qingy2024/Qwarkstar-4B
Updated
•
14
text
stringlengths 18
50k
| id
stringlengths 6
138
| metadata
dict |
|---|---|---|
Talk about chops! Comprehend the key structure like never before.
Instantly transpose any progression into each and every key. Easily
Take complexity out of learning your invaluable chord theory.
Most study of music theory begins with scales. With a little research, you'll discover that scales will often be explained in terms of notation (the lines and dots), 'whole-step/half-step' patterns (which refer to the distance between notes in the scale) or a combination of the two. Unfortunately both methods are rather complex and usually result in confusion. Let's avoid this frustration by taking a step back and looking at scales simply in terms of how chords are constructed.As you may know, a scale is simply a set of notes (usually seven) that take you from a given note through a series of other notes until it reaches the same note again an octave higher. Though the initial note is much lower in tone than the finishing note, the way our ears hear music they sound almost identical. (Think of the two "Do's" in "Do, Ra. Mi, Fa, So, La, Ti, Do.") This pattern repeats over and over, up and down. This circular behavior is common in music theory and is a principal reason the Chord Wheel is constructed the way it is.
The most fundamental scale is the Major Scale. (The "Do, Ra, Mi, etc." example we referred to earlier is the Major Scale.) Instead on concentrating on how the scale is devised, let's focus on how chords are constructed from a scale. A 'C Major Scale' starts from a C note and climbs through seven of the eleven notes in-between until it reaches the C an octave higher. By assigning numbers starting with the first note and moving upwards counting, we reach the octave C where it would be counted '8' (or another '1' if we consider it to be starting over).The most basic chord is called a triad as it contains three notes. Triads are put together by first taking any note in a scale as a starting point (called the root of the chord). From the root we skip a note in the scale and add the next note instead. So if we were starting from the first note in the C Major Scale (which, of course, is a C and we earlier numbered as '1') we would add the note numbered '3.' By once again skipping a note (the one we numbered '4') and adding the '5' note we have three notes and our first chord!Our three notes, the '1, 3 & 5' notes of the scale provide us with the first chord of the Family of Chords for the C Major Scale. By constructing chords from each note in the scale (using the 'every other note' pattern) we come up with seven basic triads (i.e. three note chords). Whenever we use the C Major Scale and the chords we have derived from it, we are said to be working in the 'Key of C.'Confused yet? If you are, fear not! Just read on, follow along and read it again later on. Like any good theory, it just takes getting used to. The beauty of actually using the Chord Wheel is that much of this knowledge will be just handed to you. However, it's important to have access to the actual theory behind what the Chord Wheel will make so easy for you to practice.
© Copyright 1995 – 2010, by The Chord Wheel. All rights reserved. | <urn:uuid:4f1cf921-e8d2-4d58-afb9-caf31a9927e8> | {
"date": "2016-06-30T19:20:21",
"dump": "CC-MAIN-2016-26",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783399117.38/warc/CC-MAIN-20160624154959-00100-ip-10-164-35-72.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9506641030311584,
"score": 4.34375,
"token_count": 722,
"url": "http://www.chordwheel.com/index.php?option=com_content&view=article&id=15:scales&catid=7:music-theory-basics&Itemid=18"
} |
- Historic Sites
The Sparck Of Rebellion
Badly disguised as Indians, a rowdy group of patriotic vandals kicked a revolution into motion
Winter 2010 | Volume 59, Issue 4
On September 5 representatives from every colony except Georgia met in Philadelphia at what came to be known as the first Continental Congress. Radicals called for Samuel Adams’s trade embargo, while moderates led by John Jay of New York and Joseph Galloway of Pennsylvania supported a strongly worded protest. All agreed that some form of action had to be taken. On behalf of the radicals, Joseph Warren of Massachusetts introduced the Suffolk Resolves, declaring the Intolerable Acts to be in violation of the colonists’ rights as English citizens and urging the creation of a revolutionary colonial government. Much to his surprise, the resolves passed, if just barely. George III was infuriated at the whole business. To him, the very calling of the Continental Congress was proof of perfidy. “The New England governments are in a state of rebellion,” he told North.
Gen. Thomas Gage, the British commander and now Massachusetts governor, received orders to strike a blow at the New England rebels. Gage learned of their whereabouts and sent troops to seize them and then destroy the supply facility at Concord. But that night Boston silversmith Paul Revere rode the 20 miles to Lexington to warn the radical leaders and everyone else along the way that the British were coming. When British troops reached the town on April 19, 1775, they encountered an armed force of 70, some of them “minutemen,” a local militia formed by an act of the provincial congress the previous year. Tensions were high and tempers short on both sides, but as the Lexington militia’s leader, Capt. John Parker, would state after the battle, he had not intended to “make or meddle” with the British troops. In fact it was the British who were advancing to form a battle line when a shot rang out—whether it was from a British or colonial musket, no one knows to this day. At the time neither side realized it was the first blast of the American Revolution.
The lone shot was followed by volleys of bullets that killed eight and wounded 10 minutemen before the British troops marched on to Concord and burned the few supplies the Americans had left there. But on their march back to Boston, the British faced the ire of local farmers organized into a well-trained embryo army, which outnumbered the British five to one and shot at them from every house, barn, and tree. By nightfall total casualties numbered 93 colonists and 273 British soldiers, putting a grim twist on Samuel Adams’s earlier exclamation to John Hancock, “What a glorious morning for America is this!”
A declaration of independence was no longer a pipe dream but a revolutionary plan in the making. From the Tea Party to the bloody fields of Concord, the thirteen colonies had proved that direct action was the surest way to free themselves from British tyranny. | <urn:uuid:74c8d0c5-65e6-4f2a-8f30-4b8ccc27ee7a> | {
"date": "2015-05-25T04:45:17",
"dump": "CC-MAIN-2015-22",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207928350.51/warc/CC-MAIN-20150521113208-00324-ip-10-180-206-219.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9701483249664307,
"score": 3.734375,
"token_count": 623,
"url": "http://www.americanheritage.com/content/sparck-of-rebellion?page=2&nid=61942"
} |
Researchers from the School of Biological Sciences discover first new chlorophyll in 60 years
20 August 2010
Found by accident in stromatolites from Western Australia's Shark Bay, the new pigment named chlorophyll f can utilise lower light energy than any other known chlorophyll.
The historic study published online in Science, challenges our understanding of the physical limits of photosynthesis - revealing that small-scale molecular changes to the structure of chlorophyll allows photosynthetic organisms to survive in almost any environment on Earth.
The new chlorophyll was discovered deep within stromatolites - rock-like structures built by photosynthetic bacteria, called cyanobacteria - by lead author Dr Min Chen from the University of Sydney.
A team of interdisciplinary scientists, including Dr Martin Schliep and Dr Zhengli Cai (University of Sydney); Associate Professor Robert Willows (Macquarie University); Professor Brett Neilan (University of New South Wales) and Professor Hugo Scheer (University of Munich), characterised the absorption properties and chemical structure of chlorophyll f, making it the fifth known type of chlorophyll molecule on Earth.
Chlorophyll is the essential molecule in oxygenic photosynthesis - the process that enables plants, algae and some bacteria to convert carbon dioxide into sugar and oxygen by using free energy from sunlight. Until recently, oxygenic photosynthesis was thought only to occur in light that is visible to human eyes, between 400nm to 700nm, as chlorophyll was strictly limited to absorbing light in this range.
This was overturned in 1996 when scientists found a cyanobacterium that could photosynthesise using light just outside the visible spectrum - at 710nm, in the infrared region - using a modified chlorophyll molecule, named chlorophyll d. Since this discovery, scientists around the world have been puzzled by how chlorophyll d is able to get enough energy from infrared light for photosynthesis.
Now the rules of photosynthesis need to be rewritten again, with the discovery of a new chlorophyll that can absorb light of even lower photon energy - 720nm - making it the most red-shifted chlorophyll to date.
In ecological terms, chlorophyll f allows cyanobacteria living deep within stromatolites to photosynthesise using low-energy infrared light, the only light able to penetrate into the structure, which challenges further our understanding of the physical limits of photosynthesis.
Dr Chen explains: "Finding the new chlorophyll was totally unexpected - it was one of those serendipitous moments of scientific discovery.
"I was actually looking for chlorophyll d, which we knew could be found in cyanobacteria living in low light conditions. I thought that stromatolites would be a good place to look, since the bacteria in the middle of the structures don't get as much light as those on the edge."
After obtaining a sample of stromatolite from Hamelin Pool, Dr Chen looked for chlorophyll d by culturing the cyanobacterial sample in infrared light of 720nm. This ensured only the survival of cyanobacteria that had chlorophylls able to absorb and use infrared light.
High performance liquid chromatography of the cultured sample performed six months later revealed not only trace amounts of chlorophyll d, but also a new chlorophyll not seen before.
Testing the optical absorption spectrum of the new chlorophyll revealed that it could absorb much longer wavelengths of light than any other known chlorophyll - 10nm longer than chlorophyll d and more than 40nm longer than chlorophyll a.
Sequential mass spectral analysis revealed the molecular weight of the new pigment to be 906 mass units. Then nuclear magnetic resonance (NMR) spectroscopy was performed to determine the chemical structure of the chlorophyll. Results indicated that chlorophylls a, b, d and f have very similar chemical structures, differing only in the position of a substitution. Yet these tiny differences in structure give the chlorophylls very different spectral properties, and hence can function in very different light environments.
"Discovering this new chlorophyll has completely overturned the traditional notion that photosynthesis needs high energy light," Dr Chen said.
"It is amazing that this new molecule, with a simple change to its chemical structure, can absorb extremely low energy light. This means that photosynthetic organisms can utilise a much larger portion of the solar spectrum than we previously thought and that the efficiency of photosynthesis is much greater than we ever imagined.
"Chlorophyll f, and its ability to absorb infrared light, can have numerous applications to industries like plant biotechnology and bioenergy.
"For us, the next challenge is to work out the function of this new chlorophyll in photosynthesis.
"Is its job to capture additional red light and pass it on to another chlorophyll, like chlorophyll a, in the reaction centre for photosynthesis?
"Or is it the only chlorophyll responsible for photosynthesis in the cyanobacterium? And if it is, then we will be speechless wondering how this molecule can get enough energy from infrared light to make oxygen from water.
"Whatever happens next, the fact that we have discovered a cyanobacterium that exploits a tiny modification in its chlorophyll molecule to photosynthesise in light that we cannot see, opens our mind to the seemingly limitless ways that organisms adapt to survive in their environment."
Media enquiries: Rachel Gleeson, 0403 067 342, 9351 4312, [email protected]
Contact: Carla Avolio
Phone: 02 9351 4543 | <urn:uuid:6050aa05-9ebd-4161-bcf2-981eda8dd9fb> | {
"date": "2014-09-21T06:57:57",
"dump": "CC-MAIN-2014-41",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657134620.5/warc/CC-MAIN-20140914011214-00341-ip-10-234-18-248.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9197516441345215,
"score": 3.515625,
"token_count": 1178,
"url": "http://sydney.edu.au/news/sobs/1699.html?newsstoryid=5469"
} |
Ever heard that “you’re born with all the brain cells you’ll ever have”? It turns out that could be a good thing – if it were true. A new study shows that at least in some circumstances, neurogenesis actually impairs memory performance.
To understand why this might be the case, consider that adults are constantly generating new neurons in a long-term memory structure – the hippocampus. This region requires a large number of neurons to store episodic memories accumulated over a lifetime (and understandably so!). Similarly, to be able to store experiences that may have occurred very quickly, neurons in the hippocampus learn very fast. Finally, to reduce interference from other similar memories, neural activity there is very sparse and highly non-overlapping (so that the hippocampus doesn’t accidentally confuse two similar experiences).
These characteristics make this region particularly useful for quickly learning relationships between stimuli. Accordingly, damage to the hippocampus, and disruption of neurogenesis in particular, will usually result in impaired learning. However, Saxe et al. showed that disrupting neurogenesis actually improves performance on some of the more complex learning tasks we know of!
Saxe et al. placed mice in the middle of an asterisk-shaped “radial arm maze” with 8 arms; at the end of each arm was a morsel of food. Then, one arm was opened. After mice had retrieved the food from the end of that arm, the same thing was repeated with another arm. Finally, after mice had retrieved food from the second arm, the mice were given a 30- second delay, and then given the option to return to the first arm or to an adjacent arm. In this case, returning to the first arm would reflect a memory failure – that arm had already been cleared of food.
Mice without neurogenesis (as dirsupted by X-Rays and a genetic manipulation) actually did better on this task than normal mice – they were more likely to go to the adjacent arm in search of food, rather than return to the original arm from which they had already cleared the food!
Mice with and without neurogenesis performed equally on a task where the delay was reduced to 5-seconds. A similar pattern emerged on another version of the task in which mice were only tested on the first arm vs. an adjacent arm, demonstrating that the memory advantage of disrupting neurogenesis is not due simply to less interference from the memory of the second arm.
To explain this thoroughly confusing result, the authors suggested two explanations:
1) Mice are less likely to suffer from the “interfering memory” of visiting the first arm because having less neurogenesis results in worse short-term memory – there is no memory to cause interference! However, this explanation seems unlikely both because of previous results (showing no short-term memory defiicts resulting from hippocampal irradiation) and because short-term memory could just as easily help performance as hurt it in the way they imply.
The second explanation is far more interesting, though somewhat counterintuitive:
2) Less neurogenesis has no effect on short-term memory, but reduces interference by making overlap between cells less likely.
“But wait, didn’t you say the purpose of neurogenesis was to create less overlap, by increasing the total number of available neurons?”
Ultimately, new neurons do result in less overlap between representations, but young neurons are quite different from older neurons in that they are far more excitable. In other words, very young neurons will fire to almost anything, and will cause increased interference from previous memories.
So less really can be more: an abundance of new neurons can result in overexcitability in the hippocampus, a region that requires very sparse, low-level patterns of activity in order to store memories without the threat of interference. Over the long term, the young neurons become less excitable and do ultimately increase the capacity of memory (indeed, they may do so even earlier for highly distinct memories). An interesting question for future research will be the kind of cues that are required to recruit non-overlapping regions of the hippocampus to make a memory “highly distinct” and thus resistant to the initially deleterious effects of hippocampal neurogenesis. | <urn:uuid:abd2c97e-55ae-4552-a32c-1c3bbf0e34eb> | {
"date": "2013-05-20T02:48:51",
"dump": "CC-MAIN-2013-20",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368698207393/warc/CC-MAIN-20130516095647-00001-ip-10-60-113-184.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9642428755760193,
"score": 3.5,
"token_count": 885,
"url": "http://scienceblogs.com/developingintelligence/2007/04/24/less-really-is-more-fewer-neur/"
} |
×
Get Full Access to Elementary Statistics - 12 Edition - Chapter 10.6 - Problem 9ccq
Get Full Access to Elementary Statistics - 12 Edition - Chapter 10.6 - Problem 9ccq
×
# 9 CQQ The exercises are based on the following sample data
ISBN: 9780321836960 18
## Solution for problem 9CCQ Chapter 10.6
Elementary Statistics | 12th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Elementary Statistics | 12th Edition
4 5 1 345 Reviews
11
1
Problem 9CCQ
?9 CQQ The exercises are based on the following sample data obtained from different second-year medical students who took blood pressure measurements of the same person (based on data from Marc Triola, MD). S y s t o l i c D i a s t o l i c If you had computed the value of the linear correlation coefficient to be 3.335, what should you conclude?
Step-by-Step Solution:
Solution 9 CQQ Answer : Step 1 : The value of correlation coefficient must lies in between -1 to 1. If the value correlation is -1, then it is negatively correlated, if it is 1 then we can say that positively correlated, if the value is o then there is no correlation. The correlation coefficient value be 3.335, we can say that there must be error in computation, the correlation value does not exceed more than one on the positive side.
Step 2 of 1
##### ISBN: 9780321836960
This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. Since the solution to 9CCQ from 10.6 chapter was answered, more than 400 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 9CCQ from chapter: 10.6 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. This full solution covers the following key subjects: based, data, marc, computed, conclude. This expansive textbook survival guide covers 121 chapters, and 3629 solutions. Elementary Statistics was written by and is associated to the ISBN: 9780321836960. The answer to “?9 CQQ The exercises are based on the following sample data obtained from different second-year medical students who took blood pressure measurements of the same person (based on data from Marc Triola, MD). S y s t o l i c D i a s t o l i c If you had computed the value of the linear correlation coefficient to be 3.335, what should you conclude?” is broken down into a number of easy to follow steps, and 68 words.
#### Related chapters
Unlock Textbook Solution | crawl-data/CC-MAIN-2021-39/segments/1631780056120.36/warc/CC-MAIN-20210918002951-20210918032951-00069.warc.gz | null |
188.8.131.52 Floods and droughts
Documented trends in floods show no evidence for a globally widespread change. Although Milly et al. (2002) identified an apparent increase in the frequency of ‘large’ floods (return period >100 years) across much of the globe from the analysis of data from large river basins, subsequent studies have provided less widespread evidence. Kundzewicz et al. (2005) found increases (in 27 cases) and decreases (in 31 cases) and no trend in the remaining 137 cases of the 195 catchments examined worldwide. Table 1.3 shows results of selected changes in runoff/streamflow, lake levels and floods/droughts. Other examples of changes in floods and droughts may be found in Table SM1.2.
Globally, very dry areas (Palmer Drought Severity Index, PDSI ² -3.0) have more than doubled since the 1970s due to a combination of ENSO events and surface warming, while very wet areas (PDSI ³ +3.0) declined by about 5%, with precipitation as the major contributing factor during the early 1980s and temperature more important thereafter (Dai et al., 2004). The areas of increasing wetness include the Northern Hemisphere high latitudes and equatorial regions. However, the use of PDSI is limited by its lack of effectiveness in tropical regions. Table 1.3 shows the trend in droughts in some regions. Documented trends in severe droughts and heavy rains (Trenberth et al., 2007, Section 3.8.2) show that hydrological conditions are becoming more intense in some regions, consistent with other findings (Huntington, 2006). | <urn:uuid:138d490c-d7a9-423d-9d4b-c575c2ba4a97> | {
"date": "2016-10-28T08:20:11",
"dump": "CC-MAIN-2016-44",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988721595.48/warc/CC-MAIN-20161020183841-00459-ip-10-171-6-4.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.931334376335144,
"score": 3.53125,
"token_count": 364,
"url": "http://www.ipcc.ch/publications_and_data/ar4/wg2/en/ch1s1-3-2-2.html"
} |
# How Many Candy Corns Are in a Bag? Exploring the Sweet Science Behind the Perfect Proportion
## Introduction
Candy corn is one of the most beloved treats during the Halloween season. But have you ever wondered just how many candy corns are in a bag? If so, you’ve come to the right place! In this article, we’ll explore the sweet science behind calculating the exact amount of candy corns that fit into a bag. We’ll cover topics such as estimating the volume of a bag, calculating candy corn density, measuring candy corn size, understanding the mechanics and physics of candy corns, and tips on counting up the perfect number. So grab a bag of candy corn and let’s get started!
## Calculating How Many Candy Corns Fit in a Bag
The first step in determining how many candy corns fit in a bag is to estimate the volume of the bag. This can be done by measuring the length, width, and height of the bag and multiplying those three numbers together. For example, if the bag measures 10 inches long, 5 inches wide, and 3 inches tall, then the volume of the bag would be 150 cubic inches.
The next step is to calculate the density of candy corn. This can be done by dividing the weight of the candy corn by its volume. For example, if a bag of candy corn weighs 1 pound and has a volume of 150 cubic inches, then the density of the candy corn would be 6.67 pounds per cubic inch.
