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theaquatics.net | 1,701,247,960,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100057.69/warc/CC-MAIN-20231129073519-20231129103519-00043.warc.gz | 669,367,884 | 25,217 | # The Astonishing Value of 10 Gallons
: How Much is 10 Gallons?
Hey there! Are you curious about how much 10 gallons actually is? Well, let me break it down for you. Imagine you have a big container that can hold 10 gallons of liquid. That’s quite a lot, isn’t it? Just to give you some perspective, picture a standard bathtub filled with water. Can you visualize it? Great! Now, imagine that bathtub being filled almost twice with water. That’s approximately how much 10 gallons is. Impressive, right?
To put it into more everyday terms, 10 gallons is equivalent to around 37.9 liters. So, imagine filling up a large bucket with almost 38 liters of liquid. It’s quite a substantial amount! You could use it to water your plants, fill up your fish tank, or even wash your car. The possibilities are endless!
In terms of fuel, 10 gallons can take you quite a distance depending on the fuel efficiency of your vehicle. On average, a car can travel approximately 300 to 350 miles with 10 gallons of gasoline. That’s like driving from one city to another or going on a road trip with your friends. So, it’s safe to say that 10 gallons of fuel can get you pretty far.
Now, let’s talk about something a bit more delicious – 10 gallons of ice cream! If you’re a fan of this frozen treat, you’ll be delighted to know that 10 gallons of ice cream is equivalent to around 320 scoops. Yes, you heard that right! You could have a scoop of ice cream every day for nearly a year and still have some left over. That’s a whole lot of sweetness!
In conclusion, 10 gallons is a significant amount when it comes to liquids. Whether it’s water, fuel, or even ice cream, it’s definitely more than meets the eye. So, next time you come across a measurement of 10 gallons, you’ll have a better understanding of just how much it truly is. Pretty fascinating, isn’t it?
In an informal style, let’s dive into the question “What’s the price of 10 gallons?” and find out the answer. Keep in mind that the following information is provided for general knowledge and may vary depending on the location and specific circumstances.
When it comes to determining the price of 10 gallons, it is essential to consider what exactly you are referring to. The price can vary based on various factors, such as the type of product or service you are referring to.
If you are talking about the price of gasoline (petrol) in a particular region, it is influenced by several factors, including crude oil prices, taxes, and transportation costs. Gasoline prices are usually measured per gallon, so knowing the price of 10 gallons would simply mean multiplying the cost per gallon by 10.
However, if you are referring to the price of another liquid or product, it is important to specify the context. For example, if you are considering the price of milk, water, or any other liquid, it may vary depending on the brand, location, and container size. Hence, it is crucial to be specific about the product in question.
Furthermore, if you are referring to a different commodity altogether, such as paint or cleaning solution, the price can vary based on the brand, quality, and quantity. It is always recommended to check the prices at local stores or online retailers to get accurate and up-to-date information.
Therefore, to determine the price of 10 gallons accurately, it is necessary to specify the product or service in question. Once you have the specific information, you can refer to local stores, online retailers, or contact suppliers for the most accurate and current pricing details.
Remember, prices can fluctuate over time, so it is always a good idea to double-check the information before making any purchasing decisions. Take into account any additional charges, taxes, or discounts that may apply to get a clear understanding of the total cost.
## How much is 10 gallons?
10 gallons is a unit of measurement commonly used to measure liquids. In simple terms, it is equivalent to approximately 37.85 liters or 3,785 milliliters. To give you a better idea, it is roughly the same as filling up a large bucket or four standard-sized water cooler bottles.
It’s important to note that the actual volume may vary depending on the substance being measured, as different liquids have different densities. For example, 10 gallons of water will have a different weight compared to 10 gallons of gasoline.
So, to summarize, 10 gallons is equal to about 37.85 liters or 3,785 milliliters. It can be visualized as a large bucket or four standard-sized water cooler bottles. However, the specific weight or volume may vary depending on the liquid being measured.
Thank you for reading, and see you next time!
## How Much Is 10 Gallons
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## Poker Odds Teacher
• App Store Info
### Description
Don't just memorize the odds, *understand* them ... This app teaches you a general method for estimating the odds for just about any Texas Hold'em scenario in your head in real time.
*** Pay once and enjoy forever ***
No in-app purchases/fees.
Continual improvement (7+ years).
*****
Notes:
(1)
This is a challenging app. You need to be pretty good at basic math operations. For example, what's the value of:
16% * 20%
We’ve added a “Math” help button to show you how to calculate this. One way to do it is, 20% means one-fifth. One-fifth of 16% is (16% / 5), which is a little more than 3%. 3.2% to be exact. Or, convert to decimals, multiply, then convert back to %age's.
(2)
This app uses specific hand situations, and some people (like Magicman74 on the U.S. App Store) don’t see the value in this because in real life play we don’t know our opponents’ cards. There’s still value in specific hand situations, because they’re the building blocks that we can use to put together and estimate the odds for more complex hand range situations. For example every good player should know the odds for specific hand situations like AJ vs. QQ pre-flop, or FlushDraw vs. TopPair on the flop. Our other apps e.g. PokerCruncher handle both specific hands and hand ranges.
*****
*****
"... helps players learn some of the math behind poker ... teaches a general method of estimating odds for different Texas Hold'em situations."
-- PokerSoftware.com
Many more great reviews on our TwoPlusTwo forum thread, and by poker experts, pros, and coaches.
(See our website.)
*****
Poker trainers, odds tables, etc. are useful tools but do they help you learn why the odds are what they are or are you just memorizing specific situations? And you can't use these tools or odds calculators in the middle of a live hand (would you really want to?; could be a “fish” tell :)). This app takes a new and different approach to poker odds - understanding instead of memorizing, so you can handle any situation that comes your way.
Our odds estimation method is a pretty simple 3-step procedure. First you count your outs and estimate your odds of improving (using for example the "Rule of 4&2"; this app shows you how). Then you consider your opponent's counter-outs, then put the two together to come up with a good odds estimate.
We've provided about 20 common and important pre-flop and post-flop scenarios for you to practice the method on. This app will take you through the 3 steps for each practice scenario, one screen per step, and will make sure that you're on track towards a good odds estimate each step of the way, correcting you if needed. We've provided info/help buttons for each sub-step.
You might be thinking, how can you cover a game as complex as Texas Hold'em in 20 scenarios? Well, we agree, you can't! But this app is about learning a general method not many specific situations, and we picked these 20 scenarios to cover a good cross section of common and important pre-flop and post-flop situations. If you understand how to compute the odds for say one generic TopPair vs TopPairTopKicker scenario or one representative Pair vs LowerPair pre-flop scenario, you'll be able to apply your knowledge to other similar situations easily.
Now, we're not saying that you're going to master this in 5 or 10 minutes. You're going to have to practice the method more than a few times, think about the outs and counter-outs, and do simple math operations in your head - basic +, -, *, /; let's not let our math teachers down :). We hope you feel that acquiring this odds skill and knowledge is more than worth the learning/practice time.
*****
See our website for our strong 7+ year free app update history.
*****
App Store reviews are greatly appreciated, thank you.
*****
Also please check out our companion apps Hold’em Odds Quizzer and PokerCruncher.
*****
### What's New in Version 5.2
Update this app’s App Store screenshots to the latest size/format. No changes to the app itself.
===== App Reviews =====
App Store reviews help Poker Odds Teacher succeed and get even better, and are greatly appreciated, thank you!
-RJ, PokerCruncher, LLC | 1,066 | 4,772 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-22 | latest | en | 0.879141 |
https://www.itl.nist.gov/div898/handbook/mpc/section3/mpc372.htm | 1,718,865,057,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00650.warc.gz | 713,487,185 | 2,508 | 2. Measurement Process Characterization
2.3. Calibration
2.3.7. Instrument control for linear calibration
## Table of t* distribution
Critical values of t* distribution
for testing the output of a linear calibration line at 3 points
Significance level alpha = 0.05
```
1. 37.544 61. 2.455
2. 7.582 62. 2.454
3. 4.826 63. 2.453
4. 3.941 64. 2.452
5. 3.518 65. 2.451
6. 3.274 66. 2.450
7. 3.115 67. 2.449
8. 3.004 68. 2.448
9. 2.923 69. 2.447
10. 2.860 70. 2.446
11. 2.811 71. 2.445
12. 2.770 72. 2.445
13. 2.737 73. 2.444
14. 2.709 74. 2.443
15. 2.685 75. 2.442
16. 2.665 76. 2.441
17. 2.647 77. 2.441
18. 2.631 78. 2.440
19. 2.617 79. 2.439
20. 2.605 80. 2.439
21. 2.594 81. 2.438
22. 2.584 82. 2.437
23. 2.574 83. 2.437
24. 2.566 84. 2.436
25. 2.558 85. 2.436
26. 2.551 86. 2.435
27. 2.545 87. 2.435
28. 2.539 88. 2.434
29. 2.534 89. 2.434
30. 2.528 90. 2.433
31. 2.524 91. 2.432
32. 2.519 92. 2.432
33. 2.515 93. 2.431
34. 2.511 94. 2.431
35. 2.507 95. 2.431
36. 2.504 96. 2.430
37. 2.501 97. 2.430
38. 2.498 98. 2.429
39. 2.495 99. 2.429
40. 2.492 100. 2.428
41. 2.489 101. 2.428
42. 2.487 102. 2.428
43. 2.484 103. 2.427
44. 2.482 104. 2.427
45. 2.480 105. 2.426
46. 2.478 106. 2.426
47. 2.476 107. 2.426
48. 2.474 108. 2.425
49. 2.472 109. 2.425
50. 2.470 110. 2.425
51. 2.469 111. 2.424
52. 2.467 112. 2.424
53. 2.466 113. 2.424
54. 2.464 114. 2.423
55. 2.463 115. 2.423
56. 2.461 116. 2.423
57. 2.460 117. 2.422
58. 2.459 118. 2.422
59. 2.457 119. 2.422
60. 2.456 120. 2.422
``` | 1,087 | 2,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-26 | latest | en | 0.151052 |
https://www.jiskha.com/display.cgi?id=1365346738 | 1,516,097,255,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886397.2/warc/CC-MAIN-20180116090056-20180116110056-00511.warc.gz | 913,741,104 | 3,791 | # Precalculus
posted by .
Find all solutions to 2 cos(theta) = - 1 on the interval 0 less than or equal to theta less than 2 pi
Give your answers as exact values, as a list separated by commas.
• Precalculus -
cosine is negative in II or III
and I know cos 60° = cosØ/3 = 1/2
so for your cosØ = -1/2
Ø = 180-60 =120° or Ø = 180+60 = 240°
Ø = 2π/3 or Ø = 4π/3
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1. Let (-7, 4) be a point on the terminal side of (theta). Find the exact values of sin(theta), csc(theta), and cot(theta). 2. Let (theta) be an angle in quadrant IV such that sin(theta)=-2/5. Find the exact values of sec(theta) and …
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https://robomateplus.com/video-lectures/iit-jee-main-advanced-video-lectures-online/iit-jee-main-advanced-xii/iit-jee-main-advanced-xii-physics/jee-main-advance-12th-pcm-physics-electrostatics-2-demo-videos/ | 1,722,717,957,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640377613.6/warc/CC-MAIN-20240803183820-20240803213820-00059.warc.gz | 408,779,489 | 59,059 | Home/Video Lectures/IIT - JEE Main & Advanced/IIT - JEE Main & Advanced-XII/IIT - JEE Main & Advanced XII - Physics/JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics module-2 Demo Videos
## JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics module-2 Demo Videos
Welcome back, students.
So, students, till now we have been dealing with examples of field and cases of field due to discrete charges. There was a charge or two point charges or multiple charges placed at different, different points. Now, let us discuss the case of continuous charge systems and to start with we will be finding electric field to a line charge at axial position. Now, students, we will be following a certain strategy to solve these questions or these cases. Number one, we need to choose the element whose field is already known to us. For example if you talk of this line charge. Now, this line charge, we can, we can imagine this line charge is made of small, small point charges and since we know the field to point charge. Therefore we can consider a very small length which will be just like a point charge. Similarly, for this ring, if we closely observe and take a very small part on this particular ring, it will also be like a point charge whose field is known to us. And so if we talk of this disc and this disc, if we closely analyse, it will be made up of small, small rings and if we know the field to a ring then we should be taking our element as the ring. So, first step is to choose an element. Step 2 is to find field to this element and next step is to integrate. So, let’s apply that in this particular scenario wherein we have to find electric field due to a finite line charge, at any point along the line charge. So, what do we have? We have a line charge and we need to find field at point P and what is given to us? It is given all these conditions that is length, distance of this point from one of the edges and we are given that this line charge has a linear charge density Lamda and length L. So, step one will be to choose a very small element whose field is known to us. So, let’s consider a small element whose field is known to us. Now, what will that element be? Remember, for line charges the elements will be a very small length which will be like a point charge. Now, if we closely observe this line charge is made up of small charge plus small charge plus small charge and so on and out of these small charges if we closely observe and then take a very small portion and say it is at a distance of x from O and has a thickness of dx then this element will be just like a point charge and so it qualifies to be an element. So our element is a point charge at distance x from origin having length dx and having a charge dq which will be equal to Lamda dx. Why Lamda dx, because linear charge density Lambda implies charge on 1 unit length is Lambda. So, charge on dx unit length will be Lambda dx. So, we know charge on element is Lambda dx. Now, what is our step 2? Step 2 is to find field due to the small element that is find dE vector. Now, what is dE in magnitude? Remember field to a point charge is k charge upon distance square. So, the charge of this element is dq and what is the distance, distance will be total length, I mean, L plus a minus x. So, hence magnitude of the field will be kdq upon L plus a minus x square. Now, as far as direction is concerned. Since this dq charge is positive. So, field due to dq charge will be in this direction that is this d vector will be kdq in this direction. We can rewrite it as kdq upon L plus a minus x square I caron. So, now we know field due to our element. We have to find field at P, what is our step 3? Step 3 is to integrate. Now, before we integrate remember, students, our all the terms should be in same variables. Now, since dq is Lambda dx, we can rewrite this equation as dE vector is equal to k Lambda dx upon a plus L minus x square I caron. Now, we can integrate now. So, integrating we need to apply limits. Now, what will be the limits be? If we closely observe limits of x will be from 0 to x equal to L, because if we closely observe line charge starts from x is equal to 0 and goes till x is equal to L. So, hence this limit will become 0 to L. Now, what is the integration of this expression? Remember, integration of this expression L plus a minus x raise to power n dx is as shown. Provided n is not equal to 1. That is integral is L plus a minus x raise to power n plus 1 upon n plus 1 and there is a minus sign outside it because in this term there is coefficient of x is minus 1. So, let’s integrate we get E vector as minus k Lambda into a plus L minus x raise to power minus 1 upon minus 1 into minus 1 I caron, and if we apply limits we will get this result. I hope you know how to apply limits. Value of expression at x is equal to upper limit minus value of expression at x is equal to lower limit. So, in this case we will get this expression. On further solving we will get the result to be k lambda L upon a plus L into a I caron.
Students, if you feel that I have been fast here please pause re-watch this particular thing because you need repetitions to understand it nicely. So, we will get back in next module till then students, thank you.
### Increase your scores by Studying with the BEST TEACHERS – Anytime and anywhere you want
2017-04-18T04:56:07+00:00 Categories: IIT - JEE Main & Advanced XII - Physics||0 Comments
Hello | 1,287 | 5,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-33 | latest | en | 0.942648 |
http://math.stackexchange.com/questions/156602/evaluate-lim-x-to-infty-frac1x-int-x4x-cos-left-frac1t-righ?answertab=oldest | 1,462,231,322,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860117914.56/warc/CC-MAIN-20160428161517-00097-ip-10-239-7-51.ec2.internal.warc.gz | 186,302,985 | 20,716 | # Evaluate $\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$
Evaluate $$\lim_{x \to \infty} \frac{1}{x} \int_x^{4x} \cos\left(\frac{1}{t}\right) \mbox {d}t$$
I was given the suggestion to define two functions as $g(x) = x$ and $f(x) = \int_x^{4x}\cos\left(\frac{1}{t}\right)dt$ so then if I could prove that both went to $\infty$ as $x$ went to $\infty$, then I could use L'Hôpital's rule on $\frac{f(x)}{g(x)}$; but I couldn't seem to do it for $f(x)$.
I can see that the limit is 3 if I just go ahead and differentiate both functions and take the ratio of the limits, but of course this is useless without finding my original intermediate form.
How do I show that $\frac{f(x)}{g(x)}$ is in intermediate form? or how else might I evaluate the original limit?
-
$f'(x) = \cos \left( \frac{1}{4x} \right) (4x)' - \cos \left( \frac{1}{x} \right) (x)'$ check this article on wikipedia - en.wikipedia.org/wiki/Fundamental_theorem_of_calculus – qoqosz Jun 10 '12 at 18:43
I am not having trouble taking the derivative, I can evaluate the limit using L'Hôpital's just fine, but I never proved that $f(x)$ goes to $\infty$ in order to be able to use it in the first place. – stariz77 Jun 10 '12 at 18:46
stariz77 right, sorry than :) – qoqosz Jun 10 '12 at 18:47
@stariz77: That the integral blows up is clear, since $\cos(1/t)$ is nearly $1$ for large $t$. – André Nicolas Jun 10 '12 at 20:07
We are only interested in behaviour when $x$ is large. For example, let $x \ge 10$. If $t \ge x$, then $0.99 \lt \cos(1/t)\lt 1$, so $2.97 x \lt \int_x^{4x}\cos(1/t)\,dt\lt 3x$. In particular, integral has infinite limit, L'Hospital's Rule very applicable. – André Nicolas Jun 11 '12 at 3:49
For other methods of solving the limit you could use mean value theorem:
$$\frac{1}{x} \int_x^{4x} \cos \frac{1}{t} \; dt = \frac{3x \cos \frac{1}{c}}{x}$$ for some $c \in (x,4x)$. Now when $x \to +\infty$ by squeeze theorem we get $3$ as a result.
-
How did you get the $3x\cos(\frac{1}{c})$ term? – stariz77 Jun 10 '12 at 18:56
In general we have: $$f'(c) = \frac{f(b) - f(a)}{b-a}$$ for some $c \in (a,b)$ if a<b. Now rewrite it as $(b-a) f'(c) = f(b) - f(a)$ and let $f(x) = \int^x \cos \frac{1}{t} \, dt$. In your case we have: $$(4x - x) \cos \frac{1}{c} = \int_x^{4x} \cos \frac{1}{t} \, dt$$ – qoqosz Jun 10 '12 at 19:00
@stariz77 btw by using $3x \cos \frac{1}{c}$ you can also determine that symbol for $f$ is $\infty$ :) – qoqosz Jun 10 '12 at 19:14
Hint: When $t \to + \infty$, $\cos(1/t) \to ?$
-
Let $y = 1/t$. Then the integral becomes \begin{align} I & = \lim_{x \rightarrow \infty} \dfrac1x \int_x^{4x} \cos(1/t) dt = \lim_{x \rightarrow \infty} \dfrac1x \int_{1/x}^{1/(4x)} \cos(y)\dfrac{-dy}{y^2}\\ & = \lim_{x \rightarrow \infty} \dfrac1x \int_{1/(4x)}^{1/x} \dfrac{\cos(y)}{y^2} dy \end{align} Now use Taylor series for $\cos(y)$ and use DCT to swap limit and integral. Or equivalently, you can write $\cos(y) = 1 + \mathcal{O}(y^2)$ and then proceed. \begin{align} I & = \lim_{x \rightarrow \infty} \dfrac1x \int_{1/(4x)}^{1/x} \dfrac{dy}{y^2} + \lim_{x \rightarrow \infty} \dfrac1x \int_{1/(4x)}^{1/x} \mathcal{O}(1) d y = \lim_{x \rightarrow \infty} \left(\dfrac1x \left. \left( - \dfrac1y \right \rvert_{1/(4x)}^{1/x} \right) + \mathcal{O}(1/x^2) \right)\\ & = \lim_{x \rightarrow \infty} \dfrac1x \left( -x + 4x\right) = 3 \end{align}
-
For $x\ge\dfrac2\pi$, Dominated Convergence says \begin{align} \lim_{x\to\infty}\frac1x\int_x^{4x}\cos\left(\frac1t\right)\,\mathrm{d}t &=\lim_{x\to\infty}\int_1^4\cos\left(\frac1{xt}\right)\,\mathrm{d}t\\ &=\int_1^41\,\mathrm{d}t\\[9pt] &=3 \end{align}
- | 1,441 | 3,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2016-18 | latest | en | 0.789274 |
https://makezine.com/article/home/fun-games/aha-puzzle-this-11/ | 1,718,797,371,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00809.warc.gz | 328,649,227 | 33,597 | # Aha! Puzzle This
### Point of Gnome Return
One hundred very smart garden gnomes are snatched from their homes by an evil wizard. He tells them he is going to line them all up in a row, and place a red or blue hat on each of their heads. They won’t be able to see the color of their own hats or anyone’s behind them, but they will be able to see the hats of the gnomes in front of them. The wizard will start at the back of the line and ask each gnome to guess the color of his own hat. Each gnome can only answer either “red” or “blue.” If he gives the wrong answer, he will be led off to work on the wizard’s landscaping for the rest of eternity. If he answers correctly, he will be returned to his own garden. Then the wizard will move on to the next gnome in line.
All of the gnomes will be able to hear the answers of the gnomes behind them, but they will not know if they were led off to forced labor or if they answered correctly and were set free. The gnomes are allowed to consult and agree on a strategy beforehand (while the wizard listens in) but after being lined up, they will not be able to communicate in any other way besides their answer of “red” or “blue” (in other words, they won’t be able to change the pitch of their voice or give any other clues to the other gnomes once they are in line and the hats are placed on their heads).
What strategy should the gnomes use to maximize the total number of gnomes that will be set free? Hint: They can do pretty well, even if the wizard hears their plan and puts the hats on in such a way to thwart whatever idea they come up with.
### Human Race
Five makers (Ben, Mark, Jason, Ruth, and Joel) raced in an extravaganza for alternative transportation. They competed in five separate races: a Segway race, a bicycle race, a hybrid car race, a skateboard race, and a roller skate race. Points were given as follows: 5 points for first place, 4 points for second, 3 points for third, 2 points for fourth, and 1 point for last place. Overall scores were determined by adding up the individual scores from each race.
Ben got the highest number of points: 24. Jason finished in the same place in four out of the five races. Joel came in first in the skateboard race, and third in the roller skate race.
The final total rankings for the races were: Ben, Mark, Jason, Ruth, and then Joel. There were no ties in any individual races, and no two racers had the same total score.
What place did Mark come in during the skateboard race?
Discuss this article with the rest of the community on our Discord server!
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### Michael H. Pryor
Michael H. Pryor is the co-founder and president of Fog Creek Software. He runs a technical interview site at techinterview.org
View more articles by Michael H. Pryor | 646 | 2,771 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-26 | latest | en | 0.966965 |
https://www.doubtnut.com/qna/12015841 | 1,718,884,557,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00537.warc.gz | 643,242,592 | 37,586 | # An α-particle and a proton are accelerated from rest through the same potential difference V. Find the ratio of de-Broglie wavelength associated with them.
Video Solution
Text Solution
Verified by Experts
## The kinetic energy gained by a charged particle of charge q, mass m, moving with velocity v, when acceleration through the potential difference V is K.E. =12mv2=qVormv=[2mqV]1/2 De-Broglie wavelength of the particle is λ=hmv=h[2mqV]1/2 so λ∝1√mq (for the same value of V) ∴λαλp=√mpqpmαqα=√mp×e4mp×2e=1√8=12√2
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## From Davisson-Germer experiment an α particle and a proton are accelerated through the same pdV. Find the ratio of the de Broglie wavelengths associated with them
A1:22
B22:1
C1:2
D2:1
• Question 2 - Select One
## An electron and a proton are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths will be
A(mpme)1/2
Bme/mp
Cmp/me
D1
• Question 3 - Select One
## An α -particle and a singly ionized 4Be8 atom are accelerated through the same potential difference. Ratio of de-broglie wavelength-
A1:2
B2:1
C4:1
D1:1
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# Sparing question with multiple drive sizes
New Member
## Sparing question with multiple drive sizes
Hello,
I have an EVA6000 with 64 drive bays populated and double drive failure protection on a single disk group.
In the 64 drive bays are:
16 x 600GB FC
16 x 450GB FC
32 x 300GB FC
I have recently retired another EVA that was populated with 146GB FC drives and would like to use those and bring my 6000 to the full 112 disk capacity.
48 x 146GB FC
I know that in reality that the 146GB drives are in fact 136.73GB. 48 x 136-73 = giving a raw capacity of 6.563TB. However some space on each drive will be required for sparing capacity.
Is my calculation below correct:
600GB (The largest drive in the group) which is actaully like 550GB (estimating) * 2 = 1100GB * 2 = 2200GB. So 2.2TB are required to be reserved for double failure protection. Spread that across all 112 drives = about 20GB per drive lost as reserved for protection leaving me 116GB to use on these 146GB drives.
Or is the protection space reserved on each drive relational to the size of the drive so the 600GB drives might set aside 80GB for spare protection and the 146GB drives set aside 5GB each? | 326 | 1,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | latest | en | 0.926151 |
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04 Jun 2009, 22:02
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Hey everyone.
I'm new to the GMAT, but thought I would join the community here!
I enjoyed reading everyone's GMAT experience, so I thought I would post mine as I went along.
I started studying 5/26 and so far have only put in about 8 hours.
The first thing I did was take the Diagnostic Test in the OG 11 to see where I stood.
My scores were:
Problem Solving - (15/24) Average
Data Sufficiency - (11/24) Average
Critical Reasoning - (16/17) Excellent
Sentence Correction - (12/18) Above Average
I'm currently a senior in college and I was surprised how hard even the simple math questions were. I finished Calculus my freshman year and really haven't had any math other than Data Analysis and Finance since.
I think for the most part the difficulty was due to the fact that I didn't recall the vocabulary.
Also I wasn't very sure how to go about answering the Data Sufficiency questions and fell victim to a few yes/no pitfalls, where I assumed if the answer was no, then it wasn't sufficient.
All in all I'm not too disappointed with my initial performance. Reviewing my mistakes was obviously critical, and I was able to learn some of the tricks, mainly with data sufficiency. Also I had pretty much forgotten all of the ratios for geometry, like the isosceles triangle hah, which was an easy fix with a quick review.
I plan to take the GMAT late July or early August, but haven't registered yet. Also I'm shooting for a 700+ score
Looking forward to working together with everyone here!
cheers!
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### Show Tags
04 Jun 2009, 22:59
Welcome to GMAT Club Jolt.
You sound rather mature for being a senior in college - most folks think of parties and to get out of school, not to get in. Keep it up and you will gain respect quickly around here.
What are you plans? Take GMAT and then wait a few years before an MBA or perhaps a Master's or PhD?
-
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05 Jun 2009, 12:54
Thanks! but I'm not sure I'm completely deserving, I'm still a typical college student for the most part hah
The reason I'm looking at getting right back into school is that I'm most interested in entrepreneurship and it would be great to find some business partners with similar interests.
I'm also applying to a teaching position abroad and considering deferring my MBA for a year if I'm accepted.
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### Show Tags
05 Jun 2009, 19:04
hey buddy, welcome to the forums! I'm in a similar situation and recently graduated in the past couple of weeks. I'm planning on taking the GMAT the end of the summer but finding it really hard to focus when studying. Haha... oh well, better get it down now than try to study for it during work. Good luck.
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### Show Tags
06 Jun 2009, 10:15
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Expert's post
FYI - it is hard to judge the performance from a paper test since
If you take a simulated computer test - any of them (all are listed here: all-gmat-cat-practice-tests-links-prices-reviews-77460.html)
Then you can use GMAT Score Estimator/Calculator to get your GMAT scores and measure your level of performance
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### Show Tags
15 Jun 2009, 01:54
well i just took GMATPREP tonight for the first time (12:30AM-3:30AM)
My scores:
Q - 32
V - 37
Total 580
Before taking it I had read through the Math Review in the OG the day before, which helped to refresh some of the concepts. Also I was able to keep an eye out for mistakes I had made in the Diagnostic Test that I took 2 weeks ago.
When reviewing my mistakes on the Quant section I noticed 5 VERY obvious mistakes on questions I should have easily gotten right, but I'm not sure how much that would have affected my score. Also I missed the first 6 questions. I didn't see any clear area that was to blame for the majority of my mistakes.
By the time I started the Verbal section I was very tired, but I don't think it affected my score much, mainly just forced me to reread a few times or take a break to close my eyes. On the Verbal section I didn't make many careless mistakes, but I did notice 80% of the questions I got wrong were SC. Luckily I have MGMAT SC for that.
I was aiming for a 590-600 for my first GMATPREP test, so I missed my goal by 10-20points.
I will take another GMATPREP test in 1-2weeks depending on when I can find time.
Should I take GMATPREP Test 2 or retake Test 1?
My goal for my next test will be 620.
cheers!
and good luck to the rest of you!
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Concentration: Entrepreneurship, Strategy
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### Show Tags
15 Jun 2009, 10:22
1
KUDOS
jolt wrote:
well i just took GMATPREP tonight for the first time (12:30AM-3:30AM)
My scores:
Q - 32
V - 37
Total 580
Before taking it I had read through the Math Review in the OG the day before, which helped to refresh some of the concepts. Also I was able to keep an eye out for mistakes I had made in the Diagnostic Test that I took 2 weeks ago.
When reviewing my mistakes on the Quant section I noticed 5 VERY obvious mistakes on questions I should have easily gotten right, but I'm not sure how much that would have affected my score. Also I missed the first 6 questions. I didn't see any clear area that was to blame for the majority of my mistakes.
By the time I started the Verbal section I was very tired, but I don't think it affected my score much, mainly just forced me to reread a few times or take a break to close my eyes. On the Verbal section I didn't make many careless mistakes, but I did notice 80% of the questions I got wrong were SC. Luckily I have MGMAT SC for that.
I was aiming for a 590-600 for my first GMATPREP test, so I missed my goal by 10-20points.
I will take another GMATPREP test in 1-2weeks depending on when I can find time.
Should I take GMATPREP Test 2 or retake Test 1?
My goal for my next test will be 620.
cheers!
and good luck to the rest of you!
Congratulations on starting your journey. Now that you've got an accurate baseline score, you should be able to identify a few key areas for improvement.
Missing the first six questions on the GMAT is very damaging to your overall score. The GMAT utilizes the first 10-15 questions in each section to determine your approximate ability. The remainder of the test is primarily used for fine tuning your score. Missing the first six questions gives the CAT a very low approximation of your ability, making it impossible for you to climb back up to a top quant score. Another way of viewing this problem is that you lose more points when you miss an easy question. By default, you start the test with a medium difficulty question. Missing the first six questions means you missed the six easiest questions you're going to see for the rest of the exam. This is far worse than missing six evenly spaced questions, or even the last six questions.
If you need to brush up on the math fundamentals, I strongly recommend Kaplan Math Workbook and MGMAT Number Properties. The combination of these two books should put you in the 43-46 range for Quant.
You should be in good hands with the MGMAT SC book. I'm currently reading it for the first time and it's helping me quite a bit.
For the sake of "saving" the GMATPrep #2, I would recommend re-taking GMATPrep #1 once you've studied a bit more. You should wait at least two weeks though to make the test worthwhile. Although I'm not a huge fan of the MGMAT CAT's, you get six for free with any book that you buy and they really are the best alternative to GMATPrep.
Good luck and welcome to GMAT Club
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15 Jun 2009, 20:45
Manhattan series is the best.... worth every penny...
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### Show Tags
08 Jul 2009, 14:07
Yesterday I took MGMAT CAT #1
My Scores
Q - 45
V - 34
Total 650
Over the past couple weeks I studied quant and worked my way through the MGMAT Number Properties book, but surprisingly I found I did the worst on Number Properties questions when I was reviewing my results. But I felt much more confident throughout the Quant section this time around.
I wasn't surprised by my Verbal score because I still haven't had the time to start studying for it and answered only 5/15 SC questions correctly. I plan to read through MGMAT SC before my next CAT.
Overall I was very pleased with the results, my goal was a 620 for this CAT
Good luck to the rest of you fellow gmatters!
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### Show Tags
21 Jul 2009, 23:43
MGMAT CAT #2
My Scores
Q - 44
V - 40
Total 700
Recently I studied the Manhattan GMAT SC Guide and made it about halfway through. I also went through about 200 of MGMAT's Quant flash cards, these were really helpful to get some quick review when I didn't really feel like studying. The SC guide helped and I was able to analyze the questions more effectively and efficiently. On MGMAT 1 I missed 10/15 SC questions. On MGMAT 2 I missed only 6/15 SC questions
cheers!
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### Show Tags
24 Jul 2009, 11:02
GMAT PREP 1 RETAKE
Q - 44
V - 41
Total 700
Director
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Schools: Michigan (Ross) - Class of 2013
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### Show Tags
24 Jul 2009, 19:08
Great job, jolt. Your test scores are coming together! If you can increase your quant score to ~47, that should give you a little bit of slack to ensure that you stay above 700 on test day.
Good luck with the remainder of your studies!
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### Show Tags
04 Aug 2009, 17:09
MGMAT CAT #3 (taken July 31)
Q - 42
V - 38
Total 660
Obviously it's never good to see your scores going down, especially after two 700s in a row. But what went wrong? Well, In the beginning of the quant section I was doing really well and by question 10-15 I was getting really hard questions. I felt like I was getting behind on time and tried to guess a few that I didn't know how to attack right away. Unfortunately I guessed on a 300-500lvl remainder question that was REALLY easy hah, after missing a few 700-800s in a row. When I got to the end of my questions I was getting really easy ones and as a result got the remaining questions right.
As for Verbal I'm not really sure what happened, I got two RC passages in a row which threw me off i know. and missed a couple more CR that I usually do.
Overall though I'm glad my score decreased. It motivated me to study more. I have put a lot more quality time into studying. I re-read the entire MGMAT Number Properties book, because the first time I didn't finish it and I could tell I had started to forget some of the NP that I had learned a month or more ago. I also started an error log for the first time. And retraced my CAT tests and wrote down all of the Q questions I had missed. However, I didn't log the V questions I missed, do any of you do this?
I set my GMAT date for August 12, so between now and then GMAT will comprise most of my time.
Any suggestions on what to study going into the final week before my test will be appreciated!
Thank you GMATClub for all the help and support so far!
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04 Aug 2009, 18:51
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I agree. A dip in your score (not a drastic plunge) is never a bad thing while studying. It definitely motivated me to study for the days after the test and I'm sure you'll be energized as well. Since you have 1 week to study, you should concentrate on specific sections rather than sections. I would go for SC since it is probably the easiest to improve in the shortest amount of time. You should also do many DS questions since you seem weak in that area. I would make sure to go through the error log multiple times. When you're sick of 'formal' studying, hop onto gmatclub and do some practice SCs and CRs. good luck!
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### Show Tags
05 Aug 2009, 00:31
MGMAT CAT #4
Q - 45 77%
V - 38 85%
Total 690 91%
One thing that is frustrating is that I got almost 100% of the 300-700 level Quant questions right, but a small percentage of the (21!) 700-800 questions right. On MGMAT CAT #1, I only received (9!) 700-800 level questions and also got a Q - 45, is it better to miss some easier questions on purpose so you can assure that you answer more easier questions right???
also i only missed 4 SC questions! ....but 5 RC somehow
anyways, back to studying!
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### Show Tags
05 Aug 2009, 18:36
Study List for Final Week!
(* indicates importance and will likely be reviewed first)
V
CR - Strengthen Conclusion
Analyze Arg Structure
Evaluate a Conclusion
RC - Tone *
Specific Detail *
SC - Pronouns *
Verbs *
Modifiers *
Idioms
Comparisons *
Clarity of Meaning
Q
WT - Probability *
Combinatorics *
Statistics
Inequalities
Variables in Choice *
Geometry - Triangles and Diagonals *
Coordinate Plane *
NP - Odds and Evens *
Divisibility and Primes
FDPs - Percents
Digits
Cheers!
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06 Aug 2009, 12:17
jolt,
dude this ia an awesome post. your details keep me checked on my own preparation. thanks for sharing and goodluck!
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06 Aug 2009, 14:21
first of all thanks whatthehell! glad to see im helping someone!
now for today's results.
Today I decided to wake up and take a mock GMAT, basically do everything I planned on doing the day of the real GMAT and see how my score turns out.
Woke up, showered, ate breakfast, and relaxed for a bit. Then I did a 15min work-out to make sure I was awake. I also tried not to eat any sugar or caffeine! I didn't want to crash during the test and I didn't think I would need it for the Essays anyway.
Drove to the testing center, which is conveniently <1min from my house, unfortunately I couldn't go in! haha
Came back home and started GMAT PREP #2 for the first time
This is the first time I have ever taken the entire GMAT including essays. As expected I did the essays comfortably in the allotted time at likely a 5.5-6 level. I reviewed the Official GMAT guide directions and sample answers as practice.
After the essays I was feeling great and had a boost of confidence. I ate a banana and drank half of my starbucks bottled coffee for energy! the math section is where I wanted the caffeine! In the beginning the math section was going great, I could tell I was solving questions that had been very hard for me in the past rather quickly. But by questions 20 I was getting tired and spent too long on probably 2 questions. By question 30 I could tell I was behind on time. However, I was able to focus and manage to read each question once through to gather necessary information and answered each question in 1min, with 1 guess. By the last question I had 3 min left and took my time to solve it.
Before starting the Verbal section I took a break and finished my coffee and did some stretches, but unfortunately I did not get the energy boost I received from before the math section! I'm thinking I might need to use an energy drink next time (RedBull) instead to ensure that extra energy boost. In any case I managed to pump myself up for about the first 7 questions, but after the first RC passage, which was short and confusing, I had run out of energy. The remainder of V questions became a grind and it was taking all of my energy to stay focused. This lasted until about question 28 or 29 where I pumped up by the realization that I was almost finished! I finally made it to the last question with 7-8 min remaining, which I probably could have used to better evaluate some RC or SC questions. But in any case I was done!
clicked through next, next, next...
