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https://uk.mathworks.com/matlabcentral/cody/problems/44457-triangle-of-numbers/solutions/1386326 | 1,603,500,254,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107881551.11/warc/CC-MAIN-20201023234043-20201024024043-00599.warc.gz | 584,068,756 | 17,832 | Cody
# Problem 44457. Triangle of numbers
Solution 1386326
Submitted on 19 Dec 2017 by shimon zusmanovsky
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
filetext = fileread('triangle.m'); assert(isempty(strfind(filetext, 'regexp')),'regexp hacks are forbidden')
2 Pass
n = 0; mat_correct = []; assert(isequal(triangle(n),mat_correct))
a = 1 b = 1 mat = []
3 Pass
n = 1; mat_correct = 1; assert(isequal(triangle(n),mat_correct))
a = 1 b = 1 mat = [] mat = 1 a = 2 b = 1
4 Pass
n = 6; mat_correct = [1 0 0; 2 3 0; 4 5 6]; assert(isequal(triangle(n),mat_correct))
a = 1 b = 1 mat = [] mat = 1 a = 2 b = 1 mat = 1 2 b = 2 mat = 1 0 2 3 a = 3 b = 1 mat = 1 0 2 3 4 0 b = 2 mat = 1 0 2 3 4 5 b = 3 mat = 1 0 0 2 3 0 4 5 6 a = 4 b = 1
5 Pass
n = 12; mat_correct = [1 0 0 0; 2 3 0 0; 4 5 6 0; 7 8 9 10; 11 12 0 0]; assert(isequal(triangle(n),mat_correct))
a = 1 b = 1 mat = [] mat = 1 a = 2 b = 1 mat = 1 2 b = 2 mat = 1 0 2 3 a = 3 b = 1 mat = 1 0 2 3 4 0 b = 2 mat = 1 0 2 3 4 5 b = 3 mat = 1 0 0 2 3 0 4 5 6 a = 4 b = 1 mat = 1 0 0 2 3 0 4 5 6 7 0 0 b = 2 mat = 1 0 0 2 3 0 4 5 6 7 8 0 b = 3 mat = 1 0 0 2 3 0 4 5 6 7 8 9 b = 4 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 a = 5 b = 1 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 11 0 0 0 b = 2 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 11 12 0 0 b = 3
6 Pass
n = 50; mat_correct = [1,zeros(1,8); 2:3,zeros(1,7); 4:6,zeros(1,6); 7:10,zeros(1,5); 11:15,zeros(1,4); 16:21,zeros(1,3); 22:28,0,0; ; 29:36,0; 37:45; 46:50,zeros(1,4)]; assert(isequal(triangle(n),mat_correct))
a = 1 b = 1 mat = [] mat = 1 a = 2 b = 1 mat = 1 2 b = 2 mat = 1 0 2 3 a = 3 b = 1 mat = 1 0 2 3 4 0 b = 2 mat = 1 0 2 3 4 5 b = 3 mat = 1 0 0 2 3 0 4 5 6 a = 4 b = 1 mat = 1 0 0 2 3 0 4 5 6 7 0 0 b = 2 mat = 1 0 0 2 3 0 4 5 6 7 8 0 b = 3 mat = 1 0 0 2 3 0 4 5 6 7 8 9 b = 4 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 a = 5 b = 1 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 11 0 0 0 b = 2 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 11 12 0 0 b = 3 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 11 12 13 0 b = 4 mat = 1 0 0 0 2 3 0 0 4 5 6 0 7 8 9 10 11 12 13 14 b = 5 mat = 1 0 0 0 0 2 3 0 0 0 4 5 6 0 0 7 8 9 10 0 11 12 13 14 15 a = 6 b = 1 mat = 1 0 0 0 0 2 3 0 0 0 4 5 6 0 0 7 8 9 10 0 11 12 13 14 15 16 0 0 0 0 b = 2 mat = 1 0 0 0 0 2 3 0 0 0 4 5 6 0 0 7 8 9 10 0 11 12 13 14 15 16 17 0 0 0 b = 3 mat = 1 0 0 0 0 2 3 0 0 0 4 5 6 0 0 7 8 9 10 0 11 12 13 14 15 16 17 18 0 0 b = 4 mat = 1 0 0 0 0 2 3 0 0 0 4 5 6 0 0 7 8 9 10 0 11 12 13 14 15 16 17 18 19 0 b = 5 mat = 1 0 0 0 0 2 3 0 0 0 4 5 6 0 0 7 8 9 10 0 11 12 13 14 15 16 17 18 19 20 b = 6 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 a = 7 b = 1 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 22 0 0 0 0 0 b = 2 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 22 23 0 0 0 0 b = 3 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 22 23 24 0 0 0 b = 4 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 22 23 24 25 0 0 b = 5 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 22 23 24 25 26 0 b = 6 mat = 1 0 0 0 0 0 2 3 0 0 0 0 4 5 6 0 0 0 7 8 9 10 0 0 11 12 13 14 15 0 16 17 18 19 20 21 22 23 24 25 26 27 b = 7 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 a = 8 b = 1 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 0 0 0 0 0 0 b = 2 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 30 0 0 0 0 0 b = 3 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 30 31 0 0 0 0 b = 4 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 30 31 32 0 0 0 b = 5 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 30 31 32 33 0 0 b = 6 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 30 31 32 33 34 0 b = 7 mat = 1 0 0 0 0 0 0 2 3 0 0 0 0 0 4 5 6 0 0 0 0 7 8 9 10 0 0 0 11 12 13 14 15 0 0 16 17 18 19 20 21 0 22 23 24 25 26 27 28 29 30 31 32 33 34 35 b = 8 mat = 1 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 4 5 6 0 0 0 0 0 7 8 9 10 0 0 0 0 11 12 13 14 15 0 0 0 16 17 18 19 20 21 0 0 22 23 24 25 26 27 28 0 29 30 31 32 33 34 35 36 a = 9 b = 1 mat = 1 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 4 5 6 0 0 0 0 0 7 8 9 10 0 0 0 0 11 12 13 14 15 0 0 0 16 17 18 19 20 21 0 0 22 23 24 25 26 27 28 0 29 30 31 32 33 34 35 36 37 0 0 0 0 0 0 0 b = 2 mat = 1 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 4 5 6 0 0 0 0 0 7 8 9 10 0 0 0 0 11 12 13 14 15 0 0 0 16 17 18 19 20 21 0 0 22 23 24 25 26 27 28 0 29 30 31 32 33 34 35 36 37 38 0 0 0 0 0 0 b = 3 mat = 1 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 4 5 6 0 0 0 0 0 7 8 9 10 0 0 0 0 11 12 13 14 15 0 0 0 16 17 18 19 20 21 0 0 22 23 24 25 26 27 28 0 29 30 31 32 33 34 35 36 37 38 39 0 0 0 0 0 b = 4 mat = 1 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 4 5 6 0 0 0 0 0 7 8 9 10 0 0 0 0 11 12 13 14 15 0 0 0 16 17 18 19 20 21 0 0 22 23 24 25 26 27 28 0 29 30 31 32 33 34 35 36 37 38 39 40 0 0 0 0 b = 5 mat = 1 0 0 0 0 0 0 0 2 3 0 0 0 0 0 0 4 5 6 0 0 0 0 0 7 8 9 10 0 ...
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Consumer%20Demand%20Theory[1]
# Consumer%20Demand%20Theory[1] - Consumer Demand Theory...
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Consumer Demand Theory: Utility and Indifference Curve Approaches The next two questions refer to the following table. Number of Apples Consumed Total Utility 0 0 1 9 2 19 3 28 4 36 5 43 6 40 Marginal utility increases for the *a. First two apples b. First three apples c. First four apples d. First five apples The consumer begins to encounter diminishing marginal utility when she consumes the a. First apple b. Second apple *c. Third apple d. Fourth apple Glenn Hall is maximizing his satisfaction by consuming two goods, A and B. If the marginal utility of A is half that of B, what is the price of A if the price of B is \$2? a. \$0.50 *b. \$1.00 c. \$2.00 d. \$3.00 An indifference curve is a curve that shows the various combinations of two goods that: a. Have identical prices b. have differing prices c. Provide a consumer equal marginal utilities *d. Provide a consumer equal total utilities In indifference curve analysis, the consumer will be in equilibrium (maximizes total utility) at the point where the a. Indifference curve intersects the budget line *b. Indifference curve is tangent to the budget line
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c. Indifference curve lies below the budget line d. Indifference curve lies above the budget line If the indifference curve appears “L” shaped, the two produ cts in question are a. Perfect substitutes *b. Perfect complements c. Independent goods d. “bads” If an indifference curve is a downward sloping straight line, the two products in question are *a. Perfect substitutes b. Perfect complements c. Independent goods d. “bads” The next two refer to the following diagram. Assume the price of good B has declined. The substitution effect is denoted by: a. oa b. ab *c. bc d. ac The income effect is denoted by: a. Oa *b. ab c. bc d. ac
At all points along a budget line, a consumer: a. Has equal utility. b. Is buying the same quantity of each good. c. Is spending all his or her money. *d. Is spending a constant amount of money. Answer the next 3 questions from the following diagram: Goods A and B represented by the indifference curves in the diagram are: a. Prefect substitutes *b. Imperfect substitutes c. Perfect complements d. Not identified by the degree of substitution If the consumer’s budget is \$100, the price of A is \$10, and the price of B is \$20, the consumer will maximize satisfaction for this budget by purchasing that mix of goods represented by point: *a. a b. b c. c d. d Assuming the same prices of A and B as in the previous question and an \$80 budget, the consumer would maximize utility by purchasing that mix of goods represented by point: a. a *b. b c. c d. d
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A parallel shift in the budget line out or away from the origin could be the result of: a. An increase in the price of both goods b. A decrease in the price of one good, while the price of the other good remained constant c. A decrease in the consumer’s money income *d. An increase in the consumer’s money income assuming that the prices of both goods remain constant.
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http://en.wikipedia.org/wiki/Studentized_range | 1,430,900,854,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430458431586.52/warc/CC-MAIN-20150501053351-00020-ip-10-235-10-82.ec2.internal.warc.gz | 66,907,042 | 11,991 | Studentized range
In statistics, the studentized range is the difference between the largest and smallest data in a sample measured in units of sample standard deviations.
The studentized range, q, is called after William Sealy Gosset (who wrote under the pseudonym "Student"), and was initially evoked by him (1927).[1] The concept was later presented by a number of actual students, Newman (1939) [2] and Keuls (1952) [3] and John Tukey in some unpublished notes. q is the basic statistic for the studentized range distribution, which is used for multiple comparison procedures, such as the single step procedure Tukey's range test , the Newman–Keuls method, and the Duncan's step down procedure, and establishing confidence intervals that are still valid after data snooping has occurred.[4]
Description
The value of the studentized range is most often represented by the variable q.
The studentized range computed from a list x1, ..., xn of numbers is given by the formulas
$q _{n,\nu}= \frac{\max\{\,x_1,\ \dots \ x_n\,\} - \min\{\,x_1,\ \dots\ x_n\}}{s} = \max_{i,j=1, \dots, n}\left\{\frac{x_i - x_j}{s}\right\}$
where
$s^2 = \frac{1}{n - 1}\sum_{i=1}^n (x_i - \overline{x})^2,$
is the sample variance, an unbiased estimator of the population variance and the square of the sample standard deviation s, and
$\overline{x} = \frac{x_1 + \cdots + x_n}{n}$
is the sample mean. The critical value of q is based on three factors:
1. α (the probability of rejecting a true null hypothesis)
2. n (the number of observations or groups)
3. v (the degrees of freedom used to estimate the sample variance)
Distribution (Normal Data) and Applications
If X1, ..., Xn are independent identically distributed random variables that are normally distributed, the probability distribution of their studentized range is what is usually called the studentized range distribution. Note that the definition of "q" does not depend on the expected value or the standard deviation of the distribution from which the sample is drawn, and therefore its probability distribution is the same regardless of those parameters. tables of the distribution quantiles are available here.
The Studentized range distribution has applications to hypothesis testing and multiple comparisons procedures. For example, Tukey's range test and Duncan's new multiple range test (MRT), in which the sample x1, ..., xn is a sample of means and q is the basic test-statistic, can be used as post-hoc analysis to test between which two groups means there is a significant difference (pairwise comparisons) after rejecting the null hypothesis that all groups are from the same population (i.e. all means are equal) by the standard analysis of variance.[5]
When only the equality of the two groups means is in question (i.e whether μ1 = μ2), the studentized range distribution is similar to the Student's t distribution, differing only in that the first takes into account the number of means under consideration, and the critical value is adjusted accordingly. The more means under consideration, the larger the critical value is. This makes sense since the more means there are, the greater the likelihood that at least some differences between pairs of means will be significally large due to chance alone..
Studentized data
Generally, the term studentized means that the variable's scale was adjusted by dividing by an estimate of a population standard deviation (see also studentized residual). The fact that the standard deviation is a sample standard deviation rather than the population standard deviation, and thus something that differs from one random sample to the next, is essential to the definition and the distribution of the Studentized data . The variability in the value of the sample standard deviation contributes additional uncertainty into the values calculated. This complicates the problem of finding the probability distribution of any statistic that is studentized.
Notes
1. ^ Student (1927). "Errors of routine analysis". Biometrika. 19 (1/2): 151–164.
2. ^ Newman D. (1939). "The Distribution of Range in Samples from a Normal Population Expressed in Terms of an Independent Estimate of Standard Deviation". Biometrika 31: 20–30. doi:10.1093/biomet/31.1-2.20.
3. ^ Keuls M. (1952). "The Use of the "Studentized Range" in Connection with an Analysis of Variance". Euphytica 1: 112–122. doi:10.1007/bf01908269.
4. ^ John A. Rafter (2002). "Multiple Comparison Methods for Means". SIAM 44 (2): 259–278. doi:10.1137/s0036144501357233.
5. ^ Pearson & Hartley (1970, Section 14.2)
References
• Pearson, E.S.; Hartley, H.O. (1970) Biometrika Tables for Statisticians, Volume 1, 3rd Edition, Cambridge University Press. ISBN 0-521-05920-8 | 1,123 | 4,735 | {"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": 3, "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-2015-18 | latest | en | 0.91923 |
https://www.overclockers.com/forums/threads/fan-noise-how-much-is-too-much.12714/ | 1,723,181,955,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640759711.49/warc/CC-MAIN-20240809044241-20240809074241-00611.warc.gz | 716,552,930 | 17,023 | # Fan noise... How much is too much?
#### e_storm
##### Member
Now, I know I have seen this question in old posts, but I can't find them.
As far as noise level of a fan goes, what dba number equates to what? I remember that over 40dba is pretty loud, but what about 30... or 20? What are silent fans rated at usually?
Also, if you go with, say 2-8cm fans instead of 1-12cm fan, does the dba ratings of the 2-8cms get combined (ie. 2-8cm fans @ 35dba each = 70dba noise total?)
Quiet system = ;D
Noisy sytem = ;p
To loud is when it bothers you or the people you live with.
Sound is kinda weird, it depends on the frequency and the distance between the sources and the reciever. Sometimes they do add up sometimes they cancel each other out. Most of the time 2 fans at 35 db each = 35 db total or real close.
one of my fans has a 40 something db rating. i can hear my comp running all the way downstairs. but i dont really care about noise.
Ridenow (May 09, 2001 10:49 a.m.):
To loud is when it bothers you or the people you live with.
Sound is kinda weird, it depends on the frequency and the distance between the sources and the reciever. Sometimes they do add up sometimes they cancel each other out. Most of the time 2 fans at 35 db each = 35 db total or real close.
Actually it will just be SLIGHTLY more, but hardly any. If you have say thirty fans going at 35db apiece, the total sound would be around 37-38. It is just the way db is equated. Also, remember that db are not linear, i forgot approximately how many db is twice as much as the other, but it graphs into some kinda curve, not a strait line.
If that's the case, why doesn't everyone go with a whole bunch of small, low noise fans instead of a couple of big, high noise ones?
I guess maybe noise doesn't bother some people as much as me. I gotta live in the same room with it.
My system got so loud (to me anyway) that I put all 4 case fans on separate switches and mounted the switches on the outside of the box so when it got hot or if I overclocked it a lot, I could just switch them on one at a time until things got cool 8)
Anyone have a cool, but real quiet system? Water cooling excluded.
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559 | 610 | 2,277 | {"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-2024-33 | latest | en | 0.959929 |
https://dojo.domo.com/discussion/53243/total-row-sum-instead-of-doing-the-calculation | 1,642,731,561,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302715.38/warc/CC-MAIN-20220121010736-20220121040736-00460.warc.gz | 283,651,894 | 18,429 | # Total row sum % instead of doing the % calculation
⚪️
Hi,
I use one beastmode to give me the MTD % of achievement vs the MTD Target.
On one card it work with this rule :
sum(`product_A`)/(SUM(DISTINCT `product_a_Target_dly`)*DAY(CURDATE()))
So it's giving me the MTD % even in the total row.
I use the same formula on another card but with product_B.... and the total row show me a sum of all the percentage instead of doing the calculation.
I don't find what I'm doing bad.... is there a way for the total row to apply the calculation and not summing it ?
PS : If i put over () at the end... it work but then the % if wrong for all the store. Only the total row become good.
Thanks :)
• Indiana 🥷
the total row is calculated on the card after the beast mode is calculated so it’s adding your percentages instead of calculating your total percentage.
if you want a total row you need to use an ETL to group your data together and sum all your columns using a group by tile then using an add constant tile put the value of total in your label field then append this row to your original dataset. Now your beast mode will work for the “total” row without telling your card to do the total row on it (it’s now included in your dataset)
• ⚪️
oh I see !
Will do that :)
Thank you !
• Budapest / Portland, OR 🟤
@Adrien most days of the week if you're doing SUM(DISTINCT) you've designed your data ... poorly. usually this happens b/c people JOIN their Actuals to Daily_Forecast and then use SUM(DISTINCT) to avoid double counting targets.
If you continue the road you're on, if you had two separate products with different daily forecast OR if you had the same forecast across two days, you'd misrepresent your forecast. | 414 | 1,736 | {"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-2022-05 | latest | en | 0.901947 |
https://www.beatthegmat.com/search.php?author_id=427678&sr=posts&sid=051681f789323b5483f9840c7bc80cb7 | 1,638,384,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964360881.12/warc/CC-MAIN-20211201173718-20211201203718-00139.warc.gz | 735,506,845 | 10,052 | ## Search found 13 matches
#### What is the $$LCM$$ of $$x$$ and $$12?$$
##### Re: What is the $$LCM$$ of $$x$$ and $$12?$$
Statement 1: 45 = $${3^2}$$*5 9 = $${3^2}$$ Hence, x = 5*$${3^a}$$ where a = 0, 1 ,2 x can be 5, 15, 45 Hence, this statement is insufficient. Statement 2: 20 = $${2^2}$$*5 4 = $${3^2}$$ Hence, x = 5*$${2^a}$$ where a = 0, 1 ,2 x can be 5, 10, 20 Hence, this statement is insufficient. Combining both...
by terminator12
Thu Jul 30, 2020 7:20 pm
Forum: Data Sufficiency
Topic: What is the $$LCM$$ of $$x$$ and $$12?$$
Replies: 1
Views: 330
#### What is the perimeter of the triangle above?
##### Re: What is the perimeter of the triangle above?
tan $${30^o}$$ = $$\frac{1}{\sqrt{3}}$$ = $$\frac{1}{base}$$
Hence, base = $$\sqrt{3}$$
Applying pythagoras theorem,
hypoteneuse = $$\sqrt{1\ +\ \left(\sqrt{3}\right)^2}$$ = 2
Hence, perimeter is 3 + $$\sqrt{3}$$
by terminator12
Thu Jul 30, 2020 7:12 pm
Forum: Problem Solving
Topic: What is the perimeter of the triangle above?
Replies: 2
Views: 331
#### Is $$x > 0 ?$$
##### Re: Is $$x > 0 ?$$
Statement 1: xy > 0 That can mean that either both are positive or both are negative. Hence, insufficient. Statement 2: x + y > 0 That can mean: i) both are positive ii) x is positive, y is less negative iii) x is negative, y is more positive Hence, this is insufficient. Combining both statements: W...
by terminator12
Thu Jul 30, 2020 7:08 pm
Forum: Data Sufficiency
Topic: Is $$x > 0 ?$$
Replies: 1
Views: 246
#### By how many years is Jason older than Allison?
##### Re: By how many years is Jason older than Allison?
Let the current ages of Jason, David and Allison be j, d and a respectively. Statement 1: j = 6 + 2d No information of a, hence clearly insufficient. Statement 2: a + 8 = 2d No information of j, hence clearly insufficient. Combining both statements, we get: j - a = 14 Hence, Jason is 14 years older ...
by terminator12
Thu Jul 30, 2020 6:05 pm
Forum: Data Sufficiency
Topic: By how many years is Jason older than Allison?
Replies: 1
Views: 253
#### What is the $$X$$ intercept of non-horizontal line $$m?$$
##### Re: What is the $$X$$ intercept of non-horizontal line $$m?$$
Consider a line: y = mx + c Slope = m X intercept (when y = 0) = $$\frac{-c}{m}$$ Y intercept (when x = 0) = c Statement 1: m = 4*c We see $$\frac{-c}{m}$$ = $$\frac{-1}{4}$$ Hence, this is sufficient. Statement 2: c = -2 We have no information on m Hence, this is insufficient. Hence, answer is A.
by terminator12
Thu Jul 30, 2020 6:00 pm
Forum: Data Sufficiency
Topic: What is the $$X$$ intercept of non-horizontal line $$m?$$
Replies: 2
Views: 441
#### $$\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=$$
##### Re: $$\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=$$
Let 1.701 be x
Hence, the equation becomes:
$$\frac{\left(4x\right)^6\cdot\left(2x\right)^7}{2^{19}\cdot x^{13}}$$
We see the powers of 2 and x cancel off in numerator and denominator and we remain with 1
by terminator12
Thu Jul 30, 2020 5:53 pm
Forum: Problem Solving
Topic: $$\dfrac{(6.804)^6\cdot(1.701)^{-13}}{2^{19}\cdot(3.402)^{-7}}=$$
Replies: 2
Views: 397
#### What is the value of $$x?$$
##### Re: What is the value of $$x?$$
Statement 1: We see 2 important points here: -2 and 2 let's make ranges about these points: 1) x < -2 - x - 2 = 2(-x + 2) x = 6 2) -2 < x < 2 x + 2 = 2(-x + 2) x = $$\frac{2}{3}$$ 3) x + 2 = 2(x - 2) x = 6 Hence, we have 2 different values of x. Hence, insufficient. Statement 2: Says x > 2 Clearly i...
by terminator12
Thu Jul 30, 2020 5:48 pm
Forum: Data Sufficiency
Topic: What is the value of $$x?$$
Replies: 1
Views: 224
#### A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =
##### Re: A bag contains x blue chips and y red chips. If the probability of selecting a red chip at random is 3/7, then x/y =
Probability of selecting red chip = $$\frac{y}{x+\ y}$$ = $$\frac{3}{7}$$
Solving this, we get $$\frac{x}{y}$$ = $$\frac{4}{3}$$
#### Percentage
##### Re: Percentage
Good question! Initially, the mileage was 20. That means pure petrol gives mileage of 20 So, kerosene gives mileage of 4 Let the % of petrol filled by the motorist be x, and % of kerosene filled be (1-x) 20*x + 4*(1-x) = 16 We get x = 75% Old cost to the gas station (C) = 100%*p New cost to the gas ...
by terminator12
Wed Jul 29, 2020 8:25 pm
Forum: GMAT Math
Topic: Percentage
Replies: 3
Views: 6476
#### Divisibility
##### Re: Divisibility
The way to solve these kinds of questions is: Divide 1,000,000 by 43 and find the remainder. Then subtract the remainder from 43, and add the result to 1,000,000 1,000,000 = 43 x 23,255 + 35 43 - 35 = 8 Adding 8 to 1,000,000 gives 1,000,008 Hence, answer is B You may also notice that: 1,000,008 = 43...
by terminator12
Tue Jul 28, 2020 9:50 pm
Forum: GMAT Math
Topic: Divisibility
Replies: 2
Views: 4919
#### Data Sufficiency
##### Re: Data Sufficiency
So the number x is of the format nnnn 1) Sum of the digits = 4n (which is divisible by 4, hence even) 2) Product of the digits = $$n^4$$ If n=even, product will be even. If n=odd, product will be odd. So, we can't be sure of this one. 3) To be divisible by 12, the number must be divisible by both 3 ...
by terminator12
Tue Jul 28, 2020 9:21 pm
Forum: GMAT Math
Topic: Data Sufficiency
Replies: 2
Views: 4975 | 1,892 | 5,401 | {"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 | 4 | CC-MAIN-2021-49 | latest | en | 0.838443 |
http://math.stackexchange.com/questions/103043/topology-exercise-compactness-circle-projective-space | 1,469,563,570,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.22/warc/CC-MAIN-20160723071025-00239-ip-10-185-27-174.ec2.internal.warc.gz | 157,835,186 | 16,961 | # topology exercise. compactness circle projective space.
Is the circle compact in $\mathbb{P}_{2}(\mathbb{C})$?
Here what I did: I considered the circle in $\mathbb{C}^2$ is $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$. The projective closure in $\mathbb{P}_{2}(\mathbb{C})$ is $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}$. The points at infinity are $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}\cap\{{x}_{0}=0\}=\{[0,1,i],[0,1,-i]\}$. So for every open cover ${\{{A}_{i}\}}_{i\in I}$ there will be a $j$ and a $k$ in $I$ fow which $[0,1,i]\in{A}_{j}$ and $[0,1,-i]\in{A}_{k}$. Therefore ${\{{A}_{i}\}}_{i\in I-\{j,k\}}$ is an open cover of the affine part $\{{{x}_{1}}^2+{{x}_{2}}^2-{{x}_{0}}^2=0\}$ and (if what i wrote is correct) the circle in $\mathbb{P}_{2}(\mathbb{C})$ is compact iff the circle $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$ is compact in $\mathbb{C}^2$. A subspace is compact in $\mathbb{C}^2$ iff it's close and limitated. Considering the continuous function $f:\mathbb{C}^2\to\mathbb{C}$ defined by $(x,y)\mapsto x^2+y^2$, the circle is ${f}^{-1}(1)$ so it's closed ($\mathbb{C}$ is T1, the points are closed). But it's limitated? Seems to me it is not.
Can anyone help me. Is what i wrote correct? Is $\{(x,y)\in\mathbb{C}^2|x^2+y^2=1\}$ limitated in $\mathbb{C}^2$?
-
$\{(x,y)\in\mathbb{C}^2\ | \ x^2+y^2=1\}$ is not compact (or bounded="limitated"), that's why there are points at infinity in the projective closure. for instance, given any $x$, you have solutions $(x,\pm\sqrt{1-x^2})$ so take some sequence of $x$'s going off to infinity to see it is unbounded (or not "limitated") – yoyo Jan 27 '12 at 19:05
Here is the first error I found: If $A_j$ and $A_k$ contain your two points at infinity, it does NOT follow that $\{A_i\}_{i\in I\setminus\{j,k\}}$ is an open cover for the affine part. For example, you could take the cover containing only the trivial open set. You can repair this problem by removing the points at infinity from each of the open sets in your cover to get an open cover of the affine part. – Aaron Jan 27 '12 at 19:10
Thank you very much... really usefull. So my reasoning is correct (with the correction Aaron made). Thanks again for your time and disponibility. P.S : And yoyo sorry for the word limitated, I've wrote the question quickly and I've wrongly translated the word from Italian. – Lorenzo Rossi Jan 27 '12 at 20:49 | 842 | 2,370 | {"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} | 3.46875 | 3 | CC-MAIN-2016-30 | latest | en | 0.841962 |
https://www.learnermath.com/multiplying-rational-expressions-examples | 1,725,980,722,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651255.81/warc/CC-MAIN-20240910125411-20240910155411-00813.warc.gz | 823,811,783 | 14,686 | # Multiplying Rational Expressions Examples,and Dividing
When you encounter multiplying rational expressions examples in Math, the general process conveniently follows the same approach as multiplying fractions that contain numbers.
Which we can look at a recap of.
## Multiplying Number Fractions:
\bf{\frac{2}{3} \space {\scriptsize{\times}} \space \frac{5}{7}} = \bf{\frac{2 \space \times \space 5}{3 \space \times \space 7}} = \bf{\frac{10}{21}}
General case:
\frac{a}{b} \space {\scriptsize{\times}} \space \frac{c}{d} = \frac{a \space \times \space c}{b \space \times \space d}
### Multiplying Rational Expressions:
If we have rational expressions \frac{Q(x)}{R(x)} and \frac{S(x)}{T(x)}.
Then, \frac{Q(x)}{R(x)} \space {\scriptsize{\times}} \space \frac{S(x)}{T(x)} = \frac{Q(x) \space \times \space S(x)}{R(x) \space \times \space T(x)}.
Example
\frac{2}{3x} {\scriptsize{\times}} \frac{x^2}{5x} = \frac{2}{3x} {\scriptsize{\times}} \frac{x^2}{5} = \frac{2 \space \times \space x^2}{3x \space \times \space 5} = \frac{2x^2}{15x} = \frac{2x}{15}
### How to Multiply Rational Expressions:
1) Look to factor the numerator and/or denominator if possible.
2) Eliminate any common factors between the numerator and the denominator.
3) Carry out any necessary multiplication and factor/simply further if possible.
## Multiplying Rational Expressions Examples
(1.1)
3 × \frac{x + 4}{9}
Solution
3 × \frac{x + 4}{9} = \frac{3}{1} {\scriptsize{\times}} \frac{x + 4}{9} = \frac{3(x + 4)}{9} = \frac{x + 4}{3}
(1.2)
\frac{5x^3}{4} {\scriptsize{\times}} \frac{3}{8x}
Solution
\frac{5x^3}{4} {\scriptsize{\times}} \frac{3}{8x} = \frac{5x^3 \space \times \space 3}{4 \space \times \space 8x} = \frac{15x^3}{32x} = \frac{15x^3}{32}
(1.3)
\frac{4a}{5a^2} {\scriptsize{\times}} \frac{10}{11a}
Solution
\frac{4a}{5a^2} {\scriptsize{\times}} \frac{10}{11a} = \frac{4}{5a} {\scriptsize{\times}} \frac{10}{11a} = \frac{4 \space \times \space 10}{5a \space \times \space 11a} = \frac{40}{55a^2} = \frac{8}{11a^2}
(1.4)
\frac{x^2 \space + \space 6x \space + \space 8}{3x^2 \space + \space x \space {\text{--}} \space 4} {\scriptsize{\times}} \frac{5x^2}{x \space + \space 4}
Solution
With multiplying rational expressions examples such as this.
We look to factor first where possible.
\frac{x^2 \space + \space 6x \space + \space 8}{3x^2 \space + \space x \space {\text{--}} \space 4} {\scriptsize{\times}} \frac{5x^2}{x \space + \space 4} = \frac{(x \space + \space 2)(x \space + \space 4)}{(3x \space {\text{--}} \space 2)(x \space + \space 2)} {\scriptsize{\times}} \frac{5x^2}{x \space + \space 4}
= \frac{(x \space + \space 4)}{(3x \space {\text{--}} \space 2)} {\scriptsize{\times}} \frac{5x^2}{x \space + \space 4} = \frac{5x^2(x \space + \space 4)}{(3x \space {\text{--}} \space2)(x \space + \space 4)} = \frac{5x^2}{3x \space {\text{--}} \space2}
(1.5)
\frac{6a^2b}{5a} {\scriptsize{\times}} \frac{3b}{2a}
Solution
\frac{6a^2b}{5a} {\scriptsize{\times}} \frac{3b}{2a} = \frac{18a^2b^2}{10a^2} = \frac{18b^2}{10} = \frac{9b^2}{5}
## Dividing Rational Expressions
Dividing rational expressions follows almost the same method as we use in multiplying rational expressions examples.
Just like with dividing fractions involving whole numbers, we multiply once we flip one of the fractions present in the division sum.
If we had, \bf{\frac{1}{4} {\scriptsize{\div}} \frac{3}{7}}.
To do this division sum we just perform the multiplication, \bf{\frac{1}{4} {\scriptsize{\times}} \frac{7}{3} \space {\scriptsize{=}} \space \frac{7}{12}}.
We follow the same approach with rational expressions.
Example
(2.1)
\frac{2x}{5} \space {\scriptsize{\div}} \space \frac{x \space {\text{--}} \space 2}{x \space + \space 1}
Solution
\frac{2x}{5} \space {\scriptsize{\div}} \space \frac{x \space {\text{--}} \space 2}{x \space + \space 1} = \frac{2x}{5} {\scriptsize{\times}} \frac{x \space + \space 1}{x \space {\text{--}} \space 2} = \frac{2x(x \space + \space 1)}{5(x \space {\text{--}} \space 2)}
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4. › Multiply/Divide Rational Expressions | 1,547 | 4,222 | {"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} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.381247 |
https://www.indcareer.com/saat-syllabus | 1,527,248,946,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867085.95/warc/CC-MAIN-20180525102302-20180525122302-00619.warc.gz | 778,930,464 | 18,472 | # SAAT Syllabus
Siksha ‘O’ Anusandhan University Admission Test (SAAT),a National Level Entrance Test is conducted by Siksha ‘O’ Anusandhan University every year to select bright students from different parts of our country. The test is being conducted for different programs offered in different constituent institutes / schools of the university.
==TOC==
## SAAT Syllabus
### MCA Syllabus
#### Unit-I : Mathematics
Logic: Statement, negation, implication, converse, contra positives, conjuction, disjunction, truth Table.
Algebra of Sets : Set operations, union, intersection, difference, symmetric difference, complement, Venn diagram, cartesian products of sets, relation and function, composite function, inverse of a function, equivalence relation, kinds of function.
Number Systems : Real numbers (algebraic and other properties, rational and irrational numbers), complex numbers, algebra of complex numbers, conjugate and square root of a complex number, cube roots of unity, De-Moivre’s theorem with simple application. Permutation and combinations and their simple applications, mathematical induction, binomial theorem. Determinants upto third order, minors and cofactors, properties of determinants. Matrices upto third order, types of matrices. Algebra of matrices, adjoint and inverse of a matrix. Application of determinants and matrices to the solution of linear equations (in three unknowns).
Trigonometry : Compound angles, multiple and sub-multiple angles, solution of trigonometric equations, properties of triangles, inverse circular functions.
Co-ordinate Geometry of two dimensions : Straight lines, pairs of straight lines, circles, equations of tangents and normals to a circle. Equations of parabola, ellipse and hyperbola, ellipse and hyperbola in simple forms and their tangents (focus, directix, eccentricity and latus rectum in all cases).
Co-ordinate Geometry of Three Dimensions : Distance and division formulae, direction cosines and direction ratios. Projections, angles between two planes, angle between a line and a plane, distance of a point from a line and plane. equations of a sphere-general equation.
Vectors: Fundamentals, dot and cross product of two vectors, scalar triple product, simple applications (to geometry, work and moment).
Differential Calculus: (Concept of limit, continuity, derivation of standard functions, successive differentiation (simple cases, Leibnitz theorem, partial differentiation (simple cases, derivatives as rate measure, maxima and minima indeterminate forms, geometrical applications such as tangents and normals to plane curves.
Integral calculus: Standard methods of integration (substitution, by parts, by partial fractions etc.). Definite integrals and properties of definite integrals, areas under plane curves, differential equations (only simple cases). (i) dy/dx = f(x) (ii) dy/dx = f(x) g(y) (iii) d2 y/dx2 = f(x) and application to motions in a straight line with constant acceleration. Probability and Statistics: Averages (mean, median and mode), dispersion (standard deviation and variance). Definition of probability; mutually exclusive events. independent events, addition theorem.
#### Unit-II : Computer Awareness
Introduction to Computer: Brief history of computers, components of a computer, computer related general knowledge, application of computers, classification of computers, simple DOS commands.
Computer Arithmetic: Number system with general base, number base conversion, elementary arithmetic operation.
BASIC Language Programming: Flow charts, algorithms, constants, variables, arithmetic and logical expression, elementary BASIC statements, writing simple programs (using sequence, repetition and control structures), ‘subscripted variables, matrix operations function and subroutines, concept of Files.
### Syllabus for BHMCT
Complete Bachelor in Hotel Management & Categoring Technology (BHMCT) syllabus
Section No. of Questions
Reasoning 20
Service Aptitude 20
General English 20
General Knowledge 20
General Science 20
Numerical Aptitude 20
### Syllabus for MBA / MHA / MHTM / MHMCT
Section No. of Questions
Verbal Reasoning 40
Analytical Reasoning 40
General Knowledge 10
Comprehension 20
Computer and Management Fundamentals 20
### Syllabus For M.Pharm
#### Unit-I
Micromeretics & powder rheology, various dispersion systems (viz. colloidal, suspensions & emulsions) kinetics & drug stability, polymorphism, surface & interfacial phenomenon. Liquid dosage forms, pharmaceutical aerosols, ophthalmic preparations. Cosmetic preparations, tablets, capsules & microencapsulation technology. Parenteral products, GMP and quality assurance, biopharmaceutics & pharmacokinetics, compartment model and kinetics, clearance concept, bioavailability & bioequivalence, formulation design of various controlled released drug delivery systems.
#### Unit - II
Stereo isomerism, conformational analysis of alkanes and cycloalkanes, relative stabilities of cycloalkanes, study of various heterocyclic compounds & polynuclear aromatic hydrocarbons, study of Amino acids, proteins, carbohydrates & lipids, brief concept on QSAR, steps involved to synthesize various categories of drugs mentioned in I.P and their SAR studies, structure elucidation of some important drugs under the category of vitamins, antibiotics and alkaloids.
#### Unit-III
Various limit tests, acid-base titrations, common ion effect, solubility product, theory & choice of indicators, precipitation
titrations, non-aqueous titrations, complexometirc & gravimetirc titrations, various oxidation & reduction titrations, RIAS, principle, basic instrumentation, elements of interpretation of spectra and applications of the following analytical tools. UV-Visible spectrophotometer IR & flame photometer NMR spectroscope including 13 CNMR & mass spectrometer, study of various chromatographic instruments viz TLC, column chromatography, HPLC, GLC & HPTLC.
#### Unit-IV
Pathophysiology of common diseases, viz. rheumatoid arthritis, gout, epilepsy, psychosis, depression, mania etc. Pharmacology of various drugs acting on peripheral nervous system, central nervous system, on cardiovascular system, on haemopoetic system, urinary system, respiratory system, GI tract system & endocrine system. Principles of chemotherapy and toxicology, pharmacology of autacoids, various bioassay of drugs/hormones & biological standardization. Basic concepts on hospital and clinical pharmacy.
#### Unit-V
Study of cultivation, collection, chemical constituents, adulterants and uses of various drugs obtained from natural sources with special emphasis on alkaloids, steroids, cardiac glycosides, terpenes, flavonoids and volatile oils. General techniques of biosynthetic studies and basic metabolic pathways. Brief introduction to biogenesis of secondary metabolites of pharmaceutical importance. Study of Plant tissue culture techniques.
#### Unit-VI
Pharmaceutical Biotechnology: Immunology & immunological preparations, genetic code & protein synthesis, genetic recombination, gene cloning and its applications, development of hybridoma for monoclonal antibodies. Microbial transformation, enzyme immobilization, study of drugs produced by biotechnology. blood products and plasma substitutes.
### Syllabus For M.Sc. Nursing
#### Unit-I : Nursing Foundations
Nursing as a Profession & it’s concept and Practice, Nursing trends and theories, Hospital Admission & discharge, Nursing Process, Intra personal relationship, Recording & reporting, vital signs, Health assessment, Equipment and linen management, Meeting needs of patient, Urinary & Bowel elimination, Psychosocial needs, Infection control & Microbiology, Administration of medication & it’s management, Care of terminally ill patients
#### Unit-II : Medical Surgical Nursing
New trends and concepts, Role & responsibilities of a Nurse in Client & family, Anatomy & Physiology related to each system, Management of patients having disorders in respiratory, cardiovascular & cardiothoracic, digestive, genitourinary, reproductive, endocrine, integumentary, musculoskeletal and immunological system etc., nursing management of patient with disorders in eye, ENT, and neurological system, management of burns and reconstructive and cosmetic surgery, management of oncological conditions, emergency and disaster management, care of elderly, management of patients C.C.U, management of occupational and industrial disorders.
#### Unit-III : Community Health Nursing
Determinants of health, epidemiology and nursing management of communicable and non-communicable diseases, demography, population and its control, health planning policies and problems, delivery of community health services, concepts and approach of community health nursing, national health and family welfare programmes, national and international health agencies, waste management at home and community, medical care follow up.
#### Unit–IV : Child Health Nursing And Neonatology
Modern concept of child care, care of neonate, IMNCI, management of behavioural and social problems in children, milestones, pediatrics medicine and surgery ICU management
#### Unit -V : Mental Health Nursing
Concepts and principles, therapeutic communication, treatment modalities and therapies used in mentally disorder, nursing management of patient with schizophrenia, psychotic and neurotic, somatisation disorder, substance used disorders, stress related disorders, personality, childhood and adolescent disorders, psychiatric emergencies and crisis intervention, legal issues in mental health nursing
#### Unit-VI : Gynaecology And Obstetrical Nursing
Anatomy & Physiology of reproductive system, assessment and management of antenatal, intranatal and postnatal mothers (normal & abnormal cases), assessment and management of normal neonates, high risk pregnancy assessment, abnormal labour management, management of high risk new borns, pharmaco-therapeutics in obstetrics, family welfare programme
#### Unit-VII : Administration And Education In Nursing
Management in Nursing, management process and nursing services in the hospital, community and educational institutions, organizational behavior and human relation, in service education and professional advancement, guidance and counseling, method of teaching educational media, I.E.C for Health, legal aspects in Nursing, professional advancement.
#### Unit–VIII : Nursing Research And Nursing Statistics
Research process, research problems and questions, review of literature, research approaches and designs, sampling and data collection, introduction to statistics(mean, median, mode, frequency distribution, coefficient, correlation statistical package and its application), biostatistics , communication and utilization of research.
### Syllabus For P.B.B.Sc. Nursing
#### Unit-I : Anatomy & Physiology
Organization of body cell, tissue, organs and different systems.
#### Unit-II : Microbiology
Immunity, infection, sterilization & micro-organism.
#### Unit-III : Psychology & Sociology
Human behaviour, learning, attention, perception, intelligence and personality.
#### Unit-IV : Fundamentals Of Nursing
Basic needs and care of the patients, therapeutic nursing care, nursing process, ethics in nursing, First-Aid
#### Unit-V : Medical Surgical Nursing
Management of patients with pain, undergoing surgery, respiratory, gastrointestinal, urinary, cardiovascular disorders nursing & Communicable diseases, assessment altered immune response, Emergency management, nursing care of elderly people. Oncology nursing.