The third step is to measure the size of each candy corn. This can be done by measuring the length, width, and height of a single candy corn. For example, if the candy corn measures 0.5 inches long, 0.25 inches wide, and 0.125 inches tall, then the volume of the candy corn would be 0.015625 cubic inches.
## The Sweet Science Behind How Many Candy Corns are in a Bag
Now that we know the volume of the bag, the density of the candy corn, and the size of each candy corn, it’s time to dive into the sweet science behind how many candy corns fit into a bag. To do this, we need to understand the mechanics, physics, and chemistry of candy corn.
The mechanics of candy corn involve understanding the forces that act upon it when it is placed in a bag. These forces include gravity, friction, and air pressure. Gravity causes the candy corn to settle at the bottom of the bag while friction causes the candy corn to stick together. Air pressure also plays a role in how the candy corn is distributed in the bag.
The physics of candy corn involve understanding the properties of the material itself. Candy corn is made up of sugar, water, corn syrup, and various flavorings. When these ingredients are mixed together, they create a substance with certain physical properties such as viscosity, surface tension, and elasticity. All of these factors play a role in how the candy corn behaves when placed in a bag.
Finally, the chemistry of candy corn involves understanding the chemical reactions that occur when the ingredients are mixed together. Different ingredients will react differently, creating different flavors and textures. The type of ingredients used in the candy corn will affect how it looks, tastes, and behaves in a bag.
## A Guide to Knowing the Ideal Amount of Candy Corns for Your Bag
Now that we understand the sweet science behind how many candy corns fit in a bag, it’s time to figure out the ideal amount of candy corn for your bag. The first step is to identify your needs. Do you want a bag of candy corn for a party or for yourself? Do you want a bag that’s full or one that’s half-full? Once you’ve identified your needs, you can move on to the next step.
The next step is to find the right size bag. The size of the bag will determine how much candy corn you can fit in it. If you’re planning on buying a large bag, then you’ll need to account for the extra space. On the other hand, if you’re buying a small bag, then you’ll need to make sure there’s enough room for the candy corn.
The final step is to determine the optimal proportion of candy corn. This can be done by dividing the volume of the bag by the volume of a single candy corn. The result will give you the ideal number of candy corns for your bag. For example, if you have a bag that’s 150 cubic inches and each candy corn is 0.015625 cubic inches, then the optimal proportion would be 9,600 candy corns.
## Estimating the Perfect Proportion of Candy Corns for a Bag
Once you’ve determined the optimal proportion of candy corn for your bag, it’s time to estimate the perfect proportion. There are several factors that can affect the perfect proportion including the shape of the bag, the size of the candy corn, and the type of candy corn. For example, if the bag is square, then the optimal proportion may be higher than if the bag is round. Similarly, if the candy corn is larger, then the optimal proportion may be lower than if the candy corn is smaller.
There are also several techniques that can be used to estimate the perfect proportion. One technique is to fill the bag with a test batch of candy corn and measure the amount that fits. This method is best for estimating the perfect proportion for a specific type of candy corn. Another technique is to calculate the exact volume of the bag and divide it by the volume of a single candy corn. This method is best for estimating the perfect proportion for any type of candy corn.
Finally, there are some tips that can be used to ensure an accurate measurement when estimating the perfect proportion. For example, it’s important to make sure the bag is completely filled with candy corn and that all of the pieces are the same size. Additionally, it’s important to make sure the measurements are precise and that the measurements are taken from the same point in the bag.
## Counting Up the Perfect Number of Candy Corns for a Bag
Once you’ve estimated the perfect proportion of candy corn for your bag, it’s time to count up the exact number. There are several strategies that can be used to count the exact number of candy corn. One strategy is to count each piece individually. This method is best for small bags where the pieces can be easily seen. Another strategy is to use a scale to weigh the bag and then divide the weight by the weight of a single candy corn.
It’s also important to keep track of the number of candy corn. One method is to mark off each piece as it’s counted. This method works best for small bags where the pieces can be easily seen. Another method is to use a tally system to keep track of the number of pieces. This method works best for larger bags where the pieces can’t be easily seen.
Finally, there are some tricks that can be used to ensure an accurate count. For example, it’s important to make sure all of the pieces are the same size and to count each piece twice to make sure the count is correct. Additionally, it’s important to count the pieces in batches to reduce the chance of missing pieces.
## Conclusion
In conclusion, determining the exact number of candy corns that fit in a bag is not as straightforward as it may seem. It requires an understanding of the mechanics, physics, and chemistry of candy corn as well as an understanding of the size of the bag and the size of the candy corn. By following the steps outlined in this article, you should now be able to calculate the exact number of candy corns that fit in a bag. So grab a bag of candy corn and start counting!
To summarize, this article explored the science behind how many candy corns fit in a bag. We discussed topics such as estimating the volume of a bag, calculating candy corn density, measuring candy corn size, understanding the mechanics and physics of candy corns, and tips on counting up the perfect number. Armed with this knowledge, you should now be able to calculate the exact number of candy corns that fit in a bag. | crawl-data/CC-MAIN-2024-10/segments/1707947476413.82/warc/CC-MAIN-20240304033910-20240304063910-00081.warc.gz | null |
# Testing
Students of high school have 10 points for each good solved task. The wrong answer is deducted by 5 points. After solving 20 tasks, student Michael had 80 points. How many tasks did he solve correctly and how many wrong?
Correct result:
a = 12
b = 8
#### Solution:
a+b = 20
10a - 5b = 80
a+b = 20
10•a - 5•b = 80
a+b = 20
10a-5b = 80
a = 12
b = 8
Our linear equations calculator calculates it.
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
Tips to related online calculators
Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation?
## Next similar math problems:
• Trip cost
In September, the trip cost CZK 12,000. How many crowns did the trip cost in June of the same year, when they have since reduced the price by a quarter and by another CZK 1,200?
Susan thought, "If I read 15 pages a day, I will read the whole book in 8 days. “How many pages would she have to read a day if she wanted to finish the book on the 6th day from the start of reading? And how many pages does the book have?
• Worker salary
The worker had a salary of CZK 18,000. During the year, his salary was increased by a quarter. He earned a total of CZK 247,500 for the whole year. From which month was his salary increased?
If I read 15 pages a day, I will read the whole book in 18 days. How long will it take to read a book if I only read 9 pages every day?
• Sum of seven
The sum of seven consecutive odd natural numbers is 119. Determine the smallest of them.
• The sum
The sum of five consecutive odd numbers is 75. Find out the sum of the second and fourth of them.
Suzan reads a book. If she read half an hour a day, she would read it in nine days. How many minutes does he have to read a day if he wants to read it three days earlier?
• The tourist
The tourist walked a quarter of the way on the first day, a third of the rest on the second day, and 20 km on the last day. How many km did he walk in three days?
• What time is it?
What time is it, when there is 4 times less time left until midnight than the time that has elapsed since noon? What time is it, when one-fifth of the hours that have passed since midnight is equal to one-third of the hours that are missing by 12 o'clock?
• Second side
Calculate the length of the other side of the rectangle if its circumference is 60 cm and one side is 10 cm long.
• Work together
Two bricklayers plastering a wall. The first would plaster it in 8 hours, the second in 12 hours. How many hours will they be done with the work if they work together?
• Diggers
The excavation was carried out in 4 days and 5 workers worked on it for 7 hours a day. Determine how long it would take to excavate if 7 workers worked on it 8 hours a day.
• Tourist route
How long is the tourist route when tourists crossed four-sevenths of the way on foot, crossed the bus twice less than on foot and passed the last 14 kilometers by boat.
• The farmer
The farmer calculated that the supply of fodder for his 20 cows was enough for 60 days. He decided to sell 2 cows and a third of the feed. How long will the feed for the rest of the peasant's herd last?
• Job applicants
Job applicants: three-fourths of applicants had experience for a position. The number that did not have prior experience was 36. How many people applied for the job?
• Library
New books were purchased for the library. Five-eighths were professional books, one-fifth were encyclopedias, and 231 books were dictionaries. How many professional books were there?
• Large family
The average age of all family members (children, mother, father, grandmother, grandfather) is 29 years. The average age of parents is 40 years, grandparents 66 years and all children are 5 years. How many children are there in this family? | crawl-data/CC-MAIN-2021-04/segments/1610704798089.76/warc/CC-MAIN-20210126042704-20210126072704-00079.warc.gz | null |
"It's a really amazing-quality genome," says David Reich of Harvard Medical School in Boston. "It's as good as modern human genome sequences, from a lot of ways of measuring it."
The pinky belonged to a girl who lived tens of thousands of years ago. Scientists aren't sure about the exact age. She is a member of an extinct group of humans called Denisovans. The name comes from Denisova cave in Siberia, where the pinky was found.
Two years ago, scientists at the Max Planck Institute for Evolutionary Anthropology in Leipzig reported that they had been able to get just enough DNA from the fossil to make a rough sequence of her DNA. But Matthias Meyer developed a far more efficient way of recovering ancient DNA, so he went back to the tiny amount of DNA left over from the first effort, and reanalyzed it.
"And from this little leftover, we were able to determine the sequence of the Denisova genome 30-fold over," says Meyer.
What that means is they were able to look at every single location along all of this girl's chromosomes 30 times to be absolutely certain that they had the right DNA letter in the right spot. The new results appear in the online edition of the journal Science.
The high-quality sequence gives scientists valuable new data for studying ancient humans. Researchers have begun, for example, to explore which modern human populations may have inherited genes from Denisovans. | <urn:uuid:eb2bc49b-0e75-4612-9305-2c6c83b1f97d> | {
"date": "2014-04-20T23:30:43",
"dump": "CC-MAIN-2014-15",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00460-ip-10-147-4-33.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9689256548881531,
"score": 3.546875,
"token_count": 297,
"url": "http://www.richarddawkins.net/news_articles/next_article?article=660&category=&videos="
} |
I'm a huge believer in hands-on education. But you have to have the right tools. If I'm going to teach my daughter about electronics, I'm not going to give her a soldering iron. And similarly, she finds prototyping boards really frustrating for her little hands. So my wonderful student Sam and I decided to look at the most tangible thing we could think of: Play-Doh. And so we spent a summer looking at different Play-Doh recipes. And these recipes probably look really familiar to any of you who have made homemade play-dough — pretty standard ingredients you probably have in your kitchen. We have two favorite recipes — one that has these ingredients and a second that had sugar instead of salt. And they're great. We can make great little sculptures with these. But the really cool thing about them is when we put them together. You see that really salty Play-Doh? Well, it conducts electricity. And this is nothing new. It turns out that regular Play-Doh that you buy at the store conducts electricity, and high school physics teachers have used that for years. But our homemade play-dough actually has half the resistance of commercial Play-Doh. And that sugar dough? Well it's 150 times more resistant to electric current than that salt dough. So what does that mean? Well it means if you them together you suddenly have circuits — circuits that the most creative, tiny, little hands can build on their own. (Applause) And so I want to do a little demo for you. So if I take this salt dough, again, it's like the play-dough you probably made as kids, and I plug it in — it's a two-lead battery pack, simple battery pack, you can buy them at Radio Shack and pretty much anywhere else — we can actually then light things up. But if any of you have studied electrical engineering, we can also create a short circuit. If I push these together, the light turns off. Right, the current wants to run through the play-dough, not through that LED. If I separate them again, I have some light. Well now if I take that sugar dough, the sugar dough doesn't want to conduct electricity. It's like a wall to the electricity. If I place that between, now all the dough is touching, but if I stick that light back in, I have light. In fact, I could even add some movement to my sculptures. If I want a spinning tail, let's grab a motor, put some play-dough on it, stick it on and we have spinning. (Applause) And once you have the basics, we can make a slightly more complicated circuit. We call this our sushi circuit. It's very popular with kids. I plug in again the power to it. And now I can start talking about parallel and series circuits. I can start plugging in lots of lights. And we can start talking about things like electrical load. What happens if I put in lots of lights and then add a motor? It'll dim. We can even add microprocessors and have this as an input and create squishy sound music that we've done. You could do parallel and series circuits for kids using this. So this is all in your home kitchen. We've actually tried to turn it into an electrical engineering lab. We have a website, it's all there. These are the home recipes. We've got some videos. You can make them yourselves. And it's been really fun since we put them up to see where these have gone. We've had a mom in Utah who used them with her kids, to a science researcher in the U.K., and curriculum developers in Hawaii. So I would encourage you all to grab some Play-Doh, grab some salt, grab some sugar and start playing. We don't usually think of our kitchen as an electrical engineering lab or little kids as circuit designers, but maybe we should. Have fun. Thank you. (Applause) | Hands-on science with squishy circuits | null |
# 2/3 times 5 in fraction form
2/3 x 5 = 10/3.
2/3 times 5 is equal to 10/3 in fraction and 3.3333 in decimal form.
2/3 times 5 in fraction form:
2/3 x 5 = ?
Arrange the fractions in the product expression form as like the below:
= 2/3 x 5/1
= (2 x 5)/(3 x 1)
Check the numerator and denominator and cancel if anything cancelled each other:
= 10/3
2/3 x 5 = 10/3
Hence,
2/3 times 5 as a fraction is equal to 10/3.
2/3 times 5 as a decimal:
2/3 x 5 = 10/3
2/3 x 5 = 3.3333
2/3 times 5 as a decimal is 3.3333
where,
2/3 is the multiplicand,
5 is the multiplier,
10/3 is the simplest form of 2/3 times 5,
3.3333 is the decimal form of 2/3 times 5.
Important Notes: 2/3 * 5
All the following questions represent 2/3 times 5 in fraction form, so it's very much important to observe the different variations of this question.
1. what is 2/3 times 5 in fraction form?
2. what is 5 times 2/3 as a fraction?
3. 2/3 x 5 = ?
4. 5 x 2/3 = ?
5. 2/3 * 5 = ?
6. 5 * 2/3 = ?
For values other than 2/3 times 5, use this below tool: | crawl-data/CC-MAIN-2023-23/segments/1685224653183.5/warc/CC-MAIN-20230606214755-20230607004755-00485.warc.gz | null |
This is a 109-page pack that includes digitally designed anchor charts for 3rd grade topics/concepts. You can print these and compile them into a notebook or you can hang them in your room when you're covering a certain topic/concept.
There are four subject areas covered:
- Language Arts/Writing
- Science/Social Studies
Here are the concepts covered in each area:
* Reading *
what readers do, where readers read, just right books, characters, setting, problem and solution, main idea and details, text connections, summary, theme vs. plot, fiction vs. nonfiction, reference materials, fluency, accuracy, comprehension, expression, point of view, narrator, sequence of events, author and illustrator, title page, table of contents, index, glossary, transition words, cause and effect
* Language Arts/Writing *
where writers write, the writing process, types of writing, synonyms and antonyms, rhyming words, blends, conjunctions, figurative language, capitalization, punctuation, commas, ABC order, subjects and predicates, types of sentences, simple vs. compound sentences, nouns, adjectives, verbs, possessive nouns, common vs. proper nouns, comparative and superlative adjectives, adverbs, prefixes, base words, suffixes, Latin and Greek roots, homophones, multiple meaning words, syllables, quotation marks
* Mathematics *
addition/subtraction with regrouping, rounding to the nearest ten/hundred, place value, odd and even numbers, standard form, expanded form, word problems, multiplication strategies, division strategies, missing/unknown factor, probability/reasonableness, linear measurement, mass/volume, area and perimeter, fractions, money, graphs, shapes, time, AM/PM, elapsed time, number patterns, lines, angles, triangles
* Science/Social Studies *
Paul Revere, Susan B. Anthony, Frederick Douglass, Mary McLeod Bethune, Franklin D. Roosevelt, Eleanor Roosevelt, Lyndon B. Johnson, Cesar Chavez, map grids, compass/directions, latitude and longitude, Equator and Prime Meridian, branches of national government, habitats, pollution, conservation, rocks, minerals, soil, fossils, magnets, heat, economics, landforms, timelines
The final page is a huge THANK YOU to all the fabulous artists who create clip art and fonts for all of us to use! Thank you to KPM Doodles, Ginger Snaps, Fancy Dog Studio, and First Grade a la Carte for their talents!!!
***Printing - if you're worried about using a bunch of color ink, try printing the pages in black and white. Print them on colored paper! That will save you TONS on colored ink! :)
I hope you and your students enjoy these anchor charts. They are a fabulous resource for review and mastery. | <urn:uuid:c1b6b20b-3dbf-4d56-969e-4e974ec17d5d> | {
"date": "2014-09-19T19:51:43",
"dump": "CC-MAIN-2014-41",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657132007.18/warc/CC-MAIN-20140914011212-00274-ip-10-196-40-205.us-west-1.compute.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.8612520694732666,
"score": 3.546875,
"token_count": 604,
"url": "http://www.teachersnotebook.com/product/3rdgradegridiron/anchor-charts-for-3rd-grade-all-subjects-ccss-aligned"
} |
## MSTAR Interventions
### Common Misconceptions and How to Prevent Them
#### Examples for Preventing or Correcting
Some students believe ratios always compare a part to a whole, like fractions.
Provide an example in which students compare a part to a part (e.g., 1 green apple to 3 red apples). Explain that ratios can compare part to part, not only part to whole. The ratio 1 to 3 compares part of the apples to another part of the apples.
Some students struggle to simplify ratios.
Demonstrate an example of simplification, explicitly describing the steps:
1. Check for common factors. (The easiest thing to remember is to use the prime numbers 2, 3, 5, 7, and 11.)
2. Divide each part of the ratio by the common factor.
Example:
Show students 8 red and 4 yellow counters.
Ask, “What is the ratio of red to yellow counters?” ()
Ask, “How do I know whether this ratio will simplify?” (Both numbers must divide by a common factor.)
Ask, “What divides into 8 and 4 evenly?” (2 and 4)
Say, “When 2 factors divide evenly, to simplify, choose the largest factor.” (4)
Say, “When I divide by 1 whole, the ratio stays the same. Because 4 is a common factor, I need to divide by , which is 1 whole."
Ask, “What is 8 divided by 4? What is 4 divided by 4?” (2, 1)
Say, “So becomes when it is simplified.”
Ask, “What is the common factor for the ratio 10 to 15?” (5)
Some students believe that ratios and rates can always be simplified to a mixed number or a whole number.
Remind students that ratios and rates must compare 2 quantities; therefore, you often cannot simplify a ratio or a rate to a mixed number. Examples: 15 boys to 5 bats becomes 3 boys to 1 bat, not 3. 14 girls to 4 boys becomes 7 girls to 2 boys, not 3 . | crawl-data/CC-MAIN-2019-43/segments/1570986675409.61/warc/CC-MAIN-20191017145741-20191017173241-00146.warc.gz | null |
### GRE : Quantitative Comparison Problem
Directions
This is the Quantitative Comparison Problem-1.
Each of the following questions consists of two quantities, one in Column A and one in Column B. You are to compare these two quantities and choose
A if the quantity in Column A is greater
B if the quantity in Column B is greater
C if the two quantities are equal
D if relationship cannot be determined from the information given
Since there are only four choices, never mark E for these type of questions.
This is the Quantitative Comparison Problem-1.Understand the basic mathematical principles behind this problem and other problems.
Example-1
J – K = 2
K – 6 = 4
Column A : J
Column B : 10
Form the Second equation, we get K = 6 + 4 = 10. Substituting this value in the first equation, we get J = 2 + K = 2 + 10 = 12. Since J (12) is greater than 10, we can choose A as the answer.
This is the Quantitative Comparison Problem-1.Understand the basic mathematical principles behind this problem and other problems.