My Score
Q - 47 81%
V - 41 93%
Total 710! 93%
This is probably the most ideal situation I could imagine on the actual test day!
I missed 12 Q questions, 2 of which were Geometry
and I only missed 7 V questions, 5SC and 2RC and 0CR!
Hopefully I an replicate this on the test day next week!
Cheers!
Senior Manager
Joined: 11 Dec 2008
Posts: 479
Location: United States
GMAT 1: 760 Q49 V44
GPA: 3.9
Followers: 40
Kudos [?]: 205 [0], given: 12
### Show Tags
06 Aug 2009, 14:45
Awesome job. You should probably focus on SC intensely for the next few days since your other areas seem very strong. Also I would recommend doing at least another full length simulation so that you build stamina for test date. Keep it up and chances are pretty good for a 700+
Intern
Joined: 04 Jun 2009
Posts: 48
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### Show Tags
08 Aug 2009, 16:07
Today I took another Full CAT, including essays (really hate writing essays for no reason hah)
during the past two days I did some review of my error log, but most of my time was spent reading through the entire MGMAT SC guide. I also put some thought into a new RC passage strategy.
GMAT PREP CAT 1
Q - 46 79%
V - 44 97%
Total - 720 95%
My quant was going well and I focused much better this time. Retaking the test over and over has really made me accustomed to the length and it seems to breeze by now, actually finished with 10 minutes left, but I missed the last 5 out of 6 questions. Ultimately missed 14 questions total. Also missed 1st question
Verbal I was doing great, even though I missed the first question and knew it right after hitting next. I missed 4 total questions, all 4 SC, which is 1 better then before I read the entire SC guide! hah
But the major point that I found was I could cut substantial time off of each question by reading the passage twice.... but why? When I was VERY familiar with the passage I didn't have to keep referring to the passage when deciding between two likely answers; furthermore, if I ever did need to refer back to a passage I knew exactly where to look!
I recommend those of you having trouble with accuracy on RC and CR, especially native speakers, to try this strategy
Now if only I can get a little better with SC!
Cheers!
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# Chapter 9
9.1 Under what noise conditions will signal averaging fail to improve the SNR?
Ans.
Signal averaging has decreased SNR improvement if signal and noise are
uncorrelated.
## 9.2 In a signal averaging application, the amplitude of uncorrelated noise is
initially 16 times as large as the signal amplitude. How many sweeps must
be averaged to give a resulting signal-to-noise ratio of 4:1?
mS/(m)^0.5N = mS/(m^0.5)*16*S=(m)^0.5/16 =S/N
Ans.
4096 sweeps
9.3 After signal averaging 4096 EEG evoked responses, the signal-to-noise ratio
is 4. Assuming that the EEG and noise sources are uncorrelated, what was
the SNR before averaging?
Ans.
1
SNR =
16
## 9.4 In a signal averaging application, the noise amplitude is initially 4 times as
large as the signal amplitude. How many sweeps must be averaged to give a
! resulting signal-to-noise ratio of 4:1?
Ans.
256 sweeps
9.5 In a signal averaging application, the signal caused by a stimulus and the
noise are slightly correlated. The frequency spectra of the signal and noise
overlap. Averaging 100 responses will improve the signal-to-noise ratio by
what factor?
Ans.
Less than 10.
50
Chapter 10
10.2 Given the following data: {15, 10, 6, 7, 5, 3, 4, 7, 15, 3}, produce the data
points that are stored using the TP algorithm.
Ans.
{15, 6, 7, 3, 15}
10.9 Does the TP algorithm (a) produce significant time-base distortion over a
very long time, (b) save every turning point (i.e., peak or valley) in a signal,
(c) provide data reduction of 4-to-1 if applied twice to a signal without
violating sampling theory, (d) provide for exactly reconstructing the original
signal, (e) perform as well as AZTEC for electroencephalography (EEG)?
Ans.
(a) No, only local distortion.
(b) No.
(c) Yes, if the final sampling rate still is higher than twice the highest
frequency present in the signal.
## (d) No, it is a lossy algorithm.
(e) Better than AZTEC for an EEG. Since the EEG is a randomly varying
signal, straight line approximations cannot effectively represent the
signal.
## 10.10 Which of the following are characteristic of a Huffman coding algorithm?
(a) Guarantees more data reduction on an ECG than AZTEC; (b) Cannot
perfectly reconstruct the sampled data points (within some designated error
range); (c) Is a variable-length code; (d) Is derived directly from Morse
code; (e) Uses ASCII codes for the most frequent A/D values; (f) Requires
advance knowledge of the frequency of occurrence of data patterns;
(g) Includes as part of the algorithm self-correcting error checks.
Ans.
(c), (f)
51
10.11 After application of the TP algorithm, what data sequence would be saved if
the data sampled by an analog-to-digital converter were:
## (a) {20, 40, 20, 40, 20, 40, 20, 40}
(b) {50, 40, 50, 20, 30, 40}
(c) {50, 50, 40, 30, 40, 50, 40, 30, 40, 50, 50, 40}
(d) {50, 25, 50, 25, 50, 25, 50, 25}
Ans.
(a) {20, 40, 20, 40}
(b) {50, 40, 20}
(c) {50, 40, 30, 50, 30, 50}
(d) {50, 25, 50, 25}
10.12 After application of the TP algorithm on a signal, the data points saved are
{50, 70, 30, 40}. If you were to reconstruct the original data set, what is the
data sequence that would best approximate it?
Ans.
{50, 60, 70, 50, 30, 35, 40}
52
10.13 The graph below shows a set of 20 data points sampled from an analog-to-
digital converter. At the top of the chart are the numerical values of the
samples. The solid lines represent AZTEC encoding of this sampled signal.
## 0 1 1 0 -1 10 20 40 50 20 -1 -30 -20 -10 0 1 0 0 1 -1
(a) List the data array that represents the AZTEC encoding of this signal.
(b) How much data reduction does AZTEC achieve for this signal?
(c) Which data points in the following list of raw data that would be saved if
the Turning Point algorithm were applied to this signal?
## (d) If this signal were encoded with a Huffman-type variable-bit-length code
with the following four bit patterns as part of the set of codes, indicate
which amplitude value you would assign to each pattern.
(e) How much data reduction does each algorithm provide (assuming that no
coding table needs to be stored for Huffman coding)?
53
Ans.
(a) {4, 0, 4, 50, 3, 30, 3, 0, 5, 0}
(b) 2 : 1
(d)
Amplitude
Code value
1 0
01 1
001 1
0001 20
## (e) All are the same 2 : 1
10.14 AZTEC encodes a signal as {2, 50, 4, 30, 4, 50, 4, 30, 4, 50, 2, 50}.
How many data points were originally sampled?
Ans.
20
10.15 After applying the AZTEC algorithm to a signal, the saved data array is
{2, 0, 3, 80, 3, 30, 3, 0, 3, 0}. Draw the waveform that AZTEC would
reconstruct from these data.
Ans.
80
40
0
30
54
10.16 AZTEC encodes a signal from an 8-bit analog-to-digital converter as
{2, 50, 4, 30, 6, 50, 6, 30, 4, 50, 2, 50}. (a) What is the amount of data
reduction? (b) What is the peak-to-peak amplitude of a signal reconstructed
from these data?
Ans.
(a) 2 : 1
(b) 20
## 10.17 AZTEC encodes a signal from an 8-bit analog-to-digital converter as
{3, 100, 5, 150, 5, 50, 5, 100, 2, 100}. The TP algorithm is applied to the
same original signal. How much more data reduction does AZTEC achieve
on the same signal compared to TP?
Ans.
Same amount of data reduction.
10.18 The graph below shows a set of 20 data points of an ECG sampled with an
8-bit analog-to-digital converter.
65
60
55
50
45
40
35
Amplitude
30
25
20
15
10
5
0
-5
-10
-15
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Sample time
55
(a) Draw a Huffman binary tree similar to the one in Figure 10.9(b)
including the probabilities of occurrence for this set of data.
(b) (5 points) From the binary tree, assign appropriate Huffman codes to the
numbers in the data array.
Number Huffman
code
10
0
10
20
60
(c) Assuming that the Huffman table does not need to be stored, how much
data reduction is achieved with Huffman coding of this sampled data
set? (Note: Only an integral number of bytes may be stored.)
(d) Decode the following Huffman-coded data and list the sample points
that it represents:
01010110001
56
Ans.
(a)
1.0
1 0
0.50 0.50
1 0
0
0.30 0.20
1 0
20
0.15 0.15
1 0
10
0.10 0.05
10 60
(b)
Data Huffman Number of Total
point code points number of
bits
10 0101 2 8
0 1 10 10
10 011 3 9
20 00 4 8
60 0100 1 4
Total = 39 bits
57
39 bits
= 5 bytes
8 bits per byte
## (c) Assuming that all initial numbers were byte-length, 20 : 5 = 4 : 1
! (d) 10 10 20 Last two bits have bit error-no code.
58
Chapter 11
## 11.5 A 100-Hz-bandwidth ECG signal is sampled at a rate of 500 samples/s. (a)
Draw the approximate frequency spectrum of the new digital signal obtained
after sampling, and label important points on the axes. (b) On the same
graph, draw the approximate spectrum that would be averaged from a set of
normal QRS complexes.
Ans.
(a)
f
100 400 500 600 900 1000 1100
(b)
f
5 10 15
59
Chapter 12
## 12.9 Experiments to determine the frequency characteristics of the average QRS
complex have shown that the largest spectral energy of the QRS complex
occurs at approximately what frequency?
Ans.
10 Hz
2
12.10 A filter with the difference equation, y(nT ) = [ y(nT " T )] + x(nT ) , is best
described as what traditional filter type?
Ans.
! with the linear mathematics that we are using.
It is nonlinearnot definable
12.11 The center frequency of the optimal QRS bandpass filter is not at the
location of the maximal spectral energy of the QRS complex. (a) What
function is maximized for the optimal filter? (b) What is the center
frequency of the optimal QRS filter for cardiotachometers? (c) If this filter
has the proper center frequency and a Q = 20, will it work properly? If not,
why not?
Ans.
(a) Signal-to-noise ratio
(b) 17 Hz
(c) No, too high a Q factor leads to excessive ringing in the output.
12.12 In addition to heart rate information, what QRS parameter is provided by the
QRS detection algorithm that is based on the first and second derivatives?
Ans.
QRS width
12.13 The derivative algorithm used in a real-time QRS detector has the difference
equation: y(nT ) = 2x(nT ) + x(nT " T ) " x(nT " 3T ) " 2x(nT " 4T ) . (a) Draw
its block diagram. (b) What is its output sequence in response to a unit step
input? Draw the output waveform.
!
60
Ans.
(a)
(b) {2, 3, 3, 2, 0, 0, }
3 3
2 2
-2T -T 0 T 2T 3T 4T 5T time
12.14 Write the equations for the amplitude and phase responses of the derivative
algorithm used in a real-time QRS detector that has the transfer function
## "2z "2 " z "1 + z1 + 2z 2
H(z) =
8
Ans.
2 sin (2"T ) + sin ("T )
(a) H("!
T) =
4
\$
(b) "H(#T ) =
2
!
12.15 A moving window integrator integrates over a window that is 30 samples
wide and has an overall amplitude scale factor of 1/30. If a unit impulse (i.e.,
! 1, 0, 0, 0, ) is applied to the input of this integrator, what is the output
sequence?
61
Ans.
"1 1 1 % 1
# , ,, ,,0,0,0,& Starts with total of 30 outputs of value,
\$ 30 30 30 ' 30
12.16 A moving window integrator is five samples wide and has a unity amplitude
scale factor (i.e., N = 1). A pacemaker pulse is described by the!sequence:
! (1, 1, 1, 1, 0, 0, 0, 0, ). Application of this pulse to the input of the moving
window integrator will produce what output sequence?
Ans.
{1,2,3,4,4,3,2,1,0,0,0,}
12.17 The transfer function of a filter used in a real-time QRS detection algorithm
is
!
(1" z "6 )2
H(z) =
(1" z "1 )2
For a sample rate of 200 sps, this filter eliminates input signals of what
frequencies?
!
Ans.
33.3, 66.7, and 100 Hz
62
Chapter 13
13.1 In the modern version of the portable arrhythmia monitor, the arrhythmia
analysis is based on mapping what two variables into two-dimensional
space?
Ans.
RR interval and QRS duration
## 13.2 Current real-time QRS detection algorithms developed at the UW can
correctly detect approximately what percentage of the QRS complexes in a
standard 24-hour database?
Ans.
More than 99.3%
13.3 In arrhythmia analysis, the RR interval and QRS duration for each beat are
mapped into a two-dimensional space. How is the location of the center of
the box marked Normal established?
Ans.
It is an average of the RR interval and QRS duration for the most-recent past
8 beats recognized as normal.
13.4 Which of the following best describe the portable arrhythmia monitor
developed at UW: (a) is a distributed processing approach, (b) selects
important signals and stores them on magnetic tape for subsequent playback
to a central computer over the telephone, (c) stores RR intervals and QRS
durations in its memory so that a 24-hour trend plot can be made for these
variables, (d) uses ST-segment levels as part of the arrhythmia analysis
algorithm, (e) saves 30 16-second ECG segments in its memory, (f)
transmits over the telephone using a separate modem that fits in a shirt
pocket, (g) currently uses an CMOS 8088 microprocessor but will be
updated soon, (h) always stores the ECG segment that preceded an alarm,
(i) is being designed as a replacement for a Holter recorder, (j) uses the new
medical satellite network to send its data to the central computer by
telemetry, (k) has a built-in accelerometer for monitoring the patient's
activity level, (l) includes 256 kbytes of RAM to store ECG signals, (m) uses
two features extracted from the ECG in the arrhythmia analysis, (n) does
near-optimal QRS detection so it will be produced commercially by a com-
63
pany early next year, (o) saves all the sampled two-channel ECG data for 24
hours, (p) stores the single ECG segment that caused an alarm, (q) analyzes
Ans.
(a), (i), and (p)
13.6 Describe the QRS detection technique that is used most in high-performance
commercial arrhythmia monitors such as in the intensive care unit.
Ans.
Template matching
13.7 Explain how you would approach the problem of writing software to do 12-
lead ECG interpretation by computer so that it would be commercially
accepted.
Ans.
The commercially-accepted software mimics the decision logic of the
physician. Thus physicians can see how the interpretation is arrived at.
13.8 What are some other techniques of measuring the ST-segment level? Give
method.
Ans.
Normally ST-level is measured at a fixed interval after a fiducial point such
as the R-peak or the J-point.
64 | 3,662 | 12,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-35 | latest | en | 0.920209 |
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teralitre femtolitre
How many teralitre in 1 femtolitre? The answer is 1.0E-27.
We assume you are converting between teraliter and femtoliter.
You can view more details on each measurement unit:
teralitre or femtolitre
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 1.0E-9 teralitre, or 1.0E+18 femtolitre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between teraliters and femtoliters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of teralitre to femtolitre
1 teralitre to femtolitre = 1.0E+27 femtolitre
2 teralitre to femtolitre = 2.0E+27 femtolitre
3 teralitre to femtolitre = 3.0E+27 femtolitre
4 teralitre to femtolitre = 4.0E+27 femtolitre
5 teralitre to femtolitre = 5.0E+27 femtolitre
6 teralitre to femtolitre = 6.0E+27 femtolitre
7 teralitre to femtolitre = 7.0E+27 femtolitre
8 teralitre to femtolitre = 8.0E+27 femtolitre
9 teralitre to femtolitre = 9.0E+27 femtolitre
10 teralitre to femtolitre = 1.0E+28 femtolitre
## ››Want other units?
You can do the reverse unit conversion from femtolitre to teralitre, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Teraliter
The SI prefix "tera" represents a factor of 1012, or in exponential notation, 1E12.
So 1 teraliter = 1012 liter.
## ››Definition: Femtoliter
The SI prefix "femto" represents a factor of 10-15, or in exponential notation, 1E-15.
So 1 femtoliter = 10-15 liter.
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Prev Next | 968 | 4,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.842594 |
https://www.doubtnut.com/qna/646341659 | 1,721,372,870,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00723.warc.gz | 653,808,208 | 36,048 | # Following figure shows cross-sections through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor B will fit snugly within conductor A, and conductor C will fit snugly within conductor B. Relationship between their end to end resistance is
A
RA=RB=RC
B
RA>Rb>RC
C
RA<RB<R
D
Information is not sufficient
Text Solution
Verified by Experts
## All the conductors have equal lengths. Area of cross-section of A is {(√3a)2−(√2a)2}=a2 Similarly area of cross-section of B = Area of crosssection of C=a2 Hence according to formula R=p1A, resistances of all the conductors are equal i.e., RA=RB=RC
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## Following figure shows cross-section through three long conductors of the same length and material, with square cross-section of edge lengths as shown. Conductor B will snugly within conductor B. Relationship between their end to end resistance is
ARA=RB=RC
BRA>RB>RC
CRA<RB<RC
DInformation is not sufficient
• Question 2 - Select One
## How is electrical conductance of a conductor related with length and area a cross section of the conductor
AG=l.a.k1
BG=k.i.a1
CG=k.a.i1
DG=k.i.a2
• Question 3 - Select One
## How is electrical conductance of a conductor related with length and area of cross section of the conductor ?
AG=l.a.κ1
BG=κ.l.a1
CG=κ.a.l1
DG=κ.l.a2
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# Using SolidWorks Flow Simulation to Determine Which Fan Blade is Better
## Using SolidWorks Flow Simulation to Determine Which Fan Blade is Better
(OP)
Hey Everybody,
I am using SolidWorks Flow Simulation to look at the internal flow of air passing through a tube. Inside the tube is a small fan blade (approximate .700" OD). I am simulating the fan blade running at 2000 RPM.
I have several versions of the fan blade, and I am trying to figure out which "metric" to use to make an apples-to-apples comparison between the 4 different fan blades that I have designed.
I do not see MASS FLOW RATE as one of the options for simulation results. MASS FLOW RATE is however one of my goals for simulation convergence. When I look at the GOAL PLOT, the MASS FLOW RATE is positive and negative...like a transient response...until the MASS FLOW RATE converges. Is the converged value for MASS FLOW RATE the number that I should be looking at?
I thought that I would have to somehow integrate the velocity over the cross-sectional area and then multiply by the density (assuming constant density for air for simplification).
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Register now while it's still free! | 513 | 2,354 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-38 | latest | en | 0.889373 |
https://stats.stackexchange.com/questions/225277/difference-between-descriptive-and-predictive-modelling/381788 | 1,627,252,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046151866.98/warc/CC-MAIN-20210725205752-20210725235752-00429.warc.gz | 547,524,412 | 40,521 | Difference between descriptive and predictive modelling
I was wondering if anyone could help me clear up the difference between descriptive and predictive modelling. I am trying to build a model to predict where house prices will go up. To do this I need to examine historical data to identify predictors of house price increase, and how they relate to predict an increase.
To me this seems like it fits the description of descriptive modelling and predictive modelling. I am looking at historical data and trying to find the set of rules that summarise how we get from the variables to the current house price, so that I can use the same rules to predict from current conditions to future unknown house prices.
However, this also seems like predictive modelling because we are finding rules to predict an outcome.
So my question is, what is this: descriptive or predictive? And how can I determine whether something is descriptive or predictive modelling?
• hi, don't you want to point out the answer? – nickolay Feb 29 '20 at 18:22
The best graphic I've seen summarizing this is from here:
The bigger focus is on whether or not the statistical model you're building is for statistical inference, or predictive accuracy. Breiman has a good paper explaining the differences, Statistical Modeling - A Tale of Two Cultures. In Breiman's paper, "data modeling" is equivalent to models built for statistical inference, whereas "algorithmic modeling" is closer to models built for predictive accuracy.
In the comment below, @boscovich mentions G. Shmueli's paper, To Explain or Predict? The paper looks to speak directly to your original question (Sections 1.2 and 1.3), and offers a trove of information beyond, for example: model evaluation and selection is also covered (Section 2.6), and two examples are provided (Section 3). This paper is probably a much more digestible starting point than Breiman's.
The biggest difference IMO is related to model fit and assessment. With models built for statistical inference, you are looking at in-sample fit (i.e. the entire sample population). With models built for predictive accuracy, you are looking at out-of-sample fit (i.e. the dataset which represents your sample population is split into a training and test set, and you judge the predictive accuracy by a measure of error - such as MSE, RMSE, MASE, etc. - on the test set). Rob Hyndman has a good paper covering measures of error, Another Look at Measures of Forecast Accuracy.
One thing that gets a bit confusing is that the terms in-sample fit and out-of-sample fit are often relative terms. Suppose your goal is a statistical model built for predictive accuracy. You randomly sample your dataset, splitting it as a 70/30 training-test set. Your model is built on the 70% training set, and scored (judged) by performance on the 30% test set. The training-test set approach is used to avoid overfitting or misspecification. Ultimately, you may want to deploy your model to the population, or even to a validation set. These can also be considered "out-of-sample". The validation set is data that are (read: should be) from the population, but that you never see during the model building process (either in training or test; for all you know it may not exist), but ultimately one where your model is deployed and scored on. A validation set approach is often used in Kaggle competitions. You build your model on the training set, score on the test set, and then it is scored again on the validation set.
Both model types are making "predictions" but it is the domain of which those predictions are judged that make a difference. For statistical inference, the fit is assessed by examining residuals, p-values at specific levels of alpha, etc. For predictive accuracy, you tend to be less concerned with those, and are more aligned with answering the question "Does this model make accurate predictions out-of-sample?"
• Maybe you can edit the Breiman paper into your answer, now that you have more points. projecteuclid.org/download/pdf_1/euclid.ss/1009213726 – Mark L. Stone Jul 23 '16 at 16:34
• @MarkL.Stone, thank you! I just added it in and expanded on the answer. Greatly appreciated. – J.M. Jul 23 '16 at 16:36
• This is another excellent paper on this subject: G.Shmueli, To Explain or to Predict?, Statistical Science 2010 stat.berkeley.edu/~aldous/157/Papers/shmueli.pdf – boscovich Jul 23 '16 at 19:55
• @boscovich thank you! I revised the reply and added the paper - it looks to be a much more digestible starting point than Breiman's. – J.M. Jul 23 '16 at 20:10
You are doing predictive modeling. You know that you're doing predictive modeling when you don't care much (or at all) about how your model is arriving at its predictions as long as its predictions are as accurate as possible without overfitting. Easy examples of this include using any of the following models: gradient boosted trees, random forests, or neural networks.
These models are so complex that even though you could look at their weights and splits if you really wanted to, the amount of information would be too overwhelming to make sense of. It's straight forward enough to make sense of one decision tree but not thousands. The same can be said of neural networks, particularly deep learning for object detection in images. A deep neural network can have millions of weights.
1. Descriptive Analytics, which use data aggregation and data mining to provide insight into the past and answer: “What has happened?”
2. Predictive Analytics, which use statistical models and forecasts techniques to understand the future and answer: “What could happen?”
3. Prescriptive Analytics, which use optimization and simulation algorithms to advice on possible outcomes and answer: “What should we do?”
• This is a somewhat narrow answer; the question could be answered in much broader terms. It would also help the credibility if a quote or source could be included. Additionally it is a very old question that did not seem to generate much interest. – cherub Dec 13 '18 at 9:49 | 1,309 | 6,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-31 | latest | en | 0.942437 |
https://electronics.stackexchange.com/questions/410091/how-to-find-voltage-on-zener-diode-while-there-is-a-current-source | 1,716,859,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00854.warc.gz | 190,206,820 | 41,132 | # How to find voltage on zener diode while there is a current source
simulate this circuit – Schematic created using CircuitLab
I'm confused about whether zener is on or off due to not being able to know voltage on V0. I tried to find by applying KCL & KVL, but current on the diode (Id) and the resistor (Ir) is not equal to each other. I'm using the shockley diode equation.(Id=Is(e^(Vd/Vt))). This current source being connected to zener diode instead of a voltage source makes something complicated. Which way should I follow to find V0? Hope one can give me a trick.
• I don't see Id or Ir marked on your schematic so I can't follow your reasoning. Also, please show all of the work you have done so far. Dec 2, 2018 at 21:08
• What would Vo be if you assume that the Zener is not conducting? Is it higher than the Zener voltage? Dec 2, 2018 at 21:12
• What I'm trying to do is that if I find the current flowing through resistor (here I use Id current for each branch one by one because Ir and Id are the same, but to find Id , Vd has to be known. The given Vd values is for constant voltage drop model that requires relatively higher voltage than ,in this circuit, V0 , therefore I am not sure whether to use it) I would find voltage on resistor so then the V0. Dec 2, 2018 at 21:48
• @Klementayn Assume the zener is off. This just means that you have two resistors in parallel, or $\frac23$k, times 3 mA. This works out to 2 V. Add 600 mV for your diodes and this would account for the entire current. But at 2.6 V, the zener isn't going to conduct current (leakage only.) So the assumption is supported. Do you need to be more detailed than this?
– jonk
Dec 2, 2018 at 21:49
There are only two options here the zener is ether conducting or it isn't
If it is:
$$\dfrac{Vo-Vz}{R1}+\dfrac{Vo-Vd}{R2}+\dfrac{Vo-Vd}{R3} = I1$$
If it isn't:
$$\dfrac{Vo-Vd}{R2}+\dfrac{Vo-Vd}{R3} = I1$$
Where Vo is the voltage a junction of the diodes, Vz is the zener voltage and Vd is forward diode drop (0.6V).
What do these say? Does only one make sense?
Treat the current source as the variable. At what current does the diode cut on? This is solved by assuming the diodes current is zero, and solving for the current needed to raise the zener diode voltage to 3V. Now that you know what current is going to turn on the diode, is 3mA greater or less than this current? With the state of the zerer solved, the rest of the problem is simple.
• Other way around. You don't know if there is sufficient voltage to turn allow current in the Zener diode. So, ignore the Zener initially and see what voltage is present at node V0. Dec 3, 2018 at 5:07
• Both methods work. I think the current first method is more deterministic as it involves no hypotheticals. Dec 3, 2018 at 5:28
• That doesn't make sense to me. The value of current is fixed and specified. It is trivial to calculate the voltage at node V0 when the Zener diode is disconnected. It is then trivial to re-calculate the voltage at Node V0 should there be sufficient voltage to allow the Zener diode to conduct. Dec 3, 2018 at 5:43
• Did you try to solve it my way to see how hard it is? Both methods are trivial. This is a trivial problem that can be solved by inspection alone, however, if the student had a more complex problem in the future, with 5 legs with zeners, than the difference in the methods might be more obvious. I was trying to teach a method that will be useful in the future. Dec 6, 2018 at 2:48
• why not post an answer where you solve the problem using the method that you suggest. Dec 6, 2018 at 16:49 | 980 | 3,582 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-22 | latest | en | 0.930377 |
https://www.hyperfinecourse.org/forums/topic/magentic-moment-and-g-factor/ | 1,657,028,802,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104576719.83/warc/CC-MAIN-20220705113756-20220705143756-00660.warc.gz | 870,917,080 | 7,516 | # Magentic moment and g-factor
homepage Forums g-factor calculations Magentic moment and g-factor
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Carlos M Fajardo Z
Participant
The magnetic moment of the ground state of 111Cd is -0.5940 nm, while its spin is 1/2. Since g=\mu/I, the g-factor is -1.188.
The free electron g factor can be expressed as g = (\mu/\mu_B) / (S/\hbar). Since S = m_s\hbar and \mu = m\mu_B, g = 10^{-3}/0.5 = 0.02.
For the neutron we can no longer use the Bohr magneton but the whole expression in terms of the mass of the neutron. \mu = -g (e\hbar / 2m_n) (S/\hbar). Rearranging this expression, we get that g is = 54.345m\mu_B.
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• You must be logged in to reply to this topic. | 247 | 748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-27 | longest | en | 0.82897 |
https://dsp.stackexchange.com/questions/64147/frequency-modulation-in-matlab | 1,718,933,228,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00466.warc.gz | 185,674,192 | 40,462 | # Frequency modulation in MATLAB
I am trying to frequency modulate a cosine signal(for assignment). I have written the following code but not getting the required output.
dt = 0.0002;
t = 0:dt:0.05;
fm = 50;
fc = 1000;
beta1 = 1;
beta2 = 2;
%making sine and m(t)
m_t = 5.*cosd(2.*pi.*fm.*t);
sin_fm = sind(2.*pi.*fm.*t); %integral of m_t. A_m adjusted in beta
%fmmod signal
fmmod1 = cosd(2.*pi.*fc.*t + beta1.*sin_fm);
fmmod2 = cosd(2.*pi.*fc.*t + beta2.*sin_fm);
%frequency_deviation
kf_1 = (beta1*fm)/5; %5 is amplitude of message signal(Am)
kf_2 = (beta2*fm)/5;
%frequency modulated signal of matlab
fmmod1_m = fmmod(m_t,fc,1/dt,kf_1);
fmmod2_m = fmmod(m_t,fc,1/dt,kf_2);
figure(1);
plot(t,fmmod1);
title('Without using fmmod beta=3');
figure(2);
plot(t,fmmod2);
title('Without using fmmod beta=5');
figure(3);
plot(t,fmmod1_m);
title('With using fmmod beta=3');
figure(4);
plot(t,fmmod2_m);
title('With using fmmod beta=5');
I have used sind in place of sin in the code because for earlier code of Amplitude Modulation sin was not giving the required waveform but when I used sind it was giving the required waveform.
AM code
dt1 = 0.0001;
fc1 = 500*1000;
fm1 = 5*1000;
t1 = 0:dt1:0.1;
%making cosine for message and carrier for higher frequency
cos_fm1 = cosd(2*pi*fm1.*t1);
cos_fc1 = cosd(2*pi*fc1.*t1);
%making Udsb_am signal for higher frequecny
Udsb_am2h = 2.*cos_fc1 + cos_fc1.*cos_fm1;
Udsb_am1h = 1.*cos_fc1 + cos_fc1.*cos_fm1;
Udsb_am_half_h = (0.5).*cos_fc1 + cos_fc1.*cos_fm1;
figure(1);
plot(t1,Udsb_am2h);
title('AM for Ac=2');
figure(2);
plot(t1,Udsb_am1h);
title('AM for Ac=1');
figure(3);
plot(t1,Udsb_am_half_h);
title('AM for Ac=0.5');
What I want to ask from the question? When should I use sind and sin?(I know sind is for degree and sin is for radian).Asking question about sind and sin because when I saw the code on internet they have used sin not sind.
What is wrong in the code of FM?
@A Q. To make your life easier, I suggest you stop using the cosd() command and only use MATLAB's cos() command. For your AM code, if variable dt1 is measured in seconds then your Fs sampling rate is 10,000 samples per second. But your carrier frequency fc1 is set to 500,000. If your fc1 = 500,000 is measured in Hz (cycles/second) then your fc1 value violates the Nyquist criterion. To continue your AM modeling, try these commands (notice the 0.3 "modulation index"):
fc1 = 1000;
fm1 = 20;
cos_fm1 = cos(2*pi*fm1.*t1);
cos_fc1 = cos(2*pi*fc1.*t1);
Udsb_am2h = 2.*cos_fc1 + .3*cos_fc1.*cos_fm1;
Udsb_am1h = 1.*cos_fc1 + .3*cos_fc1.*cos_fm1;
Udsb_am_half_h = (0.5).*cos_fc1 + 0.3*cos_fc1.*cos_fm1;
The difference is the expected input. The cosd function expects the input to be expressed in degrees, and the cos function expects the input to be in radians. So you have cosd(theta)==cos(deg2rad(theta)). This is just from the MATLAB documentation page.
Are you using these correctly? For example, you do cosd(2*pi*F*t). So we have $$\frac{2\pi \text{ radians}}{\text{cycle}}.\frac{F \text{ cycles}}{\text{second}}.[\text{seconds}]=2\pi F \text{ radians}$$, as the input. You should be using cos not cosd because your input is expressed in radians.
BTW, I do not check the correctness of the actual code. That is on you.
• If $2\pi F$ is in radians then why does the AM code does not work for sin and cos but works for sind and cosd?
– A Q
Commented Feb 26, 2020 at 15:22
• BTW, I do not check the correctness of the actual code. That is on you. Commented Feb 26, 2020 at 16:03
• Only checking the code can solve the issue. Main code is only 4 to 6 lines.
– A Q
Commented Feb 26, 2020 at 16:05 | 1,210 | 3,638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.707035 |
https://www.itl.nist.gov/div898/software/dataplot/refman2/auxillar/papppf.htm | 1,555,916,207,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578544449.50/warc/CC-MAIN-20190422055611-20190422081408-00025.warc.gz | 736,456,825 | 3,856 | Dataplot Vol 2 Vol 1
# PAPPPF
Name:
PAPPPF (LET)
Type:
Library Function
Purpose:
Compute the Polya-Aeppli percent point function.
Description:
The formula for the Polya-Aeppli probability mass function is
with and p denoting the shape parameters.
The cumulative distribution function is computed using the following recurrence relation (from page 379 of Johnson, Kemp, and Kotz)
The percent point function is computed by computing the cumulative distribution function until the appropriate probability is obtained.
Syntax:
LET <y> = PAPPPF(<x>,<theta>,<p>)
<SUBSET/EXCEPT/FOR qualification>
where <x> is a variable, number, or parameter in the interval (0,1);
<theta> is a positive number or parameter that specifies the first shape parameter;
<p> is a positive number or parameter that specifies the second shape parameter;
<y> is a variable or a parameter where the computed Polya-Aeppli ppf value is stored;
and where the <SUBSET/EXCEPT/FOR qualification> is optional.
Examples:
LET A = PAPPPF(0.95,3,0.5)
LET Y = PAPPPF(P,2,0.3)
PLOT PAPPPF(P,2,0.3) FOR P = 0 0.01 0.99
Default:
None
Synonyms:
None
Related Commands:
PAPCDF = Compute the Polya-Aeppli cumulative distribution function. PAPPDF = Compute the Polya-Aeppli probability mass function. LPOPDF = Compute the Lagrange-Poisson percent point function. BTAPDF = Compute the Borel-Tanner probability mass function. LOSPDF = Compute the lost games probability mass function. POIPDF = Compute the Poisson probability mass function. HERPDF = Compute the Hermite probability mass function. BINPDF = Compute the binomial probability mass function. NBPDF = Compute the negative binomial probability mass function. GEOPDF = Compute the geometric probability mass function. INTEGER FREQUENCY TABLE = Generate a frequency table at integer values with unequal bins. COMBINE FREQUENCY TABLE = Convert an equal width frequency table to an unequal width frequency table. KS PLOT = Generate a minimum chi-square plot. MAXIMUM LIKELIHOOD = Perform maximum likelihood estimation for a distribution.
References:
Douglas (1980), "Analysis with Standard Contagious Distributions", International Co-operative Publishing House, Fairland, MD.
Evans (1953), "Experimental Evidence Concerning Contagious Distributions in Ecology", Biometrika, 40, pp. 186-211.
Johnson, Kotz, and Kemp (1992), "Univariate Discrete Distributions", Second Edition, Wiley, pp. 378-382.
Applications:
Distributional Modeling
Implementation Date:
2006/6
Program:
```
title size 3
tic label size 3
label size 3
legend size 3
height 3
multiplot scale factor 1.5
x1label displacement 12
y1label displacement 17
.
multiplot corner coordinates 0 0 100 95
multiplot scale factor 2
label case asis
title case asis
case asis
tic offset units screen
tic offset 3 3
title displacement 2
x1label Probability
y1label X
.
xlimits 0 1
major xtic mark number 6
minor xtic mark number 3
.
multiplot 2 2
.
title Theta = 0.5, P = 0.5
plot papcdf(x,0.5,0.5) for x = 0 0.01 0.99
.
title Theta = 1, P = 0.5
plot papcdf(x,1,0.5) for x = 0 0.01 0.99
.
title Theta = 2.5, P = 0.5
plot papcdf(x,2.5,0.5) for x = 0 0.01 0.99
.
title Theta = 5, P = 0.5
plot papcdf(x,5,0.5) for x = 0 0.01 0.99
.
end of multiplot
.
justification center
move 50 97
text Percent Point for Polya-Aeppli
```
Date created: 6/20/2006
Last updated: 6/20/2006 | 980 | 3,337 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-18 | longest | en | 0.679165 |
https://help.openstreetmap.org/questions/22126/how-do-i-define-a-boundary-for-a-suburb-correctly | 1,620,527,818,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00394.warc.gz | 310,444,611 | 8,201 | # How do I define a boundary for a suburb correctly?
1 Hi everyone, I defined a boundary for the suburb I live in, because the current placement of suburb places in the map causes my street to be found in the wrong suburb. I did a multipolygon definition with the relevant ways in it and set the existing node for the suburb with the same name as label for the multipolygon. However the multipolygon does not appear as the boundary for the suburb. If I do a search for "Harleshausen" only the place-node is found not the boundary. On the contrary if I do a search for "Prenzlauer Berg" there is a result for the place-node as well as for the boundary. I tried to do the definition of my suburb according to the settings used for the "Prenzlauer Berg". Any hints how to get it right? What do I miss? Regards, AndrĂ© asked 06 May '13, 09:17 AnSc_de 16●1●1●1 accept rate: 0%
3 I assume that you are reffering to this boundary relation: relation/2911422 In the moment when writing this answer, the geometry of the relation is broken. Click the link above to see it ... before someone other is repairing it ... you can see this if you are listed there as the last editor. Solution: There are two positions where the minor suburb boundary ways meet the bigger boundaries. The bigger boundary ways have to be splitted at these points. So all ways in the relation need to form a closed circle. I really recommend to do that splitting and fixing of the relation in the offline editor JOSM, and not in the online editor Potlatch2 that you used so far. Search the OSM wiki for more details and help. And be aware: the Nominatim database needs some days or hours to be synchronized with the main OSM database. According to the tags of the relation: this is looking fine on the first view. answered 06 May '13, 17:14 stephan75 12.6k●4●55●209 accept rate: 6% 1 Thanks for pointing that out. I did not recognize the requirements for multipolygons from the wiki entry. I thought it enough that the path is closed not that the endpoints of two way must meet. (06 May '13, 18:49) AnSc_de
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question asked: 06 May '13, 09:17
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last updated: 06 May '13, 18:49 | 632 | 2,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-21 | longest | en | 0.942595 |
http://www.physicsforums.com/showthread.php?t=184139 | 1,369,180,571,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368700871976/warc/CC-MAIN-20130516104111-00018-ip-10-60-113-184.ec2.internal.warc.gz | 663,741,083 | 7,060 | Torque calculation, is this correct?