#### Unit-VI : Psychiatric Nursing
Mental health assessment, disorder and nursing & crisis intervention, legal aspects.
#### Unit-VII : MID-Wifery & Gyneacologial Nursing
Embryology and fiddle development, nursing management of pregnant women, labour, nursing management of baby at birth and mother during puriterium. Hi-risk pregnancy, obstetric operation
#### Unit-VIII : Pediatric Nursing
New Born, healthy child, sick child, behavioural disorder and common health problems during childhood, children with congenital defect, welfare of children.
#### Unit-IX : Professional Trends
Nursing as a profession, professional ethics, personal and professional growth and development, legislation in nursing, professional
and related organisation.
#### Unit -X : Administration and Management
Hospital and ward.
### B.A. LLB / B.B.A. LLB Syllabus
Section No. of Questions
English including comprehension 20
General Knowledge / Current Affairs 20
Elementary Numerical Aptitude 20
Legal Aptitude / Legal Awareness 30
Logical Reasoning 30
### LLM Syllabus
Section No. of Questions
Legal Theory 20
Constitutional Law 30
Criminal Law 30
Law of Torts 20
Law of Contracts 20 | 2,570 | 12,778 | {"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} | 3.078125 | 3 | CC-MAIN-2018-22 | latest | en | 0.835755 |
https://www.calculatestudy.com/probability-calculator | 1,726,623,194,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00584.warc.gz | 644,299,516 | 12,216 | ## Probability Calculator
A Probability Calculator is a tool used to calculate the probability of an event or outcome occurring in a given situation.
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# Exploring Probability: Your Guide to Understanding and Using Probability Calculators
## Introduction: Unraveling the World of Probability Calculators
Imagine being able to easily anticipate results, comprehend uncertainty, and make wise judgments with the aid of a tool. Greetings from the probability calculator universe! These calculators are an excellent resource for anybody interested in statistics, dice probability in games, or seeking to understand concepts such as conditional probability or implied chances.
## What is a Probability Calculator? Understanding the Basics
A probability calculator is fundamentally your mathematical crystal ball. It's a computer program that calculates the probability of an event occurring under specific circumstances. Do you want to know the likelihood of rolling a particular number on the dice? A dice probability calculator can be useful in this situation.
## Probability Formulas and Definitions
The probability of an event happening can be calculated using the following formula:
$$P(A) = \frac{n(A)}{n(S)}$$
Where:
• $$P(A)$$ is the probability of event $$A$$.
• $$n(A)$$ is the number of favorable outcomes for event $$A$$.
• $$n(S)$$ is the total number of possible outcomes in the sample space.
For example, when rolling a fair six-sided die, the probability of getting a 4 would be:
$$P(4) = \frac{1}{6}$$
Another important concept is conditional probability, which is the probability of an event occurring given that another event has already occurred. It is calculated using the formula:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$
Where:
• $$P(A|B)$$ is the probability of event $$A$$ occurring given that event $$B$$ has occurred.
• $$P(A \cap B)$$ is the probability of both events $$A$$ and $$B$$ occurring.
• $$P(B)$$ is the probability of event $$B$$ occurring.
This formula helps in understanding the likelihood of an event happening under certain conditions. For instance, the probability of drawing two aces in a deck of cards given that the first card drawn was an ace can be calculated using conditional probability.
## how to calculate probability?
Probability measures the likelihood of an event happening. It's calculated as:
$$\text{Probability} (P) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}$$
Here's a step-by-step guide to calculating probability:
1. Determine the Event: Identify the specific event you want to calculate the probability for.
2. Count Favorable Outcomes: Determine the number of outcomes favorable to the event.
3. Count Total Possible Outcomes: Find the total number of outcomes in the sample space.
4. Apply the Formula: Use the formula $$P = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}$$ to compute the probability.
5. Express as a Fraction or Decimal: The probability can be represented as a fraction, decimal, or percentage.
For example, when rolling a fair six-sided die:
• Favorable Outcomes: The number of outcomes resulting in a 4 is 1.
• Total Possible Outcomes: There are 6 possible outcomes (numbers 1 through 6) on the die.
Therefore, the probability of rolling a 4 is:
$$P(4) = \frac{1}{6} \approx 0.167 \text{ (or } 16.7\% \text{)}$$
Remember, probability ranges from 0 (impossible event) to 1 (certain event). It's a powerful tool for analyzing uncertainties and predicting outcomes.
## Probability Calculator Examples and Solutions
### Example 1: Coin Toss Probability
Consider a fair coin. What is the probability of getting heads?
Solution:
$$P(\text{Heads}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{2} = 0.5$$
### Example 2: Dice Probability
If you roll a six-sided die, what is the probability of rolling an even number?
Solution:
$$P(\text{Even number}) = \frac{\text{Number of even outcomes}}{\text{Total possible outcomes}} = \frac{3}{6} = \frac{1}{2} = 0.5$$
### Example 3: Card Probability
You draw a single card from a standard deck. What is the probability of drawing a heart?
Solution:
$$P(\text{Heart}) = \frac{\text{Number of hearts}}{\text{Total cards}} = \frac{13}{52} = \frac{1}{4} = 0.25$$
### Example 4: Bag Probability
In a bag, there are 4 red, 3 blue, and 5 green balls. What is the probability of drawing a blue ball?
Solution:
$$P(\text{Blue ball}) = \frac{\text{Number of blue balls}}{\text{Total balls}} = \frac{3}{12} = \frac{1}{4} = 0.25$$
## Types of Probability Calculator
There are several types of probability calculators designed to address different scenarios and calculations. Here's an explanation of some common types:
### 1. Basic Probability Calculator:
• Calculates the probability of a single event occurring based on the ratio of favorable outcomes to the total possible outcomes.
### 2. Dice Probability Calculator:
• Specifically designed for dice-based scenarios, determining probabilities of various outcomes when rolling one or more dice.
### 3. Card Probability Calculator:
• Calculates probabilities related to decks of cards, such as the chance of drawing specific cards from a deck.
### 4. Conditional Probability Calculator:
• Computes the probability of an event occurring given that another event has already occurred, dealing with dependent events.
### 5. Implied Probability Calculator:
• Commonly used in betting scenarios, it calculates the implied probability from given betting odds.
### 6. Multivariate Probability Calculator:
• Handles situations involving multiple variables or events and computes the probability of their combined occurrence.
### 7. Labor Probability Calculator:
• Helps estimate the probability of labor starting within a certain timeframe during pregnancy.
### 8. Probability Calculator with Mean and Standard Deviation:
• Utilizes statistical measures like mean and standard deviation to calculate probabilities in datasets.
### 9. Joint Probability Calculator:
• Focuses on computing probabilities of multiple events occurring simultaneously.
### 10. Bayesian Probability Calculator:
• Incorporates Bayes' theorem to update probabilities based on new evidence or information.
These calculators cater to diverse scenarios, offering users the ability to compute probabilities relevant to their specific situations or fields of interest.
## Dice Probability Calculator: Rolling the Odds in Your Favor
Have you ever pondered how often it is to roll two dice and get a double? These intricate dice rolls may be reduced to easily understood odds with a dice probability calculator. It's like having a board game cheat code that makes it easy to plan and predict results.
## Navigating Z-Score Probability Calculator: Decoding Standard Deviation
Dive into the world of statistics with a z-score probability calculator. It's your guide to understanding deviations from the mean and figuring out the probability of a score occurring within a standard distribution. This tool unlocks insights, making statistical analysis less daunting.
Moving beyond the basics, advanced concepts such as the Z-score probability calculator, conditional probability calculator, and implied probability calculator delve into nuanced scenarios. The Z-score probability calculator, for instance, evaluates probabilities concerning standard deviations from the mean in a dataset.
## Multi-Event Probability Calculation
When dealing with multiple events, a probability calculator for three events becomes handy. In specific domains like healthcare, a labor probability calculator aids in assessing childbirth outcomes based on various factors.
## Conditional Probability Calculator: When Events Depend on Each Other
Probability isn't always straightforward. Sometimes, one event's outcome hinges on another. Enter the conditional probability calculator. It's like peeking into the future by considering the relationship between different events, making predictions more accurate.
## Labor Probability Calculator: Navigating Pregnancy Probabilities
Expecting parents often seek insights into labor probabilities. This specialized calculator estimates the likelihood of labor starting within a certain timeframe, offering a comforting sense of preparedness.
## Probability Calculator with Mean and Standard Deviation
Understanding Data Spread When dealing with datasets, understanding the spread of values is crucial. A probability calculator incorporating mean and standard deviation aids in comprehending how data points are dispersed around the average.
## Probability with Statistical Parameters
Integrating statistical parameters like mean and standard deviation into a probability calculator elevates its functionality. This advanced feature empowers users to compute probabilities within a specified range, offering a more nuanced analysis.
## Real-World Applications of Probability Calculators
Probability calculators find widespread applications across diverse fields. In business and finance, these tools aid in risk assessment, decision-making, and forecasting. Moreover, in scientific research, they assist in hypothesis testing and predicting experimental outcomes.
## Conclusion
Probability calculators serve as indispensable tools, offering insights and predictions crucial for decision-making across industries. Embracing their functionality enables users to navigate uncertainty with informed calculations and strategic foresight.
How do probability calculators work?
An online application called a probability calculator is used to figure out the chances of an event or result happening. It aids in comprehending the likelihood of various events occurring.
How should I operate the probability calculator?
The Probability Calculator is easy to use. When finished, click the "Calculate" button after entering the total number of outcomes and the number of favourable circumstances. The likelihood of the event will be shown by the calculator immediately.
Can the Probability Calculator handle compound events?
Yes, the Probability Calculator can handle compound events, where two or more independent events are involved. It calculates the combined probability of these events.
Can I use the calculator for conditional probability?
Yes, the Probability Calculator can calculate conditional probability. Simply input the number of outcomes and favorable events for each scenario to find the conditional probability. | 2,093 | 10,562 | {"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": 2, "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} | 4.5625 | 5 | CC-MAIN-2024-38 | latest | en | 0.902384 |
https://www.convertunits.com/from/pfund/to/hyl | 1,656,488,009,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00205.warc.gz | 749,318,739 | 12,673 | ## ››Convert pfund [Denmark, Germany] to hyl
pfund hyl
How many pfund in 1 hyl? The answer is 19.6133.
We assume you are converting between pfund [Denmark, Germany] and hyl.
You can view more details on each measurement unit:
pfund or hyl
The SI base unit for mass is the kilogram.
1 kilogram is equal to 2 pfund, or 0.10197162129779 hyl.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pfund [Denmark, Germany] and hyl.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pfund to hyl
1 pfund to hyl = 0.05099 hyl
10 pfund to hyl = 0.50986 hyl
20 pfund to hyl = 1.01972 hyl
30 pfund to hyl = 1.52957 hyl
40 pfund to hyl = 2.03943 hyl
50 pfund to hyl = 2.54929 hyl
100 pfund to hyl = 5.09858 hyl
200 pfund to hyl = 10.19716 hyl
## ››Want other units?
You can do the reverse unit conversion from hyl to pfund, or enter any two units below:
## Enter two units to convert
From: To:
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 443 | 1,482 | {"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-2022-27 | latest | en | 0.809355 |
http://www.cplusplus.com/forum/beginner/109601/ | 1,500,971,822,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425117.42/warc/CC-MAIN-20170725082441-20170725102441-00000.warc.gz | 421,226,995 | 5,421 | ### variable char problem in switch loop
I have to write a sort of blackjack program using a switch loop, I have most of it done but it needs to take letters for the face cards upper and lower case which all = 10, and have a condition for the ace being 1 or 11 (which ever helps the user more). The book gave a hint that the cards should be var. type char, but everything i tried doesn't work. So my question is how do I implement this? Do I state it some how in each case or at the start of my program?
Thank you for your help.
``12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970717273747576777879808182`` `````` #include using namespace std; int main( ) { char ans, NumberOfCards; int Sum, CardValue1, CardValue2, CardValue3, CardValue4, CardValue5; do { cout << "Please input the number of cards 2, 3, 4 or 5\n"; cin >> NumberOfCards; switch (NumberOfCards) { case '2': cout << "First Card" " "; cin >> CardValue1; cout << "Second Card" " "; cin >> CardValue2; Sum = CardValue1 + CardValue2; cout << "Your Score is" " " << Sum << " \n"; break; case '3': cout << "First Card" " "; cin >> CardValue1; cout << "Second Card" " "; cin >> CardValue2; cout << "Third Card" " "; cin >> CardValue3; Sum = CardValue1 + CardValue2 + CardValue3; cout << "Your Score is" " " << Sum << " \n"; break; case '4': cout << "First Card" " "; cin >> CardValue1; cout << "Second Card" " "; cin >> CardValue2; cout << "Third Card" " "; cin >> CardValue3; cout << "Forth Card" " "; cin >> CardValue4; Sum = CardValue1 + CardValue2 + CardValue3 + CardValue4; cout << "Your Score is" " " << Sum << " \n"; break; case '5': cout << "First Card" " "; cin >> CardValue1; cout << "Second Card" " "; cin >> CardValue2; cout << "Third Card" " "; cin >> CardValue3; cout << "Forth Card" " "; cin >> CardValue4; cout << "Fifth Card" " "; cin >> CardValue5; Sum = CardValue1 + CardValue2 + CardValue3 + CardValue4 + CardValue5; cout << "Your Score is" " " << Sum << " \n"; break; default: cout << "Incorrect Input of Cards\n"; } if (Sum > 21) { cout << "Busted\n"; } else { cout << "Winner\n"; } cout << "would you like to play another hand?\n"; cout << "Press y or n\n"; cin >> ans; } while(ans == 'y' or ans == 'Y'); system("Pause"); return 0; }``````
I ask this out of curiosity: do you know how to nest loops and use arrays?
I have covered nesting but not arrays. Im pretty sure a nested if else in each case is the answer but not sure how to make it happen.
Last edited on
Study this:
``1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192`` ``````#include using namespace std; int convert(char); int main() { char ans; do { cout << "Enter the number of cards: " << endl; int numberOfCards; cin >> numberOfCards; char* cards = new char[numberOfCards]; for (int i = 0; i < numberOfCards; i++) { cout << "Enter card " << i << ": "; cin >> cards[i]; } int sum = 0; for (int i = 0; i < numberOfCards; i++) { sum += convert(cards[i]); } cout << "Your total is: " << sum << endl; if (sum > 21) { cout << "Busted..." << endl; } else { cout << "Winner!" << endl; } cout << "Would you like to play another hand?" << endl; cout << "Type y or Y to continue: "; cin >> ans; } while(ans == 'y' || ans == 'Y'); delete [] cards; cin.get(); return 0; } int convert(char c) { switch (c) { case '2': return 2; case '3': return 3; case '4': return 4; case '5': return 5; case '6': return 6; case '7': return 7; case '8': return 8; case '9': return 9; case 't': case 'T': return 10; case 'j': case 'J': return 11; case 'q': case 'Q': return 12; case 'k': case 'K': return 13; case 'a': case 'A': return 14; default: return 0; } }``````
It uses arrays and loops and lets you input any number of cards you want (because of this).
Thank you for you help, I certainly will study this. It uses some things i haven't covered in my class yet but get the gist. I think the main thing i was missing was the int convert char with individual cases for each input.
Topic archived. No new replies allowed. | 1,249 | 4,154 | {"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-2017-30 | latest | en | 0.748381 |
https://www.mapleprimes.com/questions/235402-Example-For-Finding-All-Solutions-To | 1,721,155,282,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00596.warc.gz | 762,541,080 | 26,180 | # Question:Example for finding all solutions to a system of equations
## Question:Example for finding all solutions to a system of equations
This is a training example for finding all solutions to a system of equations. If you look at the graph, you can count 36 solutions, but I managed to find 20 relatively good approximations. No attempts to get more solutions, which are also visible as intersections of graphs, did not lead to success. Therefore, there is a suspicion that there are only 20 solutions.
Is it so?
``` restart: with(plots):
a:=8.:
f1 := x1^4-1.999*x1^2*x2^2+x2^4-1;
f2 := tan(x1+x2)-x2*sin(x1);
implicitplot([f1, f2], x1 = -a .. a, x2 = -a .. a, numpoints = 25000, scaling = constrained, color = [red, blue], thickness = 1);```
``` 1, (3.192246883291975), (-3.0395187374365404)
2, (3.0952031367176476), (-3.2447717313041897)
3, (0.5881900748267959), (-1.160066226905079)
4, (-0.936866718243322), (-1.3700058362814254)
5, (-2.555853694651265), (-2.7399958564861953)
6, (-3.2556241416421168), (-3.3964651254113774)
7, (-3.583319843955091), (-3.7077839724189228)
8, (-5.364827188794712), (-5.401998918608201)
9, (-5.398295356665546), (-5.360818510223991)
10, (-3.769206506106412), (-3.6477855329362683)
11,(-1.3978806247566642), (-0.9772190664843745)
12, (-1.192159295544978),(0.6492335177657542)
13, (-3.0867255059416623),(2.927375855548188)
14, (-3.18519036357835), (3.329801919022179)
15, (2.0108268901120754), (2.243492422396739)
16, (3.1133812329649766), (3.261937603184373)
17, (4.0265558604742715), (4.130826167761226)
18, (4.124539552922121), (4.019977762680433)
19, (3.172338340844501), (3.018365965761908)
20, (2.1945695320368097), (1.9558412553082192)
```
| 638 | 1,692 | {"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} | 3.078125 | 3 | CC-MAIN-2024-30 | latest | en | 0.65808 |
http://convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=fluid+oz | 1,656,485,806,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00564.warc.gz | 17,669,088 | 3,861 | Partner with ConvertIt.com
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Please read our Help Page and FAQ Page then post a message or send e-mail. Thanks! | 476 | 1,733 | {"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-2022-27 | latest | en | 0.643987 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=55612&p=205776 | 1,582,664,318,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00469.warc.gz | 429,790,796 | 11,572 | ## Writing Equilibrium Constant Labels
Bryce Ramirez 1J
Posts: 84
Joined: Sat Aug 24, 2019 12:16 am
### Writing Equilibrium Constant Labels
When writing the products and reactants for the equilibrium constant, why do we add P before the formula that is part of the chemical equation? Does this represent the partial pressure?
sarahforman_Dis2I
Posts: 98
Joined: Sat Aug 17, 2019 12:18 am
### Re: Writing Equilibrium Constant Labels
Bryce Ramirez 1J wrote:When writing the products and reactants for the equilibrium constant, why do we add P before the formula that is part of the chemical equation? Does this represent the partial pressure?
I believe that you are correct. If all of the substances in the equilibrium reaction are in the gas phase, that means that to determine when the system is at equilibrium we will use P, the partial pressure to calculate K. If the products and reactants are all in an aqueous solution, then concentration (the estimate for activity) would be used in the equilibrium expression.
Megan Vu 1J
Posts: 91
Joined: Thu Jul 25, 2019 12:15 am
### Re: Writing Equilibrium Constant Labels
P does means partial pressure because you are using a gas throughout the entire equilibrium. The equilibrium constant would also be denoted as Kp. You can often use the equation, PV = nRT in order to get the concentration of the whole equilibrium if you were asked to find it.
Both Kp and Hc have the same system where products are over reactants in a ratio.
Amy Pham 1D
Posts: 88
Joined: Fri Aug 09, 2019 12:15 am
### Re: Writing Equilibrium Constant Labels
Little clarification, even when all the products and reactants are given in the gas phase you don't always have to use partial pressures to calculate the K value. If you are given concentrations, then using those and calculating K_c is fine as well.
Tiffany_Chen 2K
Posts: 91
Joined: Fri Aug 30, 2019 12:15 am
### Re: Writing Equilibrium Constant Labels
Kc is used for concentrations (M), which can be gases or aqueous molecules, Kp uses partial pressures of gases. Question should be clear on which one to use. | 510 | 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} | 3.09375 | 3 | CC-MAIN-2020-10 | latest | en | 0.907676 |
https://www.brightstorm.com/math/trigonometry/pythagorean-theorem/equation-of-a-circle/ | 1,675,485,772,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00621.warc.gz | 701,646,764 | 26,888 | ###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
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# Equation of a Circle - Concept
Brian McCall
###### Brian McCall
Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
Share
A circle can't be represented by a function, as proved by the vertical line test. However, we can obtain an equation that describes the full circle by using the distance formula between the given center coordinates and any point along the circumference of the circle. Once we have derived this equation of a circle, we can apply it to any other circle you may come across in a coordinate plane.
In geometry it's helpful to come up with
an equation that will describe a
circle that's centered at some point
H and K with a radius R.
Well, before we come up with this,
let's do a little backtrack.
How can I calculate the distance
of R from point HK to XY?
Well, we said our distance formula between
any two points A and B is equal to
the square root of the differences of
the Xs squared plus the differences
of the Ys squared.
So let's apply that to this problem.
AB is actually the radius of this problem.
That's what we're trying to find.
So the radius is equal to the square root.
If I subtract my Xs, I see that I have X
as this point and my center of my circle
is at H. So we're going to say this
is X minus H quantity squared.
Now let's look at our Ys.
Our Y is Y, excuse me, our Y2 is Y. And
I'm going to subtract K. Because
K is the Y coordinate of
the center of my circle.
And I'm going to square that.
So for any point on this circle, what I'm
going to do is I'm going to square
both sides of this equation.
So that will give us any point
above or below that X axis.
Because right now if we just look at R,
we're going to be given half of this circle.
Because if you remember from algebra,
your vertical line test would fail of
the circle. The vertical line test, remember, says if
you can draw a vertical line anywhere
on your graph and it intersects your
function more than once it's not a function.
So we know a circle is not a function.
So we're going to square both sides
so we get the full function.
So R squared is equal to X minus H quantity
squared plus Y minus K quantity squared.
So the equation of a circle with radius
R is this equation right here.
Where H is the X coordinate of your center
and K is your Y coordinate of your
center.
So you're going to be substituting in for
HK and for R and that will give you | 657 | 2,768 | {"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} | 4.46875 | 4 | CC-MAIN-2023-06 | longest | en | 0.95951 |
https://www.handprint.com/HP/WCL/perspect4.html | 1,701,622,108,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.42/warc/CC-MAIN-20231203161435-20231203191435-00394.warc.gz | 909,468,787 | 21,784 | ## three point perspective
At this point it's customary to explore the capabilities of 2PP in a variety of specific drawing problems. I want to keep the momentum and look at three point perspective, which allows you to construct a form in any orientation (from any viewpoint).
Three point perspective is often illustrated with aerial views of Manhattan, looking down on a skyline bristling with skyscrapers. But artists will find 3PP equally useful in still life or figure paintings — where the view downward onto a table of objects or a piece of furniture can be just as steep — and in landscape views up toward soaring cliffs or a stand of tall trees.
The 3PP perspective problems and construction methods are complex, and it may seem we lose more in clarity than we gain in drawing power. Many artists have come to the same conclusion, and avoid 3PP for simpler approaches, including freehand modification of drawings blocked out in 2PP, or the expedient of tracing photos.
I won't disagree with those solutions; they can be convenient and effective. They fall short, however, if you must add new forms around the primary form — for example, if you have traced the photograph of an existing building, and want to insert new or different buildings around it — or if you want to show the building from a different point of view, or require more precision than freehand perspective can provide. For these common situations, 3PP is invaluable.
### three point perspective
As we add vanishing points, we remove aspects of perspective that we can take for granted. In 1PP or central perspective, the relationship of the vanishing points and horizon line to the direction of view are taken for granted. In 3PP both the vanishing point locations and the relationship between in the direction of view and the ground plane (horizon line) must be specified.
Defining Features of Three Point Perspective. The diagram shows the simplest 3PP situation: a cube centered in view but first rotated 45° to one side and then downward until all front faces appear of equal size. In all three point perspective views there are no faces or edges parallel with the picture plane.
In particular, because the direction of view is still assumed to be perpendicular to the image plane, the direction of view is no longer parallel to the ground plane when the primary forms are constructed as buildings are, with walls perpendicular to the ground.
The canonical view places the three front edges of the cube in a 54.7° angle to the direction of view, so that all three vanishing points are outside the circle of view. The planes of the three front faces are at a 35.3° angle to the direction of view, with vanishing lines defined by the triangle of three vanishing points.
three point perspective: the basic geometry
The three vanishing points (vp1, vp2 and vp3) control the recession of all lines parallel to the edges of the cube. This means the outline of each face is determined by two vanishing points, rather than one as in 2PP.
Connecting the vanishing points are three vanishing lines, which control recession of all planes parallel with each front and matching back face of the cube and all planes parallel to them. Each vanishing line also contains the vanishing points for all lines parallel to their respective planes, including the diagonal vanishing points (dvp1, dvp2 and dvp3) for the planes.
A vanishing line perpendicular to the viewer's vertical orientation (parallel to the ground plane) is typically the horizon line in architectural or landscape uses of perspective. It is the vanishing line for all planes parallel to the ground plane, and contains all vanishing points for lines parallel to the ground plane (perspective rules 13 and 14).
Each vanishing line is connected to the vanishing point opposite to it by an auxiliary horizon line (shown in orange in the figure). These are the vanishing lines for measure points for each of the three dimensions of the cube. In 2PP, the horizon line was a vanishing line for both the vanishing points and measure points, but in 3PP these functions can be separated.
The auxiliary horizon lines always intersect at the direction of view (the principal point) — that is, they link the vanishing points of the object to the vanishing point of the viewer's central recession (perspective gradient). Therefore the principal point is always inside the vp triangle formed by the three vanishing lines: if it is not, then the primary form does not define right angled vanishing points (it is a pyramid or a lopsided cube).
The measure points become significantly more complex in the 3PP orientation: two vanishing points define the edges of each face, and each edge requires its own measure point. So we have in all six measure points (mp1 to mp6) — two for each vanishing point in relation to the two faces it governs.
Finally, with the visual ray method we had a simple way to rotate the vanishing points in 2PP, but this also becomes significantly more complex in 3PP. In 2PP we just had to rotate two faces joined in one right angle, which we could easily diagram in two dimensions as two lines joined in one angle. In 3PP we must rotate three faces joined in three right angles, and that complicates the visual ray approach to a perspective solution.
Direction of View & Horizon Line. A 3PP construction allows the direction of view to be oblique to the ground plane, so that we are looking down or up on objects rather than looking at them directly from one side. Consequently in 3PP it is necessary to distinguish between (1) the object geometry (the vanishing points defined by the edges of the primary form), (2) the central recession defined by the direction of view, and (3) recession on the ground plane, for example in the visual texture of forests, grassy plains, deserts or bodies of water.
For example, we can redraw the cube illustrated above in two point perspective so that it has exactly the same angular size in the field of view (using a measure bar), and is positioned below the direction of view so that we look down on its upper face at a 35° angle. This locates the top front corner on the 71° circle of view and the bottom front corner just in front of the ground line (diagram, below).
the 3PP canonical view in two point perspective
Because both the angular size of the cube and the angle of its faces to the direction of view are identical, we are viewing it from exactly the same location in physical space. All we have done is shift our gaze from the object itself to the horizon line behind it. This keeps the same visual angle between the front corner of the cube and the horizon line. But changing the direction of view in 3PP means that:
(1) the horizon line no longer must intersect the principal point, and in fact may no longer be within 90° circle of view; and
(2) the geometrical relationship between any two vanishing points (the size and shape of the triangle the vanishing points define on the image plane) depends on the location of the third vanishing point and the location of the direction of view (the orientation of the image plane to the perspective problem).
### the perspective sketch construction method
The solution is basically to draw the form first, so you can locate the vanishing points and measure points which will produce that perspective view. You then use these to reconstruct the primary form in accurate perspective, and to add objects around the primary form within the same perspective space.
Why not draw the primary form by freehand perspective alone? Because, as we've already seen in 2PP, inaccurate placement of vanishing points results in a distorted perspective view; even small distortions can be obvious in a finished drawing. There is a better way.
You start with a freehand perspective sketch or scaled down perspective drawing at the center of a fairly large piece of paper (a 3' section from a roll of wrapping paper or white butcher paper is ideal).
three point perspective: perspective sketch of the primary form
Your drawing or photograph of the primary form should be small enough to fit all perspective points on the sheet of paper, yet large enough to work with accurately — usually a drawing about 10cm or 4 inches on its longest side is practical.
Take your time with the freehand drawing, and try to capture the relative proportions of the dominant edge angles and faces as accurately as you can. Don't worry about extraneous features (such as doors, windows or domes): you want to capture the basic perspective shape as it recedes in three directions. Be sure to define the edges and corner points clearly.
You can also start with a drawing or photograph of a building or monument that presents clear vanishing lines in its edges or surfaces, in the perspective orientation you want to duplicate. This photograph is only used to specify the approximate perspective view of the primary form in the drawing, so it does not have to look anything like the primary form you actually want to draw.
Once the drawing is finished to your satisfaction, or you have taped your photograph to the sheet of paper, you draw prospecting lines from the edges of the front planes to find the three vanishing points. Using a ruler or yardstick, extend the outer edges of the form until these prospecting lines intersect at three separate points. In a cube you have three edges tending to each vanishing point; use these in combination to reconcile discrepancies and find the point that gives all three the best definition.
The Vanishing Line Triangle. Next, connect these three vanishing points with three vanishing lines. You have defined the vanishing line triangle that will define (and usually contain) the primary form.
three point perspective: the vanishing line triangle
This is the point to look at the overall placement of the vanishing points in relation to the primary form and the space around it that will appear in the finished drawing. You can block in the format outline, or sketch other large forms around the primary form, to make sure you will get the effect you want.
Constructing Auxiliary Horizon Lines. Next, draw the three auxiliary horizon lines through each vanishing point and perpendicular to the opposite vanishing line. There are two ways to do this. The quicker is to use a large carpenter's square, laying one side against each vanishing line and sliding it back and forth along the line until the other arm is exactly on the vanishing point. Then draw the line.
three point perspective: constructing the auxiliary horizon lines
A more accurate method in large drawings is to construct the perpendiculars using a long piece of fishing line, hemp (not stretchable cotton) string or strip of cardboard as a compass measure. With your thumb, a tack or a piece of tape, fix one end of the measure at the vanishing point, and with the other end scribe a wide pencil arc across the opposite vanishing line. (Put the tip of the pencil through a loop in the string or a small hole in the cardboard strip.) The arc must intersect the vanishing line at two widely spaced points. Then either scribe two intersecting arcs centered on each of these new points, or measure with a ruler 1/2 the distance between them.
In the figure, the two arcs have been scribed around vp1 and vp2, and through vp3, to define the new points X and Y. Intersecting arcs drawn from X, Y and vp3 create the new points P1 and P2; lines to these points from the corresponding vanishing points create two auxiliary horizon lines. The direction of view (dv) is always at the intersection of all three auxiliary horizon lines, so the third line can simply be drawn from vp3 through dv to the opposite vanishing line. You end up with a vanishing line triangle similar to the one shown above.
Didn't I say elsewhere that the freehand placement of vanishing points leads to distortions? No: it's the clumsy scaling of drawing size in relation to the distance between the vanishing points that introduces distortions. If your three auxiliary horizon lines are at right angles (perpendicular) to their vanishing lines, if they meet in a single point (dv), and if this point is inside the vanishing line triangle, then the triangle defines a valid (physically possible) perspective space for a rectilinear solid.
Constructing the Circle of View. Now we insert the 90° circle of view. This requires you to (1) find the midpoint of any of the three vanishing lines (connecting two vanishing points), (2) draw a semicircle of Thales over the vanishing line, (3) extend to the semicircle the auxiliary horizon line that intersects the vanishing line, (4) construct a line parallel to the vanishing line, and finally (5) draw a second arc back to this parallel line. The intersection of this arc with the parallel line defines the radius of the 90° circle of view around dv.
three point perspective: constructing the circle of view
In the traditional solution, the artist uses either a ruler or the method of intersecting arcs to find the midpoint M on the vanishing lines between two vanishing points. In the diagram, I've chosen the vanishing line between vp2 and vp3. When arcs of equal radius are inscribed across the vanishing line from the two vanishing points, they intersect at two points, x and y. (1) A line through these points defines the midpoint M of the vanishing line.
(2) From point M the artist constructs a semicircle of Thales between the two vanishing points, then (3) extends to the semicircle the auxiliary horizon line that intersects the inscribed vanishing line at P. This defines a new point C. (For visual clarity, the semicircle is shown outside the perspective triangle, but to save space it can just as well be drawn to intersect the interior auxiliary horizon line.)
(4) Next, the artist constructs a line through dv that is parallel to the vanishing line.
(5) Finally, the artist inscribes an arc from P with radius equal to PC, the extended segment of the auxiliary horizon line. This intersects the line parallel to the vanishing line at either H1 or H2, depending on where it is more convenient to construct the arc.
(6) The line segments dv-H1 or dv-H2 are equivalently the radius of the 90° circle of view. The artist draws this circle from H1 (H2) with dv as its center.
It is often useful to include the 60° circle of view, which is a second circle with a radius equal to 0.58 (58%) of the radius of the 90° circle of view. This completes the perspective space.
Locating Measure Points. The last step is locating the measure points. Six are required if they are marked along the vanishing lines, but only three if you locate them on the auxiliary horizon lines.
Auxiliary Horizon Line Measure Points. To find the measure points on the auxiliary horizon lines, use a protractor or architect's triangle (or the traditional method for constructing a perpendicular) to construct finish perpendiculars on each auxiliary horizon line, from dv to the circle of view: the intersection with the circle of view defines three new points, C1, C2 and C3. Draw arcs from each of these C points back to the auxiliary horizon line perpendicular to it, using the vanishing point on that auxiliary horizon line as the center of the arc.
three point perspective: finding the measure points
This completes the perpective space at a reduced scale. I find that this entire procedure, starting with a blank sheet of paper and ending with the finished perspective space, requires about 20 minutes to complete. Once you understand how to do it, the work goes quickly and smoothly.
You must carefully make seven measurements on this drawing (using a metric ruler) to rescale it to full size: (1) the longest distance between any two vanishing points (in the example, vp3 to vp2), (2) the distance from one of these vanishing points to the intersection with the auxiliary horizon line (vp3 to h), (3) the length of this auxiliary horizon line (h to vp1), (4) the length to the direction of view (h to dv), and finally (5-7) the distance from dv to each of the three measure points.
Divide the radius of the circle of view you want in the full sized drawing (say, 160cm) by the radius of the circle of view in your perspective sketch: multiply all the measurements by this number. This gives you the full scale perspective space. Your perspective work surface needs to be at least as long as the longest vanishing line and as wide as the 90° circle of view. In the example drawing, assuming a 3m circle of view, this would be roughly 5m by 3m.
On a surface large enough to accommodate these distances (a very large table, or a clean hardwood or linoleum floor, or a clean, flat patio, garage floor or driveway), measure out the longest vanishing line (in the figure, vp2 to vp1), and the auxiliary horizon line to vp3. Connect the three vanishing points to define the vanishing line triangle. Measure the distance from the vanishing line to dv, and draw the remaining two auxiliary horizon lines from the vanishing points through dv. Finally, mark the three mp's on each auxiliary horizon line, measured from dv.
Use the drawing scale shown in the distance to size table to compute the drawing scale — the percentage of the actual object size (for a given viewing distance) that the drawing of the primary form should have. On a piece of paper, make a rough sketch of the primary form at this size, and lay the sketch on the format (size of support) you intend to use, to make sure the proportions work.
Vanishing Line Measure Points. The 3PP method of using three measure points is convenient, but it fails when the anchor point for measurements is close to the direction of view (dv). In this case, you may want to use the vanishing line points instead.
The construction of the circle of view required a semicircle of Thales drawn around one of the vanishing lines, centered on M and intersecting the vanishing points at either end of the vanishing line, then extended the auxiliary horizon line to intersect the semicircle in a point h'. This is all you need to define the measure points on that vanishing line. (Note that you can save steps and work space by intersecting the auxiliary horizon line inside the perspective triangle, to define interior h', and construct the measure points from there.)
three point perspective: alternate method to define measure points
The point h' will always define a 90° angle with the two vanishing points on the vanishing line. That is, it is equivalent to the viewpoint in a 2PP rotation of vanishing points. So you can draw two arcs from this point back to the vanishing line, using each vanishing point as the center of an arc, to define the measure points for the vanishing line — just as you would in two point perspective.
Confusion about the choice of vanishing line measure points is usually dispelled by the following two criteria:
• The controlling vanishing point is the vanishing point for the convergence of the edges that are being sized by the measure bar. Thus, edges converging to the right side vanishing point (vp2) are controlled by that vanishing point.
• The measure point to use was defined by an arc from the controlling vanishing point. Thus, mp4 was defined by an arc centered on vp2, so mp4 is the measure point to use when sizing edges that recede to that vanishing point. The height dimension is controlled by the vertical vanishing point (vp3), which was the center of the arc used to define mp3.
Measure bars to the vanishing line measure points always must be parallel to the vanishing line containing the measure point being used, not to any auxiliary horizon line as before. Note that two measure points are always available for each dimension. In the example, mp6 can be used to size the vertical edges receding to vp3, if for some reason mp3 is inconvenient to use — but in that case, the measure bar must be parallel to the vanishing line containing mp6.
The measure bars in the illustration are the same length as those used previously, and as you can see, they define the same reduction in perspective depth. You do not need to rescale or recompute the measure bars you already have; just align them parallel with the appropriate vanishing line.
Because the semicircle on M is part of the circle of view procedure, and any vanishing line can be used to define the circle of view, you should consider the location of your anchor points in the perspective space, and place the semicircle of Thales around the vanishing line where measure points will be most convenient.
For example: I had originally put the anchor point at the front bottom corner of the cube; in that location mp3 worked fine, but the other two points created badly slanting measure lines that would introduce inaccuracies. The best alternative points would be found on the top vanishing line (between vp1 and vp2), so I should have started building the circle of view by putting the first semicircle on that side.
### constructing a 3PP cube(perspective sketch method)
Once you have constructed the 3PP space, you can begin construction of the cube or primary form. This explanation excludes the procedures necessary to scale the drawing, which are developed below.
three point perspective: locating the primary form
Measure out the perspective space on the perspective drawing surface (floor, driveway, patio), and tape or tack the support to the surface, oriented with the top edge parallel to one of the vanishing lines (or to none, if the perspective view is tilted), and the dv in the correct location within the drawing. If you do not want to work directly on your watercolor paper, reconstruct the drawing on a large sheet of butcher paper or wrapping paper, and then trace or square the drawing to the format when you are done.
Mark the dv and draw the auxiliary horizon lines, the measure points, the anchor point, and the base vanishing lines through the anchor point.
The diagram shows this done on an emperor sheet, 40" x 60", located with the dv near the bottom. (If you are going to all this trouble, you may as well make the painting spectacular!) For clarity, the support outline is omitted in the next several illustrations, although it is assumed you are working with the support in place.
three point perspective: constructing measure bars
The last preparatory step is constructing the measure bars. Do this from the center of the space (dv), because each measure bar must be parallel with its corresponding auxiliary horizon line. This is easiest to do by simply drawing the measure bar on a separate sheet of paper, directly over the auxiliary horizon lines.
Draw the measure bars to the perspective length they have in space, so that you can line them up with one end against the anchor point. The length of the measure bars determines the drawing size of the primary form, so you want these to be accurate. For example, if the dimensions of a building are 150 feet long, 75 feet wide and 36 feet high, and you used the length of the building to scale the drawing size, then the proportions between the measure bars are 1.00 to 0.50 and 1.00 to 0.24. Since we are drawing a cube, all three measure bars will be of equal length, so we define them by drawing a circle around dv (shown above).
three point perspective: constructing front vertical
The rest is a piece of cake. First, align the vertical measure bar parallel to the vertical auxiliary horizon, with the bottom end on the anchor point. Draw a line from mp3 through the top end of the measure bar to the vertical vanishing line (that is, the line parallel to the measure bar you are using). This defines the front height of the cube.
Using a yardstick, string or cardboard strip aligned with the bottom vanishing point, draw a line from the anchor point to the line from the vertical measure bar to mp3. This is the front vertical. Use the yardstick, string or cardboard strip to connect the ends of this vertical to the two side vanishing points, and draw the front top and bottom edges of the form.
three point perspective: constructing left side
Next, use the second measure bar to define the depth dimension on one side (to mp1, the measure point on the auxiliary horizon line parallel to the measure bar you are using). When one end of the measure bar is aligned with the anchor point, the back corner of the cube is located where the line from the other end of the measure bar to mp1 crosses the bottom left edge of the figure. Mark this point.
Again aligning your straight edge with vp3, draw a line from this point to the top left edge line: this is the back vertical of the cube. Connect the ends of this vertical along vanishing lines to vp2. These lines define the back left upper and lower edges of the figure.
three point perspective: constructing right side
With the third measure bar, construct the opposite side, define the corners, and connect to the vanishing points as before.
Clean up the drawing as much as necessary to visually confirm the final perspective outline meets your expectations. Then go on to add any other objects in the environment around the primary form, or perspective details on its surface (doors, windows, etc.).
three point perspective: perspective distortions
Familiarity with the 3PP mechanism will help you understand how to use it effectively. The diagram (above) gives some clues about the scale, placement and cropping of forms:
• In general, distortions toward the side vanishing points are much more objectionable than those toward the bottom vanishing point: choose a vertical or square format whenever feasible.
• Forms can be placed below the 90° border — a 90° angle placed to intersect the two side vanishing points (red line) — to emphasize height or vertical scale, but forms should not be placed near the border on either side. (All possible locations of the right angled corner of this border are defined by a circle of Thales constructed below the horizon line.)
• The same circle of view rules apply in order to reduce perspective distortions, but the circle can be displaced downward from the direction of view, as if pulled away from the horizon line by the vertical depth. It is better to think in terms of a column of view centered on the principal point and extending from below the 90° border to above the horizon line (where cloud layers in perspective can enhance distance depth to balance the vertical depth). Any format that fits within a 40° to 60° column will produce a handsome image.
When you have finished with the perspective elements, carefully release the drawing surface from the table, floor or patio, and lay it out on your painting surface to erase the guidelines, measure points, and other extraneous elements, or to transfer the perspective outline to the actual painting surface. When the drawing is fully cleaned, add by freehand any additional outlines or guidelines necessary before you begin to paint.
three point perspective: finished drawing
The diagram shows the finished perspective form, once again within the monumental 40"x60" emperor format. In this reduced diagram, the primary form appears to be little changed from the original perspective sketch. But in practice, despite all the work invested, you will be quite pleased with the increased perspective accuracy and "weight" of the finished drawing in comparison to anything you could manage by freehand methods alone.
### the horizon line construction method
Two significant problems with the perspective sketch method are that it establishes the angles of the the primary form to the viewpoint approximately, through a sketch, and that it cuts the 3PP methods loose from the procedures for scaling the drawing within the circle of view. The actual perspective angles and scale of the circle of view are derived from the drawing, rather than given at the start. An alternative method is to start with the circle of view, and from there construct the vanishing points. This method starts by specifying the location of the horizon line (a horizontal vanishing line above or below the direction of view), so I refer to it as the horizon line method of 3PP construction, though the circle of view method is also apt.