From Quantitative Comparison Problem-1 to HOME PAG | crawl-data/CC-MAIN-2021-17/segments/1618038082988.39/warc/CC-MAIN-20210415005811-20210415035811-00191.warc.gz | null |
Diabetes is a common condition that can affect adults & children. It occurs when the body
cannot produce enough insulin or stops producing insulin completely. Insulin helps control
the level of glucose (sugar) in the blood.
There are 2 types of diabetes – Type 1 and Type 2. Symptoms of diabetes include
increased thirst, passing urine more often – especially at night, tiredness, weight loss, thrush
and blurred vision. Some people may not experience any symptoms.
What is the difference between Type 1 & Type 2 Diabetes?
Type 1 Diabetes:
In Type 1 Diabetes, the body stops producing insulin. Type 1 Diabetes usually develops in
children and young adults, although it can occur at any age.
The management of Type 1 Diabetes requires treatment with insulin, combined with healthy
eating and exercise.
Type 2 Diabetes:
In Type 2 Diabetes, the body still produces insulin, but not enough for the body’s needs and
the insulin it does produce is not used effectively (this is known as insulin resistance).
Over three quarters of people with diabetes have Type 2 Diabetes. There is a strong link
between this type of diabetes and being overweight.
A simple blood test will indicate if you are at risk of developing diabetes and will also help
confirm a diagnosis.
Should you have any concerns please make an appointment to see our practice nurse who
will assess your individual case and follow up with an appointment in our diabetic clinic if
Taking control of your Diabetes
If you discover that you have diabetes, you may be worried about what the future will hold
and what changes you have to make to your life. Our team are here to give you the advice
and support that you need.
You play a very important role in the management and control of your diabetes. Weight loss,
changes to your diet, regular physical activity, and medication if required, will all help control
the glucose level in your blood and keep it as near normal as possible.
Poorly controlled diabetes can, over time, cause damage to your eyes, nerves, kidneys,
blood vessels and heart. To reduce the risk of damage here are some of the steps you can
take: Diet: Eat a healthy balanced diet
Weight: It is important to keep your weight at the right level for your height.
Smoking: You need to stop smoking
Foot care: Take care of your feet and check them daily.
Physical activity: Regular activity promotes food health. it helps to lower your blood glucose, lose weight and reduce blood pressure.
Monitoring: You will play an important role in monitoring your blood glucose levels.
Education: Learn as much as possible about your condition as this will help you to manage it.
It is important that you take these measures to manage your diabetes. | <urn:uuid:b77857c1-e5cc-47ad-a2e6-582a41210bb0> | {
"date": "2018-04-23T05:28:20",
"dump": "CC-MAIN-2018-17",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945793.18/warc/CC-MAIN-20180423050940-20180423070940-00016.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9314719438552856,
"score": 3.71875,
"token_count": 576,
"url": "http://drmc.ie/diabetic-testing-and-monitoring/"
} |
During this crucial period, the United States pursued a policy of expansion based on “manifest destiny,” the ideology that Americans were in fact destined to extend their nation across the continent.
Illustration of “Manifest Destiny” ( John Gast, 1872)
The United States even proved to be willing to go to war to secure new territories. While it managed to negotiate an agreement with Great Britain to secure the Oregon territory, acquiring the valuable territory south of it—including California and its important Pacific harbors—required the use of force, and, in 1845, the United States embarked on its first offensive war by invading Mexico.
In addition to advancing westward, the United States also continued to expand economically through investment in foreign markets and international trade. With these growing commercial interests, came a larger navy and increased international presence. The United States began to turn to the Pacific for new economic opportunities, establishing a presence in China, and opening Japan and Korea to western commercial interests. | <urn:uuid:e90951dc-3883-41ad-ac9d-2bfd56c6ccdf> | {
"date": "2014-11-26T17:14:48",
"dump": "CC-MAIN-2014-49",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416931007301.29/warc/CC-MAIN-20141125155647-00223-ip-10-235-23-156.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9514349699020386,
"score": 4.28125,
"token_count": 203,
"url": "http://history.state.gov/milestones/1830-1860"
} |
From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. [Take π = 3.14.]
Height of solid cylinder = h = 8 cm
Radius of solid cylinder = r = 6 cm
Volume of solid cylinder = πr2h
= 3.14 × 6 × 6 × 8 cm3
= 904.32 cm3
Curved Surface area of solid cylinder = 2πrh
Height of conical cavity = h = 8 cm
Radius conical cavity = r = 6 cm
Volume of conical cavity = 1/3 πr2h
= 1/3 × 3.14 × 6 × 6 × 8 cm3
= 301.44 cm3
Let l be the slant height of conical cavity
l2 = r2 + h2
l2 = (62 + 82) cm2
l2 = (36 + 64) cm2
l2 = 100 cm2
l = 10 cm
Curved Surface area of conical cavity = πrl
Since, conical cavity is hollowed out from solid cylinder, so, volume and total surface area of remaining solid will be found out by subtracting volume and total surface area of conical cavity from volume and total surface area of solid cylinder.
Volume of remaining solid = Volume of solid cylinder – Volume of conical cavity
Volume of remaining solid = 904.32 cm3 – 301.44 cm3
= 602.88 cm3
Total surface area of remaining solid = Curved Surface area of solid cylinder + Curved Surface area of conical cavity + Area of circular base
Total surface area of remaining solid = 2πrh + πrl + πr2
= πr × (2h + l + r)
= 3.14 × 6 × (2 × 8 + 10 + 6) cm2
= 3.14 × 6 × 32 cm2
= 602.88 cm2
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better. | crawl-data/CC-MAIN-2020-16/segments/1585370510846.12/warc/CC-MAIN-20200403092656-20200403122656-00459.warc.gz | null |
# How do you find the derivative of [x+(x+sin^2x)^3]^4?
$\frac{d}{\mathrm{dx}} {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4}$
$= 4 {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left[1 + 3 {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + \sin 2 x\right)\right]$
Take note: $\sin 2 x = 2 \cdot \sin x \cdot \cos x$
#### Explanation:
From the given
${\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4}$
$\frac{d}{\mathrm{dx}} {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4} =$
$4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + {\left(x + {\sin}^{2} x\right)}^{3}\right)$
$= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{3 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + {\sin}^{2} x\right)\right)$
$= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + 2 \cdot \sin x \cdot \cos x\right)\right)$
$= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + \sin 2 x\right)\right)$
God bless....I hope the explanation is useful. | crawl-data/CC-MAIN-2024-33/segments/1722640476479.26/warc/CC-MAIN-20240806032955-20240806062955-00834.warc.gz | null |
Quercus robur (English oak)
Unrivalled king of the forest in Britain, English oak (pedunculate oak) is synonymous with strength, size and longevity.
Acorns and leaves of Quercus robur (Photo: Wolfgang Stuppy)
Quercus robur L.
English oak, pedunculate oak, common oak
Least Concern (LC) according to IUCN Red List criteria.
Timber, tanning leather, food for livestock.
Tannic acid in the leaves is poisonous to horses if consumed in excess, damaging the kidneys. Acorns are poisonous to horses and cattle, though swine can consume them safely in moderation.
About this species
English oak is probably the most well-known and best-loved of the tree species native to Britain. This king of the forest can live for more than a millennium according to some sources, and grow up to 40 m high. Mature specimens are usually home to many species of wildlife. Since the Druids, and probably long before, the oak has played an important role in British culture. Couples were still wed under ancient oaks as late as Oliver Cromwell's time and the Yule log, kept from one year to another to warm the Christmas celebrations, was traditionally cut from oak.
Geography and distribution
Quercus robur is native to Asia Minor (an area corresponding to the western two-thirds of Turkey), North Africa, the Caucasus (a geopolitical region at the border of Europe and Asia) and Europe.
One of English oak's most recognisable characteristics is the shape of its leaves. Pale green in colour, they have four or five lobes on each side and are attached to the branches with almost no stalk. In contrast, the acorns are borne on long stalks known as peduncles, hence the name ‘pedunculate oak’.
Large, old oaks often have dead branches at the top. This 'stag head' is not necessarily a sign that the tree is dying. When water is short the oak simply stops supplying its upper extremities, and as a result extends its lifespan.
For such a huge, long-lived and widespread tree, the oak is surprisingly bad at reproducing naturally. First, it can be a full 50 years before the first crop of acorns (seeds) is produced. Second, most of the tens of thousands of acorns dropped are eaten by animals, or simply rot. It is therefore left to the buried acorns of forgetful or dying squirrels or jays to ensure the continued lifecycle of this giant of the countryside.
Threats and conservation
Oakwoods have diminished greatly in the UK over the last few hundred years, but the Forestry Commission is now actively promoting woodland planting with native tree species, such as the English oak. A fungal disease known as ‘sudden oak death’ has recently arrived in the UK, apparently from Europe and although the disease affects many other species including Rhododendron, Viburnum, beech, sweet chestnut and holm oak, at present native oaks appear to be relatively resistant to it. However, it may pose a threat to oaks in the future – particularly if the trees become stressed by climate change. Another, more worrying ‘disease’ known as ‘acute oak decline’ has recently been identified in the UK. This does kill our native oaks and its causes have yet to be understood.
English oak's role in biodiversity
Quercus robur supports a wealth of organisms which benefit from the food, support and shelter it supplies. Its acorns are rich in starch and provide food for birds such as the Eurasian jay (Garrulus glandarius) and small mammals such as squirrels. An abundance of insects live on the leaves, buds and bark, and even inside the acorns. In autumn the soft leaves break down easily to form a rich leaf mould beneath the tree, which supports a wealth of invertebrates and fungi (such as the edible beefsteak fungus Fistulina hepatica). The trunk and branches offer physical support to mosses, ferns and lichens, and the open leaf canopy lets light through to the woodland floor, meaning flowers such as violets, bluebells and primroses can thrive.
The green oak moth (Tortrix viridana) feeds on oak leaves in its larval (caterpillar) form and then rolls itself up in a leaf to pupate into the adult. This in turn is parasitised by at least four species of ichneumon wasps, which lay their eggs inside the defenceless pupae. Gall wasps lay their eggs inside the oak leaves, where their larvae then secrete a chemical causing the leaf to mutate and form a gall (oak apple), providing them with protection and shelter – though not enough to prevent them from being parasitised by other species of ichneumon wasp! In years when populations of caterpillar or moth larvae are high almost all the leaves on a tree can be eaten, yet the English oak is able to survive this trauma.
Birds such as great-spotted woodpeckers (Dendrocopos major) come to feed on insects in the bark. Acorns that have fallen to the ground provide food for rooks, wood pigeons and mice, which in turn attract birds of prey such as owls, buzzards and sparrowhawks.
Fungi, such as the orange oak bolete (Leccinum quercinum), form symbiotic associations with English oak roots, the fungi helping the tree to extract nutrients from the soil whilst in turn benefiting from sugars provided by the oak. Even after its death the English oak continues to support biodiversity. The decaying wood can host a vast array of fungi and the hollow trunks provide roosting sites for owls and hibernation sites for bats.
Quercus robur was named for its robust or sturdy nature (robur means strength in Latin), and since iron tools were first made, people have been felling this mighty tree for its strong and durable timber. It can take as long as 150 years before an oak is ready for use in construction, but it is well worth the wait. Until the middle of the 19th century, when iron became the material of choice for building ships, thousands upon thousands of oaks were felled every year. Oak has many other uses; oak bark has been used in leather tanning and in dyeing, insect galls have been used to make black ink, and the acorns are valued as food for livestock.
Pig grazing rights in oak woods, known as 'pannage', have been jealously guarded through the centuries. In the New Forest in southern England these rights are still exploited by Commoners (local people with rights to the forest’s resources), though on a smaller scale than in the past. Acorns have also been roasted to produce a coffee substitute, apparently quite inferior to the real thing.
The majority of the oaks propagated at Kew are grown from seed. This method is preferred for the scientific collections as it is the easiest and safest means of collecting oaks in the wild without damaging their populations. Acorns are collected green and sown directly into the Arboretum Nursery field during the autumn. The seeds germinate the following spring. Oak seed does not store well and quickly becomes non-viable as it dries out. Mice and squirrels present a threat to the seeds, and netting or wire mesh is used to keep them at bay during autumn and winter. Rare oak specimens are sown in pots in a frost-free glasshouse for added protection.
Occasionally, young trees are also brought in from nurseries, for example to give instant impact in heritage plantings on the vistas. These are grown in a reputable nursery for 12 months where they can be inspected for pests and diseases before they are transplanted to Kew.
The fresh foliage of young oaks often suffers from mildew. Trees affected in this way are not treated because once they are planted out the mildew no longer presents a problem. It is often necessary to stake young trees. Tony Hall, who manages the oak collection at Kew, says that the Arboretum is treated as an extension of the nursery, in the sense that young trees there are carefully monitored during their early years.
The ‘Coubertin oak’ at Kew
The ‘Coubertin oak’ at Kew is one of a ‘ribbon’ of 40 English oaks (Quercus robur) planted between the Olympic Park in London and Much Wenlock in Shropshire, to celebrate the London 2012 Olympic Games.
It was grown from one of 40 acorns collected by students in Much Wenlock. The acorns were from a tree planted in 1890 in honour of Baron Pierre de Coubertin who was invited to visit the Wenlock Olympian Games, a local version of the Ancient Greek Olympic Games, by William Brookes. Brookes hoped to establish a modern version of the Games and following this meeting, the Baron went on to set up the International Olympic Committee and the first modern Olympics in Greece in 1896.
Students at William Brookes School grew the ‘Coubertin oak’ acorns into young trees which were then moved to the Arboretum Nursery at Kew where they stayed for two years to develop their root systems. They were then transferred to a nursery in Cambridgeshire for a further two years before the planting in April 2012.
English oak and Turner's oak at Kew
Quercus robur is by far the most numerous species of tree at Kew and well over 1,000 specimens grow around the Gardens.
The hybrid Quercus × turneri (Turner's oak) can also be seen at Kew – a cross between the English oak (Quercus robur) and the holm oak (Quercus ilex). The semi-deciduous Turner's oak has been growing at Kew since the 18th century and was raised by Mr Turner (a nurseryman from Essex) in 1783. It was then planted in the original five-acre arboretum at Kew in 1798, where it stands today by the Princess of Wales Conservatory.
Compared to some oaks, Quercus × turneri is relatively small, growing to around 17 m. It generally develops a low, domed shape, with its twisting, horizontal branches sometimes barely above the ground. This gives an open feel to the tree, and creates a perfect opportunity for the enthusiastic tree climber!
Although the leaves of this hybrid are similar in colour to those of other oaks, they have a different shape. Instead of the familiar rounded leaf lobes of the English oak, those of Turner's oak form almost sharp, forward-pointing teeth.
The hurricane-like storm which ravaged the south of England, as well as parts of northern France, on the night of 15 October 1987 brought wind speeds of up to 185 km/h. It caused extensive damage to buildings and blew down around 15 million trees.
At Kew the entire root plate of the historic Turner's oak lifted, before settling back into the ground. Far from damaging the tree, this upheaval seemed to rejuvenate it. Previously it had been showing signs of stress and decline due to compaction of the root plate where so many people had taken shelter beneath the tree in the past. Indeed, Kew have learned from this and now routinely de-compact the root plates of the mature trees in the arboretum.
Kennedy, C. E. J. & Southwood, T. R. E. (1984). The number of species of insects associated with British trees: a re-analysis. Journal of Animal Ecology 53: 455–478.
Logan, W. B. (2005). Oak: the Frame of Civilization. W.W. Norton, New York/London.
Tansley, A. G. (1952). Oaks and Oak Woods. Methuen, London
Kew Science Editor: William Milliken
Kew contributors: Tony Hall
Copyediting: Emma Tredwell
Although every effort has been taken to ensure that the information contained in these pages is reliable and complete, notes on hazards, edibility and suchlike included here are recorded information and do not constitute recommendations. No responsibility will be taken for readers’ own actions. | <urn:uuid:87c580be-d1ca-4bef-a122-d04a971985c1> | {
"date": "2014-12-20T17:55:53",
"dump": "CC-MAIN-2014-52",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1418802770071.75/warc/CC-MAIN-20141217075250-00134-ip-10-231-17-201.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9602325558662415,
"score": 3.546875,
"token_count": 2611,
"url": "http://www.kew.org/science-conservation/plants-fungi/quercus-robur-english-oak"
} |
Butterflies are flying colored wing insects that vary in color and pattern from individual to another individual. It has wings covered with overlapping rows of scales. Most of butterflies have developed mechanisms to avoid predators making disguise coloration blending like leaf or bark of the tree. Some releases chemicals as a defense mechanism wherein butterfly evolved to have toxic chemicals. But recent finding due to extreme weather events and trend linked to ongoing anthropogenic climate change species shifts its dynamics. Droughts occur more often in larger spatial scale which has an effect on insects. Generally, drier and warmer climatic conditions have an impact either positive or negative to insect populations. The aim of this research is to address the knowledge gap using multi-decadal dataset of 163 butterfly species. All of this butterflies experienced millennium-scale drought.
Impacts of droughts on Butterflies
To know the faunal dynamics, investigation of phenology, species richness and diversity with its elevation gradient has been conducted. In which linear model used to understand differential sensitivity of butterflies to climate change at low and high elevation. A decade of dataset of 163 butterfly species across elevational gradient in Northern California has been considered. Results showed that a prolonged shift towards spring flight during drought years and change in phenology is evident across elevations. It also happened that the total flight window expanded at lower elevations while at higher elevation shifted and compressed. This leads the notion that fewer overall flight days at higher sites.
The millennium drought in California created across site with elevation-specific changes in flight windows and species richness. This resiliency reveals that lowest elevations are less detrimental than biotic-abiotic association at higher elevations. Most of the researchers hypothesized a mismatch between trophic levels as a result of climate change. But, results of butterflies from low elevation would suggest that at consumer trophic level need not always have negative impacts. Additionally, species at lowest elevations have access to agricultural lands though irrigation does not correlate the population dynamics during drought. Thus, there is a possibility that low elevation population buffered by irrigated crops or agricultural margin during drought.
Indeed, that at high elevation butterflies declined in number and become sensitive to dry years with warmer temperatures. Contrary to the theory that mountains offer microclimatic refugia and adapt species for climatic changes. It has been known that high latitude environments are warming faster with negative consequences to several species. But positive or have a neutral effect for other species. Consequently, this research suggests more thorough investigation about organismal responses to extreme weather. As well as on the extent wherein different habitat type may or may not buffer species populations against climate change.
Source: Prepared by Joan Tura from Springer BMC Climate Changes Responses
Volume 5:3 26 January 2018 | <urn:uuid:4ed3a773-34c5-486a-bbc0-51b021e3592e> | {
"date": "2018-12-12T20:44:16",
"dump": "CC-MAIN-2018-51",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376824119.26/warc/CC-MAIN-20181212203335-20181212224835-00096.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9432355165481567,
"score": 4.21875,
"token_count": 556,
"url": "https://www.biology-online.org/tag/extreme-weather/"
} |
April 27th, 2009
Members of the University at Albany Community:
As many of you are aware the White House has declared a public health emergency due to concerns of a new influenza strain originating from pigs – referred to as the “Swine Flu.” This virus is a new strain of the Influenza A virus which causes typical flu-like symptoms.
In Mexico, at least 1,400 people have been clearly diagnosed with this virus, though it is believed many more may have been infected but were not diagnosed. A number of people have died.
There have been 20 confirmed cases of the Swine Flu in the U.S. In none of these cases did the infected individual have anything more than a mild illness. None of these patients required any treatment other than the routine treatment for seasonal (regular) influenza. None were hospitalized.
Confirmed cases of Swine Flu have been reported in Mexico, New York City, Ohio, Texas, Kansas and throughout Canada.