I'm experimenting some ideas and needed a verification of my formula.
There is a torque arm of 15" with a force of 6000 lbs perpendicular to the axis of rotation. So that calculates to 7500 ft/lbs (6000*15/12). So on the same axis, there is a torque arm of 5" the must oppose this force equally, I need that force required. I calculated 18000 lbs of opposing force with 7500 ft/lbs at the axis (7500/5*12).
Is this valid?
No its not a homework question :) I need to know how powerful a single disc brake on a truck is. | 145 | 559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2013-20 | latest | en | 0.926976 |
https://studysoup.com/tsg/228537/precalculus-5-edition-chapter-8-1-problem-4 | 1,604,168,914,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922411.94/warc/CC-MAIN-20201031181658-20201031211658-00003.warc.gz | 536,802,812 | 10,816 | ×
×
# Solved: In Exercises 35, solve each system of equations
ISBN: 9780321837349 160
## Solution for problem 4 Chapter 8.1
Precalculus | 5th Edition
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2
Problem 4
In Exercises 35, solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination cx - 2y + z = 0y - 3z = -12y + 5z = -2
Step-by-Step Solution:
Step 1 of 3
Step 2 of 3
Step 3 of 3
#### Related chapters
Unlock Textbook Solution
Solved: In Exercises 35, solve each system of equations
× | 214 | 716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-45 | latest | en | 0.789168 |
http://blog.keenessays.com/2018/03/for-the-nonprofit-ballet-company-in-problem-17-of-chapter-9-assume-following-pe/ | 1,529,651,822,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00301.warc.gz | 41,382,972 | 9,200 | ## For the nonprofit ballet company in Problem 17 of Chapter 9, assume following pe
For the nonprofit ballet company in Problem 17 of Chapter 9, assume following percentages of donors and gift levels: The company has set a financial goal of \$150,000. How many prospective donors must they contact for donations at the \$100 level or below to have a 95% chance of meeting this goal? Assume that the number of gifts at each level follow a discrete uniform distribution or a uniform distribution for the percentage of solicitations at the \$100 and \$50 levels. | 118 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-26 | longest | en | 0.900487 |
https://www.freezingblue.com/flashcards/print_preview.cgi?cardsetID=33427 | 1,513,218,868,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948537139.36/warc/CC-MAIN-20171214020144-20171214040144-00098.warc.gz | 738,712,538 | 4,836 | # Chemistry Chapter 1
Home > Preview
The flashcards below were created by user christavictoria on FreezingBlue Flashcards.
1. what is a mixture?
a kind of matter that can be seperated by physical means into other kinds of matter.
2. what are the two types of mixtures?
• homogeneous: non-uniform throughout
• heterogeneous: uniform throughout.
3. what is the difference between a law and a theory?
A LAW IS A CONCISE STATEMENT OR MATHEMATICAL EQUATION DESCRIBING A FUNDAMENTAL RELATIONSHIP OR REGULARITY OF NATURE does not OFFER AN EXPLANATION, WHILE A THEORY IS A TESTED EXPLANATION, CANNOT BE PROVE, AND MAY BE MODIFIED.
4. What is the law of conservation of mass?
The total mass remains constand during a chemical change (reaction) because mass cannot be created or destroyed.
5. what is precision?
the CLOSENESS to the set of calues obtained from repeated measurement of the same quantity. Not neccessarily accurate.
6. what is accuracy?
the closeness of a single measurement to its true value.
7. mega
10^6
8. kilo
10^3
9. deci
10^-1
10. centi
10^-2
11. mili
10^-2
12. micro
10^-6
13. nano
10^-9
14. pico
10^-10
15. Temperature:
the measure of "hotness". It is related to how energetic the molecules are in a particular sample.
16. Heat:
A quantity of energy transferred in a chemical reaction. Heat flows from an area of high temp. to an area od lower temp.
17. the formula for going from degrees celcuis to degrees KELVIN
C + 273.15K = C
18. Celcius to Farenheight
F = (C x 1.8) +32F
19. Farenheight to Celcuis
C= F - 32F / 1.8
20. Kelvin to Celcius
C = (K-273.15K)
21. Chemical Formula:
a notation that uses atomic symbols with numerical subscripts to convey the relative proportions of atoms of elements uin a substance.
22. molecular compound:
a definite group of atoms that are chemically bonded together.
23. Molecular formula:
a formula giving the exact number of different atoms iof an atoms in a molecule.
24. Ion:
electrically charged particle obtained from an atom or group of atoms by adding or removing one or more electrons.
## Card Set Information
Author: christavictoria ID: 33427 Filename: Chemistry Chapter 1 Updated: 2010-09-08 16:01:33 Tags: Chem ch1 Folders: Description: Chemistry chapter 1 Show Answers:
Home > Flashcards > Print Preview | 611 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2017-51 | longest | en | 0.862589 |
https://academicpros.net/owlv2-online-teaching-and-learning-resource-from-cengage-learning-microsoft-edge-a/ | 1,721,255,110,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514809.11/warc/CC-MAIN-20240717212939-20240718002939-00168.warc.gz | 72,395,132 | 13,974 | # OWLv2 | Online teaching and learning resource from Cengage Learning – Microsoft Edge A
Please see the attached picture for the question that I am needing help with.
OWLv2 | Online teaching and learning resource from Cengage Learning – Microsoft EdgeA https://sjc.cengagenow.com/ilm/takeAssignment/takeCovalentActivity.do?locator=assignment-take&takeAssignmentSessionLocator=assignment-takeChapter 10[Review Topics][References]1. Mole Ratio1 ptsMUse the References to access important values if needed for this question.2. Stoichiometry: Mole/Mole (Intege… 1 pts 2reqExpressing amounts of energy in different energy units is necessary to solve many chemistry problems. Forpractice, complete the following table3. Stoichiometry: Mole/Mole1 pts 2reqThe Joule (J) is the SI unit of energy. 1 calorie (cal) – 4 184 J1. Stoichiometry: Mass/Mass1 pts 2reqJcalKJ5. Stoichiometry: Mass/Mole1 pts 2req6955. Percent Yield: Calculate1 pts 2req101Percent Yield: Applicationspts 2req0.6348. Limiting Reactants1 pts 2reqSubmit AnswerRetry Entire Group9 more group attempts remaining9. Energy: Unit Conversions1 pts 2reqQuestionOQuestionQuestion10. Thermochemical Equations: En…1 pts 2req11. Thermochemical Equations: Wr… 1 pts 2req12. Thermochemical Stoichiometry1 pts 2reqType here to searchea6:59 PM5/11/2019hp
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 671 | 2,690 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.759551 |
https://www.quantconnect.com/terminal/cache/embedded_backtest_54c5e8425b6f4f6f142d8c2a64fc0c6d.html | 1,669,486,644,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446708046.99/warc/CC-MAIN-20221126180719-20221126210719-00522.warc.gz | 1,034,930,349 | 40,727 | Overall Statistics Total Trades40Average Win132.52%Average Loss-39.82%Compounding Annual Return32.097%Drawdown68.300%Expectancy1.164Net Profit2821.538%Sharpe Ratio0.947Loss Rate50%Win Rate50%Profit-Loss Ratio3.33Alpha0.199Beta0.917Annual Standard Deviation0.283Annual Variance0.08Information Ratio0.823Tracking Error0.235Treynor Ratio0.292Total Fees\$487.72
```namespace QuantConnect
{
/*
* QuantConnect University: Full Basic Template:
*
* The underlying QCAlgorithm class is full of helper methods which enable you to use QuantConnect.
* We have explained some of these here, but the full algorithm can be found at:
* https://github.com/QuantConnect/Lean/tree/master/Algorithm
*/
/// <summary>
/// Basic template algorithm simply initializes the date range and cash
/// </summary>
public class LRP : QCAlgorithm
{
//the leverage for each holding
decimal leverage = 3.8m;//Config.GetValue<decimal>("leverage", 3.5m);
//the days interval to perform rebalance
int days = 391;//Config.GetInt("days", 30);
DateTime rebalanced;
decimal tlt = 0.3m;//Config.GetValue<decimal>("tlt", 0.5m);
decimal spy = 0.5m;//Config.GetValue<decimal>("spy", 0.4m);
decimal gld = 0.1m;
/// <summary>
/// Initialise the data and resolution required, as well as the cash and start-end dates for your algorithm. All algorithms must initialized.
/// </summary>
public override void Initialize()
{
SetStartDate(2004, 11, 19); //Set Start Date
SetEndDate(2017, 1, 1); //Set End Date
SetCash(100000); //Set Strategy Cash. Should be 3 month T Bills.
//using etf rather than futures
gld = 1 - tlt - spy;
//minimum of 0.1 gld
if (gld < 0.1m || gld + tlt + spy > 1)
{
Quit();
}
rebalanced = this.Time;
Schedule.On(DateRules.EveryDay(), TimeRules.AfterMarketOpen("SPY"), () =>
{
{
rebalanced = this.Time;
SetHoldings("TLT", tlt * leverage);
SetHoldings("SPY", spy * leverage);
SetHoldings("GLD", gld * leverage);
Debug("Rebalance");
}
});
}
public override void OnMarginCall(List<SubmitOrderRequest> requests)
{
//if (!LiveMode)
//{
// this.Quit();
//}
}
/// <summary>
/// OnData event is the primary entry point for your algorithm. Each new data point will be pumped in here.
/// </summary>
/// <param name="data">Slice object keyed by symbol containing the stock data</param>
public override void OnData(Slice data)
{
if (!Portfolio.Invested)
{
SetHoldings("TLT", tlt * leverage);
SetHoldings("SPY", spy * leverage);
SetHoldings("GLD", gld * leverage);
Debug("Purchased Stock");
}
}
}
}``` | 695 | 2,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2022-49 | latest | en | 0.592177 |
https://has3ong.github.io/boj/boj-16509/ | 1,670,190,327,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710980.82/warc/CC-MAIN-20221204204504-20221204234504-00133.warc.gz | 328,773,276 | 5,907 | # BOJ - 16509 - 장군
Updated:
``````import sys
from collections import deque
dx = [(-1, -2, -3), (-1, -2, -3), (0, -1, -2), (0, -1, -2), (0, 1, 2), (0, 1, 2), (1, 2, 3), (1, 2, 3)]
dy = [(0, -1, -2), (0, 1, 2), (-1, -2, -3), (1, 2, 3), (-1, -2, -3), (1, 2, 3), (0, -1, -2), (0, 1, 2)]
def moveCheck(i, x, y, ex, ey):
nx, ny = x, y
for j in range(3):
nx, ny = x + dx[i][j], y + dy[i][j]
if nx < 0 or nx >= 10 or ny < 0 or ny >= 9:
return -1, -1
if j < 2 and nx == ex and ny == ey:
return -1, -1
return nx, ny
def BFS(maps, sx, sy, ex, ey):
dq = deque()
dq.append((sx, sy))
maps[sx][sy] = 0
while dq:
x, y = dq.popleft()
if x == ex and y == ey:
return maps[x][y]
for i in range(8):
nx, ny = moveCheck(i, x, y, ex, ey)
if nx != -1 and maps[nx][ny] == -1:
dq.append((nx, ny))
maps[nx][ny] = maps[x][y] + 1
return -1
def solution():
maps = [[-1] * 9 for _ in range(10)]
print(BFS(maps, sx, sy, ex, ey))
solution()
``````
https://www.acmicpc.net/problem/16509
Categories:
Updated: | 455 | 984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2022-49 | latest | en | 0.382849 |
https://www.jiskha.com/questions/1474640/you-push-a-1kg-block-against-a-spring-with-spring-constant-k-100-nmon-a-horizontal-table | 1,627,841,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.63/warc/CC-MAIN-20210801154943-20210801184943-00685.warc.gz | 861,943,889 | 5,014 | # Physics
You push a 1kg block against a spring with spring constant k = 100 Nmon a horizontal table top. The spring is compressed to 0.2m away from its relaxed length, and then the block is released. If the horizontal table top has a
coefficient of kinetic friction of 0.2, how much further past the end of the spring does the block go before coming to rest?
1. 👍
2. 👎
3. 👁
1. energy stored in spring = (1/2) k x^2
= (1/2)(100)(.04)
work done by friction
= mu m g *d
so
50*.04 = .2 * 1 *9.81* d
solve for d
1. 👍
2. 👎
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A 0.500 kg block is sitting on a horizontal, frictionless surface. The block is connected to a horizontal spring with a force constant of 124 N/m. The other end of the horizontal spring rests against a wall. When a 100.0 g arrow
4. ### physics
A block of mass m = 3.53 kg is attached to a spring which is resting on a horizontal frictionless table. The block is pushed into the spring, compressing it by 5.00 m, and is then released from rest. The spring begins to push the
1. ### physics
A 0.250 kg block on a vertical spring with a spring constant of 5.00 103 N/m is pushed downward, compressing the spring 0.120 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above
2. ### Physics
A 2.9 kg block slides with a speed of 2.1 m/s on a friction less horizontal surface until it encounters a spring. (a) If the block compresses the spring 5.6 cm before coming to rest, what is the force constant on the spring? (b)
3. ### physics again it didnt work.
A moving 1.60 kg block collides with a horizontal spring whose spring constant is 295 N/m. The block compresses the spring a maximum distance of 3.50 cm from its rest position. The coefficient of kinetic friction between the block
4. ### Physics
1. A 5 kg weight is hung from a vertical spring. The spring stretches by 5 cm. How much mass should be hung from the spring so it will stretch by 10 cm from its original length? A. 15 kg B. 10 kg C. 20 kg D. 25 kg 2. What is the
1. ### physics
A block of ice of mass 4.10 kg is placed against a horizontal spring that has force constant k = 210 N/m and is compressed a distance 2.60×10−2 m. The spring is released and accelerates the block along a horizontal surface. You
2. ### physics
A 1kg mass is suspended from a vertical spring with a spring constant of 100 N/m and the equilibrium position is noted. The spring is then pushed upward (compressed) a distance x=10 cm before the mass is released from rest. How
3. ### physics
The drawing shows three identical springs hanging from the ceiling. Nothing is attached to the first spring, whereas a W = 4.20-N block hangs from the second spring. A block of unknown weight hangs from the third spring. From the
4. ### physics
You push a 2.0 kg block against a horizontal spring, compressing the spring by 15 cm. Then you release the block, and the spring sends it sliding across a tabletop. It stops 75 cm from where you released it. The spring constant is | 918 | 3,488 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-31 | latest | en | 0.888713 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-4-proportions-percents-and-solving-inequalities-4-4-inequalities-problem-set-4-4-page-169/59 | 1,560,746,054,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00382.warc.gz | 776,293,177 | 13,201 | Elementary Algebra
$x\lt-\dfrac{9}{4}$
Using the properties of inequalities, the solution to the given inequality, $-x-4-3x\gt5 ,$ is \begin{array}{l}\require{cancel} -x-3x\gt5+4 \\\\ -4x\gt9 \\\\ x\lt\dfrac{9}{-4} \\\\ x\lt-\dfrac{9}{4} .\end{array} | 104 | 251 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-26 | latest | en | 0.579905 |
https://thepackaginginsider.com/ask-and-learn/how-long-is-1-second-in-light-speed | 1,686,369,341,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656963.83/warc/CC-MAIN-20230610030340-20230610060340-00057.warc.gz | 647,251,099 | 13,787 | # How long is 1 second in light speed?
The speed of light traveling through a vacuum is exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second — a universal constant known in equations as "c," or light speed.
## How long is a light-second in time?
A light-second is a unit of length equal to the distance light in empty space travels in one second, and is exactly 299,792,458 meters. The light-second is used in telecommunications, astronomy, and physics.
## How long is a minute at light speed?
A light minute equals 60 light seconds.
## What happens if you travel at the speed of light for 1 second?
No, humans cannot survive travelling at the speed of light. You see, if an object travels at the speed of light, its mass will increase exponentially! Consider this… the speed of light is 300,000 kilometers per second (186,000 miles per second) and when an object moves at this speed, its mass will become infinite.
## What is 1% the speed of light?
While 1% of anything doesn't sound like much, with light, that's still really fast – close to 7 million miles per hour! At 1% the speed of light, it would take a little over a second to get from Los Angeles to New York. This is more than 10,000 times faster than a commercial jet.
## Is The dark Faster Than Light?
Nothing's faster than the speed of light. Except the speed of dark. That might sound like the tagline of a grim and gritty movie that's trying way too hard, but it also happens to be true.
## Can we travel 20% the speed of light?
Nothing can travel faster than 300,000 kilometers per second (186,000 miles per second). Only massless particles, including photons, which make up light, can travel at that speed. It's impossible to accelerate any material object up to the speed of light because it would take an infinite amount of energy to do so.
## How fast is the speed of dark?
Darkness travels at the speed of light. More accurately, darkness does not exist by itself as a unique physical entity, but is simply the absence of light. Any time you block out most of the light – for instance, by cupping your hands together – you get darkness.
## Do you age at the speed of light?
Five years on a ship traveling at 99 percent the speed of light (2.5 years out and 2.5 years back) corresponds to roughly 36 years on Earth. When the spaceship returned to Earth, the people onboard would come back 31 years in their future--but they would be only five years older than when they left.
## Do you age slower at the speed of light?
Changes to time and distance
Perhaps one of the most famous effects of special relativity is that for a human moving near the speed of light, time slows down. In this scenario, a person moving at near light speed would age more slowly. This effect is called time dilation.
## How much is 1 year light?
Light-year is the distance light travels in one year. Light zips through interstellar space at 186,000 miles (300,000 kilometers) per second and 5.88 trillion miles (9.46 trillion kilometers) per year.
## How fast is 100 light speed?
The speed of light in vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics. The speed of light c is exactly equal to 299,792,458 metres per second (approximately 300,000 kilometres per second; 186,000 miles per second; 671 million miles per hour).
## Why can't we go faster than light?
The real reason you cannot travel through space faster than light is because you are always traveling through spacetime at the speed of light.
## What is 1 second from light?
The light-second is a unit of length useful in astronomy, telecommunications and relativistic physics. It is defined as the distance that light travels in free space in one second, and is equal to exactly 299792458 metres (983571056 ft).
## Is light faster than a second?
Light travels approximately 1 foot per nanosecond or 186 miles per millisecond or 300,000 kilometers per second. Light from distant objects such as stars can take millions of years to make the journey to our eyes so that we see them as they were in the distant past rather than how they are right now.
## How many light-years are there in 1 second?
Solution : 3.26 light years make 1 par sec.
## Do pilots age slower?
Chou did the math, and it turns out that frequent fliers actually age the tiniest bit more quickly than those of us with both feet on the ground. Planes travel at high enough altitudes that the weak gravitational field speeds up the tick rate of a clock on board more than the high speeds slow it down.
## Does time freeze at the speed of light?
If you were able to travel at the speed of light, all of your motion would be wrapped up in getting you to travel at the maximum speed through space, and there would be none left to help you travel through time — and, for you, time would stop. At the speed of light, there is no passage of time.
## What can age you faster?
Exposure to light is a top cause of premature aging: Sun exposure causes many skin problems. Ultraviolet (UV) light and exposure to sunlight age your skin more quickly than it would age naturally. The result is called photoaging, and it's responsible for 90% of visible changes to your skin.
## What's the fastest thing in the universe?
So light is the fastest thing. Nothing can go faster than that. It's kind of like the speed limit of the universe.
## Is sound faster than light?
No, sound travels much slower than the speed of light. The speed of sound in air is about 340 metres per second. Whereas, the light will travel through the vacuum at a speed of 300 million metres per second.
300,000 km/s. So when I say your shadow can travel faster than light, I don't blame you if your skeptic senses are tingling. But humour me, because shadows are one thing that can travel faster than light.
## Can we go 99% the speed of light?
Albert Einstein's special theory of relativity states that photons—or particles of light—travel at a constant speed of 670,616,629 miles per hour. As far as we know, nothing can travel faster than this. But across the universe, particles are often accelerated to 99.99 percent the speed of light.
## Why is light so fast?
That's because all massless particles are able to travel at this speed, and since light is massless, it can travel at that speed.
## Can we reach 99% the speed of light?
As you attain more and more speed, the amount of energy needed to accelerate you closer to the speed of light increases exponentially. Simply put, the amount of energy needed to get from 0%-1% of light speed is not the same as 98%-99%. Because of this, you can never really reach the speed of light. | 1,524 | 6,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-23 | latest | en | 0.945931 |
http://mdct.ru/future-value/future-value-of-a-growing-annuity-calculator/ | 1,529,335,847,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267860570.57/warc/CC-MAIN-20180618144750-20180618164750-00431.warc.gz | 218,129,049 | 13,534 | Future Value of a Growing Annuity Calculator
What does it do?
This future value of a growing annuity calculator works out the future value (FV) of a regular sum of money (Pmt) which is growing or declining at a constant rate (g) each period. The payments are made at the end of each period for n periods, and a discount rate i is applied.
Formula
The calculator uses the future value of a growing annuity formula as shown below:
`FV = Pmt x ( (1 + i)n - (1 + g)n ) / (i - g)`
Instructions
The Excel future value of a growing annuity calculator, available for download below, is used to compute the future value by entering details relating to the regular payment, growth rate, discount rate and the number of periods. The calculator is used as follows:
Step 1
Enter the regular payment amount (Pmt). The regular payment is the amount received at the end of each period for n periods. The amount must be the same for each period.
Step 2
Enter the growth rate (g). The growth rate is the rate at which the original payment (Pmt) is growing each period. The rate should be for a period, so for example, if the period is a year, then the rate should be the yearly growth rate.
Step 3
Enter the discount rate (i). The discount rate is the rate used to discount each payment amount back from the end of the period in which is was made, to the beginning of period 1 (today). The rate should be for a period, so for example, if the period is a year, then the rate should be the yearly rate.
Step 4
Enter the number of periods (n). The number of periods is entered. A period can be any term (month, year etc), but must be consistent with the discount rate provided (see step 2).
Step 5
The future value of a growing annuity calculator works out the future value (FV). The answer is the value at the end of period n of an a regular sum of money growing at a constant rate (g) each period, received at the end of each of the n periods, and discounted at a rate of i. It is the future value of a growing annuity.
The future value of a growing annuity calculator is one type of tvm calculator used in time value of money calculations, discover another at the links below.
Notes and major health warnings
Users use this free annuity calculator at their own risk. We make no warranty or representation as to its accuracy and we are covered by the terms of our legal disclaimer, which you are deemed to have read. This is an example of a growing annuity future value calculator that you might use when considering how to calculate the future value of a growing annuity. It is purely illustrative of a future value of an annuity calculator. This is not intended to reflect general standards or targets for any particular business, company or sector. If you do spot a mistake in this future value growing annuity calculator, please let us know and we will try to fix it.
Future Value of a Growing Annuity Calculator November 6th, 2016
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https://forum.posit.co/t/analysis-help-choosing-method-of-analysis/20079/2 | 1,718,562,363,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00694.warc.gz | 229,631,181 | 5,646 | # analysis help--choosing method of analysis
The message often indicates that many of the terms used in the calculation are very small, due to an insufficiently large number of observations. Beyond that it's hard to say without a FAQ: What's a reproducible example (`reprex`) and how do I do one? and the characteristics of your data in terms of number of rows and variables, if you have them in a data frame and the type of each variable.
More generally in doing this sort of problem you want to identify an `outcome`, conventionally called y as a function of some other set of variables, x_i ... x_n. How each is encoded has a big influence on your choice of tools.
For example y may be binary yes/no, TRUE/FALSE, 1,0. A binary variable is an example of a categorical variable that can take on only one of two value. On the other hand, it may be continuous numerically. For example if you are measuring attitudes toward premarital sex by gathering data on the number of sexual partners a subject reports before a first marriage, if any, you may have a range of numbers ranging from 0 to 32, say (depending on the person, the culture, and other intangibles). Other categorical data may take one several different values, say flavors of ice cream.
Take a look at one of the built in datasets
``````data(mtcars)
str(mtcars)
'data.frame': 32 obs. of 11 variables:
\$ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
\$ cyl : num 6 6 4 6 8 6 8 4 4 6 ...
\$ disp: num 160 160 108 258 360 ...
\$ hp : num 110 110 93 110 175 105 245 62 95 123 ...
\$ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
\$ wt : num 2.62 2.88 2.32 3.21 3.44 ...
\$ qsec: num 16.5 17 18.6 19.4 17 ...
\$ vs : num 0 0 1 1 0 1 0 1 1 1 ...
\$ am : num 1 1 1 0 0 0 0 0 0 0 ...
\$ gear: num 4 4 4 3 3 3 3 4 4 4 ...
\$ carb: num 4 4 1 1 2 1 4 2 2 4 ...
``````
Think about which of these are continuous and which are categorical and which are binary. In this set of automobile data, miles per gallon, mpg, is a continuous variable as is horsepower, hp. You might ask if mpg is affected by hp, and choose a linear regression model
``````fit <- lm(mpg ~ hp, data = mtcars)
summary(fit)
Call:
lm(formula = mpg ~ hp, data = mtcars)
Residuals:
Min 1Q Median 3Q Max
-5.7121 -2.1122 -0.8854 1.5819 8.2360
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 30.09886 1.63392 18.421 < 2e-16 ***
hp -0.06823 0.01012 -6.742 1.79e-07 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.863 on 30 degrees of freedom
Multiple R-squared: 0.6024, Adjusted R-squared: 0.5892
F-statistic: 45.46 on 1 and 30 DF, p-value: 1.788e-07
``````
and you'd conclude that there is a big effect that's has less than a one in 17 million probability of being due solely to chance.
Try
``````fit <- lm(carb ~ gear, data = mtcars)
summary(fit)
``````
and think about what kind of data (continuous or categorical) the number of carburetors and gears represent.
Then do the same with your data in understanding how it's represented.
1 Like | 1,048 | 3,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-26 | latest | en | 0.834307 |
http://www.docstoc.com/docs/109650025/03-Depreciation---Declining-Balance | 1,430,103,570,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246656965.63/warc/CC-MAIN-20150417045736-00247-ip-10-235-10-82.ec2.internal.warc.gz | 456,662,152 | 37,103 | # 03 Depreciation - Declining Balance by xiaoyounan
VIEWS: 1 PAGES: 2
• pg 1
``` 6B – Driving – Lesson 03
Running Costs & Depreciation – Declining Balance Depreciation.
Name: _________________
1) How does declining balance depreciation differ from straight line depreciation?
2) A ute is bought for \$34500 and depreciates at 20% p.a. [100% - 20% = 80% =
0.80]. The ute will be sold when the value falls below \$10000. After how many
years will it be sold?
Year 0 Value
Year 1 \$24500 x 0.80
Year 2
Year 3
Year 4
Year 5
Year 6
3) A RV is bought for \$130,000. If it depreciates at 15% per annum [100% - 15% =
85% = 0.85]. How long before its value falls below \$40,000 and is sold.
Year Value Year Value
Year 0 \$130,000 Year 5
Year 1 Year 6
Year 2 Year 7
Year 3 Year 8
Year 4 Year 9
4) What do we mean by Salvage value?
1
n
5) Calculate the salvage value, using the formula S = V0(1 – r) .
a. A motorcycle costing \$4500 depreciates at 25% p.a. for 7 years.
n
S = V0(1 – r)
7
= \$4500 x ( 1 - 0.25) = \$________
b. A computer bought for \$2700 is depreciated at 15% p.a. for 5 years.
n
S = V0(1 – r)
5
= \$2700 x ( 1 - 0.15) = \$________
c. A car purchased for \$24000 is depreciated at 10%/annum for 11 years.
6) How many years will it take for a computer costing \$1800 depreciate by 20% p.a.
to a salvage value of \$500 ?
n Try n = 5 Answer < 500 T or F
S = V0(1 – r)
Try n = __ Answer < 500 T or F
Try n = __ Answer < 500 T or F
n
= \$1800x( 1 - 0.20) = \$500 Try n = __ Answer < 500 T or F
7) How many years will it take for a Car costing \$52,000 depreciate by 10% p.a. to a
salvage value of \$20,000 ?
2
```
To top | 619 | 1,833 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2015-18 | longest | en | 0.820923 |
https://physics.stackexchange.com/questions/121598/walking-swinging | 1,660,490,492,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572043.2/warc/CC-MAIN-20220814143522-20220814173522-00185.warc.gz | 428,272,707 | 65,165 | # Walking & Swinging
How can I show that the most convenient way to move the arms while walking is swinging them back and forth, alternatively?
To pose the question in another way: can I prove, starting from the conservation of momentum and angular momentum, that given a rigid solid body moving at constant speed and with two appendices on the side, this appendices will move as our arms do when we walk "freely"?
• The mechanics of walking are quite complex. A Google should find you articles on the subject. Basically, swinging the arms in opposition to the legs reduces the torque on the body and therefore makes it twist less. Jun 24, 2014 at 11:37
• @JohnRennie minor nitpick: I'd prefer to say "complicated" (not "complex") so as to avoid suggesting the solution requires going into the complex plane. Jun 24, 2014 at 11:45
• @JohnRennie "...reduces the torque on the body and therefore makes it twist less". Is there a way to show this formally, without relying on experimental data? Jun 24, 2014 at 12:16
This is a responce to usumdelphini's comment:
Is there a way to show this formally, without relying on experimental data?
I'm not going to attempt a deep analysis of walking, but it's fairly straightforward to show that using your arm reduces twisting. Suppose you're looking down on the person walking from above, and suppose they're a cylinder $^1$. Then you'd see something like:
When you move your right leg forward this takes some force $F$, and by Newton's third law there is an equal and opposite force on the point where the leg is attached to your body (the green dot). This causes a torque on your body of:
$$T = Fr$$
and as a result your body will tend to twist clockwise.
Now suppose you move your left arm forward at the same time:
Moving your arm creates a torque of $T_{arm} = fr$ in the same way, but the torque is in the opposite direction. So the net torque on your body is:
$$T = Fr - fr = (F - f)r$$
and the result is that your body will twist less.
$^1$ traditionally we assume everything is spherical, but I'll make an exception in this case | 492 | 2,091 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2022-33 | latest | en | 0.924772 |
https://www.bloodraynebetrayal.com/suzanna-escobar/how-to-write-better/is-the-negative-terminal-of-a-battery-0-volts/ | 1,660,896,088,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573630.12/warc/CC-MAIN-20220819070211-20220819100211-00053.warc.gz | 594,608,419 | 31,287 | ## Is the negative terminal of a battery 0 volts?
The negative battery terminal is 0. Some electronic equipment ( stereo’s, surround sound receivers etc.) do neither, they have an internal power supply that generates both positive and negative voltages with the common (0 volts) grounded.
## Can battery voltage negative?
If the poles of your voltmeter are correctly connected, there is a possibility that the battery suffered a phenomenon called “polarity reversal”. In this case there is the possibility that battery B will get a negative voltage. This can happen when there is a mixed use of old and new batteries together.
Is voltage 0 at the end of a circuit?
In fact, all points starting from 5,6,7 and 8 are the same point in the circuit (represent the same point). This means that there is no voltage between thus points. So, the answer is 0V, no difference in potential between those two points. They are at the same electric potential.
How can voltage be negative?
Negative voltage in a circuit is voltage that is more negative in polarity than the ground of the circuit. The first circuit is oriented one way and the other circuit is oriented the same way but with the voltage source flipped around, so that is has opposite polarity orientation than the first.
### How do you know if a voltage is negative?
In the case when a voltage source has negative voltage, it just means the negative terminal of the battery is connected to the positive side of the circuit and the positive terminal of the battery is connected to the negative side of the circuit .
### Can power be negative?
Power is the rate of change in the energy of a system, so if energy leaving the system is positive, then energy entering would mean that the power is negative. Yes.
What happens when voltage is zero?
If there were no pressure or voltage, then when you opened the valve or closed the circuit there would be no flow. Because voltage is like pressure and current is the rate at which electricity is flowing around a complete circuit.
Why is DC voltage negative?
It is the highest, more safe that compromise voltage to pass over long wires. The negative voltage is used, so the current leaked due moisture does not electroplate away the copper, which is present in the cables.
#### Is 0 positive or negative zero?
So let’s say 0 is positive, just for the minute. Take the rule, “When multiplying, if the signs are different, the result is negative.” So then (positive) zero times negative 3 is (negative) zero. Like this:
#### How do I know if my car battery is positive or negative?
When you open the hood and find the battery, on the top left corner will be the minus sign, and on the top right will be the plus sign. The minus sign represents negative, and the plus sign represents positive. When your car dies, the most important thing to do is to stay calm.
What is the difference between positive and negative potentials in batteries?
The two electrodes of a battery or accumulator have different potentials. The electrode with the higher potential is referred to as positive, the electrode with the lower potential is referred to as negative.
What does the Plus and minus sign mean on a battery?
As you might guess, the plus sign indicates the positive battery terminal, while the minus sign indicates the negative battery terminal. Most batteries also have a positive and negative sign stamped into the case. In many cases, the battery cable itself will also be red in color. | 710 | 3,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2022-33 | latest | en | 0.94846 |
https://www.vikschaatcorner.com/what-is-bel-and-db-explain/ | 1,701,472,136,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100308.37/warc/CC-MAIN-20231201215122-20231202005122-00861.warc.gz | 1,200,698,855 | 18,122 | # What is Bel and dB explain?
## What is Bel and dB explain?
The decibel is defined as one tenth of a bel where one bel represents a difference in level between two intensities I1, I0 where one is ten times greater than the other. Thus, the intensity level is the comparison of one intensity to another and may be expressed: Intensity level = 10 log10 (I1 /I0) (dB)
## How is dB calculated?
The most basic form for deciBel calculations is a comparison of power levels. As might be expected it is ten times the logarithm of the output divided by the input. The factor ten is used because deciBels rather than Bels are used.
What is dB measurement?
decibel (dB), unit for expressing the ratio between two physical quantities, usually amounts of acoustic or electric power, or for measuring the relative loudness of sounds. One decibel (0.1 bel) equals 10 times the common logarithm of the power ratio.
What does 0 decibels mean?
The lowest hearing decibel level is 0 dB, which indicates nearly total silence and is the softest sound that the human ear can hear.
### Is decibel scale logarithmic?
In decibels, this ratio is expressed by taking its logarithm (to base 10) and multiplying by 10. Because the decibel scale is a logarithmic one, doubling the gain does not double the decibel value.
### Why is dB used?
The decibel (dB) is a logarithmic unit used to measure sound level. It is also widely used in electronics, signals and communication. The dB is a logarithmic way of describing a ratio. The ratio may be power, sound pressure, voltage or intensity or several other things.
How do you convert dB to ratio?
dB to power ratio conversion The power P2 is equal to the reference power P1 times 10 raised by the gain in GdB divided by 10. P2 is the power level. P1 is the referenced power level. GdB is the power ratio or gain in dB.
Why do we use dB?
## How do you read decibels?
On the decibel scale, the quietest audible sound (perceived near total silence) is 0 dB. A sound 10 times more powerful is 10 dB. A sound 100 times more powerful than near total silence is 20 dB. A sound 1,000 times more powerful than near total silence is 30 dB, 40 dB and so on. | 524 | 2,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-50 | longest | en | 0.916067 |
http://teacherlink.ed.usu.edu/tlnasa/reference/ImagineDVD/Files/starchild/docs/StarChild/questions/calculation.html | 1,544,647,208,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824119.26/warc/CC-MAIN-20181212203335-20181212224835-00249.warc.gz | 289,795,853 | 2,603 | ## Making the Sun Shine by Nuclear Fusion
All of these calculations are done in scientific notation to make it easier to write! Briefly, in scientific notation, the exponent of the 10 is equal to the number of places you move the decimal. For example, 1.0 x 105 is 100,000. Negatives work the same way but move the decimal in the opposite direction. So 1.0 x 10-5 is 0.00001.
The mass of a proton is 1.67 x 10-27 kg The reaction chain requires four protons as the mass input. A helium nucleus has a mass of 6.6326 x 10-27 kg and the 2 positrons produced during the reaction chain have a mass of 9.1139x10-31 kg each, making a total of 6.6344 x 10-27 kg total output mass. So after the reaction, there is less mass than there was before. Where did this mass deficit of 4.7227 x 10-29 kg go? The most famous equation in the world (E=mc2) supplies the answer! How much energy will 4.7227 x 10-29 kg be converted into? (If m is in kg and c is in m/sec, E will be in units of Joules, J)
Now consider this:
We can measure the power output of the Sun to be L=3.9 x 1026 J/s. If each reaction in the proton-proton cycle produces 4.3 x 10-12 Joules, how many reactions must take place each second to power the Sun?
# reactions = 3.9 x 1026 / 4.3 x 10-12 = 9 x 1037
Remember we use up 4 protons in each reaction, so each second, we use up 3.6 x 1038 protons! How many protons are there in the Sun? We can get a quick estimate by calculating
# protons in the Sun ~ (total mass of Sun) / (mass of proton)
~ 2.0 x 1030 kg / 1.67 x 10-27 kg
~ 1.0 x 1057
If we have 1057 protons to use and we use 1038 each second, we have enough protons to last for about 1019 seconds…or over 100,000,000,000 years! It is important to note that the Sun will in fact use up only about 10% of its protons during its lifetime, however.