A discussion of the 3PP geometry will clarify how this method works. Because all parallel lines converge to the same (single) vanishing point (perspective rule 6), and the 3PP vanishing points define visual rays at right angles to each other, the 3PP vanishing points are equivalently defined by the three right angled edges of a cube that can be turned or rotated around a front corner fixed on the direction of view (diagram, right).
These edges converge to the three right angle vanishing points at the vanishing lines for the three planes defined by the three front faces of the cube (perspective rule 14). Therefore the vanishing lines between the pairs of vanishing points will be parallel to the line intersections of the three front faces of this cube with the image plane (green, corollary to perspective rule 11). As a result, we have reduced the geometry of the 3PP vanishing points to the geometry of a three sided pyramid thrust through the image plane in any arbitrary angle and rotation.
As explained earlier, the circle of view framework provides a method to specify exactly the location of any vanishing point as a line rotated to the required angle around the viewpoint folded into the image plane. What we require is a way to perform this folding for elements of the 3PP "pyramid".
This is done by moving the fixed corner of the cube forward until it coincides with the viewpoint. In that position its three edges define three visual rays to the vanishing points (magenta lines, diagram above right). More important: the altitude of the pyramid is now equal to the viewing distance and therefore to the radius of the 90° circle of view (diagram, below).
folding a pyramid right triangle into the image plane
Two kinds of folding operations are possible in this 3PP geometry. First are the auxiliary line folds that define the interior angle between a pyramid edge, or the pyramid face perpendicular to it, and the direction of view. These are found by folding into the image plane a vertical section of the pyramid defined by an auxiliary horizon line, for example the interior triangle PVC defined by the auxiliary horizon line PC in the diagram above. This triangle contains the two triangles VdvC and VdvP, each containing a right angle at dv. The fold brings line Vdv into the image plane as x'dv. Because the edge Cdv is continuous with edge Pdv, the right angle at dv is preserved. And the image edges Cx' = CV and Px' = PV. Therefore, by triangular equalities, the image angle 1' equals the interior angle 1, the angle between the direction of view and the face ABV.
This fold also identifies (at Cx'dv) the angle between the vanishing point C and the direction of view, so this folding down of an interior section of the perspective pyramid is geometrically identical to the folding of the viewpoint into the circle of view that is used to rotate vanishing points to the direction of view.
The second kind of folding operations are the vanishing line folds that define an exterior angle of one face of the perspective pyramid (angle 2) as a "plan view" of the angle in the image plane (angle 2'). This is the angle, on the face of the 3PP pyramid, between the edge of triangle ABV and its altitude PV. The fold is achieved by constructing a line (ab) that intersects the direction of view parallel to the vanishing line (AB). This line intersects the circle of view at x'. Because Vdv equals x'dv, the line Px' equals line PV, the altitude of ABV. Therefore an arc constructed on P with radius Px' intersects the auxiliary horizon line at x, and Px = PV. Therefore the right triangle ABx is the perpendicular view of the foreshortened triangle ABV, and x is the auxiliary viewpoint for the horizon line AB.
three right triangles folded out of the 3PP pyramid
The diagram (above) shows the three possible vanishing line folds and auxiliary viewpoints (x, y and z) constructed from a 3PP vanishing line triangle. Study this diagram carefully until you understand how each fold has been done.
The geometry of triangles is efficient: defining any one side with its two adjacent angles, or any two sides with their common angle, defines the rest of the triangle. Therefore only two folding operations are necessary to define the image of a 3PP vanishing line triangle: one auxiliary horizon line fold and one vanishing line fold. This is sufficient to define the location of all three vanishing points and vanishing lines in relation to the direction of view and circle of view.
Finally, the 3PP construction releases the direction of view from its parallel position to the ground plane, and this creates several novel features in the perspective geometry which affect in particular the scaling of the 3PP drawing. For now I only want to describe this geometry and define a few new terms (diagram, below).
elevation view of 3PP geometry
In this example we assume the perspective view is downward in relation to the ground plane: it can just as well be upward (as the top of a skyscraper viewed from the ground) or tilted (as a city viewed from a turning airplane), a problem I leave for the reader. In the downward view case:
• The image plane is oblique to the ground plane, as is the direction of view. As a result the direction of view does not terminate in a vanishing point, but in a fixation point, some physical point on the ground. This fixation distance is typically different from the object distance from the station point to the primary form.
• The station point S is still directly under the viewpoint, but now the station point appears on the image plane, where it is the image s equivalent to the vertical vanishing point (vp1).
• The horizon line is now located above the direction of view in the circle of view, which means the principal point, the vanishing point for the viewer's central recession (at p), is no longer the same as the orthogonal vanishing point (at h), the vanishing point of ground plane recession.
• The primary form appears in rotation foreshortening — the vertical and horizontal dimensions are in a different scale. Foreshortening is corrected by using the measure points; measure bars parallel to the image plane may be rotated in the image plane to any other angle. However, it is sometimes useful to estimate the amount of vertical foreshortening, for example when planning the image layout. This is found by a cosine correction for foreshortening:
where θ is the horizon angle. Because the angle of view to the ground plane is only equal to the horizon angle at the fixation point, a measure bar established at any other point must be calculated with the correct angle of view to that point on the ground plane.
• An object's angular size or image size is determined by the sight line distance from the viewpoint, which is simply the hypotenuse of the right triangle formed by the object distance and viewing height.
These points need to be understood in order to apply the correct distance & size calculations when scaling the drawing in a 3PP construction.
### constructing a 3PP drawing(horizon line method)
The horizon line method builds on the assumption that most three dimensional perspective problems concern a viewer whose line of sight is not parallel to the ground plane. Either the viewer is looking upward, toward the top of a tower, building, mountain or cliff; or the viewer is looking downward, from a vantage at the top of a tower, building, mountain or cliff.
Approximate Horizon Line Method. In this approach the artist places the horizon line and vanishing points by judgment or whim, but uses the pyramid folds to make these landmarks consistent with each other.
The first step is the placement of the horizon line in relation to the principal point: either above (for a downward direction of view) or below (for an upward direction of view). Then, using a drafting triangle, the artist finds the 90° angle at one of the diagonal vanishing points, and extends this line until it meets the median line below the circle of view: this is the vertical vanishing point (vp1).
three point perspective: rotating the horizon line
It is useful to bisect this angle to find the diagonal view (45° from either the horizon line or vp1 visual rays), as this projects in depth the viewing height above the ground plane.
Next the second vanishing point (vp2) is located on the horizon line somewhere to the left of the median line. A ruler laid from dv to this point will show the angle to view of a cubic form in perspective space at the direction of view.
three point perspective: approximately placing vp2
Once the location of the point is completed, draw the vanishing line between the two vanishing points. Then you must construct a perpendicular line from this vanishing line through the direction of view (dv), as described here.
The steps are: (1) draw a circular arc around dv that intersects the vanishing line at two widely spaced points, a and b; (2) draw an arc from each point with a radius greater than half the segment length between them; (3) draw a line through the double intersection of the arcs to define the normal point c; (4) draw a line from c through dv until it intersects the horizon line on the opposite side of the circle of view. This is the auxiliary horizon line for the constructed vanishing line; it locates vp3.
three point perspective: completed "approximate" perspective triangle
Construct the third vanishing line and its auxiliary horizon line from vp2 through dv.
Find the internal or external altitude points on the auxiliary horizon lines, and from these locate the six measure points on the vanishing lines. (The diagram above shows one internal altitude point, h' and the two measure points constructed from it.) This completes the three point perspective triangle.
Exact Horizon Line Method. In some cases (illustrated below) it is desirable to locate the three vanishing points precisely. In this case the pyramid folds are precisely defined with a protractor or using the tangent ratio for the required angle, applied to the radius length of the circle of view that is perpendicular to the viewpoint.
Required is one auxiliary horizon line fold along the vertical auxiliary horizon line (median line) to establish the tilt of the horizon line and the location of the vertical vanishing point (vp1), and one vanishing line fold along the horizon line to establish the left/right placement of vp2 and vp2.
The diagram (below) shows these operations to provide an exact 25° downward angle of view to the ground plane (upward horizon angle of 25°), and a placement of vp2 55° to the left of the median line. This places vp3 35° to the right of the median line.
three point perspective: exact rotation of vanishing points
The vanishing lines are added as before; the auxiliary vanishing lines can be drawn directly, as lines from the vanishing points through dv to the opposite vanishing line, because the vanishing points have already been precisely located.
Measure points have been added using the "alternative" horizon line method described above.
three point perspective: completed "exact" perspective triangle
Although this 3PP triangle is very similar to the one constructed from an approximate judgment of the correct angles, here all the perspective landmarks are exactly placed from given values established in advance. This is especially important when the goal is a 3PP view of a specific primary form from a specific location — as is typical in architectural renderings or historical reconstructions — or when a certain arrangement of key forms within the image is required.
Diagonal Vanishing Points. It is usually very useful to take the extra step and establish the diagonal vanishing points on the horizon line. Once this is done a unit dimension on the station line can be projected across the ground plane, using the method of projecting a unit dimension in depth from the diagonal vanishing points.
three point perspective: locating the central dvp's
In fact, no separate rotation is required to define the diagonal vanishing points from the auxiliary viewpoint: they are already located at the intersection of the arc used to define the auxiliary viewpoint with the horizon line (diagram, above). The method of rotating the diagonal vanishing points around the auxiliary viewpoint A, so that a 90° angle is bisected by the vertical auxiliary horizon line (median line), is shown simply to confirm this.
Once these diagonal vanishing points have been established, ovp serves as the orthogonal vanishing point, the convergence for recession in depth parallel to the ground plane (the ground plane central recession); but dv remains the principal point, the convergence for recession parallel to the direction of view (the viewer's central recession). The depth of transversals across orthogonals to ovp are found by diagonals to dvp1 and dvp2; the depth of transversals across orthogonals to dv are found by vanishing lines to points on the circle of view.
Scaling the 3PP Drawing. This task is more complex than it is in one or two point perspective, but I outline it here because I have not seen it discussed in any other source. A minimal reliance on trigonometry is required, both to validate the basic principles and to provide calculation shortcuts or remedies to complex construction problems.
Construction Methods. Three drawing scale guides are already available: (1) the circle of view and the many visual angles that can be computed within it; (2) the viewing height in depth, added when the horizon line was rotated; and (3) a ground line scale, which is used in combination with the orthogonal vanishing point (ovp) to project a unit dimension in depth to approximately locate objects in depth and scale their image size. Provided image scale and perspective accuracy are not critically important, these are almost always adequate to scale the 3PP drawing.
three point perspective: ground plane recession
Two scaling approaches can be used. In the first example (above), an arbitrary unit dimension of 50 cm measured along the station line image is projected into perspective space by orthogonals drawn to the orthogonal vanishing point (ovp). These indicate that the viewing height in depth is about 7.2 times the viewing height (measured from the station line) and that the fixation point is about 16 units away.
Dividing the viewing height by the depth units yields the ground plane width of the unit dimension. If the viewing height is 300 meters, then the unit dimension represents about 300/7.2 = 42 meters, and the fixation point is about 672 meters from the station line — all distances measured on the ground plane rather than along the line of sight. If the viewing height is 3 feet, then the unit dimension represents 36/7.2 = 5 inches and the fixation point is 80 inches away.
The orthogonals define this unit dimension at any depth; a transversal established at the base of the primary form creates a measure bar in unit dimensions at that distance. In the diagram (above), a measure bar is shown at the fixation point that is 3 unit dimensions long. If the viewing height is 300 meters, then the measure bar defines a width of 126 meters at a distance of 672 meters. This image bar is used to measure out the size of the primary form image along and above the base transversal.
three point perspective: key scaling dimensions
The second method is to establish an exact unit dimension. This is done by (1) drawing a line from one of the dvp's through the viewing height in depth (vhd) and extending this line until it intersects the station line image, then (2) dividing the distance from this intersection to the station point (vp1) by an appropriate number of units. In the example this procedure yields a station line length of 354 cm, which is conveniently divided into six 59 cm units. If the viewing height is 270 meters, then this unit dimension represents exactly 45 meters on the ground plane.
As explained in the discussion of scaling the drawing in the 1PP context, the location of the format requires the artist to decide the appropriate size and location of the primary form image. The principal scaling restriction is the horizon line rule: the horizon line intersects all forms at a height above their intersection with the ground plane that is equal to the viewing height, or causes the forms to appear below the horizon line by an equivalent added proportion of their total height. Following the procedure explained for central perspective, the artist finds the ratio between the viewing height and object height, then places the object so that the horizon line divides or stands above the object image by this ratio. This rule holds regardless of the angle of the direction of view to the ground plane.
In the example I will develop, I want to render a primary form that is 300 meters high and 125 meters wide. I want the upper portion of the form to cut the horizon line, so that its top portion is silhouetted against the sky. Therefore I plan on approximately 30 meters of the form appearing above the horizon line and the remaining 270 meters below the horizon line. This means the viewing height will also be 270 meters (according to the horizon height rule), and 30/300 or 10% of the primary form image will be above the horizon line.
To minimize perspective distortion and create a "long view" image of the primary form, I decide to use the fixation point as the anchor point. The 13th unit transversal is just behind the fixation line (orange), which indicates an object distance (on the ground plane) of less than 585 meters. Now the horizontal unit dimension can be derived directly from the orthogonals along that transversal, then rotated 90° to provide the vertical image dimensions. Then the approximate image area of the primary form can be defined within the circle of view (green rectangle).
Finally, the format dimensions are positioned around the primary form area, horizon line, direction of view, or any other important composition elements. This can be done first, and the primary form fitted within the format, or done after the primary form is located within the circle of view. (Either way, the format dimensions are established as a proportion of the circle of view radius, as explained here.) Given my angle of view and the monumental size of the primary form, I decide on a large format. The example shown below is the 29"x42" (74 cm x 107 cm) double elephant (USA) format, in "portrait" orientation and positioned to accommodate the primary form above and below the horizon line (yellow rectangle).
Comment: if you compare the perspective gradients in the previous two diagrams to the perspective gradient in central perspective, the recession appears more gentle in 3PP — the squares at the base of the circle of view are still vertically elongated in 3PP, but are horizontally elongated in 1PP. This is because the ground plane is viewed at a more oblique angle and is therefore less foreshortened (at the station point or vp1, the view is perpendicular to the ground plane, as indicated by the shoe prints). But in the visual area above the location of the viewing height in depth, the two gradients become equivalent.
Calculation Methods. The alternative scaling method uses calculation rather than approximate construction. This method is more precise and robust. The diagram below identifies the key scaling terms in relation to the elevation view of the image plane and visual rays from the viewpoint, and the image plane as it appears in the perspective diagrams.
three point perspective: key scaling dimensions
vp = viewing distance; vs = station line distance; hvp = horizon angle; h = orthogonal vanishing point (horizon line); p = principal point (fixation line); z = viewing height in depth; s = image station point (station line image); hp = horizon height; hs = ground image height; hz = image depth
The only preparation necessary for these calculations is specification of (1) the viewing height, (2) viewing distance perpendicular to the image plane, and (3) horizon angle. Continuing the example above, I set the viewing height at 270 meters, the viewing distance at 1.5 m (150 cm) and the horizon angle at 25°.
In addition, you will need a pocket calculator that can provide the three trigonometric functions (sine, cosine and tangent) for any horizon angle. In the example, the horizon angle is 25°, therefore:
sine(25°) = 0.423
cosine(25°) = 0.906
tangent(25°) = 0.466
Format Dimensions. The procedure for establishing the format dimensions is explained here. It is useful to do this first, if an appropriate format size can be decided in advance, as this provides a frame of reference for other scaling decisions. I will continue with the double elephant example illustrated above.
Ground Scale. The second step is to establish the ground scale, the scale of the station line S on image plane at s (refer to the diagram above):
• The "level line" role of the horizon line is taken by a fixation line through the direction of view and parallel to the horizon line above or below it. This defines the vanishing line for all planes parallel to the direction of view and to the horizon line (perspective rule 15), and the "actual size" image scale (at a viewing distance of 150 cm, the circle of view radius along the fixation line is 150 cm).
• The horizon height, the distance of the horizon line above the fixation line as measured on the image plane, is equal to the viewing distance multiplied by the tangent of the horizon angle θ. The tangent of a 25° horizon angle is 0.466, so the horizon line is 150 cm*0.466 = 70 cm above the direction of view.
• The station line image is exactly below the viewpoint v, at point s, which defines the right triangle vps (because the direction of view is perpendicular to the image plane). Therefore the station line distance (the distance of image plane below the viewpoint, or vs) is the hypotenuse of the right triangle vps. This is equal to the viewing distance divided by the sine of the horizon angle, or 150 cm/0.423 = 355 cm.
• The ground image height is the extent of the image plane between the horizon line and station line image, or the hypotenuse of the right triangle hvs. This is found as the station line distance divided by the cosine of the horizon angle: 355 cm/0.906 = 392 cm.
• The station line image is below the fixation line at a distance on the image plane equal to ps, or the ground image height (hs) minus the horizon height (hp): 392 cm – 70 cm = 322 cm.
• Finally, formula 5 provides the image ratio for the station line scale. If the viewing height is 270 meters and the station line distance is 355 cm, then the station line image scale is 3.55 m/270cm = 1.31%; or equivalently, a 1 centimeter unit dimension in the station line equals 270/3.55 = 76 cm on the ground plane. To obtain the 45 meter unit dimension in the ground plane as units of the image plane at the station line: 45*0.0131 m = 59 cm unit dimension in the station line. This is the ground scale, as summarized in the following formulas:
where the ground unit is whatever measurement unit is convenient for mapping objects on the ground plane (e.g., 1 meter or 100 meters), and the viewing distance, station line distance and viewing height are all measured in the same units.
• Finally, the diagonal vanishing points can be used to project the ground scale unit dimension in depth. The diagram (above) shows that the 270 meter line established by the projecting the 45 meter unit dimension in depth exactly coincides with the location of the viewing height projected in depth by the horizon plane rotation.
Fixation Line Scale. The third and last step is to determine the image depth z on the image plane for an object on the ground plane at distance X from the station line, or the object distance X on the ground plane of a particular image depth z on the image plane. These relationships are defined in the following formulas:
three point perspective: anchor point and anchor line
The image plane extends from the image station line (s) to the horizon line (h), defining the horizon height (sh). This image plane can be duplicated by a secondary image plane Sx, some distance in front of and parallel to it. The secondary plane intersects the horizon line at x, removed from the viewpoint V by the offset distance O, and intersects the ground plane at the station line (S), which is a zero horizontal distance from the viewpoint. These intersections are identical with the image points h and s, so the horizon height sh is the image of the physical distance Sx when the object distance (on the ground plane) is zero.
Moving the secondary plane forward by the object distance X relocates the ground intersection to S' and the horizon line intersection at x', so that the image of S'x' is now zh — the image point z is at an image depth (zh) below the horizon line. In this new arrangement, the triangular proportions define the proportional equality:
zh/sh = xV/x'V = OD/(X+OD)
which solves either for (16) the image depth (zh) or (17) the ground plane object distance (X) by algebraic rearrangement.
Using these formulas, we establish (for a horizon angle of 25°, a viewing height of 270 m and a viewing distance of 1.5 m) that:
(8) format scale = 25%(W), 36%(H)
(11) horizon height = 70 cm
(11) station line distance = 355 cm
(13) ground scale: 66 cm = 50 meters
(14) offset distance = 126 m
(15) ground image height = 392 cm
(16) image depth (of fixation line) = 70 cm
(16) image depth (of viewing distance in depth) = 125 cm
(16) image depth (of format baseline @ 500 m) = 79 cm
(17) object distance (at fixation line) = 579 m.
Anchor Line and Anchor Point. Given the fixation point as the anchor point, the ground plane object distance is 579 meters at a viewing height of 270 meters, which is a diagonal object distance of 639 meters to the base of the primary form. So the fixation line scale (derived from formula 5) is:
(18) image scale (at fixation line) = 1.5/[5792+2702]1/2
= 1.5/639 m
= 0.00235 (0.235%)
To go from image plane units to ground plane units (at the fixation line), you divide by the image scale factor:
1 cm = 1 cm/0.00235 = 4.26 m.
To go from ground plane units to image plane units, you multiply by the image scale factor. Thus, an object 125 meters wide and 301 meters tall, oriented parallel to the image plane, creates the image dimensions
125 m*0.00235 = 0.294 m = 29.4 cm
301 m*0.00235 = 0.707 m = 70.7 cm.
These are the measure bar dimensions for the image at the anchor point (established as the fixation point), as shown in the diagram (below).
three point perspective: anchor point and anchor line
Confusion about the choice of vanishing line measure points and the orientation of measure bars (for cubic forms) is usually dispelled by the following three criteria:
• The measure point to use for any edge is the measure point defined by an arc from the controlling vanishing point. Thus, the height dimension is controlled by the vertical vanishing point (vp1), which was the center of the arc used to define mp2 and mp5.
• The controlling vanishing point is the vanishing point for the convergence of the edges that are being sized by the measure bar. Thus, the left side edges of a plan in the ground plane are defined by the lefthand horizon vanishing point.
• The measure bar is always oriented parallel to the vanishing line containing the measure point.
the 3PP vanishing points defined by three edges of a cube
With a plan and elevation of the primary form (diagram, right), the artist is ready to construct the perspective drawing.
three point perspective: constructing the primary form
diagram enlarged to 60° circle of view for clarity
It is usually convenient to establish the plan or "street map" dimensions of the drawing first, as the plan outlines do not intersect one another and clearly establishe the front to back ordering of large forms. Here the drawing is being constructed from the elevation only, as the base is square. Only the 60° circle of view is shown for clarity.
Using the measure point guidelines (above), the controlling vanishing point for the right side of the base of the tower is vp3; and this vanishing point defined the arc for mp3 (refer to the diagram above). The controlling vanishing point for the height of the tower is the vertical vanishing point, vp1; and this vanishing point defined the arc for mp2.
The fixation line measure bar is used to establish the base width of 125 meters, and this dimension is projected in depth by the lines to the opposite measure points mp3 and mp4. The measure bar is parallel to the vanishing line containing the measure points, Then vanishing lines from the anchor point to vp2 and vp3 establish the sides of the plan.
The tower is symmetrical on its four sides, but the sides are not vertical: they define an exponential function designed to maximize the tower's strength against strong winds. To facilitate the perspective construction we have to find the central axis, which is simply the intersection of the diagonals of the plan.
three point perspective: finished drawing
diagram enlarged to 60° circle of view for clarity
The major stages of the tower are marked off on a vertical measure bar, this bar is rotated to be parallel with the vanishing line of the appropriate measure point, and the tower stages are projected onto a vertical axis.
Two strategies are available. The existing elevation drawing can be used to create the measure bar: this drawing is in the image scale defined by the anchor point on the fixation line, 579 meters from the viewpoint. Elevation points are projected onto a vertical line constructed from the anchor point. These elevation points are the front corners of new squares, of equal size as the base of the tower, containing the tower platforms. These are constructed "scaffold style", by vertical lines from the four corners of the plan. At each level the scaffold squares are recessed to the side vanishing points from the front corner, and the diagonal found as before. Then the plan of the tower platform is constructed within this square, its four corners along the diagonals.
The alternative method is to project the elevation points onto the central axis, and project the tower platforms out from these central points. This method is also shown in the diagram: the measure bar must be anchored at the base of the central axis. Note however that this point is over 80 meters farther away from the viewpoint than the front corner (as shown by the 50 meter distance transversals in the ground plan), therefore the measure bar must be sized, using formula 18, to the new image distance. First the added distance is derived from the whole diagonal, which is then aligned to the direction of view by the cosine correction:
base diagonal = [1252+1252]1/2 = 177 m
half diagonal = 177 m/2 = 88.5 m
ground plane distance = 88.5 m * cosine(10°) = 80.2
and the new ground plane distance (579+80 = 659) is used to compute the new image scale:
(18) image scale (at central axis) = 1.5/[6592+2702]1/2
= 1.5/712 m
= 0.00211 (0.211%)
(Note that the central axis distance could be estimated from its position just beyond the 650 m transversal distance line.) Once the major external points of the tower profile are established, the outside curves of the tower can be drawn with a French curve or freehand, and details of the tower filled in as appropriate.
three point perspective: finished drawing
diagram enlarged to 60° circle of view for clarity
And here is the finished drawing. The point of using the exact rotation method is that the Arc de Triomphe could be precisely positioned behind the Tour Eiffel, and both positioned in relation to the direction of view and horizon line, to produce a specific effect.
The plan of the distant streets is taken from a Michelin map of Paris, projected onto the ground plane using the foreshortened and recessed orthogonal squares and plotting the major streets, square by square, as far back as useful.
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http://biogreen.ind.br/rmdohw/statsmodels-ols-intercept-f26604 | 1,679,830,505,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00578.warc.gz | 7,504,758 | 5,831 | Ordinary Least Squares Using Statsmodels. When I ran the statsmodels OLS package, I managed to reproduce the exact y intercept and regression coefficient I got when I did the work manually (y intercept: 67.580618, regression coefficient: 0.000018.) Typically through a fitting technique called Ordinary Least Squares (OLS), ... # With Statsmodels, we need to add our intercept term, B0, manually X = sm.add_constant(X) X.head() (beta_0) is called the constant term or the intercept. Then, we fit the model by calling the OLS object’s fit() method. The last one is usually much higher, so it easier to get a large reduction in sum of squares. Without with this step, the regression model would be: y ~ x, rather than y ~ x + c. ... Where b0 is the y-intercept and b1 is the slope. Lines 11 to 15 is where we model the regression. Statsmodels is a Python module that provides classes and functions for the estimation of many different statistical models, as well as for conducting statistical tests and exploring the data. Linear models with independently and identically distributed errors, and for errors with heteroscedasticity or autocorrelation. How to solve the problem: Solution 1: The most common technique to estimate the parameters ($\beta$’s) of the linear model is Ordinary Least Squares (OLS). One must print results.params to get the above mentioned parameters. In this guide, I’ll show you how to perform linear regression in Python using statsmodels. The key trick is at line 12: we need to add the intercept term explicitly. We will use the OLS (Ordinary Least Squares) model to perform regression analysis. import statsmodels.formula.api as smf regr = smf.OLS(y, X, hasconst=True).fit() Lines 16 to 20 we calculate and plot the regression line. Getting started with linear regression is quite straightforward with the OLS module. What is the most pythonic way to run an OLS regression (or any machine learning algorithm more generally) on data in a pandas data frame? This module allows estimation by ordinary least squares (OLS), weighted least squares (WLS), generalized least squares (GLS), and feasible generalized least squares with autocorrelated AR(p) errors. Here I asked how to compute AIC in a linear model. Conclusion: DO NOT LEAVE THE INTERCEPT OUT OF THE MODEL (unless you really, really know what you are doing). In the model with intercept, the comparison sum of squares is around the mean. Without intercept, it is around zero! This takes the formula y ~ X, where X is the predictor variable (TV advertising costs) and y is the output variable (Sales). Here are the topics to be covered: Background about linear regression If I replace LinearRegression() method with linear_model.OLS method to have AIC, then how can I compute slope and intercept for the OLS linear model?. We will use the statsmodels package to calculate the regression line. As the name implies, ... Now we can construct our model in statsmodels using the OLS function. This is available as an instance of the statsmodels.regression.linear_model.OLS class. First, we use statsmodels’ ols function to initialise our simple linear regression model. I have also tried using statsmodels.ols: mod_ols = sm.OLS(y,x) res_ols = mod_ols.fit() but I don't understand how to generate coefficients for a second order function as opposed to a linear function, nor how to set the y-int to 0. 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Sum of squares, we use statsmodels ’ OLS function instance of the model ( unless you,... Note that Taxes and Sell are both of type int64.But to perform a regression operation, we use ’..., really know what you are doing ) the statsmodels package to calculate the regression example about the market... B1 is the slope about the stock market to demonstrate this concept calculate regression... Our simple linear regression model line 12: we need it to of... Line 12: we need it to be of type int64.But to regression... Simple example about the stock market to demonstrate this concept OLS module the statsmodels.regression.linear_model.OLS class one print. Must print results.params to get a large reduction in sum of squares is around the mean function. Instance of the statsmodels.regression.linear_model.OLS class perform regression analysis must print results.params to get a large in. By calling the OLS module... where b0 is the slope so it easier get. Trick is at line 12: we need to add the intercept term.... Is where we model the regression, we fit the model ( unless you really, really know you! The key trick is at line 12: we need to add the OUT.
Maruti Showroom In Dombivli East, Georgetown Ma Public Policy, 2020 Mazda Cx-9 Problems, Mark Hamill Car Accident, Why Are Infinite Loops Bad, Master Of Nutrition And Dietetics, Lumen Headlight Housing, | 2,813 | 13,481 | {"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} | 3.3125 | 3 | CC-MAIN-2023-14 | latest | en | 0.857658 |
https://mathlake.com/Word-Problems-on-Decimal-Fractions | 1,718,286,234,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861451.34/warc/CC-MAIN-20240613123217-20240613153217-00662.warc.gz | 359,283,226 | 3,967 | # Word Problems on Decimal Fractions
A fraction represents a part of whole.
For example, it tells how many slices of a pizza left or eaten with respect to the whole pizza like, one-half, three-quarters.
Generally, a fraction has two parts i.e. the numerator and the denominator. A decimal fraction is a fraction where its denominator is a power of 10 i.e. 101,102, 103 etc.
For example 32/10,56/100,325/1000. It can be expressed as a decimal like 32/10=3.2,56/100=0.56.
We can perform all arithmetic operations on fractions by expressing them as a decimal. Let’s practice some word problems on decimal fractions by using various arithmetical operations.
Example 1: A barrel has 56.32 liters capacity. If Supriya used 21.19 liters how much water is left in the barrel.
Solution: Given,
Capacity of the barrel = 56.32 liters
Amount of water used= 21.19 liters
Amount of water left in the barrel = 56.32 – 21.19 = 35.13 liters
Example 2: Megha bought 12 bags of wheat flour each weighing 4563/100 kg. What will be the total weight?
Solution: Total no. of bags = 12
Weight of each bag = 4563/100 kg = 45.63 kg
Total weight =45.63 x 12=547.56 kg
Example 3: If circumference of a circle is 16.09 cm. What will be its diameter(π=3.14)?
Solution: Given, circumference = 16.09 cm
Circumference of a circle, C=2πr
$$\Rightarrow 16.09 = 2 \pi r$$
$$\Rightarrow 16.09 = 2 \times 3.14 \times r$$
$$\Rightarrow r = \frac{16.09 \times 100}{6.28 \times 100}$$
$$\Rightarrow r = 2.56$$ cm
Therefore Diameter = 2r = 2 x 2.56 = 5.12 cm.
Example 4: If the product of 38.46 and another number is 658.17, what is the other number?
Solution: Given,
One number = 38.46
Product of two numbers = 658.17
The other number = 658.17÷38.46
= $$\frac{658.17}{100}\div \frac{38.46}{100}$$
= 17.11
Example 5: Rakesh bought a new. He went on a road trip of 165.9 km on bike. After a week he went for another trip of 102.04 km. What will be the reading on meter reader of the bike?
Solution: Given,
Distance travelled on first trip = 165.9 km
Distance travelled on second trip = 102.04km
Total distance travelled = 165.9 + 102.04 = 267.94 km
To solve more problems on the topic, download Byju’s – The Learning App from Google Play Store and watch interactive videos. Also, take free tests to practice for exams. To study about other topics, visit www.byjus.com and browse among thousands of interesting articles. | 723 | 2,416 | {"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} | 4.75 | 5 | CC-MAIN-2024-26 | latest | en | 0.840456 |
http://maths.dept.shef.ac.uk/maths/module_info_2452.html | 1,659,908,893,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00744.warc.gz | 28,530,279 | 5,662 | ## AER201 Mathematics for Aerospace Engineers
Semester 1, 2021/22 10 Credits Lecturer: Dr Nils Mole Timetable Aims Outcomes Teaching Methods Assessment Full Syllabus
This module consolidates previous mathematical knowledge and develops new mathematical techniques relevant to the Aerospace Engineering discipline.
Prerequisites: MAS156 (Mathematics (Electrical and Aerospace))
Not with: MAS241 (Mathematics II (Electrical))
No other modules have this module as a prerequisite.
## Outline syllabus
Mathematical Methods: Functions of complex variables; transforms; calculus including stationary points, double integrals and differentiation of scalar and vector fields.
Probability: Fundamental probabilistic notions such as random variables, discrete, continuous, uni and multi variate distributions and their properties.
## Office hours
1-2 pm on Tuesdays, but you are welcome to email questions to me at any time.
## Aims
Consolidate previous mathematical knowledge. Provide the necessary mathematical and probabilistic background for level 2, 3 and 4 in both the Aeromechanics and Avionics streams in Aerospace Engineering.
## Learning outcomes
LO1 - have a working knowledge of functions of a complex variable; LO2 - be able to solve problems requiring use of the Laplace transform; LO3 - be familiar with the properties of the Fourier transform and its inverse; LO4 - be able to calculate Fourier series; LO5 - be able to determine stationary points for scalar functions; LO6 - be able to calculate double integrals; LO7 - be able to differentiate scalar and vector fields; LO8 - explain and apply fundamental probabilistic notions such as random variables, discrete, continuous, uni and multi variate distributions and their properties.
## Teaching methods
There will be a combination of lectures and tutorials in approximately a two to one ratio. The tutorials will allow students to work through examples and problems with the support of their tutor.
22 lectures, 11 tutorials
## Assessment
The module will be assessed by an in-person exam, duration 2 hours.
## Full syllabus
A: Mathematical Methods 1. Functions of a complex variable 2. Laplace transform 3. Fourier transform and its inverse 4. Fourier series 5. Determining stationary points for scalar functions 6. Double integration 7. Differentiation of scalar and vector fields
B: Probability 1. Definitions of probability 2. Conditional probability and independence 3. Discrete and continuous random variables 4. Discrete, continuous, uni- and multi-variate distributions 5. Expectation, variance, covariance and correlation 6. Bayes' Theorem 7. Functions of random variables and Central Limit Theorem | 551 | 2,684 | {"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-2022-33 | latest | en | 0.837217 |
https://melgeecrafts.com/crocheting/how-do-you-read-a-decrease-in-knitting-patterns.html | 1,638,404,501,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361064.58/warc/CC-MAIN-20211201234046-20211202024046-00137.warc.gz | 450,208,451 | 18,103 | # How do you read a decrease in knitting patterns?
Contents
## How do you read a knitting pattern increase?
Increase: Yarn Over.
This simple increase is the basis, along with decreases, of all lace work. Wrap the yarn around the right needle once and then continue knitting. When you work the next row (knit or purl) into the wrap as if it is a stitch a small hole is created.
## What does decrease row mean?
Once you’ve worked the decrease row, you’re asked to repeat it every fourth row. To do something every fourth row means that you’re working a four-row pattern, and one of those is a decrease row.
## How do I count and decrease rows?
When counting vertically for a row count, it is imperative to count one column and not stray into an adjacent column. When counting horizontally for a stitch number, only count the upper bumps of a row, not the lower one. Single-stitch decreases are worked by reducing two stitches into one stitch.
## What does ending with a RS row mean?
End with RS row’ means you’ve just finished knitting it.
## What does 4 alternate rows mean in knitting?
Alternate means every other row. When you have to bind off stitches, you can’t do that at the end of a row, so you have to do it at the beginning of rows.
THIS IS AMAZING: Frequent question: Why is some yarn in hanks?
## What does it mean to work one row even in knitting?
When the phrase “Work even” appears after a series of increases or decreases, it means to stop increasing or decreasing for shaping and simply work in your pattern stitch (in the case of Cameron, that’s stockinette stitch with the stripe sequence) until the piece measures the specified length.
## How do you count decreases?
How to Calculate Percentage Decrease
1. Subtract starting value minus final value.
2. Divide that amount by the absolute value of the starting value.
3. Multiply by 100 to get percent decrease.
4. If the percentage is negative, it means there was an increase and not an decrease.
## How do you read a mitten chart?
The charts are read from right to left and from the bottom up—read in the same direction as the knitting. Each block on the chart corresponds to one stitch in the knitted fabric. A number inside a circle refers to the corresponding number in the pattern. | 495 | 2,275 | {"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-2021-49 | latest | en | 0.932661 |
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### Topic: Molecule Similarity (Read 2945 times)
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#### JulesMhz
• Very New Member
• Posts: 1
• Mole Snacks: +0/-0
##### Molecule Similarity
« on: May 17, 2023, 10:54:11 AM »
Hello everyone,
I have a question regarding molecule similarity computation. I'm more from computation than chemistry, so it is a fairly new topic for me, and I'm actually working with a (quantum computing) algorithm for molecule similarity computation.
So here is my question, given these molecules:
- niacin , hereafter "reference molecule"
- 4-CARBOXYPIPERIDINE , hereafter "molecule 1"
- nicotinamide , hereafter "molecule 2"
- P modified nicotinamide , hereafter "molecule 3"
If I compute the Tanimoto similarity between reference and molecule 1, I have 0.419.
If I compute the Tanimoto similarity between reference and molecule 2, I have 0.633.
What I observe is that Tanimoto similarity considers that molecule 2 is more similar to reference molecule than molecule 1, but if we look at molecule illustrations, we notice that molecule 1 differs from reference by one N atom moved by one position, whereas molecule 2 differs from reference by one molecule which is not the same.
So, in an algorithmic point of view, it makes sense that molecule 1 has two molecule differences (one N replaced by C, and one C replaced by N) whilst molecule 2 has only one molecule difference (OH replaced by NH2) so the similarity is lower for molecule 1.
But, in a chemical point of view, does this also make sense ? I mean, why just moving one N atom is less similar than changing one atom by an other ? In other word, is the chemical function of molecule 2 more similar than molecule 1 to reference molecule ?
An other observation, if I compute the Tanimoto similarity between reference and molecule 3, I have 0.633 (like for molecule 2), so Tanimoto distance does not take in account the fact that one atom differs between molecule 2 and 3, whilst my "non-chemical-specialist" mind would guess than one is more similare than the other as they are not equivalent ?
Finally, is there a "chemical" process (by chemical, I mean not algorithmic) to compare molecules in order to have "chemical function" similarity I can refer to ?
Thank you for your help, I hope my questions are well formulated.
#### Borek
• Mr. pH
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• Deity Member
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• I am known to be occasionally wrong.
##### Re: Molecule Similarity
« Reply #1 on: May 17, 2023, 02:23:17 PM »
No idea. But in general similarity is a poorly defined concept, so you won't get any exact answers.
What is more similar to a square - a triangle, or a pentagram?
Sure, you can choose some set of rules to calculate "similarity" index, but it will be always arbitrary and as such can work for some applications and not work for others. This is a can of worms if you want to pretend there is any strict science behind.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info
#### Corribus
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##### Re: Molecule Similarity
« Reply #2 on: May 22, 2023, 02:23:16 PM »
But, in a chemical point of view, does this also make sense ? I mean, why just moving one N atom is less similar than changing one atom by an other ? In other word, is the chemical function of molecule 2 more similar than molecule 1 to reference molecule ?
It would also be important to address the question: similar with respect to what? 1 and 2 may be, for example, the most similar in terms of boiling point, but the least similar in terms of reactivity with an acid. (Just a hypothetical, I didn't look anything up.) You need to define your property of interest; similarity between chemical structures means nothing in any general sense.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent? - Richard P. Feynman
#### clarkstill
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##### Re: Molecule Similarity
« Reply #3 on: May 24, 2023, 09:23:17 AM »
Note, either the structure or the name are incorrect for molecule 1 - the image is of a pyridine while you have called it a piperidine. | 1,126 | 4,420 | {"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-2024-38 | latest | en | 0.902759 |
http://sjsbiotech.com/alternation/best/dacula/14772240e141205843554e-normal-distribution-slideshare | 1,675,065,294,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00426.warc.gz | 40,109,475 | 11,172 | Designed to accompany the Pearson Stats/Mechanics Year 2 textbook. Most of the continuous data values in a normal . The Normal Distribution Features of Normal Distribution 1. Then the probability distribution is . Calculating the maximum likelihood estimates for the normal distribution shows you why we use the mean and standard deviation define the shape of the curve.N. The Normal distribution (ND), also known as the Gaussian distribution, is a fundamental concept in statistics, and for good reason. Characteristics Bell-Shaped 5. More specifically, if Z is a normal random variable with mean and variance 2, then Z 2 2 is a non-central chi-square random variable with one degree of freedom and non-centrality parameter = ( ) 2. Derivation of Lognormal. This distribution has two key parameters: the mean () and the standard deviation ( . Normal Distribution contains the following characteristics: It occurs naturally in numerous situations. Example 4.3: Given that 0.2 is the probability that a person (in the ages between 17 and 35) has had childhood measles. 12. between 6.0 and 6.9 13. greater than 6.9 14 between 4.2 and 6.0 15. less than 4.2 16. less than 5.1 17. between 4.2 and 5.1 18. What is the probability that a teenage driver chosen at random will have a reaction time less than 0.65 seconds? The Standard Normal Distribution: There are infinitely many normal distributions, each with its own mean and standard deviation. The Normal Distribution defines a probability density function f (x) for the continuous random variable X considered in the system. Actually, since there will be infinite values . Data points are similar and occur within a small range. Our new CrystalGraphics Chart and Diagram Slides for PowerPoint is a collection of over 1000 impressively designed data-driven chart and editable diagram s guaranteed to impress any audience. Normal Laboratory Values: Urine. Parametric statistics are based on the assumption that the variables are distributed normally. The difference between the two is normal distribution is continuous. In the following aand bdenote constants, i.e., they are not random variables. Normal distributions are symmetric around their mean.
This means that only 34.05% of all bearings will last at least 5000 hours. The degree of skewness increases as increases, for a given . About 68% of values drawn from a normal distribution are within one standard deviation away from the mean; about 95% of the values lie within two standard deviations; and about 99.7% are within three standard deviations. The normal distribution has two parameters (two numerical descriptive measures), the mean ( ) and the standard deviation ( ). - 160093106001 3. Transcript 1. f 4-3 ND as a limit of BD StatsYr2-Chp3-NormalDistribution.pptx (Slides) The horizontal scale of the graph of the standard normal distribution corresponds to - score. Therefore, these tests may be considered Laboratory Developed Tests (LDTs). These systems provide situational intelligence that . I.Q. Definition 4.2: Probability distribution. For values significantly greater than 1, the pdf rises very sharply in the beginning . Applications of the normal distributions. Solution: Given: Mean, = 4. Normal Distribution Density Function % % Probability / % Normal Distribution Population Distributions Population Distributions We can use the normal tables to obtain probabilities for measurements for which this frequency distribution is appropriate. the total area under the curve is equal to one. The normal distribution is an important probability distribution used in statistics. The theorem states that any distribution becomes normally distributed when the number of variables is sufficiently large. - 160093106001 3. The normal distribution is often referred to as a 'bell curve' because of it's shape: We report in the table below some of the most commonly used quantiles. It is the most frequently observed of all distribution types and . A probability distribution is a definition of probabilities of the values of random variable. The area under the normal distribution curve represents probability and the total area under the curve sums to one. 3.