Please be advised that anyone with a history of travel to these areas in the past seven to ten days with symptoms of a respiratory illness may be infected with Swine Flu and should be tested. The University’s Health Center is able to collect specimens for testing which are then sent to an outside lab for analysis and will provide further advice and treatment as warranted.
The symptoms of the Swine Flu are exactly the same as a typical flu and include sore throat, nasal congestion, cough, fever and muscle aches. Even the mildest form of the flu, including Swine Flu, will result in symptoms similar to a simple cold with the possibility of fever and muscle aches in severe forms.
At this point, the University is monitoring the situation with the assistance of the Albany County Department of Health, the New York State Department of Health, the Centers for Disease Control and the World Health Organization.
As with all respiratory infections the best way to minimize risk of illness is by avoiding the virus. The flu is spread when an infected person coughs or sneezes and other people breathe in the infected air droplets.
To avoid catching or spreading the flu, you are encouraged to take the following steps:
- Avoid close contact with people who are sick.
When you are sick, keep your distance from others (minimum three to six feet) to protect them from getting sick too.
- Seek medical attention when you are sick.
This will help prevent others from catching your illness.
- Cover your mouth and nose with a tissue when coughing or sneezing.
It may prevent those around you from getting sick. Coughing or sneezing into the bend of your elbow, when tissues are not available, directs the cough/sneeze downward and is helpful.
- Clean your hands often.
Washing your hands often will help protect you from germs. Twenty seconds of hand washing with hot water and soap or use of a disinfectant hand lotion is recommended.
- Avoid touching your eyes, nose or mouth.
Germs are often spread when a person touches something that is contaminated with germs and then touches his or her eyes, nose, or mouth.
Information from the Center for Disease Control on Swine Flu is available at: http://www.cdc.gov/swineflu/investigation.htm
I will update you should additional information become available. If you have any questions not answered by the above resources, please send them to [email protected].
We will do our best to answer your questions in a timely manner.
Peter Vellis, DO
University Health Center | <urn:uuid:4801065e-0ada-4e02-ab72-d4184249a83e> | {
"date": "2016-06-27T18:36:13",
"dump": "CC-MAIN-2016-26",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783396106.25/warc/CC-MAIN-20160624154956-00180-ip-10-164-35-72.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9560803174972534,
"score": 3.625,
"token_count": 746,
"url": "http://www.albany.edu/h1n1/april27-09.shtml"
} |
# Which counting numbers are also whole numbers?
A number is a mathematical value used for measuring, counting, and performing different arithmetic calculations such as addition, subtraction, division, multiplication. A number system is characterized as an arrangement of writing to express numbers. It is the mathematical notation for addressing numbers. It provides a unique representation of every number. Numbers have different categories such as natural numbers, Whole numbers, Integers, Rational Numbers, Irrational numbers, Real numbers, Complex Numbers. Examples are 1, 50, -50, 0.25, 8.9999 etc.
### Natural Numbers
Counting Numbers are called Natural Numbers. These are a part of the number system, including all the positive integers from 1 to infinity. They are called counting numbers because they do not include zero or negative numbers. Examples of natural numbers are 1, 2 , 9 ,1023 etc.
A whole number is a set of numbers that includes all positive integers and 0. These are a part of real numbers Which do not include fractions, decimals, or negative numbers. Natural Numbers along with 0 are called whole numbers. Examples of whole numbers are 0, 1, 2 , 9 ,1023 etc.
### Which counting numbers are also whole numbers?
All the counting Numbers such as 1, 2, 3, 4 . . . . . are whole numbers. But all whole numbers are not counting numbers. Whole Numbers excluding 0 can be called counting numbers. In order to put it in the simplest way, all natural numbers are counting numbers and all whole numbers are natural numbers except 0, therefore, apart from 0, all whole numbers can be considered as counting numbers.
N = W – {0}
### Sample Problems
Question 1: Choose Natural Numbers from the following set of numbers,
11, 0, 25, 0.387, 1.20, 2506.
All counting numbers are natural numbers, hence 11, 25, 2506 are natural numbers here.
Question 2: Choose whole Numbers from the following set of numbers?
1.1, 0, 125, 0.387, 1.20, 256, 69
All counting numbers along with 0 are whole numbers, hence 0, 125, 256, 69 are whole numbers here.
Question 3: Can all decimal numbers be counting numbers?
Numbers including decimals are not counting numbers.
Question 4: Which one of the given numbers is a natural number?
0, 2.3, 22/7, 9.87, 12, 11.2 | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00041.warc.gz | null |
KS3: On entering the school students are able to study French or Spanish. They will study this in two periods per week.
Classes are set based on mixed ability and the teachers ensure that work is set at an appropriately challenging level for the ability of each student. All students should by a bilingual dictionary for the language they will be studying as this will enable each students to produce at home a better independent learning homework.
The objectives of the Year 7 languages course are to enable all students to have a basic vocabulary and knowledge of grammar in order to be able to use familiar language in familiar contexts up to National Curriculum Level 4a on a series of topics.
Year 8 and Year 9 follow similar lesson patterns and the aim is for students to consolidate their knowledge of grammar to be able to independently manipulate language to create paragraphs in the future and past tenses. The most able will also include complex features and tenses.
The teaching of Languages at GAC is very good and the department is staffed by very enthusiastic, experienced linguists. We devise our own Schemes of Work designed to make language learning enjoyable and engaging, whilst developing a strong linguistic foundation in knowledge of grammar. We carry out projects including researching areas and traditions of the French and Spanish speaking worlds, film analysis and story writing alongside the traditional topic areas of Home and Town, Family, Hobbies, Healthy Living.
GAC believes in the importance of studying languages to enhance cultural awareness, to extend literacy in our native language and to build communication and problem solving skills.
A very small number of students who would benefit from baseline Literacy and Numeracy support are withdrawn from MFL classes but this is at the discretion of the school. Students who make a real effort and work well in class and at home achieve very strong results.
GCSE: We currently use the Edexcel examination board and build on the sure vocabulary and grammar foundation built at Key Stage 3. The GCSE course is begun in the final term of Year 9 and after students have made their options. We aim that all students can produce short passages from memory, accurately applying grammar rules and knowledge about a range of key subjects laid down by the examination boards.
Students studying a language at GCSE level will have to produce two pieces of written coursework and two spoken pieces. A rough outline is given below and letters will be sent detailing the dates and titles for full GCSE assessment pieces in advance of each test taking place. Information regarding the new GCSE structure will be added at a later stage.
As at Key Stage 3, the department devises its own Schemes of Work for Key Stage 4 but supplements this heavily with the examination board endorsed Textbooks ‘Edexcel GCSE for French’ and ‘Edexcel GCSE for Spanish’. | <urn:uuid:4faff41d-6902-4bd8-b187-23de583c2664> | {
"date": "2018-05-20T11:50:04",
"dump": "CC-MAIN-2018-22",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794863410.22/warc/CC-MAIN-20180520112233-20180520132233-00416.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9544017314910889,
"score": 3.65625,
"token_count": 571,
"url": "https://coventry.graceacademy.org.uk/subjects/gcse/mfl/"
} |
Yesterday I mentioned that the IceCube Neutrino observatory doesn’t detect neutrinos directly, but instead detects the flash of light that occurs after a neutrino collides with a water molecule. As you can imagine, the light emitted with a single neutrino collides with a single molecule is pretty faint. So how do we observe such faint light? We use a photomultiplier tube (or PMT), some of which are so sensitive they can detect the light of a single photon.
A PMT is based on something known as the photoelectric effect. Early observations of the photoelectric effect date back to the late 1800s, but it was made famous by Einstein’s 1905 paper, for which he won the Nobel prize. The basic idea of the photoelectric effect is that a negatively charged metallic plate can emit electrons when ultraviolet light is shined on it. Einstein demonstrated that this was due to photons striking the electrons and knocking them free of the metal.
Of course a single photon can only knock out one electron, which would be very difficult to observe. So we need to use another effect known as secondary emission. In this case, fast moving electron striking a metal plate can knock free several slower moving electrons, kind of how a bowling ball can knock over several pins.
In a PMT, a photon strikes a negatively charged metal plate called the photocathode, which releases the first electron. This is then accelerated toward another metal plate (known as a dynode) which it strikes, releasing a few electrons. These are then accelerated to another dynode, releasing more electrons. The result is a kind of cascade effect (seen in the image above), where a single photon causes an avalanche of electrons to reach the final metal plate (the anode) where an electric current is created that we can measure easily.
So a photomultiplier tube allows faint light to trigger an electric current strong enough for us to measure. Different types of PMTs have different light sensitivities, depending on the type of photocathode they use. Some are sensitive to visible wavelengths, while others are sensitive to infrared or ultraviolet.
Under the right conditions, we really can detect the light of a single photon. | <urn:uuid:1b86f8e9-830f-4089-8735-260c94d3c34d> | {
"date": "2021-09-24T11:14:48",
"dump": "CC-MAIN-2021-39",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057524.58/warc/CC-MAIN-20210924110455-20210924140455-00697.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9429926872253418,
"score": 3.9375,
"token_count": 457,
"url": "https://briankoberlein.com/blog/yop/"
} |
There are three types of veins in the human body: superficial veins, perforating veins and deep veins. Deep veins are connected to the vena cava, which is the largest vein in the body and is directly connected to the lungs and heart. Deep vein thrombosis (DVT) occurs when a blood clot forms in a deep vein. DVT is most likely to occur in the calf, pelvis or thigh, though it can occur in other areas such as the chest or arm.
While half of patients with DVT will not experience symptoms, the other half is likely to experience:
- Sudden swelling or pain
- Skin warmth
- Leg pain that worsens during walking or standing
- Red or blue skin discoloration
DVT is a particularly dangerous condition because it can result in a pulmonary embolism. This occurs when a blood clot in one of the deep vein breaks away and travels to the lungs. Once there, the clot can block blood flow, straining the lungs and heart. A pulmonary embolism is considered an emergency and can quickly be fatal if the clot is large enough.
The most common symptom of a pulmonary embolism is shortness of breath. Other symptoms may include:
- Chest pain that might reach into the shoulder, arm, jaw and neck
- Rapid breathing or heartbeat
- Restlessness and anxiety
- Coughing up blood
- Lightheadedness and fainting
DVT can result from a problem with the body's clotting system. In this case, a small clot causes inflammation that then allows larger clots to form. Additionally, poor blood flow in the leg veins increases the risk for DVT, especially during periods of prolonged motionlessness. Because of this, DVT has been called Economy Class Syndrome, since clots may form during long airplane trips where there is limited leg room. The truth, however, is that long flights rarely cause DVT. Most often, the condition is found in hospitalized, bedbound patients.
Other factors increasing the risk for DVT include:
- History of stroke, heart attack or congestive heart failure
- Inflammatory bowel syndrome | <urn:uuid:ca81e415-eec4-4622-8326-e08d8d7a99fa> | {
"date": "2018-06-20T15:12:06",
"dump": "CC-MAIN-2018-26",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267863650.42/warc/CC-MAIN-20180620143814-20180620163814-00297.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9422197937965393,
"score": 3.796875,
"token_count": 443,
"url": "https://heart.templehealth.org/patient-care/conditions-we-treat/deep-vein-thrombosis"
} |
# Is f(x)=(1-e^(2x))/(2x-4) increasing or decreasing at x=-1?
Jan 5, 2016
Increasing.
#### Explanation:
Find the sign of the derivative at $x = - 1$.
To find $f ' \left(x\right)$, use the quotient rule.
$f ' \left(x\right) = \frac{\left(2 x - 4\right) \frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) - \left(1 - {e}^{2 x}\right) \frac{d}{\mathrm{dx}} \left(2 x - 4\right)}{2 x - 4} ^ 2$
The derivative of each part:
The first requires the chain rule:$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$
$\frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) = - 2 {e}^{2 x}$
$\frac{d}{\mathrm{dx}} \left(2 x - 4\right) = 2$
Thus,
$f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(- 2 {e}^{2 x}\right) - 2 \left(1 - {e}^{2 x}\right)}{2 x - 4} ^ 2$
$f ' \left(x\right) = \frac{- 4 x {e}^{2 x} + 10 {e}^{2 x} + 2}{2 x - 4} ^ 2$
Find $f ' \left(- 1\right)$ If it's greater than $0$, the function is increasing at that point. If it's less than $0$, the function is decreasing at that point.
$f ' \left(- 1\right) = \frac{- 4 \left(- 2\right) {e}^{-} 2 + 10 {e}^{-} 2 + 2}{- 2 - 4} ^ 2 = \frac{\frac{18}{e} ^ 2 + 2}{36}$
Since this is $> 0$, the function is increasing when $x = - 1$.
graph{(1-e^(2x))/(2x-4) [-5.93, 6.56, -2.596, 3.647]} | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00511.warc.gz | null |
# For what values of x, if any, does f(x) = 1/((12x+3)sin(pi+(6pi)/x) have vertical asymptotes?
Aug 7, 2017
$x = - \frac{1}{4}$ and $x = \frac{6}{k}$ for integer $k$
#### Explanation:
$12 x + 3 = 0$ at $x = - \frac{1}{4}$
sin(pi+(6pi)/x) = 0# where
$\pi + \frac{6 \pi}{x} = k \pi$ for integer $k$
And that happens where $\frac{6 \pi}{x} = k \pi$ for integer $k$
Which is at $x = \frac{6}{k}$ for integer $k$ | crawl-data/CC-MAIN-2022-21/segments/1652663011588.83/warc/CC-MAIN-20220528000300-20220528030300-00246.warc.gz | null |
GMAT Club Forumhttps://gmatclub.com:443/forum/ In the figure above, QRS is a straight line and QR = PR. Ishttps://gmatclub.com/forum/in-the-figure-above-qrs-is-a-straight-line-and-qr-pr-is-143843.html Page 1 of 1
Author: JJ2014 [ 09 Dec 2012, 11:43 ] Post subject: In the figure above, QRS is a straight line and QR = PR. Is Attachment: Screen shot 2012-12-09 at 1.40.06 PM.png [ 19.42 KiB | Viewed 24232 times ] In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?(1) Length PQ = Length PR(2) Line TR bisects angle PRS
Author: mikemcgarry [ 09 Dec 2012, 15:58 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is In the figure, QRS is a straight line. QR=PR. Are TR and PQ parallel?1) Length PQ = Length PR2) Line TR bisects angle PRS From the prompt, we know that triangle QPR is isosceles, with QR = PR. By the Isosceles Triangle theorem, we know that angle Q = angle P. Statement #1 PQ = PR.This is enough to guarantee that triangle QPR is equilateral, but we don't know anything about ray RT, so we have no idea whether that is parallel to anything else. This statement, alone and by itself, is not sufficient. Statement #2 Line TR bisects angle PRSA fascinating statement. Let's think about this. We already know angle Q = angle P. Call the measure of that angle M. Look angle QRP --- call the measure of that angle K. Clearly, within triangle QPR, M + M + K = 180, but Euclid's famous theorem. Now, look at angle PRS. This is what is known as the "exterior angle" of a triangle, and there's a special theorem about this. The Remote Interior Angle Theorem:If the exterior angle of a triangle is adjacent to the angle of the triangle, then the measure of the exterior angle is equal to the sum of the two "remote" interior angles of the triangle --- that is, the two angles of the triangle which the exterior angle is not touching. If you think about this, it has to be true, because(angle Q) + (angle P) + (angle PRQ) = 180, because they're the three angles in a triangle(angle PRQ) + (exterior angle PRS) = 180, because they make a straight lineSubtract (angle PRQ) from both sides of both equations:(angle Q) + (angle P) = 180 - (angle PRQ) (exterior angle PRS) = 180 - (angle PRQ) Since the two things on the left are equal to the same thing, they are equal to each other. (angle Q) + (angle P) = (exterior angle PRS) Now, going back to the letters we were using ---- if (angle Q) = (angle P) = M, this means (exterior angle PRS) = 2M. If we bisect exterior angle PRS, each piece will have a measure of M. Thus, (angle PRT) = (angle TRS) = MWell, now we know that (angle Q) = (angle TRS) = M. If corresponding angles are congruent, then the lines must be parallel. PQ must be parallel to TR. This statement, alone and by itself, is sufficient to answer the prompt question. Answer = BDoes all this make sense?Mike
Author: shanmugamgsn [ 10 Dec 2012, 17:53 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is Hi Mike well explained,I do have a doubt ! now we know that (angle Q) = (angle TRS) = M. If corresponding angles are congruent, then the lines must be parallel. PQ must be parallel to TR. This statement, alone and by itself, is sufficient to answer the prompt question. Correspoding angle means the angle made on its side? angle Q ==> side PQangle TRS ==> side TRvery basic question but i understood complex things
Author: mikemcgarry [ 10 Dec 2012, 18:09 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is shanmugamgsn wrote:I do have a doubt ! Corresponding angle means the angle made on its side? No. "Corresponding angles" is a technical term from Euclidean Geometry. It's the name of a particular pair of angles formed when a transversal crosses a pair of parallel lines.Attachment: parallel line diagram.JPG [ 22.96 KiB | Viewed 23480 times ] The following pairs are corresponding angles1 & 52 & 63 & 74 & 8Corresponding angles are congruent if and only if the lines are parallel. The following pairs are alternate interior angles3 & 64 & 5Alternate interior angles are congruent if and only if the lines are parallelThe following pairs are alternate exterior angles1 & 82 & 7Alternate exterior angles are congruent if and only if the lines are parallelThe following pairs are same side interior angles3 & 54 & 6Same side interior angles are supplementary if and only if the lines are parallelThe following pairs are same side exterior angles1 & 72 & 8Same side exterior angles are supplementary if and only if the lines are parallelThose are all the names relating pairing an angle at one vertex with an angle at the other vertex, when a transversal intersects a pair of parallel lines. Does all this make sense?Mike
Author: priyamne [ 11 Dec 2012, 04:40 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is JJ2014 wrote:Attachment:Screen shot 2012-12-09 at 1.40.06 PM.pngIn the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?(1) Length PQ = Length PR(2) Line TR bisects angle PRSAns: For lines TR and PQ to be parallel angle PQR= angle TRS. From statement 1 we get angle PQR=x=60 but nothing about angle TRS.From statement 2 we get PRQ=180-2X , therefore PRS=180-(180-2x)=2x and TR bisects it so angle TRS=x which is equal to PQR. Therefore the answer is (B).
Author: jlgdr [ 22 Apr 2014, 08:36 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is Let's solve. So we need to know if TR is parallel to PQ. Now then, let's hit the first statement. We are told that PQ=QR. Now we know that PQR is an equilateral triangle but still no info on TR. Therefore, insufficient. Statement 2, we have that TR bisects PRS. Now let's see. So we know that QR=PR from the question stem. Hence angle QRP is 180-2x, 'x' being the angles P and Q respecively. Therefore angle PRS will be 2x since QRS is a straight line with total measure of 180 degrees. Now if TRS bisects then angle TRS is x only. Which means that the angles Q and R are equal and thus PQ // TR. B standsCheers!J
Author: BrainLab [ 22 Dec 2015, 04:29 ]
Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is
JJ2014 wrote:
Attachment:
The attachment Screen shot 2012-12-09 at 1.40.06 PM.png is no longer available
In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?