The StarChild site is a service of the High Energy Astrophysics Science Archive Research Center (HEASARC), Dr. Alan Smale (Director), within the Astrophysics Science Division (ASD) at NASA/GSFC.
StarChild Authors: The StarChild Team
StarChild Graphics & Music: Acknowledgments
StarChild Project Leader: Dr. Laura A. Whitlock
Responsible NASA Official: Phil Newman | 634 | 2,178 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-51 | latest | en | 0.877649 |
https://www.coursehero.com/file/209727/Syllabus2007/ | 1,516,242,265,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887054.15/warc/CC-MAIN-20180118012249-20180118032249-00298.warc.gz | 891,234,041 | 64,393 | Syllabus2007
# Syllabus2007 - SYLLABUS ECONOMICS 501.01 INTERMEDIATE...
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S YLLABUS E CONOMICS 501.01: I NTERMEDIATE M ICROECONOMICS Professor Patricia B. Reagan C ONTACT I NFORMATION E-mail: Telephone: 292-0192 or 442-7385 Office Address: 450 Arps Hall (1945 North High) Office Hours: Thursday 11:30-1:15 and by appointment Virtual Office Hours: [email protected] The web site for the course is: http://economics.sbs.ohio-state.edu/reagan/501 . Homework assignments and supplemental handouts will be posted at that site. G RADING Your grade for this course will be based on six problem sets, one mid-term exam, and a final exam. The fraction of the points allocated to each is shown below. The problem set with the lowest score will be dropped. Thus, your score on the problem sets will be calculated from your score on your five best problem sets. An alternative grade breakdown is provided for students who do poorly on the midterm but who do better on the final. Your score will be calculated using both point allocations, and the highest score will be chosen automatically. Grade Breakdown Alternative Grade Breakdown Best 5 Problem Sets (6% each) 30% 30% Midterm 30% 20% Final Exam 40% 50% You are responsible for all material in the classes and book. The midterm will cover chapters 1, 2, 3, 4, 6, 7, and 8. The final exam will be comprehensive with an emphasis on new material (chapters 9, 10, 11, 14, 16, 17, 18). Exams will consist of short answer written questions. Notes and calculators are not permitted during the exams. M AKE U P P OLICY No make up exams will be provided for the midterm. Students missing the midterm for unavoidable and formally verifiable reasons will be graded on the alternative grade breakdown (out of a total of 80 points, i.e., homework has a weight of 3/8 and the final has a weight of 5/8). Students missing the final for unavoidable and formally verifiable reasons will be given a make up exam. Exams missed for avoidable or for unverifiable reasons will be assigned a grade of 0. P
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Ask a homework question - tutors are online | 639 | 2,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-05 | latest | en | 0.879156 |
https://www.sophia.org/tutorials/12-writing-equations-from-graphs | 1,527,177,707,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866511.32/warc/CC-MAIN-20180524151157-20180524171157-00325.warc.gz | 838,602,921 | 13,236 | ### Online College Courses for Credit
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## Determine an equation from a graph by SVMathTeach
Theory and examples to help in the solution of the problems in the Summer Assignment "Getting Ready for AP Calculus -12. Writing Equations from Graphs"
## Alg II - Quadratics - Find Equation Given Graph by ToriBall
Theory and examples to help in the solution of the problems in the Summer Assignment "Getting Ready for AP Calculus -12. Writing Equations from Graphs"
## Finding the Equation of a Circle in Standard Form by mattemath
Theory and examples to help in the solution of the problems in the Summer Assignment "Getting Ready for AP Calculus -12. Writing Equations from Graphs"
## Finding the Equation given a graph of an absolute value function by William Lindsay
Theory and examples to help in the solution of the problems in the Summer Assignment "Getting Ready for AP Calculus -12. Writing Equations from Graphs"
## Modeling the equation of a piecewise defined function from its graph, by Krista King | CalculusExpert.com
Source: Krista King | CalculusExpert.com: https://www.youtube.com/watch?v=UvYj2E8cghs | 427 | 2,052 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-22 | latest | en | 0.878058 |
https://applet-magic.com/chargeEnergy.htm | 1,631,889,595,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055645.75/warc/CC-MAIN-20210917120628-20210917150628-00657.warc.gz | 146,525,829 | 3,531 | The Energies of the Electrical Fields of Charged Particles and their Charge Radii
San José State University
applet-magic.com
Thayer Watkins
Silicon Valley
U.S.A.
The Energies of the Electrical Fields of
Charged Particles and their Charge Radii
## Background
A motionless, spherically-shaped charge of radius R and magnitude Q establishes an electric field throughout space with an intensity E given by
#### E = KQ/s² for s≥R
where K is a constant and s is the distance to the center of the charge. In the MKS system K is equal to 8.988×109 N·m²/C². Often the constant K is expressed in the form of
#### K = 1/(4πε0)
where ε0 is the permittivity of free space and its value is 8.854x10−12 farads per meter.
The intensity for s<R depends upon the distribution of the charge within the radius R.
The energy density H per unit volume of an electric field is given by
#### H = ½εE²
where ε is the permittivity of the medium in which the field exists. The permittivity can be taken to be that of free space. Thus the energy contained in a spherical shell of radius s and thickness ds is
#### ½ε0(KQ/s²)0K²Q²/s4
The total enegy contained in the field for s≥R is then
#### H = ∫R∞[½εE = 2εK²Q²π ∫R∞ds/s² = 2εK²Q²π[1/R]
Taking into account that K=1/(4πε0) this reduces to
## Point Charges?
For a charged point particle R is 0 and thus the energy of a charged point particle is infinite. When an electron and a positron annihilate each other the fields of both disappear, but amount of energy released is not infinite. Therefore electrons and positrons are not point particles. There are similar annihilation of other particles and their anti-particles.
Thus point charges do not and cannot exist.
The energy released upon the annihilation of a particle is mc² where m is the particle mass and c is the speed of light. This means that
#### R = Q²/[8πε0mc²]
It is notable that the size of a particle is inversely proportional to its mass.
## The Electron
The charge of an electron is 1.60217657×10-19 coulombs and its mass is 9.10938291×10-31 kilograms. This implies that the charge radius of an electron is
#### R = (1.602×10-19)²/[8*(3.1416)*(8.854x10−12)(9.10938×10-31)(3×108)²] which reduces to 1.40674×10-15 m = 1.40674 fermi .
This is a reasonable value given the scale of atoms. The radius of the lowest orbit of an electron in a hydrogen atom is 5.2918×10−11 m. This is over 37 thousand times the above value for the radius of an electron.
## The Proton
The mass of a proton is 1836 times that of an electron. If its mass-energy is entirely due to that of its electric field then its charge radius would have to be 1/1836 of the radius of an electron. The conventional estimate of the charge radius of a proton based upon scattering experiments usually electron beams is 0.8775 fermi. This means that a proton's electric field accounts for only 1.6 times the mass-energy of an electron. The rest of the 1836 ratio involves proton mass-energy from a different source.
(To be continued.) | 809 | 3,013 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-39 | latest | en | 0.907693 |
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# Past JEE Paper Analysis - JEE Mains 2013 Math Section
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This presentation is provides a brief analysis and overview of the JEE Mains 2013 Math section. Understand which topics to devote more time to and what to expect from the JEE Mains 2014 Math section.
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### Past JEE Paper Analysis - JEE Mains 2013 Math Section
1. 1. JEE Mains 2013 Analysis Math Section
2. 2. Importance of this analysis Mathematics is like a nightmare for a lot of students. Some students even doubt themselves whether they will be able to clear the cutoff or not. But remember that you can’t be an engineer if you fear mathematics. Mathematics is one of the most interesting and beautiful subjects an engineer has to deal with. Once you know how to approach maths it can give you the edge to score a good rank as maths generally has a low cutoff. The next few slides contain some important points from last year’s JEE Mains mathematics section so as to equip you with the trends prevailing currently. This will definitely refine your preparation and last minute study approach. www.spanedea.com
3. 3. Overall Analysis JEE mains (previously called AIEEE) 2013 mathematics section was of 120 marks. Each question carried four marks for each correct and negative marking of one mark for each wrong answer. Some Key Features of JEE Mains 2013 Maths section: 1. Paper was easy compared to AIEEE 2012 paper. 2. Only 10% of the paper could be categorized as ‘tough’. 3. Covered almost every topic: Paper was well balanced as it covered every concept. 4. Concepts: Students with proper understanding of concept scored quite comfortably. 5. 11th and 12th Portion: Marks were equally distributed in 11th (64) and 12th (56). 6. Two questions, one from vector and one three dimensional geometry, were repeated from last to last year paper. 7. Some non-IIT topics were also significant. 8. Maths was the easiest compared to Physics and Chemistry www.spanedea.com
4. 4. Major Topic Analysis ● ● ● ● ● Algebra as a whole is very big so it’s obvious for it to make 40% of the paper, but algebra of 11th is more important. Calculus is the second most important. One of the main reason why integral calculus is more important is that it has applications in every engineering field. Integral calculus is very interesting and easy. Once integration is learned topics such as area under curve and differential equations are just easy to cover. Coordinate geometry was well covered in-terms of marks and it is justified as portion-wise it is large too. So my advice is to choose which topics to go for in between coordinate geometry itself. There is not much to analyse about the paper as it had followed the basic trend covering each topic and concept and also kept its share equal in both 11th and 12th. So right thing is to cover that topics well which you have already covered, not to pick anything new, at least before your revision is done. After your revision, cover new topics in accordance weightage ratio. www.spanedea.com
5. 5. www.spanedea.com
6. 6. Sub-Topic Analysis One can quickly grasp the situation from the chart shown but let’s see some of the highlights and comparisons to make the picture more clear: 1. Straight line should be given equivalent time in comparison to circle and parabola. 2. Ellipse and hyperbola are not asked deeply so going beyond tangents and normals in detail is not a smart thing. 3. Mathematical reasoning and statistics (not part of JEE Advanced syllabus) also had a good share. 4. Complete algebra was covered but no topic had more than 4 marks. So if you missed out on one or two topics, then don’t worry too much. 5. Paper was uniform covering all concepts. So if you are covering some topic well enough but not others, then hear the alarm ringing! 6. Definite integral should be done as thoroughly as it is important from the JEE Advanced point of view also. 7. Vectors and three dimensional geometry has weightage so don’t take them causally. 8. The mark distribution was more or less uniform with most sub-topics receiving 4 marks. www.spanedea.com
7. 7. Topic Wise Analysis Algebra 11th Coordinate Geometry Algebra 12th Differential Calculus Integral Calculus Vector and Three dimensional geometry General Mathematics www.spanedea.com
8. 8. Thank You! What is Spanedea? Spanedea is an online education platform that allows students preparing for the IIT JEE to take oneto-one classes, over the Internet, with an experienced IIT JEE tutor through Spanedea’s virtual classroom! ● ● ● ● ● ● 300+ IIT JEE tutors - IITians, IIT Alumni & experienced faculty Doubt clearing, customized topic-wise, and full-length courses Specialized subject classes for students already attending other coaching institutes. Live, interactive and one-to-one classes. Extremely reliable Parent involvement - in our sessions, parents can keep track of their child’s progress, and attend the class with them. ‘Apply for a Free 1 Hour Trial Class’ Email: learn@spanedea.net Landline: +91-22-4023-8349; Mobile: +91-80800-25432; Headquarters: Openedex India Pvt. Ltd., 703, C-Wing, Mangalya, Andheri (E), Mumbai-400059, India www.spanedea.com | 1,305 | 5,484 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-39 | latest | en | 0.922476 |
http://www.jiskha.com/members/profile/posts.cgi?name=dean&page=8 | 1,386,361,535,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163052469/warc/CC-MAIN-20131204131732-00053-ip-10-33-133-15.ec2.internal.warc.gz | 398,873,429 | 3,015 | Friday
December 6, 2013
# Posts by dean
Total # Posts: 156
life science
how do fungi liv e without water and in extreme cold?
how children learn
which of the following questions ia a thinker/doer most likely to ask?
skills for the effective teacher aide
Acc to the SMCR model, problems in communication can arise when the speaker fails to do what?
math,
3/4 is= to something over 12
math
Theres 26 black cards in a deck of cards. Half of the deck of cards will remain red/black. 1/2 of the cards are shown as black cards.
math
Well, there is 52 cards, and there is only 4 aces, aswell as 9's. What is the probability that you will get those cards?
Pre-Algebra
Well, sorry -.- Thanks for the answer, lol. I thought it would be 28.
Pre-Algebra
Can I have someone check this please? I wouldn't want to do the whole page, then find out I had done them all wrong. :( [Percents of Change] A \$100 digital watch is now \$72 Decrease, 72/100 = negative percent. 100/72.00 = .72 .72 - 1 = 30%? Thanks. 30% is my final answer, ...
math213
Chocolate + singles Chocolate + doubles Chocolate + triples Vanilla + singles Vanilla + doubles Vanilla + triples Strawberry + singles Strawberry + doubles Strawberry + triples There are a total of how many different ways? 3+3+3=? Find it, and thats your final answer.
math
Are you needing to reduce this fraction? Divide?
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(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \sqrt2 \qquad \textbf{(C)}\ 1/2 \qquad \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$
## Solution
Solution by e_power_pi_times_i
Denote the side of one square as $s$. Then the diagonal of the second square is $s$, so the side of the second square is $\dfrac{s\sqrt{2}}{2}$. The area of the second square is $\dfrac{1}{2}s^2$, so the ratio of the areas is $\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{\textbf{(A) } 2}$. | 243 | 713 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 6, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-25 | latest | en | 0.801457 |
http://www.scout.com/college/nebraska/story/192246-a-return-to-nasty-the-blackshirts-reborn | 1,481,157,276,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542288.7/warc/CC-MAIN-20161202170902-00025-ip-10-31-129-80.ec2.internal.warc.gz | 705,776,091 | 42,120 | "> ">
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You want to know the biggest difference between last year's defense and this one that is amongst the very best in all the country? It's almost elementary. The difference between how this year's defense operates versus last year is the difference between, 2+2-6+4-5+6+7=10 and 5+5=10. Taking the complexity out of the side of the ball where being complex is almost contradictory to the very reason defenses are successful in the first place.
Also, if I knew the right keystroke, I would add a little 10 on the side of the sum to the far right, if only to illustrate not just the simplicity, but the tenth power representing the effort gained from letting players play instead of thinking about nuances on a side of the ball where nuances are ultimately self-defeating.
Yes, this defense has came a long ways in a very short time, the statistical difference from a year ago almost staggering. While the season has yet to be completed, you don't need to analyze the rest of the year to try and find a fair comparison to what the NU defense did all of last year. The disparity is such, heck, it doesn't even matter. Night and Day, wet and dry, to or fro, so incredible are the differences, comparisons don't even apply, but for the sake of giving you a slight perspective, we will anyway and keep in mind when looking at the difference, less is better.
Nebraska '02 Category Nebraska '03 Difference 23.9 Opp. pts. per game 13.0 -10.9 361.9 Opp. yds. per game 255.6 -106.3 3.7 Opp. yds. per rush 3.1 -.6 6.6 Opp. yds. per pass 5.0 -1.6 215.3 Opp. yds. pass. per game 154.1 -61.2 146.6 Opp. yds. rush per game 101.4 -45.2 5.0 Opp. yards per play 4.0 -1.0 37% Opp. 3rd dwn. cnvers. % 30% -7% 44% Opp. 4th dwn. cnvers. % 33% -11% 13 (14 games) Total Interceptions 16 (7 games) +3 (+1.37 per gm) 8 (14 games) Total Opp. fumbles lost 13 (7 games) +5 (+1.29 per gm) 1 (14 games) Total TDs scored on "D" 3 (7 games) +2 (+.36 per gm)
Any questions?
How about some more? This year, Demorrio Williams leads the team in sacks with 6 1/2 and all of last year he had 1. Barrett Ruud is already a half a sack higher than his total from last year as well.
Ok, so we've seen the results, but some might just look at this complete reversal of performance and speculate as to how any one or group of coaches can make that much difference. Well, I think it's obvious that they can.
For a second, let's just forget that Jimmy Williams played over double-digit years in the pros. And, let's put aside Marvin Sanders turning down a job with the St. Louis Rams to return to Lincoln to coach the secondary. Finally, let's chuck the entire NFL resume' of one, Bo Pelini. What you have left then isn't the credentials, but the attitude and that's the biggest difference between this year's team and last.
Sure, technique is different, schemes are simplified and people with some impressive backgrounds usually make great impacts from their experiences, but if you ask Bo Pelini, this arrival wasn't the arrival of a system, rather a mind-set in how you approach the game. That approach being "full out, four quarters".
Relentless, intense and complete effort every single play of the game. That's what Pelini preached to his unit when he got here. Effort first, everything else second and when he was satisfied that the idea was imprinted, that's when players were given the tools to play formations, schemes and systems.
While Craig Bohl might have been lauded last year during practice for his ability to recall schemes and defensive plays down to the myriad gaps, assignments and formations that made up his ultra-complex methodology, Pelini would probably be remembered more for the F-bombs he drops from time to time because someone wasn't close to hyperventilating due to the effort after each and every play.
One thinks, one does and that's really the main reason for what's going on with the blackshirts.
No longer are players hamstrung by a variety of responsibilities. Now, they have one. Get to the ball. No longer are players thinking ad-nauseam about gaps, slots and assignments. Now, it's get to the ball. And, while last year, the defense seemed to be on it's heels, folding like a wet blanket at times when they got down, now, it's "pick your ass up" and oh yeah, get to the damn ball.
That's the reason for the turnovers. That's the reason for the short-fields the offense often has and that's the reason people are smiling now when even a fake grin was hard to come by in recent years at NU.
Of course, nobody on the defensive staff will admit to having accomplished anything. As far as Pelini is concerned, "We are no where near to the finished product" is the quote you will hear every time someone asks how pleased he is with this group. "It's a process", "We're making progress" or "We've got a long ways to go" add to the long list of what are now cliches for college football's most valuable assistant.
The irony here is that the appreciation everyone has shown for the "simplified" defense has more than overshadowed just how smart this guy is. Except for that hiccup in the fourth quarter against Missouri which was actually an entire team collapse, Pelini has shown brilliance in being able to adjust to some very complex schemes and some players that are awfully hard to scheme against.
While Utah State didn't have the players, they had an offensive system about as complex as you can get. Shifts, motions, players just moving all over the place, I thought I was watching a game out of arena football. Pelini adjusted, putting Demorrio on the line, Williams getting 3 sacks and a forced fumble, all in about five minutes. Thus the end of the Aggies.
Against Oklahoma State, a team that has one of the best trios on offense around the entire country, you might assume that you have to pick who hurts you because you simply can't stop them all. Well, Bell didn't get a hundred yards on the ground, Josh Fields didn't even get 100 yards in the air and Mr. Everything, Heisman-hopeful, Rashaun Woods, well, he didn't even get 50 yards worth of catches.
And, most recently against the gifted but young, Reggie McNeal, McNeal might have nightmares about that game until he's old and grey. 5 turnovers from the Aggie standout, 3 in the air and 2 he put on the ground, Texas A and M went from moving the ball very well in the first half to being all but stifled in the second.
You would have to give the Aggies credit though, because they did something against this Pelini defense that hasn't all year. They scored in the 3rd quarter. Yeah, well, it was only a field goal, but still, the season-long futility for teams right after the second half, it was ended, even if it was futile in the end.
How about that though? Last year, I am sure you were actually getting used to Nebraska's defense getting run down like deer on I-80 as they gave up one scoring drive after another when the start of the second half ensued. The infamous half-time adjustments that NU used to make pre-Bohl were a matter of waxing nostalgic, but it would seem NU is back to their old (and good) ways.
It's a return in some many ways, really. A year like last season can make you forget so many of the things NU did right at one point, but Bo Pelini, Marvin Sanders, Jimmy Williams and Jeff Jamrog have helped to remind NU fans just what that was like.
It's a return to flying around the field. It's a return to intimidating offenses rather than just trying to slow them down. It's a return to wide-eyed quarterbacks, running-backs who can't find holes and offensive coordinators that cringe when they see "Nebraska" as their up-coming game. | 1,886 | 7,804 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-50 | longest | en | 0.934105 |
https://fr.mathworks.com/matlabcentral/cody/problems/31-remove-all-the-words-that-end-with-ain/solutions/1847185 | 1,566,762,657,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027330800.17/warc/CC-MAIN-20190825194252-20190825220252-00089.warc.gz | 469,711,209 | 15,472 | Cody
# Problem 31. Remove all the words that end with "ain"
Solution 1847185
Submitted on 12 Jun 2019 by Zizhuo Pang
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
s1 = 'The rain in Spain falls mainly on the plain'; s2 = 'The in falls mainly on the '; assert(strcmp(remAin(s1),s2));
2 Pass
s1 = 'The pain from my migraine makes me complain'; s2 = 'The from my migraine makes me '; assert(strcmp(remAin(s1),s2));
3 Pass
s1 = 'I had to explain that "ain" is not a word'; s2 = 'I had to that "" is not a word'; assert(strcmp(remAin(s1),s2)); | 202 | 668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2019-35 | longest | en | 0.892362 |
https://cs.stackexchange.com/questions/28031/syntax-checking-expressed-mathematically | 1,561,550,101,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000306.84/warc/CC-MAIN-20190626114215-20190626140215-00100.warc.gz | 406,183,277 | 34,720 | # Syntax checking expressed mathematically
1) Set B is the finite set of all rules for a particular programming language.
2) Set A is the infinite set of all valid strings of that language and is defined by Set B.
I'm having a problem defining expressing that mathematically.
My so far unsuccessful idea:
• Define a set C - the set of all possible strings.
• set A is a finite union between infinite sets A_i such that for every rule B_i
$B_i ( C ) = A_i$
can It be defined with an implication ? for every rule B_i of B applied to every string c_i of C, if $B_i ( c_j ) \implies c_j$
• To make this easier to read use latex formatting around your functions( put a dollar sign on each side of the function A_i becomes $A_i$ ) – lPlant Jun 27 '14 at 15:38
• Also initially you say $A$ is the set of rules and $B$ is a set of strings, but then refer to rule $B_i$ applying a string to it to get a new rule which does not make sense. – lPlant Jun 27 '14 at 15:48
• Can you explain in words the meaning of what you write formally. What are your rules supposed to look like and be used. What does it mean and/or produce when you apply a rule to a string. What is the meaning of defining something with an implication. Math is to some extent a Lego game, but you have to understand the shapes of the bricks to fit them together meaningfully. I am asking what shape you see to the bricks you propose to use, and how you propose to assemble them. – babou Jun 27 '14 at 17:44 | 373 | 1,473 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-26 | longest | en | 0.941746 |
https://gmatclub.com/forum/q-regarding-specific-work-experience-13134.html | 1,506,204,699,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689779.81/warc/CC-MAIN-20170923213057-20170923233057-00410.warc.gz | 651,781,922 | 38,881 | It is currently 23 Sep 2017, 15:11
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# Q regarding specific work experience
Author Message
Manager
Joined: 07 Oct 2004
Posts: 99
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Q regarding specific work experience [#permalink]
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12 Jan 2005, 22:09
I'm curious how a particular position might be viewed by various ADCOms of top tier programs. I'm working as a financial analyst for small, regional HMO. The work is much more biased toward operations finance as opposed to capital budgeting, etc. I know a lot comes down to what you make of the position, but what would be the first honest impression of the application reviewer?
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CIO
Joined: 31 Oct 2003
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Location: Ann Arbor
### Show Tags
17 Jan 2005, 14:52
Depends on what you're stating as your career goal, what school you're applying too etc.
Generally speaking, a lot of people tend to believe that if they don't have a big name behind their career experience, that they don't have a chance. That's not the case. As to your question about first impression. It would probably be something along the lines of "hmm don't recognize the name. Well, let's see what types of experiences eastcoaster has had. Is eastcoaster a leader? A team player? Does eastcoaster have the types of experiences from which his/her classmates can benefit?
_________________
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Manager
Joined: 07 Oct 2004
Posts: 99
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### Show Tags
04 Feb 2005, 12:18
Thanks for the response, but I'm thinking the first impression would be something more like .."Certainly have never heard of that company! What's wrong with this guy that he couldn't land a more reputable job? Well, let's take a look at the rest of the application, but there better be something damn impressive!" Please tell me I'm wrong.
I'm utterly confused as to what constitutes quality work experience. Are folks that worked as teachers/artists/ski instructors automatically out of the mix? Thanks
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CIO
Joined: 31 Oct 2003
Posts: 539
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Location: Ann Arbor
### Show Tags
06 Feb 2005, 11:36
Since the tone of your post shows that you don't believe me, I'll past Stanford's response to your question:
Do I have to work for one of the top five companies in my field in order to be a competitive candidate for admission to the Stanford MBA Program?
We make a concerted effort to attract and admit students from a broad range of professional backgrounds and experiences. Therefore, there are no particular industries or sets of companies that put an applicant at a competitive advantage in the admission process. Although investment banks and management consulting firms are represented at Stanford, we also draw students from non-profit organizations, multi-national corporations, family businesses, entrepreneurial ventures, and the public sector.
And yes all of the career types that you pointed to can get an MBA. One of my most impressive study group members was a middle school teacher before HBS. Not trying to plug our book too much, but your skepticism leads me to believe that you would benefit greatly from its insights!
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# Help!!
0
97
1
In the figure, AC = 26 units, AD = x = BE, DB = 9 units, DE = 13 units and measure of angle BDE = measure of angle BCA. What is the value of x?
Dec 3, 2021
#1
0
x / 13 = (9 + x) / 26
26x = 13(9 + x)
13x = 117
x = 9
Dec 3, 2021 | 117 | 255 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2022-27 | latest | en | 0.879538 |
https://www.gradesaver.com/textbooks/math/algebra/algebra-1/chapter-9-quadratic-functions-and-equations-9-8-systems-of-linear-and-quadratic-equations-practice-and-problem-solving-exercises-page-586/33 | 1,721,811,087,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518198.93/warc/CC-MAIN-20240724075911-20240724105911-00677.warc.gz | 676,350,485 | 14,195 | ## Algebra 1
Published by Prentice Hall
# Chapter 9 - Quadratic Functions and Equations - 9-8 Systems of Linear and Quadratic Equations - Practice and Problem-Solving Exercises - Page 586: 33
#### Answer
The student didn't see that the determinant is zero (and thus there are no intersecting points between the two equations).
#### Work Step by Step
$x^2+2x+4=y$ $y=x+1$ $x^2+2x+4=x+1$ $x^2+2x+4-x-1=x+1-x-1$ $x^2+x+3 = 0$ $b^2-4ac$ $1^2-4*1*3$ $1-12=-11$ Since the determinant is negative, then there are no real-number solutions.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 217 | 698 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-30 | latest | en | 0.852655 |
https://www.physicsforums.com/threads/mathematica-plotting-with-a-slightly-complicated-relationship-between-variables.653125/ | 1,508,312,805,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822822.66/warc/CC-MAIN-20171018070528-20171018090528-00273.warc.gz | 967,622,111 | 14,655 | # Mathematica plotting with a slightly complicated relationship between variables
1. Nov 18, 2012
### McLaren Rulez
Hi,
So here's my latest problem as I get to grips with Mathematica. I will give a simpler version of the one I am actually doing.
I started with a closed set of differential equations. Consider the following equations
$$a'(t)=P*b(t)$$
$$b'(t)=P^{2}*a(t)$$
with known initial conditions and P is a constant. I solved it for a specific value of P using s=NDSolve[...] with the equations and initial conditions inside the NDSolve argument. I can get a plot of a(t) against t using Plot[Evaluate[a(t)/.s], {t, 0, Infinity}]. So far, so good.
Now, I want to get a feel for how the system behaves for an arbitrary P. I tried to upgrade s to a function that takes in argument p. That is, s[P_]=NDSolve[...] and this works as well. So now, I can input s[5] and get the solution of my DEs for P=5, say. Note that now, I have to use Plot[Evaluate[a(t)/.s[5], {t, 0, Infinity}] to get my plot of a(t) against t.
Similarly, I can get a plot of a[t] against P for a given t. That is, Plot[Evaluate[a(1)/.s[P], {P, 0, Infinity}] works too. But for some reason, this takes a very long time. It takes a good two or three minutes for each value of t. Why is it so slow?
I eventually want to get a 3D plot of a(t) against t and P. I'm not sure how to go about it and even if I did, if it takes two or three minutes for one value of t, it will never be able to do a 3D plot of all t and all P, will it?
Thank you for your help.
2. Nov 18, 2012
### Bill Simpson
You have asked Mathematica to give you a detailed accurate plot where p ranges to infinity. To do that it needs to evaluate the function lots of places. Every evaluation requires doing a complete NDSolve again. You can put in a bit of diagnostic code that will tell you where or how many times NDSolve is being evaluated to verify this.
Mathematica has no idea that your function might be relatively smooth and only doing a handful of NDSolve might be enough. Perhaps you know enough that you could tell Mathematica to only do NDSolve for a fixed list of values of p and then do a ListPlot and connect those points together. Perhaps that would let you accomplish this with a dozen or two NDSolve.
Then when you want to do 3D perhaps you could accomplish that with a hundred or two NDSolve.
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http://faculty.kfupm.edu.sa/COE/ashraf/RichFilesTeaching/COE022_200/OnlineQuizes/Quiz1/quiz.html | 1,501,203,407,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436316.91/warc/CC-MAIN-20170728002503-20170728022503-00693.warc.gz | 124,044,715 | 2,710 | Copyright © Dean Johnson 2000 (johnson@wmich.edu)
## Quiz#1: Boolean Algebra
The following questions are practice problems associated with the lecture material on the subject of Boolean Algebra. These "bigger problems" have appeared on past exams in this course; variation of them will be found on your first exam in a few weeks. Is best to solve these algebracic derivations with paper and pencil, rather than by guessing and pushing buttons, by the way! (Go back to homepage.) Which of the following relationships represents the dual of the Boolean property x + x' = 1? (Note: * = AND, + = OR) An equivalent representation for the Boolean expression A' + 1 is Simplification of the Boolean expression (A+B+C)(D+E)' + (A+B+C)(D+E) yields which of the following results? Given the function F(A,B,X,Y) = AB + X'Y, the most simplified Boolean representation for F' is An equivalent representation for the Boolean expression A + A' is Simplification of the Boolean expression AB + ABC + ABCD + ABCDE + ABCDEF yields which of the following results? Given the function F(X,Y,Z) = XZ + Z(X'+ XY), the equivalent, most simplified Boolean representation for F is Given that F = A'B'+ C'+ D'+ E', which of the following represent the correct expression for F'? Simplification of the Boolean expression (A+B+C) + (A+B+C)'(D+E) yields which of the following results? Which of the following Boolean functions is algebraically complete? Which of the following relationships represents the dual of the Boolean property x + x'y = x + y Choose the best answer. Simplification of the Boolean expression (A + B)'(C + D + E)' + (A + B)' yields which of the following results? Given that F = (A + B'+ C)(D + E), which of the following represents the only correct expression for F'? Simplification of the Boolean expression AB + A(BC)' yields which of following results? Simplifing (AB)' + A'B to most basic form yields which of the following expressions? Which of the following Boolean functions is not algebraically complete? | 465 | 2,009 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2017-30 | longest | en | 0.905228 |
https://electric-skateboard.builders/t/great-battery-deal-for-aus/24361 | 1,611,480,726,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703547475.44/warc/CC-MAIN-20210124075754-20210124105754-00199.warc.gz | 316,008,274 | 6,696 | # GREAT battery deal for AUS
While browsing hobbyking, I found this deal on a Lipo. It’s 6S 5000mAh and 20-30C.
Unfortunately only in Australia warehouse, but 50% off. When stacking these, you could do a 12S4P (10Ah) for ~100 bucks!
EDIT: I meant 12S4P if you think of 2500mah a parallel group. Sure, when you count one of these batteries as a parallel, you will get either 6S4P or 12S2P
4 Likes
Would be a great deal if it were in the US
yeah I cry… EU would be so nice man. for real a 12S4P for 100 dollars? you can’t beat that
Where are you getting 12S4P for \$100? 4 of those packs would give you a 6S4p or a 12S2P correct? am I missing something?
1 Like
12S4P, using 5Ah 20C continuous cells, means 4P x 20C x 5Ah = 400A continous at 3.6V x 12S = 43.2V
So 43.2V x 400A continuous means over 17 kilowatts -> what are we powering again?
1 Like
Thanks for the tip
1 Like
Thanks @Davey - I saw this deal too. Just grabbed 4
1 Like
So lucky man. I wish I could buy some too…
No. Well I meant that normally you count 2500mah as one Parallel group. You would get a 12S 10Ah battery out of four. But sure if you take 5000mah as one Parallel group, then you are totally right.
I think threy are using the fact they the capacity is almost about double that of li-ion so with 4 you get a 10Ah.
1 Like
No you got me wrong. 2P of these batteries (25Ah=10Ah) is equal to a normal 4P li-ion (42,5Ah=10Ah). This battery 12S 10Ah and 20C discharge rate can output 200A, not 400A.
Ahh ok, I was a bit confused, you re counting each 6s pack as a 6S2P. Gotcha .
1 Like | 509 | 1,575 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-04 | latest | en | 0.931809 |
https://lottomatic.live/66/how-to-win | 1,713,125,551,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816893.9/warc/CC-MAIN-20240414192536-20240414222536-00771.warc.gz | 375,135,715 | 49,152 | # How to win Quebec 49
In order to be able to win the Quebec 49, you should first know how this particular lottery behaves. Our comprehensive analysis shows that each lottery has certain behaviors that are unique to them.
The data on this page is based on the latest one hundred results and can be used to develop a playing strategy to help you win significant amounts or even that big jackpot we all dream of.
These facts have been extracted from the Quebec 49 matrix, one of the most comprehensive Quebec 49 dataset available online, and are being updated after the next drawing so you may want to bookmark this page for further reference.
Let's take a look at what could help you win the Quebec 49.
## Five outstanding Quebec 49 facts
1. Down trend: all winning numbers in one draw were equal or lower than those in the previous draw a number of 18 times.
2. It's been now 7 drawings since no up or down trend took place.
3. Two of the winning numbers were contained by the previous 2 drawings a number of 24 times.
4. The sum of the digits that make up the winning numbers was 41 a number of 8 times.
5. 14 times, the greatest difference between two adjacent winning numbers was 14.
### Other facts you may want to consider
• The sum of the winning numbers increased between 0% and 10% compared to the previous results a number of 12 times.
• The sum of the digits that made up the winning numbers increased between 0% and 10% compared to the previous results a number of 22 times.
• The range of the winning numbers (that is the difference between the highest number and the lowest number drawn) increased between 0% and 10% compared to the previous results a number of 18 times.
• While two consecutive numbers have been drawn 49 times, another 47 times there were no consecutive numbers drawn at all. The other 4 drawings had more than two consecutive numbers.
### How to use all this
The past winning numbers facts mentioned above can be used to hugely decrease the number of tickets you generate for the next draw using our Quebec 49 generator.
You can just as well use them with any other lottery tool you may have or simply to review the numbers you fill on your tickets by hand.
If you're in a lottery place willing to play and find yourself uninspired, come back to this page, look at the latest outstanding facts. They may just help you win the Quebec 49.
### Further tools to help you win
Feel free to use the lottery tools and the Quebec 49 statistics linked in the menu on this page (three horizontal lines icon in the upper right corner of your screen). | 560 | 2,585 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-18 | latest | en | 0.9655 |
https://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=29&t=8315&start=790 | 1,596,616,404,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735916.91/warc/CC-MAIN-20200805065524-20200805095524-00065.warc.gz | 712,329,229 | 28,953 | PSLE marks the graduation of Primary school students and their entry into Secondary schools as teenagers. Discuss all issues about Secondary schooling here.
mathtuition88
BlackBelt
Posts: 901
Joined: Thu Apr 25,
Total Likes:16
### Re: Secondary School Mathematics
the above method is very nice.
mathtuition88
BlackBelt
Posts: 901
Joined: Thu Apr 25,
Total Likes:16
### Expansion and Factorisation, HCI IP Sec 2 Revision Pack
Does your child need practice on Expansion and Factorisation, especially the more challenging questions? It is a key technique that needs to be honed for the O Levels.
Practice Hwa Chong Institution Expansion and Factorisation (Sec 2, IP) here.
http://mathtuition88.wordpress.com/2013 ... orisation/
PiggyLalala
KiasuGrandMaster
Posts: 4053
Joined: Tue May 24,
Total Likes:23
### Re: Secondary School Mathematics
Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
mathtuition88
BlackBelt
Posts: 901
Joined: Thu Apr 25,
Total Likes:16
### Re: Secondary School Mathematics
PiggyLalala wrote:Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
Dear Sir/Mdm,
Best wishes.
PiggyLalala
KiasuGrandMaster
Posts: 4053
Joined: Tue May 24,
Total Likes:23
### Re: Secondary School Mathematics
mathtuition88 wrote:
PiggyLalala wrote:Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
Dear Sir/Mdm,
Best wishes.
very much for the solution. Was using the wrong approach- change of base n made a big mess... haha..
I did not think of letting y = the log expression n change it into its exponential form.
Thank you so much.
mathtuition88
BlackBelt
Posts: 901
Joined: Thu Apr 25,
Total Likes:16
### Re: Secondary School Mathematics
PiggyLalala wrote:
mathtuition88 wrote:
PiggyLalala wrote:Need help to solve the following question.
Log (base 9) a = log (base 12) b =log (base 16) (a+b), find the value of a/b.
TIA
Dear Sir/Mdm,
Best wishes.
very much for the solution. Was using the wrong approach- change of base n made a big mess... haha..
I did not think of letting y = the log expression n change it into its exponential form.
Thank you so much.
You are welcome, glad to help.