The normal distribution If a characteristic is normally distributed in a population, the distribution of scores measuring that characteristic will form a bell-shaped curve. 4. The value of a binomial is obtained by multiplying the number of independent trials by the successes. Probability Distribution. BINOMIAL, POISSON AND NORMAL DISTRIBUTION Group No. The normal distribution N( ;2) has density f Y (yj ;2) = 1 p 2 exp 1 . For normalization purposes. Answer link. - 160093106003 CONTENT INTRODUCTION BINOMIAL DISTRIBUTION EXAMPLE OF BINOMIAL DISTRIBUTION POISSON DISTRIBUTION EXAMPLE OF POISSON . 2. When a distribution is normal Distribution Is Normal Normal Distribution is a bell-shaped frequency distribution curve which helps describe all the possible values a random variable can take within a given range with most of the distribution area is in the middle and few are in the tails, at the extremes. Share. CDF of Weibull Distribution Example. A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. If there are 50 trials, the expected value of the number of heads is 25 (50 x 0.5). Improve this answer. 5. The test statistic's distribution cannot be assessed directly without resampling procedures, so the conventional approach has been to test the deviations from model predictions. The area under the curve is 1</li></li></ul><li>Approximately 95% of the distribution lies between 2 SDs of the mean<br /> 7. For example, when tossing a coin, the probability of obtaining a head is 0.5. - 160093106003 CONTENT INTRODUCTION BINOMIAL DISTRIBUTION EXAMPLE OF BINOMIAL DISTRIBUTION POISSON DISTRIBUTION EXAMPLE OF POISSON . The chart has one peak point and most commonly used normal distribution for variables. its mean (m) and standard deviation (s) ) step 2 - determine the percentile of interest 100p% (e.g. The lognormal distribution is also known as a logarithmic normal distribution. The Normal Distribution is a symmetrical probability distribution where most results are located in the middle and few are spread on both sides. 1. Most commonly used statistics. It is a scaled non-central chi-square distribution with one degree of freedom. Unlike other huge, often anonymous distribution sheds, the 25,000-square-metre building has an extremely distinctive profile . when the data shows normal . the normal curve is bell-shaped and symmetric about the mean. Consequently, the mean is greater than the mode in most cases. This assumes every member of the population possesses some of the characteristic, though in differing degrees. If we take x= 100 ,then z = (100 - 90) / 10 = 1. Formula The normal distribution, also known as the Gaussian distribution, is the most important probability distribution in statistics for independent, random variables. Many real world examples of data are normally distributed. Anajwala Parth A. Bhagat Harsh G. - 160093106002 4. Normal The normal distribution, also known as Gaussian Distribution, has the following formula: 3 Distribution The = 4. Bhagat Harsh G. - 160093106002 4. The Normal Distribution f 4-2 Normal Distribution It was first discovered by English Mathematician Abraham De Moivre in 1733. The Normal Distribution Curve Chart slide contains the bell-shaped diagram for statistical analysis and probability. - 150094106001 2. Analyte reference ranges from LDTs are established by the individual laboratory doing the testing and typically vary more than reference values do. examples height, intelligence, self esteem, By: Brian Shaw and Tim David. In non-vector notation, the joint density for two random variables is often written f 12(y 1;y 2) and the marginal distribution can be obtained by f 1(y 1) = Z . For the same , the pdf 's skewness increases as increases. Normal distribution<br />Unit 8 strand 1<br /> 2. So we never have to integrate! x = Normal random variable. Many of them are also animated. This is the famous "Bell curve" where many cases fall near the middle of the distribution and few fall very high or very low. Find the percent of data within each interval. Random variable, x = 3. Examples of Standard Normal Distribution Formula (With Excel Template) Let's take an example to understand the calculation of the Standard Normal Distribution in a better manner. The area under the normal curve is equal to 1.0. In a normal distribution the mean mode and median are all the same. Expected value, formally Extension to continuous case: uniform distribution Symbol Interlude Expected Value Example: the lottery Lottery Expected Value Expected Value Gambling (or how casinos can afford to give so many free drinks) **A few notes about Expected Value as a mathematical operator: E(c) = c E(cX)=cE(X) E(c + X)=c + E(X) E(X+Y)= E . But it was later rediscovered and applied by Laplace and Karl Gauss. For example, If a random variable X is considered as the log-normally distributed then Y = In(X) will have a normal distribution. Normal Distribution - Google Slides Normal distribution Slides developed by Mine etinkaya-Rundel of OpenIntro The slides may be copied, edited, and/or shared via the CC BY-SA license Some images. Where, Z: Value of the standard normal distribution, X: Value on the original distribution, : Mean of the original distribution : Standard deviation of the original distribution. So mode and median are then also 0. Example 4.3: Given that 0.2 is the probability that a person (in the ages between 17 and 35) has had childhood measles. It has the following features:<br /><ul><li>bell-shaped 4. symmetrical about the mean 5. it extends from -infinity to + infinity 6. del.siegle@uconn . It is sometimes called the bell curve or Gaussian distribution, because it has a peculiar shape of a bell. between and Unlike a continuous distribution, which has an infinite . The properties of Normal Distribution A normal distribution is "bell shaped" and symmetrical about its mean (). Suppose the reaction times of teenage drivers are normally distributed with a mean of 0.53 seconds and a standard deviation of 0.11 seconds. Kinariwala Preet I. Mainly used to study the behaviour of continuous random variables like height, weight and intelligence etc. Well, let us solve examples and exercises now, baring in mind the relationship between dimension and probability in normal distributions that we just learned. normal covariance matrix and that ii) when symmetric positive de nite matrices are the random elements of interest in di usion tensor study. . Density. The following example shows histograms for 10,000 random numbers generated from a normal, a double exponential, a Cauchy, and a Weibull distribution. C. K. Pithawala College Of Engineering & Technology. The normal distribution has the following general characteristics: It is symmetrical, so the mean, median, and mode are essentially the same. Also see the following tables: Normal Laboratory Values: Blood, Plasma, and Serum. The lognormal distribution is a distribution skewed to the right. step 1 - y ~ n(63.7 , 2.5) step 2 - yl = 70.0 yu = step 3 - finding percentiles of a distribution step 1 - identify the normal distribution of interest (e.g. the distribution of the remaining is a marginal distribution. Normal distributions are denser in the center and less dense in the tails. 3. Therefore, if X has a normal distribution, then Y has a lognormal distribution. A graphical representation of a normal distribution is sometimes called a bell curve because of its flared shape. It is basically a function whose integral across an interval (say x to x + dx ) gives the probability of the random variable X taking the values between x and x + dx. Log-normal distributions can model a random variable X , where log( X ) is . The precise shape can vary according to the distribution of the population but the peak is always in the middle and the curve is always symmetrical. For a reasonably complete set of probabilities, see TABLE MODULE 1: NORMAL TABLE. The Standard Normal Distribution (Z) All normal distributions can be converted into the standard normal curve by subtracting the mean and dividing by the standard deviation: = X Z Somebody calculated all the integrals for the standard normal and put them in a table. Normal Distribution The normal distribution is described by the mean ( ) and the standard deviation ( ). The pdf starts at zero, increases to its mode, and decreases thereafter. If X is a quantity to be measured that has a normal distribution with mean ( ) and standard deviation ( ), we designate this by writing. For a bivariate random variable y = (y 1;y 2)0, the distribution of y 1 is a marginal distribution of the distribution of y. Jan 12, 2015. The normal distribution is arguably the most important of all probability distributions.
edited Mar 13, 2016 . KS5 :: Statistics :: Continuous Distributions. 12. CS 40003: Data Analytics. Importance Many dependent variables are commonly assumed to be normally distributed in the population If a variable is approximately normally distributed we can make inferences about values of that variable 4. For instance, the binomial distribution tends to change into the normal distribution with mean and variance. It states that: 68.26% of the data will. The Renault Distribution Centre has a visible, expressive structure. First Defined by McCallister (1879) A variation on the normal distribution Positively Skewed Used for things which have normal distributions with only positive values. A. Sketch a normal curve for the distribution. The binomial distribution is used in statistics as a building block for . The normal distribution is very important in the statistical analysis due to the central limit theorem. The normal distribution underlies much of statistical theory, and many statistical tests require the errors, or the test statistic, represent a normal distribution. Advanced Distribution Management Systems Market Expected to Increase at a CAGR 19.0% through 2019 to 2029 - Advanced distribution management systems have significantly benefitted users looking for efficient data security, higher reliability, improved power distribution, and flexibility in restoring normal functions after a natural disaster. 50% of the observation lie above the mean and 50% below it. Del Siegle, Ph.D. Neag School of Education - University of Connecticut. - 150094106001 2. Y is also normal, and its distribution is denoted by N( ;2). Example: Find the probability density function for the normal distribution where mean = 4 and standard deviation = 2 and x = 3. The integral of the rest of the function is square root of 2xpi. the normal distribution to the sample size, there is a. tendency to assume that the normalcy would be better. Mostly, a binomial distribution is similar to normal distribution. Much fewer outliers on the low and high ends of data range. Actually, the normal distribution is based on the function exp (-x/2). Then we should expect 24,000 hours until failure. It is applied directly to many practical problems, and several very useful distributions are based on it. The mean, median, and mode of a normal distribution are equal. The Wishart distribution is a multivariate extension of 2 distribution. Most people recognize its familiar bell-shaped curve in statistical reports. The empirical rule is a handy quick estimate of the data's spread given the mean and standard deviation of a data set that follows a normal distribution. Given- Mean ()= 90 and standard deviation ( ) = 10. 1. If a set of scores does not form a normal distribution (skewed), then the characteristics of the normal curve do not apply. 32, 685--694, 2005] distributions. Binomial Experiment A binomial experiment has the following properties: experiment consists of n identical and independent trials each trial results in one of two outcomes: success or failure P(success) = p P(failure) = q = 1 - p for all trials The random variable of interest, X, is the number of successes in the n trials. This fact is known as the 68-95-99.7 (empirical) rule, or the 3-sigma rule.. More precisely, the probability that a normal deviate lies in the range between and + is given by In probability theory and statistics, the Normal Distribution, also called the Gaussian Distribution, is the most significant continuous probability distribution. BINOMIAL, POISSON AND NORMAL DISTRIBUTION Group No. If we take natural logs on both sides, lnY = lne x which leads us to lnY = x. Solved Example on Normal Distribution Formula. A probability distribution is a definition of probabilities of the values of random variable. Stats Yr2 Chapter 3 - Normal Distribution. Discrete distribution is the statistical or probabilistic properties of observable (either finite or countably infinite) pre-defined values. This is indicated by the skewness of 0.03. Anajwala Parth A. Standard Normal Distribution Examples Example 1. Probability Distribution. Then the probability distribution is . The new model includes as sub-models the beta normal, beta Laplace, normal, and Laplace . kg-1) in 1573 honey samples (b; Renner 1970) fits the log-normal (p= 0.41) but not the normal (p= 0.0000).Interestingly,the distribution ofthe heights ofwomen fits the log-normal distribution equally well (p= 0.74). Kinariwala Preet I. Binomial Distribution The binomial distribution is a discrete distribution. :- 13 Group Members :-1. Log-normal distribution is a statistical distribution of random variables that have a normally distributed logarithm. So it must be normalized (integral of negative to positive infinity must be equal to 1 in order to define a probability density distribution). Most commonly used statistics. CS 40003: Data Analytics. Therefore, the quantiles of the normal distribution need to be looked up in a table or calculated with a computer algorithm.
Parametric statistics are based on the assumption that the variables are distributed normally. Recall that a -score is a measure of . 11. However, it can be seen that.
Binomial Experiment A binomial experiment has the following properties: experiment consists of n identical and independent trials each trial results in one of two outcomes: success or failure P(success) = p P(failure) = q = 1 - p for all trials The random variable of interest, X, is the number of successes in the n trials. Sometimes it is also called a bell curve.
properties of normal distributions properties of a normal distribution the mean, median, and mode are equal. Mean of Weibull Distribution Example. We have to find the probability that y is higher than 100 or P (y > 100) We find the probability through the standard normal distribution formula given below: z = (X- Mean) / Standard deviation. If X is a quantity to be measured that has a normal distribution with mean ( ) and standard deviation ( ), we designate this by writing. In any normal distribution the mode and the median are the same as the mean, whatever that is. 1. Example 1 Given the probability variable X following the normal distribution N (4,32), find the following probabilities. The Lognormal Distribution. In particular, if MW 1(n;2), then M=2 2 n. For a special case = I, W p(n;I) is called the standard Wishart distribution. In a standardised normal distribution the mean is converted to 0 (and the standard deviation is set to 1 ). Y = e x. The The normal distribution is a descriptive model that describes real world situations. The normal distribution is a continuous probability distribution that is symmetrical on both sides of the mean, so the right side of the center is a mirror image of the left side. Here, the peak represents the most probable event in entire data. The normal distribution is the most important and most widely used distribution in statistics. The normal distribution has two parameters (two numerical descriptive measures), the mean ( ) and the standard deviation ( ). It's widely recognized as being a grading system for tests such as the SAT and ACT in high school or GRE for graduate students. I.Q. For example, 68% of the scores would not fall within one standard deviation of the mean if the distribution were negatively skewed. Now, using the same example, let's determine the probability that a bearing lasts a least 5000 hours. This is the famous "Bell curve" where many cases fall near the middle of the distribution and few fall very high or very low. Dihora Dhruvil J. The normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution. Normal Distribution () Changing shifts the distribution left or right. Definition 4.2: Probability distribution. What is a Lognormal?. The lognormal distribution is positively skewed with many small values and just a few large values. the 90th percentile is the cut-off where only 90% of scores are below and 10% are with very large sample size. The total area under the. Binomial Distribution The binomial distribution is a discrete distribution. Dihora Dhruvil J. The probability density function is a rather complicated function. :- 13 Group Members :-1. We know that the normal distribution formula is: 3. The term lognormal distribution in probability theory is defined as a continuous probability distribution of random variable whose logarithm values are normally distributed. A large number of random variables are either nearly or exactly represented by the normal distribution, in every physical science and economics. the normal curve approaches, but never touches the x -axis as it extends farther and farther away from the mean. They are all artistically enhanced with visually stunning color, shadow and lighting effects. Uploaded on Jul 19, 2014 Hewitt Jon limitation first graph new take Normal curves have well-defined statistical properties. Normal Distribution The first histogram is a sample from a normal distribution.
Normal curves have well-defined statistical properties. C. K. Pithawala College Of Engineering & Technology. Standard deviation, = 2. A generalized normal distribution, \emph{Journal of Applied Statistics}. The probability density function is a rather complicated function. 1 Univariate Normal (Gaussian) Distribution Let Y be a random variable with mean (expectation) and variance 2 >0. The normal distribution is a symmetric distribution with well-behaved tails. Name of quantile Probability p Quantile Q(p) First millile: 0.001-3.0902: Fifth millile: 0.005-2.5758: First percentile: 0.010 It has the shape of a bell and can entirely be described by its mean and standard deviation. The t-distribution is used as an alternative to the normal distribution when sample sizes are small in order to estimate confidence or determine critical values that an observation is a given distance from the mean.It is a consequence of the sample standard deviation being a biased or underestimate (usually) of the population standard deviation. | 4,896 | 23,051 | {"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-2023-06 | latest | en | 0.90192 |
https://learnchannel-tv.com/en/drives/3-phase-inductance-motor/reactive-power-compensation/ | 1,726,541,360,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00090.warc.gz | 337,276,381 | 21,109 | ## Compensation of three-phase induction motors
### Why compensate?
Because large inductive loads strain the power supply system, It is recommended that big induction motors should not be compensated. Therefore, capacítors are added to improve the power factor PF or cos φ.
As a reference value motors above 5 kvar should be compensated. Calculate the reactive power consumption in rated operation for the motor shown below and decide whether this motor must be compensated or not.
Motor plate 3 Phase Induction motor
Ql = √3 U * I * sin φ | auxiliary calculation: PF = cos φ = 0,85 => φ ≈ 31,7888 => sin φ ≈ 0,52678
Ql = √3 * 400V * 24A * 0,52678 = 8,763 kvar => The motor should be compensated.
In practice, you will not compensate all the reactive power that occurs at nominal load. The reason is: At low load (the extreme case would be no-load), lower reactive currents will flow and you would have overcompensated the motor, which is undesirable.
Either a target power factor is specified or the capacítor power can be taken from a table.
According to the specification the motor should be compensated at cos φ2 = 0.98. The power triangle with and without compensation you can take from the following sketch:
Power vector diagram
Note:
active power P in kW
apparent power S in kVA
reactive power in kvar
### Determine the required capacitive reactive power Qbc to get the new active power factor:
Qc ges = Pzu (tan φ1 - tan φ2 | φ1 before compensation; φ2 after compensation
In our case:
P = √3 U * I * cos φ = √3 * 400V * 24A * 0,85 = 14,133 kW
Before compensation: cos φ1 = 0,85 => φ1 ≈ 31,79°
After compensation: cos φ2 = 0,98 => φ2 ≈ 11,48°
=> Qc total = 14,133 kW (tan 31,79° - tan11,48°) = 5,889 kvar
### Determinate the capacity of each capacitor:
First of all, it is possible to connect the capacítors in star or delta:
Capacítors for compensation
The total reactive power of our motor is Qc total = 5.889 kvar. Whether in star or delta, 1/3 of the reactive power now takes a single capacítor:
Qc = 1/3 Qc total = 1/3 * 5.889 kvar = 1.963 kvar
To show how the capacítive reactive resístance is related to the reactive power, we make a "bridge" to the ohmic resístance:
P = U2 compared with QC = U2 … Eq. (1)
. R XC
and XC = 1 / 2π f C … Eq. (2)
… Eq.(2) in …Eq.(1) results in: QC = U2 = U2 2π f C
. 1 / 2π f C
=> C = QC = QC …Eq.(3) | ω = 2π f
U2 2π f U2 ω
From equation ...(3) it can be seen that to determine the capacítor size it is important to know whether they are connected in star or delta connection. Why? If the capacítors are connected in star, the capacítor voltage (phase voltage) is reduced by the factor √3 to 230 V (400 V line voltage), i.e. the capacítance of the capacítors is increased three times.
We check this statement:
Capacítors connected in delta:
C = QC = 1963 var = 3,905 * 10-5 F ≈ 39 µF
. U2 2π f (400V)2 2π50s-1
Capacítors connected in star:
C = QC = 1963 var = 1,181 * 10-4 F ≈ 118 µF
. U2 2π f (230V)2 2π50s-1 | 998 | 3,334 | {"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} | 3.84375 | 4 | CC-MAIN-2024-38 | latest | en | 0.780123 |
https://mathematica.stackexchange.com/questions/33000/variable-partitioning-of-the-data-in-the-form-of-list-of-lists | 1,720,806,448,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00542.warc.gz | 310,298,512 | 44,330 | # Variable partitioning of the data in the form of "list of lists"
I am looking for a way for variable partitioning of the data in the form of a "list of lists" for interpolating the imported data from excel. The data looks like:
data = {
{{5., 9.53333, 0.057735}, {5., 19.3333, 0.057735}, {5.,29.2667, 0.057735},
{5., 39.0667, 0.11547}, {5., 49., 0.}, {10., 11., 0.}, {10., 22.4, 4.35117*10^-15},
{10., 33.6667, 0.057735}, {10., 45.1, 0.}, {10., 56.3333, 0.11547},
{15., 12.4667, 0.057735}, {15., 25.2667, 0.057735}, {15., 38., 0.1},
{15., 50.9333, 0.057735}, {15., 63.7667, 0.057735}, {20., 13.8333, 0.057735},
{20., 28.1, 0.1}, {20., 42.3667, 0.057735}, {20., 56.5333, 0.057735},
{20., 70.8667, 0.152753}}
}
Now I want to interpolated this data using Interpolation function but the problem is that Interpolation needs the data in the form:
{{5., 9.53333}, 0.057735}, {{5., 19.3333}, 0.057735},...}
f = Interpolation[data[[1]], InterpolationOrder -> 2]
it interpolates but produces a warning as well:
Interpolation::udeg: Interpolation on unstructured grids is currently only
supported for InterpolationOrder -> 1 or InterpolationOrder -> All.
Order will be reduced to 1.
I would have been happy if the 3D plot of f[x,y] would have been a bit better than this (I hate spikes in the plot):
Overall, I think that the problem can be resolved by supplying "structured grids" to the Interpolation command and that's where I have got stuck. I need to arrange the data as it is required in the Interpolation command. I have also checked Mr Wizard's solution for [dynamic partitioning](Partitioning with varying partition size"list manipulation - Partitioning with varying partition size") but it dose not work (or I must say - I could not make it work) in this case. I have got a feeling that some modification in that will do the job but I do not know the workaround for that.
So the question is how to do that?
There are many ways to repartition the data. Here's one:
data2 = Flatten[data /. {x_, y_, z_} -> {{x, y}, z}, 1]
Then you can interpolate as before:
f = Interpolation[data2, InterpolationOrder -> 1]
Note that I have changed the order to 1 -- this is what the warning was telling you -- with data that is not on a regular grid, you cannot use order greater than 1. The jagged edges are then caused (not by the interpolation) but by the number of points used to plot. So changing the PlotPoints option makes it smoother:
Plot3D[f[x, y], {x, 5, 20}, {y, 9.5, 70}, PlotPoints -> 200]
As Mr Wizard points out, it might be safer to make sure that the replacement rule only applies to the numerical data, hence:
data2 = Flatten[data /. {x_?NumericQ, y_?NumericQ, z_?NumericQ} -> {{x, y}, z}, 1]
would be safer.
• I wouldn't recommend this. You could get a false match with the pattern {x_, y_, z_} causing a nasty bug. You should make the replacement at a specific level or use _?NumericQ. Please see: (6543), (7688), (17497). Commented Sep 25, 2013 at 12:56
• Elegant and perfect. I never used Flatten command like that. Great !! Don't mind it but I can't +1 the reply. Sorry to You, Mr Wizard and Artes. Thanks a lot to all of you. Cheers !! :) Commented Sep 25, 2013 at 13:00
• Thanks for the caution Mr Wizard. Checking out the possibility !! Commented Sep 25, 2013 at 13:03
• Now you get my +1. Commented Sep 25, 2013 at 13:14
• @Om. Have you seen these?: (8458), (20334) Commented Sep 26, 2013 at 12:26
I believe all you need is this:
newdata = {{#, #2}, #3} & @@@ data[[1]]
{{{5., 9.53333}, 0.057735}, {{5., 19.3333}, 0.057735}, {{5., 29.2667}, 0.057735}, . . .
If your data contains multiple sets you can use the long form of Apply and the appropriate levelspec, here {2}:
newdata = Apply[{{#, #2}, #3} &, data, {2}]
Reference:
• I must say that was quick. May be it was easy for you. Thanks a ton. c :) Commented Sep 25, 2013 at 12:56
• @Om. You're welcome. Yes, it was easy, but I spend way too much time playing with Mathematica syntax. :-) I think you will find a lot of power in Slot (#, #2, etc.) which is used in Function (short form &), and Apply. I use these functions all the time. Sometimes they are not optimal due to unpacking but they are convenient and concise, therefore often my first approach. Commented Sep 25, 2013 at 13:02
d = {{#1, #2}, #3} & @@@ Flatten[data, 1]
or
d = {Most @ #, Last @ #}& /@ Flatten[data, 1]
e.g.
{Most@#, Last@#} & /@ Flatten[data, 1] // Short
{{{5., 9.53333}, 0.057735}, <<18>>, {{20., 70.8667}, 0.152753}}
Edit
Given data allow only for InterpolationOrder -> 1
f = Interpolation[ d, InterpolationOrder -> 1];
To visualize data we can make use of various options of Plot3D. To get a better resolution we rescaled z-axis using appropriate values of BoxRatios.
Plot3D[ f[x, y], {x, 5, 20}, {y, 14, 70}, PlotPoints -> 150, MaxRecursion -> 4,
MeshFunctions -> {#3 &}, Mesh -> 30, ClippingStyle -> None,
ColorFunction -> "DeepSeaColors", BoxRatios -> {15, 60, 5},
ViewPoint -> {-1, -3, 1}] | 1,599 | 4,990 | {"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} | 2.671875 | 3 | CC-MAIN-2024-30 | latest | en | 0.873825 |
https://www.nuomiphp.com/eplan/977.html | 1,601,351,119,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00642.warc.gz | 890,671,351 | 11,048 | # 其他 - 用python pandas 装箱列
46.5
44.2
100.0
42.12
bins = [0, 1, 5, 10, 25, 50, 100]
[0, 1] bin amount
[1, 5] etc
[5, 10] etc
......
Night Walker
20
jezrael 2017-07-24 14:31
bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = pd.cut(df['percentage'], bins)
print (df)
percentage binned
0 46.50 (25, 50]
1 44.20 (25, 50]
2 100.00 (50, 100]
3 42.12 (25, 50]
bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
df['binned'] = pd.cut(df['percentage'], bins=bins, labels=labels)
print (df)
percentage binned
0 46.50 5
1 44.20 5
2 100.00 6
3 42.12 5
bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = np.searchsorted(bins, df['percentage'].values)
print (df)
percentage binned
0 46.50 5
1 44.20 5
2 100.00 6
3 42.12 5
...然后value_countsor groupby和合计size
s = pd.cut(df['percentage'], bins=bins).value_counts()
print (s)
(25, 50] 3
(50, 100] 1
(10, 25] 0
(5, 10] 0
(1, 5] 0
(0, 1] 0
Name: percentage, dtype: int64
s = df.groupby(pd.cut(df['percentage'], bins=bins)).size()
print (s)
percentage
(0, 1] 0
(1, 5] 0
(5, 10] 0
(10, 25] 0
(25, 50] 3
(50, 100] 1
dtype: int64
Series像这样的方法Series.value_counts()将使用所有类别，即使数据中不存在某些类别，也可以使用categorical 操作 | 606 | 1,322 | {"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} | 2.703125 | 3 | CC-MAIN-2020-40 | longest | en | 0.365692 |
https://edurev.in/test/6950/Inequalities-MCQ-4 | 1,686,179,893,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224654016.91/warc/CC-MAIN-20230607211505-20230608001505-00159.warc.gz | 254,986,585 | 38,584 | LR > Inequalities MCQ - 4
# Inequalities MCQ - 4
Test Description
## 20 Questions MCQ Test | Inequalities MCQ - 4
Inequalities MCQ - 4 for LR 2023 is part of LR preparation. The Inequalities MCQ - 4 questions and answers have been prepared according to the LR exam syllabus.The Inequalities MCQ - 4 MCQs are made for LR 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Inequalities MCQ - 4 below.
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Inequalities MCQ - 4 - Question 1
### Study the following information to answer the given questions A©B means A is either smaller than or equal to B A\$B means A is neither greater than nor smaller than B A@B means A is either greater than or equal to B A*B means A is greater than B A%B means A is smaller than B Statements: M @ R, R % T, T \$ K Conclusions: I. K *M II. T * M
Inequalities MCQ - 4 - Question 2
### Study the following information to answer the given questions A©B means A is either smaller than or equal to B A\$B means A is neither greater than nor smaller than B A@B means A is either greater than or equal to B A*B means A is greater than B A%B means A is smaller than B Statements: K * T, T @ B, B © M Conclusions: I. M% T II. K © B
Inequalities MCQ - 4 - Question 3
### Study the following information to answer the given questions A©B means A is either smaller than or equal to B A\$B means A is neither greater than nor smaller than B A@B means A is either greater than or equal to B A*B means A is greater than B A%B means A is smaller than B Statements: F % J, B © J, W @ R Conclusions: I. F \$ J II. J % F
Inequalities MCQ - 4 - Question 4
Study the following information to answer the given questions
A©B means A is either smaller than or equal to B
A\$B means A is neither greater than nor smaller than B
A@B means A is either greater than or equal to B
A*B means A is greater than B
A%B means A is smaller than B
Statements:
A © N, N*V, V \$ J
Conclusions:
I. J @ N
Inequalities MCQ - 4 - Question 5
Study the following information to answer the given questions
A©B means A is either smaller than or equal to B
A\$B means A is neither greater than nor smaller than B
A@B means A is either greater than or equal to B
A*B means A is greater than B
A%B means A is smaller than B
Statements:
D \$ M, M % W, W @ R
Conclusions:
I. R * W
Inequalities MCQ - 4 - Question 6
Statements:
A ≥ B, B > C, C< D, D ≤ E
Conclusions: I. D > B II. E >C
Inequalities MCQ - 4 - Question 7
Statements:
P = S, P < Q, R ≤ Q, R ≤ T
Conclusions:
I. Q > S
II. Q = T
Inequalities MCQ - 4 - Question 8
Statements:
X < Y, Y = Z, Z > A, A ≥ B
Conclusions:
I.Z > B
II.A < Y
Inequalities MCQ - 4 - Question 9
Statements:
A > N, K ≥ N, K > M, R > M
Conclusions:
I. R≥A
II. M = N
Inequalities MCQ - 4 - Question 10
Statements:
P ≤ Q, P > R, S = R, S ≤ T
Conclusions:
I. Q ≥ P
II.Q = T
Inequalities MCQ - 4 - Question 11
P & Q – P is neither smaller than nor equal to Q
P@Q – P is neither greater than nor equal to Q
P*Q – P is not smaller than Q
P\$Q – P is not greater than Q
P%Q – P is neither greater than nor smaller than Q
Statement A*B, B&E, F&E, F*D
Conclusion 1) A&E 2) A&D
Inequalities MCQ - 4 - Question 12
P & Q – P is neither smaller than nor equal to Q
P@Q – P is neither greater than nor equal to Q
P*Q – P is not smaller than Q
P\$Q – P is not greater than Q
P%Q – P is neither greater than nor smaller than Q
Statement
A%B, B&C, C*D, D\$E
Conclusion :
1) A & C
2) B * D
Inequalities MCQ - 4 - Question 13
P & Q – P is neither smaller than nor equal to Q
P@Q – P is neither greater than nor equal to Q
P*Q – P is not smaller than Q
P\$Q – P is not greater than Q
P%Q – P is neither greater than nor smaller than Q
Statement
A % B, C & B, C * D, E@D
Conclusions :
1) C % D
2) C & D
Inequalities MCQ - 4 - Question 14
P & Q – P is neither smaller than nor equal to Q
P@Q – P is neither greater than nor equal to Q
P*Q – P is not smaller than Q
P\$Q – P is not greater than Q
P%Q – P is neither greater than nor smaller than Q
Statement
A&B, B @ C, C \$ D, E &D
Conclusion :
1) E&A
2) B*E
Inequalities MCQ - 4 - Question 15
P & Q – P is neither smaller than nor equal to Q
P@Q – P is neither greater than nor equal to Q
P*Q – P is not smaller than Q
P\$Q – P is not greater than Q
P%Q – P is neither greater than nor smaller than Q
Statement
A*B, B\$C, C&D, D*E
Conclusion:
1) A&D
2) E@C
Inequalities MCQ - 4 - Question 16
What should be placed in place of question mark in equation A >= B > E? F>=D to make A> D always true.
Detailed Solution for Inequalities MCQ - 4 - Question 16
A>=B>E>F>=D and A>=B>E>=F>=D (Both > and >= makes A>D true)
Inequalities MCQ - 4 - Question 17
What should be placed in place of question mark in equation K = M? N>=R>S to make K> N always false.
Detailed Solution for Inequalities MCQ - 4 - Question 17
K=M=R>S and K=M=N>= R>S (In both cases K>N are always false
Inequalities MCQ - 4 - Question 18
Which of the following should be place in the blank spaces of the expression A_D_C_B_E from left to right in order to make A>D and C>E always true.
Detailed Solution for Inequalities MCQ - 4 - Question 18
A>DB>=E (A>D and C>E always follow)
Inequalities MCQ - 4 - Question 19
Which of the following should be place in the blank space of the expression F _ P_U_M_C to make Expression F> M and U > C always false.
Detailed Solution for Inequalities MCQ - 4 - Question 19
FM and U >C doesn’t follow for all values of F,M and U,C )
Inequalities MCQ - 4 - Question 20
What should be placed in place of question mark in the expression P = Q =T to make R>T always true.
Detailed Solution for Inequalities MCQ - 4 - Question 20
P = Q S >= T (R>S>=T, so R>T directly follows )
Information about Inequalities MCQ - 4 Page
In this test you can find the Exam questions for Inequalities MCQ - 4 solved & explained in the simplest way possible. Besides giving Questions and answers for Inequalities MCQ - 4, EduRev gives you an ample number of Online tests for practice | 1,995 | 6,521 | {"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-2023-23 | latest | en | 0.915413 |
https://space.stackexchange.com/questions/4093/how-can-i-predict-what-an-objects-orbital-state-vectors-will-be-in-the-future | 1,721,098,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514726.17/warc/CC-MAIN-20240716015512-20240716045512-00583.warc.gz | 471,280,519 | 42,044 | # How can I predict what an object's orbital state vectors will be in the future?
I have a position and velocity vector for an orbiting object and can calculate the orbital elements. From a simple two-body situation, how can I predict where in the orbit the object will be at some future time? What are the mathematical steps involved?
There are two ways. One is to numerically integrate from the current time to the future time. That is conceptually the easiest, as well as the simplest to implement (given a good integration routine). On the other hand, it is the most compute intensive, most prone to accumulated numerical error depending on how far you have to go, and provides no insight into the orbit.
For situations that are not your two-body problem, that is really the only option.
The second is to derive the six orbital elements from the six components of position and velocity (assuming some convention for $t=0$). That tells you a great deal about the future of the object, such as the size, shape, and orientation of the orbit, and the orbit period. From the elements you can compute the position and velocity at any future time with a few simple formulas, without having to integrate through the times in between. Again, assuming your two-body problem.
To answer the question in the comments, the position and velocity can be computed using the orbital elements thusly. In the plane of the orbit, where $\mu$ is the $GM$, $a$ is the semi-major axis, $e$ is the eccentricity, and $\tau$ is the eccentric anomaly (more on that one in a bit):
$x=a\left(\cos\tau-e\right)$
$y=a\sqrt{1-e^2}\sin\tau$
$z=0$
$v_x=-\sqrt{\mu\over a}{\sin\tau\over 1-e\cos\tau}$
$v_y=\sqrt{\mu\over a}{\sqrt{1-e^2}\cos\tau\over 1-e\cos\tau}$
$v_z=0$
You can then rotate those coordinates to the actual plane of the orbit by applying an Euler rotation with the angles $\Omega$, $i$, and $\omega$, where those are the longitude of the ascending node, the inclination, and the argument of periapsis respectively.
The eccentric anomaly goes from $0$ to $2\pi$ over one orbit, and is a convenient substitute for time. It is related to time by:
$t=\sqrt{a^3\over\mu}\left(\tau-e\sin\tau\right)$
There is no closed form solution to get $\tau$ from $t$, so you need to solve that numerically. Or if you are, for example, plotting, you can calculate the coordinates and $t$ all from $\tau$ and plot parametrically.
This all assumes the convention that $t=0$ and $\tau=0$ is at periapsis. You will need to offset $t$ for your solution, since I expect that you will want $t=0$ to be at your initial condition, which is probably not at periapsis. That is where the sixth orbital element for the anomaly comes into play.
This will give you the answer in a fake coordinate system for the two-body problem that is pretty darned close to the real coordinate system if the body being orbited is much, much heavier than the object orbiting. However if that's not the case, then you need to convert the coordinates back to the real situation for both bodies, since they are both orbiting their combined center of mass. An example of that case is the Moon orbiting the Earth.
• The equations for position and velocity based on a, e and tau, are these just for elliptical orbits or would they work for hyperbolic orbits as well? Commented Sep 21, 2017 at 2:13
• No, but there are similar equations for hyperbolic orbits using hyperbolic trig functions. Then $\tau$ goes from $-\infty$ to $\infty$, with the closest approach at $\tau=0$. Commented Sep 24, 2017 at 3:15
@Stu, I'll add some information from the equations of motion perspective. These comments are more geared towards artificial satellite orbiting Earth problem--additional bodies and perturbations throws a lot of this out the window, but some fundamentals remain. Apologies in advance if you already know this.
We make some assumptions on the relative masses of the two-bodies involved, and assume the central body has a far greater mass. We also assume an inertial reference frame, and that the bodies are point masses. Using these we can develop the 2BP equation of motion where $$r$$ is the position vector from the central body center of mass, to the second body
$$\ddot{\vec{r}}=-\frac{\mu}{r^2} \frac{\vec{r}}{r}$$
Now, the fact that we have an $$r^3$$ in the denominator shouldn't worry you - This is the inverse square gravity law, and the additional $$r$$ scales $$\vec{r}$$ to produce a unit vector. This looks a little nasty to solve in closed form, since we have a non-linear, ordinary differential equation (which is also coupled due to the $$r$$ term). There are 3 second-order ODEs, which can be expressed as 6 first-order ODEs. If you have access to a numerical integrator, all you need now are initial conditions on your position and velocity vectors (the 6 states), set your desired time, and away you go. You can propagate this problem backward in time as easily as forward in time.
With respect to orbital element and position velocity conversions, the unperturbed two-body problem essentially has one degree of freedom, that being your choice of anomaly in the plane. With one anomaly specified, you can calculate any of the other two. Given the mean anomaly $$M$$, you can iterate to find the eccentric anomaly $$E$$, and from there determine the true anomaly $$\nu$$ (with care to avoid quadrant ambiguity with the arctangent function). $$M=E-e\sin{E}$$
$$\tan{\frac{\nu}{2}} = \sqrt{\frac{1+e}{1-e}}\tan{\frac{E}{2}}$$
Propagating the anomalies allows one to more easily visualize the orbit, versus observing position/velocity vectors. There exist numerous open-source routines to easily convert (typically named rv2oe or rv2coe) between position/velocity and orbital elements. This is a good website for Vallado's collection of software. | 1,397 | 5,821 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-30 | latest | en | 0.888536 |
https://www.teacherspayteachers.com/Product/I-SPY-SUBTRACT-to-20-NO-PREP-MATH-Puzzles-REVIEW-Subtraction-Gr-1-2-CORE-1027055 | 1,537,755,595,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267160085.77/warc/CC-MAIN-20180924011731-20180924032131-00028.warc.gz | 874,822,305 | 21,747 | # I SPY SUBTRACT to 20 ... NO PREP MATH Puzzles REVIEW Subtraction | Gr 1-2 CORE
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REVIEW Subtraction: I SPY SUBTRACT to 20 . . . NO PREP . . . MATH Puzzles
I SPY SUBTRACT to 20 is a fun way to reinforce subtraction facts. Grades 1-2 CORE Math
Subtraction puzzles work just like a word search puzzle except using numbers.
. . . NO PREP . . . PRINT . . . USE . . . NO PREP . . .
"Always something EXTRA" to help TEACH / REVIEW / PRACTICE skills in FUN ways.
Answers are found from left to right and from top to bottom.
Students search for the basic subtraction facts listed at the bottom of each puzzle.
Students then circle the minuends, subtrahends, and differences on the grid and put in the subtraction sign (–) and equal sign (=) in the appropriate places.
An answer sheet is provided for each I SPY SUBTRACT puzzle so students can
check their own work as they finish each puzzle.
♥ ♥ ♥ Always, samples of BEST SELLING and NEW products ♥ ♥ ♥
Use I SPY SUBTRACT to 20 as:
morning warm-up activities
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A note from the authors:
In creating the puzzles and answer keys, we noticed we were repeating the subtraction patterns over and over, 11 - 3 = 8, 11 - 3 = 8; 18 - 6 = 12, 18 - 6 = 12, and realized how this repetition will help students reinforce the basic addition facts as they search the puzzles.
© Jean VanDerford and Linda Schwartz for Teaching Stuff Place
Tiger by Theresa Cardinali
OTHER BASIC FACTS PRACTICE FOR YOUR STUDENTS
< SPIN in SPACE ADDITION DISCS
Click here for SPIN in SPACE ADDITION DISCS 10 Space Crafts, each with a disc, wheel, gives practice of 11 mental addition facts in order and difficulty. That's 110 mental problems.
< CREEPY CRAWLY SUBTRACTION DISCS
Click here for CREEPY CRAWLY SUBTRACTION DISCS SUBTRACTION FACTS on wheels with numbers to 19 on 10 discs.
< ADDITION and SUBTRACTION in DESIGNS
Click here for ADDITION and SUBTRACTION in DESIGNS Addition and Subtraction facts on 10 designs to add and subtract with numbers to 20. Mental Addition and Subtraction to 20 will complete colorful geometric designs. The worksheets reinforce addition and subtraction as students copy the problems from each design.
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< RHYMING WORDS in DESIGNS
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Enjoy, Jean VanDerford
Teaching and Learning are Fun with Creative and Engaging Activities
Our goal is to make learning fun for kids. ♥ ♥ ♥
Please take time to review I SPY SUBTRACT to 20
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at Teaching Stuff Place
Teaching Stuff Place
I SPY SUBTRACT to 20
Puzzles
COMMON CORE MATH STANDARDS
CCSS.Math.Content.1.OA.C.6 Add and subtract within 20, demonstrating fluency for addition and subtraction within 10.
CCSS.Math.Content.2.OA.B.2 Fluently add and subtract within 20 using mental strategies.2 By end of Grade 2, know from memory all sums of two one-digit numbers.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 928 | 3,637 | {"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} | 3.796875 | 4 | CC-MAIN-2018-39 | latest | en | 0.853575 |
https://kr.mathworks.com/matlabcentral/answers/278393-impedance-calculation-from-experimental-data?s_tid=srchtitle | 1,624,306,901,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00311.warc.gz | 323,363,876 | 25,565 | # Impedance calculation from experimental data
조회 수: 69(최근 30일)
Gisela 2016년 4월 11일
댓글: Ruben Díaz Granero 2021년 5월 13일
I have some experimental data (V and I) that I want to use to calculate the impedance Z using matlab. So far I've calculated the Fourier transform for both voltage and current and to calculate the impedance I want to determine the ratio between the two of them. This however gives a vector for each frequency and in order to obtain the Nyquist plot I need to use only one of the values of the vector. Yet I don't know which of the values I have to choose.
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### 채택된 답변
Star Strider 2016년 4월 11일
I do not understand what your data are. Do you have a vector or matrix?
Ideally, you would either have the voltage and current measured at each frequency, or the impulse response of which you took the fft. Impedance Z(jω) is by definition a complex quantity, so you would simply calculate Z=V./I to get the complex impedance.