(1) Length PQ = Length PR
(2) Line TR bisects angle PRS
(1) this tells us that all 3 sides are equal = Equilateral Triangle. Not sufficient, as there is no info about TR
(2) Now look at the attachment, I'll use the property of the "Exterior Angles of a Triangle" Angle Y is the sum of the angles on the both sides of TR
Exterior angle of a triangle is equal to the sum of the opposite interior angles
Angle Y=2X and if TR bisects angle Y it means that Angle TRS = $$\frac{Y}{2}=\frac{2X}{2}=X$$ and when this two angles of to different triangles are equal then QP and TR are parallel. | crawl-data/CC-MAIN-2023-14/segments/1679296948708.2/warc/CC-MAIN-20230327220742-20230328010742-00671.warc.gz | null |
# How to Convert 10 MG to ML?
Milligrams (mg) and milliliters (ml) are two common units of measurement used in medicine and science. Milligrams are used to measure the mass of a substance, while milliliters are used to measure the volume of a liquid.
When working with medications, it’s important to understand the conversion between milligrams and milliliters, as many medicines are prescribed in milligrams but administered in milliliters.
The conversion between milligrams and milliliters can be done using the conversion factor of 1 mg = 0.001 ml. This means that one milligram is equivalent to 0.001 milliliters. To convert milligrams to milliliters, multiply the number by 0.001.
### Let’s understand 10 MG to ML conversion:
For example, to convert 10 milligrams to milliliters, you would perform the following calculation:
1 mg = 0.001 ml
10 mg = 10 * 0.001 ml = 0.01 ml
So, 10 mg is equivalent to 0.01 ml
## Know more about how to Convert 10 MG to ML?
It’s important to note that the conversion factor between MG to ML will vary depending on the measured substance, as different meanings have different densities. However, the conversion factor of 1 mg = 0.001 ml can be used as a general guideline in most cases.
In conclusion, converting 10 mg to ml is a simple mathematical calculation that can be done using the conversion factor of 1 mg = 0.001 ml. Understanding the conversion between 10 mg to ml is vital in many fields, including medicine and science, as it allows for the accurate measurement and administration of substances.
### FAQs:
• #### What is the conversion factor from milligrams to milliliters?
The conversion factor from milligrams (mg) to milliliters (ml) is 1 mg = 0.001 ml.
• #### How do I convert 10 MG to ML?
To convert 10 milligrams to milliliters, multiply 10 by 0.001: 10 mg * 0.001 = 0.01 ml. So, 10 mg is equal to 0.01 ml.
• #### Can 1 mg = 0.001 ml conversion factor be used for all substances?
The conversion factor of 1 mg = 0.001 ml is a general guideline and can be used for many substances, but it may not be accurate for all senses due to differences in densities.
• #### Why is it important to understand the conversion between milligrams and milliliters?
It is important to understand the conversion between milligrams and milliliters in fields such as medicine and science, as many medications are prescribed in milligrams but administered in milliliters. Accurate measurement and administration of substances are crucial in these fields.
• #### What happens if I use the wrong conversion factor when converting milligrams to milliliters?
Using the wrong conversion factor when converting milligrams to milliliters can result in an incorrect measurement and administration of a substance. This can lead to adverse effects and should be avoided. It is important to use the correct conversion factor, or to consult a professional, to ensure accurate measurements. | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00321.warc.gz | null |
The word shock can be used in a range of ways, but when used in a first aid context it describes a physical condition that results from a loss of circulating body fluid. It should not be confused with emotional shock that might occur, say, when a person has received bad news (although the external signs are very similar).
What happens in cases of shock
A severe loss of body fluid will lead to a drop in blood pressure. Eventually the blood's circulation around the body will deteriorate and the remaining blood flow will be directed to the vital organs such as the brain. Blood will therefore be directed away from the outer areas of the body, so the casualty will appear paler than previously and the skin will feel cold and clammy. As blood flow slows, so does the amount of oxygen reaching the brain.
The casualty may appear to be confused, weak and dizzy, and may eventually deteriorate into unconsciousness. To try to compensate for this lack of oxygen, the heart and breathing rates both speed up, gradually becoming weaker, and may eventually cease.
Potential causes of shock include: severe internal or external bleeding; burns; severe vomiting and diarrhoea, especially in children and the elderly; problems with the heart.
First Aid Treatment
Keep the casualty warm but do not allow her to get overheated. If you are outside, try to get something underneath the casualty if you can do so easily. Wrap blankets and coats around her, paying particular attention to the head, through which much body heat is lost.
Maintain a careful eye on the casualty's airway and be prepared to turn her into the recovery position if necessary, or even to resuscitate if breathing stops. Try to clear back bystanders and loosen tight clothing to allow maximum air to the casualty.
Keep the casualty still and preferably sitting or lying down. If the casualty is very giddy, lay her down with her legs raised to ensure that maximum blood and therefore maximum oxygen is sent to the brain.
Reassure the casualty but keep your comments realistic. Do not say that everything is going to be fine when it is obvious that there is something seriously wrong. Let the casualty know that everything that can be done is being done and that help has been called for. If she has other worries then try to resolve these.
Treat the cause of the shock and aim to prevent further fluid loss.
Ensure that appropriate medical help is on the way.
Signs and symptoms
Shock kills, so it is vital that you can recognise these signs and symptoms. With internal bleeding in particular, shock can occur some time after an accident, so if a person with a history of injury starts to display these symptoms coupled with any of the symptoms of internal bleeding, advise her to seek urgent medical attention, or take or send her to hospital.
(c) Health-care-clinic.org All rights reserved
Disclaimer: Health-care-clinic.org website is designed for educational purposes only. It is not intended to treat, diagnose, cure, or prevent any disease. Always take the advice of professional health care for specific medical advice, diagnoses, and treatment. We will not be liable for any complications, or other medical accidents arising from the use of any information on this web site. Please note that medical information is constantly changing. Therefore some information may be out of date. | <urn:uuid:db3e04d9-7d61-4074-b1d7-9542f970cce1> | {
"date": "2017-03-26T22:50:05",
"dump": "CC-MAIN-2017-13",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189313.82/warc/CC-MAIN-20170322212949-00472-ip-10-233-31-227.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9405887722969055,
"score": 4.1875,
"token_count": 684,
"url": "http://www.health-care-clinic.org/first-aid/shock.html"
} |
The use of solar power became very popular in the 1970s, but has fallen in and out of favour since depending on the potential savings when compared with fossil-fuel energy costs.
The photovoltaic effect is when photo cells convert sunlight directly into electricity - this has been used for sometime to power certain calculators, for example. Photovoltaic cells (PV's) can be used as roof tiles. They cover the roof of a house and take advantage of the light coming from the Sun. This is trapped by the cell and turned into electricity.
Another way to take advantage of the energy from the sun is to design buildings so they can collect the heat. They do this by designing the building sensibly and facing it in a way where it can use the sun to the maximum benefit. Large glass windows help with this, especially during the winter when the Sun is very low. In the summer, balconies and trees protect the building from getting too much heat.
More common method in the UK is using the benefits of the sun to heat our water pipes. Painting the thin pipes black and putting them in a 'greenhouse' type insulator can heat our water supply and therefore reduce the cost of using electricity to heat it.
As well as the fact that energy from the Sun is readily available, there are many other benefits. By locating photovoltaic cells on top of houses, no extra land space is needed and they can also be situated in urban areas. The technology now needed is about 90% cheaper than it was in the 1970s. Houses with solar roof tiles can in fact generate more electricity than is needed at certain times in the day, and can sell this back to local electricity companies.
However the UK is behind many other countries in Europe and the rest of the World when it comes to using solar power technologies. In Japan and the USA, billions has been spent on developing PV over a number of years, and more recently, Germany has started to push lots of money into the development of it for projects there. China has pledged to throw its economic might behind a national solar plan and the government has committed to raising its share of renewable energy to 6%, from the current 1.5%.
Alternative energy sources: | <urn:uuid:1b834544-5924-4ba6-a037-b575e7e6da54> | {
"date": "2015-08-29T21:14:53",
"dump": "CC-MAIN-2015-35",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644064538.31/warc/CC-MAIN-20150827025424-00291-ip-10-171-96-226.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9632356762886047,
"score": 3.6875,
"token_count": 455,
"url": "http://www.bbc.co.uk/climate/adaptation/solar_power.shtml"
} |
Trigonometric Functions on Any Angle
1 / 28
# Trigonometric Functions on Any Angle - PowerPoint PPT Presentation
Trigonometric Functions on Any Angle. Section 4.4. Objectives. Determine the quadrant in which the terminal side of an angle occurs. Find the reference angle of a given angle.
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
## Trigonometric Functions on Any Angle
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
### Trigonometric Functions on Any Angle
Section 4.4
Objectives
• Determine the quadrant in which the terminal side of an angle occurs.
• Find the reference angle of a given angle.
• Determine the sine, cosine, tangent, cotangent, secant, and cosecant values of an angle given one of the sine, cosine, tangent, cotangent, secant, or cosecant value of the angle.
Vocabulary
• reference angle
• sine of an angle
• cosine of an angle
• terminal side of an angle
• initial side of an angle
• tangent of an angle
• cotangent of an angle
• secant of an angle
• cosecant of an angle
Reference Angle
A reference angle is the smallest distance between the terminal side of an angle and the x-axis.
All reference angles will be between 0 and π/2.
continued on next slide
Reference Angle
There is a straight-forward process for finding reference angles.
Step 1 – Find the angle coterminal to the given angle that is between 0 and 2π.
continued on next slide
Reference Angle
There is a straight-forward process for finding reference angles.
Step 2 – Determine the quadrant in which the terminal side of the angle falls.
continued on next slide
Reference Angle
There is a straight-forward process for finding reference angles.
Step 3 – Calculate the reference angle using the quadrant-specific directions.
continued on next slide
θ
Reference Angle
For quadrant I, the shortest distance from the terminal side of the angle to the x-axis is the same as the angle θ.
Thus
where the reference angle is
continued on next slide
θ
Reference Angle
For quadrant II, the shortest distance from the terminal side of the angle to the x-axis is shown in blue. This is the rest of the distance from the terminal side of the angle to π.
Thus
This distance is the reference angle.
Note: Here put subtracted the angle from π since the angle was smaller than π. This gave us the positive reference angle. If we had subtracted π from the angle, we would have needed to take the absolute value of the answer.
continued on next slide
θ
Reference Angle
This distance is the reference angle.
For quadrant III, the shortest distance from the terminal side of the angle to the x-axis is shown in blue. This is the distance from the π to the terminal side of the angle.
Thus
continued on next slide
θ
Reference Angle
Note: Here put subtracted π from the angle since the angle was larger than π. This gave us the positive reference angle. If we had subtracted the angle from π, we would have needed to take the absolute value of the answer.
continued on next slide
θ
Reference Angle
This distance is the reference angle.
For quadrant IV, the shortest distance from the terminal side of the angle to the x-axis is shown in blue. This is the rest of the distance from the terminal side of the angle to 2π.
Thus
continued on next slide
θ
Reference Angle
Note: Here put subtracted the angle from 2π since the angle was smaller than 2π. This gave us the positive reference angle. If we had subtracted 2π from the angle, we would have needed to take the absolute value of the answer.
continued on next slide
Reference Angle Summary
Step 1 – Find the angle coterminal to the given angle that is between 0 and 2π.
Step 2 – Determine the quadrant in which the terminal side of the angle falls.
Step 3 – Calculate the reference angle using the quadrant-specific directions indicated to the right.
In which quadrant is the angle?
To find out what quadrant θ is in, we need to determine which direction to go and how far. Since the angle is negative, we need to go in the clockwise direction. The distance we need to go is one whole π and 1/6 of a π further.
This red part is approximately 1/6 of a π further.
This blue part is one whole π in the clockwise direction
Now that we have drawn the angle, we can see that the angle θ is in quadrant II.
continued on next slide
What is the reference angle, , for the angle
?
Using our summary for finding a reference angle, we start by finding an angle coterminal to θ that is between 0 and 2π. Thus we need to start by adding 2π to our angle.
continued on next slide
What is the reference angle, , for the angle
?
The next step is to determine what quadrant our coterminal angle is in. We really already did this in the first question of the problem. Coterminal angles always terminate in the same quadrant. Thus our coterminal angle is in quadrant II.
continued on next slide
Finally we need to use the quadrant II directions for finding the reference angle.
Thus the reference angle is
Evaluate each of the following for .
To solve a problem like this, we want to start by finding the reference angle for θ.
Since our angle is bigger than 2π, we need to subtract 2π to find the coterminal angle that is between 0 and 2π.
continued on next slide
Our next step is to figure out
what quadrant is in. You can
see from the picture that we are in quadrant II.
Evaluate each of the following for .
To find the reference angle for an angle in quadrant II, we subtract the coterminal angle from π.
This will give us a reference angle of
continued on next slide
We will now use the basic trigonometric function values for
The only thing that we will need to change might be the signs of the basic values. Remember that the sign of the cosine and tangent functions will be negative in quadrant II. The sign of the sine will still be positive in quadrant II.
Evaluate each of the following for .
continued on next slide
Evaluate each of the following for .
Once again, we will use our reference angle to determine the basic trigonometric function value. The only difference between the basic value and the value for our angle may be the sign.
continued on next slide
Evaluate each of the following for .
Once again, we will use our reference angle to determine the basic trigonometric function value. The only difference between the basic value and the value for our angle may be the sign.
continued on next slide
Evaluate each of the following for .
Once again, we will use our reference angle to determine the basic trigonometric function value. The only difference between the basic value and the value for our angle may be the sign. | crawl-data/CC-MAIN-2018-51/segments/1544376823817.62/warc/CC-MAIN-20181212091014-20181212112514-00323.warc.gz | null |
Analysis of literature: “Black Art” By Amiri Baraka
There are much articles and literature in the American history that speaks of racism, racial identity, and race as a concept. “The Black Arts” by Amiri Baraka is a unique piece of literature that interconnects art with racial identity. The poem is well connected with the sensitivity of racism among Black Africans and the association with different forms of art. The Black Arts Movement of the 1960's which is also commonly recognized as BAM is the artistic aspect of the Black Art Movement of the mid-nineteenth century. The Black Art Movement gained popularity as a movement that promoted arts and craft for the Black community and allowed them participation to the full extent. The term gained more momentum and significance after the launch of free jazz in the year 1960s and then in 1980s for the introduction of wireless connections with musicians and the development of new musical instruments at large.
It was the period from 1990s onwards that the world witnessed significant changes on the musical front. The initial style of this music resembled the Anglo-African style of music by the name Soul. It was a rugged, tough and rockfish style of music that gained massive appeal from the music lovers. The initial developments on the musical front also included the launch of CDs by the New Yorkers live music. The emphasis of these steps was aimed at encouraging the promotion of music and other forms of art without discriminating on the basis of color and creed. These steps were also included in the Pan-Africanism movement that overlapped with the Black Arts Movement. The similarities of time and space between the followers of each movement allowed for sharing their cultural beliefs and other perspectives with each other. They also shared common intellectual spaces that contributed towards promoting intellectual values in their society in addition to harmonization of their cultural traditions. The women community also participated in this cultural exchange and demonstrated their unique skills and talents at various platforms under this movement. The women involved in each of these movements competed against each other in various art exhibitions, theatres, stage shows, and on television and radio mediums. These female leaders read literature debating the existence of a definable and tangible black existence and raised pertinent questions concerning the impact of gender on this black identity.
The Black Art Movement is the most significant revolution ... | <urn:uuid:8b8c86bc-1dc9-41f6-8c60-25ce1c0f1298> | {
"date": "2023-09-26T23:13:38",
"dump": "CC-MAIN-2023-40",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00056.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9641838073730469,
"score": 3.640625,
"token_count": 464,
"url": "https://www.researchomatic.com/Analysis-Of-Literature-black-Art-By-Amiri-Baraka-83835.html"
} |
Courtesy of NASA
Viking was the culmination of a series of missions to explore the planet Mars; they began in 1964 with Mariner 4, and continued with the Mariner 6 and 7 flybys in 1969, and the Mariner 9 orbital mission in 1971 and 1972.
Viking was designed to orbit Mars and to land and operate on the planet's surface. Two identical spacecraft, each consisting of a lander and an orbiter, were built.
NASA's Langley Research Center in Hampton, Virginia, had management responsibility for the Viking project from its inception in 1968 until April 1, 1978, when the Jet Propulsion Laboratory assumed the task. Martin Marietta Aerospace in Denver, Colorado, developed the landers. NASA's Lewis Research Center in Cleveland, Ohio, had responsibility for the Titan- Centaur launch vehicles. JPL's initial assignment was development of the orbiters, tracking and data acquisition, and the Mission Control and Computing Center.
NASA launched both spacecraft from Cape Canaveral, Florida -- Viking 1 on August 20, 1975, and Viking 2 on September 9, 1975. The landers were sterilized before launch to prevent contamination of Mars with organisms from Earth. The spacecraft spent nearly a year cruising to Mars. Viking 1 reached Mars orbit on June 19, 1976; Viking 2 began orbiting Mars on August 7, 1976.
After studying orbiter photos, the Viking site certification team considered the original landing site for Viking 1 unsafe. The team examined nearby sites, and Viking 1 landed on July 20, 1976, on the western slope of Chryse Planitia (the Plains of Gold) at 22.3°N latitude, 48.0° longitude.
The site certification team also decided the planned landing site for Viking 2 was unsafe after it examined high-resolution photos. Certification of a new landing site took place in time for a Mars landing on September 3, 1976, at Utopia Planitia, at 47.7°N latitude, and 225.8° longitude.
The Viking mission was planned to continue for 90 days after landing. Each orbiter and lander operated far beyond its design lifetime. Viking Orbiter 1 exceeded four years of active flight operations in Mars orbit.
The Viking project's primary mission ended November 15, 1976, 11 days before Mars's superior conjunction (its passage behind the Sun). After conjunction, in mid-December 1976, controllers re-established telemetry and command operations, and began extended-mission operations.
The first spacecraft to cease functioning was Viking Orbiter 2 on July 25, 1978; the spacecraft had used all the gas in its attitude-control system, which kept the craft's solar panels pointed at the Sun to power the orbiter. When the spacecraft drifted off the Sun line, the controllers at JPL sent commands to shut off power to Viking Orbiter 2's transmitter.
Viking Orbiter 1 began to run short of attitude-control gas in 1978, but through careful planning to conserve the remaining supply, engineers found it possible to continue acquiring science data at a reduced level for another two years. The gas supply was finally exhausted and Viking Orbiter 1's electrical power was commanded off on August 7, 1980, after 1,489 orbits of Mars.
The last data from Viking Lander 2 arrived at Earth on April 11, 1980. Lander 1 made its final transmission to Earth Nov. 11, 1982. Controllers at JPL tried unsuccessfully for another six and one-half months to regain contact with Viking Lander 1. The overall mission came to an end May 21, 1983.
With a single exception -- the seismic instruments -- the science instruments acquired more data than expected. The seismometer on Viking Lander 1 would not work after landing, and the seismometer on Viking Lander 2 detected only one event that may have been seismic. Nevertheless, it provided data on wind velocity at the landing site to supplement information from the meteorology experiment, and showed that Mars has very low seismic background.
The three biology experiments discovered unexpected and enigmatic chemical activity in the Martian soil, but provided no clear evidence for the presence of living microorganisms in soil near the landing sites. According to mission biologists, Mars is self-sterilizing. They believe the combination of solar ultraviolet radiation that saturates the surface, the extreme dryness of the soil and the oxidizing nature of the soil chemistry prevent the formation of living organisms in the Martian soil. The question of life on Mars at some time in the distant past remains open.
The landers' gas chromatograph/mass spectrometer instruments found no sign of organic chemistry at either landing site, but they did provide a precise and definitive analysis of the composition of the Martian atmosphere and found previously undetected trace elements. The X-ray fluorescence spectrometers measured elemental composition of the Martian soil.
Viking measured physical and magnetic properties of the soil. As the landers descended toward the surface they also measured composition and physical properties of the Martian upper atmosphere.