That question is quite creative, enjoyed solving it.
Qwertymum
YellowBelt
Posts: 11
Joined: Wed Dec 21,
### Re: Secondary School Mathematics
My son asks me a Math question from his MY Revision Paper which I need help. Find it rather challenging! Or if there is something not right? Could anyone out there kindly help to solve, thanks in advance.
" A father in his will left all his money to his children as such: \$1000 and 1/10 of what then remains to the first born; then \$2000 and 1/10 of what then remains to the second; then \$3000 and 1/10 of what then remains to the third born; and so no. When this was done each child had the same amount. How many children were there? "
jieheng
BrownBelt
Posts: 613
Joined: Thu Mar 24,
Total Likes:3
### Re: Secondary School Mathematics
Qwertymum wrote:My son asks me a Math question from his MY Revision Paper which I need help. Find it rather challenging! Or if there is something not right? Could anyone out there kindly help to solve, thanks in advance.
" A father in his will left all his money to his children as such: \$1000 and 1/10 of what then remains to the first born; then \$2000 and 1/10 of what then remains to the second; then \$3000 and 1/10 of what then remains to the third born; and so no. When this was done each child had the same amount. How many children were there? "
Father's money [-1000-][1U][-------------------9U-------------------]
Father's money [-1000-][1U][--2000--][1P][-----------9P-----------]
Father's money = 1000 + 10U
the money that 1st child will receive = 1000 + 1U
the money that 2nd child will receive = 2000 + 1P
Each child will receive the same amount of money
1000 + 1U = 2000 + 1P
1U = 1000 + 1P ------------(1)
9U = 2000 + 10P ----------(2)
(1)*9 , 9U = 9000 + 9P ---(3)
(2) = (3) ,
2000 + 10P = 9000 + 9P
1P = 7000
From (1)
1U = 1000 + 1P = 1000 + 7000 = 8000
Father's money = 1000 + 10U = 1000 + 10*8 = 81000
the amount of money that each child will receive = 1000 + 1U = 9000
No of children = 81000 / 9000 = 9
Qwertymum
YellowBelt
Posts: 11
Joined: Wed Dec 21,
### Re: Secondary School Mathematics
jieheng wrote:
Qwertymum wrote:My son asks me a Math question from his MY Revision Paper which I need help. Find it rather challenging! Or if there is something not right? Could anyone out there kindly help to solve, thanks in advance.
" A father in his will left all his money to his children as such: \$1000 and 1/10 of what then remains to the first born; then \$2000 and 1/10 of what then remains to the second; then \$3000 and 1/10 of what then remains to the third born; and so no. When this was done each child had the same amount. How many children were there? "
Father's money [-1000-][1U][-------------------9U-------------------]
Father's money [-1000-][1U][--2000--][1P][-----------9P-----------]
Father's money = 1000 + 10U
the money that 1st child will receive = 1000 + 1U
the money that 2nd child will receive = 2000 + 1P
Each child will receive the same amount of money
1000 + 1U = 2000 + 1P
1U = 1000 + 1P ------------(1)
9U = 2000 + 10P ----------(2)
(1)*9 , 9U = 9000 + 9P ---(3)
(2) = (3) ,
2000 + 10P = 9000 + 9P
1P = 7000
From (1)
1U = 1000 + 1P = 1000 + 7000 = 8000
Father's money = 1000 + 10U = 1000 + 10*8 = 81000
the amount of money that each child will receive = 1000 + 1U = 9000
No of children = 81000 / 9000 = 9
Wow! That's fast. Thanks a lot.
alwaysLovely
YellowBelt
Posts: 19
Joined: Wed Jan 14,
### Re: Cubic Equations
Solving cubic equations is taught in Secondary 3 Additional Maths.
There are three methods:
1) Synthetic division
2) Comparing coefficients
3) Long division
I've two videos on demonstrating how to solve a cubic equation using the first two methods.
I believe this post will be useful for you.
http://www.singaporeolevelmaths.com/201 ... nts-video/
Cheers! | 1,871 | 6,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-34 | latest | en | 0.882318 |
https://www.rdocumentation.org/packages/psych/versions/1.8.12/topics/fa.lookup | 1,558,406,399,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256215.47/warc/CC-MAIN-20190521022141-20190521044141-00165.warc.gz | 920,617,608 | 7,418 | # fa.lookup
0th
Percentile
##### A set of functions for factorial and empirical scale construction
When constructing scales through rational, factorial, or empirical means, it is useful to examine the content of the items that relate most highly to each other (e.g., the factor loadings of fa.lookup of a set of items) , or to some specific set of criteria (e.g., bestScales). Given a dictionary of item content, these routines will sort by factor loading or criteria correlations and display the item content.
Keywords
multivariate, models
##### Usage
lookup(x,y,criteria=NULL)
fa.lookup(f,dictionary=NULL,digits=2,cut=.0,n=NULL,sort=TRUE)
item.lookup(f,m, dictionary,cut=.3, digits = 2)
keys.lookup(keys.list,dictionary)
##### Arguments
x
A data matrix or data frame depending upon the function.
y
A data matrix or data frame or a vector
criteria
Which variables (by name or location) should be the empirical target for bestScales and bestItems. May be a separate object.
f
The object returned from either a factor analysis (fa) or a principal components analysis (principal)
keys.list
A list of scoring keys suitable to use for make.keys
cut
Return all values in abs(x[,c1]) > cut.
n
Return the n best items per factor (as long as they have their highest loading on that factor)
dictionary
a data.frame with rownames corresponding to rownames in the f\$loadings matrix or colnames of the data matrix or correlation matrix, and entries (may be multiple columns) of item content.
m
A data frame of item means
digits
round to digits
sort
Should the factors be sorted first?
##### Details
fa.lookup and lookup are simple helper functions to summarize correlation matrices or factor loading matrices. bestItems will sort the specified column (criteria) of x on the basis of the (absolute) value of the column. The return as a default is just the rowname of the variable with those absolute values > cut. If there is a dictionary of item content and item names, then include the contents as a two column matrix with rownames corresponding to the item name and then as many fields as desired for item content. (See the example dictionary bfi.dictionary).
lookup is used by bestItems and will find values in c1 of y that match those in x. It returns those rows of y of that match x. Suppose that you have a "dictionary" of the many variables in a study but you want to consider a small subset of them in a data set x. Then, you can find the entries in the dictionary corresponding to x by lookup(rownames(x),y) If the column is not specified, then it will match by rownames(y).
fa.lookup is used when examining the output of a factor analysis and one wants the corresponding variable names and contents. The returned object may then be printed in LaTex by using the df2latex function with the char option set to TRUE.
Similarly, given a correlation matrix, r, of the x variables, if you want to find the items that most correlate with another item or scale, and then show the contents of that item from the dictionary, bestItems(r,c1=column number or name of x, contents = y)
item.lookup combines the output from a factor analysis fa with simple descriptive statistics (a data frame of means) with a dictionary. Items are grouped by factor loadings > cut, and then sorted by item mean. This allows a better understanding of how a scale works, in terms of the meaning of the item endorsements.
##### Value
bestItems returns a sorted list of factor loadings or correlations with the labels as provided in the dictionary.
lookup is a very simple implementation of the match function.
fa.lookup takes a factor/cluster analysis object (or just a keys like matrix), sorts it using fa.sort and then matches by row.name to the corresponding dictionary entries.
##### Note
Although empirical scale construction is appealing, it has the basic problem of capitalizing on chance. Thus, be careful of over interpreting the results unless working with large samples. Iteration and bootstrapping aggregation (bagging) gives information on the stability of the solutions. See bestScales
To create a dictionary, create an object with row names as the item numbers, and the columns as the item content. See the link{bfi.dictionary} as an example. The bfi.dictionary was constructed from a spreadsheet with multiple columns, the first of which was the column names of the bfi. See the first (not run) example.
##### References
Revelle, W. (in preparation) An introduction to psychometric theory with applications in R. Springer. (Available online at https://personality-project.org/r/book).
fa, iclust,principal, bestScales and bestItems
• lookup
• fa.lookup
• item.lookup
• keys.lookup
##### Examples
# NOT RUN {
#Tne following shows how to create a dictionary
#first, copy the spreadsheet to the clipboard
# rownames(bfi.dictionary) <- bfi.dictionary[1] #the first column had the names
# bfi.dictionary <- bfi.dictionary[-1] #these are redundant, drop them
f5 <- fa(bfi,5)
m <- colMeans(bfi,na.rm=TRUE)
item.lookup(f5,m,dictionary=bfi.dictionary[2])
fa.lookup(f5,dictionary=bfi.dictionary[2]) #just show the item content, not the source of the items
# }
Documentation reproduced from package psych, version 1.8.12, License: GPL (>= 2)
### Community examples
Looks like there are no examples yet. | 1,184 | 5,323 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-22 | latest | en | 0.746942 |
https://se.mathworks.com/matlabcentral/cody/problems/1804-fangs-of-a-vampire-number/solutions/304159 | 1,610,739,062,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00454.warc.gz | 628,761,687 | 17,430 | Cody
# Problem 1804. Fangs of a vampire number
Solution 304159
Submitted on 14 Aug 2013
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 1; y = 1; assert(~are_fangs(x,y))
2 Pass
%% x = 21; y = 60; assert(are_fangs(x,y))
3 Fail
%% x = randi(9,1); y = randi([10 99],1); assert(~are_fangs(x,y))
Error: Assertion failed.
4 Pass
%% x = 15; y = 93; assert(are_fangs(x,y))
5 Pass
%% x = randi(9,1)*10; y = randi(9,1)*10; assert(~are_fangs(x,y))
6 Fail
%% x = 2; y = 50006; assert(~are_fangs(x,y))
Error: Assertion failed.
7 Pass
%% x = 150; y=930; assert(~are_fangs(x,y))
8 Fail
%% x = 300; y = 501; assert(are_fangs(x,y))
Error: Assertion failed.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 309 | 921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-04 | latest | en | 0.533905 |
https://investortonight.com/purchasing-power-parity-theory/ | 1,721,566,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00654.warc.gz | 262,899,676 | 62,686 | ## What is Purchasing Power Parity?
Purchasing power parity (PPP) is an economic theory and a technique used to determine the relative value of currencies, estimating the amount of adjustment needed on the exchange rate between countries in order for the exchange to be equivalent to (or on par with) each currency’s purchasing power.
PPP theory specify a precise relationship between relative inflation rates of two countries and their exchange rates. PPP theory suggests that the equilibrium exchange rate will adjust the same magnitude as the differential in inflation rates between two countries.
## What is Purchasing Power Parity Theory?
The purchasing power parity theory (PPP) of exchange rate determination states that the exchange rate between any two currencies equals the ratio of their price levels. The PPP theory focuses on the inflation-exchange rate relationships.
This theory is said to have been originally formulated by Wheatley in 1802 and later by Blake in 1810. However, according to Krugman and Obstfeld, ‘The basic idea of PPP was put forth in the writings of 19th-century British economists, among them David Ricardo (the originator of the theory of comparative advantage).’
Whoever might be the originator of the basic idea of the PPP theory, it was Gustav Cassel, a Swedish economist, who reformulated the PPP theory and developed the concept of an equilibrium rate of exchange in 1920.
This theory asserts that the relative values of different currencies correspond to the relation between the real-purchasing power of each currency in its own country.
In other words, under a free and inconvertible paper currency system, the rate of exchange between any two currencies is determined by their purchasing power in the respective currencies. This theory is called the purchasing power theory of the rate of exchange.
The PPP theory has two versions:
• The absolute purchasing power parity theory
• The relative purchasing power parity theory
## Absolute Purchasing Power Parity Theory
According to the absolute version of purchasing power parity theory, the equilibrium exchange rate is determined in terms of the ratio of the absolute price levels in any two countries.
### Absolute PPP Theory Formula
The exchange rate under this version of the PPP theory is given as,
ER = PA / PB
where ER = exchange rate, PA = price level in country A, and PB = price level in country B.
For example, suppose a basket of goods can be bought in India for ₹100 and in the US for \$2. In that case, the exchange rate between the Indian rupee (INR) and the US\$ will be determined as follows:
### Assumption of PPP Theory
The absolute version of the PPP theory is based on the following assumptions:
• There are no transportation costs.
• There are no tariffs on imports and subsidies on exports.
• There is free trade between nations.
### Criticism of Absolute PPP theory
The absolute version of the PPP theory has certain serious shortcomings which make its validity questionable.
• It ignores the effect of transportation costs which affect the final price paid by traders
• Many countries impose tariffs on imports and subsidize exports. These factors affect the price and, hence, the real purchasing power of the currency
• It takes into account only traded goods and services and ignores the price levels of non-traded goods which are part and parcel of the general price level. Also, the PPP theory ignores capital account transactions which do matter in exchange rate determination
## Relative Purchasing Power Parity Theory
The relative purchasing power parity theory is a modified version of its absolute version. While the absolute version assumes price level to remain constant, in reality, price levels do not remain constant. And a change in price levels is bound to change the exchange rate.
The relative PPP theory gives a measure of the change in the exchange rate under the conditions of changes in relative prices.
The relative PPP theory states that the relative change in the exchange rate over time is proportional to the change in the relative price level over a period of time.
### Relative PPP Theory Formula
The formula for relative PPP theory is given as,
ERN = (PAN / PA0 ) / (PBN / PB0 ) = ER0
where:
ERN and ER0 are the exchange rates in year N and base year 0, respectively
PAN and PA0 are the price levels in country A in year N and base year 0, respectively
PBN and PB0 are the price levels in country B year N and base year 0, respectively
In fact, the basic purpose of the relative PPP theory is to determine the equilibrium exchange rate under conditions of changing price levels. Also, the relative PPP theory can be used to measure the change in the exchange rate owing to a change in the price levels.
For example, suppose there are two counties A and B and their respective price levels in the base year are given as PA0 = 100 and PB0 = 100. In this case, the exchange rate between the currencies of the two countries can be expressed as,
ERN = PA0 / PB0 = 100/100 = 1
Now let the price levels in the two countries in year N change to PAN= 150 and PBN = 200, respectively. In that case, the relative exchange rate (RER) in year N between the two countries can be measured as follows:
PERN = (PAN / PA0 ) / (PBN / PB0 ) = (150 / 100) / (200 / 100)
This measure of relative PPP shows that the currency in country B has depreciated by 0.25 or by 25 per cent. The above calculation shows that if the price index in country A increases in year N from 100 to 150, i.e., 50 per cent, and the price level in country B increases from 100 to 200, the currency in country B depreciates by 25 per cent in year N due to a higher rise in its price level.
### Criticism of Relative PPP theory
Most of the deficiencies of the absolute PPP theory apply also to the relative PPP theory of exchange rate. The relative PPP theory has been criticized also on the following grounds:
• It is based on the general price index of a country. This index included prices of both traded and non-traded goods and services whereas the exchange determination pertains only to internationally traded goods and services. Therefore, the relative PPP theory yields a misleading exchange rate.
• The base year and weightage of goods services used in the construction of the price index varies from country to country depending on the nature and structure of production.
Therefore, the price index number does not reflect the relative price levels in the different countries. The variation in the exchange rate under this condition based on the relative price structure does not reflect the actual purchasing power.
• Apart from non-traded goods and services, certain kinds of services, such as banking, insurance, consultancy services, etc., are considered part of foreign transactions but they are not included in the price index number. Therefore, changes in relative prices do not reflect changes in the purchasing power of a currency.
• A large amount of capital transfers take place between nations, which affect the demand for foreign exchange. The change in the demand for foreign exchange does affect the exchange rate. But this kind of a change in exchange rate is not accounted for in the relative PPP theory.
• As Haberler has pointed out, the imposition of an embargo or tariffs, and the provision of subsidies, causes a change in the actual purchasing power of a currency. But such factors are not accounted for by the relative purchasing power theory.
• A change in the exchange rate depends, by and large, on the elasticities of reciprocal demand for foreign exchange but the PPP theory only recognizes a change in the exchange rate owing to changes in relative prices.
• The relative PPP theory postulates that relative commodity prices are the sole determinants of international transactions and that a change in relative prices is the sole determinant of the exchange rate.
But in reality, changes in the exchange rate are also because of disequilibrium in BOP caused by capital transfers, service payments, and changes in real income. | 1,645 | 8,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-30 | latest | en | 0.94178 |
https://integers.info/544 | 1,720,890,972,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514510.7/warc/CC-MAIN-20240713150314-20240713180314-00593.warc.gz | 268,198,554 | 4,592 | # integers.info
Five hundred and forty-four
## 544 in other numeral systems
Binary: Octal: Hex: Roman:
25 × 17
## Divisors of 544
1, 2, 4, 8, 16, 17, 32, 34, 68, 136, 272 and 544
## Is 544 in the Fibonacci number sequence?
No.
Its nearest Fibonacci number neighbors are 377 and 610. | 99 | 292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.777388 |
https://www.mersenneforum.org/showthread.php?s=1056632db7ed7f7ac1cd2376f276db49&t=21778&goto=nextnewest | 1,660,851,745,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573399.40/warc/CC-MAIN-20220818185216-20220818215216-00336.warc.gz | 755,442,956 | 8,104 | mersenneforum.org Are these polynomials being generated correctly?
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2016-12-02, 19:01 #1 Sam Kennedy Oct 2012 2·41 Posts Are these polynomials being generated correctly? I'm just double checking that the polynomials in my upcoming implementation of the MPQS are being generated correctly: n = 1287871475892167 First prime is near sqrt(sqrt(2*N)/M), where M is the sieve size, in this case: 500,000 Polynomials generated (Ax + B)^2 - N: A: 121 B: 109 A: 169 B: 51 A: 361 B: 345 A: 1849 B: 533 A: 2209 B: 1655 A: 2809 B: 2757 This is how I'm generating them: After finding a prime Q close to sqrt(sqrt(2*N)/M), which is a quadratic residue mod N, I set: A = Q^2 h = N^((Q+1)/4) mod Q v = (2h)^-1 * ((N - h^2)/Q) mod Q B = h + vQ mod Q^2 If B is even, then B = A - B. In the trial division stage, values of X are plugged into the polynomial (Ax + B)^2 - N, and divided by A, before being divided over the factor base. Is this process correct? Are the polynomials above correct for the given input? Thank You Last fiddled with by Sam Kennedy on 2016-12-02 at 19:02
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https://encyclopediaofmath.org/index.php?title=Homological_algebra&printable=yes | 1,675,432,619,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500056.55/warc/CC-MAIN-20230203122526-20230203152526-00210.warc.gz | 252,811,501 | 9,852 | # Homological algebra
The branch of algebra whose main study is derived functors on various categories of algebraic objects (modules over a given ring, sheaves, etc.).
One of the origins of homological algebra is the (singular) homology theory of topological spaces. In this, to each topological space $X$ is associated a sequence of abelian groups ${H_{n}}(X)$ (homology groups), while to a continuous mapping $f: X \to Y$ of spaces is associated a set of homomorphisms $f_{n}: {H_{n}}(X) \to {H_{n}}(Y)$ of homology groups. Each $n$-dimensional singular simplex $T$ of a topological space $X$ has a boundary consisting of singular simplices of dimension $n - 1$. If $K_{n}$ is the free abelian group generated by all the $n$-dimensional simplices, then the function $\partial$ that assigns to each $T$ the alternating sum $\partial T$ of its boundary simplices defines a homomorphism $\partial: K_{n} \to K_{n - 1}$ so that $\cdots \stackrel{\partial}{\longrightarrow} K_{n - 1} \stackrel{\partial}{\longrightarrow} K_{n - 1} \stackrel{\partial}{\longrightarrow} K_{n - 2} \stackrel{\partial}{\longrightarrow} \cdots$ is a complex of abelian groups, and a continuous mapping of spaces induces homomorphisms of their respective complexes. Certain properties of the space $X$ or the mappings $f: X \to Y$ may be deduced from the properties of the homology groups $H_{n}$ of this complex, or from the corresponding homomorphisms $f_{n}$ of these homology groups. This makes it possible, in a number of cases, to reduce the study of topological objects to the study of certain algebraic objects, as is done in analytic geometry (with the difference that the transition from geometry to algebra in homology theory is irreversible).
In algebra, in turn, in the theory of groups (cf. extension of a group), one studies, in fact, the first and the second homology and cohomology groups. Extensive preparatory material was developed in the theory of associative algebras, the theory of Lie algebras, the theory of finite-dimensional algebras, the theory of rings and the theory of quadratic forms.
The language of homological algebra arose mainly from the process of studying homology groups. There appeared arrows as symbols for mappings and commutative diagrams (if, in a diagram, any two paths with a common beginning and end give rise to the same composite mapping, then the diagram is said to be commutative). Sequences of homomorphisms in which the kernel of each outgoing homomorphism coincides with the image of the incoming one were encountered; such sequences are called exact. It became customary to specify mathematical objects together with their mappings; the correspondences most preferred were those between objects that preserve the mappings. These correspondences became known as functors. The principal advantages of this language — the amount of information conveyed, naturalness and clarity — were soon recognized. For example, the language of homological algebra was employed ([5]) in the axiomatic exposure of the fundamentals of algebraic topology. Nowadays, this language is used in numerous studies, including those not employing homological methods.
By the mid-1940’s, homological algebra had become an independent branch of algebra. The principal domain of application of homological algebra is the category of modules over a ring. Most of the results known for modules may be applied to abelian categories with certain restrictions (this is because such categories are embeddable into categories of modules). In the most fruitful extension of the domain of application of homological algebra ([4]), the latter was extended so as to apply to arbitrary abelian categories with enough injective objects, and became applicable to arithmetical algebraic geometry and to the theory of functions in several complex variables (cf. Grothendieck category).
The principal functors of homological algebra are $\operatorname{Hom}(A,B)$ (the group of homomorphisms of a module $A$ to a module $B$) and $A \otimes B$ (the tensor product of $A$ and $B$). The base of the theory is the study of derived functors, which may be constructed as follows. An arbitrary module $A$ can be represented as a quotient module of a free module $F_{0}$, after which a similar representation $F_{1}$ is considered for the kernel of the preceding representation, etc. One thus obtains an exact sequence: $$\cdots \longrightarrow F_{n} \longrightarrow \cdots \longrightarrow F_{1} \longrightarrow F_{0} \longrightarrow A \longrightarrow 0.$$ An exact sequence $$\cdots \longrightarrow P_{n} \longrightarrow \cdots \longrightarrow P_{0} \longrightarrow A \longrightarrow 0,$$ where all the modules $P_{i}$ are projective, is called a projective resolution of the module $A$. The application of a covariant additive functor $T$ yields a complex, whose homology groups are known as the left derived functors of $T$; they are denoted by $L_{n} T$. A dual construction (for a contravariant functor) — or by using injective modules and injective resolutions (for a covariant functor) — is performed to obtain the right derived functors $R^{n} T$. In a certain sense, the derived functors are a measure of the deviation of the functor from exactness. They are not affected by the arbitrariness involved in the construction of a resolution. To each exact sequence $$0 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0$$ correspond two infinite exact sequences of derived functors: $$\cdots \longrightarrow {L_{n + 1} T}(C) \longrightarrow {L_{n} T}(A) \longrightarrow {L_{n} T}(B) \longrightarrow {L_{n} T}(C) \longrightarrow {L_{n - 1} T}(A) \longrightarrow \cdots$$ and $$\cdots \longrightarrow {R^{n - 1} T}(C) \longrightarrow {R^{n} T}(A) \longrightarrow {R^{n} T}(B) \longrightarrow {R^{n} T}(C) \longrightarrow {R^{n + 1} T}(A) \longrightarrow \cdots.$$
The following notation is accepted for the derived functors of basic functors: $${L_{n}}(A \otimes_{R} B) = {\operatorname{Tor}^{R}_{n}}(A,B), \qquad R^{n} {\operatorname{Hom}_{R}}(A,B) = {\operatorname{Ext}_{R}^{n}}(A,B).$$ Both of these functors are functors of two arguments, $A$ and $B$, and for this reason, the construction of the derived functor as described above is not directly applicable to them. In such a case, one can fix one argument and construct a resolution for the other, or, having constructed resolutions of both arguments, one can construct a binary complex. The same result will be obtained in all cases. The group ${\operatorname{Ext}_{R}^{1}}(A,B)$ is isomorphic to the group of extensions of the module $B$ by the module $A$ (and in this form, it has been studied for a long time). The establishment of the new relations considerably extended and advanced the theory of extensions of modules. The group ${\operatorname{Tor}^{\mathbb{Z}}_{1}}(A,\mathbb{Q} / \mathbb{Z})$ is the torsion part of a group $A$. The generalization of this observation resulted in the development of the general theory of torsion.
The homology theory of algebraic systems forms part of the general scheme of derived functors. Thus, let $\Lambda = \mathbb{Z} G$ be the group ring (cf. group algebra) of a multiplicative group $G$ over the ring $\mathbb{Z}$ of integers, and let $A$ be a left and $B$ a right $\Lambda$-module. The study of the groups $${H^{n}}(G,A) = {\operatorname{Ext}_{\mathbb{Z} G}^{n}}(\mathbb{Z},A) \qquad \text{and} \qquad {H^{n}}(G,B) = {\operatorname{Tor}^{\mathbb{Z} G}_{n}}(B,\mathbb{Z}),$$ where $\mathbb{Z}$ is considered as a trivial left $\mathbb{Z} G$-module, is the theory of homology and cohomology of groups. Let $L$ be a Lie algebra over a field $\mathbb{k}$, let $U L$ be its universal enveloping algebra, and let $A$ be a $U L$-module. The study of the groups $${H^{n}}(L,A) = {\operatorname{Ext}_{UL}^{n}}(\mathbb{k},A),$$ where $\mathbb{k}$ is considered as a trivial $U L$-module, is the theory of cohomology of Lie algebras. Appropriate cohomology and homology groups of monoids, abelian groups, algebras, graded algebras, rings, etc., are defined in a similar manner. The guideline in each case is the fact that the second cohomology group is the group of extensions for the type of algebraic systems under consideration.
In turn, the homology groups of algebraic systems form the subject of study of relative homological algebra.
In concrete cases, derived functors of functors are usually computed by means of an explicit resolution. The resolution may be finite (e.g., the length of the resolution of an arbitrary abelian group does not exceed $1$). There has long been interest in the length of the shortest resolution (this length is called the homological dimension). The first significant result in this direction is the Hilbert syzygy theorem (appearing at the end of the 19th century). Homological-dimension theory is one of the actively-developing branches of homological algebra. The transition from modules with various restrictions of finiteness to the general case is often performed with the aid of the inductive limit and projective limit functors, $\lim_{\rightarrow}$ and $\lim_{\leftarrow}$. Thus, any group is the inductive limit of its finitely-generated subgroups. Every compact totally-disconnected group is representable as the projective limit of its finite quotient groups. Interest in these groups stems from their connection with Galois theory. The derived functors of these functors are used in homological dimension theory.
Derived functors for non-additive functors have been studied (e.g., functors that relate an abelian group to its group ring or to its symmetric algebra).
The principal means of computations in homological algebra, other than the resolutions already mentioned, are spectral sequences and the homology product. The former, which are a most powerful tool in the study of derived functors, approximate the homology groups of a group by the homology groups of a subgroup and a quotient group of it. The homology product relates to the study of homomorphisms of the type $$L_{n} T \times L_{m} T \to L_{n + m} T,$$ which combine derived functors with one another.
Methods of homological algebra are now extensively employed in very different branches of mathematics, like functional analysis, the theory of functions of a complex variable, differential equations, etc. Branches of algebra such as algebraic $K$-theory, algebraic geometry or algebraic number theory would be unthinkable without homological algebra.
#### References
[1] H. Cartan, S. Eilenberg, “Homological algebra”, Princeton Univ. Press (1956). MR0077480 Zbl 0075.24305 [2] S. MacLane, “Homology”, Springer (1963). Zbl 0818.18001 Zbl 0328.18009 [3] H. Bass, “Algebraic $K$-theory”, Benjamin (1968). MR249491 [4] A. Grothendieck, “Sur quelques points d’algèbre homologique”, Tohôku Math. J., 9 (1957), pp. 119–221. MR0102537 [5] S. Eilenberg, N.E. Steenrod, “Foundations of algebraic topology”, Princeton Univ. Press (1952). MR0050886 Zbl 0047.41402 [6] Itogi Nauk. Ser. Mat., Algebra 1964 (1966), pp. 203–236. [7] N.E. Steenrood (ed.), Reviews of papers in algebraic and differential topology, topological and homological algebra, 2, Amer. Math. Soc. (1968), pp. 1174–1364.
#### References
[a1] P.J. Hilton, U. Stammbach, “A course in homological algebra”, Springer (1971). MR0346025 Zbl 0238.18006
How to Cite This Entry:
Homological algebra. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Homological_algebra&oldid=41199
This article was adapted from an original article by V.E. GovorovA.V. Mikhalev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article | 2,948 | 11,675 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2023-06 | latest | en | 0.866935 |
https://blog.plover.com/2019/04/ | 1,719,296,026,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865560.33/warc/CC-MAIN-20240625041023-20240625071023-00726.warc.gz | 106,786,211 | 8,089 | # The Universe of Discourse
Fri, 26 Apr 2019
What is the shed in “watershed”? Is it a garden shed? No.
I guessed that it meant a piece of land that sheds water into some stream or river. Wrong!
The Big Dictionary says that this shed is:
The parting made in the hair by combing along the top of the head.
This meaning of “shed” fell out of use after the end of the 17th century.
This week I learned that there are no fewer than seven fanfics on AO3 that concern the Complaint letter to Ea-Nasir, a 3750-year-old Babylonian cuneiform tablet from an merchant angry at the poor-quality copper ingots he was sold. Truly, we live in an age of marvels.
I've said here before that I don't usually find written material funny, with very rare exceptions. But this story, Pay me Baby, Treat me Right, was a rare exception. I found it completely sidesplitting.
(Caution: sexual content.)
[ Addendum: However, I still demand to know: Where the hell is my Sonar Taxlaw fanfic? Fanfic writers of the world, don't think this gets you off the hook! ]
[ Addendum 20200824: Only 16 months later, there are now eleven works on AO3. ]
This is definitely the worst thing I learned this month. It's way worse than that picture of Elvis meeting Nixon.
Nobel Laureate and noted war criminal Henry Kissinger is also an honorary member of the Harlem Globetrotters.
As Maciej Cegłowski said, “And yet the cruel earth refuses to open and swallow up everyone involved.”
Thu, 25 Apr 2019
Katara was given the homework exercise of rationalizing the denominator of $$\frac1{\sqrt2+\sqrt3+\sqrt5}$$ which she found troublesome. You evidently need to start by multiplying the numerator and denominator by !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, obtaining $$\frac1{(\sqrt2+\sqrt3+\sqrt5)}\cdot \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{-\sqrt2 + \sqrt 3 + \sqrt 5} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{(-2 +3 + 5 + 2\sqrt{15})} = \frac{-\sqrt2 + \sqrt 3 + \sqrt 5}{6 + 2\sqrt{15}}$$ and then you go from there, multiplying the top and bottom by !!6 - 2\sqrt{15}!!. It is a mess.
But when I did it, it was much quicker. Instead of using !!-\sqrt2 + \sqrt 3 + \sqrt 5!!, I went with !!\sqrt2 + \sqrt 3 - \sqrt 5!!, not for any reason, but just at random. This got me: $$\frac1{\sqrt2+\sqrt3+\sqrt5}\cdot \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{\sqrt2 + \sqrt 3 - \sqrt 5} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{(2 +3 - 5 + 2\sqrt{6})} = \frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}}$$
with the !!2+3-5!! vanishing in the denominator. Then the next step is quite easy; just get rid of the !!\sqrt6!!: $$\frac{\sqrt2 + \sqrt 3 - \sqrt 5}{2\sqrt{6}}\cdot \frac{\sqrt6}{\sqrt6} = \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}$$
which is correct.
I wish I could take credit for this, but it was pure dumb luck.
It's often pointed out that women, even famous and accomplished women, are often described in newspaper stories as being someone's wife, but that the reverse rarely occurs. The only really well-known exception I could think of was Pierre Curie, who was a famous, prizewinning scientist (1903 Nobel Laureate, yo), but is often identified as having been the husband of Marie Skłodowska Curie (also 1903 Nobel Laureate).
But last week brought another example to my attention. There ware a great many news articles reporting that Salma Hayek's husband had pledged money to help rebuild Notre Dame cathedral. His name is François-Henri Pinault, and he is a billionaire. And the husband of Salma Hayek.
For example:
“Billionaire Francois Pinault and his son, Francois-Henri Pinault, who is married to actress Salma Hayek, said…”
Notre Dame fire – Salma Hayek’s French billionaire husband Francois-Henri Pinault pledges £86million
Salma Hayek’s Billionaire Husband Pledges More Than \$110 Million to Rebuild Paris' Notre Dame Cathedral
(etc.)
[ Addendum 20190808: Walt Mankowski brings up the excellent example of Sir Max Mallowan, CBE, a famous archaeologist and one of the original excavators of Ur. However, he is better known for having been the husband of Dame Agatha Christie from 1930 until he death in 1976. ]
[ Addendum 20230425: I am proud to announce that I have achieved this distinction! In this article from Slate, I am referred to only as “Lorrie Kim's husband”. ]
[ Addendum 20231215: Antoine Veil is interred in the Panthéon, but only because he was married to Simone Veil ] | 1,227 | 4,353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-26 | latest | en | 0.932463 |
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The risk event value/impact is an estimate of the gain or loss that will be incurred if the risk event should take place. This value will apply to all elements of the project including budget, scope, quality and schedule. To assess the consequences and severity of the risk events, the amount at stake and the criticality of each must be determined. Note that these two variables may vary with time depending on the stage in the project life cycle. In most cases, the amount at stake and criticality can be derived by a simple examination of the available data and some subjective judgment. In complex situations however, it may be necessary to develop some form of mathematical model and to construct and run a series of computer-generated analyses. After identifying risks, the project manager will create a \\\"Risk Register\\\" that will be appended to the project management plan. The Risk Register will contain the following information: -List of identified risks -List of potential responses -Root causes of the risks -Updated risk categories During the qualitative phase of risk analysis, the Risk Register will also include the following information: -Priority list of project risks -Risks grouped by category -Risks requiring near-term responses -Risks that need more analysis -Watch list for low priority risks -Trends in qualitative analysis risk source..
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Estimating the Risk Event Value/Impact
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Estimating the Risk Event Value/Impact
The extreme notion of price movements in assets is currently implicit in the practice of management. The adequacy of capital assumes a threshold classifying observable changes in the business risk factor either as ordinary or extreme. At first a probability that will be able to measure the extent that an event will influence a particular portfolio must be chosen. In addition, this probability has to determine the proper threshold of the risk (Fonkych, Taylor, & Rand Corporation.2005).
This underlying approach is the limit centre of the theorem, which produces a normal asymptotic distribution for the risk factor being considered. It is easy to get an asymptotic distribution for risk factor values possible. This paper uses extreme distribution model to calculate risk value and produce some remarkable results. The sample information illustrates that the theory of extreme distribution performs remarkably well to capture the extent of extreme events and the rate of occurrence in the financial market. In fact, the theory of statistics of extreme looks more natural and ...
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http://piping-designer.com/index.php/properties/fluid-mechanics/1981-force | 1,591,039,837,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347419593.76/warc/CC-MAIN-20200601180335-20200601210335-00539.warc.gz | 93,812,280 | 9,736 | # Force
Written by Jerry Ratzlaff on . Posted in Classical Mechanics
Force, abbreviated as F, is the push or pull of an object resulting in a change from rest or motion. So when you apply force to an object the velocity changes, the change in velocity is acceleration. Force is a vector quantity having magnitude and direction, some of these include acceleration, displacement, drag, lift, momentum, thrust, torque, velocity, and weight.
## Formulas that use Force
$$\large{ F = m \; a }$$ $$\large{ F = \frac {\Delta p}{\Delta t} }$$ $$\large{ F = \frac {I}{\Delta t} }$$ (impulse) $$\large{ F = \frac{M}{l} }$$ (moment) $$\large{ F = \frac{P}{v} }$$ (power) $$\large{ F = \frac{P \; t}{d} }$$ (power) $$\large{ F = p \; A }$$ (pressure) $$\large{ F = -k_s \; \left( x - x_0 \right) }$$ (spring constant) $$\large{ F = \sigma \; A_c }$$ (stress) (yield strength) $$\large{ F = \frac {\tau}{l} }$$ (torque) $$\large{ F = \frac {\tau}{d} }$$ (torque) $$\large{ F = \frac{W}{d} }$$ (work)
### Where:
$$\large{ F }$$ = force
$$\large{ a }$$ = acceleration
$$\large{ A }$$ = area
$$\large{ A_c }$$ = area cross-section
$$\large{ d }$$ = displacement
$$\large{ x }$$ = distance from equilibrium
$$\large{ I }$$ = impulse
$$\large{ l }$$ = length, moment arm
$$\large{ m }$$ = mass
$$\large{ M }$$ = moment
$$\large{ \Delta p }$$ = momentum differential
$$\large{ P }$$ = power
$$\large{ p }$$ = pressure
$$\large{ k_s }$$ = spring constant
$$\large{ x_0 }$$ = spring equilibrium position
$$\large{ \sigma }$$ (Greek symbol sigma) = stress
$$\large{ t }$$ = time
$$\large{ \Delta t }$$ = time differential
$$\large{ \tau }$$ (Greek symbol tau) = torque
$$\large{ v }$$ = velocity
$$\large{ W }$$ = work
$$\large{ \sigma }$$ (Greek symbol sigma) = yield strength
## General Three-dimensional Force
All other combinations of nonconcurrent, nonparallel and noncoplanar forces.
## Force types
All forces can be divided into two basic types of forces:
• Contact force
• Air resistance force
• Applied force
• Friction force
• Normal force
• Spring force
• Tension force
• Non-contact force
• Gravity force
• Electric force
• Magnetic force
### Air Resistance Force
Air resistance force, abbreviated as $$F_{ar}$$, is the force in opposition to a moving object through the air.