##### 댓글 수: 3표시숨기기 이전 댓글 수: 2
Ruben Díaz Granero 2021년 5월 13일
Hi! Have you solved your problem? I've exactly the same problem. An Input current data with a voltage response from a modeling circuit. I(t) --> H(s) --> V(t). The teorical solution is applying the FFT to each temporal sine wave and you will obtain I(w) and V(w). Then the ratio V(w)/I(w) it is supposed to be the impedance Z(jw) but I can't reach this point.
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Translated by | 428 | 1,542 | {"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-2021-25 | latest | en | 0.869025 |
http://tug.org/pipermail/metapost/2005-January/000087.html | 1,547,679,532,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657907.79/warc/CC-MAIN-20190116215800-20190117001800-00541.warc.gz | 251,243,199 | 2,539 | # [metapost] Re: all intersections between two paths *
Laurence Finston lfinsto1 at gwdg.de
Tue Jan 11 12:47:21 CET 2005
```On Tue, 11 Jan 2005, Larry Siebenmann wrote:
>
> > How does one arrive at the value 9 for the maximum number of intersections
> > of other `paths' of length 1? Is there a (relatively) simple proof, or
> > can I look this up somewhere?
>
> Given a generic bezier degree n planar path t |--> b(t), there
> is an "implicitization" process (that goes back to Euler they
> say) [...]
Thank you for your explanation.
>
> > Given the manipulations possible with connectors, I think it may be
> > difficult to filter out `paths' with infinitely many intersections.
>
> I am optimistic that it can be done in some practical sense. Have
> you a specific challenge?
My specific challenge is implementing routines for finding
intersections of arbitrary three-dimensional Metafont-like
`paths' in GNU 3DLDF. Currently, this is not possible. I
plan to reimplement them as NURBs because of the property of
projective invariance, which the latter possess.
I am not, however, bound by the arithmetical limitations
built into MF, since I haven't tried to implement whole
number arithmetic and just use `floats' or `doubles',
depending on the value of a preprocessor macro.
>
> I assume you mean by connectors the short paths that metafont seems
> to insert to link paths that almost but not quite chain together.
> (?)
No, that's not what I meant, but I expressed myself poorly.
I should have said "given the manipulations possible with
control points." What I meant by "connectors" was actually
"path joins", e.g., ".. tension a and b .." and "direction
specifiers", which, when present, are ultimately used to
find control points.
>
> By pairs of `paths' with infinitely many intersections I imagine
> you include paths that remain very near to one another for a
> 'noticeable' stretch?
Yes. I also think that under certain circustances, the
dimensions of the `pens' used for drawing the `paths' should
be taken into account, since the drawings of two `paths' might
intersect or become tangent although the `paths' themselves
do not. This opens another can of worms, though.
In discussions about intersections in GNU 3DLDF,
Martijn van Manen has pointed out that the case of objects
getting very close without actually becoming tangent or
intersecting are also problematical.
Laurence
```
More information about the metapost mailing list | 608 | 2,472 | {"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} | 2.578125 | 3 | CC-MAIN-2019-04 | latest | en | 0.942033 |
https://www.infocomm.ky/choke-price-definition/ | 1,723,641,558,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00375.warc.gz | 637,439,063 | 11,479 | Choke Price Definition.
A choke price is the highest price that a consumer is willing to pay for a good or service. In other words, it is the price point at which the consumer "chokes" on the price and is unwilling to pay any more. This concept is important to businesses because it helps them to determine the optimal price point for their goods and services.
What are the 5 types of elasticity of supply?
1. Price Elasticity of Supply: This measures the responsiveness of quantity supplied to changes in price. A good is said to have price elasticity of supply if an increase in price leads to an increase in quantity supplied and vice versa.
2. Income Elasticity of Supply: This measures the responsiveness of quantity supplied to changes in income. A good is said to have income elasticity of supply if an increase in income leads to an increase in quantity supplied and vice versa.
3. Cross Elasticity of Supply: This measures the responsiveness of quantity supplied of one good to changes in the price of another good. A good is said to have cross elasticity of supply if an increase in the price of another good leads to an increase in quantity supplied of the first good and vice versa.
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What are the 3 types of elasticity of demand?
The three types of elasticity of demand are: price elasticity of demand, income elasticity of demand, and cross elasticity of demand.
Price elasticity of demand is a measure of how much the quantity demanded of a good or service changes in response to a change in its price. Income elasticity of demand is a measure of how much the quantity demanded of a good or service changes in response to a change in consumers' incomes. Cross elasticity of demand is a measure of how much the quantity demanded of a good or service changes in response to a change in the price of a related good or service.
What are the types of price elasticity?
Price elasticity refers to how sensitive demand is to changes in price. There are four main types of price elasticity:
1. Perfectly inelastic: demand does not change at all in response to price changes.
2. Inelastic: demand changes very little in response to price changes.
3. Elastic: demand changes significantly in response to price changes.
4. Perfectly elastic: demand changes infinitely in response to price changes. Why do prices increase when supply decreases? When the supply of a good decreases, the prices of the good will increase because there is less of the good available for purchase. The law of supply and demand dictates that when the demand for a good is high and the supply is low, the prices of the good will increase. This is because there are more people who want to purchase the good than there are available units of the good, so the price of the good goes up in order to ration the good among the people who want to purchase it. | 796 | 3,905 | {"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} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.942979 |
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Quadratic function to standard form solver Arithmetic word problems Website that gives you answers to math problems Steps to solve algebra problems 4th grade math word problems with answers | 798 | 4,019 | {"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} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.955692 |
https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=59&t=40613&p=138370 | 1,606,791,602,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141542358.71/warc/CC-MAIN-20201201013119-20201201043119-00648.warc.gz | 360,234,081 | 11,248 | ## Problem 12.45 (6th edition)
Acidity $K_{a}$
Basicity $K_{b}$
The Conjugate Seesaw $K_{a}\times K_{b}=K_{w}$
pamcoronel1H
Posts: 45
Joined: Fri Sep 28, 2018 12:25 am
### Problem 12.45 (6th edition)
I am confused about the pattern of strength described in the solution. In my lecture notes I have it that the stronger the acid, the weaker it conjugate base, so that would be that a higher pKa would mean a lower kb and therefore a weaker conjugate base, right?
The solutions manual says that the stronger the pKa the stronger the conjugate base, so now I am confused on which pattern is right. Could someone explain the relationship between pKa and the strength of its conjugate base?
Thanks!
(By the way, for my answer I put d>a>b>c, which was the complete opposite of the answer)
Eshwar Venkat 1F
Posts: 32
Joined: Fri Sep 28, 2018 12:22 am
### Re: Problem 12.45 (6th edition)
The trend is that the higher the pKa value, the weaker the acid, and the stronger the base, the weaker its conjugate acid will be. As a result, the base strengths for this problem correspond to the increasing pKa values for the conjugate acids.
Vicky Lu 1L
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am
### Re: Problem 12.45 (6th edition)
The trend is that the a smaller pKa/larger the Ka, the stronger the acid and the larger the pKb/smaller the Kb, the stronger the acid. Likewise, the smaller the pKb/larger the Kb, the stronger the base and the larger the pKA/smaller Ka, the stronger the base. | 422 | 1,489 | {"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": 3, "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} | 2.859375 | 3 | CC-MAIN-2020-50 | latest | en | 0.891105 |
https://mycbseguide.com/blog/cbse-question-paper-class-12-physics/ | 1,601,189,574,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400265461.58/warc/CC-MAIN-20200927054550-20200927084550-00339.warc.gz | 520,811,276 | 53,289 | # CBSE Question Paper class 12 Physics
## myCBSEguide App
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## Previous Year Question Papers – Download Free PDF
CBSE Question Paper class 12 Physics conducted by Central Board of Secondary Education, New Delhi in the month of March 2018-19. CBSE previous year question papers with solution are available in myCBSEguide mobile app and cbse guide website. The Best CBSE App for students and teachers is myCBSEguide which provides complete study material and practice papers to cbse schools in India and abroad.
CBSE Question Paper 2018 Physics
## Class 12 Physics chapters wise list
1. Electric Charges and Fields
2. Electrostatic Potential and Capacitance
3. Current Electricity
4. Moving Charges and Magnetism
5. Magnetism and Matter
6. Electromagnetic Induction
7. Alternating Current
8. Electromagnetic Waves
9. Ray Optics and Optical Instruments
10. Wave Optics
11. Dual Nature of Radiation and Matter
12. Atoms
13. Nuclei
14. Semiconductor Electronic: Material, Devices and Simple Circuits
15. Communication Systems
## CBSE Question Paper class 12 Physics
Time allowed: 3 hours
Maximum Marks: 70
General Instructions :
1. All questions are compulsory. There are 26 questions in all.
2. This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
4. There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions.
5. You may use the following values of physical constants wherever necessary :
$c = 3 \times {10^8} m/s$
$h = 6.63 \times {10^{ - 34}} Js$
$e = 1.6 \times {10^{ - 19}}C$
${\mu _0} = 4\pi \times {10^{ - 7}}\; Tm{A^{ - 1}}$
${\varepsilon _0} = 8.854 \times {10^{ - 12}}{C^2}{N^{ - 1}} {m^{ - 2}}$
$\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9} N{m^2}{C^{ - 2}}$
Mass of electron (me) $= 9.1 \times {10^{ - 31}}kg$
Mass of neutron $= 1.675 \times {10^{ - 27}}kg$
Mass of proton $= 1.673 \times {10^{ - 27}}kg$
Avogadro’s number $= 6.023 \times {10^{23}}$ per gram mole
Boltzmann constant = $= 1.38 \times {10^{ - 23}}J{K^{ - 1}}$
### Section A
1. A proton and an electron traveling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?
Ans.
Electron
2. Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery.
Ans.
(a) Ultra violet rays (b) Ultra violet rays / Laser
3. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity.
Ans.
4. Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two – the parent or the daughter nucleus – would have higher binding energy per nucleon?
Ans.
Daughter nucleus
5. Which mode of propagation is used by short wave broadcast services?
Ans.
Skywave propagation
### Section B
1. Two electric bulbs P and Q have their resistances in the ratio of 1: 2. They are connected in series across a battery. Find the ratio of the power dissipation in these bulbs.
Ans.
Power = I2R
The current, in the two bulbs, is the same as they are connected in series.
$\therefore \frac{{{P_1}}}{{{P_2}}} = \frac{{{I^2}{R_1}}}{{{I^2}{R_2}}} = \frac{{{R_1}}}{{{R_2}}}$
2. A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 $\Omega$ as shown in the figure. Find the value of current in the circuit.
ORIn a potentiometer arrangement for determining the emf of a cell, the balance point of the cell in open circuit is 350 cm. When a resistance of 9 $\Omega$ is used in the external circuit of the cell, the balance point shifts to 300 cm. Determine the internal resistance of the cell.
Ans.
By Kirchoff‟s law, we have, for the loop ABCD, +200 – 38i– 10 = 0
$\therefore i = \frac{{190}}{{38}}A = 5A$
Alternatively:
The two cells being in ‘opposition’,
$\therefore$ net $\varepsilon mf$ = (200 – 10)V = 190 V
Now $I = \frac{V}{R}$
$\therefore I = \frac{{190V}}{{38\Omega }} = 5A$
[Note: Some students may use the formulae,$\frac{\varepsilon }{r} = \frac{{{\varepsilon _1}}}{{{r_1}}} + \frac{{{\varepsilon _2}}}{{{r_2}}}$ and $r = \frac {{r_1}{r_2}} {{r_1} + {r_2}}$
For two cells connected in parallel
They may then say that r = 0;
$\varepsilon$ is indeterminate and hence
I is also indeterminate
Award full marks(2) to students giving this line of reasoning.]ORWe have $r = \left( {\frac{{{l_1}}}{{{l_2}}} - 1} \right)R$$= \left( {\frac{{{l_1} - {l_2}}}{{{l_2}}}} \right)R$
$\therefore r = \left( {\frac{{350 - 300}}{{300}}} \right) \times 9\Omega$$= \frac{{50}}{{300}} \times 9\Omega = 1.5\Omega$
1. Why are infra-red waves often called heat waves ? Explain.
2. What do you understand by the statement, ‘‘Electromagnetic waves transport momentum’’?
3. Ans.
1. Infrared rays are readily absorbed by the (water) molecules in most of the substances and hence increases their thermal motion.
(If the student just writes that “infrared ray produce heating effects”, award ½ mark only)
2. Electromagnetic waves can set (and sustain) charges in motion. Hence, they are said to transport momentum.
(Also accept the following: Electromagnetic waves are known to exert „radiation pressure‟. This pressure is due to the force associated with rate of change of momentum. Hence, EM waves transport momentum)
4. If light of wavelength 412·5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
MetalWork Function
Na1.92
K2.15
Ca3.20
Mo4.17
Ans. The energy of a photon of incident radiation is given by
$E = \frac{{hc}}{\lambda }$
$\therefore E = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{(412.5 \times {{10}^{ - 9}}) \times (1.6 \times {{10}^{ - 19}})}}eV$
$\cong 3.01eV$
Hence only Na and K will show photoelectric emission
[Note: Award this ½ mark even if the student writes the name of only one of these metals] Reason: The energy of the incident photon is more than the work function of only these two metals.
5. A carrier wave of peak voltage 15 V is used to transmit a message signal. Find the peak voltage of the modulating signal in order to have a modulation index of 60%.
Ans.
We have
$\mu = \frac{{{A_m}}}{{{A_c}}}$
Here $\mu = 60\% = \frac{3}{5}$
$\therefore {A_m} = \mu {A_c} = \frac{3}{5} \times 15V$
= 9V
These are first 10 questions only. To view and download complete question paper with solution
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## Last Year Question Paper Class 12 Physics 2018
Download class 12 Physics question paper with solution from best CBSE App the myCBSEguide. CBSE class 12 Physics question paper 2018 in PDF format with solution will help you to understand the latest question paper pattern and marking scheme of the CBSE board examination. You will get to know the difficulty level of the question paper. CBSE question papers 2018 for class 12 Physics have 26 questions with solution
## Previous Year Question Paper for class 12 in PDF
Question papers 2019, 2018, 2017, 2016, 2015, 2014, 2013, 2012, 2011, 2010, 209, 2008, 2007, 2006, 2005 and so on for all the subjects are available under this download link. Practicing real question paper certainly helps students to get confidence and improve performance in weak areas.
To download CBSE Question Paper class 12 Accountancy, Chemistry, Physics, History, Political Science, Economics, Geography, Computer Science, Home Science, Accountancy, Business Studies, and Home Science; do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through the best app for CBSE students and myCBSEguide website.
## Test Generator
Create Papers with your Name & Logo | 2,368 | 8,585 | {"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": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 32, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.4375 | 3 | CC-MAIN-2020-40 | longest | en | 0.852893 |
https://www.geeksforgeeks.org/find-all-angles-of-a-triangle-in-3d/?ref=rp | 1,652,988,676,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662529658.48/warc/CC-MAIN-20220519172853-20220519202853-00118.warc.gz | 931,445,792 | 29,108 | # Find all angles of a triangle in 3D
• Last Updated : 21 Jul, 2021
Given coordinates of 3 vertices of a triangle in 3D i.e. A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3). The task is to find out all the angles of the triangle formed by above coordinates.
Examples:
Input:
x1 = -1, y1 = 3, z1 = 2
x2 = 2, y2 = 3, z2 = 5
x3 = 3, y3 = 5, z3 = -2
Output:
angle A = 90.0 degree
angle B = 54.736 degree
angle C = 35.264 degree
Approach:
For finding angle A find out direction ratios of AB and AC :
direction ratios of AB = x2-x1, y2-y1, z2-z1
direction ratios of AC = x3-x1, y3-y1, z3-z1
then angle A =
For finding angle B find out direction ratios of BA and BC :
direction ratios of BA = x1-x2, y1-y2, z1-z2
direction ratios of BC = x3-x2, y3-y2, z3-z2
then angle B =
For finding angle C find out direction ratios of CB and CA :
direction ratios of CB = x2-x3, y2-y3, z2-z3
direction ratios of CA = x1-x3, y1-y3, z1-z3
then angle C =
Below is the implementation of above approach:
## C++
//CPP program for finding all angles of a triangle#include#includeusing namespace std; // function for finding the anglefloat angle_triangle(int x1, int x2, int x3, int y1, int y2, int y3, int z1, int z2, int z3){ int num = (x2-x1)*(x3-x1)+(y2-y1)*(y3-y1)+(z2-z1)*(z3-z1) ; float den = sqrt(pow((x2-x1),2)+pow((y2-y1),2)+pow((z2-z1),2))*\ sqrt(pow((x3-x1),2)+pow((y3-y1),2)+pow((z3-z1),2)) ; float angle = acos(num / den)*(180.0/3.141592653589793238463) ; return angle ;} // Driver codeint main(){int x1 = -1;int y1 = 3;int z1 = 2;int x2 = 2;int y2 = 3;int z2 = 5;int x3 = 3;int y3 = 5;int z3 = -2;float angle_A = angle_triangle(x1, x2, x3, y1, y2, y3, z1, z2, z3);float angle_B = angle_triangle(x2, x3, x1, y2, y3, y1, z2, z3, z1);float angle_C = angle_triangle(x3, x2, x1, y3, y2, y1, z3, z2, z1);cout<<"Angles are :"<
## Java
//Java program for finding all angles of a triangle class GFG{// function for finding the anglestatic double angle_triangle(int x1, int x2, int x3, int y1, int y2, int y3, int z1, int z2, int z3){ int num = (x2-x1)*(x3-x1)+(y2-y1)*(y3-y1)+(z2-z1)*(z3-z1) ; double den = Math.sqrt(Math.pow((x2-x1),2)+ Math.pow((y2-y1),2)+Math.pow((z2-z1),2))* Math.sqrt(Math.pow((x3-x1),2)+ Math.pow((y3-y1),2)+Math.pow((z3-z1),2)) ; double angle = Math.acos(num / den)*(180.0/3.141592653589793238463) ; return angle ;} // Driver codepublic static void main(String[] args){int x1 = -1;int y1 = 3;int z1 = 2;int x2 = 2;int y2 = 3;int z2 = 5;int x3 = 3;int y3 = 5;int z3 = -2;double angle_A = angle_triangle(x1, x2, x3, y1, y2, y3, z1, z2, z3);double angle_B = angle_triangle(x2, x3, x1, y2, y3, y1, z2, z3, z1);double angle_C = angle_triangle(x3, x2, x1, y3, y2, y1, z3, z2, z1);System.out.println("Angles are :");System.out.println("angle A = "+angle_A+" degree");System.out.println("angle B = "+angle_B+" degree");System.out.println("angle C = "+angle_C+" degree");}}// This code is contributed by mits
## Python3
# Python Code for finding all angles of a triangleimport math # function for finding the angledef angle_triangle(x1, x2, x3, y1, y2, y3, z1, z2, z3): num = (x2-x1)*(x3-x1)+(y2-y1)*(y3-y1)+(z2-z1)*(z3-z1) den = math.sqrt((x2-x1)**2+(y2-y1)**2+(z2-z1)**2)*\ math.sqrt((x3-x1)**2+(y3-y1)**2+(z3-z1)**2) angle = math.degrees(math.acos(num / den)) return round(angle, 3) # driver code x1 = -1y1 = 3z1 = 2x2 = 2y2 = 3z2 = 5x3 = 3y3 = 5z3 = -2angle_A = angle_triangle(x1, x2, x3, y1, y2, y3, z1, z2, z3)angle_B = angle_triangle(x2, x3, x1, y2, y3, y1, z2, z3, z1)angle_C = angle_triangle(x3, x2, x1, y3, y2, y1, z3, z2, z1)print("Angles are :")print("angle A = ", angle_A, "degree")print("angle B = ", angle_B, "degree")print("angle C = ", angle_C, "degree")
## C#
// C# program for finding all// angles of a triangleusing System; class GFG{ // function for finding the anglestatic double angle_triangle(int x1, int x2, int x3, int y1, int y2, int y3, int z1, int z2, int z3){ int num = (x2 - x1) * (x3 - x1) + (y2 - y1) * (y3 - y1) + (z2 - z1) * (z3 - z1); double den = Math.Sqrt(Math.Pow((x2 - x1), 2) + Math.Pow((y2 - y1), 2) + Math.Pow((z2 - z1), 2)) * Math.Sqrt(Math.Pow((x3 - x1), 2) + Math.Pow((y3 - y1), 2) + Math.Pow((z3 - z1), 2)); double angle = Math.Acos(num / den) * (180.0/3.141592653589793238463); return angle ;} // Driver codepublic static void Main(){ int x1 = -1, y1 = 3, z1 = 2; int x2 = 2, y2 = 3, z2 = 5; int x3 = 3, y3 = 5, z3 = -2; double angle_A = angle_triangle(x1, x2, x3, y1, y2, y3, z1, z2, z3); double angle_B = angle_triangle(x2, x3, x1, y2, y3, y1, z2, z3, z1); double angle_C = angle_triangle(x3, x2, x1, y3, y2, y1, z3, z2, z1); Console.WriteLine("Angles are :"); Console.WriteLine("angle A = " + angle_A + " degree"); Console.WriteLine("angle B = " + angle_B + " degree"); Console.WriteLine("angle C = " + angle_C + " degree");}} // This code is contributed by 29AjayKumar
## PHP
## Javascript
Output:
Angles are :
angle A = 90.0 degree
angle B = 54.736 degree
angle C = 35.264 degree`
My Personal Notes arrow_drop_up | 2,158 | 5,906 | {"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} | 3.765625 | 4 | CC-MAIN-2022-21 | latest | en | 0.601925 |
https://math.answers.com/other-math/Are_all_numbers_divisible_by_3_also_divisible_by_6 | 1,721,568,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517701.96/warc/CC-MAIN-20240721121510-20240721151510-00845.warc.gz | 347,259,147 | 48,449 | 0
# Are all numbers divisible by 3 also divisible by 6?
Updated: 4/28/2022
Wiki User
14y ago
Talking in Integers only, then no.
Example, 3 is not divisible by 6.
Talking all natural numbers... then yes. 3 / 6 = 0.5
Wiki User
14y ago | 79 | 241 | {"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-2024-30 | latest | en | 0.874085 |
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First, research your chosen agency’s (The Reedy Creek Fire Department) reach on their social media platforms and then choose one of the options below to complete: Option 1: Based on your previous examination of your community demographics, do you feel they successfully leverage the digital tools they have chosen? Support your assessment
### THANK YOU LETTER PLEASE SEE BELOW FOR DETAILS OF THE CLASS MGT 1007 – Business and Career Dynamics This course is designed to familiarize students with the
THANK YOU LETTER PLEASE SEE BELOW FOR DETAILS OF THE CLASS MGT 1007 – Business and Career Dynamics This course is designed to familiarize students with the reality of today’s workplace, human reso THANK YOU LETTER PLEASE SEE BELOW FOR DETAILS OF THE CLASS MGT 1007 – Business and Career
### COSTCO CASE STUDY PART 2 For this assignment, you will provide a 6–8 slide presentation for the executive team. The presentation will include the
COSTCO CASE STUDY PART 2 For this assignment, you will provide a 6–8 slide presentation for the executive team. The presentation will include the situational analysis, problem, alternatives, and marketing strategy recommendation. Follow the instructions below as you conduct your analysis. Include content from Part 1 Week 2 as it
### THANK YOU LETTER PLEASE SEE BELOW FOR DETAILS OF THE CLASS MGT 1007 – Business and Career Dynamics This course is designed to familiarize students with
THANK YOU LETTER PLEASE SEE BELOW FOR DETAILS OF THE CLASS MGT 1007 – Business and Career Dynamics This course is designed to familiarize students with the reality of today’s workplace, human resource management issues, and such lifetime advancement and management strategies as reinventing oneself, building relationships in a culturally
### This assignment asks you to review and analyze the “Costco” case study on pp. C-18-42 of your textbook and respond to the following questions using both
This assignment asks you to review and analyze the “Costco” case study on pp. C-18-42 of your textbook and respond to the following questions using both theory and practical managerial thinking. For t This assignment asks you to review and analyze the “Costco” case study on pp. C-18-42 of your
### “I Manger Reflection” In this reflection you should think about-and address-the following points: 1) Personal Profile: How do you see yourself a
“I Manger Reflection” In this reflection you should think about-and address-the following points: 1) Personal Profile: How do you see yourself as a manager, given your generational/cultural facts, abilities, personal values, and assessment feedback -give supportive details. Take one additional assessment that interests you in the Take
### What did international businesses do during/after each of the below-mentioned crisis to build resilience against global supply chains disruptions. You will
What did international businesses do during/after each of the below-mentioned crisis to build resilience against global supply chains disruptions. You will need to list out individually what was done after – 1. Pandemic 2. Russia-Ukraine War 3. Suez Canal Crisis 4. The Middle East Crisis One page for each topic.
### Overview In today’s information age, customers have access to various digital platforms to access and share information. Customers also use these platforms
Overview In today’s information age, customers have access to various digital platforms to access and share information. Customers also use these platforms to communicate with service providers, share opinions, and provide feedback. Therefore, it is important for product owners and service providers to be ready with a response plan for | 1,840 | 9,251 | {"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-2024-18 | latest | en | 0.951939 |
http://jacobsphysics.blogspot.com/2011/05/brightness-of-bulb-quantitative.html | 1,472,193,658,000,000,000 | text/html | crawl-data/CC-MAIN-2016-36/segments/1471982295358.50/warc/CC-MAIN-20160823195815-00169-ip-10-153-172-175.ec2.internal.warc.gz | 135,092,283 | 16,284 | Buy that special someone an AP Physics prep book: 5 Steps to a 5: AP Physics 1
Visit Burrito Girl's handmade ceramics shop, The Muddy Rabbit: Yarn bowls, tea sets, dinner ware...
## 09 May 2011
### Brightness of a bulb -- quantitative demonstration
Pop quiz, hotshot: The brightness of a light bulb depends on the bulb's
(A) voltage
(B) current
(C) resistance
(D) power
Sure, for *identical* bulbs, voltage and/or current will kind of relate to brightness. For a fixed resistance, the equations I2R and V2/R show that with the same resistance, power, and therefore brightness, will increase as current or voltage increases. However, you'll eventually get into trouble assuming that a bulb carrying more current is necessarily brighter than one carrying less current.
It's pretty straightforward to show experimentally that voltage does not necessarily correlate with a bulb's brightness. Just get a bunch of miniature flashlight bulbs and holders from Radio Shack or Harbor Freight or somewhere. Be sure to get bulbs with different voltage and current ratings, so that their resistances are different. Connect them in parallel to a battery -- they will take the same voltage, but will NOT be just as bright as one another.
Michael Gray, a veteran of my 2010 AP Summer Institute and a frequent contributor to this column, came up with a much cleverer and more subtle demonstration. He showed with a single light bulb that the bulb's brightness depends on the power, not the voltage or current. How? He measured the bulb's brightness directly with a Vernier light sensor. Of course! Brilliant.
He connected a bulb to a battery. He showed that, by the equation V2/R, doubling the bulb's voltage should not just double the bulb's brightness, but quadruple the brightness. He darkened the room, and placed a light sensor a fixed distance from the bulb. He zeroed the sensor for the ambient light, and turned the bulb on. When he doubled the bulb's voltage, the sensor reading quadrupled. Physics works.
I will certainly use this demonstration next year. Thanks, Michael! | 465 | 2,091 | {"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} | 3.4375 | 3 | CC-MAIN-2016-36 | latest | en | 0.889806 |
https://es.mathworks.com/matlabcentral/answers/64328-series-summation-issue-conversion-to-double-error | 1,627,766,763,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154126.73/warc/CC-MAIN-20210731203400-20210731233400-00294.warc.gz | 247,045,010 | 25,939 | # Series summation issue - conversion to double error
6 views (last 30 days)
Jon on 20 Feb 2013
Hi, I'm trying to write a series summation code and I'm getting the "conversion to double" error when I plot.
Here's the part of the code with trouble:
syms m
rhs = (1./(1+(2.*(t- m.*p))./tc)).* exp((-2.*r^2)./(r^2.*(1+ (2.*(t-m.*p)./tc))));
summation1 = symsum(rhs,0,1.89E5);
Where tc, r, and p are constants and t is a timespace with 36 spaces.
Is the problem that I'm asking it to sum from 0 to 1.89e5 with those 36 time spaces?
Thanks
EDIT: http://i.imgur.com/mz2KV5h.jpg Link is of the sum
bym on 21 Feb 2013
At some point, Matlab gives up & returns the indefinite summation. You can try this:
clc;clear
syms m
tc = 3;
p = 3;
t = 1:10;
r =2;
rhs = (1./(1+(2.*(t- m.*p))./tc)).* exp((-2.*r^2)./(r^2.*(1+ (2.*(t-m.*p)./tc))));
summation1 = symsum(rhs);
f = matlabFunction(summation1);
f(189000)
f(10)
ans =
1.0e-005 *
Columns 1 through 8
-0.2646 -0.2646 -0.2646 -0.2646 -0.2646 -0.2646 -0.2646 -0.2646
Columns 9 through 10
-0.2646 -0.2646
ans =
Columns 1 through 8
-0.0608 -0.0634 -0.0662 -0.0692 -0.0725 -0.0762 -0.0802 -0.0847
Columns 9 through 10
-0.0897 -0.0954
Jon on 23 Feb 2013
Thank you, this will work for now while I have a dozen or so points, but it gets quite complex when I want to extend my time.
Many thanks!
Walter Roberson on 20 Feb 2013
You could try
double(summation1)
If that gives you the same error then your summation1 must still contain some symbolic variable. Try
symvar(summation1)
##### 2 CommentsShowHide 1 older comment
Jon on 21 Feb 2013
Here's the sum for completeness | 590 | 1,601 | {"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} | 3.15625 | 3 | CC-MAIN-2021-31 | latest | en | 0.782892 |
https://www.questia.com/read/120532445/credit-risk-pricing-measurement-and-management | 1,448,409,103,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398444047.40/warc/CC-MAIN-20151124205404-00264-ip-10-71-132-137.ec2.internal.warc.gz | 897,275,068 | 19,975 | Credit Risk: Pricing, Measurement, and Management
By David Lando | Go to book overview
Appendix D
Stochastic Calculus for Jump-Diffusions
This appendix gives an overview of some basic tools involving processes with jumps. The reader is assumed to be familiar with diffusion processes and the Itô formula for such processes. This appendix gives the extension of Itô’s formula to processes with a finite number of jumps on finite intervals. It also describes how the distribution of jump-diffusions can change under an equivalent change of measure. We define the concepts in a framework which is more general, namely within a class of processes known as special semimartingales. Working within this class gives us a way of defining what we mean by a jump-diffusion and it also carries additional dividends. For example, special semimartingales represent the most general class of processes within which we can formulate “beta” relationships, i.e. equations which describe the excess returns on processes in terms of their local covariation with a state price density process (which then in turn can be related to the return on a market portfolio). We look at how this can be formulated.
D.1 The Poisson Process
The Poisson process is the fundamental building block of jump processes. It is just as essential to modeling jump processes as Brownian motion is to diffusion modeling. Given a filtered probability space (Ω, F, P, F), a Poisson process N with intensity λ ≥ 0 is the unique process satisfying the following properties:
i. N0 = 0; ii. N has independent increments; and iii. P(Nt–Ns = k) = ((λ(t–s))k/k!) exp(–λ(t–s)) for k ∈ N0 and t > s ≥ 0
Just as one needs to prove the existence of a Brownian motion (after all, the defining properties could be self-contradictory), one also has to “construct” a Poisson process. This is a lot easier than for Brownian motion. The construction is simply done by letting ε1, ε2, … be a sequence of independent, identically, and exponentially distributed random variables with
-275-
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Your work will be lost once you leave this Web page. | 994 | 4,240 | {"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-2015-48 | latest | en | 0.907227 |
https://blog.csdn.net/styshoo/article/details/54743057 | 1,556,224,682,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578733077.68/warc/CC-MAIN-20190425193912-20190425215912-00189.warc.gz | 361,072,331 | 27,753 | # 原题链接
https://leetcode.com/problems/number-complement/
# 原题
Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.
Note:
The given integer is guaranteed to fit within the range of a 32-bit signed integer.
You could assume no leading zero bit in the integer’s binary representation.
Example 1:
Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.
Example 2:
Input: 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.
# 解法
public int findComplement(int num) {
int valid = 0; // 最高位为1的位数
int tmp = num;
while(tmp > 0) {
tmp /= 2;
valid++;
}
return ~num & ((1 << valid) - 1);
}
public int findComplement(int num) {
return ~num & ((Integer.highestOneBit(num) << 1) - 1);
}
# 测试用例：
public static void main(String[] args) {
Solution s = new Solution();
assert(s.findComplement(5) == 2);
assert(s.findComplement(1) == 0);
assert(s.findComplement(0) == 0);
} | 321 | 1,118 | {"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} | 3.5625 | 4 | CC-MAIN-2019-18 | latest | en | 0.649769 |
https://www.studypug.com/sg/sg-gce-n(a)-level-a-maths/solve-quadratic-equations-by-completing-the-square | 1,656,333,480,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103331729.20/warc/CC-MAIN-20220627103810-20220627133810-00521.warc.gz | 1,077,600,150 | 25,344 | # Solving quadratic equations by completing the square
0/1
##### Introduction
###### Lessons
1. Solve by completing the square: $2{x^2} - 12x + 10 = 0$
0/3
##### Examples
###### Lessons
1. Solving a quadratic equation with TWO REAL SOLUTIONS
Solve by completing the square: $x^2+10x+6=0$
1. Solving a quadratic equation with ONE (REPEATED) REAL SOLUTION
Solve by completing the square: $9x^2+25=30x$
1. Solving a quadratic equation with TWO COMPLEX SOLUTIONS
Solve by completing the square: $-3x^2-24x=49$
###### Free to Join!
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###### Topic Notes
When a quadratic equation cannot be factorized, we can use the method of completing the square to solve the equation.
4-step approach:
1. isolate X's on one side of the equation
2. factor out the leading coefficient of $X^2$
3. "completing the square"
• X-side: inside the bracket, add (half of the coefficient of $X)^2$
• Y-side: add [ leading coefficient $\cdot$ (half of the coefficient of $X)^2$ ]
4. clean up
• X-side: convert to perfect-square form
• Y-side: clean up the algebra | 561 | 2,314 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "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} | 3.5 | 4 | CC-MAIN-2022-27 | latest | en | 0.889783 |
https://ko.webqc.org/balance.php?reaction=Ag2S%28s%29+%2B+H2%28g%29+%3D+Ag%28s%29+%2B+H2S%28g%29+ | 1,571,405,533,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00373.warc.gz | 568,065,084 | 5,979 | #### 온라인 화학반응식 계산기
균형 방정식:
Ag2S(s) + H2(g) = 2 Ag(s) + H2S(g)
반응 방식: single replacement
Reaction stoichiometry 한계 반응물
화합물계수몰 질량무게
Ag2S(s)1247.8014
H2(g)12.01588
Ag(s)2107.8682
H2S(g)134.08088
단위: 몰 질량 - g/mol, 질량 - g.
무료로 제공되는 이 화학 소프트웨어를 친구들에게 알리세요!
이 균형 방정식에 직접 링크:
화학 방정식을 균형에 대한 지침 :
• Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
• Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
• To enter an electron into a chemical equation use {-} or e
• To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
• Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
• Compound states [like (s) (aq) or (g)] are not required.
• If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.
• Reaction stoichiometry could be computed for a balanced equation. Enter either the number of moles or weight for one of the compounds to compute the rest.
• Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents.
완전한 비례값이 계산된 화학 방정식의 예 : 화학 방정식 시약 (전체 방정식을 권장)의 예 : 저희 화학반응식 계산기에 만족하셨다면 만족도 평가를 남겨주세요
오늘 계산된 화학반응식
온라인 화학 도구 메뉴로 돌아가기 | 555 | 1,563 | {"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-2019-43 | latest | en | 0.434475 |
https://www.convertunits.com/from/zettapascal/to/terabar | 1,657,002,830,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104514861.81/warc/CC-MAIN-20220705053147-20220705083147-00024.warc.gz | 781,112,469 | 17,320 | ## ››Convert zettapascal to terabar
zettapascal terabar
How many zettapascal in 1 terabar? The answer is 0.0001.
We assume you are converting between zettapascal and terabar.
You can view more details on each measurement unit:
zettapascal or terabar
The SI derived unit for pressure is the pascal.
1 pascal is equal to 1.0E-21 zettapascal, or 1.0E-17 terabar.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between zettapascals and terabars.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of zettapascal to terabar
1 zettapascal to terabar = 10000 terabar
2 zettapascal to terabar = 20000 terabar
3 zettapascal to terabar = 30000 terabar
4 zettapascal to terabar = 40000 terabar
5 zettapascal to terabar = 50000 terabar
6 zettapascal to terabar = 60000 terabar
7 zettapascal to terabar = 70000 terabar
8 zettapascal to terabar = 80000 terabar
9 zettapascal to terabar = 90000 terabar
10 zettapascal to terabar = 100000 terabar
## ››Want other units?
You can do the reverse unit conversion from terabar to zettapascal, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Zettapascal
The SI prefix "zetta" represents a factor of 1021, or in exponential notation, 1E21.
So 1 zettapascal = 1021 pascals.
The definition of a pascal is as follows:
The pascal (symbol Pa) is the SI unit of pressure.It is equivalent to one newton per square metre. The unit is named after Blaise Pascal, the eminent French mathematician, physicist and philosopher.
## ››Definition: Terabar
The SI prefix "tera" represents a factor of 1012, or in exponential notation, 1E12.
So 1 terabar = 1012 bars.
The definition of a bar is as follows:
The bar is a measurement unit of pressure, equal to 1,000,000 dynes per square centimetre (baryes), or 100,000 newtons per square metre (pascals). The word bar is of Greek origin, báros meaning weight. Its official symbol is "bar"; the earlier "b" is now deprecated, but still often seen especially as "mb" rather than the proper "mbar" for millibars.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 730 | 2,612 | {"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} | 3.640625 | 4 | CC-MAIN-2022-27 | longest | en | 0.78126 |
http://www.advogato.org/person/ssp/diary.html?start=27 | 1,455,248,891,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163421.31/warc/CC-MAIN-20160205193923-00285-ip-10-236-182-209.ec2.internal.warc.gz | 264,264,173 | 16,084 | Older blog entries for ssp (starting at number 27)
The Radix Heap is a priority queue that has better caching behavior than the well-known binary heap, but also two restrictions: (a) that all the keys in the heap are integers and (b) that you can never insert a new item that is smaller than all the other items currently in the heap.
These restrictions are not that severe. The Radix Heap still works in many algorithms that use heaps as a subroutine: Dijkstra’s shortest-path algorithm, Prim’s minimum spanning tree algorithm, various sweepline algorithms in computational geometry.
Here is how it works. If we assume that the keys are 32 bit integers, the radix heap will have 33 buckets, each one containing a list of items. We also maintain one global value last_deleted, which is initially MIN_INT and otherwise contains the last value extracted from the queue.
The invariant is this:
The items in bucket $k$ differ from last_deleted in bit $k - 1$, but not in bit $k$ or higher. The items in bucket 0 are equal to last_deleted.
For example, if we compare an item from bucket 10 to last_deleted, we will find that bits 31–10 are equal, bit 9 is different, and bits 8–0 may or may not be different.
Here is an example of a radix heap where the last extracted value was 7:
As an example, consider the item 13 in bucket 4. The bit pattern of 7 is 0111 and the bit pattern of 13 is 1101, so the highest bit that is different is bit number 3. Therefore the item 13 belongs in bucket $3 + 1 = 4$. Buckets 1, 2, and 3 are empty, but that’s because a number that differs from 7 in bits 0, 1, or 2 would be smaller than 7 and so isn’t allowed in the heap according to restriction (b).
Operations
When a new item is inserted, it has to be added to the correct bucket. How can we compute the bucket number? We have to find the highest bit where the new item differs from last_deleted. This is easily done by XORing them together and then finding the highest bit in the result. Adding one then gives the bucket number:
bucket_no = highest_bit (new_element XOR last_deleted) + 1
where highest_bit(x) is a function that returns the highest set bit of x, or $-1$ if x is 0.
Inserting the item clearly preserves the invariant because the new item will be in the correct bucket, and last_deleted didn’t change, so all the existing items are still in the right place.
Extracting the minimum involves first finding the minimal item by walking the lowest-numbered non-empty bucket and finding the minimal item in that bucket. Then that item is deleted and last_deleted is updated. Then the bucket is walked again and all the items are redistributed into new buckets according to the new last_deleted item.
The extracted item will be the minimal one in the data structure because we picked the minimal item in the redistributed bucket, and all the buckets with lower numbers are empty. And if there were a smaller item in one of the buckets with higher numbers, it would be differing from last_deleted in one of the more significant bits, say bit $k$. But since the items in the redistributed bucket are equal to last_deleted in bit $k$, the hypothetical smaller item would then have to also be smaller than last_deleted, which it can’t be because of restriction (b) mentioned in the introduction. Note that this argument also works for two-complement signed integers.
We have to be sure this doesn’t violate the invariant. First note that all the items that are being redistributed will satisfy the invariant because they are simply being inserted. The items in a bucket with a higher number $k$ were all different from the old last_deleted in the $(k-1)$th bit. This bit must then necessarily also be different from the $(k-1)$th bit in the new last_deleted, because if it weren’t, the new last_deleted would itself have belonged in bucket $k$. And finally, since the bucket being redistributed is the lowest-numbered non-empty one, there can’t be any items in a bucket with a lower number. So the invariant still holds.
In the example above, if we extract the two ‘7’s from bucket 0 and the ‘8’ from bucket 4, the new heap will look like this:
Notice that bucket 4, where the ‘8’ came from, is now empty.
Performance
Inserting into the radix heap takes constant time because all we have to do is add the new item to a list. Determining the highest set bit can be done in constant time with an instruction such as bsr.
The performance of extraction is dominated by the redistribution of items. When a bucket is redistributed, it ends up being empty. To see why, remember that all the items are different from last_deleted in the $(k - 1)$th bit. Because the new last_deleted comes from bucket $k$, the items are now all equal to last_deleted in the $(k - 1)th$ bit. Hence they will all be redistributed to a lower-numbered bucket.
Now consider the life-cycle of a single element. In the worst case it starts out being added to bucket 31 and every time it is redistributed, it moves to a lower-numbered bucket. When it reaches bucket 0, it will be next in line for extraction. It follows that the maximum number of redistributions that an element can experience is 31.
Since a redistribution takes constant time per element distributed, and since an element will only be redistributed $d$ times, where $d$ is the number of bits in the element, it follows that the amortized time complexity of extraction is $O(d)$. In practice we will often do better though, because most items will not move through all the buckets.
Caching performance
Some descriptions of the radix heap recommend implementing the buckets as doubly linked lists, but that would be a mistake because linked lists have terrible cache locality. It is better to implement them as dynamically growing arrays. If you do that, the top of the buckets will tend to be hot which means the per-item number of cache misses during redistribution of a bucket will tend to be $O(1/B)$, where $B$ is the number of integers in a cache line. This means the amortized cache-miss complexity of extraction will be closer to $O(d/B)$ than to $O(d)$.