The two landers continuously monitored weather at the landing sites. Weather in the Martian midsummer was repetitious, but in other seasons it became variable and more interesting. Cyclic variations appeared in weather patterns (probably the passage of alternating cyclones and anticyclones). Atmospheric temperatures at the southern landing site (Viking Lander 1) were as high as -14°C (7°F) at midday, and the predawn summer temperature was -77°C (-107°F). In contrast, the diurnal temperatures at the northern landing site (Viking Lander 2) during midwinter dust storms varied as little as 4°C (7°F) on some days. The lowest predawn temperature was -120°C (-184°F), about the frost point ofcarbon dioxide. A thin layer of water frost covered the ground around Viking Lander 2 each winter.
Barometric pressure varies at each landing site on a semiannual basis, because carbon dioxide, the major constituent of the atmosphere, freezes out to form an immense polar cap, alternately at each pole. The carbon dioxide forms a great cover of snow and then evaporates again with the coming of spring in each hemisphere. When the southern cap was largest, the mean daily pressure observed by Viking Lander 1 was as low as 6.8 millibars; at other times of the year it was as high as 9.0 millibars. The pressures at the Viking Lander 2 site were 7.3 and 10.8 millibars. (For comparison, the surface pressure on Earth at sea level is about 1,000 millibars.)
Martian winds generally blow more slowly than expected. Scientists had expected them to reach speeds of several hundred miles an hour from observing global dust storms, but neither lander recorded gusts over 120 kilometers (74 miles) an hour, and average velocities were considerably lower. Nevertheless, the orbiters observed more than a dozen small dust storms. During the first southern summer, two global dust storms occurred, about four Earth months apart. Both storms obscured the Sun at the landing sites for a time and hid most of the planet's surface from the orbiters' cameras. The strong winds that caused the storms blew in the southern hemisphere.
Photographs from the landers and orbiters surpassed expectations in quality and quality. The total exceeded 4,500 from the landers and 52,000 from the orbiters. The landers provided the first close-up look at the surface, monitored variations in atmospheric opacity over several Martian years, and determined the mean size of the atmospheric aerosols. The orbiter cameras observed new and often puzzling terrain and provided clearer detail on known features, including some color and stereo observations. Viking's orbiters mapped 97 percent of the Martian surface.
The infrared thermal mappers and the atmospheric water detectors on the orbiters acquired data almost daily, observing the planet at low and high resolution. The massive quantity of data from the two instruments will require considerable time for analysis and understanding of the global meteorology of Mars. Viking also definitively determined that the residual north polar ice cap (that survives the northern summer) is water ice, rather than frozen carbon dioxide (dry ice) as once believed.
Analysis of radio signals from the landers and the orbiters -- including Doppler, ranging and occultation data, and the signal strength of the lander-to-orbiter relay link -- provided a variety of valuable information.
Other significant discoveries of the Viking mission include:
Space History Viking Spacecraft Introduction
Copyright © 1997-2000 by Calvin J. Hamilton. All rights reserved. Privacy Statement. | <urn:uuid:faee3679-cf33-40d2-9291-137ad996f31e> | {
"date": "2018-06-24T11:05:13",
"dump": "CC-MAIN-2018-26",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00578.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9340593814849854,
"score": 3.71875,
"token_count": 1755,
"url": "http://solarviews.com/germ/vikingfs.htm"
} |
# Calculator search results
Formula
Factorize the expression
$x ^{ 4 } +3x ^{ 3 } -7x ^{ 2 } -27x-18$
$\left ( x - 3 \right ) \left ( x + 1 \right ) \left ( x + 2 \right ) \left ( x + 3 \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 27 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 }$
Do factorization
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 } \right )$
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 } \right )$
Do factorization
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$
Factorize to use the polynomial formula of sum and difference
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$
Sort the factors
$\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
Sort the factors
$\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
Sort the factors
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$
Solution search results | crawl-data/CC-MAIN-2022-21/segments/1652662558015.52/warc/CC-MAIN-20220523101705-20220523131705-00407.warc.gz | null |
GeeksforGeeks App
Open App
Browser
Continue
## Related Articles
• NCERT Solutions for Class 9 Maths
# Class 9 NCERT Solutions- Chapter 5 Introduction to Euclid’s Geometry – Exercise 5.1
### Question 1: Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There is an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In Given fig, if AB = PQ and PQ = XY, then AB = XY.
Solution:
(i) False
Reason: If we mark a point O on the surface of a paper. Using pencil and scale, we can draw infinite number of straight lines passing through O.
(ii) False
Reason: Through two distinct points there can be only one line that can be drawn. Hence, the statement mentioned above is False.
(iii) True
Reason: A line that is terminated can be indefinitely produced on both sides as a line can be extended on both its sides infinitely. Hence, the statement mentioned above is True.
(iv) True
Reason: The radii of two circles are equal when the two circles are equal. The circumference and the centre of both the circles coincide; and thus, the radius of the two circles should be equal. Hence, the statement mentioned above is True.
(v) True
Reason: According to Euclid’s 1st axiom- “Things which are equal to the same thing are also equal to one another”. Hence, the statement mentioned above is True.
### Question 2: Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them?
(i) Parallel lines
(ii) Perpendicular lines
(iii) Line segment
(v) Square
Solution:
Yes, there are other terms which need to be defined first, so that we understand better:
• Plane: Flat surfaces in which geometric figures can be drawn are known are plane. A plane surface is a surface which lies evenly with the straight lines on itself.
• Point: A dimensionless dot which is drawn on a plane surface is known as point. A point is that which has no part.
• Line: A collection of points that has only length and no breadth is known as a line. And it can be extended on both directions. A line is breadth-less length.
(i) Parallel lines: Two lines l and m in a plan are said to be parallel if they have a no common point and we can write them as l || m.
(ii) Perpendicular lines: Two lines A and B are said to be perpendicular if the form a right angle and we can write them as A ⊥ B.
(iii) Line Segment: A line segment is a part of line and having a definite length. It has two end-points. In the figure, a line segment is shown having end points P and Q. Write it as arrow over PQ.
(iv) Radius of Circle: The distance from the centre to a point on the circle is called the radius of the circle. In the figure, l is centre and m is a point on the circle, then lm is the radius of the circle.
(v) Square: A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a Square
In the given figure ABCD is a Square.
### Question 3: Consider two ‘postulates’ given below:
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist at least three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent?
Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms such as ‘Point and Line’. Undefined terms in the postulates are:
• There are many points that lie in a plane. But, in the postulates as given here, the position of the point C is not given, as of whether it lies on the line segment joining AB or it is not joining line segment.
• On top of that, there is no information about whether the points are in same plane or not.
And
Yes, these postulates are consistent when we deal with these two situations:
• Point C is lying on the line segment AB in between A and B.
• Point C does not lie on the line segment AB.
No, they don’t follow from Euclid’s postulates. They follow the axioms i.e “Given two distinct points, there is a unique line that passes through them.”
### Question 4: If a point C lies between two points A and B such that AC = BC, then prove that AC = 12 AB, explain by drawing the figure.
Solution:
AC = BC (Given)
As we have studied in this chapter “If equals are added to equals then there wholes are also equal”.
Therefore, AC + BC = BC + AC
⇒ 2AC = BC+AC
As we have studied in this chapter, we know that,
BC+AC = AB (as it coincides with line segment AB)
∴ 2 AC = AB (If equals are added to equals, the wholes are equal).
⇒ AC = (½)AB.
### Question 5: In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Solution:
Let, AB be the line segment as given in the Question.
Assume that points C and D are the two different mid points of line segment AB.
Therefore, C and D are the midpoints of AB.
Now, as C and D are mid points of Ab we have,
AC = CB and AD = DB
CB+AC = AB (as it coincides with line segment AB)
Now,
Adding AC to the L.H.S and R.H.S of the equation AC = CB
We get, AC+AC = CB+AC (If equals are added to equals, the wholes are equal.)
⇒ 2AC = AB — (i)
Similarly,
2 AD = AB — (ii)
From equation (i) and (ii), Since R.H.S are same, we equate the L.H.S we get,
2 AC = 2 AD (Things which are equal to the same thing are equal to one another.)
⇒ AC = AD (Things which are double of the same things are equal to one another.)
Thus, we conclude that C and D are the same points.
This contradicts our assumption that C and D are two different mid points of AB.
Thus, it is proved that every line segment has one and only one mid-point.
Hence, Proved.
### Question 6: In Figure, if AC = BD, then prove that AB = CD.
Solution:
According to the question, AC = BD
From the given figure we can conclude that,,
AC = AB+BC
BD = BC+CD
⇒ AB+BC = BC+CD (AC = BD, given)
As we have studied, according to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal.
Subtracting BC from the L.H.S and R.H.S of the equation AB+BC = BC+CD, we get,
AB+BC-BC = BC+CD-BC
AB = CD
Hence Proved.
### Question 7: Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate).
Solution:
Euclid’s fifth axiom states that “the whole is greater than the part.”
For Example: A cake. When it is whole or complete, assume that it measures 2 pounds but when a part from it is taken out and measured, its weight will be smaller than the previous measurement. So, the fifth axiom of Euclid is true for all the materials in the universe. Hence, Axiom 5, in the list of Euclid’s axioms, is considered a ‘universal truth’.
My Personal Notes arrow_drop_up
Related Tutorials | crawl-data/CC-MAIN-2023-23/segments/1685224655027.51/warc/CC-MAIN-20230608135911-20230608165911-00432.warc.gz | null |
# What is rounded to the nearest 100?
## What is rounded to the nearest 100?
The rule for rounding to the nearest hundred is to look at the tens digit. If it is 5 or more, then round up. If it is 4 or less, then round down. Basically,in each hundred, all numbers up to 49 round down and numbers from 50 to 99 round up to the next hundred.
What is the greatest 3 digit number that round off to 100?
999
100 is the smallest 3-digit number and 999 is the greatest 3-digit number.
What is 246 to the nearest 100?
200
To round a number to the nearest hundred, you make it into the hundred that is closest. Take the number 246, for example, which is between 200 and 300. It is closer to 200. So 246 rounded to the nearest hundred is 200.
### What number rounded to the nearest 100 is 500?
4. 509 is between 500 and 600 and rounded to the nearest 100 is 500 .
What is the smallest number that rounds to 100?
The smallest whole number that if we rounded to the nearest hundred will give us a result of 95,200 is the number 95,150.
What is 739 rounded to the nearest hundred?
700
739 rounded to the nearest hundred is 700. When you round to the nearest hundred, your answer is always going to be an even hundred.
#### How do you round 1000?
Rounding to the nearest 1000 To round a number to the nearest 1000, look at the hundreds digit. If the hundreds digit is 5 or more, round up. If the hundreds digit is 4 or less, round down. The hundreds digit in 4559 is 5.
What does 750 round to?
750 is exactly half-way between 700 and 800, and 791 is between 750 and 800. So 791 is closer to 800 than 700 on the number line and it rounds up to 800.
How do you round off numbers to the nearest 100?
This makes the question easier to understand. The easiest way to learn about rounding off numbers is to use place value and a number line. Click here to learn about Place Value. Click here for Rounding Off to the Nearest 100. The first step is to make sure your child knows how to count by 5s.
## What is the smallest possible number that will round off to 700?
From the number line we can see that the smallest possible number that will round off to 700 when rounded to the nearest ten is 695. To check our answer, we round off 694 to the nearest ten. We will get 690 not 700. So 695 is the correct answer.
What is 7089 rounded to the nearest hundred?
When your child has understood the process of rounding a number, we can use a short cut. I will use the number 7089 to illustrate. Underline the last 3 digits – 7 089. The digit 0 is in the hundreds place, so 7089 will become either 7 000 or 7 100 after rounding off to the nearest hundred.
What does it mean to round up to the nearest integer?
Rounding up, sometimes referred to as “taking the ceiling” of a number means rounding up towards the nearest integer. For example, when rounding to the ones place, any non-integer value will be rounded up to the next highest integer, as shown below: 5.01. ⇒. | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00117.warc.gz | null |
A new study has found that racial disparities in incidence of diabetes have more to do with living conditions than just genetics.
Researchers from the Hopkins Centre for Health Disparities Solutions and Case Western Reserve University School of Medicine showed that when African Americans and whites live in similar environments and have similar incomes, their diabetes rates are similar.
In recent decades the United States has seen a sharp increase in diabetes prevalence, with African Americans having a considerably higher occurrence of type 2 diabetes and other related complications compared to whites.
"While we often hear media reports of genes that account for race differences in health outcomes, genes are but one of many factors that lead to the major health conditions that account for most deaths in the United States," said Dr Thomas LaVeist, director of the Hopkins Centre for Health Disparities Solutions and lead author of the study.
"I don't mean to suggest that genetics play no role in race differences in health, but before we can conclude that health disparities are mainly a matter of genetics we need to first identify a gene, polymorphism or gene mutation that exists in one race group and not others.
"And when that gene is found we need to then demonstrate that that gene is also associated with diabetes," LaVeist added.
The study's authors said their findings support the need for future health disparities research and creative approaches to examining health disparities within samples that account for socioeconomic and social environmental factors.
The study appears in the Journal of General Internal Medicine. | <urn:uuid:d8af6c25-ae3e-4c28-b41f-eb7e628ab2ce> | {
"date": "2017-06-29T11:28:52",
"dump": "CC-MAIN-2017-26",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128323908.87/warc/CC-MAIN-20170629103036-20170629123036-00097.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9599934816360474,
"score": 3.515625,
"token_count": 298,
"url": "http://www.medindia.net/news/Racial-Disparities-in-Diabetes-Incidence-Linked-to-Living-Conditions-58427-1.htm"
} |
One such complex issue deals with harvesting natural gas from West Virginia’s abundant Marcellus Shale reserves. The Marcellus Shale is a formation of sedimentary rocks that lies under a 95,000-square-mile area that includes part of southwestern New York, most of western Pennsylvania, eastern Ohio and nearly all of West Virginia. Marcellus gas reserves lie between 3,000 feet and 9,000 feet under the land’s surface.
Formed in the Appalachian Basin over 300 million years ago, the Marcellus Shale formation has recently become an economically viable source of natural gas due to technological advances in horizontal drilling and the hydraulic fracturing process. Hydraulic fracturing, or “fracking,” involves injecting a mixture of water and chemicals under high pressure to release the natural gas reserves contained in the shale.
While this industry could potentially provide significant economic benefits to the state, lawmakers are currently grappling with a myriad of issues to ensure that this is done in a balanced manner beneficial to all citizens and stakeholders.
According to a recently published study produced by West Virginia University and the Oil and Natural Gas Association, between 2002 and 2008, West Virginia led the nation in the number of gas drilling permits issued. More than 2,800 permits were issued for new drilling in 45 of the state’s 55 counties.
The industry-funded study focuses on the economic impacts of our growing efforts to extract natural gas from the Marcellus.
In 2009, the study shows, West Virginia’s natural gas industry generated more than $12 billion in business, created more than 24,000 jobs in the state and paid more than $550 million in wages.
The report also notes that it is possible that Marcellus development created between 7,600 and 8,500 additional jobs in West Virginia in 2010. According to the report, by the year 2015, West Virginia could see 19,000 more jobs because of Marcellus development and related activities.
With any new industry the positive economic outlook must be balanced with proper care for the environment and a respect for all citizens’ rights. This industrial process has created many questions for lawmakers to consider with regard to both the environment and landowner rights.
Other advocates contend that tapping into the Marcellus shale field comes at a high cost, ranging from contamination of ground water to damage to local roadways from moving heavy equipment to the drilling sites to infringement on landowners’ rights.
The Legislature is considering how to regulate it responsibly as the U.S. Environmental Protection Agency is examining the fracking process to determine whether it endangers supplies of drinking water.
In order to address some of these concerns through regulation the “Hydraulic Fracturing and Horizontal Drilling Act” was introduced in both the Senate (SB 258) and the House (HB 2878) in late January. This bill came out of extensive work by interim committees during the months leading up to session. The bill is now being considered in the Judiciary Committee of each body. It is a comprehensive bill that will regulate Marcellus Shale development in areas including land use and surface owner rights, water quality and quantity, disclosure of chemicals used in the hydraulic fracturing process and waste management and disposal.
Clearly this is a very complex, detailed and highly important issue. Lawmakers will be hard at work during the second half of this legislative session and beyond to come up with real solutions that strike the appropriate age-old balance between industry, environment and citizens’ rights.
Senate Bill 200 will correct the names of state institutions of higher education. The bill will correctly name West Liberty State University as West Liberty University.
Senate Bill 342 will appropriate $8 million for the purpose of financing the special elections to fill a vacancy in the office of Governor.
House Bill 2853 will provide for a primary and special election to fill the vacancy in the office of the Governor. The bill would call for the primary election to be held on May 14 and the general election to be held on Oct. 4.
Senate Bill 78 would require parental consent and accompaniment for a minor using a tanning device. The bill would require minors between the ages of 14 to 17 to present a parental consent form, and minors under the age of 14 must present the consent form, and also be accompanied by a parent or legal guardian.
Senate Bill 186 would establish the West Virginia State Police as the entity that is authorized to issue administrative subpoenas to Internet service providers in cases of suspected child pornography. This bill gives the West Virginia State Police the authority to define offenses and set penalties and fees.
Senate Bill 195 would adjust the requirements for an individual to become a magistrate. The bill would require magistrates to possess a bachelor’s degree, an associate’s degree in criminal justice or at least four years of prior experience as a magistrate. The bill would go in effect in 2014.
Senate Bill 254 would make a supplementary appropriation of federal funds to the Development Office and the Division of Human Services - Energy Assistance.
Senate Bill 255 would make supplementary appropriations of remaining moneys to various accounts, such as the Governor’s Office, DHHR, Division of Rehabilitation Services, amoung other agencies.
Senate Bill 256 would require sex offenders to verify their e-mail and other online identities in the same way as they register their physical address.
Senate Bill 281 would make it a crime for a person to put certain types of invasive software such as, spyware or a virus, on mobile devices of another person without his/her consent or knowledge.
Senate Bill 349 would make it a requirement for a bittering agent to be placed in certain engine coolants and antifreezes to prevent personal injury or death of human beings and animals. If implemented, violation of this bill would be considered a misdemeanor.
Senate Bill 438 would change the election process for magistrates to be elected by division.
Senate Bill 265 would allow certain offenders to have contact with children, but only when the court finds it in the best interest of the child.
Senate Bill 419 would create the “Health Care Choice Act” by seeking to increase the availability of health insurance coverage by allowing insurers authorized to sell insurance in Kentucky, Ohio, Maryland, Pennsylvania and Virginia to issue accident and sickness policies in West Virginia.
Senate Bill 429 would provide educational scholarships for the children of war veterans. The scholarship would include tuition, institutional fees and standard room and board allowance. The bill would provide that the West Virginia Division of Veterans’ Affairs administer the scholarship program.
House Bill 2013 would increase the training that a dispatcher must have to work at a 911 center, adding a new 40 hour course in “emergency medical dispatch.” The bill further would require each dispatch center to develop protocols for dispatching their emergency medical calls.
House Bill 2368 would require the Board of Barbers and Cosmetologists to establish an apprenticeship program. This bill would update the existing system for training and procedures relating to the practice of beauty care.
House Bill 2503 would authorize the Board of Barbers and Cosmetologists to require government identification to be presented prior to issuance of licenses. The bill also authorizes the board to retain all information regarding licensees.
House Bill 2562 would move the State Athletic Commission under the Lottery Commission to assist in administrative functions. The bill would also make mixed martial arts a licensed sport with regulations set forth by the State Athletic Commission.