### Applied Force
Applied force, abbreviated as $$F_a$$, can come from different types of forces, one of them could be Newton's Second Law. There really is no one formula.
### Centrifugal force
Centrifugal force is when a force pushes away from the center of a circle, but this does not really exist. When an object travels in a circle, the object always wants to go straight, but the centripetal force keeps the object traveling along an axis of rotation.
### Centripetal force
The direction of the centripetal force is always toward the center of the circle and perpendicular to the velocity. Any combination of forces causing a uniform circular motion can be called centripetal force.
### Collinear Force
Collinear forces all share the same line of action.
### Concurrent Force
Concurrent forces are all acting at the same point.
### Coplannar Parallel Force
Coplannar parallel forces can be in the same or opposite direction and are on the same plane.
### Electric Force
An electric force, abbreviated as $$F_e$$, is an attraction or repulsion force between any two charged objects.
### Friction Force
Friction force, abbreviated as $$F_f$$, is the force an object exerts on the surface of an object as it slides across.
### Gravity Force
Gravity force is the force exerted between two masses.
### Magnetic Force
A magnetic force, abbreviated as $$F_m$$, is an attraction or repulsion force between electrically charged particles moving in a magnetic field.
### Non Coplannar Force
All non coplannar forces lines of action lie on different planes.
### Non Coplannar Concurrent Force
All non coplannar concurrent forces act at the same point but their lines of action lie on different planes.
### Non Coplannar Non Concurrent Force
All non coplannar non concurrent forces do not meet and their lines do not lie on the same plane.
### Normal Force
Normal force, abbreviated as $$F_n$$, is the supported force between two contacting objects.
### Spring Force
Spring force, abbreviated as $$F_s$$, is the force that compresses (a repulsive force) or stretches (an attractive force) the spring.
### Tension Force
Tension force, abbreviated as $$F_t$$, is the force that is exerted through a rope, string or wire when pulled from opposite directions.
Tags: Equations for Force | 1,238 | 4,616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2020-24 | latest | en | 0.72864 |
http://tailieu.vn/tag/arithmetic-mean.html | 1,524,411,605,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945604.91/warc/CC-MAIN-20180422135010-20180422155010-00551.warc.gz | 297,095,573 | 15,144 | # Arithmetic mean
Xem 1-12 trên 12 kết quả Arithmetic mean
• ### MATHEMATICS TEST 30 Minutes — 30 Questions
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• ### MATHEMATICS TEST 60 Minutes — 60 Questions
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In the question, you started with six terms. Let a through f equal those six terms: 19 5 a1b1c1d1e1f 6 114 5 a 1 b 1 c 1 d 1 e 1 f f 5 114 2 ~a 1 b 1 c 1 d 1 e! Letting f 5 the number removed, here’s the arithmetic-mean formula, applied to the remaining five numbers: 21 5 a1b1c1d1e 5 105 5 a 1 b 1 c 1 d 1 e Substitute 105 for (a 1 b 1 c 1 d 1 e) in the first equation: f 5 114 2 105 f59 Step 5:...
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This book is partly based on lectures I gave at NUI Galway and Trinity College Dublin between 1998 and 2000. It is by no means a comprehensive guide to all the mathematics an engineer might encounter during the course of his or her degree. The aim is more to highlight and explain some areas commonly found difficult, such as calculus, and to ease the transition from school level to university level mathematics, where sometimes the subject matter is similar, but the emphasis is usually different.
• ### Lecture Statistical techniques in business and economics - Chapter 3: Describing Data: measures of central tendency
When you have completed this chapter, you will be able to: Calculate the arithmetic mean, the weighted mean, the median, the mode, and the geometric mean of a given data set; identify the relative positions of the arithmetic mean, median and mode for both symmetric and skewed distributions; point out the proper uses and common misuses of each measure; explain your choice of the measure of central tendency of data; explain your choice of the measure of central tendency of data.
• ### Ecological Informatics Scope, Techniques and Applications 2nd Edition
The Fuzzy Set Theory developed by L. Zadeh (Zadeh 1965) as a possible way to handle uncertainty is particularly useful for the representation of vague expert knowledge and processing uncertain or imprecise information. The Fuzzy Set Theory is based on an extension of the classical meaning of the term "set" and formulates specific logical and arithmetical operations for processing information defined in the form of fuzzy sets and fuzzy rules. | 1,328 | 5,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2018-17 | latest | en | 0.873845 |
https://scicomp.stackexchange.com/questions/37561/automatic-finite-differences | 1,723,599,852,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00458.warc.gz | 408,943,360 | 45,994 | # Automatic finite differences
Given numbers $$x, y \in \mathbb{R}$$ where $$\frac{|y-x|}{|x|}$$ is small, and code that implements the function $$f$$ with a sequence of arithmetic operations, I would like to compute to high accuracy the finite difference $$\frac{f(y) - f(x)}{y-x},$$ which is plagued by cancellation. Is there a variant of automatic differentiation that lets me do it (without working in higher precision, possibly)? At a first sight it looks like the basic ideas of AD transfer to this similar problem: start from the pair $$(x, y-x)$$, and apply the same sequence of operations to it in "forward mode". But I can't get anything useful with a search for "automatic finite differences".
• Probably also have a look at the complex step derivative (which basically is automatic differentiation in disguise). Commented Jun 10, 2021 at 6:24
• @davidhigh thanks, but I don't really see how that would help in my case. How is $f(x+iz)$ related to an incremental ratio? Could you expand, if you have something in mind? Commented Jun 10, 2021 at 6:32
• @FredericoPoloni: nope, nothing in mind. It was just a reflex based on the words AD and Finite Difference. But as you stated in the comment section of the other answer, your question is a different setup. Commented Jun 10, 2021 at 21:44
This is what Griewank et al. call "Piecewise linearization in secant mode", see for instance https://opus4.kobv.de/opus4-zib/files/6164/newton_secant_approx_paper.pdf. The aim of that research was to capture the kinks of absolute value operations with the same precision a tangent or a secant captures the local behavior of a smooth function, with an implementation as an extension of the forward mode of automatic/algorithmic differentiation (Adol-C and home-brew python classes). But as part of that it of course also computes secants of smooth functions.
The secant slopes are computed using the available tricks for the elementary operations, for instance if $$v(x)=\sin(u(x))$$, then $$S_v=\frac{v(y)-v(x)}{y-x}=2\frac{\sin\frac{u(y)-u(x)}{2}}{y-x}·\cos\frac{u(y)+u(x)}2 \\ =2\cos\frac{u(y)+u(x)}2·\frac{\sin\left(\frac{S_u}2·(y-x)\right)}{y-x}$$ where $$S_u$$, $$S_v$$ are the slopes of the secants of $$u$$ and $$v$$ for the given pair of points.
The numerical evaluation of the sine close to zero is usually implemented in a way that the quotient $$\frac{\sin\epsilon}{\epsilon}$$ is correctly evaluated. If one is unsure about that, one could of course approximate that quotient for values below some threshold using the Taylor series $$1-\frac16\epsilon^2+O(\epsilon^4)$$.
This basic procedure for elementary operations gets extended to the full function by propagating it along the computational graph, the same way as the forward mode of AD. And as there, variants exist in the implementation philosophy, such as reading out the function into a tape and using a "tape machine" for the secant computation, or transforming the computational graph/AST directly into a tree data structure where the secant computation proceeds in node operations, or by propagating a secant "number" type through the original (overloaded) function without reading out the function structure.
Potentially related and useful (I gave these resources to my students when teaching Intro to Computational Mathematics, kinda useful pedagogically too):
"Automatic Source-to-Source Error Compensation of Floating-Point Programs" by Laurent Thévenoux, Philippe Langlois and Matthieu Martel : https://hal.archives-ouvertes.fr/hal-01158399/document
Herbie: Automatically Improving Floating Point Accuracy (https://herbie.uwplse.org/) by Programming Languages and Software Engineering group of the University of Washington (https://uwplse.org/): the main publication is "Automatically Improving Accuracy for Floating Point Expressions" by Pavel Panchekha, Alex Sachez-Stern, James R. Wilcox and Zachary Tatlock which can be found at https://herbie.uwplse.org/pldi15-paper.pdf
There is an analogue of automatic differentiation for divided differences. Here is a minimal sample implementation in C++ that computes the divided difference of $$f(x)=x^3+x^2$$ but which can be extended in a number of directions. Implementations will look similar in Haskell, Python and other languages:
#include <iostream>
template<typename T>
struct FD
{
T a, b, c;
FD operator+(const FD& other)
{
return FD{a + other.a, b + other.b, c + other.c};
}
FD operator*(const FD& other)
{
return FD{a * other.a,
b * other.b,
c * other.b + a * other.c};
}
};
template<typename T>
float FiniteDifference(T (*)(T), T a, T b)
{
return f(FD{a, b, 1.f}).c;
}
template<typename T>
T f(T x)
{
return x * x * x + x * x;
}
int main()
{
std::cout << FiniteDifference(f, 3.f, 8.f) << std::endl;
}
There are a number of ways of looking at this:
(1) if we define $$M=\pmatrix{a&1\\0&b}$$ we can extend analytic functions $$f$$ to act on matrices and so compute $$f(M)$$. The top right entry is the divided difference. The type FD represents the matrix $$\pmatrix{a&c\\0&b}$$ and gives an efficient way to compute products and sums. Note that many methods for optimizing computations over reals carry over to matrices. Eg. we can compute finite differences of $$f(x)=x^n$$ by repeated squaring and the same method can be applied here. For related methods see Kahan & Fateman.
(2) Another point of view is that the variable c is using a modified chain rule to track the differences as functions like + and * are applied. The rule for * is a modified form of the Leibniz rule.
Perhaps what @davidhigh has in mind is the following
$$\frac{f(y)-f(x)}{y-x} = \frac{f(x+h) - f(x)}{h} = f'(x) \,\, \text{ as } h\rightarrow{0}\,\, \text{substituting} \,\, y-x=h$$
And you can get $$f'(x)$$ by complex step differentiation.
• Thanks, but I am still missing something; in this question I am not interested in $f'(x)$: I want to compute the incremental ratio exactly given $x$ and $y$ (close to each other but distinct), not its limit. This is a different setup than the classical one of algorithmic differentiation. Commented Jun 10, 2021 at 9:54
• Well, it seems that I am missing something. Wouldn't your ratio be equal to the derivative when $y-x$ is small?
– NNN
Commented Jun 10, 2021 at 9:56
• No; its limit is equal to the derivative, but the ratio itself is not. Commented Jun 10, 2021 at 9:57
• Take $f(x) = \sin(x)$ for example. Its derivative is $\cos(x)$, but Herbie (mentioned in my answer) suggests a much different function to implement $(\sin(y)-\sin(x))/(y-x)$ if $|h| = |y-x|<10^{-3}$: $\left(\cos x - h \cdot \left(\sin x \cdot 0.5 + h \cdot \left(\cos x \cdot 0.16666666666666666\right)\right)\right) + 0.041666666666666664 \cdot \left(\sin x \cdot {h}^{3}\right)$. This function reduces the average error from 46.2 bits to 0 bits in the vicinity of zero. (Also if $h=0$ you get $\cos(x)$ back which is nice too). Commented Jun 10, 2021 at 14:52
• Thanks everyone. For some reason, this reminds me of 'modified equations'. ocw.mit.edu/courses/mathematics/…
– NNN
Commented Jun 11, 2021 at 6:45 | 1,886 | 7,064 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 21, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.91303 |
https://developer.aliyun.com/article/90734 | 1,643,078,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304749.63/warc/CC-MAIN-20220125005757-20220125035757-00420.warc.gz | 257,548,063 | 17,763 | # 《Python极客项目编程 》——2.3 代码
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### 2.3.1 Spiro构造函数
# a class that draws a Spirograph
class Spiro:
# constructor
def __init__(self, xc, yc, col, R, r, l):
# create the turtle object
1 self.t = turtle.Turtle()
# set the cursor shape
2 self.t.shape('turtle')
# set the step in degrees
3 self.step = 5
# set the drawing complete flag
4 self.drawingComplete = False
# set the parameters
5 self.setparams(xc, yc, col, R, r, l)
# initialize the drawing
6 self.restart()
### 2.3.2 设置函数
# set the parameters
def setparams(self, xc, yc, col, R, r, l):
# the Spirograph parameters
1 self.xc = xc
self.yc = yc
2 self.R = int(R)
self.r = int(r)
self.l = l
self.col = col
# reduce r/R to its smallest form by dividing with the GCD
3 gcdVal = gcd(self.r, self.R)
4 self.nRot = self.r//gcdVal
self.k = r/float(R)
# set the color
self.t.color(*col)
# store the current angle
5 self.a = 0
### 2.3.3 restart()方法
# restart the drawing
def restart(self):
# set the flag
1 self.drawingComplete = False
# show the turtle
2 self.t.showturtle()
# go to the first point
3 self.t.up()
4 R, k, l = self.R, self.k, self.l
a = 0.0
5 x = R*((1-k)*math.cos(a) + l*k*math.cos((1-k)*a/k))
y = R*((1-k)*math.sin(a) - l*k*math.sin((1-k)*a/k))
6 self.t.setpos(self.xc + x, self.yc + y)
7 self.t.down()
### 2.3.4 draw()方法
draw()方法用连续的线段绘制该曲线。
# draw the whole thing
def draw(self):
# draw the rest of the points
R, k, l = self.R, self.k, self.l
1 for i in range(0, 360*self.nRot + 1, self.step):
2 x = R*((1-k)*math.cos(a) + l*k*math.cos((1-k)*a/k))
y = R*((1-k)*math.sin(a) - l*k*math.sin((1-k)*a/k))
self.t.setpos(self.xc + x, self.yc + y)
# drawing is now done so hide the turtle cursor
3 self.t.hideturtle()
### 2.3.5 创建动画
update()方法展示了一段一段绘制曲线来创建动画时所使用的绘图方法。
# update by one step
def update(self):
# skip the rest of the steps if done
1 if self.drawingComplete:
return
# increment the angle
2 self.a += self.step
# draw a step
R, k, l = self.R, self.k, self.l
# set the angle
x= self.R*((1-k)*math.cos(a) + l*k*math.cos((1-k)*a/k))
y = self.R*((1-k)*math.sin(a) - l*k*math.sin((1-k)*a/k))
self.t.setpos(self.xc + x, self.yc + y)
# if drawing is complete, set the flag
4 if self.a >= 360*self.nRot:
self.drawingComplete = True
# drawing is now done so hide the turtle cursor
self.t.hideturtle()
### 2.3.6 SpiroAnimator类
SpiroAnimator类让我们同时绘制随机的螺线。该类使用一个计时器，每次绘制曲线的一段。这种技术定期更新图像，并允许程序处理事件，如按键、鼠标点击，等等。但是，这种计时器技术需要对绘制代码进行一些调整。
# a class for animating Spirographs
class SpiroAnimator:
# constructor
def __init__(self, N):
# set the timer value in milliseconds
1 self.deltaT = 10
# get the window dimensions
2 self.width = turtle.window_width()
self.height = turtle.window_height()
# create the Spiro objects
3 self.spiros = []
for i in range(N):
# generate random parameters
4 rparams = self.genRandomParams()
# set the spiro parameters
5 spiro = Spiro(*rparams)
self.spiros.append(spiro)
# call timer
6 turtle.ontimer(self.update, self.deltaT)
### 2.3.7 genRandomParams()方法
# generate random parameters
def genRandomParams(self):
width, height = self.width, self.height
1 R = random.randint(50, min(width, height)//2)
2 r = random.randint(10, 9*R//10)
3 l = random.uniform(0.1, 0.9)
4 xc = random.randint(-width//2, width//2)
5 yc = random.randint(-height//2, height//2)
6 col = (random.random(),
random.random(),
random.random())
7 return (xc, yc, col, R, r, l)
### 2.3.8 重新启动程序
# restart spiro drawing
def restart(self):
for spiro in self.spiros:
# clear
spiro.clear()
# generate random parameters
rparams = self.genRandomParams()
# set the spiro parameters
spiro.setparams(*rparams)
# restart drawing
spiro.restart()
### 2.3.9 update()方法
def update(self):
# update all spiros
1 nComplete = 0
for spiro in self.spiros:
# update
2 spiro.update()
# count completed spiros
3 if spiro.drawingComplete:
nComplete += 1
# restart if all spiros are complete
4 if nComplete == len(self.spiros):
self.restart()
# call the timer
5 turtle.ontimer(self.update, self.deltaT)
update()方法使用一个计数器nComplete来记录已画的Spiro对象的数目。在1行初始化后，它遍历Spiro对象的列表，在2行更新它们，如果一个Spiro完成，就在3行将计数器加1。
### 2.3.10 显示或隐藏光标
# toggle turtle cursor on and off
def toggleTurtles(self):
for spiro in self.spiros:
if spiro.t.isvisible():
spiro.t.hideturtle()
else:
spiro.t.showturtle()
### 2.3.11 保存曲线
# save drawings as PNG files
def saveDrawing():
# hide the turtle cursor
1 turtle.hideturtle()
# generate unique filenames
2 dateStr = (datetime.now()).strftime("%d%b%Y-%H%M%S")
fileName = 'spiro-' + dateStr
print('saving drawing to %s.eps/png' % fileName)
# get the tkinter canvas
3 canvas = turtle.getcanvas()
# save the drawing as a postscipt image
4 canvas.postscript(file = fileName + '.eps')
# use the Pillow module to convert the postscript image file to PNG
5 img = Image.open(fileName + '.eps')
6 img.save(fileName + '.png', 'png')
# show the turtle cursor
7 turtle.showturtle()
turtle程序采用tkinter创建的用户界面（UI）窗口，在3和4行，利用tkinter的canvas对象，将窗口保存为嵌入式PostScript（EPS）文件格式。由于EPS是矢量格式，你可以用高分辨率打印它，但PNG用途更广，所以在5行用Pillow打开EPS文件，并在6行将它保存为PNG文件。最后，在7行，取消隐藏海龟光标。
### 2.3.12 解析命令行参数和初始化
1 parser = argparse.ArgumentParser(description=descStr)
help="The three arguments in sparams: R, r, l.")
# parse args
3 args = parser.parse_args()
# set the width of the drawing window to 80 percent of the screen width
1 turtle.setup(width=0.8)
# set the cursor shape to turtle
2 turtle.shape('turtle')
# set the title to Spirographs!
3 turtle.title("Spirographs!")
# add the key handler to save our drawings
4 turtle.onkey(saveDrawing, "s")
# start listening
5 turtle.listen()
# hide the main turtle cursor
6 turtle.hideturtle()
# check for any arguments sent to --sparams and draw the Spirograph
1 if args.sparams:
2 params = [float(x) for x in args.sparams]
# draw the Spirograph with the given parameters
col = (0.0, 0.0, 0.0)
3 spiro = Spiro(0, 0, col, *params)
4 spiro.draw()
else:
# create the animator object
5 spiroAnim = SpiroAnimator(4)
# add a key handler to toggle the turtle cursor
6 turtle.onkey(spiroAnim.toggleTurtles, "t")
# add a key handler to restart the animation
7 turtle.onkey(spiroAnim.restart, "space")
# start the turtle main loop
8 turtle.mainloop()
20911 0
【HEVC学习与研究】38、HEVC编码过程中的块分割结构
【本文主要分为前后两部分，前半部分基本是Vivienne Sze、Madhukar BudagaviGary和J. Sullivan所编著的《High Efficiency Video Coding (HEVC) ——Algorithms and Architectures》的第三章前半部分的笔记，后半部分是在HM-10.0中对Intra预测时块分割相应的代码研究。
1306 0
DL之RNN：人工智能为你写代码——基于TF利用RNN算法实现生成编程语言代码(C++语言)、训练&测试过程全记录（二）
DL之RNN：人工智能为你写代码——基于TF利用RNN算法实现生成编程语言代码(C++语言)、训练&测试过程全记录
29 0
numpy基础代码操练
In [20]: b[0,:,1] Out[20]: array([1, 5, 9]) In [21]: b[0,:,1] Out[21]: array([1, 5, 9]) In [22]: b[0,:,-1] Out[22]: array([ 3, 7, 11]) ...
929 0
05.Java网络编程（代码实践）
786 0
+关注
12049
0
《2021云上架构与运维峰会演讲合集》
《零基础CSS入门教程》
《零基础HTML入门教程》 | 2,496 | 7,261 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-05 | latest | en | 0.279752 |
http://support.ptc.com/help/creo/ced_modeling_express/r20.3.0.0/en/ced_modeling/OSDM_Annotation/am_hatch_subpatterns.html | 1,642,607,701,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00366.warc.gz | 61,279,587 | 4,058 | > > > Define hatch sub-patterns
Define hatch sub-patterns
Sub-patterns are the basic elements that comprise the total hatch pattern. They are simple patterns of individually defined lines, which together form the whole pattern. The appearance of each line is determined by a number of settings, such as color and line type.
The sub-patterns of existing hatching can be changed. The procedure for defining sub-patterns is similar for both new and existing hatch patterns. In the Sub Pattern dialog box, you specify the following attributes:
Color Sets the color of the simple pattern. Line Type Sets the line type of the simple pattern. Distance Determines the repetition width of the simple pattern within the whole pattern. The repetition width is calculated as: repetition width = distance x distance of the hatch. You can enter any positive value for the distance. Offset Determines the starting point of the simple pattern within the whole pattern. The starting point is calculated as: starting point = offset x distance of the hatch. You can enter any value between 0 and less than 1 for the offset. Angle Determines the drawing angle of the simple pattern. The drawing angle is calculated as: drawing angle = angle + angle of the hatch.
To create a new sub-pattern,
1. Click Annotation and then, in the Annotate group, click the arrow next to Properties.
2. Click Hatch. The Modify Hatch dialog box opens.
3. Click the Subpattern button in the general section. The Edit Subpattern dialog box opens.
4. Click Insert in the Edit Subpattern dialog box. The dialog box displays a new sub-pattern with default attributes (a solid white line with distance 1 and offset and angle values of 0).
5. Enter the desired changes.
6. Click to complete the operation.
Note that the new sub-pattern is inserted after the current one. To delete the current sub-pattern, click Delete in the Edit Subpattern dialog box.
The number of the current sub-pattern and the total number of existing sub-patterns is shown as a fraction in the menu. Click the right or left arrows, when applicable, to move forward or backward among the sub-patterns. | 436 | 2,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-05 | latest | en | 0.800773 |
http://www.mathconcentration.com/profile/DikshaMaheshwari?xg_source=activity | 1,527,397,530,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868003.97/warc/CC-MAIN-20180527044401-20180527064401-00494.warc.gz | 407,647,329 | 19,158 | One Stop Shop for Math Teacher Resources
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## Notes
### Figure This Challenge #56
• Complete Solution will be given on May 17, 2015
Complete Solution:
…
Continue
Created by Wanda Collins May 10, 2015 at 1:56pm. Last updated by Wanda Collins May 10, 2015.
## Math Homework Help Online
Professional math homework help get your math solved today.
Do you need help with math homework? Our reliable company provides only the best math homework help.
# Math Limerick
Question: Why is this a mathematical limerick?
( (12 + 144 + 20 + 3 Sqrt[4]) / 7 ) + 5*11 = 92 + 0 .
A dozen, a gross, and a score,
plus three times the square root of four, divided by seven, plus five times eleven, is nine squared and not a bit more.
---Jon Saxton (math textbook author)
Presentation Suggestions:
Challenge students to invent their own math limerick!
The Math Behind the Fact:
It is fun to mix mathematics with poetry.
Resources:
Su, Francis E., et al. "Math Limerick." Math Fun Facts.
funfacts
• View All | 440 | 1,748 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-22 | latest | en | 0.826436 |
http://www.ask.com/question/how-to-convert-hz-to-khz | 1,397,864,884,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00493-ip-10-147-4-33.ec2.internal.warc.gz | 284,284,681 | 16,556 | # How to Convert HZ to KHz?
One Hz, or hertz, is equal to 0.001 kHz, or kilohertz. So, to convert Hz to kHz, you multiply the total number of Hz by 0.001. For example, 100 Hz times 0.001 equals 0.1 kHz.
Q&A Related to "How to Convert HZ to KHz"
1. Download a copy of the audio-editing program GoldWave (see Resources) Run the installation program and then open it. 2. Click on the "Open" icon, located in the top-right http://www.ehow.com/how_5161846_convert-wav-khz.ht... | 146 | 474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-15 | latest | en | 0.809087 |
http://mathhelpforum.com/calculus/37101-area-triangle-using-cross-product-print.html | 1,502,936,939,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102819.58/warc/CC-MAIN-20170817013033-20170817033033-00644.warc.gz | 263,077,125 | 2,583 | # Area of triangle using cross product
• May 4th 2008, 08:37 AM
paushali
Area of triangle using cross product
Find the area of the triangle if the vertices are:
A=(2,1)
B=(4,-1)
C=(6,3) using vector cross product.
• May 4th 2008, 08:47 AM
galactus
$P_{1}(2,1), \;\ P_{2}(4,-1), \;\ P_{3}(6,3)$
$\overrightarrow{P_{1}P_{2}}=(2,-2)$
$\overrightarrow{P_{1}P_{3}}=(4,2)$
$\overrightarrow{P_{1}P_{2}}\times\overrightarrow{P _{1}P_{3}}=(0,0,12)$
$\frac{1}{2}||\overrightarrow{P_{1}P_{2}}\times\ove rrightarrow{P_{1}P_{3}}||=?$
Can you finish?. | 229 | 543 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-34 | longest | en | 0.482094 |
https://learn.careers360.com/school/question-provide-solution-for-r-d-sharma-maths-class-12-chapter-19-definite-integrals-exercise-19-point-1-question-30-maths-textbook-solution/?question_number=30.0 | 1,717,064,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059632.19/warc/CC-MAIN-20240530083640-20240530113640-00272.warc.gz | 293,050,141 | 40,192 | #### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 30 Maths Textbook Solution.
Answer: $\frac{-\pi }{4}$
Hint: Use indefinite formula then put the limit to solve this integral
Given: $\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$
Solution:
$\int_{0}^{\frac{\pi}{2}} x^{2} \cos 2 x d x$
Integrating by parts then
\begin{aligned} &=\left[x^{2} \int \cos 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left\{\frac{d\left(x^{2}\right)}{d x} \int \cos 2 x d x\right\} d x \\ &=\left[x^{2} \frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2 x \frac{\sin 2 x}{2} \end{aligned} \quad\left[\int \cos a x d x=\frac{\sin a x}{a}\right]
$=\frac{1}{2}\left[x^{2} \sin 2 x\right]-\frac{2}{2} \int_{0}^{\frac{\pi}{2}} x \sin 2 x d x$
Again using integrating on by parts method
$=\frac{1}{2}\left[\left(\frac{\pi}{2}\right)^{2} \sin 2 \times \frac{\pi}{2}-0^{2} \sin 0\right]-\left[x \int \sin 2 x d x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(\frac{d(x)}{d x} \int \sin 2 x d x\right) d x$
\begin{aligned} &=\frac{1}{2}\left[\frac{\pi^{2}}{4} \sin \pi-\sin 0 \times 0\right]-\left[x\left(\frac{-\cos 2 x}{2}\right)\right]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}\left(\frac{-\cos 2 x}{2}\right) d x \\ &=\frac{1}{2} \times 0+\frac{1}{2}[x \cos 2 x]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \frac{\cos 2 x}{2} d x & {\left[\begin{array}{l} \int \sin a x d x=\frac{-\cos a x}{a} \\ \sin \pi=0, \sin 0=0 \end{array}\right]} \end{aligned}
$=\frac{1}{2}\left[\frac{\pi}{2} \cos 2 \times \frac{\pi}{2}-0 \times \cos 2 \times 0\right]-\frac{1}{2}\left[\frac{\sin 2 x}{2}\right]_{0}^{\frac{\pi}{2}} \quad\left[\int \cos a x d x=\frac{\sin a x}{a}\right]$
\begin{aligned} &=\frac{1}{2}\left[\frac{\pi}{2} \cos \pi-0\right]-\frac{1}{2 \times 2}[\sin 2 x]_{0}^{\frac{\pi}{2}} \\ &=\frac{1}{2}\left[\frac{\pi}{2}\right](-1)-\frac{1}{4}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right] \end{aligned} $\left [ \cos \pi =-1 \right ]$
\begin{aligned} &=\frac{1}{2}\left(\frac{-\pi}{2}\right)-\frac{1}{4}[\sin \pi-\sin 0] \\ &=-\frac{\pi}{4}-\frac{1}{4}[0-0] \end{aligned} $\left [ \sin \pi =\sin 0=0 \right ]$
$=\frac{-\pi }{4}$ | 1,005 | 2,364 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-22 | latest | en | 0.32633 |
https://www.aqua-calc.com/one-to-one/density/long-ton-per-liter/milligram-per-cubic-foot/228 | 1,610,923,135,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703513194.17/warc/CC-MAIN-20210117205246-20210117235246-00059.warc.gz | 708,400,008 | 9,784 | # 228 long tons per liter [long tn/l] in milligrams per cubic foot
## long tons/liter to milligram/foot³ unit converter of density
228 long tons per liter [long tn/l] = 6 559 843 756 032 milligrams per cubic foot [mg/ft³]
### long tons per liter to milligrams per cubic foot density conversion cards
• 228
through
252
long tons per liter
• 228 long tn/l to mg/ft³ = 6 559 843 756 032 mg/ft³
• 229 long tn/l to mg/ft³ = 6 588 615 000 576 mg/ft³
• 230 long tn/l to mg/ft³ = 6 617 386 245 120 mg/ft³
• 231 long tn/l to mg/ft³ = 6 646 157 489 664 mg/ft³
• 232 long tn/l to mg/ft³ = 6 674 928 734 208 mg/ft³
• 233 long tn/l to mg/ft³ = 6 703 699 978 752 mg/ft³
• 234 long tn/l to mg/ft³ = 6 732 471 223 296 mg/ft³
• 235 long tn/l to mg/ft³ = 6 761 242 467 840 mg/ft³
• 236 long tn/l to mg/ft³ = 6 790 013 712 384 mg/ft³
• 237 long tn/l to mg/ft³ = 6 818 784 956 928 mg/ft³
• 238 long tn/l to mg/ft³ = 6 847 556 201 472 mg/ft³
• 239 long tn/l to mg/ft³ = 6 876 327 446 016 mg/ft³
• 240 long tn/l to mg/ft³ = 6 905 098 690 560 mg/ft³
• 241 long tn/l to mg/ft³ = 6 933 869 935 104 mg/ft³
• 242 long tn/l to mg/ft³ = 6 962 641 179 648 mg/ft³
• 243 long tn/l to mg/ft³ = 6 991 412 424 192 mg/ft³
• 244 long tn/l to mg/ft³ = 7 020 183 668 736 mg/ft³
• 245 long tn/l to mg/ft³ = 7 048 954 913 280 mg/ft³
• 246 long tn/l to mg/ft³ = 7 077 726 157 824 mg/ft³
• 247 long tn/l to mg/ft³ = 7 106 497 402 368 mg/ft³
• 248 long tn/l to mg/ft³ = 7 135 268 646 912 mg/ft³
• 249 long tn/l to mg/ft³ = 7 164 039 891 456 mg/ft³
• 250 long tn/l to mg/ft³ = 7 192 811 136 000 mg/ft³
• 251 long tn/l to mg/ft³ = 7 221 582 380 544 mg/ft³
• 252 long tn/l to mg/ft³ = 7 250 353 625 088 mg/ft³
• 253
through
277
long tons per liter
• 253 long tn/l to mg/ft³ = 7 279 124 869 632 mg/ft³
• 254 long tn/l to mg/ft³ = 7 307 896 114 176 mg/ft³
• 255 long tn/l to mg/ft³ = 7 336 667 358 720 mg/ft³
• 256 long tn/l to mg/ft³ = 7 365 438 603 264 mg/ft³
• 257 long tn/l to mg/ft³ = 7 394 209 847 808 mg/ft³
• 258 long tn/l to mg/ft³ = 7 422 981 092 352 mg/ft³
• 259 long tn/l to mg/ft³ = 7 451 752 336 896 mg/ft³
• 260 long tn/l to mg/ft³ = 7 480 523 581 440 mg/ft³
• 261 long tn/l to mg/ft³ = 7 509 294 825 984 mg/ft³
• 262 long tn/l to mg/ft³ = 7 538 066 070 528 mg/ft³
• 263 long tn/l to mg/ft³ = 7 566 837 315 072 mg/ft³
• 264 long tn/l to mg/ft³ = 7 595 608 559 616 mg/ft³
• 265 long tn/l to mg/ft³ = 7 624 379 804 160 mg/ft³
• 266 long tn/l to mg/ft³ = 7 653 151 048 704 mg/ft³
• 267 long tn/l to mg/ft³ = 7 681 922 293 248 mg/ft³
• 268 long tn/l to mg/ft³ = 7 710 693 537 792 mg/ft³
• 269 long tn/l to mg/ft³ = 7 739 464 782 336 mg/ft³
• 270 long tn/l to mg/ft³ = 7 768 236 026 880 mg/ft³
• 271 long tn/l to mg/ft³ = 7 797 007 271 424 mg/ft³
• 272 long tn/l to mg/ft³ = 7 825 778 515 968 mg/ft³
• 273 long tn/l to mg/ft³ = 7 854 549 760 512 mg/ft³
• 274 long tn/l to mg/ft³ = 7 883 321 005 056 mg/ft³
• 275 long tn/l to mg/ft³ = 7 912 092 249 600 mg/ft³
• 276 long tn/l to mg/ft³ = 7 940 863 494 144 mg/ft³
• 277 long tn/l to mg/ft³ = 7 969 634 738 688 mg/ft³
• 278
through
302
long tons per liter
• 278 long tn/l to mg/ft³ = 7 998 405 983 232 mg/ft³
• 279 long tn/l to mg/ft³ = 8 027 177 227 776 mg/ft³
• 280 long tn/l to mg/ft³ = 8 055 948 472 320 mg/ft³
• 281 long tn/l to mg/ft³ = 8 084 719 716 864 mg/ft³
• 282 long tn/l to mg/ft³ = 8 113 490 961 408 mg/ft³
• 283 long tn/l to mg/ft³ = 8 142 262 205 952 mg/ft³
• 284 long tn/l to mg/ft³ = 8 171 033 450 496 mg/ft³
• 285 long tn/l to mg/ft³ = 8 199 804 695 040 mg/ft³
• 286 long tn/l to mg/ft³ = 8 228 575 939 584 mg/ft³
• 287 long tn/l to mg/ft³ = 8 257 347 184 128 mg/ft³
• 288 long tn/l to mg/ft³ = 8 286 118 428 672 mg/ft³
• 289 long tn/l to mg/ft³ = 8 314 889 673 216 mg/ft³
• 290 long tn/l to mg/ft³ = 8 343 660 917 760 mg/ft³
• 291 long tn/l to mg/ft³ = 8 372 432 162 304 mg/ft³
• 292 long tn/l to mg/ft³ = 8 401 203 406 848 mg/ft³
• 293 long tn/l to mg/ft³ = 8 429 974 651 392 mg/ft³
• 294 long tn/l to mg/ft³ = 8 458 745 895 936 mg/ft³
• 295 long tn/l to mg/ft³ = 8 487 517 140 480 mg/ft³
• 296 long tn/l to mg/ft³ = 8 516 288 385 024 mg/ft³
• 297 long tn/l to mg/ft³ = 8 545 059 629 568 mg/ft³
• 298 long tn/l to mg/ft³ = 8 573 830 874 112 mg/ft³
• 299 long tn/l to mg/ft³ = 8 602 602 118 656 mg/ft³
• 300 long tn/l to mg/ft³ = 8 631 373 363 200 mg/ft³
• 301 long tn/l to mg/ft³ = 8 660 144 607 744 mg/ft³
• 302 long tn/l to mg/ft³ = 8 688 915 852 288 mg/ft³
• 303
through
327
long tons per liter
• 303 long tn/l to mg/ft³ = 8 717 687 096 832 mg/ft³
• 304 long tn/l to mg/ft³ = 8 746 458 341 376 mg/ft³
• 305 long tn/l to mg/ft³ = 8 775 229 585 920 mg/ft³
• 306 long tn/l to mg/ft³ = 8 804 000 830 464 mg/ft³
• 307 long tn/l to mg/ft³ = 8 832 772 075 008 mg/ft³
• 308 long tn/l to mg/ft³ = 8 861 543 319 552 mg/ft³
• 309 long tn/l to mg/ft³ = 8 890 314 564 096 mg/ft³
• 310 long tn/l to mg/ft³ = 8 919 085 808 640 mg/ft³
• 311 long tn/l to mg/ft³ = 8 947 857 053 184 mg/ft³
• 312 long tn/l to mg/ft³ = 8 976 628 297 728 mg/ft³
• 313 long tn/l to mg/ft³ = 9 005 399 542 272 mg/ft³
• 314 long tn/l to mg/ft³ = 9 034 170 786 816 mg/ft³
• 315 long tn/l to mg/ft³ = 9 062 942 031 360 mg/ft³
• 316 long tn/l to mg/ft³ = 9 091 713 275 904 mg/ft³
• 317 long tn/l to mg/ft³ = 9 120 484 520 448 mg/ft³
• 318 long tn/l to mg/ft³ = 9 149 255 764 992 mg/ft³
• 319 long tn/l to mg/ft³ = 9 178 027 009 536 mg/ft³
• 320 long tn/l to mg/ft³ = 9 206 798 254 080 mg/ft³
• 321 long tn/l to mg/ft³ = 9 235 569 498 624 mg/ft³
• 322 long tn/l to mg/ft³ = 9 264 340 743 168 mg/ft³
• 323 long tn/l to mg/ft³ = 9 293 111 987 712 mg/ft³
• 324 long tn/l to mg/ft³ = 9 321 883 232 256 mg/ft³
• 325 long tn/l to mg/ft³ = 9 350 654 476 800 mg/ft³
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• 327 long tn/l to mg/ft³ = 9 408 196 965 888 mg/ft³
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1 Digital logic fundamentals Question Bank Subject Name : Digital Logic Fundamentals Subject code: CA102T Staff Name: R.Roseline Unit I 1. What is Number system? 2. Define binary logic. 3. Show how negative numbers are stored in a digital computer? 4. Convert decimal number (300.55)10 to octal and hexadecimal number? 5. State and Prove De Morgan s theorem. 6. Write short notes on Boolean laws and theorem? 7. What are the basic gates? Show along with the truth table. 8. Convert binary number to decimal and octal number. 9. Convert a. (450.12)8 to hexadecimal b. (C2AE)16 to octal 10. What are the Universal gates? Brief any one. 11. Brief binary number system with example? 12. Brief the Huntington postulates of Boolean algebra? 13. State and Prove demorgan s double inversion theorem. 1. Convert decimal ( )10 to binary, octal and hexadecimal. 2. Explain the various Boolean laws with logic diagrams. 3. Explain the logic gates in detail? 4. Define the duality principle in Boolean algebra? 5. State and prove the absorption theorem using postulates? 6. Show the theorems of Boolean algebra with correct derivations. 7. Define the term logic gates. Explain the various basic gates with their symbols and truth table.