In a regular binary heap, both insertion and extraction require $\Theta(\log n)$ swaps in the worst case, and each swap (except for those very close to the top of the heap) will cause a cache miss.
In other words, if $d = \Theta(\log n)$, extraction from a radix heap will tend to generate $\Theta(\log n / B)$ cache misses, where a binary heap will require $\Theta(\log n)$.
Syndicated 2013-05-25 00:00:00 from Søren Sandmann Pedersen
Fast Multiplication of Normalized 16 bit Numbers with SSE2
If you are compositing pixels with 16 bits per component, you often need this computation:
uint16_t a, b, r;
r = (a * b + 0x7fff) / 65535;
There is a well-known way to do this quickly without a division:
uint32_t t;
t = a * b + 0x8000;
r = (t + (t >> 16)) >> 16;
Since we are compositing pixels we want to do this with SSE2 instructions, but because the code above uses 32 bit arithmetic, we can only do four operations at a time, even though SSE registers have room for eight 16 bit values. Here is a direct translation into SSE2:
a = punpcklwd (a, 0);
b = punpcklwd (b, 0);
a = pmulld (a, b);
b = psrld (a, 16);
a = psrld (a, 16);
a = packusdw (a, 0);
But there is another way that better matches SSE2:
uint16_t lo, hi, t, r;
hi = (a * b) >> 16;
lo = (a * b) & 0xffff;
t = lo >> 15;
hi += t;
t = hi ^ 0x7fff;
if ((int16_t)lo > (int16_t)t)
lo = 0xffff;
else
lo = 0x0000;
r = hi - lo;
This version is better because it avoids the unpacking to 32 bits. Here is the translation into SSE2:
t = pmulhuw (a, b);
a = pmullw (a, b);
b = psrlw (a, 15);
b = pxor (t, 0x7fff);
a = pcmpgtw (a, b);
a = psubw (t, a);
This is not only shorter, it also makes use of the full width of the SSE registers, computing eight results at a time.
Unfortunately SSE2 doesn’t have 8-bit variants of pmulhuw, pmullw, and psrlw, so we can’t use this trick for the more common case where pixels have 8 bits per component.
Exercise: Why does the second version work?
Syndicated 2013-05-16 05:00:56 from ssp
Sysprof 1.1.8
A new version 1.1.8 of Sysprof is out.
This is a release candidate for 1.2.0 and contains mainly bug fixes.
Syndicated 2013-05-16 05:00:56 from ssp
Gamma Correction vs. Premultiplied Pixels
Pixels with 8 bits per channel are normally sRGB encoded because that allocates more bits to darker colors where human vision is the most sensitive. (Actually, it’s really more of a historical accident, but sRGB nevertheless remains useful for this reason). The relationship between sRGB and linear RGB is that you get an sRGB pixel by raising each component of a linear pixel to the power of $1/2.2$.
A lot of graphics software does alpha blending directly on these sRGB pixels using alpha values that are linearly coded (ie., an alpha value of 0 means no coverage, 0.5 means half coverage, and 1 means full coverage). Because alpha blending is best done with premultiplied pixels, such systems store pixels in this format:
[ alpha, alpha * red_s, alpha * green_s, alpha * blue_s ]
where alpha is linearly coded, and (red_s, green_s, blue_s) are sRGB coded. As long as you are happy with blending in sRGB, this works well. Also, if you simply discard the alpha channel of such pixels and display them directly on a monitor, it will look as if the pixels were alpha blended (in the sRGB space) on top of a black background, which is the desired result.
But what if you want to blend in linear RGB? If you use the format above, some expensive conversions will be required. To convert to premultiplied linear, you have to first divide by alpha, then raise each color to 2.2, then multiply by alpha. To convert back, you must divide by alpha, raise to $1/2.2$, then multiply with alpha.
The conversions can be avoided if you store the pixels linearly, ie., keeping the premultiplication, but coding red, green, and blue linearly instead of as sRGB. This makes blending fast, but the downside is that you need deeper pixels. With only 8 bits per pixel, the linear coding loses too much precision in darker tones. Another problems is that to display these pixels, you will either have to convert them to sRGB, or if the video card can scan them out directly, you have to make sure that the gamma ramp is set to compensate for the fact that the monitor expects sRGB pixels.
[ alpha, alpha_s * red_s, alpha_s * green_s, alpha_s * blue_s ]
That is, the alpha channel is stored linearly, and the color channels are stored in sRGB, premultiplied with the alpha value raised to 1/2.2. Ie., the red component is now
(red * alpha)^(1/2.2),
where before it was
alpha * red^(1/2.2).
It is sufficient to use 8 bits per channel with this format because of the sRGB encoding. Discarding the alpha channel and displaying the pixels on a monitor will produce pixels that are alpha blended (in linear space) against black, as desired.
You can convert to linear RGB simply by raising the R, G, and B components to 2.2, and back by raising to $1/2.2$. Or, if you feel like cheating, use an exponent of 2 so that the conversions become a multiplication and a square root respectively.
This is also the pixel format to use with texture samplers that implement the sRGB OpenGL extensions (textures and framebuffers). These extensions say precisely that the R, G, and B components are raised to 2.2 before texture filtering, and raised to 1/2.2 after the final raster operation.
Syndicated 2013-05-16 05:00:56 from ssp
Over is not Translucency
The Porter/Duff Over operator, also known as the “Normal” blend mode in Photoshop, computes the amount of light that is reflected when a pixel partially covers another:
The fraction of bg that is covered is denoted alpha. This operator is the correct one to use when the foreground image is an opaque mask that partially covers the background:
A photon that hits this image will be reflected back to your eyes by either the foreground or the background, but not both. For each foreground pixel, the alpha value tells us the probability of each:
$a \cdot \text{fg} + (1 - a) \cdot \text{bg}$
This is the definition of the Porter/Duff Over operator for non-premultiplied pixels.
But if alpha is interpreted as translucency, then the Over operator is not the correct one to use. The Over operator will act as if each pixel is partially covering the background:
Which is not how translucency works. A translucent material reflects some light and lets other light through. The light that is let through is reflected by the background and interacts with the foreground again.
Let’s look at this in more detail. Please follow along in the diagram to the right. First with probability $a$, the photon is reflected back towards the viewer:
$a \cdot \text{fg}$
With probability $(1 - a)$, it passes through the foreground, hits the background, and is reflected back out. The photon now hits the backside of the foreground pixel. With probability $(1 - a)$, the foreground pixel lets the photon back out to the viewer. The result so far:
\begin{align*} &a\cdot \text{fg} \\ +&(1 - a) \cdot \text{bg} \cdot (1 - a) \end{align*}
But we are not done yet, because with probability $a$ the foreground pixel reflects the photon once again back towards the background pixel. There it will be reflected, hit the backside of the foreground pixel again, which lets it through to our eyes with probability $(1 - a)$. We get another term where the final $(1 - a)$ is replaced with $a \cdot \text{fg} \cdot \text {bg} \cdot (1 - a)$:
\begin{align*} &a\cdot \text{fg} \\ +&(1 - a) \cdot \text{bg} \cdot (1 - a)\\ +&(1 - a) \cdot \text{bg} \cdot a \cdot \text{fg} \cdot \text{bg} \cdot (1 - a) \end{align*}
And so on. In each round, we gain another term which is identical to the previous one, except that it has an additional $a \cdot \text{fg} \cdot \text{bg}$ factor:
\begin{align*} &a\cdot \text{fg} \\ +&(1 - a) \cdot \text{bg} \cdot (1 - a)\\ +&(1 - a) \cdot \text{bg} \cdot a \cdot \text{fg} \cdot \text{bg} \cdot (1 - a)\\ +&(1 - a) \cdot \text{bg} \cdot a \cdot \text{fg} \cdot \text{bg} \cdot a \cdot \text{fg} \cdot \text{bg} \cdot (1 - a) \\ +&\cdots \end{align*}
or more compactly:
$\displaystyle a \cdot \text{fg} + (1 - a)^2 \cdot \text{bg} \cdot \sum_{i=0}^\infty (a \cdot \text{fg} \cdot \text{bg})^i$
Because we are dealing with pixels, both $a$, $\text{fg}$, and $\text{bg}$ are less than 1, so the sum is a geometric series:
$\displaystyle \sum_{i=0}^\infty x^i = \frac{1}{1 - x}$
Putting them together, we get:
$\displaystyle a \cdot \text{fg} + \frac{(1 - a)^2 \cdot bg}{1 - a \cdot \text{fg} \cdot \text{bg}}$
I have sidestepped the issue of premultiplication by assuming that background alpha is 1. The calculations with premultipled colors are similar, and for the color components, the result is simply:
$\displaystyle r = \text{fg} + \frac{(1 - a_\text{fg})^2 \cdot \text{bg}}{1 - \text{fg}\cdot\text{bg}}$
The issue of destination alpha is more complicated. With the Over operator, both foreground and background are opaque masks, so the light that survives both has the same color as the input light. With translucency, the transmitted light has a different color, which means the resulting alpha value must in principle be different for each color component. But that’s not possible for ARGB pixels. A similar argument to the above shows that the resulting alpha value would be:
$\displaystyle r = 1 - \frac{(1 - a)\cdot (1 - b)}{1 - \text{fg} \cdot \text{bg}}$
where $b$ is the background alpha. The problem is the dependency on $\text{fg}$ and $\text{bg}$. If we simply assume for the purposes of the alpha computation that $\text{fg}$ and $\text{bg}$ are equal to $a$ and $b$, we get this:
$\displaystyle r = 1 - \frac{(1 - a)\cdot (1 - b)}{1 - a \cdot b}$
which is equal to
$\displaystyle a + \frac{(1 - a)^2 \cdot b}{1 - a \cdot b}$
Ie., exactly the same computation as the one for the color channels. So we can define the Translucency Operator as this:
$\displaystyle r = \text{fg} + \frac{(1 - a)^2 \cdot \text{bg}}{1 - \text{fg} \cdot \text{bg}}$
for all four channels.
Here is an example of what the operator looks like. The image below is what you will get if you use the Over operator to implement a selection rectangle. Mouse over to see what it would look like if you used the Translucency operator.
Both were computed in linear RGB. Typical implementations will often compute the Over operator in sRGB, so that’s what see if you actually select some icons in Nautilus. If you want to compare all three, open these in tabs:
Over, in sRGB
Translucency, in linear RGB
Over, in linear RGB
And for good measure, even though it makes zero sense to do this,
Translucency, in sRGB
Syndicated 2013-05-16 05:00:56 from ssp
Sysprof 1.2.0
A new stable releasenew stable release of Sysprof is now available. Download version 1.2.0.
Syndicated 2013-05-16 05:00:56 from ssp
Big-O Misconceptions
In computer science and sometimes mathematics, big-O notation is used to talk about how quickly a function grows while disregarding multiplicative and additive constants. When classifying algorithms, big-O notation is useful because it lets us abstract away the differences between real computers as just multiplicative and additive constants.
Big-O is not a difficult concept at all, but it seems to be common even for people who should know better to misunderstand some aspects of it. The following is a list of misconceptions that I have seen in the wild.
But first a definition: We write
$f(n) = O(g(n))$
when $f(n) \le M g(n)$ for sufficiently large $n$, for some positive constant $M$.
Misconception 1: “The Equals Sign Means Equality”
$f(n) = O(g(n))$
is a widespread travestry. If you take it at face value, you can deduce that since $5 n$ and $3 n$ are both equal to $O(n)$, then $3 n$ must be equal to $5 n$ and so $3 = 5$.
The expression $f(n) = O(g(n))$ doesn’t type check. The left-hand-side is a function, the right-hand-side is a … what, exactly? There is no help to be found in the definition. It just says “we write” without concerning itself with the fact that what “we write” is total nonsense.
The way to interpret the right-hand side is as a set of functions:
$O(f) = \{ g \mid g(n) \le M f(n) \text{ for some $$M > 0$$ for large $$n$$}\}.$
With this definition, the world makes sense again: If $f(n) = 3 n$ and $g(n) = 5 n$, then $f \in O(n)$ and $g \in O(n)$, but there is no equality involved so we can’t make bogus deductions like $3=5$. We can however make the correct observation that $O(n) \subseteq O(n \log n)\subseteq O(n^2) \subseteq O(n^3)$, something that would be difficult to express with the equals sign.
Misconception 2: “Informally, Big-O Means ‘Approximately Equal’"
If an algorithm takes $5 n^2$ seconds to complete, that algorithm is $O(n^2)$ because for the constant $M=7$ and sufficiently large $n$, $5 n^2 \le 7 n^2$. But an algorithm that runs in constant time, say 3 seconds, is also $O(n^2)$ because for sufficiently large $n$, $3 \le n^2$.
So informally, big-O means approximately less than or equal, not approximately equal.
If someone says “Topological Sort, like other sorting algorithms, is $O(n \log n)$", then that is technically correct, but severely misleading, because Toplogical Sort is also $O(n)$ which is a subset of $O(n \log n)$. Chances are whoever said it meant something false.
If someone says “In the worst case, any comparison based sorting algorithm must make $O(n \log n)$ comparisons” that is not a correct statement. Translated into English it becomes:
“In the worst case, any comparison based sorting algorithm must make fewer than or equal to $M n \log (n)$ comparisons”
which is not true: You can easily come up with a comparison based sorting algorithm that makes more comparisons in the worst case.
To be precise about these things we have other types of notation at our disposal. Informally:
$O()$: Less than or equal, disregarding constants $\Omega()$: Greater than or equal, disregarding constants $o()$: Stricly less than, disregarding constants $\Theta()$: Equal to, disregarding constants
and some more. The correct statement about lower bounds is this: “In the worst case, any comparison based sorting algorithm must make $\Omega(n \log n)$ comparisons. In English that becomes:
“In the worst case, any comparison based sorting algorithm must make at least $M n \log (n)$ comparisons”
which is true. And a correct, non-misleading statement about Topological Sort is that it is $\Theta(n)$, because it has a lower bound of $\Omega(n)$ and an upper bound of $O(n)$.
Misconception 3: “Big-O is a Statement About Time”
Big-O is used for making statements about functions. The functions can measure time or space or cache misses or rabbits on an island or anything or nothing. Big-O notation doesn’t care.
In fact, when used for algorithms, big-O is almost never about time. It is about primitive operations.
When someone says that the time complexity of MergeSort is $O(n \log n)$, they usually mean that the number of comparisons that MergeSort makes is $O(n \log n)$. That in itself doesn’t tell us what the time complexity of any particular MergeSort might be because that would depend how much time it takes to make a comparison. In other words, the $O(n \log n)$ refers to comparisons as the primitive operation.
The important point here is that when big-O is applied to algorithms, there is always an underlying model of computation. The claim that the time complexity of MergeSort is $O(n \log n)$, is implicitly referencing an model of computation where a comparison takes constant time and everything else is free.
Which is fine as far as it goes. It lets us compare MergeSort to other comparison based sorts, such as QuickSort or ShellSort or BubbleSort, and in many real situations, comparing two sort keys really does take constant time.
However, it doesn’t allow us to compare MergeSort to RadixSort because RadixSort is not comparison based. It simply doesn’t ever make a comparison between two keys, so its time complexity in the comparison model is 0. The statement that RadixSort is $O(n)$ implicitly references a model in which the keys can be lexicographically picked apart in constant time. Which is also fine, because in many real situations, you actually can do that.
To compare RadixSort to MergeSort, we must first define a shared model of computation. If we are sorting strings that are $k$ bytes long, we might take “read a byte” as a primitive operation that takes constant time with everything else being free.
In this model, MergeSort makes $O(n \log n)$ string comparisons each of which makes $O(k)$ byte comparisons, so the time complexity is $O(k\cdot n \log n)$. One common implementation of RadixSort will make $k$ passes over the $n$ strings with each pass reading one byte, and so has time complexity $O(n k)$.
Misconception 4: Big-O Is About Worst Case
Big-O is often used to make statements about functions that measure the worst case behavior of an algorithm, but big-O notation doesn’t imply anything of the sort.
If someone is talking about the randomized QuickSort and says that it is $O(n \log n)$, they presumably mean that its expected running time is $O(n \log n)$. If they say that QuickSort is $O(n^2)$ they are probably talking about its worst case complexity. Both statements can be considered true depending on what type of running time the functions involved are measuring.
Syndicated 2013-05-16 05:00:56 from ssp
Porter/Duff Compositing and Blend Modes
In the Porter/Duff compositing algebra, images are equipped with an alpha channel that determines on a per-pixel basis whether the image is there or not. When the alpha channel is 1, the image is fully there, when it is 0, the image isn’t there at all, and when it is in between, the image is partially there. In other words, the alpha channel describes the shape of the image, it does not describe opacity. The way to think of images with an alpha channel is as irregularly shaped pieces of cardboard, not as colored glass. Consider these two images:
When we combine them, each pixel of the result can be divided into four regions:
One region where only the source is present, one where only the destination is present, one where both are present, and one where neither is present.
By deciding on what happens in each of the four regions, various effects can be generated. For example, if the destination-only region is treated as blank, the source-only region is filled with the source color, and the ‘both’ region is filled with the destination color like this:
The effect is as if the destination image is trimmed to match the source image, and then held up in front of it:
The Porter/Duff operator that does this is called “Dest Atop”.
There are twelve of these operators, each one characterized by its behavior in the three regions: source, destination and both. The ‘neither’ region is always blank. The source and destination regions can either be blank or filled with the source or destination colors respectively.
The formula for the operators is a linear combination of the contents of the four regions, where the weights are the areas of each region:
$A_\text{src} \cdot [s] + A_\text{dest} \cdot [d] + A_\text{both} \cdot [b]$
Where $[s]$ is either 0 or the color of the source pixel, $[d]$ either 0 or the color of the destination pixel, and $[b]$ is either 0, the color of the source pixel, or the color of the destination pixel. With the alpha channel being interpreted as coverage, the areas are given by these formulas:
$A_\text{src} = \alpha_\text{s} \cdot (1 - \alpha_\text{d})$
$A_\text{dst} = \alpha_\text{d} \cdot (1 - \alpha_\text{s})$
$A_\text{both} = \alpha_\text{s} \cdot \alpha_\text{d}$
The alpha channel of the result is computed in a similar way:
$A_\text{src} \cdot [\text{as}] + A_\text{dest} \cdot [\text{ad}] + A_\text{both} \cdot [\text{ab}]$
where $[\text{as}]$ and $[\text{ad}]$ are either 0 or 1 depending on whether the source and destination regions are present, and where $[\text{ab}]$ is 0 when the ‘both’ region is blank, and 1 otherwise.
Here is a table of all the Porter/Duff operators:
$[\text{s}]$ $[\text{d}]$ $[\text{b}]$ Src $s$ $0$ s Atop $0$ $d$ s Over $s$ $d$ s In $0$ $0$ s Out $s$ $0$ $0$ Dest $0$ $d$ d DestAtop $s$ $0$ d DestOver $s$ $d$ d DestIn $0$ $0$ d DestOut $0$ $d$ $0$ Clear $0$ $0$ $0$ Xor $s$ $d$ $0$
And here is how they look:
Despite being referred to as alpha blending and despite alpha often being used to model opacity, in concept Porter/Duff is not a way to blend the source and destination shapes. It is way to overlay, combine and trim them as if they were pieces of cardboard. The only places where source and destination pixels are actually blended is where the antialiased edges meet.
Blending
Photoshop and the Gimp have a concept of layers which are images stacked on top of each other. In Porter/Duff, stacking images on top of each other is done with the “Over” operator, which is also what Photoshop/Gimp use by default to composite layers:
Conceptually, two pieces of cardboard are held up with one in front of the other. Neither shape is trimmed, and in places where both are present, only the top layer is visible.
A layer in these programs also has an associated Blend Mode which can be used to modify what happens in places where both are visible. For example, the ‘Color Dodge’ blend mode computes a mix of source and destination according to this formula:
$\begin{equation*} B(s,d)= \begin{cases} 0 & \text{if $$d=0$$,} \\ 1 & \text{if $$d \ge (1 - s)$$,} \\ d / (1 - s) & \text{otherwise} \end{cases} \end{equation*}$
The result is this:
Unlike with the regular Over operator, in this case there is a substantial chunk of the output where the result is actually a mix of the source and destination.
Layers in Photoshop and Gimp are not tailored to each other (except for layer masks, which we will ignore here), so the compositing of the layer stack is done with the source-only and destination-only region set to source and destination respectively. However, there is nothing in principle stopping us from setting the source-only and destination-only regions to blank, but keeping the blend mode in the ‘both’ region, so that tailoring could be supported alongside blending. For example, we could set the ‘source’ region to blank, the ‘destination’ region to the destination color, and the ‘both’ region to ColorDodge:
Here are the four combinations that involve a ColorDodge blend mode:
In this model the original twelve Porter/Duff operators can be viewed as the results of three simple blend modes:
Source: $B(s, d) = s$ Dest: $B(s, d) = d$ Zero: $B(s, d) = 0$
In this generalization of Porter/Duff the blend mode is chosen from a large set of formulas, and each formula gives rise to four new compositing operators characterized by whether the source and destination are blank or contain the corresponding pixel color.
Here is a table of the operators that are generated by various blend modes:
The general formula is still an area weighted average:
$A_\text{src} \cdot [s] + A_\text{dest} \cdot [d] + A_\text{both}\cdot B(s, d)$
where [s] and [d] are the source and destination colors respectively or 0, but where $B(s, d)$ is no longer restricted to one of $0$, $s$, and $d$, but can instead be chosen from a large set of formulas.
The output of the alpha channel is the same as before:
$A_\text{src} \cdot [\text{as}] + A_\text{dest} \cdot [\text{ad}] + A_\text{both} \cdot [\text{ab}]$
except that [ab] is now determined by the blend mode. For the Zero blend mode there is no coverage in the both region, so [ab] is 0; for most others, there is full coverage, so [ab] is 1.
Syndicated 2013-05-16 05:00:56 from ssp
17 Mar 2013 (updated 25 Mar 2013 at 13:22 UTC) »
Porter/Duff Compositing and Blend Modes
In the Porter/Duff compositing algebra, images are equipped with an alpha channel that determines on a per-pixel basis whether the image is there or not. When the alpha channel is 1, the image is fully there, when it is 0, the image isn’t there at all, and when it is in between, the image is partially there. In other words, the alpha channel describes the shape of the image, it does not describe opacity. The way to think of images with an alpha channel is as irregularly shaped pieces of cardboard, not as colored glass. Consider these two images:
When we combine them, each pixel of the result can be divided into four regions:
One region where only the source is present, one where only the destination is present, one where both are present, and one where neither is present.
By deciding on what happens in each of the four regions, various effects can be generated. For example, if the destination-only region is treated as blank, the source-only region is filled with the source color, and the ‘both’ region is filled with the destination color like this:
The effect is as if the destination image is trimmed to match the source image, and then held up in front of it:
The Porter/Duff operator that does this is called “Dest Atop”.
There are twelve of these operators, each one characterized by its behavior in the three regions: source, destination and both. The ‘neither’ region is always blank. The source and destination regions can either be blank or filled with the source or destination colors respectively.
The formula for the operators is a linear combination of the contents of the four regions, where the weights are the areas of each region:
$$A_\text{src} \cdot [s] + A_\text{dest} \cdot [d] + A_\text{both} \cdot [b]$$
Where $$[s]$$ is either 0 or the color of the source pixel, $$[d]$$ either 0 or the color of the destination pixel, and $$[b]$$ is either 0, the color of the source pixel, or the color of the destination pixel. With the alpha channel being interpreted as coverage, the areas are given by these formulas:
$$A_\text{src} = \alpha_\text{s} \cdot (1 – \alpha_\text{d})$$
$$A_\text{dst} = \alpha_\text{d} \cdot (1 – \alpha_\text{s})$$
$$A_\text{both} = \alpha_\text{s} \cdot \alpha_\text{d}$$
The alpha channel of the result is computed in a similar way:
$$A_\text{src} \cdot [\text{as}] + A_\text{dest} \cdot [\text{ad}] + A_\text{both} \cdot [\text{ab}]$$
where $$[\text{as}]$$ and $$[\text{ad}]$$ are either 0 or 1 depending on whether the source and destination regions are present, and where $$[\text{ab}]$$ is 0 when the ‘both’ region is blank, and 1 otherwise.
Here is a table of all the Porter/Duff operators:
$$[\text{s}]$$ $$[\text{d}]$$ $$[\text{b}]$$ Src $$s$$ $$0$$ s Atop $$0$$ $$d$$ s Over $$s$$ $$d$$ s In $$0$$ $$0$$ s Out $$s$$ $$0$$ $$0$$ Dest $$0$$ $$d$$ d DestAtop $$s$$ $$0$$ d DestOver $$s$$ $$d$$ d DestIn $$0$$ $$0$$ d DestOut $$0$$ $$d$$ $$0$$ Clear $$0$$ $$0$$ $$0$$ Xor $$s$$ $$d$$ $$0$$
And here is how they look:
Despite being referred to as alpha blending and despite alpha often being used to model opacity, in concept Porter/Duff is not a way to blend the source and destination shapes. It is way to overlay, combine and trim them as if they were pieces of cardboard. The only places where source and destination pixels are actually blended is where the antialiased edges meet.
Blending
Photoshop and the Gimp have a concept of layers which are images stacked on top of each other. In Porter/Duff, stacking images on top of each other is done with the “Over” operator, which is also what Photoshop/Gimp use by default to composite layers:
Conceptually, two pieces of cardboard are held up with one in front of the other. Neither shape is trimmed, and in places where both are present, only the top layer is visible.
A layer in these programs also has an associated Blend Mode which can be used to modify what happens in places where both are visible. For example, the ‘Color Dodge’ blend mode computes a mix of source and destination according to this formula:
$$\begin{equation*} B(s,d)= \begin{cases} 0 & \text{if \(d=0$$,}
\\
1 & \text{if $$d \ge (1 – s)$$,}
\\
d / (1 – s) & \text{otherwise}
\end{cases}
\end{equation*}\)
The result is this:
Unlike with the regular Over operator, in this case there is a substantial chunk of the output where the result is actually a mix of the source and destination.
Layers in Photoshop and Gimp are not tailored to each other (except for layer masks, which we will ignore here), so the compositing of the layer stack is done with the source-only and destination-only region set to source and destination respectively. However, there is nothing in principle stopping us from setting the source-only and destination-only regions to blank, but keeping the blend mode in the ‘both’ region, so that tailoring could be supported alongside blending. For example, we could set the ‘source’ region to blank, the ‘destination’ region to the destination color, and the ‘both’ region to ColorDodge:
Here are the four combinations that involve a ColorDodge blend mode:
In this model the original twelve Porter/Duff operators can be viewed as the results of three simple blend modes:
Source: $$B(s, d) = s$$ Dest: $$B(s, d) = d$$ Zero: $$B(s, d) = 0$$
In this generalization of Porter/Duff the blend mode is chosen from a large set of formulas, and each formula gives rise to four new compositing operators characterized by whether the source and destination are blank or contain the corresponding pixel color.
Here is a table of the operators that are generated by various blend modes:
The general formula is still an area weighted average:
$$A_\text{src} \cdot [s] + A_\text{dest} \cdot [d] + A_\text{both}\cdot B(s, d)$$
where [s] and [d] are the source and destination colors respectively or 0, but where $$B(s, d)$$ is no longer restricted to one of $$0$$, $$s$$, and $$d$$, but can instead be chosen from a large set of formulas.
The output of the alpha channel is the same as before:
$$A_\text{src} \cdot [\text{as}] + A_\text{dest} \cdot [\text{ad}] + A_\text{both} \cdot [\text{ab}]$$
except that [ab] is now determined by the blend mode. For the Zero blend mode there is no coverage in the both region, so [ab] is 0; for most others, there is full coverage, so [ab] is 1.
Syndicated 2013-03-17 18:50:24 (Updated 2013-03-25 13:06:40) from Søren Sandmann Pedersen
15 Oct 2012 (updated 19 Oct 2012 at 09:08 UTC) »
Big-O Misconceptions
In computer science and sometimes mathematics, big-O notation is used
to talk about how quickly a function grows while disregarding multiplicative and additive constants. When classifying algorithms, big-O notation is useful because it lets us abstract away the differences between real computers as just multiplicative and additive constants.
Big-O is not a difficult concept at all, but it seems to be common even for people who should know better to misunderstand some aspects of it. The following is a list of misconceptions that I have seen in the wild.
But first a definition: We write $$f(n) = O(g(n))$$ when $$f(n) \le M g(n)$$ for sufficiently large $$n$$, for some positive constant $$M$$.
Misconception 1: “The Equals Sign Means Equality”
The equals sign in $$f = O(g(n))$$ is a widespread travestry. If you take it at face value, you can deduce that since $$5 n$$ and $$3 n$$ are both equal to $$O(n)$$, then $$3 n$$ must be equal to $$5 n$$ and so $$3 = 5$$.
The expression $$f = O(g(n))$$ doesn’t type check. The left-hand-side is a function, the right-hand-side is a … what, exactly? There is no help to be found in the definition. It just says “we write” without concerning itself with the fact that what “we write” is total nonsense.
The way to interpret the right-hand side is as a set of functions: $$O(f) = \{ g \mid g(n) \le M f(n) \text{ for some $$M > 0$$ for large $$n$$}\}.$$ With this definition, the world makes sense again: If $$f(n) = 3 n$$ and $$g(n) = 5 n$$, then $$f \in O(n)$$ and $$g \in O(n)$$, but there is no equality involved so we can’t make bogus deductions like $$3=5$$. We can however make the correct observation that $$O(n) \subseteq O(n \log n)\subseteq O(n^2) \subseteq O(n^3)$$, something that would be difficult to express with the equals sign.
Misconception 2: “Informally, Big-O Means ‘Approximately Equal’”
If an algorithm takes $$5 n^2$$ seconds to complete, that algorithm is $$O(n^2)$$ because for the constant $$M=7$$ and sufficiently large $$n$$, $$5 n^2 \le 7 n^2$$. But an algorithm that runs in constant time, say 3 seconds, is also $$O(n^2)$$ because for sufficiently large $$n$$, $$3 \le n^2$$.
So informally, big-O means approximately less than or equal, not approximately equal.
If someone says “Topological Sort, like other sorting algorithms, is O(n log n)”, then that is technically correct, but severely misleading, because Toplogical Sort is also $$O(n)$$ which is a subset of $$O(n \log n)$$. Chances are whoever said it meant something false.
If someone says “In the worst case, any comparison based sorting algorithm must make $$O(n \log n)$$ comparisons” that is not a correct statement. Translated into English it becomes:
“In the worst case, any comparison based sorting algorithm must make fewer than or equal to $$M n \log (n)$$ comparisons”
which is not true: You can easily come up with a comparison based sorting algorithm that makes more comparisons in the worst case.
To be precise about these things we have other types of notation at our disposal. Informally:
$$O()$$: Less than or equal, disregarding constants $$\Omega()$$: Greater than or equal, disregarding constants $$o()$$: Stricly less than, disregarding constants $$\Theta()$$: Equal to, disregarding constants
and some more. The correct statement about lower bounds is this: “In the worst case, any comparison based sorting algorithm must make $$\Omega(n \log n)$$ comparisons. In English that becomes:
“In the worst case, any comparison based sorting algorithm must make at least $$M n \log (n)$$ comparisons”
which is true. And a correct, non-misleading statement about Topological Sort is that it is $$\Theta(n)$$, because it has a lower bound of $$\Omega(n)$$ and an upper bound of $$O(n)$$.
Misconception 3: “Big-O is a Statement About Time”
Big-O is used for making statements about functions. The functions can measure time or space or cache misses or rabbits on an island or anything or nothing. Big-O notation doesn’t care.
In fact, when used for algorithms, big-O is almost never about time. It is about primitive operations.
When someone says that the time complexity of MergeSort is $$O(n \log n)$$, they usually mean that the number of comparisons that MergeSort makes is $$O(n \log n)$$. That in itself doesn’t tell us what the time complexity of any particular MergeSort might be because that would depend how much time it takes to make a comparison. In other words, the $$O(n \log n)$$ refers to comparisons as the primitive operation.
The important point here is that when big-O is applied to algorithms, there is always an underlying model of computation. The claim that the time complexity of MergeSort is $$O(n \log n)$$, is implicitly referencing an model of computation where a comparison takes constant time and everything else is free.
Which is fine as far as it goes. It lets us compare MergeSort to other comparison based sorts, such as QuickSort or ShellSort or BubbleSort, and in many real situations, comparing two sort keys really does take constant time.
However, it doesn’t allow us to compare MergeSort to RadixSort because RadixSort is not comparison based. It simply doesn’t ever make a comparison between two keys, so its time complexity in the comparison model is 0. The statement that RadixSort is $$O(n)$$ implicitly references a model in which the keys can be lexicographically picked apart in constant time. Which is also fine, because in many real situations, you actually can do that.
To compare RadixSort to MergeSort, we must first define a shared model of computation. If we are sorting strings that are $$k$$ bytes long, we might take “read a byte” as a primitive operation that takes constant time with everything else being free.
In this model, MergeSort makes $$O(n \log n)$$ string comparisons each of which makes $$O(k)$$ byte comparisons, so the time complexity is $$O(k n \log n)$$. One common implementation of RadixSort will make $$k$$ passes over the $$n$$ strings with each pass reading one byte, and so has time complexity $$O(n k)$$.
Misconception 4: Big-O Is About Worst Case
Big-O is often used to make statements about functions that measure the worst case behavior of an algorithm, but big-O notation doesn’t imply anything of the sort.
If someone is talking about the randomized QuickSort and says that it is $$O(n \log n)$$, they presumably mean that its expected running time is $$O(n \log n)$$. If they say that QuickSort is $$O(n^2)$$ they are probably
talking about its worst case complexity. Both statements can be considered true depending on what type of running time the functions involved are measuring.
Syndicated 2012-10-15 09:16:39 (Updated 2012-10-19 08:11:32) from Søren Sandmann Pedersen
18 older entries... | 10,964 | 43,219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 133, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "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} | 3.1875 | 3 | CC-MAIN-2016-07 | longest | en | 0.937283 |
https://stats.stackexchange.com/questions/325855/adding-a-correlated-variable-to-the-regression | 1,701,394,021,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00585.warc.gz | 602,464,160 | 43,169 | # Adding a correlated variable to the regression?
I am running a regression like below: $Y = \beta_0+\beta_1X_1+\beta_2X_2 +\beta_3X_3+\epsilon$, where $X_1$ is a gender dummy and the result shows it has NO statistic significance.
Then I added one more variable $X_1\cdot X_3$ to the model above (with the purpose to test whether the effect of $X_3$ on Y differs by gender), but was surprised that in the new model, all the variables became statistically significant. I just think that the new variable $X_1X_3$ should have even reduced the significance of the dummy $X_1$ because these two variables have high correlation. So why could it happen?
• stats.stackexchange.com/questions/160026/… this might fit your purposes. Roughly spoken, you are changing your reference category and therefore the interpretation of the dummy variables changes. Jan 30, 2018 at 13:37
Think of something like this:
n <- 500
women <- n/2
X.women <- runif(women)
X.men <- runif(n-women)
beta3.women <- 2
beta3.men <- -beta3.women
beta0 <- 1
beta1 <- -beta0
y.women <- beta1 + beta3.women*X.women + rnorm(women, sd=.1)
y.men <- beta0 + beta3.men*X.men + rnorm(n-women, sd=.1)
y <- c(y.women,y.men)
X <- c(X.women,X.men)
plot(X.women,y.women, col="magenta", pch=19, xlab="X", ylab="y")
points(X.men,y.men, col="blue", pch=19)
women.dummy <- c(rep(1,women),rep(0,n-women))
summary(lm(y~X+women.dummy))
summary(lm(y~women.dummy+X+I(women.dummy*X)))
The slopes and intercepts of this artificial dataset are opposite in sign, so that, on average, when fitting no interaction, the regression line is about flat, with no real need for a different intercept for men and women.
Once we introduce the interaction, separate lines are fit for men and women, with evidently both significant intercepts and slopes. (This ignores $X_2$, which seems tangential to the story.)
> summary(lm(y~X+women.dummy))
Call:
lm(formula = y ~ X + women.dummy)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.07295 0.05595 1.304 0.193
X 0.01416 0.08995 0.157 0.875
women.dummy -0.02515 0.05338 -0.471 0.638
> summary(lm(y~women.dummy+X+I(women.dummy*X)))
Call:
lm(formula = y ~ women.dummy + X + I(women.dummy * X))
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.99616 0.01189 83.81 <2e-16 ***
women.dummy -1.98039 0.01763 -112.34 <2e-16 ***
X -1.98613 0.02174 -91.36 <2e-16 ***
I(women.dummy * X) 3.95259 0.03056 129.34 <2e-16 ***
Somewhat more theoretically, the OLS estimator $\hat\beta=(X'X)^{-1}X'y=(X'X/n)^{-1}X'y/n$ will tend to zero (and hence not be likely to produce significant results) if $X'y/n\to_p0$.
In the present context $$\frac{X'y}{n}=\begin{pmatrix}\overline{Y}\\ \overline{Y}_w\\ \overline{X_3y}\end{pmatrix},$$ where $\overline{Y}_w$ denotes the sample average for women and $$\overline{X_3y}=\frac{1}{n}\sum_{i=1}^nX_{3i}y_i$$ Combining the true models for men $$y_i=1-2X_{3i}+u_i$$ and women, $$y_i=-1+2X_{3i}+u_i$$ into a joint model gives, with $W_i$ the female dummy, (so that setting $W_i=1$ retrieves the model for women and setting $W_i=0$ that for men) $$y_i=1-2\cdot W_i-2X_{3i}+4X_{3i}W_i+u_i$$ Thus, $$\frac{1}{n}\sum_{i=1}^nX_{3i}y_i=\frac{1}{n}\sum_{i=1}^nX_{3i}(1-2\cdot W_i-2X_{3i}+4X_{3i}W_i+u_i)$$ As $X_{3i}$ is generated as standard uniform, $\frac{1}{n}\sum_{i=1}^nX_{3i}\to_p 0.5$ and $\frac{1}{n}\sum_{i=1}^nX_{3i}^2\to_p 1/3$. Plugging into the models for men and women then directly yields that the first two entries of $X'y/n$ tend to zero.
Next, $-2\frac{1}{n}\sum_{i=1}^nX_{3i}W_i=-2\frac{1}{n}\sum_{i\in\text{women}}X_{3i}=-2\frac{n_w}{n}\frac{1}{n_w}\sum_{i\in\text{women}}X_{3i}$. The way I generated the data, $\frac{n_w}{n}=0.5$ and, again, $\frac{1}{n_w}\sum_{i\in\text{women}}X_{3i}\to_p0.5$. Similarly, $$4\frac{1}{n}\sum_{i=1}^nX_{3i}^2W_i=4\frac{1}{2}\frac{1}{3}$$ Hence, $$\frac{X'y}{n}\to_p\begin{pmatrix}0\\0\\0\end{pmatrix}$$ Overall, we see that some specific features of the DGP were required to exhibit the result, in particular for the proportion of "men" and "women" in the data as well as for the coefficients and the distribution of the regressors to be such that the OLS estimator has a plim of zero.
There certainly are other possibilities, but playing around with the code will show that yet other specifications will not exhibit this result.
• I also added a little theoretical analysis, if that helps. Jan 30, 2018 at 15:28 | 1,584 | 4,498 | {"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": 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} | 3.0625 | 3 | CC-MAIN-2023-50 | longest | en | 0.849184 |
https://www.sscadda.com/target-ssc-cgl-10000-questions-reasoning-questions-for-ssc-cgl-day-132/ | 1,686,379,737,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657144.94/warc/CC-MAIN-20230610062920-20230610092920-00409.warc.gz | 1,103,627,047 | 120,816 | Latest SSC jobs » SSC GD Constable 2019: Check Important Notice » Target SSC CGL | 10,000+ Questions...
# Target SSC CGL | 10,000+ Questions | Reasoning Questions For SSC CGL : Day 132
This is the new year with new goals, new experiences and lots of exams to be scheduled soon. SSC CGL has recently released the exam dates so now it is time to gear up your preparations and try hard to get success. ADDA247 never fails to deliver something new and fruitful for you all. This time also we are providing you the best study plan as well as a study material. We are here going to prepare your Reasoning section for the SSC CGL. In this article, we are providing you the details that how we are going to help you to clear the examination this year. We ADDA247 is going to provide you daily tests for all the subjects. The topic-wise quiz will be done from January till May. This will help you to get a deeper knowledge of all the topics and will prepare you thoroughly.
Q1. In the following question, two statements are given each followed by two conclusions I and II. You have to consider the statement to be true even if they seem to be at variance from commonly known facts. You have to decide which of the given conclusions, if any, follows from the given statements.
Statements:
(I) All hill-stations have a sunset point.
(II) X is a hill station.
Conclusion:
(I) X has a sunset point.
(II) Place other than hill-station do not have sunset point.
(a) Conclusion I follows
(b) Conclusion II follows
(c) Neither I nor II follows
(d) Both I and II follows
Q2. A series is given with one term missing. Choose the correct alternative from the given ones that will complete the series.
BE, HK, NQ, ?
(a) ST
(b) TU
(c) TW
(d) TS
Q3. A series is given with one term missing. Choose the correct alternative from the given ones that will complete the series.
? , YV, BY, FC
(a) MN
(b) VS
(c) XU
(d) WT
Q4. ln a company all Mondays and Sundays are offs. If a month starts with a Monday and has 31 days then how many offs will be there in that month?
(a) 7
(b) 8
(c) 9
(d) 5
Q5. Arrange the given words in the sequence in which they occur in the dictionary.
i. Speaker
ii. Surreptitious
iii. Spontaneous
iv. Spurious
(a) iv, ii, i, iii
(b) iii, ii, iv, i
(c) iv, iii, i, ii
(d) i, iii, iv, ii
Q6. A word is represented by only one set of numbers as given in any one of the alternatives. The sets of numbers given in the alternatives are represented by two classes of alphabets as shown in the given two matrices. The columns and rows of Matrix-I are numbered from 0 to 4 and that of Matrix-II are numbered from 5 to 9. A letter from these matrices can be represented first by its row and next by its column, for example, ‘N’ can be represented by 21, 67 etc. and ‘R’ can be represented by 66, 57 etc. Similarly, you have to identify the set for the word ‘TOAST’.
(a) 41, 33, 02, 88, 87
(b) 55, 77, 56, 96, 00
(c) 87, 02, 11, 86, 55
(d) 00, 24, 44, 76, 41
Q7. Introducing a boy Rahul says, “He is the son of the brother of the only daughter of my maternal grandfather”. How is the boy related to the Rahul?
(a) Grandfather
(b) Son
(c) Cousin
(d) Father
Q8. If a mirror is placed on the line MN, then which of the answer figures is the right image of the given figure?