House Bill 2663 would require the Public Service Commission to be present in hearings in which it has retained the right to serve as the initial fact finder in the case. This would include any associated public protest hearings.
House Bill 2708 would remove the 12-month limitation on the length of agreements between law-enforcement agencies. Rather than expiring after a year, agreements under this proposed bill would remain in effect unless and until the agreement is changed or withdrawn by the head of one of the law-enforcement agencies.
House Bill 2750 would make the commission of sexual assault a consideration when issuing the permanent or temporary end to a parent-child relationship. This bill would allow a judge to take into consideration sexual assault or sexual abuse when deciding whether or not to remove a child from the home.
House Bill 2752 would increase the age of persons applying for appointment to a position on a police force within certain cities. The age would be increased from 35 to 40 years old. This increase will pertain to any person(s) applying for a position in a Class I or Class II city.
House Bill 2757 would provide for the evaluation of professional personnel within the public school systems. This bill would require the State Board of Education to establish a task force to address rule changes for personnel evaluations. The bill would also require the state board to report the evaluations to the Legislative Oversight Commission on Education Accountability.
House Bill 2787 would transfer the licensing of private investigators and security guards from the Secretary of the State to the Division of Justice and Community Services. All procedures, rules and regulations for these professions would be transferred to the Division of Justice.
House Bill 2864 would create a misdemeanor crime of unlawful restraint in the first or second degree. First-degree unlawful restraint would be considered intentionally restraining someone by use without proper authority. This would carry a maximum sentence of one year in jail and a $500 fee. Second-degree restraint is defined in this bill as restraint based on intimidation. This crime would carry a maximum of six months in jail and a $100 fine.
House Bill 2871 would provide that the Brownfield Economic Development districts comply with local planning laws. This compliance must take place before an application for such districts would be approved.
House Bill 2936 would change the date of the canvassing votes in a primary election. This bill would change the date from the Friday following a primary election to the Monday following a primary election. | <urn:uuid:05fa1144-70bd-45b7-8f34-9a83c05e4dbf> | {
"date": "2016-09-29T17:31:22",
"dump": "CC-MAIN-2016-40",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661905.96/warc/CC-MAIN-20160924173741-00282-ip-10-143-35-109.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9504767060279846,
"score": 3.515625,
"token_count": 2013,
"url": "http://www.legis.state.wv.us/Wrapup/2011/issue_04/wrapup.cfm"
} |
## Algebra 1
$\frac{1}{8}$
If a=-4, and b=1, in order to evaluate the expression $\frac{-b}{2a}$ all we must do is simply plug in the givens into the expression. $\frac{-b}{2a}$=$\frac{-(1)}{2(-4)}$=$\frac{-1}{-8}$=$\frac{1}{8}$ | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00708.warc.gz | null |
GMAT Question of the Day - Daily to your Mailbox; hard ones only
It is currently 18 Oct 2019, 01:43
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# When positive integer x is divided by 44, the remainder is 28
Author Message
TAGS:
### Hide Tags
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4007
When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
30 Mar 2017, 08:35
2
1
Top Contributor
2
00:00
Difficulty:
5% (low)
Question Stats:
87% (01:34) correct 13% (02:01) wrong based on 117 sessions
### HideShow timer Statistics
When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11?
A) 1
B) 3
C) 5
D) 7
E) 9
*kudos for all correct solutions
_________________
Test confidently with gmatprepnow.com
Math Expert
Joined: 02 Aug 2009
Posts: 7977
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
30 Mar 2017, 08:51
4
1
GMATPrepNow wrote:
When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11?
A) 1
B) 3
C) 5
D) 7
E) 9
*kudos for all correct solutions
Good Question Brent..
When x is divided by 44, remainder is 28...
So the number x is 44t+28, where t is integer..
When y is divided by 22, remainder is 14..
So the number y is 22s+14, where s is integer..
N=x+y = 44t+28+22s+14
So when N is divided by 11, $$\frac{44t+22s+42}{11}$$
44t and 22s are div by 11..
42=11*3+9..
So 9 is the remainder
E
_________________
##### General Discussion
VP
Joined: 05 Mar 2015
Posts: 1003
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
30 Mar 2017, 10:46
2
GMATPrepNow wrote:
When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11?
A) 1
B) 3
C) 5
D) 7
E) 9
*kudos for all correct solutions
x = 44a+28( a = any integer)
Let a =1 then x = 44+28= 72
similarly
y=22b+14 (b=any integer)
let b=1
then y= 36
thus N=x+y =72+36=108
108/11 = 9 9/11
ans E
Manager
Status: GMAT...one last time for good!!
Joined: 10 Jul 2012
Posts: 54
Location: India
Concentration: General Management
GMAT 1: 660 Q47 V34
GPA: 3.5
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
30 Mar 2017, 11:46
1
GMATPrepNow wrote:
When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11?
A) 1
B) 3
C) 5
D) 7
E) 9
*kudos for all correct solutions
x=44a+28;y=14+22b
For a=1 and b=1
x=72,y=36
n=x+5=108
remainder-9
_________________
Kudos for a correct solution
VP
Joined: 07 Dec 2014
Posts: 1222
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
30 Mar 2017, 11:54
2
GMATPrepNow wrote:
When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11?
A) 1
B) 3
C) 5
D) 7
E) 9
*kudos for all correct solutions
least values of x and y are 28 and 14 respectively
x+y=42=N
N/11 gives a remainder of 9
E
Current Student
Joined: 07 Jan 2016
Posts: 1088
Location: India
GMAT 1: 710 Q49 V36
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
30 Mar 2017, 13:09
1
remainder 28 when divided by 44
so 44a + 28
remainder 14 when divided by 22
so 22b + 14
now N = x+y
44a + 22b + 42
N/11
as we know 44 and 22 are divisible by 11 only 42 is remaining
42/11 remainder = -2 or 9
44- 2 = 42
33+9 = 42
E
Manager
Joined: 12 Jun 2016
Posts: 212
Location: India
WE: Sales (Telecommunications)
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
04 Aug 2017, 09:45
HI!
Did it by number picking
Possible value of x = 28, 72, 100...
Possible values of y = 14, 36, 58....
Since there can be only one unique solution, add to find the smallest possible value of n = 28 + 14 = 42. Remainder of 42/11 = 9
Is the Kudos for correct solution still valid?
_________________
My Best is yet to come!
Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4775
Location: India
GPA: 3.5
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink]
### Show Tags
20 Jan 2019, 09:14
GMATPrepNow wrote:
When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11?
A) 1
B) 3
C) 5
D) 7
E) 9
*kudos for all correct solutions
Let $$x = 28$$ and $$y = 14$$
So, $$N = 28 + 14$$ => $$42$$
$$\frac{N}{11} = 11*3 + 9$$, Thus Answer must be (E) 9
_________________
Thanks and Regards
Abhishek....
PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS
How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )
Re: When positive integer x is divided by 44, the remainder is 28 [#permalink] 20 Jan 2019, 09:14
Display posts from previous: Sort by | crawl-data/CC-MAIN-2019-43/segments/1570986679439.48/warc/CC-MAIN-20191018081630-20191018105130-00234.warc.gz | null |
1. ## librarian
A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all.
a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135
b) if 10 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.096
c) If 20 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.0042
any idea?
2. Originally Posted by swoopesjr01
A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all.
a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135
Binomial distribution problem.
Out of every $100$ books signed out $5$ are returned damaged and $1$ is not
returned at all. So of every $99$ books returned $5$ are damaged so the probability
that a returned book is damaged is $5/99\approx 0.0505$
Of $4$ books returned the probability that exactly $2$ are damaged is:
$P={4 \choose 2} 0.0505^2 (1-0.0505)^2$ $=\frac{4!}{2!\ 2!} 0.0505^2\ 0.9495^2\approx 0.01379$
Ooops - the above is the probability that of four returns two are damaged
irrespective of how many were signed out.
- No in retrospect this is right
I just managed to confuse myself when rereading the explanation as it is not
clear enough. There is a clearer (at least I think so) explanation in one of my
later posts.
Simulation shows that the proper probability has about $95\%$ chance of being in the interval $[0.01376,0.01396]$
Now if the question had been:
a) if 4 people each sign out one book, what is the probability that exactly 2
of the books will be returned damaged?
$P_1={4 \choose 2} 0.05^2 (1-0.05)^2=\frac{4!}{2!\ 2!} 0.05^2\ 0.95^2\approx 0.0135$
RonL
3. Originally Posted by swoopesjr01
c) If 20 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.0042
Again, this is binomial probability thus,
$\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
The other one is,
${20 \choose 2}.05^2.95^{18}=.0159$
Since this is a conjunction probability (this and that)
You need to multiply them to get,
$\approx 0.00415944$
4. Originally Posted by ThePerfectHacker
Again, this is binomial probability thus,
$\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
The other one is,
${20 \choose 2}.05^2.95^{18}=.0159$
Since this is a conjunction probability (this and that)
You need to multiply them to get,
$\approx 0.00415944$
The trouble is that the two clauses in the conjunction are not independent -
but multiplying may well be OK-ish here but I would like to be more confident
than I am.
RonL
5. Originally Posted by CaptainBlack
The trouble is that the two clauses in the conjunction are not independent -
but multiplying may well be OK-ish here but I would like to be more confident
than I am.
RonL
I was troubled by that too, I just saw that my hypothesis matched his answers. So I ignored that.
6. Originally Posted by ThePerfectHacker
I was troubled by that too, I just saw that my hypothesis matched his answers. So I ignored that.
I was distrubed by the given answers to all three questions in the original
post, that is why I stopped after doing (a) - I had to go away and think
If I get the chance (and I probably will as my research budget has been
frozen due to contract delays) I plan to see if simulation will throw any light
on the correctness of the given answers.
(Rule 1 of Applied Probability: When in doubt about the correctness of a
probability calculation - simulate)
RonL
7. Originally Posted by ThePerfectHacker
Again, this is binomial probability thus,
$\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$
The other one is,
${20 \choose 2}.05^2.95^{18}=.0159$
Typo here,
${20 \choose 2}0.01^2\ 0.99^{18}=.0159$
RonL
8. Originally Posted by CaptainBlack
I was distrubed by the given answers to all three questions in the original
post, that is why I stopped after doing (a) - I had to go away and think
If I get the chance (and I probably will as my research budget has been
frozen due to contract delays) I plan to see if simulation will throw any light
on the correctness of the given answers.
(Rule 1 of Applied Probability: When in doubt about the correctness of a
probability calculation - simulate)
RonL
Simulations done.
For b) they give p~0.00023
for c) they give p~0.00357
RonL
9. Originally Posted by swoopesjr01
A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all.
a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135
Given that a book is returned the probability that it is damaged is $0.05/0.99 \approx 0.50505$,
and the probability that it is not damages is $1-0.050505 \approx 0.949495$.
So the probability that if four books are returned from four checked out, that
exactly two are returned damaged is:
$
P_1={4 \choose 2} 0.050505^2 (1-0.050505)^2=$
$\frac{4!}{2!\ 2!} 0.050505^2\ 0.949495^2\approx 0.013798
$
RonL | crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00084-ip-10-31-129-80.ec2.internal.warc.gz | null |
There is a species of bees called “commercial” bees. These bees are kept by beekeepers to pollinate crops such as tomatoes, sweet peppers, and oilseed. This population of managed bees is coming down with “fast evolving viruses”, according to the University of Exeter in Science Daily News.
Then there are “wild” bees, free to fly around, not employed by beekeepers. The viruses that the commercial bees have are starting to spread to the wild bee population. Currently, researchers are “calling for new measures” to protect the wild pollinators, and confine the commercial, diseased population. In the article, Dr. Lena Wilfert said this can be controlled by beekeepers keeping a vigil eye and monitoring the commercial bees they own. It is their “responsibility” to do so. Also, interesingly enough, the international transport of these commercial bees must have more checks and security. They must be screened better, in order to get a better sense of how many have a disease, so they know not to set any of the commercial bees free into the wild.
The major cause of the spread is the Varroa mite. This spreads viruses, such as the Deformed Wing Virus, and may increase the power of the viral spread. It significantly weakens bees, causing their RNA to deteriorate. The article says that it has been “identified as an emerging disease in pollinators,” and there is a connection between wild bumblebees who have it, and commercial honeybees.
The poor management of the commercial bee community is the cause of this horrible break out of diseases among innocent wild bees. In the future, researchers plan to investigate which species of commercial bees are the major cause of the breakout and spread. The wild bee population is extremely important for our environment, and beekeepers need to realize that, and make sure their bee farm does not spread disastrous diseases.
*Additional information is found through the last two hyperlinks.*
*Original article is the first hyperlink.* | <urn:uuid:dcb914ec-174c-47b2-b4b2-7c15ad37f169> | {
"date": "2021-06-23T12:48:36",
"dump": "CC-MAIN-2021-25",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488538041.86/warc/CC-MAIN-20210623103524-20210623133524-00017.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9504067897796631,
"score": 3.625,
"token_count": 424,
"url": "https://jnewbio.edublogs.org/2015/01/17/the-buzzing-battle-of-the-bees/"
} |
Orodromeus was a small, herbivorous, or plant-eating, dinosaur that inhabited North America during the late Cretaceous period, about 65 to 98 million years ago. Orodromeus is classified as a member of the family Hypsilophodontidae, which was one of the most successful dinosaur families, flourishing for about 100 million years from the late Jurassic through the early Cretaceous periods. The Hypsilophodontidae belong to the order Ornithischia (the bird-hipped dinosaurs) and the suborder Ornithopoda, which consists of dinosaurs that exhibited primarily bipedal, or two-legged, locomotion.
The first fossil evidence of Orodromeus was discovered in the 1980s while a crew of paleontologists was examining nests of hadrosaur dinosaurs in Montana. At nearby sites that were later dubbed Egg Mountain and Egg Island, they discovered a large number of eggs of a new ornithopod dinosaur among the hadrosaur remains. The new dinosaur was named Orodromeus, which means “mountain runner,” because of both the location of the find and the animal’s presumed swiftness, based on its long hind limbs and slender build. The eggs and nests were remarkably well preserved; one complete nest contained 19 eggs laid in a precise spiral. After radiographic examination revealed embryonic bones inside the eggs, paleontologists carefully opened one egg and removed the preserved embryo for further study.
The condition of the nesting site, along with structural aspects of the embryo, shed light on the presumed parenting habits of Orodromeus. The eggs were less trampled than those found in nests of other dinosaur genera, suggesting that the young left the site soon after emerging from their shells—therefore not trampling the eggs. The presumption that Orodromeus and other hypsilophodonts engaged in only minimal parental care was further corroborated by the bone structure of the embryos. The joints of the hatchlings were well developed, suggesting that the young were capable of immediately leaving the site to fend for themselves. In contrast, the offspring of the hadrosaurid Maiasaura had poorly developed leg joints, making them incapable of leaving the nest soon after hatching and therefore in need of more parental care. (See also Maiasaura.)
Orodromeus grew to about 8 feet (2.4 meters) in length. Its head was small and light, with a birdlike beak. The hind legs were long and slender, and the arms ended in hands with five short, blunt-clawed fingers. Orodromeus was bipedal, meaning that it stood and walked on two legs. Along with its confamilials—that is, animals belonging to the same family—Orodromeus had effective grinding teeth and cheeks that held food during chewing. Some evidence suggests that they may have traveled in herds.
Horner, John, and Dobb, Edwin. Dinosaur Lives: Unearthing an Evolutionary Saga (HarperCollins, 1997). Lambert, David, and the Diagram Group. Dinosaur Data Book: The Definitive Illustrated Encyclopedia of Dinosaurs and Other Prehistoric Reptiles (Gramercy, 1998). Lessem, Don, and Glut, D.F. The Dinosaur Society’s Dinosaur Encyclopedia (Random, 1993). Lockley, Martin. Tracking Dinosaurs: A New Look at an Ancient World (Cambridge Univ. Press, 1991). Norell, M.A., and others. Discovering Dinosaurs in the American Museum of Natural History (Knopf, 1995). Norman, David. The Illustrated Encyclopedia of Dinosaurs (Crescent, 1985). Sattler, H.R. The New Illustrated Dinosaur Dictionary (Lothrop, 1990). Weishampel, D.B., and others, eds. The Dinosauria (Univ. of Calif. Press, 1990). Dixon, Dougal. Questions and Answers About Dinosaurs (Kingfisher, 1995). Farlow, J.O. On the Tracks of Dinosaurs (Watts, 1991). Gohier, François. 165 Million Years of Dinosaurs (Silver Burdett, 1995). Green, Tamara. Looking at: The Dinosaur Atlas (Gareth Stevens, 1997). Sokoloff, Myka-Lynne. Discovering Dinosaurs (Sadlier-Oxford, 1997). Theodorou, Rod. When Dinosaurs Ruled the Earth (Thomson Learning, 1996). Unwin, David. The New Book of Dinosaurs (Copper Beech, 1997). | <urn:uuid:8c33bfd8-3434-4382-a279-bd7d2468cf77> | {
"date": "2022-11-28T04:54:57",
"dump": "CC-MAIN-2022-49",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00616.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9248247742652893,
"score": 4,
"token_count": 944,
"url": "https://kids.britannica.com/students/article/Orodromeus/312839"
} |
The Sahara Desert and in particular its northern regions have attracted its share of attention from Atlantis investigators. However unlikely it may appear as a possible location for Atlantis it must be kept in mind that the Sahara of prehistory was very different from what we see today. Not only was it wetter at various periods in the past, but also there is clear evidence for the existence of a large inland sea extending across the borders of modern Algeria and Tunisia. This evidence is in the form of the chotts or salt flats in both countries. This proposed sea is considered by some to have been the Lake Tritonis referred to by classical writers. It is suggested that some form of tectonic/seismic activity, common in the region, was responsible for isolating this body of seawater from the Mediterranean and eventually turning it into the salt flats we see today.
An even more extensive inland sea, further south, was proposed by Ali Bey el Abbassi and based on his theory a map was published in 1802 which can be viewed online(c).
More recently, Riaan Booysen has published an illustrated paper on the ancient inland Saharan seas as indicated on the 16th century maps of Mercator and Ortelius(i). King’s College London runs The Sahara Megalakes Project which studies the Megalakes and the Saharan Palaeoclimate record(m).
A 2013 report in New Scientist magazine(d) revealed that 100,000 years ago the Sahara had been home to three large rivers that flowed northward, which probably provided migration routes for our ancestors.
Other studies(h) have shown the previous existence of a huge river system in the Western Sahara, which flowed into the Atlantic on the Mauritanian coast.
An article in the Sept. 2008 edition of National Geographic pointed out that the Saharan climate has been similar for the past 70,000 years except for a period beginning 12,000 years ago when a number of factors combined to alter this fact. A northerly shift by seasonal monsoons brought additional rain to an area the size of contiguous USA. This period of a greener Sahara lasted until around 4,500 years ago.
More recent studies claim that “there’s geologic evidence from ocean sediments that these orbitally-paced Green Sahara events occur as far back as the Miocene epoch (23 million to 5 million years ago), including during periods when atmospheric carbon dioxide was similar to, and possibly higher, than today’s levels. So, a future Green Sahara event is still highly likely in the distant future.”
More recent studies claim that “there’s geologic evidence from ocean sediments that these orbitally-paced Green Sahara events occur as far back as the Miocene epoch (23 million to 5 million years ago), including during periods when atmospheric carbon dioxide was similar to, and possibly higher, than today’s levels. So, a future Green Sahara event is still highly likely in the distant future.” (p)
Henri Lhote contributed an article to the Reader’s Digest’s, The World’s Last Mysteries , regarding the ‘green’ Sahara that existed prior to 2500 BC. An interesting question might be; what happened circa 2500 BC to cause this reversal? Some have suggested a connection between the aridification of the Sahara and the destruction of Atlantis!