2 8. What are the universal gates and explain them? 9. Prove the De Morgan s theorem with logic diagrams. Unit II 1. Convert to the canonical form F(x,y,z)= (1,3,7) 2. Draw K-map for 3-variables. 3. Write a note on Don t care condition with example. 4. Explain the karnaugh map method 5. Simplify the Boolean function F(A,B,C,D)= (0,1,2,5,8,9,10) as i. Sum of products ii. Product of sums 6. Simplify the following expression using Quine McClausky method and verify using K-map F(A,B,C,D)= (0,1,2,3,4,6,8,10,12,14) 7. Simplify the boolean function F(W,X,Y,Z)= (1,3,7,11,15) and the don t care condition d(w,x,y,z)= (0,2,5) 8. Convert the following to the other canonical form a) F(X,Y,Z)= (1,3,7) b) F(A,B,C,D)=π(0,1,2,3,4,6,12) 9. Explain Pos method with example? 10. Simplify the following using K-map f(x,y,z)= (0,2,4,5,6) 11. Find the product of maxterms for the given function F(X,Y,Z)= (0,2,4,5,6) 12. Differentiate between SOP and POS. 13. Simplify F(W,X,Y,Z)= (1,3,7,11,15) with don t care function d(w,x,y,z)= (0,2,5) 14. Find the POS for the function F(x,y,z)=π(0,1,4,5) 15. Discuss about five variable k-map. 1. Simplify using k-map F=(W,X,Y,Z)= (2,3,12,13,14,15) 2. Simplify using Quine McClausky tabulation method
3 F= (0,1,2,810,11,14,15) 3. What is SOP and POS? Explain. 4. Explain the method of Karnaugh map simplification with don t care condition. Give example. 5. Simplify the following expression using Quine McClausky tabulation method and verify using K-Map F(A,B,C,D)= (0,1,2,3,5,6,7,8,11,13) 6. Explain the steps involved in 4-variable k-map simplification? 7. Simplify using tabulation method F= (0,1,2,8,10,11,14,15) 8. Simplify the Boolean function using k-map F(W,X,Y,Z)= (1,3,4,6,9,11,12,14) 9. Find using K-map F(W,X,Y,Z)= (0,1,2,4,5,6,8,9,12,13,14) 10. Find SOP : i) F=A+B c ii)f=xy+x z 11. Explain about SOP with an example. 12. Explain Quine McClausky tabulation method with example. 13. Write down the steps to convert Boolean functions into sum of minterms and product of max terms with suitable example. Unit III 1. Discuss briefly the Full adder with the truth table and circuit 2. Explain the BCD adder with block diagram 3. Write note on the full- subtractor with block diagram and truth table 4. Write the working principle of Full- adder with its circuit and truth table. 5. Explain ROM and its types. 6. What is PLA? Draw the block diagram of PLA and explain it?
4 1. Give the working principle of BCD adder. 2. Write about the working style of half adder and full adder with the help of circuits. 3. Explain the block diagram of full adder. 4. Construct a 4- bit binary adder. 5. List the various types of ROM in detail. 6. Implement the circuit of a PLA with 3 input, 2 output and 4 product terms F1(A,B,C)= (3,5,6,7) F2(A,B,c)= (0,2,4) 7. Design a circuit using ROM for F1(A1,A0)= (1,2,3) F2(A1,A0)= (0,2) Unit IV 1. Explain the working of encoder? 2. What is decoder? Draw 1 of 16 decoder circuit and explain it. 3. Design a BCD to decimal decoder? 4. What is encoder? Give the truth table of octal to binary encoder? 5. Implement the following function with a multiplexer? F(A,B,C,D)= (0,1,3,4,8,9,15) 6. Explain about the multiplexer with a block diagram. 1. Give a detailed account on encoder. 2. Draw the gate equivalent circuit of 8 x 1 multiplexer. Explain its working? 3. What is demultiplexer? Explain the quadruple 2 to 1 multiplexer with neat logic diagram and function table.
5 4. Design a combinational circuit using ROM. The circuit accepts a 3-bit number and generates an output binary number equal to square of input number. 5. Explain about encoder and decoder. 6. Types of ROM in detail. 7. How will you implement F(A,B,C)= (1,3,5,6,7) with a multiplexer. 8. Explain in detail about ROM/PLA. 9. Design a 1 of 16 decoder with a neat diagram. UNIT V 1. What is latch? Explain. 2. What is Flip-Flop? Why we need it? 3. Explain the working of JK flip flop? 4. Write short notes on shift registers. 5. How can you make D flip flop from RS flip flop? 6. Define counter. List out its types. 7. Brief the working principle of D flip flop. 8. How does a 4 bit register work with a parallel load? 9. Draw the circuit of T flip flop and explain briefly. 10. Brief the working of RS flip flop. 1. Explain the JK flip flop with a neat diagram. 2. Explain the shift register in detail. 3. Explain the concept of BCD counter with parallel load? 4. Discuss about the Master slave flip flop 5. Explain the bidirectional shift register with parallel load? 6. Explain in detail about ripple counter.
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MULTIMEDIA COLLEGE JALAN GURNEY KIRI KUALA LUMPUR
STUDENT IDENTIFICATION NO MULTIMEDIA COLLEGE JALAN GURNEY KIRI 54100 KUALA LUMPUR SECOND SEMESTER FINAL EXAMINATION, 2013/2014 SESSION ITC2223 COMPUTER ORGANIZATION & ARCHITECTURE DSEW-E-F 1/13 18 FEBRUARY
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COURSE STRUCTURE (W.E.F. 2011 Batch Students) (Total Unit 7.5) Course Theory Unit Course Sessional Unit Code Code DCP 4001 Data Structures 1.0 DCP 4002 Data Structures Lab. 0.5 DEC 4003 Electronics Circuits
Semester: I Credits: 5. Category: MC No.of hrs/week: 5 CA PROGRAMMING IN C
Semester: I Credits: 5 Category: MC No.of hrs/week: 5 CA1505 - PROGRAMMING IN C Objective: This course aims at explaining the basic concepts of computers and an easy understanding of C Language by the | 8,961 | 32,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-21 | longest | en | 0.764338 |
http://www.khanacademy.org/math/recreational-math/math-warmup/distribution-warmup/a/bean-machines | 1,427,397,921,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131292567.7/warc/CC-MAIN-20150323172132-00032-ip-10-168-14-71.ec2.internal.warc.gz | 491,086,017 | 17,698 | Math warmups
# Distribution warmup
Take a moment and think about a sequence of random events you have experienced in the past. Such as lottery numbers, coin flips or dice rolls.
We call these independent events because the past outcomes do not influence future outcomes. If we flip a coin 3 times and it lands heads each time, we know the fourth flip is still equally likely to be heads or tails.
One interesting question is how many heads occur, on average, in a sequence of coin flips. If we flip a coin 10 times, what can we say about the likely outcome? Can we say anything at all?
Instead of flipping coins all day we can use an equivalent mechanism to speed things up. What is equivalent to a coin flip? Well, any process which results in 2 equally likely outcomes. For example, we could drop a disk directly over a peg which deflects it either to the right or left.
To model a sequence of 12 coin flips we must chain multiple collisions together. This is exactly what Francis Galton did in the middle of the 19th century with his bean machine. Below is a diagram of his device:
At the bottom of the machine are buckets which represent the total number of right vs. left deflections in the sequence. For example the middle bucket represents an equal number of right vs. left deflections. While the buckets at the far left edge represent 12 left deflections.
If we drop hundreds of balls into this machine how do you think they will distribute among the various buckets? Is one bucket more likely to full up than another? Are they all equal?
Let’s find out! Below we have a simulation of Galton's device for you to use:
What do you notice?
Next up we have an exercise to test your understanding of this phenomenon. | 368 | 1,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2015-14 | latest | en | 0.945319 |
http://pdesolutions.com/help/usage_misc_stages.html | 1,534,553,525,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213247.0/warc/CC-MAIN-20180818001437-20180818021437-00491.warc.gz | 320,748,735 | 3,311 | Sample Problems > Usage > Misc > stages
# stages
Navigation: Sample Problems > Usage > Misc >
# stages
{ STAGES.PDE
This example demonstrates the use of staging to solve a problem for a range of parameters.
We stage both the equation parameters and the solution ERRLIM.
The problem is a nonlinear test, which solves a modified steady-state Burgers equation.
}
title 'Staged Problem'
select
stages = 3 { run only the first three of the listed stages }
errlim = staged(0.01, 0.001, 0.0005)
Variables
u
definitions
scale = staged(1, 2, 4, 8) { extra value ignored }
a = 1/scale
Initial values
u = 1 - (x-1)^2 - (y-1)^2
equations
U: div(a*grad(u)) + scale*u*dx(u) +4 = 0;
boundaries
region 1
start(0,0)
value(u)=0
line to (2,0) to (2,2) to (0,2) to close
monitors
contour(u)
plots
surface(u) report scale as "Scale"
contour(u) report scale as "Scale"
histories
history(integral(u)) vs scale as "Ingegral vs Scale"
end | 293 | 954 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-34 | longest | en | 0.744838 |
http://windowssecrets.com/forums/showthread.php/85585-If-Cell-Contains-cancel-make-1-if-not-%28XP-Excell%29 | 1,503,169,449,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886105712.28/warc/CC-MAIN-20170819182059-20170819202059-00202.warc.gz | 468,130,091 | 13,442 | # Thread: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
1. ## If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
Hello All,
I'm trying to sort a stack of info to see if cells in a column contain "Cancel" within any part of the cell. The problem is Cancel may appear first or after a few words, so i can't do a "lookleft" type If. After I do the If statement I'll use a Countif to tally them up to determine how many cells needed to be canceled.
If(A1 contains "cancel",1,"")
If(A2 contains "shipped",2,"")
and so on.
2. ## Re: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
Not sure why you switch from "cancel" to "shipped" and from 1 to 2, but see if you can adapt the formula
=IF(ISERROR(FIND("cancel",A1)),0,1)
3. ## Re: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
OK, that works great. My next question is how can I make that formula look for multiple objects..
`=IF(ISERROR(FIND("cancel",B1),1,(FIND("Shipped",B 1),2,(FIND("Held",B1),0,3)
obviously it's not working the way I have it but translated I need it to be:
if it finds Cancel give 1, if it gives shipped give 2, if it gives held, give 3.
Can that all be done in one formula or will I need multiple columns for each? I've attached to sheet that I need this to apply to.
4. ## Re: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
You can use this formula in row 2:
<code>
=IF(ISERROR(FIND("Shipped",D2)),IF(ISERROR(FIND("H eld",D2)),IF(ISERROR(FIND("Cancel",D2)),0,1),2),3 )
</code>
and fill down as far as needed.
Note: calculation is set to manual in your workbook, so you'll have to select the cells and press F9 to update their values.
5. ## Re: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
Try: =IF(ISERROR(FIND("Cancel",B1)),1,IF(ISERROR(FIND(" Shipped",B1)),2,IF(ISERROR(FIND("Held",B1)),3,"")
6. ## Re: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
=IF(ISNUMBER(FIND("cancel",B1)),1,IF(ISNUMBER(FIND ("Shipped",B1)),2,IF(ISNUMBER(FIND("Held",B1)),3,0 )))
If you do not want it case sensitive then use:
=IF(ISNUMBER(SEARCH("cancel",B1)),1,IF(ISNUMBER(SE ARCH("Shipped",B1)),2,IF(ISNUMBER(SEARCH("Held",B1 )),3,0)))
Steve
7. ## Re: If Cell Contains 'cancel' make 1 , if not '' (XP Excell)
That won't work (apart from the two missing closing parentheses): if Cancel is not found, this formula will always return 1, without looking for Shipped or Held.
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• | 777 | 2,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2017-34 | latest | en | 0.804573 |
https://www.atozexams.com/mcq/quantitative-aptitude/104.html | 1,722,758,589,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640393185.10/warc/CC-MAIN-20240804071743-20240804101743-00792.warc.gz | 530,606,251 | 10,327 | # What is the year next to 1990 which will have the same calendar as that of the year 1990
1. 1992
2. 1997
3. 1995
4. 1996
4
1996
Explanation :
No Explanation available for this question
# Find prt:qsu, given that p:r=2:3, q:s=4:3, r:t=8:9, s:u=3:2 and t:q=3:4
1. 4/27
2. 8/27
3. 16/27
4. 18/27
4
16/27
Explanation :
No Explanation available for this question
# January 1, 2004 was a Thursday, what day of the week lies on January 1 2005
1. Saturday
2. Monday
3. Saturday
4. Tuesday
4
Saturday
Explanation :
No Explanation available for this question
# Find the number of five-digit multiples of 4, which can be formed using the digits from 1 to 7, using each digit at most once.
1. 720
2. 84
3. 840
4. 600
4
600
Explanation :
No Explanation available for this question
# What is the difference between local value and face value of 7 in the numerical 657903
1. 6993
2. 69993
3. 7000
4. 7
4
6993
Explanation :
No Explanation available for this question
# If the first day of a year (other than leap year) was Friday, then which was the last day of that year
1. Saturday
2. Friday
3. Tuesday
4. Monday
4
Friday
Explanation :
No Explanation available for this question
# In a company, the number of married employees is a prime number. Which of the following cannot be the ratio of the total number of employees and the number of unmarried employees
1. 173:90
2. 193:80
3. 181:32
4. 201:32
4
201:32
Explanation :
No Explanation available for this question
# 108 + 109 + 110 + ... + 202 =
1. 14615
2. 14625
3. 14715
4. 14725
4
14725
Explanation :
No Explanation available for this question
# The speeds of Aniket, Brihat and Chatur are in the ratio of 4:3:2. If Chatur takes half an hour more than Brihat to travel from P to Q, what is the total time taken by Aniket to travel from P to Q and then back from Q to P
1. 120 minutes
2. 90 minutes
3. 60 minutes
4. 45 minutes
4
90 minutes
Explanation :
No Explanation available for this question
1. Saturday
2. Thursday
3. Wednesday
4. Tuesday
4 | 694 | 2,104 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2024-33 | latest | en | 0.739275 |
https://www.physicsforums.com/threads/rolling-without-slipping-too-many-equations.609866/ | 1,534,595,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221213666.61/warc/CC-MAIN-20180818114957-20180818134957-00552.warc.gz | 972,577,402 | 16,550 | Homework Help: Rolling without slipping - too many equations
1. May 29, 2012
jason.farnon
1. The problem statement, all variables and given/known data
I think I have too many equations for unknowns for rolling without slipping. Suppose a wheel is pushed and left rolling along the ground without slipping.
2. Relevant equations
T = I*alpha, where alpha is the rotation about the wheel's centroid and I is the moment of inertia.
F = ma.
omega = r*v, where omega is the rotational speed, r is the wheel radius and v is the translational speed. The rolling without slipping condition. From this, alpha = r*a.
T = r*F, where F is friction.
3. The attempt at a solution
I*alpha = r*F, so alpha = r*F/I. For rolling without slipping, a = r^2 * F/I. But from Newton's second law, a = F/m. What am I doing wrong. Thanks.
2. May 29, 2012
azizlwl
There are 3 'F' here not one.
F force apllied
Fnet=ma
Ffriction=Iα
Fnet=F apllied±Ffriction=ma
Last edited: May 29, 2012
3. May 30, 2012
jason.farnon
What is the force applied you're referring to? I was referring to the situation after the person/external agent releases the wheel with some initial velocity, and the wheel is just rolling by itself on the ground without slipping subject to friction.
4. May 30, 2012
Staff: Mentor
So the wheel is just rolling along with a = 0, since there's no longer a force pushing it?
5. May 30, 2012
azizlwl
Without external forces, the wheel will keep on rolling forever at initial rotational and translational velocity.
6. May 30, 2012
cupid.callin
What, according to you, would happen if someone rotates the wheel clockwise and
A) there is no friction
B) there is friction. In this case what will be direction of friction?
7. May 30, 2012
azizlwl
The constraint is rolling without slipping.
Then friction as the torque for rotation.
A. Will rotate forever like the satellite.
B. The friction is anticlockwise.
8. May 30, 2012
jason.farnon
right, no one is pushing it, but there is friction. sorry if I was unclear in setting up the problem. Say I take a wheel, cylinder, ball or whatever and set it rolling on the ground, as you might pitch a bowling ball. It has a certain initial velocity but I am no longer pushing it. Friction is still present, but it is not dissipative, so there is rolling without sliding.
9. May 30, 2012
Staff: Mentor
Assuming you have met the conditions for rolling without slipping on a horizontal surface, the friction will be zero.
10. May 30, 2012
jason.farnon
I don't follow. If friction were zero it seems to me there would be no rolling at all. The object would just slide.
11. May 30, 2012
Staff: Mentor
If friction were nonzero, then it couldn't be rolling without slipping at some constant speed. Friction is needed to change the rotational speed, but the friction force becomes zero once the conditions are met for rolling without slipping.
12. May 30, 2012
truesearch
You have to be careful and distiguish between 2 types of friction....sliding friction and rolling friction. When something like a snooker ball is first struck the ball slides over the surface and sliding friction causes the balls linear speed to decrease.
The frictional force acting on the ball produces a turning effect which makes the ball rotate.
At the point of contact with the table the forward velocity will decrease to a value, v, and the (backwards) rotation velocity will increase to a velocity, v. At this point there is no relative motion between the point of contact on the ball and the point of contact on the table so sliding friction will become zero. The ball will then continue to roll with no sliding (slipping).
The link between linear velocity and angular velocity when this happens is v = ωr
There is now only rolling friction which is very small compared to sliding friction
Last edited: May 30, 2012
13. May 30, 2012
jason.farnon
Thanks, that is something I was wondering about. But, maybe I'm missing something, but I don't see that it answers my original question. Take the point that there is only rolling friction and thereafter. I don't see what is wrong with my two derivations for the acceleration of the ball, but they are contradictory.
14. May 30, 2012
Staff: Mentor
Once the ball is rolling without slipping, there is no acceleration. (Ignoring rolling friction.)
Your derivation takes "F" as some force acting on the wheel. What's that force? The applied force? But that's zero, since you stopped pushing. Friction? There is none. (I'm talking static friction, not rolling friction.)
15. May 30, 2012
jason.farnon
I was using F to refer to the "rolling friction". Why does that affect the derivation? The translational effect on the centroid of the wheel/ball/object should be analyzable separately from the spinning wheel, shouldn't it? I have encountered derivations where that appears to be the case.
16. May 30, 2012
azizlwl
Rolling friction is static friction.
Just like a block on the a rough table, the static fraction is equal or less to the force applied.
When no force applied, static friction is equal to zero.
When no force applied on rotating object without slippage then slipping
friction is zero.
17. May 31, 2012
Staff: Mentor
I don't think you are. The term "rolling friction" usually refers to a dissipative force caused by the deformation of the contact surfaces. When truesearch used the term I think he actually meant the static friction that appears during rolling motion. That static friction force is zero, once rolling without slipping is attained.
Actual rolling friction is a bit more complicated and is the force that will eventually stop the wheel as it rolls along the horizontal surface. | 1,379 | 5,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2018-34 | latest | en | 0.927403 |
http://bankersdaily.in/11647-2/ | 1,627,849,468,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154219.62/warc/CC-MAIN-20210801190212-20210801220212-00381.warc.gz | 4,461,859 | 23,107 | ## Data Interpretation For SBI PO : 64
D 1-5) Go through the data set given below and solve the questions based on it.
The pie chart shows the productions of different products in degree.
#### The total number of production = 16200
1) The production of Bovanto is what percentage more than the production of Fanta?
a) 112%
b) 114%
c) 120%
d) 111%
e) None of these
d) 111%
The production of Fanta =45/360×16200=2025
The production of Bovanto =95/360×16200 = 4275
Required percentage = (4275-2025)/2025×100 = 111%
2) The production of maaza in the year 2017 is 20% more than the production in the year 2016, the production of Pepsi in the year 2017 is 30% more than the production in the year 2016, then what is the approximate Difference between the production of Pepsi and Maaza in the year 2016?
a) 2040
b) 2030
c) 2130
d) 1940
e) None of these
b) 2030
The production of maaza in the year 2017 =15/360×16200=675 is 20% more than in the year 2016
Then, the production of maaza in the year 2016 =675/120×100 = 562.5
The production of Pepsi in the year 2017=75/360×16200 = 3375
Then, the production of Pepsi in the year 2016 =3375/130×100 = 2596
Required difference = 2596 – 562.5= 2033.5 ~2030
3) what is the Ratio of the production of Torina and Pepsi in the year 2017?
a) 13:25
b) 17:15
c) 15:13
d) 13:15
e) None of these
d) 13:15
Required Ratio =65/360×16200 :75/360× 16200
= 65:75 = 13:15
4) what is the total production of maaza, coca cola, Torina, Pepsi, and Fanta?
a) 10135
b) 10115
c) 10125
d) 11125
e) None of these
c) 10125
The production of Fanta =45/360×16200=2025
The production of Pepsi in the year 2017=75/360×16200 = 3375
The production of maaza in the year 2017 =15/360×16200=675
The production of Torina in the year 2017 =65/360×16200=2925
The production of coca cola in the year 2017 =25/360×16200=1125
Required total = 2025 + 3375 + 675 + 2925 + 1125 = 10125
5) What is the difference between the production of Pepsi and coca cola?
a) 2350
b) 2250
c) 2450
d) 2050
e) None of these
b) 2250
The production of Pepsi in the year 2017=75/360×16200 = 3375
The production of coca cola in the year 2017 =25/360×16200=1125
Required difference = 3375 – 1125 = 2250.
D.6-10) Go through the data set given below and solve the questions based on it.
The pie chart 1 shows the percentage of eggs export to different countries from Andhra Pradesh. The pie chart 2 given the details about shipment cost of various countries per egg (in Rs)
## Total number of eggs= 300000
Total shipment cost of various countries =Rs.200
6) what is the total shipment cost of Ukraine and Vanuatu?
a) 369000
b) 471400
c) 3960000
d) 470400
e) None of these
c) 3960000
Number of eggs in Ukraine=300000×18/100=54000
Number of eggs in vanuatu=300000×25/100=75000
per egg cost for ukraine=200×20/100=Rs.40
Per egg cost for Vanuatu=200×12/100=Rs.24
Total shipment cost of Ukraine and Vanuatu=54000×40+75000×24
=2160000+1800000
=Rs.3960000
7) If the total shipment cost of Kenya is 20% more than total shipment cost of virgin islands(US) and the number of eggs export to Kenya is 50 % of the number of eggs export to Uruguay, then what is the approximate shipment cost of Kenya per egg (in Rs)?
a) Rs.82
b) Rs.80
c) Rs.83
d) Rs.80.5
e) None of these
b) Rs.80
Number of eggs export to Virgin islands(US)=300000×10/100=30000
Per egg shipment cost for virgin islands (US)=200×20/100=Rs.40
Total shipment cost of virgin islands=30000×40=1200000
Total shipment cost of kenya=120/100×1200000=1440000
Number of eggs export to Uruguay=300000×12/100=36000
Number of eggs export to Kenya=50% of 36000=50/100×36000=18000
Required, the shipment cost of Kenya per egg = 1440000/18000= Rs.80
8) what is the total shipment cost of Virgin Islands (British)?
a) 2105000
b) 1200000
c) 1800000
d) 200400
e) None of these
c) 1800000
Number of eggs export to virgin islands(British)=300000×15/100=45000
per egg cost for virgin islands (British)=200×20/100=Rs.40
The Total shipment cost of Virgin Islands (British)= 45000×40=Rs.1800000
9) what is the different between the total shipment cost of Uganda and the total shipment cost of Uruguay?
a) 69000
b) 85600
c) 96000
d) 86000
e) None of these
c) 96000
Number of eggs export to Uganda=300000×20/100=60000
per egg cost of Uganda=200×10/100=20
Total shipment cost of Uganda=20×60000=1200000
Number of eggs export to Uruguay=300000×12/100=36000
Per egg cost of Uruguay=200×18/100=36
Total shipment cost of Uruguay=36×36000=1296000
Require difference=1296000-1200000=96000
10) The total shipment cost of Vanuatu is what percentage more than the total shipment cost of Virgin Islands (U.S)?
a) 45%
b) 56%
c) 55%
d) 50%
e) None of these
d) 50%
number of eggs export to Vanuatu=300000×25/100=75000
Per egg cost of Vanuatu=200×12/100=24
Total shipment cost of Vanuatu=24×75000=1800000
Number of eggs export to Virgin island (US)=300000×10/100=30000
Per egg cost of Virgin island=200×20/100=40
Total shipment cost of Virgin island=40×30000=1200000
Required percentage=600000/1200000×100=50% | 1,715 | 5,115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-31 | latest | en | 0.794832 |
https://mystifying-ramanujan.netlify.app/hat-matrix-and-leverage.html | 1,660,247,592,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571502.25/warc/CC-MAIN-20220811194507-20220811224507-00654.warc.gz | 417,266,992 | 10,168 | Hat Matrix And Leverage
The leverage h i i is a measure of the distance between the x value for the i t h data point and the mean of the x values for all n data points.
Hat matrix and leverage. It couldbe economic mobility or sightings of Bigfoot and the same design matrix willgive us the same hat matrix and leverages. Thus the HC2 and HC3 estimator differ as the values of hii are quite different. The sum of the h i i equals p the number of parameters regression coefficients including the intercept.
It is useful for investigating whether one or more observations are outlying with regard to their X values and therefore might be excessively influencing the regression results. Because the sum of the leverage values is p the number of coefficients in the regression model an observation i can be considered an outlier if its leverage substantially exceeds pn where n is the number of observations. However the points farther away at the extreme of the regressor values will have more leverage.
Each point of the data set tries to pull the ordinary least squares OLS line towards itself. Hat matrix H for the matrix or are you asking more about the importance of the matrix product on the righthand side. It is useful for investigating whether one or more observations are outlying with regard to their X values and therefore might be excessively influencing the regression results.
Hat Matrix and Leverages Basic idea. Use the hat matrix to identify outliers inX. Leverage considered large if it is bigger than.
The hat matrix provides a measure of leverage. And the estimated β i coefficients will naturally be calculated as X T X 1 X T. Begingroup Are you asking why we need to have a special namesymbol ie.
Authors Michail Tsagris R implementation and documentation. A vector with the diagonal Hat matrix values the leverage of each observation. The leverage h i i is a number between 0 and 1 inclusive.
Learn How To Leverage Your Social Presence To Drive In Store Sales With Our Free Guide Social Media Management Software Social Media Social Media Strategies
Source : pinterest.com | 414 | 2,101 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.865538 |
https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-14-partial-derivatives-section-14-8-lagrange-multipliers-exercises-14-8-page-852/2 | 1,726,347,406,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.74/warc/CC-MAIN-20240914193334-20240914223334-00817.warc.gz | 735,731,928 | 13,257 | ## Thomas' Calculus 13th Edition
$(\pm \sqrt 5, \sqrt 5)$ and $(\pm \sqrt 5,-\sqrt 5)$
Consider $f(x,y) =xy$ $g(x,y)=x^2+y^2-10=0$ ...(1) Consider the gradient equation as follows: $\nabla =\lambda \nabla g$ Here, $y=2 \lambda x$ or, $y=\pm \dfrac{1}{2}$ Now, $x=2 \lambda y$ or, $x=\pm y$ From equation (1), we have $g(x,y)=x^2+y^2-10=0$ This implies that $x=\pm \sqrt 5$ and $y= \pm \sqrt 5$ Points: $(\pm \sqrt 5, \sqrt 5)$ and $(\pm \sqrt 5,-\sqrt 5)$ | 204 | 456 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-38 | latest | en | 0.525801 |
https://math.eretrandre.org/tetrationforum/archive/index.php?thread-1188.html | 1,580,069,789,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00011.warc.gz | 549,110,995 | 2,173 | # Tetration Forum
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In the title I mentioned that this is a repost.
Indeed these ideas are not completely new and have been mentioned before.
However I Will reinvent my own wheel , by a different presentation of the/my ideas ...
The basic principle is very old : infinite Sum and telescoping.
We want to solve t(exp(x)) = e t(x).
Define
0 < a < 1.
f(x,a) = exp(x) + a x.
f^[n+1](x,a) = f( f^[n](x,a) ,a).
—-
Now consider
for Every real y there is a solution x to
f(x,a) = y.
Nomatter What a is , but still 0 < a < 1.
——
Now define the Sum from m = - Oo to Oo :
t(z,a) = ... + exp(z + 2)/f^[-2](1/2,a) + exp(z + 1)/f^[-1](1/2,a) + ...
So t(z,a) = sum_m exp(z - m)/f^[m](1/2,a).
Now let Lim a —> 0 :
z = x then :
Lim t(z,0) = t(z) = t(x)
t(exp(x)) = e t(x).
Qed
___
For practical reasons I suggest
Compute for x < 1 :
t(x) by computing t(exp(x))/e.
——
We can continue
since 1/z + 1/exp(z) + 1/exp(exp(z)) + ...
Is nowhere analytic since exp iterations are chaotic we have :
t(x) is nowhere analytic.
Also
t(exp(x)) = t(x) e.
ln t( exp(x) ) = ln t(x) + 1
Thus
ln t(x) = slog(x).
A nowhere analytic slog(x) !
——-
Due to the telescoping infinite Sum , this mzthod is probably the simplest nowhere analytic solution in the spirit of calculus.
I assume it was worthy of Being repeated.
Regards
Tommy1729 | 459 | 1,450 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2020-05 | longest | en | 0.826763 |
https://myschool.ng/classroom/chemistry/830?page=10 | 1,653,349,287,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00388.warc.gz | 469,823,449 | 14,078 | ### Which is the temperature of a given mass of a gas initially at 0°C and...
JAMB 1999
Which is the temperature of a given mass of a gas initially at 0°C and 9 atm, if the pressure is reduced to 3 atm at constant volume?
• A. 91K
• B. 182K
• C. 273K
• D. 819K
##### Explanation
From P1/T1 = P2/T2 = 9/273 = 3/T2
T2 x 9 = 273 x 3
T2 = (273 x 3)/9
T2 = 91K | 141 | 357 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2022-21 | latest | en | 0.851107 |
https://bestevew.com/6-2-in-cm/ | 1,656,882,263,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104249664.70/warc/CC-MAIN-20220703195118-20220703225118-00660.warc.gz | 180,748,937 | 16,493 | ## Convert 6’2 in cm – Bestevew
There is 187.96 cm in 6’2. Here is the right solution to fix 6’2 in cm or what is 6’2 in cm. The easiest solution is to get 6’2 to cm. Stay tuned.
## Convert 6’2 in cm
1 feet = 30.48 cm
6 feet 2 inch or 6’2 = 187.96 cm
## Definition of Feet:
A unit of length, it’s commonly used in modern United States measurements.
The foot is equal to 0.3048 meters or about 12 inches but can be converted into any other measurement with ease by dividing one number by another.
This small unit of measurement is easy to use and understand. The foot has been around since about 700 B.C. when the Egyptians measured length in terms relative to how many times their feet would fit into a walkway or room that was common for them back then.
That’s how they got measurements like cubits, elbows, paces, etc. They were used in the Roman Empire and became more widely adopted when they conquered Europe. This conversion chart is how many inches is 6 feet.
Folding feet are quite common in furniture. They provide stability and can be used for storage, but may not have the same aesthetic appeal as other types of joints like dovetail or butt hinges because they’re less sleek-looking than those styles.
The name “foot” may refer either singularly (as when we speak directly about one foot) or collectively – referring to all six appendages attached beneath your toes which help propel you forward!
## Definition of Inch:
Equivalent to 1 x 36 yards or 1 x 12 feet. Derived from the Roman uncia (“12th”), the word “inch” is often used to translate similar units in other systems of measurement, understood to be derived from the width of a person’s thumb.
## Definition of cm:
The first step in understanding the Imperial system of measurement (and any similar system) is to find out what the “units” are. These are the measures used by people to measure stuff.
There is no universal standard, and therefore no universal definition. It’s not a problem. It’s an opportunity. If you’re curious, or you think you might find it useful someday.
## What is 6’2 in cm?
6’2 is equal to 187.96 cm.
### Following This Method You can also solve this:
• What is 6’2 in cm?
• How many cm are in 6’2 ?
• How to convert 6’2 to cm?
• How to measure 6’2 to cm? | 560 | 2,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2022-27 | longest | en | 0.951568 |
https://philosophy.stackexchange.com/questions/5922/what-is-the-difference-between-dogma-and-an-axiom/5930 | 1,653,160,582,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00057.warc.gz | 520,587,348 | 69,963 | # What is the difference between dogma and an axiom?
My understanding of axioms is that they are self evident truths that require no proof, which in my mind is similar to a dogmatic belief in the sense that dogma is a set of beliefs or doctrines that are established as undoubtedly in truth.
• Taking words from your own question, what is the difference betweeen "truth" and "belief"? That should give you an answer to your question. Jan 31, 2013 at 13:39
Axiom is a statement taken to hold within a particular theory. One can combine the axioms to prove things within that theory. One may add or remove axioms to the theory to get another theory:
Euclid: ... 5. If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
Lobachevsky: ... what Euclid says except (5)
Dogmas are axioms of cultural, religious, political theories. Again, one may add or remove dogmas to get a new theory, e.g.:
Arius: ...the Son is not unbegotten ... before he was begotten, or created, or purposed, or established, he was not.
Nicean Council: ... those who say: 'There was a time when he was not;' ... are condemned by the holy catholic and apostolic Church.
The difference is that it is perfectly ok to handle different sets of axioms in, say, mathematics and prove a theorem in Euclidean geometry one day and a theorem in Lobachevskian the next - just remembering when the fifth postulate does or doesn't hold, but it's not considered acceptable to hold several sets of dogmas at once. Life of Pi provides an illustration of the controversy of such a stance (the main character is Hindu, Muslim and Catholic simultaneously and his brothers-in-dogmas don't like to share him with the competition).
I'm sure early geometers were much more religious about their axiom-dogmas than the modern mathematicians, but I have no proof.
• Axiom: Let us assume X is true. Dogma: X is true.
– SF.
Jan 31, 2013 at 20:34
• Although I don't have a concrete example at hand, I would imagine that you can find instances of pairs of dogmas that contradict each other (within the same "theory") whereas pairs of axioms should not.
– Drux
Feb 1, 2013 at 14:35
• Isn't "axioms within a theory shouldn't contradict each other" an axiom?
– artm
Feb 1, 2013 at 14:59
• @SF. Great :) Possibly the most concise explanation ever! You should post that as an answer.
– DBK
Feb 7, 2013 at 17:26
• @Drux: Axioms define the reality the theory is to describe, so if they define it in a self-conflicting way, the theory must work around that conflict. Take Lobachevsky's geometry which contains a seemingly self-conflicting axiom, and then the theory must bend the underlying space to satisfy it.
– SF.
Feb 8, 2013 at 5:36
An axiom is something that is self-evidently true; it is so obvious that there is no controversy about it. In mathematics, you just have to accept some very basic notions in order to avoid circular reasoning. These can't be proven, but they can always (and often very easily) be observed.
Example from Euclid's Elements:
Common notions:
1. Things that are equal to the same thing are also equal to one another (Transitive property of equality).
2. If equals are added to equals, then the wholes are equal.
3. If equals are subtracted from equals, then the remainders are equal.
4. Things that coincide with one another equal one another (Reflexive Property).
5. The whole is greater than the part.
Or an example from propositional logic:
If:
• A is true
• B is true
Then
• A&B is true
A dogma refers to (usually a religious) teaching that is considered undoubtedly and absolutely true. It is something you accept without any direct observation; dogmas are accepted by faith only.
I should add that some people would say that there is no difference between axioms and dogmas, because 'self-evident truths' are in some sense based on faith; that is that you accept on faith that anything that seems obvious and self-evident is true. An interesting read on this subject is Wittgenstein's On Certainty. I also want to stress that I don't mean to say that an axiom is "better" than a religious dogma (or vice versa for that matter).
• Good answer, but I'm not so sure about axioms generally being obviously self-evident. Take the axiom of choice, for example: there is huge division in mathematics as to whether it's true or not, and many proofs are written based on a by-faith acceptance (or rejection) of said axiom. In this regard I suppose I'm one of those who thinks there isn't a real difference between axioms and dogma - except maybe context. Feb 15, 2013 at 16:12
• @commando I accept your point. Not all axioms are self-evident, but I still think there is a difference between an axiom and a dogma. An axiom is only something you accept within a theory (self-evident or not); a dogma is far more pervasive. Whether I accept the axiom of choice doesn't change my life; whether I accept to dogma of Christian faith, for instance, will. Also,I don't think there have been any wars over which axiom is true, but there has been plenty of conflict over dogmas.