Q9. Identify the diagram that best represents the relationship among the given classes.
Liquid, Pizza, Milk
Q10. A piece of paper is folded and punched as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
In the last month i.e. in May daily we will provide you with a test of the previous years’ question papers, this will increase your confidence of solving the real exam and will make you familiar with the real-time exam.
S1. Ans.(a)
S2. Ans.(c)
Sol.
S3. Ans.(d)
Sol.
S4. Ans.(c)
S5. Ans.(d)
Sol. Speaker → Spontaneous → Spurious → Surreptitious
S6. Ans.(d)
Sol.
S7. Ans.(c)
Sol.
S8. Ans.(c)
S9. Ans.(d)
S10. Ans.(c)
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#### SSC CGL 2020 CAPSULE General Awareness And General Science: Free PDF | Download Now
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#### Congratulations!
General Awareness & Science Capsule PDF | 1,178 | 4,366 | {"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} | 2.78125 | 3 | CC-MAIN-2023-23 | latest | en | 0.933064 |
https://www.neuraldesigner.com/blog/statistics/ | 1,721,063,720,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00814.warc.gz | 787,321,883 | 41,291 | When building a machine learning model, knowing the ranges of all the variables is imperative; data statistics provide precious information. Indeed, they put the data set in context.
The most important statistical parameters are the minimum, the maximum, the mean, and the standard deviation. We must always perform a simple statistical analysis to check data consistency.
This way, we need to calculate statistics for each data set variable.Recall that the data matrix comprises the variables
and the samples
(columns and rows of the data set).
The values that a variable takes for each sample in the data set are
## 2. Minimum and maximum
The minimum of a variable is the smallest value of that variable in the data set.
The minimum of the variable is denoted by $v_{jmin}$, and it is defined as follows,
\begin{eqnarray}
\boxed{
v_{jmin} = \min_{i = 1, \ldots, p} d_{ij}.
}
\end{eqnarray}
A variable’s maximum is the variable’s biggest value in the data set.
Similarly, the maximum is denoted by $v_{jmax}$, and it is defined as follows,
\begin{eqnarray}
\boxed{
v_{jmax} = \max_{i=1, \ldots, p} d_{ij}.
}
\end{eqnarray}
## 3. Mean and standard deviation
The mean of a variable is the average value of that variable in the data set. The mean is denoted by $v_{jmean}$ and is defined as,
\begin{eqnarray}
\boxed{v_{jmean} = \frac{1}{p}\sum_{i=1}^{p} d_{ij}.}
\end{eqnarray}
where $d_{ij}$ is the data matrix element and $p$ is the number of samples.
The standard deviation measures how dispersed the data is about the mean. The standard deviation of a variable is denoted by $v_{jstd}$ and is defined as,
\begin{eqnarray}
\boxed{
v_{jstd} = \sqrt{\frac{1}{p}\sum_{i=1}^{p}\left(d_{ij}-v_{jmean}\right)^2}.
}
\end{eqnarray}
where $v_{jmean}$ is the mean of the variable $v_{j}$.
The graphical representation is the standard distribution curve, called the Gaussian bell.
A low standard deviation means that all the values are close to the mean.
Conversely, a high standard deviation means the values are spread out around the mean and from each other.
### Example: Predict the noise generated by airfoil blades
NASA conducts a study of the noise generated by an aircraft in order to make a model to reduce it.
The file airfoil_self_noise.csv contains the data for this example.
Here the number of variables (columns) is 6, and the number of instances (rows) is 1503.
We can calculate the basic statistics of each variable using the formulas described above.
The following table displays the minimum, maximum, mean, and standard deviation for every input variable in the data set.
Name Minimum Maximum Mean Deviation frequency 200 20000 2890 3150 angle_of_attack 0 0.22 6.78 5.92 chord_length 0.0254 0.305 0.137 0.0935 free_stream_velocity 31.7 71.3 50.9 15.6 suction_side_displacement_thickness 0.000401 0.0584 0.0111 0.0132 scaled_sound_pressure_level 103 141 125 6.9
By performing this simple statistical analysis, we can check the consistency of the data.
## 4. Conclusions
Statistics put the data set in context.
It is essential to perform a simple statistical analysis to check the consistency of the data before building the model.
This is done by calculating each variable’s most important statistical parameters, such as the minimum and maximum values, mean, and standard deviation. | 857 | 3,321 | {"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 | 4 | CC-MAIN-2024-30 | latest | en | 0.814172 |
http://www.jiskha.com/display.cgi?id=1395297054 | 1,498,345,003,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320362.97/warc/CC-MAIN-20170624221310-20170625001310-00261.warc.gz | 577,443,632 | 3,845 | # College math
posted by .
A and B agree to share the cost of an \$18 pizza based on how much each would eat. B ate 3/2 of the pizza that A ate. How much should A and B each pay?
A=
B=
Thank you
• College math -
A + 1.5 A = 18
2.5 A = 18
A = 7.20
B = 10.80
• College math -
f(x)=x^2+7 upward 8 units | 111 | 308 | {"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-2017-26 | latest | en | 0.956904 |
https://kazakhsteppe.com/questions-262 | 1,674,779,760,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494852.95/warc/CC-MAIN-20230127001911-20230127031911-00413.warc.gz | 373,916,724 | 4,503 | # Solve math problems free
These can be very helpful when you're stuck on a problem and don't know how to Solve math problems free. We will give you answers to homework.
## Solving math problems free
The best way to Solve math problems free is to eliminate as many options as possible. Word math problems are typically more challenging than arithmetic problems. This is because word problems require you to think about what you’re trying to calculate and how to get there. The good news is that you don’t need to be a math whiz to solve word math problems. All you need to know is the right formulas. Once you know how to calculate a problem, then all you need to do is multiply or divide the two sides of the equation. For example: If a man has 10 apples and 15 oranges, how many oranges does he have? To solve this problem, you first need to calculate how many apples and oranges the man has. To do this, multiply the number of apples by 5 (5 x 10 = 50) and then add 15 (15 + 5 = 20) to get 75. Finally, divide 75 by 2 (75 ÷ 2 = 37) to say that the man has 37 oranges left.
Absolute value equations can be solved with a simple formula. First, you have to know the values of both sides of the equation. The left side will always be positive, and the right side will always be negative. Then, you just subtract one value from the other and solve for the unknown. Absolute value equations are most often used in math, physics and engineering. But they can also be applied to other fields like finance and economics. For example, if you want to sell a car for \$1,000 but you paid \$1,500 for it, your sales price is \$500 too high. In this case, you need to deduct \$500 from your original price to get a realistic selling price. With absolute value equations, it's all about knowing the relationship between two sides of an equation (the left and right sides) and how to find their difference or subtraction (the unknown).
Solving for angle in a right triangle is actually quite easy, but it’s important to remember a few things. Firstly, you can never solve for the "length" of the side unless that side is a right triangle (which means all three sides are equal). Secondly, when solving for angle in a right triangle, you always need to have an initial guess as to what angle you’re looking for. Lastly, the values for angles must always be expressed in degrees. One of the most common problems with solving for angle in a right triangle is when you try to find the "perpendicular bisector" of one of the sides. When this happens, it's usually because you're trying to find *the* perpendicular bisector of the side instead of finding its length (which would give you a third angle). The easiest way to avoid this problem is to always think about which side you're looking at first and make sure that angle is always used as your starting point.
This will stop the leak. Another example is that if you want to save money on food, you could replace something that’s expensive with something that’s less expensive. For example, instead of buying expensive steak, you could buy hamburger meat or chicken. One of the main uses of substitution is to reduce waste. For example, when people have too much garbage to throw away, they might simply throw out some things and replace them with other things. Or when people have too much food to eat, they might just eat less food and replace it with other things. Solving each system by elimination
Love it!!! It's been several years since I've been any of these kinds of math problems and I have to help my children with their math all the time. what I love the most is the fact that it shows you the steps to get the answers and refreshes my memory so I can explain it to my kids. Again. I absolutely love it!!! Thank you!!!
Rosalyn Barnes
This app is helping me immensely in my exam prep. Would recommend it to everybody who has some quick syllabus completing to do and doesn't know solutions to some problems! Though I do hope that you'll add word problem solving in the near future as well. Thank you!
Belinda Reed
Beginning & intermediate algebra Trig equations solver Math homework scanner How to solve radical equations Free math help calculator with steps Find center and radius of circle solver | 923 | 4,251 | {"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-2023-06 | latest | en | 0.949836 |
https://forums.welltrainedmind.com/topic/548197-math-curriculum-questions%C3%A2%E2%82%AC%C2%A6-saxon-intermediate-3-vs-54/ | 1,642,611,959,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301475.82/warc/CC-MAIN-20220119155216-20220119185216-00502.warc.gz | 325,597,881 | 33,534 | # Math curriculum questions… Saxon Intermediate 3 vs 5/4
## Recommended Posts
The more I think about it, the more I am leaning towards moving away from MUS after we finish Gamma. I love the conceptual foundation MUS has laid for my DS8 and I think it was exactly what he needed for K-2. But as we move into 3rd grade, DS no longer needs the manipulatives and I am finding the way that MUS is taught is so different from the way I teach and understand math that I am unable to help DS (who is also very confused by some of the methods). So I am looking at some other options.
Today I am looking at Saxon Intermediate 3. He is probably ready for the concepts in Saxon 5/4 but since we have only used MUS, I thought Intermediate 3 might help with the learning curve of a new curriculum. He is sorely lacking in math fact memorization and 'real world' applications of time, money, temperature, weights and measures, etc. Cathy Duffy describes Intermediate 3 as being pretty independent which is hugely important. However, if we move on to Saxon 5/4, I have the advantage of using the Teaching Tapes (anybody have opinions on these?).
Thoughts on Saxon Inter3 and 5/4? What do you love/hate?
Other options I am researching are Teaching Textbooks and RightStart
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If he struggles with fact memorization, I would hesitate for 5/4. In 5/4, they expect you to pretty much have addition and subtraction with a few lessons of review in the beginning and then go to multiplication. The way they teach multiplication is mostly skip counting. Multiplication is also accelerated as the large focus is getting into division. Saxon 3 is when multiplication is really introduced.
We just switched from Saxon to A Beka after three years of Saxon math because their way of introducing math facts were, in my opinion, haphazard and hurting my son's retention. Take multiplication. The concept of groups of objects and faster way to do addition was fine, but it was the order of the tables that really bothered me. They would start with the easy tables of ones, tens and fives. They use skip counting to teach it. They will start the child skip counting in the daily warmup through certain extraneous stuff. The first table we learned after ones, twos, tens and fives were the times seven tables. These were drilled in by having the child count by sevens and ask how many days were in a certain number of weeks. Then they went with the twelves by asking how many months were in a year. Then it went to fours and then eights. The eights were introduced with the gallon system that they were supposed to have drilled into memory from before. The perfect squares were introduced on their own. Then whatever is left because that is when I closed the book. I thought 5/4 would be better, but the fact intro for multiplication is fewer than ten lessons and in random order again. When learning addition, they started with all these tricks and gimmicks, and whatever facts were leftover and didn't fit were called the oddball facts. Their method of memory is just page after page of black and white math facts and flash cards.
With all that said, I am not against fact memorization through drill. But Saxon was very dry and colorless and they lacked a logical order of fact presentation. We just switched to A Beka for its color and order of fact families. I plan to supplement with Multiplication the Fun Way. There are also several game ideas and colorful pages. They are similar, but in my opinion, their sequence is more logical. But I wouldnt do A Beka past elementary. I had always thought I would do Saxon in upper grades, but my son has developed such a hatred for it that I am not sure.
Many do well in Saxon, and I am not saying A Beka is right for your family. But I wanted you to understand the things you will encounter in regards to fact memorization in Saxon. I did regular Saxon 3, not the Intermediate. In fact, I have the book I am selling if you want it. :). If they still struggle with addition and subtraction, I would take the summer to work with him on the side. Addimals and Xtramath Are free apps that have been helping us this year. Getting out counters and showing all the ways to make 11 or 12, etc. also help. By third grade, most programs are not going to provide the extra help someone struggling with addition will need. You will need to supplement some extra practice with any program you do. If he doesn't have addition and subtraction down, multiplication and division are that much harder.
If you go with Saxon, I'd go with Intermediate 3 to give you more time to get those facts down and for the slowed pace of introducing multiplication.
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His math and subtraction facts are not an issue. They are not automatic but he can do them mentally relatively quickly. His multiplication facts he uses skip counting every single time. That is one of our projects this summer, to memorize the times tables themselves because the skip counting is wearing. me. out. :)
He is already doing single and double digit multiplication. So if Saxon Intermediate 3 is just introducing multiplication, that may be behind where he is. But if it would be a good intro to the *way* Saxon does things, it may be worth a little remediation...
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I would start him in 5/4. I love Saxon k-3 and 5/4+, but the format of Intermediate 3 was awful.
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Well then he could probably do 5/4. I believe Saxon has a placement test. While there are there things in 3 than just multiplication, such as fractions, they do take the entire year to present all the multiplication facts. It is a very incremental program with little bits that spiral around. If you already have that level of multiplication down, 5/4 would probably be good, but do the placement test to be sure. Many third graders are in 5/4 and start Saxon a year ahead. However, if you think he would have problems with writing out all the problems on notebook paper or the formatting of 5/4, you could look at Intermediate 4. The 5/4 and up have different authors than K-3 and the Intermediate and tend to be preferred more than the others, but only you know if your child is ready for that format yet.
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The more I think about it, the more I am leaning towards moving away from MUS after we finish Gamma. I love the conceptual foundation MUS has laid for my DS8 and I think it was exactly what he needed for K-2. But as we move into 3rd grade, DS no longer needs the manipulatives and I am finding the way that MUS is taught is so different from the way I teach and understand math that I am unable to help DS (who is also very confused by some of the methods). So I am looking at some other options.
Today I am looking at Saxon Intermediate 3. He is probably ready for the concepts in Saxon 5/4 but since we have only used MUS, I thought Intermediate 3 might help with the learning curve of a new curriculum. He is sorely lacking in math fact memorization and 'real world' applications of time, money, temperature, weights and measures, etc. Cathy Duffy describes Intermediate 3 as being pretty independent which is hugely important. However, if we move on to Saxon 5/4, I have the advantage of using the Teaching Tapes (anybody have opinions on these?).
Thoughts on Saxon Inter3 and 5/4? What do you love/hate?
Other options I am researching are Teaching Textbooks and RightStart
Has he done the Saxon placement test?
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Only 75 emoji are allowed. | 1,752 | 7,743 | {"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-2022-05 | latest | en | 0.971295 |
https://www.unitconverters.net/length/petameter-to-mil.htm | 1,659,899,703,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570692.22/warc/CC-MAIN-20220807181008-20220807211008-00414.warc.gz | 914,455,319 | 4,124 | Home / Length Conversion / Convert Petameter to Mil
# Convert Petameter to Mil
Please provide values below to convert petameter [Pm] to mil [mil, thou], or vice versa.
From: petameter To: mil
### Petameter to Mil Conversion Table
Petameter [Pm]Mil [mil, Thou]
0.01 Pm3.9370078740158E+17 mil, thou
0.1 Pm3.9370078740158E+18 mil, thou
1 Pm3.9370078740158E+19 mil, thou
2 Pm7.8740157480315E+19 mil, thou
3 Pm1.1811023622047E+20 mil, thou
5 Pm1.9685039370079E+20 mil, thou
10 Pm3.9370078740158E+20 mil, thou
20 Pm7.8740157480315E+20 mil, thou
50 Pm1.9685039370079E+21 mil, thou
100 Pm3.9370078740157E+21 mil, thou
1000 Pm3.9370078740157E+22 mil, thou
### How to Convert Petameter to Mil
1 Pm = 3.9370078740158E+19 mil, thou
1 mil, thou = 2.54E-20 Pm
Example: convert 15 Pm to mil, thou:
15 Pm = 15 × 3.9370078740158E+19 mil, thou = 5.9055118110236E+20 mil, thou | 360 | 866 | {"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-2022-33 | longest | en | 0.47542 |
https://www.teacherspayteachers.com/Product/Forms-of-a-Number-Math-Stations-2587965 | 1,510,983,075,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804610.37/warc/CC-MAIN-20171118040756-20171118060756-00562.warc.gz | 867,627,796 | 25,582 | Total:
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# Forms of a Number Math Stations
Common Core Standards
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This second grade forms of a number math station set is a no frills all action product meant to cement students' knowledge of using different forms of a number for standardized testing. All ten of the activities included were created entirely in black and white to save on colored ink, and make for easy use of the copy machine.
I use these activities in class and with my testing study group after school.
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It was created with the following standards in mind:
TEKS
2.2(B) use standard, word, and expanded forms to represent numbers up to 1,200
Common Core
CCSS.MATH.CONTENT.2.NBT.A.1 Understand that the three-digit number represents amounts of hundreds, tens, and ones; e.g., 706 equals 7 hundreds, 0 tens, and 6 ones. Understand the following as special cases:
CCSS.MATH.CONTENT.2.NBT.A.1.A 100 can be thought of as a bundle of ten tens- called a “hundred”
CCSS.MATH.CONTENT.2.NBT.A.1.B The numbers 100, 200, 300, 400, 500, 600, 700, 800, 900 refer to one, two, three, four, five, six, seven, eight, or nine hundreds (and 0 tens and 0 ones).
++++++++++++++++++++++++++++++++++++++++++++++++++++
All of the activities included are designed to be used to fill in any gaps that students may have understanding how to use forms of a number.
Each activity can be completed in a variety of settings to meet the needs of every learner in your classroom.
-Independent assessment
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Also included in this product:
Recording sheets to hold students accountable
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Forms of a Number
Comparing and Ordering Numbers to 1,200
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Generating Problem Situations Using Addition and Subtraction Within 1,000
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Or get them all in this BIG BUNDLE.
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 666 | 2,818 | {"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-2017-47 | latest | en | 0.889643 |
https://www.chemteam.info/AcidBase/pOH.html | 1,708,973,029,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00025.warc.gz | 697,048,920 | 2,554 | ### pOH and how to calculate it
Sörenson defined pH as the negative logarithm of the hydrogen ion concentration.
pH = −log [H+]
We can define the pOH in a similar way:
pOH = −log [OH¯]
In words, the pOH is the negative logarithm of the hydroxide ion concentration.
Example #1: The [OH¯] in a solution is measured to be 0.0010 M. What is the pOH?
Solution:
1) Plug the [OH¯] into the pOH definition:
pOH = −log 0.0010
2) An alternate way to write this is:
pOH = −log 10¯3
3) Since the log of 10¯3 is -3, we have:
pOH = −(−3)
pOH = 3.00
Example #2: Calculate the pOH of a solution in which the [OH¯] is 4.20 x 10¯4 M.
Solution:
pOH = −log 4.20 x 10¯4
This problem can be done very easily using your calculator. However, be warned about putting numbers into the calculator.
Enter 4.20 x 10¯4 into the calculator, press the "log" button (NOT "ln") and then the sign change button (usually labeled with a "+/-").
pOH = 3.377
I hope you took a look at the significant figures and pH discussion. If not, why don't you go ahead and do that right now. I can wait.
Comment regarding the examples below: keep in mind this equation:
pH + pOH = 14
The ChemTeam also keeps in mind that acidic pH is less than 7 and that a basic pH is greater than 7. So, if I have a pOH = 4, I know that the pH = 10 and that this is a basic solution. In a similar way, if I know the pOH is 11, then the pH is 3 and this is an acidic solution.
For the examples below, convert each hydroxide ion concentration into a pOH. Identify each as an acidic pOH or a basic pOH.
Example #3: 0.0045 M
pOH = −log 0.0045
pOH = −(−2.35)
2.35
This is a basic pOH.
Example #4: 5.0 x 10¯10 M
pOH = −log 5.0 x 10¯10
pOH = −(−9.30) = 9.30
This is an acidic pOH.
Example #5: 1.0 M
pOH = −log 1.0
pOH = −(−0.00)
pOH = 0.00
This is a basic pOH.
Yes, a pOH of zero is possible, it is just uncommon. In fact, watch out for this teacher test trick. What's the pOH when [OH¯] = 2.0 M? That's right, NEGATIVE 0.30. It is possible to have a negative pOH, it is just uncommon to see them.
Example #6: 3.27 x 10¯3 M
pOH = −log 3.27 x 10¯3 = −(−2.485) = 2.485
This is a basic pOH.
Example #7: 1.00 x 10¯12 M
pOH = −log 1.00 x 10¯12 = 12.000
This is an acidic pOH.
Example #8: 0.00010 M
pOH = −log 0.00010 = 4.0
This is a basic pOH.
Suppose you know the pOH and you want to get to the hydroxide ion concentration ([OH¯])?
Here is the equation for that:
[OH¯] = 10¯pOH
That's right, ten to the minus pOH gets you back to the [OH¯] (called the hydroxide ion concentration). This is actually pretty easy to do with the calculator.
Example #9: Calculate the [OH¯] given a pOH of 2.45.
1) The calculator technique depends on which type of calculator button you have. The following instructions assume you have a key labeled EITHER xy or yx.
(a) Enter the number "10" into the calculator. (Do NOT then press the EXP or EE key.)
(b) Press the xy (or the other, depending on what you have)
(c) Enter 2.45 and make it negative with the +/- key.
(d) Press the equals button and the calculator will do its thing.
2) The following instructions are for a calculator with a key labeled "10x."
Enter the 2.45, make it negative, then press the "10x" key. An answer appears!! Just remember to round it to the proper number of significant figures and you're on your way.
3) One more comment about the way the answer appears on the calculator. The two most common ways for the answer to appear are:
3.548133892E-3 or 0.003548133892
That E-3 means this:
x 10¯3 <--- that x is a times sign
The final answer (to the proper number of significant figures) is
[OH¯] = 3.5 x 10¯3 M or 0.0035 M
Notice the inclusion of the M for molarity. | 1,176 | 3,711 | {"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} | 3.953125 | 4 | CC-MAIN-2024-10 | latest | en | 0.921821 |
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# Rate Equations
Chemical reactions are processes in which a set of reactants are converted into products as a result of changes to their structures. These structural changes can happen at different speeds, similar to how race cars can travel at different speeds. Just like how it's important to understand how the speed of a race car can be affected, understanding how the speed…
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# Rate Equations
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Chemical reactions are processes in which a set of reactants are converted into products as a result of changes to their structures. These structural changes can happen at different speeds, similar to how race cars can travel at different speeds. Just like how it's important to understand how the speed of a race car can be affected, understanding how the speed of chemical changes can be affected is an important part of physical chemistry.
The rate equation is an expression that links the rate of a reaction to the concentration of the species involved.
• We will be looking at the rate equation.
• We'll see what it tells us about the rate of a reaction.
• Finally, we'll explore the rate constant and reaction orders, and we'll briefly touch on methods used to determine the rate equation.
## Rate equation chemistry
The rate of a reaction is how quickly a reaction occurs. But what does that mean and what does it tell us? Well, one way of looking at it is to think about how much product is made in a period of time, which will depend on how much reactant is used up. In essence, we can say that the rate of a reaction is the speed at which reactants are converted into products.
The rate of reaction is the change in concentration of reactants or products over time. It is typically measured in mol dm-3 s-1.
### Measuring rate of reaction
To calculate the rate of a reaction, we need to measure the change in the amount of reactant/product from the start of the reaction to the end.
We define the rate of a reaction as a measure of how much product is formed, or how much reactant is used, over a period of time.
We can measure this by observing things like colour change, pH change, volume of gas produced, or change in mass of a solid reactant. You should then be able to convert your data values into figures for concentration. This data is plotted on a line graph, with time on the x-axis and concentration on the y-axis. From the graph, we can find a value for the rate of reaction by working out the line's gradient. We either calculate an overall rate of reaction or an instantaneous rate of reaction. Both use the following equation:
$\mathrm{rate}\mathrm{of}\mathrm{reaction}=\frac{\mathrm{change}\mathrm{in}\mathrm{concentration}}{\mathrm{time}\mathrm{taken}}$
#### Overall rate of reaction
Calculating an overall rate of reaction is fairly straightforward. You divide the overall change in concentration of a reactant or product by the time taken. For a graph of concentration against time, this means dividing the change in y-values by the change in x-values. Here's an example.
Calculate the overall rate of reaction for the following graph.
Fig. 1 - A concentration-time graph used to calculate rate of reaction
To find the overall rate of reaction, we divide the change in concentration by the time taken.
It doesn't matter whether you measure the concentration of a product or reactant - both will give you a valid answer.
Fig. 2 - A concentration-time graph used to calculate rate of reaction
Here, the concentration starts at 40 mol dm-3 and ends at 8 mol dm-3. This is a change of 40 - 8 = 32 mol dm-3. The reaction takes 200 seconds. The rate of reaction is therefore $\frac{32}{200}=0.16\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$.
#### Instantaneous rate of reaction
Sometimes, finding an overall rate of reaction isn't that useful. You might instead want to know how the rate of reaction changes over time. To do this, you calculate instantaneous rates of reaction. This involves drawing a tangent to the curve at a particular point and finding its gradient. Again, this is given by the change in concentration divided by the time taken - in other words, the change in y-values divided by the change in x-values.
Calculate the instantaneous rate of reaction for the following graph at 60 seconds.
Fig. 3 - A concentration-time graph used to calculate rate of reaction
We first need to find the 60-second mark on the curve. We draw a tangent to the curve at this point.
Remember that a tangent is a straight line that just touches the curve at a specified point.
Next, we calculate the gradient of this tangent by dividing the change in concentration by the time taken. You do this by turning the tangent into a right-angled triangle.
Fig. 4 - A concentration-time graph used to calculate rate of reaction
Here, we can see from our right-angled triangle that the concentration starts at 22 mol dm-3 and ends at 6 mol dm-3. This is an overall change of 16 mol dm-3. This change in concentration takes place between 20 and 120 seconds, meaning it takes 100 seconds in total. The instantaneous rate of reaction is therefore $\frac{16}{100}=0.16\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-1}$
## Rate of reaction equation
Let's look at something different: the rate equation. The rate equation in chemistry is a formula that we can use to find the rate of a reaction using the concentration of species involved in the reaction. Here's what it looks like:
Fig. 5 - Rate equation
At first glance it certainly looks confusing, but once you understand what's going on it isn't all that bad.
• k is the rate constant.
• The letters A and B in the rate equation are used to represent species involved in the reaction. These could be reactants or catalysts.
• The square brackets around the letters represent concentration. So, [A] is used to show the concentration of species A.
• The letters m and n represent the order of the reaction with respect to a certain species. They show the power that the concentration of that species is raised to in the rate equation. Overall, [A]m represents the concentration of A, raised to the power of m. This means that A has the order m.
### The rate constant
k is the rate constant. It is used in the rate equation to link the concentrations of certain species to the rate of that reaction. The value of k changes depending on the reaction and reaction conditions. However, k is always constant for a certain reaction at a particular temperature. If you were to carry out the exact same reaction at different temperatures, k would change, but if you carried it out at the same temperature, k would stay the same: after all, it is a constant!
To learn more about how the rate constant relates to temperature, read The Arrhenius Equation. And if you want to find out how to calculate the rate constant, alongside its units, head over to Determining Rate Constant.
### Orders of reaction
In chemical reactions, reactants and catalysts (if there are any) have an order of reaction. The sum of the individual orders of species in a reaction equals the overall order of the equation.
In the rate equation, the order of a reaction with respect to a species is shown using a power. For example, in the rate equation we looked at above, the order of A is represented by the letter m. The order of a reaction with respect to a species tells us how the concentration of that particular species affects the reaction rate. Some species don't affect the rate whatsoever, while other species affect it dramatically.
Any non-negative number can be an order, and species can also have fractional orders like 5/2. But for the purpose of your exams, you only need to know about zero, first and second-order reactants.
#### Zero-order reactants
The concentration of a zero-order reactant doesn't affect the rate of reaction. If you double its concentration, the rate stays the same. This is because 20 = 1. Because they have no effect on the rate of reaction, zero-order reactants don't appear in the rate equation.
#### First-order reactants
The concentration of first-order reactants is directly proportional to the rate of reaction. If you double the concentration of a first-order reactant, the rate of reaction also doubles. This is because 21 = 2.
If a reactant is first-order, it appears in the rate equation raised to the power of 1. However, we don't tend to write the number 1 because raising something to the power of 1 has no effect on its value. You'll see first-order reactants in the rate equation as [A], where A represents the species.
#### Second-order reactants
The concentration of second-order reactants has an exponential effect on the rate of reaction. Doubling the concentration of a second-order reactant causes the rate of reaction to quadruple. This is because 22 = 4.
If a reactant is second-order then we put it in the rate equation raised to the power of 2; in other words, squared. You'll see it in the rate equation as [A]2.
#### Order of a reaction
The overall order of a reaction is the sum of all the individual reactant orders. Remember that the order of a reactant is the power that it is raised to in the rate equation. If you're ever asked to find the overall order of reaction, simply add together all of the powers present in the equation and you'll reach your final answer.
Understanding orders of reaction is a bit tricky, so let's look at an example to help you understand.
A reaction has the chemical equation and rate equation shown below.
$\mathrm{A}+\mathrm{B}+\mathrm{C}\to \mathrm{D}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{rate}=\mathrm{k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^{2}$
Describe the effect of doubling the concentrations of A, B, and C.
First of all, looking at the rate equation, we can see that the only species present are A and B. C does not appear at all. It must therefore be zero-order. Hence, doubling the concentration of C will have no effect on the rate of reaction.
On the other hand, A does appear in the rate equation. It looks like [A] isn't raised to any power, but as we learnt above, [A] is the same as saying [A]1. A is therefore first-order. Doubling the concentration of A will cause the rate of reaction to double, because 21 = 2.
B also appears in the rate equation. It is raised to the power of 2, meaning that it is second-order. Doubling the concentration of B will cause the rate of reaction to quadruple, because 22 = 4.
## Determining the rate equation
There are a few different methods we can use to determine the rate equation for a reaction. The basic principles come down to determining the species involved in the rate equation and then finding each of their orders. The main methods for doing this are:
• The initial rates method.
• Using rate-concentration graphs.
• Finding first-order reactants from their half-life.
• Inspecting the reaction mechanism.
We cover these methods in much more detail in Determining Reaction Order, but we'll explore them briefly now.
### Initial rates
The initial rates method involves measuring the rate of the same reaction over several experiments, each with different starting concentrations of a particular reactant. This method allows us to see numerically how the concentration of the reactant affects the rate of the reaction. We do this for each reactant, and can use the information to determine the reactant's order.
### Rate-concentration graphs
Earlier in the article, we looked at how you use graphs showing concentration of a species against time to calculate rate of reaction at a specific instant. You can then take the values for instantaneous rate of reaction and plot them against concentration to make a rate-concentration graph. These take specific shapes, depending on the order of the species involved.
• A horizontal straight line shows that the rate of reaction is unaffected by the concentration of the species. The species is therefore zero-order.
• A sloping straight line through the origin shows that the rate is directly proportional to the concentration of the species. The species is therefore first-order.
• A curved line through the origin shows that the rate is exponentially proportional to the concentration of the species. The species is second-order or higher.
Fig. 6 - Rate-concentration graphs for reactants with different orders
### Half-life equations
The half-life, ${\mathrm{t}}_{1/2}$, of a reactant is the time it takes for the concentration of that reactant to become half of what it was where you started measuring from. There's an interesting feature of first-order reactants: they have a constant half-life. This means that it takes the same amount of time to get from, say, a concentration of 1.0 to a concentration of 0.5 mol dm-3, as it does to get from a concentration of 0.8 to 0.4 mol dm-3. In both cases, the concentration has halved.
You can measure half-life using concentration-time graphs. Pick any point on the graph and look at the concentration for that time value. Then, see how long it takes to halve the concentration. Repeat this again to find multiple half-lives for a species. If all of the half-lives are the same, the species is first-order.
Fig. 7 - A concentration-time graph used to show half-life
#### Half-life and rate constant
The half-life of a first-order reactant relates to the rate constant, k, using the following equation:
$\mathrm{k}=\frac{\mathrm{ln}\left(2\right)}{{\mathrm{t}}_{1/2}}$
This means that once you know the half-life of a first-order reactant, you can easily find k.
### Reaction mechanism
Reactions can have mechanisms with one step or multiple steps. Each step happens at a different speed, and the rate of reaction is determined by the slowest step. We call the slowest step in a reaction the rate-determining step, and it gives us an idea of what the rate equation is likely to look like. This is because the rate equation is only made up of reacting species found in the steps up to and including the rate-determining step. The number of moles of each species relates to its order.
If you know the reaction mechanism and the rate-determining step of a reaction, you can predict the rate equation!
## Rate Equations - Key takeaways
• The rate of a chemical reaction is the change in concentration of reactants or products over time.
• Rate of reaction can be represented by a rate equation. Rate equations are composed of a rate constant (k), and reactant concentrations raised to the power of their respective order.
• Rate constants are constant for a particular reaction at a certain temperature.
• The order of a reaction with respect to a species tells us how the rate of reaction depends on the concentration of that species.
• The concentration of zero-order reactants has no effect on the rate of reaction.
• The concentration of first-order reactants is directly proportional to the rate of reaction.
• The concentration of second-order reactants has an exponential effect on the rate of reaction.
• The rate equation can be determined using the initial rates method, by identifying the shapes of graphs, by calculating half-lives, and by inspecting the reaction mechanism.
To calculate the rate equation, you need to find out the order of reaction with respect to each species involved in the reaction. You also need to find the rate constant, k. You can do this experimentally. Once you've formed a rate equation, you can substitute in known concentration values and find the rate of reaction at a particular instant.
Rate equations are written in the form rate = k [A]m [B]n. The rate constant, k, is a value that is always constant for a particular reaction at a particular temperature. [A] represents the concentration of A, whilst the letter m represents the order of the reaction with respect to A. Overall, [A]m means the concentration of A, raised to the power of m. To write a rate equation, you work out the rate constant and the orders of reaction with respect to each species involved, and write them in the form given above.
The rate equation tells us the rate of a reaction. This means that it tells us the rate of change of reactant or product concentration during a reaction. So, by calculating the rate equation, you can find the rate of change.
You can find the rate of a reaction by using the rate equation. The rate equation is a formula that tells us the rate of any reaction from the concentration of its reactants.
Some factors that affect the rate of a reaction include reactant concentration, surface area, temperature, and activation energy.
## Rate Equations Quiz - Teste dein Wissen
Question
Why is the Arrhenius equation useful in chemistry?
It allows us to relate the temperature of a reaction with its rate
Show question
Question
What does the letter A represent in the Arrhenius equation?
A is the Arrhenius constant
Show question
Question
What information about a reaction can the value of ex give us in the Arrhenius equation?
the number of reacting particles that have enough energy to react
Show question
Question
Rearrange the Arrhenius equation into its logarithmic form.
ln(k) = ln(A) - Ea/RT
Show question
Question
Given a graph showing an Arrhenius plot for a chemical reaction, how could you use the plotted data to determine activation energy and the rate constant for that reaction?
The activation energy would be equal to the gradient of the line.
The rate constant would be equal to the y-intercept.
Show question
Question
When drawing an Arrhenius plot, what would you label your axis?
The x-axis would be 1/T
The y-axis would be ln(K)
Show question
Question
What does the Arrhenius equation show us about the effect of temperature on the rate of a reaction?
The rate of a reaction will increase if the temperature at which the reaction occurs increases.
Show question
Question
What is a rate equation?
A mathematical equation that links the rate of reaction with the concentration of species involved in the reaction.
Show question
Question
What does the letter k represent in the rate equation?
The rate constant
Show question
Question
What do the letters m and n represent in the rate equation?
The order of the reaction with respect to a certain species.
Show question
Question
Give four ways of determining the rate equation.
• Initial rates method
• Rate-concentration graphs
• Half-life
• Reaction mechanism
Show question
Question
When using the initial rates method to determine the rate equation, you _____.
Keep the external conditions the same.
Show question
Question
How can we use half-lives to help determine the rate equation?
We can use half-lives to identify first-order reactants. First-order reactants have a constant half-life. You can see this by plotting a concentration-time graph.
Show question
Question
What is half-life?
The half-life (t1/2) of a species is the time it takes for half of the species to be used in the reaction. In other words, it is the time it takes for its concentration to halve.
Show question
Question
True or false? Second-order reactants have a constant half-life.
False
Show question
Question
Only species in _____ feature in the rate equation.
The steps up to and including the rate-determining step
Show question
Question
Which step in a chemical reaction is the rate-determining step?
The slowest step
Show question
Question
What is rate of reaction?
Change in concentration of reactants or products over time.
Show question
Question
What are the units for rate of reaction?
mol dm-3 s-1
Show question
Question
Give three ways of measuring rate of reaction.
• pH change
• Change in mass
• Volume of gas produced
• Colour change
Show question
Question
What does the letter k represent in the rate equation?
The rate constant
Show question
Question
What does [A] represent in the rate equation?
The concentration of species A, typically in mol dm-3
Show question
Question
What do the letters m and n represent in the rate equation?
The order of reaction with respect to a particular species. They show the power that the species is raised to in the rate equation.
Show question
Question
rate = k [A]2 [B]
What is the order of reaction with respect to A?
2
Show question
Question
rate = k [A]2 [B]
What is the overall order of the reaction?
3
Show question
Question
'k' is constant at different temperatures. True or false?
False
Show question
Question
What are the units of the rate constant k?
It depends
Show question
Question
rate = k [A] [B]
What are the units of k for this reaction?
mol-1 dm3 s-1
Show question
Question
How does the concentration of zero-order species affect the rate of reaction?
It has no effect on the rate of reaction.
Show question
Question
How does the concentration of first-order species affect the rate of reaction?
It is directly proportional to the rate of reaction.
Show question
Question
How does the concentration of second-order species affect the rate of reaction?
It has an exponential effect on the rate of reaction.
Show question
Question
The concentration of a second-order species doubles. Predict the effect on the rate of reaction.
Show question
Question
The concentration of a first-order species increases by a factor of 3. Predict the effect on the rate of reaction.
The rate of reaction increases by a factor of 3.
Show question
Question
Which of these do not appear in the Arrhenius equation?
ΔT
Show question
Question
What are the units of Ea in the Arrhenius equation?
J mol-1
Show question
Question
What are the units of k in the Arrhenius equation?
mol-1 dm3 s-1
Show question
Question
What does k represent in the Arrhenius equation?
The rate constant
Show question
Question
What does T represent in the Arrhenius equation and what are its units?
Temperature, K
Show question
Question
According to the Arrhenius equation, increasing the temperature _____ the rate of reaction.
Increases
Show question
Question
According to the Arrhenius equation, decreasing the activation energy _____ the rate of reaction.
Increases
Show question
Question
What is the significance of the y-value at the point where x = 0 on an Arrhenius plot?
The y-value at the point where x = 0 is equal to ln(A)
Show question
Question
What is the Arrhenius equation?
The Arrhenius equation is a mathematical formula that relates the rate constant of a reaction with the activation energy and temperature of that reaction.
Show question
Question
What goes on the y-axis on an Arrhenius plot?
ln(k)
Show question
Question
What information does the y-coordinate of the point at which x = 0 give you?
ln(A)
Show question
Question
A reaction takes place at 600K. The Arrhenius constant equals 3.8 x 1011 s-1 and the activation energy of this reaction is 240 kJ mol-1. Calculate the rate constant, k.
4.73 x 10-10 s-1
Show question
Question
A reaction takes place at 650K. The Arrhenius constant equals 5.617 x 1012 mol-1 dm3 s-1 and the rate constant equals 0.53 mol-1 dm3 s-1. Calculate the activation energy, giving your answer to three significant figures.
162000 J mol-1 (162 kJ mol-1)
Show question
Question
True or false? For reactions with an activation energy of around 50 kJ mol-1, increasing the temperature from 290K to 200K will triple the rate constant.
False
Show question
Question
What is the rate constant?
The rate constant is a proportionality constant that links the concentrations of certain species to the rate of a chemical reaction.
Show question
Question
What is the symbol for the rate constant?
k
Show question
Question
Which of the following statements is always true about the rate constant?
0
Show question
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https://www.kodytools.com/units/massflow/from/kgps/to/mgps | 1,726,537,131,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651722.42/warc/CC-MAIN-20240917004428-20240917034428-00622.warc.gz | 791,564,619 | 15,398 | # Kilogram/Second to Milligram/Second Converter
1 Kilogram/Second = 1000000 Milligram/Second
## One Kilogram/Second is Equal to How Many Milligram/Second?
The answer is one Kilogram/Second is equal to 1000000 Milligram/Second and that means we can also write it as 1 Kilogram/Second = 1000000 Milligram/Second. Feel free to use our online unit conversion calculator to convert the unit from Kilogram/Second to Milligram/Second. Just simply enter value 1 in Kilogram/Second and see the result in Milligram/Second.
Manually converting Kilogram/Second to Milligram/Second can be time-consuming,especially when you don’t have enough knowledge about Mass Flow units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Kilogram/Second to Milligram/Second converter tool to get the job done as soon as possible.
We have so many online tools available to convert Kilogram/Second to Milligram/Second, but not every online tool gives an accurate result and that is why we have created this online Kilogram/Second to Milligram/Second converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Kilogram/Second to Milligram/Second (kg/s to mg/s)
By using our Kilogram/Second to Milligram/Second conversion tool, you know that one Kilogram/Second is equivalent to 1000000 Milligram/Second. Hence, to convert Kilogram/Second to Milligram/Second, we just need to multiply the number by 1000000. We are going to use very simple Kilogram/Second to Milligram/Second conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Kilogram/Second} = 1 \times 1000000 = \text{1000000 Milligram/Second}$$
## What Unit of Measure is Kilogram/Second?
Kilogram/Second or Kilogram per Second is a unit of measurement for mass flow rate. It is defined as flow or movement of one kilogram of mass in one second.
## What is the Symbol of Kilogram/Second?
The symbol of Kilogram/Second is kg/s. This means you can also write one Kilogram/Second as 1 kg/s.
## What Unit of Measure is Milligram/Second?
Milligram/Second or Milligram per Second is a unit of measurement for mass flow rate. It is defined as flow or movement of one milligram of mass in one second.
## What is the Symbol of Milligram/Second?
The symbol of Milligram/Second is mg/s. This means you can also write one Milligram/Second as 1 mg/s.
## How to Use Kilogram/Second to Milligram/Second Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Kilogram/Second and in the first input field, enter a value.
• From the second dropdown, select Milligram/Second.
• Instantly, the tool will convert the value from Kilogram/Second to Milligram/Second and display the result in the second input field.