More recently, human activity has been blamed as a major contributory factor for the desertification of the Sahara region less than 10,000 years ago.(n)
Related to the above is a recent study of sediments off the west coast of Africa, which resulted in the discovery of what was “primarily a new “beat,” in which the Sahara vacillated between wet and dry climates every 20,000 years, in sync with the region’s monsoon activity and the periodic tilting of the Earth.” (o)
In 1868, it was proposed by D.A. Godron, the French botanist, that the Sahara was the location of Atlantis. In 2003, the non-existent archaeologist Dr.Carla Sage announced that she was hoping to lead an international expedition to the Sahara in search of Atlantis. Her contention was that “Atlantis was the capital of a vast North African empire with ports on the Gulf of Sidra”. This report is now confirmed to have been a hoax! I am indebted to Stel Pavlou for uncovering the origin of this story(e).
Gary Gilligan, the well-known catastrophist, wrote a thought-provoking article(k) on the origin of the Saharan sands, which he claims are extraterrestrial in origin and expands on the idea in his 2016 book Extraterrestrial Sands .
David Mattingly, an archaeologist at Leicester University has found that an ancient people known as the Garamantes had an extensive civilisation in the Sahara(l). He has evidence of at least three cities and twenty other settlements. The Garamantes reached their peak around 100 BC and then gradually diminished in influence as fossil water supplies reduced until in the 7th century AD they were subjected to Islamic domination. Some researchers such as Frank Joseph have identified the Garamantes as being linked with the Sea Peoples. Bob Idjennaden has published short but informative Kindle books about both the Garamantes and the Sea Peoples , without a suggestion of any connection between the two.
The discovery of the megalithic structures discovered at Nabta Playa (Nabta Lake) in the Egyptian Sahara has provided evidence for the existence of a sophisticated society in that area around 5000 BC. In the same region, near the Dakhleh Oasis, archaeologists have produced data that supports the idea that pre-Pharaonic Egypt had Desert Origins rather than being an importation from Mesopotamia or elsewhere(a).
Nabta Playa is not unique, in fact the largest megalithic ellipse in the world is to be found at Mzorah, 27 km from Lixus in Morocco(b). It appears that the construction methods employed at both Mezorah and Nabta Playa are both similar to that used in the British Isles. An even more impressive site is Adam’s Calendar in South Africa which has been claimed as 75,000-250,000 years old.
West of Cairo near the border with Libya is the Siwa Oasis, where it has now been demonstrated that “it is in fact home to one of Ancient Egypt’s astounding solar-calendar technologies– the solar equinox alignment between the Timasirayn Temple and the Temple of Amun Oracle in Aghurmi.”(j).
I think we can expect further exciting discoveries in the Sahara leading to a clearer picture of the prehistoric cultures of the region and what connections there are, if any, with Plato’s Atlantis. In the meanwhile in the Eastern Egyptian Desert, Douglas Brewer, a professor of archaeology at the University of Illinois, has discovered over 1,000 examples of rock art, including numerous depictions of boats although the sites, so far undisclosed, are remote from water.
Even more remarkable is the report(e) of March 2015 that a survey of the Messak Settafet escarpment in the central Sahara revealed that there were enough discarded stone tools in the region “to build more than one Great Pyramid for every square kilometre of land on the continent”! Coincidentally, around the same time it was reported that over a thousand stone tools had been found in the Northern Utah Desert(g). What the Utah discovery lacked in quantity was made up for in quality with the finding of the largest known Haskett point spear head, measuring around nine inches in length.
(a) Saudi Aramco World (2006, Vol. 57, No.5 p.2-11)
(d) NewScientist.com, 16 September 2013, https://tinyurl.com/mg9vcoz
(l) See: Archive 3268
The Garamantes, referred to by Herodotus, are generally considered to have been the first Libyan empire. However, attempts to link them to the Atlantis story would appear to be undermined by the fact that they flourished around the middle of the first millennium BC, which would appear to be far too recent to fit any interpretation of Plato’s 9,000 ‘years’, be they solar, lunar or seasonal.
Frank Joseph identifies the Garamantes with the Sea Peoples whom he considers to have been Atlantean refugees. However, the Garamantes are generally accepted as having developed urban centres in what is now southern Libya(a)(b), which is not what you might expect from a maritime culture. Recent satellite images(c) offer new information on the extent of the Garamantes domain.
Further information is available on the Temehu.com website(d), but perhaps the most intriguing theory is that the Garamantes originally came from the Carpatho–Danubian region of eastern Europe sometimes referred to as Dacia. It is suggested that the fair-skinned Berbers of North Africa are the descendants of those European invaders!
Bob Idjennaden, a Belgian living in Ireland, has published a short Kindle book(f) on the Garamantes. He has also authored a series of Kindle books on various aspects of ancient African history, including one about the Sea Peoples, co-authored with Taklit Mebarek.
(f) https://www.amazon.co.uk/Buried-Kingdoms-Garamantes-Forgotten-Civilisations-ebook/sim/B007Q239OE/2 (link broken Sept. 2020)< | <urn:uuid:d0407815-0a16-4c9f-a828-7192dea12b9d> | {
"date": "2021-01-20T06:06:06",
"dump": "CC-MAIN-2021-04",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519923.26/warc/CC-MAIN-20210120054203-20210120084203-00418.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9572874903678894,
"score": 3.578125,
"token_count": 2021,
"url": "https://atlantipedia.ie/samples/tag/garamantes/"
} |
XML elements can have attributes, just like HTML.
Attributes are designed to contain data related to a specific element.
XML Attributes Must be Quoted
Attribute values must always be quoted. Either single or double quotes can be used.
For a person's gender, the <person> element can be written like this:
or like this:
If the attribute value itself contains double quotes you can use single quotes, like in this example:
or you can use character entities:
XML Elements vs. Attributes
Take a look at these examples:
In the first example gender is an attribute. In the last, gender is an element. Both examples provide the same information.
There are no rules about when to use attributes or when to use elements in XML.
My Favorite Way
The following three XML documents contain exactly the same information:
A date attribute is used in the first example:
A <date> element is used in the second example:
An expanded <date> element is used in the third example: (THIS IS MY FAVORITE):
Avoid XML Attributes?
Some things to consider when using attributes are:
- attributes cannot contain multiple values (elements can)
- attributes cannot contain tree structures (elements can)
- attributes are not easily expandable (for future changes)
Don't end up like this:
to="Tove" from="Jani" heading="Reminder"
body="Don't forget me this weekend!">
XML Attributes for Metadata
Sometimes ID references are assigned to elements. These IDs can be used to identify XML elements in much the same way as the id attribute in HTML. This example demonstrates this:
<body>Don't forget me this weekend!</body>
<body>I will not</body>
The id attributes above are for identifying the different notes. It is not a part of the note itself.
What I'm trying to say here is that metadata (data about data) should be stored as attributes, and the data itself should be stored as elements. | <urn:uuid:d1e7619a-c82a-4b47-a571-fa13f71bdf00> | {
"date": "2019-05-23T07:35:44",
"dump": "CC-MAIN-2019-22",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257156.50/warc/CC-MAIN-20190523063645-20190523085645-00176.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.6791422367095947,
"score": 3.78125,
"token_count": 429,
"url": "http://welookups.com/xml/xml_attributes.html"
} |
Start a 10-Day Free Trial to Unlock the Full Review
Why Lesson Planet?
Find quality lesson planning resources, fast!
Share & remix collections to collaborate.
Organize your curriculum with collections. Easy!
Have time to be more creative & energetic with your students!
The color sight word "green" is the focus of this language arts worksheet. Learners cut and past pictures of green objects to match sentences about those green objects. A color as well as a black and white copy are provided.
3 Views 13 Downloads | <urn:uuid:d27d9a89-e3c8-4958-8fd0-134acde55188> | {
"date": "2017-03-29T01:37:08",
"dump": "CC-MAIN-2017-13",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218190134.67/warc/CC-MAIN-20170322212950-00181-ip-10-233-31-227.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.8966724276542664,
"score": 3.640625,
"token_count": 111,
"url": "https://www.lessonplanet.com/teachers/green-cut-and-paste"
} |
# Probabilities about drwing lucky cards
I wanna to calculate a probabilities of buying a box of lucky cards. However I am messed up with the equations.
The question: Let there is a box with 30 cards. and 6 of them are lucky cards.
Calculate the probabilities of buying cards one by one until 6 cards are all brought.
At which number would the probability that the chance is higher than 80%?
-
Presumably each purchase is independent and can buy each card with equal probability $\frac{1}{30}$.
Suppose $P(n,k)$ is the probability that after $n$ purchases you are missing $k$ lucky cards. Then $$P(n+1,k)=\frac{30-k}{30}P(n,k)+\frac{k+1}{30}P(n,k+1)$$ starting at $P(0,6)=1$ and $P(0,k)=0$ for $k \not = 6$. This can easily be calculated, for example using a spreadsheet.
You are interested in the smallest $n$ such that $P(n,0) \gt 0.8$. It turns out that $P(97,0) \approx 0.79402$ while $P(98,0) \approx 0.80031$.
As a check on the calculations, the expected number of cards needed to be bought to get all the lucky cards is both $\displaystyle\sum_{n=0}^\infty (1-P(n,0)) = 73.5$ and $30\left(\frac16 + \frac15 + \frac14 + \frac13 + \frac12 + \frac11\right)=73.5$.
If purchases were without replacement (meaning that at $30$ purchases you have all the cards, lucky and unlucky), the recurrence becomes
$$P(n+1,k)=\frac{30-n-k}{30-n}P(n,k)+\frac{k+1}{30-n}P(n,k+1).$$
You now find $P(29,0) = 0.8$. This is not a surprise as the probability the final card purchased is lucky is $\frac{6}{30}=0.2$.
-
Thanks for answering the question, How about if the box is fixed to 30 cards and fixed lucky cards are 6. Once the lucky card is picked and no replacement. Will the answer be different? – Kitw Dec 18 '12 at 7:17
Shall the equation become, P(n+1,k)=(30−k)/(24+k) * P(n,k)+ (k+1)/(24+k) * P(n,k+1) – Kitw Dec 18 '12 at 7:39
Followed above, the result turns a bit wield, am I miss something? <a href="picturepush.com/public/11714784"><img src="www1.picturepush.com/photo/a/11714784/img/11714784.png"; border="0" alt="Image Hosted by PicturePush - Photo Sharing" /></a> – Kitw Dec 18 '12 at 7:53
@Kitw: Not quite, instead see my addition. The fact that your rows of probabilities do not add up to 1 should suggest an issue. – Henry Dec 18 '12 at 7:55
Many thanks for solving my mess :) – Kitw Dec 18 '12 at 8:03
If you buy cards one by one, the box of 30 doesn't matter. How many total cards are there? You have the Coupon collector's problem but need to specify the conditions better before there can be a specific solution.
-
Thanks for the answer, Had just read the Coupon collector's problem, does it replace everytime? If the box size and number of cards are fixed, will the math be the same – Kitw Dec 18 '12 at 7:19
The fact is to find the expected value of how many times is the best to stop buying more. On the same condition, What is the probabilities that first 12 pick will get half cards and so on. – Kitw Dec 18 '12 at 7:22
@Kitw: In the classic Coupon collector's problem, each coupon is randomly selected from the universe, so yes, it replaces every time. You can use the classic coupon collector's problem for 6 cards and multiply by 5, because $\frac 45$ of the time you don't get a lucky card and can ignore it. You want the number of draws to complete your set of 6. – Ross Millikan Dec 18 '12 at 14:27 | crawl-data/CC-MAIN-2016-30/segments/1469257826916.34/warc/CC-MAIN-20160723071026-00137-ip-10-185-27-174.ec2.internal.warc.gz | null |
Short-sightedness – or myopia – is a vision problem in which close objects can be seen clearly but objects far away are blurred. Short-sightedness occurs if the eyeball is too long or the cornea has too much curvature. As a result, the light entering the eye can’t focus correctly and distant objects appear blurred and distorted.
Short-sightedness is a common condition and affects around 30% of the U.S. population. It usually first occurs in school-age children and, because the eye continues to grow during childhood, it will progress until about age 20.
What Causes Short-sightedness?
The exact causes of the condition are unknown, but research indicates that two factors may be primarily responsible for its development:
- Visual stress
There is significant evidence to suggest that many people inherit short-sightedness, or at least a tendency to develop it. If one or both parents are short-sighted, there is a greater likelihood their children will be short-sighted.
Even though the tendency to develop the condition may be inherited, its development may be affected by how the eyes are used. Those who spend a lot of time doing close visual work, such as reading or working in front of a computer, may be more likely to develop short-sightedness.
Symptoms of short-sightedness may also be a sign of a blood sugar condition, such as diabetes or an early indication of a developing cataract.
Diagnosis and Treatment
A comprehensive eye test, using several procedures, by your optometrist will determine if you have short-sightedness.
There are several options available for persons with short-sightedness to regain good distance vision. These are:
- Contact lenses
- LASIK surgery
- Vision therapy for persons with stress-related short-sightedness
Eyeglasses are the primary choice for correcting short-sightedness. A single vision lens is usually prescribed to provide good vision at all distances. But patients over 40, and children or adults whose short-sightedness is due to intense close vision work may have bifocal or progressive addition lenses prescribed for them. These lenses provide different strengths or powers throughout the lens to provide for distant and close vision.
Some patients will find that contact lenses offer better vision than eyeglasses and may suit them better cosmetically. Although they provide a wider field of vision, they require regular cleaning and care to safeguard eye health.
Orthokeratology involves the fitting of a series of rigid gas permeable contact lenses to reshape the cornea. The contact lenses are worn overnight during sleep, and removed for the entire day allowing patients to go about their daily activities eyewear-free. Your optometrist can advise you further about the process and whether or not it would suit your particular requirements.
Using a laser beam of light, short-sightedness can be corrected by reshaping the cornea in a process called LASIK surgery. Although this surgery has generally proved successful, it is still a developing therapy in which some problems have occurred. Speak to your optometrist to acquire full information about this process and how it is likely to affect you.
Vision therapy is an option for patients whose blurred distance vision is caused by a spasm of the muscles which control eye focusing. Various exercises are used to improve the ability to focus and regain good vision.
People with short-sightedness have a number of options to correct their vision problem. Your optometrist will help you select the treatment that best meets your visual and lifestyle needs.
Related – Medical Reasons for Wearing Sunglasses, Stevens-Johnson Syndrome, Glaucoma Information, Onchocerciasis: River Blindness, Presbyopia, Posterior Vitreous Detachment, Eye Anatomy: Important Definitions, Duane Syndrome, Helpful Glaucoma Information. 0 | <urn:uuid:d149c21d-4ed9-422f-ac60-d25f57819266> | {
"date": "2016-07-01T04:15:58",
"dump": "CC-MAIN-2016-26",
"file_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783400031.51/warc/CC-MAIN-20160624155000-00089-ip-10-164-35-72.ec2.internal.warc.gz",
"int_score": 4,
"language": "en",
"language_score": 0.9221934080123901,
"score": 4,
"token_count": 794,
"url": "http://blog.framesdirect.com/myopia-short-sightedness/"
} |
# Conditional probability regarding multiple choice
Q: A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible.
If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices. As the teacher you want the test to measure what the student knows. If the student answers a question correctly what’s the probability they knew the answer?
A: Stuck in the process of trying to answer this.
The way I thought about it was using Baye's rule and treating it as a conditional probability.
In that, $$P(answer correct | knew answer) = \frac{P(know answer | answer correct) \cdot P(answer correct)}{P(knew answer)}$$
The probability that the student knows the answer is 0.5 so that gives us the denominator. Figuring out probability for the answer being correct and knowing the answer given a correct answer is a bit more confusing for me. Would appreciate guidance.
Let A represent the event that the question is answered correctly
Let X represent the event that the student knows the correct answer
Let Y represent the event that the student is able to eliminate one choice
Let Z represent the event that the student is not able to eliminate any choice
I interpret the statement ....
the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible.
... to mean ...
$$P(X)=0.5; P(Y)=.25;p(Z)=.25$$
So ...
$$\begin{eqnarray*} P(X|A) &=& \frac{P(A \cap X)}{P(A)} \\&=& \frac{P(A \cap X)}{P(A\cap X)+P(A\cap Y)+P(A\cap Z)} \\&=& \frac{P(A |X)P(X)}{P(A |X)P(X)+P(A |Y)P(Y)+P(A |Z)P(Z)} \\&=& \frac{1 \cdot \frac 12}{1 \cdot \frac 12+\frac 14 \cdot \frac 13+\frac 14 \cdot \frac 14} \\&=& \frac {\frac 12} {\frac{24}{48}+\frac4{48}+\frac3{48}} =\frac{24}{31} \end{eqnarray*}$$
• Ah, yes, I interpreted the problem statement differently in my Answer, but I bet you have the correct interpretation! +1 – Bram28 Jul 12 '17 at 19:36
With
$A$: students answers correctly
$B$: student knew answer
what you want is $P(B|A)$, rather than $P(A|B)$
Now, we are given that:
$$P(B)= \frac{1}{2}$$
and hence
$$P(B^C)=1-\frac{1}{2}=\frac{1}{2}$$
Also:
$P(A|B)=1$ (if they knew the answer they give the correct answer)
and
$$P(A|B^C)=\frac{1}{4}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{4}=\frac{1}{12}+\frac{3}{16}=\frac{4}{48}+\frac{9}{48}=\frac{13}{48}$$
(When they don't know the answer there is a $\frac{1}{4}$ probability they can eliminate one of the answers and thus have a $\frac{1}{3}$ probability of guessing correctly between the remaining 3, and there is a $\frac{3}{4}$ probability they can't eliminate any one answer in which case they have a $\frac{1}{4}$ probability of guessing correctly)
Next, we have that:
$$P(A \cap B)=P(A|B)\cdot P(B)=1\cdot \frac{1}{2}=\frac{1}{2}$$
and
$$P(A \cap B^C)=P(A|B^C)\cdot P(B^C)=\frac{13}{48}\cdot \frac{1}{2}=\frac{13}{96}$$
and thus:
$$P(A)=P(A \cap B)+P(A \cap B^C)=\frac{1}{2}+\frac{13}{96}=\frac{61}{96}$$
So finally:
$$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{2}}{\frac{61}{96}}=\frac{48}{61}$$ | crawl-data/CC-MAIN-2019-26/segments/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00286.warc.gz | null |
End of preview. Expand
in Dataset Viewer.
Qwark Corpus
1.3B+ high quality tokens from the internet, based on HuggingFaceTB/smollm-corpus's fineweb-edu-dedup subset as well as FineMath-4+.
Filtering process:
| Step | Description | Rows |
|---|---|---|
| 1. Stream dataset until 600K samples have been selected from SmolLM Corpus | Keep only items with score >= 3.5 | 600,000 |
| 2. Remove items with length > 50,000 | Filter items exceeding 50,000 characters in length | 597,142 |
| 3. Combine with a selection of 4,000 TED transcripts | Add educational TED talk transcripts to the dataset | 601,147 |
| 4. Stream 400K samples from FineMath-4+ | Keep only items with score >= 4.0 | 1,001,147 |
| 5. Remove items with length > 50,000 | Filter items exceeding 50,000 characters in length | 999,245 |
- Downloads last month
- 21
Size of downloaded dataset files:
2.72 GB
Size of the auto-converted Parquet files:
2.72 GB
Number of rows:
999,245
Models trained or fine-tuned on qingy2024/qwark-corpus