– Ben
Feb 15, 2013 at 17:12
• That seems to be true but, at an advanced level, it's IMO wrong. For example, Newtonian mechanics is "self-evident" and may be called "axiomatic". But it's not true in 'relativistic' environments (near the speed of light, near black holes), for which you need Einsteinian mechanics. You can do calculations taking Newtonian physics as "axiomatic" even though they're false ("let's assume that Newton is correct, e.g. because relativistic effects are negligible..."). Similarly Euclidian geometry depends on "axioms" which seem to be self-evident and can be defined as true but aren't inherently true. Oct 28, 2014 at 2:42
• The 'That' which started my previous comment was referring to your opening sentence: that "An axiom is something that is self-evidently true; it is so obvious that there is no controversy about it." Instead, an "axiom" is something that's unprovable: which you need to assume is true in order to use it as a basis for going on to prove other things which assume it: "A is axiomatic, therefore B". You'll usually choose 'useful' or 'seemingly self-evident' axioms. but that's a whole other story. Oct 28, 2014 at 3:13
• I'd say "Then A&B is true" is something that defines & operator. May 5, 2018 at 11:45
In religions, some (should be majority) people accept the existence of God by belief. while some minority of them accept God by observation (thru meditation, etc). By believing, the faith can be shaken. By direct experience (the perfect balance of mindfulness and concentration), it is just there, there is no doubt about it.
• "By direct experience... it is just there, there is no doubt about it." I disagree, hallucinations are not just there. May 5, 2018 at 11:38
• An hallucination is not a direct experience. Here 'direct' should mean 'unmediated'. It should be noted that Buddhism explains the monotheistic God as misinterpreted meditative experience and thus like an hallucination, but the issue is subtle and this should not be seen as simple atheism. An experience is never false but our interpretation of it may be all over the place.
– user20253
May 5, 2018 at 11:49
• Direct exp. is difficult topic. Here I give one easy example to show what direct exp. is and that it is not hallucination : After direct experience, the world seen is just exactly the same as before direct experience, there is no flying cars, talking animals, etc. It is just that some things that have not perceived or not clearly perceived before direct exp, (simply because they are being ignored) are now perceived more clearly and understood better. When things become clearer and understood better, that is Awakening (direct experience). Therefore it is different from hallucination.
– God
May 6, 2018 at 14:58
When starting with a set of axioms, you might end up with a contradiction. This indicates that one of the axioms are wrong. You then figure out which one it is, and reject it. (This is a simplification, but sufficient in answering this question.)
When you are not allowed to reject the axiom even when you have successfully shown it to be false, the axiom you are not allowed to reject is now dogma.
• Not if the axioms are definitions. Jul 13, 2017 at 18:43
• @ChristopherE, definitions are a subset of axioms in universal form (forall X forall P(Y): X is subset P(Y)). May 5, 2018 at 11:39 | 2,086 | 8,604 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-21 | longest | en | 0.966963 |
https://projecteuler.net/problem=427 | 1,566,682,577,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00075.warc.gz | 596,952,216 | 2,769 | ## n-sequences
Published on Saturday, 11th May 2013, 10:00 pm; Solved by 230;
Difficulty rating: 95%
### Problem 427
A sequence of integers S = {si} is called an n-sequence if it has n elements and each element si satisfies 1 ≤ sin. Thus there are nn distinct n-sequences in total. For example, the sequence S = {1, 5, 5, 10, 7, 7, 7, 2, 3, 7} is a 10-sequence.
For any sequence S, let L(S) be the length of the longest contiguous subsequence of S with the same value. For example, for the given sequence S above, L(S) = 3, because of the three consecutive 7's.
Let f(n) = ∑ L(S) for all n-sequences S.
For example, f(3) = 45, f(7) = 1403689 and f(11) = 481496895121.
Find f(7 500 000) mod 1 000 000 009. | 245 | 711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2019-35 | longest | en | 0.864499 |
http://chemthermo.ddbst.com/Problems_Solutions/ch14.html | 1,720,967,595,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00875.warc.gz | 8,741,393 | 5,983 | Problems and Solutions Chapter 2
## Problems and SolutionsChapter 14 Practical Applications
### Textbook Examples:
XPS-file display is available in Internet Explorer. In case of Firefox, select Internet Explorer when asked for the software to open the file.
14.01 Pressure Build-Up in Vessel Filled with Water and Nitrogen (p.570)
14.02 Outlet Temperature of a Throttle Valve (p. 572)
14.03 Adiabatic Compression of R22 (p.575)
14.04 Choice of Optimal Refrigerant (p. 576)
14.05 Required Size of a Rupture Disk (p. 580)
14.06 Minimum Crossflow Area of a Valve (p. 582)
### Additional Problems:
P14.01 Joule-Thomson Coefficient of Nitrogen Using the Virial Equation and the SRK EOS
Calculate the Joule-Thomson coefficient of nitrogen at a temperature of 150 K and a pressure of 10 atm using the
a) virial equation truncated after the second term using a second virial coefficient
estimated via the Tsonopoulos method.
b) Soave-Redlich-Kwong equation of state.
All required parameters are given in Appendix A.
P14.02 Work and Temperature Change Upon Isentropic Compression of Oxygen
Oxygen at 25°C and a pressure of 1 bar is compressed to 10 bar. Calculate the required work and the temperature of the compressed gas assuming isentropic compression
a)
using ideal gas law
b)
using the Soave-Redlich-Kwong equation of state
All required physical property parameters are given in Appendix A.
P14.03 Reversible and Isothermal Compression of Liquid Water
A water stream (2000kg/h at 25°C and 1 bar) is compressed to 100 bar in a cooled pump. The process can assumed to be reversible and isothermal. Calculate the required work and heat duty of the cooling system. The thermal expansion coefficient α = 0.207·10-3 K-1 and compressibility coefficient χ= 0.46·10-4 bar-1 can assumed to be constant over the relevant temperature and pressure range. The molar volume of liquid water at feed conditions is 18.07 cm3/mol.
P14.04 Heat Effect Upon Mixing of Methane and Dodecane at Elevated Temperature and Pressure Using SRK
A feed stream of 1600 kg/h of methane is adiabatically mixed with 170 kg/h of n-dodecane. Both streams are at 160°C and a pressure of 20 bar. Calculate the temperature of the stream leaving the mixer using the Soave-Redlich-Kwong equation of state with a binary parameter of k12 = 0. Explain the result.
All physical property parameters for methane can be found in Appendix A. The required values for n-dodecane are: Tc = 658.8 K, Pc = 1809.7 kPa, ω = 0.562, cPid = 379.8 J/mol K.
P14.05 Required Power for R134a Compression Using a High Precision Equation of State
The refrigerant R134a is compressed from J1 = 5°C, saturated vapor, to P2 = 10 bar. The isentropic efficiency of the compressor is hth = 0.7. The mechanical efficiency is hmech = 0.7. Calculate the power of the compressor. The mass flow is 3,000 kg/h. Use a high-precision equation of state.
P14.06 Required Volume for a Gas Storage Tank for Ammonia
In a 50 m3 vessel, liquid ammonia at J1 = 50°C is stored at P1 = 100 bar. Due to a vessel failure, the ammonia is collected in a backup vessel. Which is the necessary volume of the backup vessel, if P2 = 10 bar must not be exceeded?
P14.07 Liquid Nitrogen Production Via Volume Expansion of the Compressed Gas
In a Linde plant, nitrogen at (J = -104°C, 240 bar) is let down to P = 1 bar through a valve. How much liquid nitrogen is produced?
P14.08 Required Compressor Power for Isothermal and Adiabatic Compression of a Gas Mixture (CO2, O2) Using the Ideal Gas Law
A mixed stream consisting of 1 kmol/h CO2 and 1 kmol/h O2 is compressed from T1 = 290 K, P1 = 1 bar to P2 = 5 bar. Calculate the compressor power for
a)
adiabatic compression
b)
isothermal compression
The mixture should be regarded as an ideal gas. The compression should be assumed to be reversible in both cases.
P14.09 Temperature Change Upon Ethylene Expansion in Throttle Valves Using a High Precision EOS
In an LDPE (Low Density Polyethylene) plant, ethylene is expanded from P0 = 3000 bar, T0 = 600 K to P1 = 300 bar by a throttle valve. By a second throttle valve, it is expanded to environmental pressure P2 = 1 bar. Calculate the temperatures T1 and T2 by using a high-precision equation of state. The velocity terms in the First Law should be neglected.
P14.10 Leakage Rate Change in Vacuum Distillation When Lowering the Column Pressure
In vacuum distillation columns, the leakage of ambient air into the column is always a problem and might lead to an explosive atmosphere in the condenser. How does the leakage rate rise if the column operating pressure is lowered from P1 = 400 mbar to P2 = 100 mbar? The ambient pressure shall be 1.013 bar.
P14.11 Pressure Rise In a Storage Tank Upon Heating
A vessel (1 m3) containing 500 kg propylene at J = 30°C is exposed to sun radiation. What is the initial pressure? The safety valves of the vessel actuate at P = 60 bar. Use a high-precision equation of state to calculate the respective temperature.
P14.12 Work and Temperature Change Upon Adiabatic Compression of Oxygen
Oxygen (J = 25°C, ) is compressed adiabatically from P1 = 1 bar to P2 = 10 bar. Calculate the power of the compressor and the outlet temperature of the gas using
a) the ideal gas law
b) the Soave-Redlich-Kwong equation of state
The isentropic efficiency of the compressor is hth = 0.75. The mechanical efficiency is hmech = 0.95.
P14.13 Thermodynamic Cycle Calculation Using a High-Precision EOS
A thermodynamic cycle is operated with water at the following conditions:
1. Isobaric heating to P1 = 100 bar, J1 = 350°C
2. Reversible and adiabatic expansion of the vapor in a turbine to P2 = 1 bar.
3. Isobaric condensation.
4. Isentropic compression of the liquid in a pump to P4 =100 bar.
Calculate the thermal efficiency of the process defined by
Use the high-precision equation of state.
P14.14 Refrigeration Cycle Calculation Using the Peng-Robinson EOS
A refrigerator is operated with R12 (dichlorodifluoromethane). The particular steps of the compression cycle are:
- Isobaric condensation without subcooling at J1 = 30°C.
- Adiabatic pressure relief by a throttle valve to P2 = Ps(-20°C).
- Complete isobaric evaporation of the refrigerant at J2 = J3 = -20°C
without superheating
- Isentropic compression of the saturated vapor to P4 = Ps(30°C).
Calculate the process data for the steps 1-4 using the Peng-Robinson equation of state.
P14.15 Joule-Thomson Coefficient for Methane Using the Peng-Robinson EOS
Calculate the Joule-Thomson coefficient for methane at T = 300 K and P = 30 bar using the Peng-Robinson equation of state. The critical data and the acentric factor can be taken from App. A.
P14.17 Compressor Duty and State Properties after Ammonia Compression
Gaseous ammonia (100°C, 5 bar) is compressed to P2 = 10 bar. The thermal efficiency is hth = 0.8, the mechanical efficiency is hmech = 0.9. Calculate the compressor duty and the state properties at the compressor outlet. | 1,846 | 7,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-30 | latest | en | 0.711251 |
https://html.alldatasheet.com/html-pdf/48654/AD/ADG419BN/130/5/ADG419BN.html | 1,627,279,736,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152000.25/warc/CC-MAIN-20210726031942-20210726061942-00691.warc.gz | 310,062,988 | 14,714 | Electronic Components Datasheet Search
Delete All
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## ADG419BN Datasheet(HTML) 5 Page - Analog Devices
5 / 8 pageADG419REV. A–5–Typical Performance Characteristics–TA = +25 CVDD = +12VVSS = –12VVDD = +5VVSS = –5VVDD = +15VVSS = –15VVDD = +10VVSS = –10VVS, VD – Volts50400–1515–10–50510302010Figure 1. RON as a Function of VD (VS): Dual Supply VoltageVDD = +15VVSS = –15VVL = +5V+25 C+85 C+125 CVS, VD – Volts50400–1515–10–50510302010Figure 2. RON as a Function of VD (VS) for DifferentTemperaturesVDD = +15VVSS = –15VTA = +25 CID (ON)IS (OFF)ID (OFF)VS, VD – Volts0.020.01–0.03–1515–10–505100.00–0.01– 0.02Figure 3. Leakage Currents as a Function of VS (VD)TA = +25 CVDD = +12VVSS = 0VVDD = +5VVSS = 0VVDD = +15VVSS = 0VVDD = +10VVSS = 0VVS, VD – Volts100800015510604020Figure 4. RON as a Function of VD (VS): Single SupplyVoltageVS, VD – Volts100800VDD = +12VVSS = 0VVL = +5V+125 C+85 C+25 C012369604020Figure 5. RON as a Function of VD (VS) for DifferentTemperaturesVDD = +12VVSS = 0VTA = +25 CIS (OFF)ID (OFF)ID (ON)VS, VD – Volts0.0060.004–0.004122468100.0020.000–0.0020Figure 6. Leakage Currents as a Function of VS (VD)
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https://community.powerbi.com/t5/Desktop/Calculating-Measure/td-p/133038 | 1,597,268,915,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738944.95/warc/CC-MAIN-20200812200445-20200812230445-00129.warc.gz | 262,065,195 | 92,298 | cancel
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Highlighted
Helper IV
## Calculating Measure
Hi,
I have this following situation where I want to calculate = Sum of (product sale by group)/Sum of Total product in Group
From this data, desired output is: = (599/2332) = 0.25
Product Group Product sale by Group Total product in group P1 G1 78 122 P1 G2 24 48 P1 G3 18 76 P1 G4 49 160 P2 G1 39 123 P2 G2 6 41 P2 G3 19 57 P2 G4 55 152 P3 G1 39 139 P3 G2 7 31 P3 G3 5 71 P3 G4 17 132 P4 G1 18 122 P4 G2 30 47 P4 G3 13 61 P4 G4 9 115 P5 G1 8 122 P5 G2 16 49 P5 G3 2 115 P5 G4 40 123 P6 G1 14 95 P6 G2 26 66 P6 G3 9 121 P6 G4 58 144
1 ACCEPTED SOLUTION
Accepted Solutions
Highlighted
Microsoft
## Re: Calculating Measure
This will give you the result you say you want, regardless of what/where you filter. Not sure if that is what you are after though
```New Measure = DIVIDE(
CALCULATE(
SUM(Table1[Product sale by Group]),
ALL('Table1')
)
, CALCULATE(
SUM(Table1[Total product in group]),
ALL('Table1')
)
)```
Proud to be a Datanaut!
3 REPLIES 3
Highlighted
Responsive Resident
## Re: Calculating Measure
Hi
Not sure if I got you right. The formula of your measure would be:
```=DIVIDE(
SUM(product sale by group),
SUM(Total product in Group),
0)
```
Hope this helps
JJ
Highlighted
Microsoft
## Re: Calculating Measure
This will give you the result you say you want, regardless of what/where you filter. Not sure if that is what you are after though
```New Measure = DIVIDE(
CALCULATE(
SUM(Table1[Product sale by Group]),
ALL('Table1')
)
, CALCULATE(
SUM(Table1[Total product in group]),
ALL('Table1')
)
)```
Proud to be a Datanaut!
Highlighted
Microsoft
## Re: Calculating Measure
Hi @aktripathi2506,
As the @Phil_Seamark posted, create a measure using the formula and create a card to display the expected result as follows.
If you have resolved your issue, please mark the right reply as answer, if it doesn't, please feel free to ask.
Best Regards,
Angelia
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Top Solution Authors
Top Kudoed Authors | 743 | 2,580 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-34 | latest | en | 0.793132 |
https://www.techieindoor.com/data-structure-and-algorithm/ | 1,675,772,207,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500456.61/warc/CC-MAIN-20230207102930-20230207132930-00355.warc.gz | 1,026,928,141 | 32,473 | Data Structure and algorithm
In this tutorial, We are going to learn about Data Structure and algorithm. So by using data structure and algorithm, We can collect and organise data in such a way that we can perform operations on these data in an effective way.
An algorithm is a set of instructions or logic, written in order, to accomplish a certain specific task.
Algorithms should satisfy the following given properties:
• Input
• Output
• Definiteness
• Finiteness
• Correctness
The performance of algorithm is measured by following given property:
• Time Complexity
• Space Complexity
Basic type of data structures:
• Queue
• Stack
• Tree
• graph etc.
Graph
Array
• Find the median of two sorted array
• Program to merge two sorted array in-place
Stack
`https://www.techieindoor.com/go-lang-tutorial/`
`https://en.wikipedia.org/wiki/Algorithms_%2B_Data_Structures_%3D_Programs` | 202 | 894 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2023-06 | longest | en | 0.831797 |
https://people.revoledu.com/kardi/tutorial/Regression/DataAnalysisAddIns.html | 1,582,755,252,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146562.94/warc/CC-MAIN-20200226211749-20200227001749-00308.warc.gz | 504,160,545 | 4,451 | ## Excel Data Analysis Add Ins
Microsoft Excel has excellent Data Analysis menu and one of them contain regression.
To use the data analysis menu (suppose your MS excel do not have it), you need to install the Add-ins: Go to menu Tools > Add Ins > check the Analysis Tool pack and then press OK button. Next time when you open the Tools menu again, you will see Data Analysis in the bottom of Tools menu.
To use Regression from Data Analysis:
1. Click menu Tools > Data Analysis and you will see Data Analysis dialog. Scroll down and you will see Regression. Select the Regression and click OK button.
2. Regression dialog will pop up and you can fill the Input Y range and input X range. If you have defined the name of your data (e.g. Population and Year) using menu Insert > Name >Define, you can just type the name of your data instead of searching it. When you click the OK button in the regression dialog, MS excel will automate the procedure to compute regression for you.
Suppose we have the following 5 data points and we want to predict the population data for the year 2005 using linear regression model.
We put the independent variable Year as and dependent variable Population as .
Here is the result of the Data Analysis of Regression. There are many numbers on it and I will tell you only what is important:
First, you need to examine the Coefficients. These are the intercept and the slope of the regression analysis. The slope is 0.136, intercept is -267.2 for the 5 observations data above. Thus, your regression model is
Next, your regression model needs to pass all the three criteria below
1. The R square must be bigger than 0.80
2. The significant F (from ANOVA) must be smaller than 0.05
3. The absolute value of t-statistics must be larger than 1.96 for =0.05 and must larger than 1.645 for =0.10
If you cannot pass any of the three criteria above, you should plot your data and check whether your data is really a linear model (could be model by a straight line). If your data is not linear, you may transform it to linear model or you should you non-linear regression instead. The explanation on why these criteria are important is quite long and I would rather ask you to read statistical book for the explanation.
For our example, it seems all the criteria above could be passed by our model. Using this regression model, we can predict the number of population in the city for year 2005. By inputting 2005 to the year we get population = 5.48 (x hundred thousand persons).
Preferable reference for this tutorial is
Teknomo, Kardi (2015) Regression Model using Microsoft Excel. http://people.revoledu.com/kardi/tutorial/Regression/ | 597 | 2,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-10 | longest | en | 0.857728 |
https://de.mathworks.com/matlabcentral/cody/problems/2061-narcissistic-number/solutions/2900492 | 1,604,196,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107922746.99/warc/CC-MAIN-20201101001251-20201101031251-00664.warc.gz | 285,452,874 | 17,386 | Cody
# Problem 2061. Narcissistic number ?
Solution 2900492
Submitted on 1 Sep 2020 by MtF
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
2 Pass
x = 9; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
3 Pass
x = 10; y_correct = false; assert(isequal(isnarcissistic(x),y_correct))
4 Pass
x = 99; y_correct = false; assert(isequal(isnarcissistic(x),y_correct))
5 Pass
x = 152; y_correct = false; assert(isequal(isnarcissistic(x),y_correct))
6 Pass
x = 153; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
7 Pass
x = 154; y_correct = false; assert(isequal(isnarcissistic(x),y_correct))
8 Pass
x = 371; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
9 Pass
x = 370; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
10 Pass
x = 1634; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
11 Pass
x = 8207; y_correct = false; assert(isequal(isnarcissistic(x),y_correct))
12 Pass
x = 9474; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
13 Pass
x = 9926315; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
14 Pass
x = 88593477; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
15 Pass
x = 9800817; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
16 Pass
x = 54748; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
17 Pass
x = 4679307774; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
18 Pass
x = 472335975; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
19 Pass
x = 32164049650; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
20 Pass
x = 32164049651; y_correct = true; assert(isequal(isnarcissistic(x),y_correct))
21 Pass
x = 32164049652; y_correct = false; assert(isequal(isnarcissistic(x),y_correct))
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https://www.zvab.com/9783642096082/3642096085/plp | 1,501,024,014,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425407.14/warc/CC-MAIN-20170725222357-20170726002357-00053.warc.gz | 865,524,862 | 11,356 | # Computational Acoustics of Noise Propagation in Fluids - Finite and Boundary Element Methods
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The book provides a survey of numerical methods for acoustics, namely the finite element method (FEM) and the boundary element method (BEM). It is the first book summarizing FEM and BEM (and optimization) for acoustics. The book shows that both methods can be effectively used for many other cases, FEM even for open domains and BEM for closed ones. Emphasis of the book is put on numerical aspects and on treatment of the exterior problem in acoustics, i.e. noise radiation.
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From the Back Cover:
Among numerical methods applied in acoustics, the Finite Element Method (FEM) is normally favored for interior problems whereas the Boundary Element Method (BEM) is quite popular for exterior ones.
That is why this valuable reference provides a complete survey of methods for computational acoustics, namely FEM and BEM. It demonstrates that both methods can be effectively used in the complementary cases.
The chapters by well-known authors are evenly balanced: 10 chapters on FEM and 10 on BEM. An initial conceptual chapter describes the derivation of the wave equation and supplies a unified approach to FEM and BEM for the harmonic case. A categorization of the remaining chapters and a personal outlook complete this introduction. In what follows, both FEM and BEM are discussed in the context of very different problems.
Firstly, this comprises numerical issues, e.g. convergence, multi-frequency solutions and highly efficient methods; and secondly, solutions techniques for the particular difficulties that arise with external problems, e.g. discussion of absorbing boundaries for FEM and treatment of the non-uniqueness problem for BEM. Finally, both parts on FEM and on BEM are completed by chapters on related problems, e.g. formulations for fluid-structure interaction. In addition to time-harmonic problems, transient problems are considered in some chapters. Many theoretical and industrial applications are presented.
Overall, this book is a unified review of the state-of-the-art on FEM and BEM for computational acoustics.
Review:
From the reviews: “This book contains a review of recent developments and new results for the application of finite and boundary element methods in acoustic simulations. ... Without a doubt it will become a popular reference in the acoustic community. The volume can be highly recommended to graduate students and young researchers, as well as practitioners in the industry, as a representative compilation of recent methodical developments and interesting applications in the broad field of vibroacoustic simulation.” (Frank Ihlenburg, Acta Acustica United With Acustica, Vol. 95 (4), 2009)
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## Re: Proposal to change the ranges in array trimming to be "half-open" in
From: Egeyar Bagcioglu Subject: Re: Proposal to change the ranges in array trimming to be "half-open" interval Date: Sun, 1 Nov 2020 13:40:04 +0100 User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:78.0) Gecko/20100101 Thunderbird/78.3.1
```
On 10/29/20 5:40 PM, Jose E. Marchesi wrote:
```
```Hi Mohammad.
```
```Let's talk about ranges in array trimming!
```poke
defvar a = [1, 2, 3, 4, 5];
defvar b = a[1:3];
assert (b == [2, 3, 4]);
```
The `1:3` range is an inclusive interval from `1` to `3`.
Mathematically, the interval is `[1, 3]`. A "closed interval".
IMHO, closed intervals are hard to deal with.
Let me explain a little more.
This is the scheme used by Poke for indexing elements:
1, 2, 3, 4, 5
^ ^ ^ ^ ^
| | | | |
0 1 2 3 4
But I argue that the following scheme for indexing elements is much more
powerful and easier to deal with:
1, 2, 3, 4, 5
^ ^ ^ ^ ^ ^
| | | | | |
0 1 2 3 4 5
For a sequence of `N` elements there are `N+1` positions.
Positions points **between** elements not **at** elements.
```
```Well, in an array subscript a position actually refers to the element
following the referred position, right, not the position itself :)
Reminds me of the Emacs pointer.
But yeah, I like the proposed change; using semi-open intervals makes
more sense for zero-based arrays.
And now is the time to do changes like this :)
Other opinions?
```
```
Since you asked...
I find it more natural to think about array slices in the following way:
```
The first number is the begin index, while the second number is the number of elements to be found in that array slice.
```
```
This logic gives the same results with Mohammad's suggestion for the examples he included below:
```
```poke'
assert (a[0:1] == [1]);
assert (a[0:5] == [1, 2, 3, 4, 5]);
/* a[0:0] is an empty array */
assert (a[:2] + a[2:] == a); /* instead of `a[:2] + a[2+1:]` */
```
The difference is when the begin index is non-0.
```
The example array "a" that's been used earlier in this thread is 1-based. Let me use a 0 based array "b" where b = [0, 1, 2, 3, 4].
```
What I'd prefer is the following:
b[2:3] == [2, 3, 4];
What Mohammad suggests:
b[2:3] == [2]
```
What I am suggesting here is also how we pass arrays as arguments in C. I find it much more intuitive and I feel very strongly about this approach. Having said that, I'll respect our maintainer's final decision, whatever it will be.
```
Regards
Ege
```
```This is the approached used in `STL` of `C++` (which is designed by Alex
Stepanov).
The ranges in STL are half-open intervals (mathematically denoted as `[f, l)`).
- The number of elements in a range is `l - f`.
- Dividing a range into two sub-ranges are simple:
`[f, m)` and `[m, l)`
New Poke:
```poke'
assert (a[0:1] == [1]);
assert (a[0:5] == [1, 2, 3, 4, 5]);
/* a[0:0] is an empty array */
assert (a[:2] + a[2:] == a); /* instead of `a[:2] + a[2+1:]` */
```
Regards,
Mohammad-Reza
Reference:
- [Generic Programming](https://youtu.be/iwJpxWHuZQY?t=19m02s) by Sean Parent.
```
```
```
reply via email to
[Prev in Thread] Current Thread [Next in Thread] | 1,067 | 3,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2024-26 | latest | en | 0.846201 |
http://e2e.ti.com/support/audio/f/6/t/829853?LM4950-Some-questions | 1,582,974,298,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00174.warc.gz | 42,993,360 | 18,634 | If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.
• TI Thinks Resolved
# LM4950: Some questions
Part Number: LM4950
Hi Sirs,
Sorry to bother you.
If we use LM4950, speaker is 4ohm/2w (Use SE)
Is it based on the following figure to determine the output voltage??
The AMP front-end input (AUDIO CODEC output) should be 1.2~1.1 Vrms
Will the setting of gain as shown below be correct ??(speaker is 4ohm / 2W)
• Hi, Shu-Cheng,
Please take a look at my responses below and let me know if you have additional questions or comments on it.
If we use LM4950, speaker is 4ohm/2w (Use SE) Is it based on the following figure to determine the output voltage??
Yes, figure 23 of datasheet shows the output power that you may reach for a RL = 4ohms (SE operation) with a THD+N = 10% and THD+N = 1%. You would need to select the power supply according to your power requirements.
Will the setting of gain as shown below be correct ??(speaker is 4ohm / 2W)
This resistors configuration results in a gain of ~16.3. Could you provide details about the input that you will be using and the expected output?
Best regards,
Luis Fernando Rodríguez S.
• In reply to Luis Fernando Rodriguez S.:
Hi Sirs,
LM4950_0815-Edit.pdf
The line below CN9 is anti-pops
Input spec as follows
• In reply to Shu-Cheng LIN:
Hi Sirs,
Sorry to bother you.
Have any update on here?
Thanks!!
• In reply to Shu-Cheng LIN:
Hi, Shu-Cheng,
Sorry for the delay on this.
I already reviewed the schematic and everything seems to be in order. I didn't find anything to modify.
The configured gain in this case is given by Rf / RI = (48.7K / 20K) = 2.435. If a 1.2 Vrms signal is used, you should be able to reach almost 2.9Vrms at the output. This would be enough to reach 2W output power with a 4ohms load.
Best regards,
Luis Fernando Rodríguez S. | 534 | 1,957 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2020-10 | latest | en | 0.910312 |
https://engineerscommunity.com/t/4-20ma-current-scaling-formula/6867 | 1,660,187,628,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571232.43/warc/CC-MAIN-20220811012302-20220811042302-00166.warc.gz | 249,802,248 | 5,578 | # 4-20mA Current Scaling Formula
This measurement to current converter tool will convert any linear measurement reading into the ideal current loop signal over a linear range of 4 to 20 milliamps, and display a 4-20mA conversion scale for the chosen measurement range.
### Formula
This tool uses the following formula to calculate the current output signal over a 4-20mA scale from a specified measurement input reading and range:
Linear mA out = 4 + (16 x ((Rdg – Low Limit) / (High Limit – Low Limit)))
### Parameters
#### Measurement Unit
Add the measurement parameter value in the specified measurement unit you wish to convert.
Add the measurement parameter value you wish to convert in any units.
#### Lowest Measurement
Enter the lower limit for the measurement range to be converted. For example zero would be a typical value for most instrumentation but you can also add negative (-) or positive (+) values as well. Please enter the value in the same units as specified for all other input values.
#### Highest Measurement
Enter the upper limit for the measurement range to be converted, e.g. full scale or full range. This value can be negative (-) or positive (+). Please enter the value in the same units as specified for all other input values.
#### Current Loop Signal (4-20mA) – Answer
This is the ideal converted signal for the input measurement range specified, and represents the ideal output reading excluding all measurement and instrumentation errors. The calculated result is displayed in milliamps (mA) based on a linear 4 to 20 mA scaling.
## Help
### Calculator precision
How precise is the calculation from measurement to linear 4-20mA output?
The linear 4 to 20 milliamp signal Answer is displayed to a precision of 9 significant figures.
### Measurement types
What type of measurement can I convert?
Any measurement which is linearly proportional to the 4-20mA signal can be converted, such as pressure, temperature, force, load, weight, position, movement, speed, rpm, depth, level, etc…
### Input & Output
Does it matter if the measurement is an input or a output?
No it does not matter if the measurement, the calculation is the same whether it is an input or an output.
### Bidirectional measurements
Can I convert a bidirectional measurement?
Yes, you can enter the lowest measurement as a negative value by prefixing it with a minus (-) sign.
### Compound ranges
Can I enter a compound range?
Yes, any negative range can be specified in the Lowest text box can be combine with any positive range specified in the Highest text box.
### Elevated zero point
Is it possible to convert a measurement range where the lowest value is offset in positive direction from zero?
Yes, the lowest measurement point can start from any value above zero.
### Measurement accuracy checking
Can this converter help to determine the accuracy of measurement instrument?
Yes, this converter will calculate the ideal 4-20mA signal so that you can compare it to an observed 4-20mA reading.
wow good information | 629 | 3,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-33 | longest | en | 0.754783 |
http://sciencing.com/solve-rational-expressions-4422956.html | 1,490,576,040,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189316.33/warc/CC-MAIN-20170322212949-00168-ip-10-233-31-227.ec2.internal.warc.gz | 303,915,144 | 12,526 | # How to Solve Rational Expressions
By Samantha Torrence
A rational expression is an expression whose numerator and denominator are polynomials. For example 2x+3/4+5x is a rational expression. There are many ways of solving a rational expression, one must be able to simplify the rational expression, as well as add subtract multiply and divide to obtain an answer.
## Simplifying Rational Expressions
Factor the polynomials in the numerator.
Factor the polynomials in the denominator.
Cancel out the common factors in the rational expression.
## Adding and Subtracting Rational Expressions
Find the like denominators in the rational expressions if they do not already have like denominators. To Find like denominators find the least common factors in the denominators.
Add or subtract the numerators of the rational expressions with the like denomiators.
Simplify the rational expressions into lowest terms.
## Multiply and Divide Rational Expressions
Factor the rational expressions.
Multiply the rational expressions.
Divide rational expressions by multiplying the first rational expression by the reciprocal of the second rational expression. Follow the steps for multiplying fractions. | 224 | 1,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2017-13 | latest | en | 0.900321 |
http://msgroups.net/excel.newusers/issue-writing-if-or-error-formula-types/538239 | 1,575,899,524,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540518882.71/warc/CC-MAIN-20191209121316-20191209145316-00074.warc.gz | 94,859,954 | 12,002 | Issue writing IF OR Error Formula Types
```Having trouble putting a formala together.
I have the following formula which works:
Senario : \$d\$d = either weight or unit or unknown (of an ingredient)
e8= is a quantity number ( either in grams or units, o
an ingredient. eg 100 or 1)
Raw data is in nutrition!\$b\$6:\$g\$70
The following formula works so far:
f8 ( which is the no. o
calories)=if(\$d\$d="weight",e8/100,e8)*vlookup(c8,nutrition!\$b\$6:\$g\$70,3,false)
But I also want to add to it that
If e8(quantity)=0 or blank or invalid then "qty?" should appear in th
f8 cell
And not sure what to do to avoid an error in f8 if \$d\$d = unknown??
Am completely stuck here, need help asap.
lozza
--
lozzam
```
0
10/26/2006 4:11:21 AM
excel.newusers 15348 articles. 2 followers.
1 Replies
642 Views
Similar Articles
[PageSpeed] 13
```Hazarding a wild guess ..
Perhaps you mean something like this:
=IF(E8="","qty?",IF(D8="weight",E8/100,E8*VLOOKUP(C8,nutrition!\$B\$6:\$G\$70,3,0)))
--
Max
Singapore
http://savefile.com/projects/236895
xdemechanik
---
"lozzam" <lozzam.2g9vgp@news.excelbanter.com> wrote in message
news:lozzam.2g9vgp@news.excelbanter.com...
>
> Having trouble putting a formala together.
>
> I have the following formula which works:
>
> Senario : \$d\$d = either weight or unit or unknown (of an ingredient)
> e8= is a quantity number ( either in grams or units, of
> an ingredient. eg 100 or 1)
> Raw data is in nutrition!\$b\$6:\$g\$70
>
> The following formula works so far:
>
> f8 ( which is the no. of
> calories)=if(\$d\$d="weight",e8/100,e8)*vlookup(c8,nutrition!\$b\$6:\$g\$70,3,false)
>
> But I also want to add to it that
>
> If e8(quantity)=0 or blank or invalid then "qty?" should appear in the
> f8 cell
>
> And not sure what to do to avoid an error in f8 if \$d\$d = unknown??
>
> Am completely stuck here, need help asap.
>
> lozzam
>
>
>
>
> --
> lozzam
```
0
demechanik (4694)
10/26/2006 7:14:05 AM
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Whenever I run chkdsk /r, it finds quite a few errors and corrects them. An example is a lot of orphaned files. Then, if I run it again right away, it does the same thing and finds the sa,e orphaned files. What causes this and how can I correct it? Thank you. On 14 May 2010, five256@NOwhere.com wrote in microsoft.public.windows.vista.general: > Whenever I run chkdsk /r, it finds quite a few errors and corrects > them. An example is a lot of orphaned files. Then, if I run it > again right away, it does the same thing and finds the sa,e > orphaned files. What causes... | 3,997 | 15,921 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2019-51 | latest | en | 0.775238 |
http://www.docstoc.com/docs/91380414/4 | 1,430,952,857,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430459915807.89/warc/CC-MAIN-20150501055835-00010-ip-10-235-10-82.ec2.internal.warc.gz | 387,482,778 | 38,950 | # 4 by liwenting
VIEWS: 2 PAGES: 2
• pg 1
``` Total No. of P ages: 2 6169
Register Number:
Name of the Candidate:
P.G.DIPLOMA EXAMINATION, 2009
(ACTUARIAL STATISTICS)
(PAPER-II)
120. INSURANCE AND ANNUITIES
Dec) (Time: 3 Hours
Maximum: 100 Marks
PART-A (5×8=40)
1. Explain Endowment Insurance.
2. Describe level benefit insurance.
3. Describe the insurance payable at the end of year of death.
4. Derive the formula for present value of a immediate annuity due of 1.p.a
5. Define annuity. Explain the types of annuities.
6. Derive an expression for endowment assurance.
7. Define communication function. State some of the relationships between
the commutation notations.
8. What is the value, at the end of 8 years of a immediate annuity of Rs.125
p.a for 12 years, the rate of interest being 5% p.a
PART-B (5×12=60)
9. a) Describe the different aspects involved in insurance payable at the
moment of death.
(OR)
b) Find the net single premium for a 20 year endowment assurance with
sum assured Rs.500 payable at the moment of death for a select life
aged 40 subject to the following information at 4% interest.
Given M[40]=1,904.86 D[40]=6981.60
M60 = 1,477.08 D60 = 2855.59
2
10. a) Establish the relationship between insurances payable at the moment of
death and at the end of the year of death.
(OR)
b) The following particulars are given :
x 25 26 27 28 29 30
lx 97380 97088 96794 96496 96194 95887
dx 292 294 298 302 307 313
Calculate the level annual premium for assurance given below, ignonng
interest and expenses.
i) A temporary assurance of Rs.1000 for 2 years for a person aged
25.
ii) An endowment assurance of Rs. 1000 for 4 years to a person aged
25.
11. a) Prove that m/ a n = a m+n –a m
(OR)
b) A serier of 8 annual sums of money is payable, the first payment
taking place at the end of one year from now. The first five payments
are Rs.300 each and the last three payments are Rs.200 each. Find the
present values of the eight payments @8% p.a.
12. a) Explain i) Endowment insurance
ii) Double endowment insurance
(OR)
M x M xn
b) Prove that A1n
x
Dx
13. a) Write an explanatory note on increasing life annuity.
(OR)
S x 1
b) Prove that (Ia)x=
Dx
*****
```
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