Kilogram/Second
1
Milligram/Second
1000000
# Kilogram/Second to Milligram/Second Conversion Table
Kilogram/Second [kg/s]Milligram/Second [mg/s]Description
1 Kilogram/Second1000000 Milligram/Second1 Kilogram/Second = 1000000 Milligram/Second
2 Kilogram/Second2000000 Milligram/Second2 Kilogram/Second = 2000000 Milligram/Second
3 Kilogram/Second3000000 Milligram/Second3 Kilogram/Second = 3000000 Milligram/Second
4 Kilogram/Second4000000 Milligram/Second4 Kilogram/Second = 4000000 Milligram/Second
5 Kilogram/Second5000000 Milligram/Second5 Kilogram/Second = 5000000 Milligram/Second
6 Kilogram/Second6000000 Milligram/Second6 Kilogram/Second = 6000000 Milligram/Second
7 Kilogram/Second7000000 Milligram/Second7 Kilogram/Second = 7000000 Milligram/Second
8 Kilogram/Second8000000 Milligram/Second8 Kilogram/Second = 8000000 Milligram/Second
9 Kilogram/Second9000000 Milligram/Second9 Kilogram/Second = 9000000 Milligram/Second
10 Kilogram/Second10000000 Milligram/Second10 Kilogram/Second = 10000000 Milligram/Second
100 Kilogram/Second100000000 Milligram/Second100 Kilogram/Second = 100000000 Milligram/Second
1000 Kilogram/Second1000000000 Milligram/Second1000 Kilogram/Second = 1000000000 Milligram/Second
# Kilogram/Second to Other Units Conversion Table
ConversionDescription
1 Kilogram/Second = 86400 Kilogram/Day1 Kilogram/Second in Kilogram/Day is equal to 86400
1 Kilogram/Second = 3600 Kilogram/Hour1 Kilogram/Second in Kilogram/Hour is equal to 3600
1 Kilogram/Second = 60 Kilogram/Minute1 Kilogram/Second in Kilogram/Minute is equal to 60
1 Kilogram/Second = 86400000000 Milligram/Day1 Kilogram/Second in Milligram/Day is equal to 86400000000
1 Kilogram/Second = 3600000000 Milligram/Hour1 Kilogram/Second in Milligram/Hour is equal to 3600000000
1 Kilogram/Second = 60000000 Milligram/Minute1 Kilogram/Second in Milligram/Minute is equal to 60000000
1 Kilogram/Second = 1000000 Milligram/Second1 Kilogram/Second in Milligram/Second is equal to 1000000
1 Kilogram/Second = 86400000 Gram/Day1 Kilogram/Second in Gram/Day is equal to 86400000
1 Kilogram/Second = 3600000 Gram/Hour1 Kilogram/Second in Gram/Hour is equal to 3600000
1 Kilogram/Second = 60000 Gram/Minute1 Kilogram/Second in Gram/Minute is equal to 60000
1 Kilogram/Second = 1000 Gram/Second1 Kilogram/Second in Gram/Second is equal to 1000
1 Kilogram/Second = 8.64e-11 Exagram/Day1 Kilogram/Second in Exagram/Day is equal to 8.64e-11
1 Kilogram/Second = 3.6e-12 Exagram/Hour1 Kilogram/Second in Exagram/Hour is equal to 3.6e-12
1 Kilogram/Second = 6e-14 Exagram/Minute1 Kilogram/Second in Exagram/Minute is equal to 6e-14
1 Kilogram/Second = 1e-15 Exagram/Second1 Kilogram/Second in Exagram/Second is equal to 1e-15
1 Kilogram/Second = 8.64e-8 Petagram/Day1 Kilogram/Second in Petagram/Day is equal to 8.64e-8
1 Kilogram/Second = 3.6e-9 Petagram/Hour1 Kilogram/Second in Petagram/Hour is equal to 3.6e-9
1 Kilogram/Second = 6e-11 Petagram/Minute1 Kilogram/Second in Petagram/Minute is equal to 6e-11
1 Kilogram/Second = 1e-12 Petagram/Second1 Kilogram/Second in Petagram/Second is equal to 1e-12
1 Kilogram/Second = 0.0000864 Teragram/Day1 Kilogram/Second in Teragram/Day is equal to 0.0000864
1 Kilogram/Second = 0.0000036 Teragram/Hour1 Kilogram/Second in Teragram/Hour is equal to 0.0000036
1 Kilogram/Second = 6e-8 Teragram/Minute1 Kilogram/Second in Teragram/Minute is equal to 6e-8
1 Kilogram/Second = 1e-9 Teragram/Second1 Kilogram/Second in Teragram/Second is equal to 1e-9
1 Kilogram/Second = 0.0864 Gigagram/Day1 Kilogram/Second in Gigagram/Day is equal to 0.0864
1 Kilogram/Second = 0.0036 Gigagram/Hour1 Kilogram/Second in Gigagram/Hour is equal to 0.0036
1 Kilogram/Second = 0.00006 Gigagram/Minute1 Kilogram/Second in Gigagram/Minute is equal to 0.00006
1 Kilogram/Second = 0.000001 Gigagram/Second1 Kilogram/Second in Gigagram/Second is equal to 0.000001
1 Kilogram/Second = 86.4 Megagram/Day1 Kilogram/Second in Megagram/Day is equal to 86.4
1 Kilogram/Second = 3.6 Megagram/Hour1 Kilogram/Second in Megagram/Hour is equal to 3.6
1 Kilogram/Second = 0.06 Megagram/Minute1 Kilogram/Second in Megagram/Minute is equal to 0.06
1 Kilogram/Second = 0.001 Megagram/Second1 Kilogram/Second in Megagram/Second is equal to 0.001
1 Kilogram/Second = 864000 Hectogram/Day1 Kilogram/Second in Hectogram/Day is equal to 864000
1 Kilogram/Second = 36000 Hectogram/Hour1 Kilogram/Second in Hectogram/Hour is equal to 36000
1 Kilogram/Second = 600 Hectogram/Minute1 Kilogram/Second in Hectogram/Minute is equal to 600
1 Kilogram/Second = 10 Hectogram/Second1 Kilogram/Second in Hectogram/Second is equal to 10
1 Kilogram/Second = 8640000 Dekagram/Day1 Kilogram/Second in Dekagram/Day is equal to 8640000
1 Kilogram/Second = 360000 Dekagram/Hour1 Kilogram/Second in Dekagram/Hour is equal to 360000
1 Kilogram/Second = 6000 Dekagram/Minute1 Kilogram/Second in Dekagram/Minute is equal to 6000
1 Kilogram/Second = 100 Dekagram/Second1 Kilogram/Second in Dekagram/Second is equal to 100
1 Kilogram/Second = 864000000 Decigram/Day1 Kilogram/Second in Decigram/Day is equal to 864000000
1 Kilogram/Second = 36000000 Decigram/Hour1 Kilogram/Second in Decigram/Hour is equal to 36000000
1 Kilogram/Second = 600000 Decigram/Minute1 Kilogram/Second in Decigram/Minute is equal to 600000
1 Kilogram/Second = 10000 Decigram/Second1 Kilogram/Second in Decigram/Second is equal to 10000
1 Kilogram/Second = 8640000000 Centigram/Day1 Kilogram/Second in Centigram/Day is equal to 8640000000
1 Kilogram/Second = 360000000 Centigram/Hour1 Kilogram/Second in Centigram/Hour is equal to 360000000
1 Kilogram/Second = 6000000 Centigram/Minute1 Kilogram/Second in Centigram/Minute is equal to 6000000
1 Kilogram/Second = 100000 Centigram/Second1 Kilogram/Second in Centigram/Second is equal to 100000
1 Kilogram/Second = 86400000000000 Microgram/Day1 Kilogram/Second in Microgram/Day is equal to 86400000000000
1 Kilogram/Second = 3600000000000 Microgram/Hour1 Kilogram/Second in Microgram/Hour is equal to 3600000000000
1 Kilogram/Second = 60000000000 Microgram/Minute1 Kilogram/Second in Microgram/Minute is equal to 60000000000
1 Kilogram/Second = 1000000000 Microgram/Second1 Kilogram/Second in Microgram/Second is equal to 1000000000
1 Kilogram/Second = 190479.39 Pound/Day1 Kilogram/Second in Pound/Day is equal to 190479.39
1 Kilogram/Second = 7936.64 Pound/Hour1 Kilogram/Second in Pound/Hour is equal to 7936.64
1 Kilogram/Second = 132.28 Pound/Minute1 Kilogram/Second in Pound/Minute is equal to 132.28
1 Kilogram/Second = 2.2 Pound/Second1 Kilogram/Second in Pound/Second is equal to 2.2
1 Kilogram/Second = 3047670.31 Ounce/Day1 Kilogram/Second in Ounce/Day is equal to 3047670.31
1 Kilogram/Second = 126986.26 Ounce/Hour1 Kilogram/Second in Ounce/Hour is equal to 126986.26
1 Kilogram/Second = 2116.44 Ounce/Minute1 Kilogram/Second in Ounce/Minute is equal to 2116.44
1 Kilogram/Second = 35.27 Ounce/Second1 Kilogram/Second in Ounce/Second is equal to 35.27
1 Kilogram/Second = 95.24 Ton/Day [Short]1 Kilogram/Second in Ton/Day [Short] is equal to 95.24
1 Kilogram/Second = 3.97 Ton/Hour [Short]1 Kilogram/Second in Ton/Hour [Short] is equal to 3.97
1 Kilogram/Second = 0.066138678655463 Ton/Minute [Short]1 Kilogram/Second in Ton/Minute [Short] is equal to 0.066138678655463
1 Kilogram/Second = 0.0011023113109244 Ton/Second [Short]1 Kilogram/Second in Ton/Second [Short] is equal to 0.0011023113109244
1 Kilogram/Second = 86.4 Ton/Day [Metric]1 Kilogram/Second in Ton/Day [Metric] is equal to 86.4
1 Kilogram/Second = 3.6 Ton/Hour [Metric]1 Kilogram/Second in Ton/Hour [Metric] is equal to 3.6
1 Kilogram/Second = 0.06 Ton/Minute [Metric]1 Kilogram/Second in Ton/Minute [Metric] is equal to 0.06
1 Kilogram/Second = 0.001 Ton/Second [Metric]1 Kilogram/Second in Ton/Second [Metric] is equal to 0.001 | 3,316 | 10,653 | {"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} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.923408 |
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As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
Q1. The volume of a proton is approximately:
• 1.5×10^(-30) cm^3
• 1.5×10^(-38) cm^3
• 1.5×10^(-34) cm^3
• None of these
Solution
An experimental fact.
Q2.When the frequency of light incident on a metallic plate is doubled, the KE of the emitted photoelectrons will be:
• Douled
• Halved
• Increased but more than doubled of the previous KE
• Unchanged
Solution
hv_1=work function+K∙E_1 2×hv_1=work function+K∙E_2
Q3. The wavelength of a spectral line emitted by hydrogen atom in the Lyman series is16/15R cm. What is the value of n_2?(R=Rydberg constant)
• 2
• 3
• 4
• 1
Solution
For Lyman series, 1/λ=R[1/1^2 -1/(n_2^2 )] 15R/16=R[1/1^2 -1/(n_2^2 )] 15R/16R=[(n_2^2-1)/(n_2^2 )] 15/16=(n_2^2-1)/(n_2^2 ) 15n_2^2=16n_2^2-16 n_2^2=16,n_2=4
Q4. Ground state electronic configuration of nitrogen atom can be represented as
•
•
•
•
Solution
Q5.Which diagram best represents the appearance of the line spectrum of atomic hydrogen in the visible region?
•
•
•
•
Solution
Q6. Who introduced the concept of electron spin?
• Schrodinger
• Planck
• Bohr
• Uhlenbeck and Gaudsmit
Solution
They proposed the concept of electron spin.
Q7.Which is not basic postulate of Dalton's atomic theory?
• Atoms are neither created nor destroyed in a chemical reaction
• In a given compound, the relative number and kinds of atoms are constant
• Atoms of all elements are alike, including their masses
• Each element is composed of extremely small particles called atoms
Solution
Atoms of an element are alike.
Q8.The threshold frequency of a metal is 4×10^14 s^(-1). The minimum energy of photon to cause photoelectric effect is:
• 3.06×10^(-12) J
• 1.4×10^(-18) J
• 3.4×10^(-19) J
• 2.64×10^(-19) J
Solution
E_Mini=hv_0
Q9.In photoelectric effect the number of photo-electron emitted is proportional to :
• Intensity of incident beam
• Frequency of incident beam
• Velocity of incident beam
• Work function of photo cathode
Solution
More intense beam will give out more electrons.
Q10. Which of the following elements has least number of electrons in its M-shell?
• K
• Mn
• Ni
• Sc
Solution
〖_19〗K=1s^2,2s^2 2p^6,3s^2 3p^6,4s^1 〖_25〗Mn=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5 〖_28〗Ni=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^8 〖_21〗Sc=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1 Therefore, K has least number of electrons in its M-shell (n=3)=8.
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from：http://statistics.ats.ucla.edu/stat/sas/faq/compreg3.htm
Sometimes your research may predict that the size of a regression coefficient may vary across groups. For example, you might believe that the regression coefficient of height predicting weight would differ across three age
groups (young, middle age, senior citizen). Below, we have a data file with 10 fictional young people, 10 fictional middle age people, and 10 fictional senior citizens, along with their height in inches and their weight in
pounds. The variableage indicates the age group and is coded 1 for young people, 2 for middle aged, and 3 for senior citizens.
DATA htwt;
INPUT id age height weight ;
CARDS;
1 1 56 140
2 1 60 155
3 1 64 143
4 1 68 161
5 1 72 139
6 1 54 159
7 1 62 138
8 1 65 121
9 1 65 161
10 1 70 145
11 2 56 117
12 2 60 125
13 2 64 133
14 2 68 141
15 2 72 149
16 2 54 109
17 2 62 128
18 2 65 131
19 2 65 131
20 2 70 145
21 3 64 211
22 3 68 223
23 3 72 235
24 3 76 247
25 3 80 259
26 3 62 201
27 3 69 228
28 3 74 245
29 3 75 241
30 3 82 269
;
RUN;
We analyze their data separately using the proc reg below.
PROC REG DATA=htwt;
BY age ;
MODEL weight = height ;
RUN;
The parameter estimates (coefficients) for the young, middle age, and senior citizens are shown below. below, and the results do seem to suggest thatheight is a stronger predictor of weight for seniors (3.18) than for the middle
aged (2.09). The results also seem to suggest that height does not predictweight as strongly for the young (-.37) as for the middle aged and seniors. However, we would need to perform specific significance tests to be able
to make claims about the differences among these regression coefficients.
AGE=1
Parameter Standard T for H0:
Variable DF Estimate Error Parameter=0 Prob > |T|
INTERCEP 1 170.166445 49.43018216 3.443 0.0088
HEIGHT 1 -0.376831 0.77433413 -0.487 0.6396
AGE=2
Parameter Standard T for H0:
Variable DF Estimate Error Parameter=0 Prob > |T|
INTERCEP 1 -2.397470 7.05327189 -0.340 0.7427
HEIGHT 1 2.095872 0.11049098 18.969 0.0001
AGE=3
Parameter Standard T for H0:
Variable DF Estimate Error Parameter=0 Prob > |T|
INTERCEP 1 5.601677 8.93019669 0.627 0.5480
HEIGHT 1 3.189727 0.12323669 25.883 0.0001
We can compare the regression coefficients among these three age groups to test the null hypothesis
Ho: B1 = B2 = B3
where B1 is the regression for for the young, B2 is the regression for for the middle aged, and B3 is the regression for for senior citizens. To do this analysis, we first make
a dummy variable called age1 that is coded 1 if young (age=1), 0 otherwise, and age2 that is coded 1 if middle aged (age=2), 0 otherwise. We also create age1ht that is age1 times height,
and age2ht that is age2 times height.
data htwt2;
set htwt;
age1 = . ;
age2 = . ;
IF age = 1 then age1 = 1; ELSE age1 = 0 ;
IF age = 2 then age2 = 1; ELSE age2 = 0 ;
age1ht = age1*height ;
age2ht = age2*height ;
RUN;
We can now use age1 age2 height, age1ht and age2ht as predictors in the regression equation in proc reg below. In the proc reg we use the
TEST age1ht=0, age2ht=0;
statement to test the null hypothesis
Ho: B1 = B2 = B3
This test will have two degrees of freedom because it compares among three regression coefficients.
PROC REG DATA=htwt2 ;
MODEL weight = age1 age2 height age1ht age2ht ;
TEST age1ht=0, age2ht=0 ;
RUN;
The output below shows that the null hypothesis
Ho: B1 = B2 = B3
can be rejected (F=17.29, p = 0.0000). This means that the regression coefficients between height and weight do indeed significantly differ across the 3 age groups (young, middle age, senior citizen).
Model: MODEL1
Dependent Variable: WEIGHT
Analysis of Variance
Sum of Mean
Source DF Squares Square F Value Prob>F
Model 5 69595.35464 13919.07093 220.261 0.0001
Error 24 1516.64536 63.19356
C Total 29 71112.00000
Root MSE 7.94944 R-square 0.9787
Dep Mean 171.00000 Adj R-sq 0.9742
C.V. 4.64879
Parameter Estimates
Parameter Standard T for H0:
Variable DF Estimate Error Parameter=0 Prob > |T|
INTERCEP 1 5.601677 29.48853690 0.190 0.8509
AGE1 1 164.564768 41.55490307 3.960 0.0006
AGE2 1 -7.999147 41.55490307 -0.192 0.8490
HEIGHT 1 3.189727 0.40694172 7.838 0.0001
AGE1HT 1 -3.566558 0.61316088 -5.817 0.0001
AGE2HT 1 -1.093855 0.61316088 -1.784 0.0871
Dependent Variable: WEIGHT
Test: Numerator: 1092.7718 DF: 2 F value: 17.2925
Denominator: 63.19356 DF: 24 Prob>F: 0.0001
It is also possible to run such an analysis in proc glm, using syntax as shown below. Instead of using a test statement, the contrast statement is used to test the null hypothesis
Ho: B1 = B2 = B3
The contrast statement uses the comma to join together what would have been two separate one degree of freedom tests into a single two degree of freedom test that tests the null hypothesis above.
PROC GLM DATA=htwt2 ;
CLASS age ;
MODEL weight = age height age*height / SOLUTION ;
CONTRAST 'test equal slopes' age*height 1 -1 0,
age*height 0 1 -1 ;
RUN;
If you compare the contrast output from proc glm (labeled test equal slopes found below with the output from test from proc glm above, you will see the F values and p values
are the same. This is because these two tests are equivalent.
General Linear Models Procedure
Class Level Information
Class Levels Values
AGE 3 1 2 3
Number of observations in data set = 30
General Linear Models Procedure
Dependent Variable: WEIGHT
Sum of Mean
Source DF Squares Square F Value Pr > F
Model 5 69595.354644 13919.070929 220.26 0.0001
Error 24 1516.645356 63.193557
Corrected Total 29 71112.000000
R-Square C.V. Root MSE WEIGHT Mean
0.978672 4.648794 7.9494375 171.00000
Source DF Type I SS Mean Square F Value Pr > F
AGE 2 64350.600000 32175.300000 509.15 0.0001
HEIGHT 1 3059.211075 3059.211075 48.41 0.0001
HEIGHT*AGE 2 2185.543569 1092.771784 17.29 0.0001
Source DF Type III SS Mean Square F Value Pr > F
AGE 2 1395.9046778 697.9523389 11.04 0.0004
HEIGHT 1 2597.0189017 2597.0189017 41.10 0.0001
HEIGHT*AGE 2 2185.5435689 1092.7717845 17.29 0.0001
Contrast DF Contrast SS Mean Square F Value Pr > F
test equal slopes 2 2185.5435689 1092.7717845 17.29 0.0001
T for H0: Pr > |T| Std Error of
Parameter Estimate Parameter=0 Estimate
INTERCEPT 5.6016771 B 0.19 0.8509 29.48853690
AGE 1 164.5647676 B 3.96 0.0006 41.55490307
2 -7.9991472 B -0.19 0.8490 41.55490307
3 0.0000000 B . . .
HEIGHT 3.1897275 B 7.84 0.0001 0.40694172
HEIGHT*AGE 1 -3.5665584 B -5.82 0.0001 0.61316088
2 -1.0938553 B -1.78 0.0871 0.61316088
3 0.0000000 B . . .
NOTE: The X'X matrix has been found to be singular and a generalized inverse
was used to solve the normal equations. Estimates followed by the
letter 'B' are biased, and are not unique estimators of the parameters.
You might notice that the null hypothesis that we are testing
Ho: B1 = B2 = B3
is similar to the null hypothesis that you might test using ANOVA to compare the means of the three groups,
Ho: Mu1 = Mu2 = Mu3
In ANOVA, you can get an overall F test testing the null hypothesis. In addition to that overall test, you could perform planned comparisons among the three groups. So far we have seen how to to an overall test of the equality of the three regression coefficients,
and now we will test planned comparisons among the regression coefficients. Below, we show how you can perform two such tests using the contrasta statement in proc glm. The first contrastcompares the regression
coefficients of the middle aged vs. senior.
Ho: B2 = B3
The second contrast compares the regression coefficients of the young vs. middle aged and seniors.
Ho: B1 = (B2 + B3)/2
PROC GLM DATA=htwt2 ;
CLASS age ;
MODEL weight = age height age*height ;
CONTRAST 'Mid Age vs. Sen. ' age*height 0 1 -1 ;
CONTRAST 'Yng vs (Mid & Sen)' age*height -2 1 1 ;
RUN;
The output from contrast indicates that regression coefficients for middle aged and seniors do not significantly differ (F=3.18, p=0.0871) The secondcontrast was significant (F=29.96, p=0.0000) indicating that the regression
coefficients for the young differ from the middle age and seniors combined.
General Linear Models Procedure
Class Level Information
Class Levels Values
AGE 3 1 2 3
Number of observations in data set = 30
General Linear Models Procedure
Dependent Variable: WEIGHT
Sum of Mean
Source DF Squares Square F Value Pr > F
Model 5 69595.354644 13919.070929 220.26 0.0001
Error 24 1516.645356 63.193557
Corrected Total 29 71112.000000
R-Square C.V. Root MSE WEIGHT Mean
0.978672 4.648794 7.9494375 171.00000
Source DF Type I SS Mean Square F Value Pr > F
AGE 2 64350.600000 32175.300000 509.15 0.0001
HEIGHT 1 3059.211075 3059.211075 48.41 0.0001
HEIGHT*AGE 2 2185.543569 1092.771784 17.29 0.0001
Source DF Type III SS Mean Square F Value Pr > F
AGE 2 1395.9046778 697.9523389 11.04 0.0004
HEIGHT 1 2597.0189017 2597.0189017 41.10 0.0001
HEIGHT*AGE 2 2185.5435689 1092.7717845 17.29 0.0001
Contrast DF Contrast SS Mean Square F Value Pr > F
Mid Age vs. Sen. 1 201.1146303 201.1146303 3.18 0.0871
Yng vs (Mid & Sen) 1 1893.2074903 1893.2074903 29.96 0.0001
We can do the exact same analysis in proc reg by coding age1 and age2 like the coding shown in the contrast statements above We will create age1that will be:
0
for young
1
for middle age
-1
for senior
and we will create age2 that will be:
-2
for young
1
for middle age
1
for senior
The significance tests in proc reg below for age1ht and age2ht will correspond to the contrast statements we used in proc glm above.
data htwt3;
set htwt;
age1 = . ;
age2 = . ;
IF age = 1 then age1 = 0;
IF age = 2 then age1 = 1;
IF age = 3 then age1 = -1;
IF age = 1 then age2 = -2;
IF age = 2 then age2 = 1;
IF age = 3 then age2 = 1;
age1ht = age1*height ;
age2ht = age2*height ;
RUN;
PROC REG DATA=htwt3 ;
MODEL weight = age1 age2 height age1ht age2ht ;
RUN;
The results below correspond to the proc reg results above except that the proc glm results are reported as F values and the proc reg results are reported as t values. We can square the t values to make them
comparable to the F values. Indeed, for the comparison of Middle age vs. Seniors, the t value of -1.784 when squared becomes 3.183, the same as the F value from proc glm. Likewise, for the comparison of Young vs. middle & Senior the t value
from proc reg is 5.473 and when squared becomes 29.954, the same as the F value from proc glm.
Model: MODEL1
Dependent Variable: WEIGHT
Analysis of Variance
Sum of Mean
Source DF Squares Square F Value Prob>F
Model 5 69595.35464 13919.07093 220.261 0.0001
Error 24 1516.64536 63.19356
C Total 29 71112.00000
Root MSE 7.94944 R-square 0.9787
Dep Mean 171.00000 Adj R-sq 0.9742
C.V. 4.64879
Parameter Estimates
Parameter Standard T for H0:
Variable DF Estimate Error Parameter=0 Prob > |T|
INTERCEP 1 57.790217 16.94450462 3.411 0.0023
AGE1 1 -3.999574 20.77745154 -0.192 0.8490
AGE2 1 -56.188114 11.96726393 -4.695 0.0001
HEIGHT 1 1.636256 0.25524084 6.411 0.0001
AGE1HT 1 -0.546928 0.30658044 -1.784 0.0871
AGE2HT 1 1.006544 0.18389498 5.473 0.0001
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• 罗布麻和罗布麻 物种，不同测序技术和比对工具之间的基因组比较，以了解它们如何影响叶绿体基因序列组装。
• 我们知道对于数组的比较来说,实际比较的是它们中的每一个对应位置上元素. 所以最终都是要比较对象的. 我们还知道对于Swift中的类来说,要实现==操作符,需要遵守Equatable协议,并实现==方法. 比如对于类A来说: ...
本文简单介绍了Swift中派生与不派生自NSObject的类,在等于比较时表现出的不同行为;还顺带讨论了创建大数组时效率的问题.
等于或不等于
我们知道对于数组的比较来说,实际比较的是它们中的每一个对应位置上元素.
所以最终都是要比较对象的.
我们还知道对于Swift中的类来说,要实现==操作符,需要遵守Equatable协议,并实现==方法.
比如对于类A来说:
class A:NSObject{
var name:String
var id:Int
init(name:String,id:Int){
self.name = name
self.id = id
}
convenience override init() {
self.init(name: "", id: 0)
}
}
要实现如下方法:
static func ==(lhs:A,rhs:A)->Bool{
if lhs.name == rhs.name,lhs.id == rhs.id{
return true
}
return false
}
派生自NSObject的时候
一般情况下,上面的讨论无疑是正确,但同时却是不严谨的!
why???
因为当A继承自NSObject时,在做等于比较时不会调用==方法!
它会调用另一个方法:isEqual()
它的实现很简单:
override func isEqual(_ object: Any?) -> Bool {
let other = object as! A
return self == other
}
如何有效率的创建大数组
为了验证上面的结论,我们创建2个数组来比较,数组元素类型自然是A了:
let COUNT = 1000
var ary0 = [A]()
var ary1 = [A]()
for i in 0..<COUNT{
ary0.append(A(name: "A\(i)", id: i))
ary1.append(A(name: "A\(i)", id: i))
}
print("ary0 equ ary1 : \(ary0 == ary1)")
我们先让A不继承自NSObject,然后再继承自NSObject.
两种情况下比较都是相等的!!!
不过这里有一个小问题:你会发现哪怕只是创建1000个元素,也会耗时将近6秒钟的时间,这可不算快!
这是因为:Swift数组的添加操作非常费时,千万不要append方法创建大数组啊!
相反,我们可以用赋值方法创建数组!
let COUNT = 1000
var ary0:[A] = []
var ary1:[A] = []
Timing().timing {
ary0 = Array<A>(repeating: A(), count: 1000)
for i in 0..<COUNT{
ary0[i] = A(name: "A\(i)", id: i)
}
}
Timing().timing {
for i in 0..<COUNT{
ary1.append(A(name: "A\(i)", id: i))
}
}
如上,现在ary0是用赋值方法来创建的,我们来看看会快多少:
ary0平均执行时间 : 0.17282307147979736
ary1平均执行时间 : 5.686642050743103
is equ : true
答案显而易见!
快的不是一点哦!
最后ary0和ary1是相等的,这也进一步验证了我们上面的讨论! ?
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... | 9,153 | 21,099 | {"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} | 3.28125 | 3 | CC-MAIN-2021-21 | latest | en | 0.736439 |
https://www.jiskha.com/display.cgi?id=1193277189 | 1,503,407,775,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886110774.86/warc/CC-MAIN-20170822123737-20170822143737-00609.warc.gz | 945,013,428 | 4,221 | # algebra-i post the words that i need to fit toget
posted by .
What catastrophes makes ur kitten happy?
I need to fit all this words together:
LET, CHAIN, THE , HAVE, CAT, CRY, ONE,OUT, OVER , LITTLE, DOG, SPILT, NEW, OUTSIDE, MILK, OLD.
• algebra-i post the words that i need to fit toget -
You've posted this question three times. And three of us have tried to answer it. If you'd given us complete information from the start, we might have been able to help you. But now it's your turn. How do you think these words fit together?
And what does it have to do with math?
• Algebra 1 -
When You Cry Over Spilt Milk.
## Similar Questions
1. ### spanish
can u help me translate some words? First of aLL, you need to post the words here!
2. ### TAKE A GUESS
WHAT CATASTROPHE MAKES YOUR KITTEN HAPPY?
3. ### algebra-what catastrophe makes a kitten happy?
what catastrophe makes a kitten happy? OPTION : LET, CHAIN, THE , HAVE, CAT, CRY, ONE,OUT, OVER , LITTLE, DOG, SPILT, NEW, OUTSIDE, MILK, OLD.
4. ### english
Their fears proved ___ as the tower is a triumph of design what word could fit?
5. ### Socials
I am doing a crossword puzzle and I need help on two words. The first one is: Royal documents that set out terms and permission.( 3 words) There are 15 letters and the 7th letter is S. I thought it would be LettersPatent, but that …
6. ### English
need some advice! My learners will work on a text next week.In the text are 7 new words. The aim of the lesson is to learn those words and practice them in a few ways. What can I do to practice those new words. Any interesting ideas …
7. ### English
Can you help with writing a paragraph using all these words?
8. ### poetry
Is this a good Poem for a senior in creative writing?
9. ### English
Please help me find these words in the book hatchet. These words are located in chapter 13-16 and all I need to knnow is where these words are located. I have found the book online and I have posted it underneath. I have done a couple …
10. ### Behavioral Science
A man has a dog, cat, and mouse on one side of the river. He needs to get all of the animals over to the other side of the river. He can only take one animal at a time. If he leaves the cat alone with the mouse the cat will eat the …
More Similar Questions | 588 | 2,288 | {"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-2017-34 | latest | en | 0.932217 |
https://ubraintv-jp.com/the-molar-volume-of-a-gas-at-stp-occupies/ | 1,638,119,455,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358570.48/warc/CC-MAIN-20211128164634-20211128194634-00023.warc.gz | 674,847,990 | 5,300 | ## The Mole and also the Volume that Gas
It is fairly tricky to discover the variety of moles of a gas by weighing its mass. Chemists determine the variety of moles of any gas by measuring its volume. However, this cannot be done for solids and also liquids.It is found that under the exact same temperature and pressure, equal quantities of every gases save on computer the same number of particles. Therefore, chemists introduced the ide of molar volume.
You are watching: The molar volume of a gas at stp occupies
Molar volume of a gas is characterized as the volume that one mole the the gas.Thus, the molar volume is additionally the volume lived in by 6.02 x 1023 particles of gas.The molar volume of any kind of gas is 22.4 dm3 mol-1 in ~ STP or 24 dm3 mol-1 at room conditions.Note: STP refers to standard temperature the 0°C and also pressure the 1 atmosphere. Room problems refer come the temperature the 25°C and the push of 1 atmosphere.This means that one mole of any type of gas rectal the same volume at STP, i beg your pardon is 22.4 dm3. Under room conditions, one mole of any kind of gas rectal 24 dm3.
number above Each of these balloons contains 6.02 x 1023 gas molecules.The adhering to relationship shows exactly how the volume that a gas can be converted to the number of moles and vice versa.
In calculations, make certain that the volume the gas and also the molar volume are of the same unit, the is, both are in cm3 or both space in dm3. Remember, 1 dm3 = 1000 cm3
## The Mole and the Volume the Gas problems with Solutions
1. What is the volume of 0.4 mole that carbon dioxide gas at STP?Solution:Given the variety of moles the carbon dioxide, CO2 = 0.4 molTherefore, the volume of CO2 = variety of moles that CO2 x molar volume at STP= 0.4 x 22.4 dm3 = 8.96 dm3
2. Uncover the number of moles of ammonia gas had in a sample that 60 cm3 that the gas in ~ room conditions. Solution:
## The Relationship Between Mole, variety of particles, Mass and Volume
The complying with shows the relationships in between the variety of moles, variety of particles, mass and volume of gases.
In many calculations, we an initial convert various other quantities such together the number of particles, massive or volume come the variety of moles (refer to Table).
See more: The Excretory System That Removes Solid And Liquid Wastes From The Body?
Table: summary of procedures in calculations involving the number of moles
Conversion Steps From mass to volume Mass → variety of moles → volume From volume to mass Volume → number of moles → mass From volume to number of particles Volume → number of moles → number of particles From number of particles to volume Number of corpuscle → number of moles → volume
## The Relationship Between Mole, variety of particles, Mass and also Volume Problems with Solutions
1. What is the volume of 12 g of methane at STP? Solution:
2. A sample the 120 cm3 that carbon dioxide is accumulated at room conditions in an experiment. Calculate the mass of the sample that carbon dioxide.Solution:Given the volume the carbon dioxide, CO2 = 120 cm3 = 120/1000 dm3 = 0.12 dm3
Therefore, the mass of CO2 = variety of moles that CO2 × molar fixed of CO2 = 0.005 × <12 + 2(16)>= 0.005 × 44= 0.22 g
3. How countless hydrogen molecules room there in 6 dm3 the hydrogen gas at room conditions? Solution:
4. Find the volume of nitrogen gas in cm3 in ~ STP that consists of 2.408 × 1023 nitrogen molecules. Solution:
Filed Under: Chemistry Tagged With: The Mole and also the Volume of Gas, The Mole and the Volume the Gas difficulties with Solutions, The Relationship between Mole number of particles Mass and also Volume | 909 | 3,672 | {"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-2021-49 | latest | en | 0.920023 |
http://www.go4expert.com/forums/hi-im-pointers-c-post38030/ | 1,406,785,426,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1406510272584.13/warc/CC-MAIN-20140728011752-00115-ip-10-146-231-18.ec2.internal.warc.gz | 556,390,215 | 4,781 | A quick lesson on pointers. If a function has to change something that is passed in, then you need to pass in a pointer to that thing. For example,
Code:
```void add1(int x)
{
x=x+1;
}
int main()
{
int var;
add1(var);
}```
won't increment var (although it will increment the local copy, which is held in x and discarded when the function returns), you need to do
Code:
```void add1(int *x)
{
*x=*x+1;
}
int main()
{
int var;
add1(&var);
}```
instead. The & operator in this context takes the address of the variable it precedes; &var means "the address of var", and in add1() the * operator means the computer will treat the variable as a pointer and access memory at the location contained in the variable; suppose you have int foo; int *baz=&foo; then to access foo through the baz pointer you use *baz, and you can increment foo with *baz=*baz+1;. This is known a DEREFERENCING the pointer - a very important term.
The same is true for pointers, if you need to modify a pointer, then you have to pass in the address of that pointer. So:
Code:
```void incr_ptr(int *x)
{
x=x+1;
}
int main()
{
int x[2];
int *x_ptr=&x[0];
incr_ptr(x_ptr);
}```
won't increment x_ptr to point to x[1], although as before it WILL increase the local variable x, which is discarded when the function returns. If you want to increase x_ptr itself, then just as before you have to pass in a pointer and dereference it:
Code:
```void incr_ptr(int **x)
{
*x=*x+1;
}
int main()
{
int x[2];
int *x_ptr=&x[0];
incr_ptr(&x_ptr);
}```
So your function change() only changes the local variable x, then discards that change. It leaves xx untouched. To modify xx you need to pass in a pointer to xx. xx itself is a pointer to int, so the x parameter to change() needs to be a pointer to a pointer to int. So:
Code:
```void change(int **x,int *y)
{
*x = y;
};
int main()
{
int *xx, *yy;
yy=malloc(sizeof int);
change (&xx,yy);
//etc```
should work. Note that y isn't declared as int** because you don't need to modify yy. This is a neat little example of using different levels of INDIRECTION (another important term - it's "indirect" because you don't go directly to the variable. int x would be direct. x is that number. int *ptr2x; is indirect; to get at x you have to go indirectly, via ptr2x, with *ptr2x).
Despite the simplicity of the program there are still two bugs worth a mention:
The value displayed by the program will be unpredictable, because *yy is uninitialised. Also, the program leaks memory, because the malloc() is not matched by a free(). Both these are major bugs; a memory leak in a long running program will mean that the program will eventually stop working because there's no memory left, and uninitialised variables will mean any behaviour dependent on the value of those variables will be unpredictable.
So some good habits for you to develop immediately are:
- ALWAYS initialise variables. Initialise them to zero if you can't think of a suitable value.
- ALWAYS ALWAYS ALWAYS initialise pointers. Initialise them to zero.
- ALWAYS write "free" somewhere, ideally in such a way as to cause a compiler error, when you write malloc(), because that will force you to deal with the free at the most important time (same is true for new/delete/new[]/delete[] in C++). Leaving it will mean that you forget it, which means (a) you'll get a leak and (b) you'll have to spend hours trying to find what is leaking. ALWAYS write resource allocation and resource deallocation simultaneously, or as simultaneously as possible.
- Remember that allocating memory simply gives you somewhere to stuff a value. It doesn't do any value stuffing for you. int **yy=malloc(sizeof int); initialises yy to point to an int, but the int it points to is NOT DEFINED.
jose_peeterson like this | 922 | 3,766 | {"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} | 2.6875 | 3 | CC-MAIN-2014-23 | longest | en | 0.858954 |
https://www.abebooks.com/9780324225532/Microsoft-Excel-Companion-Business-Statistics-0324225539/plp | 1,508,394,805,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823229.49/warc/CC-MAIN-20171019050401-20171019070401-00379.warc.gz | 850,348,410 | 14,649 | ## A Microsoft Excel Companion for Business Statistics
### David Eldredge
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This manual/workbook provides step-by-step instructions for using Excel to solve most of the problems found in introductory business statistics. Numerous visual examples of screen captures show data input, pull-down menus, dialog boxes, and statistical results. The workbook introduces the statistical capabilities of Excel to students who have little or no prior experience with Microsoft Windows and/or Excel. Chapter 1 includes a brief introduction to the Windows environment and an introduction to Excel.
"synopsis" may belong to another edition of this title.
David L. Eldredge is Professor of Computer Science in the College of Business and Public Policy at Murray State University. He earned a Bachelor of Science degree in aerospace engineering from Iowa State University, and an M.S. and a Ph.D. in operations research from the Ohio State University. Professor Eldredge's current teaching responsibilities include management science for Murray State University's MBA program and business statistics for the undergraduate business administration program. He served as Dean of the College of Business for eight years. Dave Eldredge has had over fifteen books and papers published primarily in the areas of simulation and gaming, and the use of Microsoft Excel for data analysis. He also has ten years of industrial experience with four different Fortune 500 firms. Dave Eldredge and his wife Judy are parents of three daughters -- Kristi, Lynn and Tracey - and have seven grandchildren. The Eldredges share a love for camping and travel, they have visited all 50 states and many other countries.
Review:
1. Introduction to Statistics with Excel. 2. Descriptive Charts and Graphs. 3. Descriptive Numerical Measures. 4. Probability and Sampling Distributions. 5. Statistical Inference for Population Means. 6. Statistical Inference for Population Proportions. 7. Statistical Inference for Population Variances. 8. Analysis of Variance. 9. Applications of the Chi-Square Statistic. 10. Regression Analysis. 11. Time Series Forecasting. 12. Statistical Quality Control Charts. Appendix A. Excel Data Analysis Tools. Appendix B. Excel Statistical Functions. Appendix C. Procedures for Constructing a Box Plot.
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## 1.A Microsoft Excel Companion for Business Statistics
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## 2.A Microsoft Excel Companion for Business Statistics
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Destination, Rates & Speeds | 745 | 3,258 | {"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} | 2.875 | 3 | CC-MAIN-2017-43 | latest | en | 0.903018 |
https://math.icalculator.com/inequalities/quadratic-inequalities.html | 1,716,521,394,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00704.warc.gz | 319,184,090 | 5,945 | # Math Tutorial 10.2 - Quadratic Inequalities
Please provide a rating, it takes seconds and helps us to keep this resource free for all to use
In this Math tutorial, you will learn:
• How to identify whether a given number is a root of a quadratic inequality or not?
• How to write a quadratic inequality in the standard form?
• What happens to the sign of a quadratic inequality when the discriminant is positive? Zero? Negative?
• How to study the sign of a quadratic inequality?
• How to find the solution set(s) of a quadratic inequality?
## Introduction
In the previous tutorial, we explained how to solve linear inequalities in one or two variables. Now, we will explain how to solve quadratic inequalities, i.e. inequalities that contain one of the variables in the second power. Obviously, such inequalities are more complicated to solve compared to linear ones. Therefore, we are dedicating this entire tutorial only to quadratic inequalities.
Please select a specific "Quadratic Inequalities" lesson from the table below, review the video tutorial, print the revision notes or use the practice question to improve your knowledge of this math topic.
Inequalities Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
Lesson IDMath Lesson TitleLessonVideo
Lesson
10.2.3Solving Quadratic Inequalities by Studying the Sign
## Whats next?
Enjoy the "Quadratic Inequalities" math tutorial? People who liked the "Quadratic Inequalities" tutorial found the following resources useful:
1. Math tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
2. Inequalities Video tutorial: Quadratic Inequalities. Watch or listen to the Quadratic Inequalities video tutorial, a useful way to help you revise when travelling to and from school/college
3. Inequalities Revision Notes: Quadratic Inequalities. Print the notes so you can revise the key points covered in the math tutorial for Quadratic Inequalities
5. Check your calculations for Inequalities questions with our excellent Inequalities calculators which contain full equations and calculations clearly displayed line by line. See the Inequalities Calculators by iCalculator™ below.
6. Continuing learning inequalities - read our next math tutorial: Graphing Inequalities
## Help others Learning Math just like you
Please provide a rating, it takes seconds and helps us to keep this resource free for all to use | 510 | 2,450 | {"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} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.841069 |
http://imechanica.org/taxonomy/term/128?page=5 | 1,550,295,092,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479885.8/warc/CC-MAIN-20190216045013-20190216071013-00360.warc.gz | 126,095,527 | 15,092 | # education
## Looking for further learning opportunities
I am a PhD candidate at Shanghai Jiao Tong University, China.
My research intrest is Digital Image Correlation (DIC).
I am looking for the opportunities for further study.
My email: libangjian#sjtu.edu.cn
Thank you for considering my request.
All the best to you.
## Discrete Element Modelling (DEM)
Choose a channel featured in the header of iMechanica:
Can I use DEM using Abaqus 6.14?? If yes how??
Thank you
## Large Deformation - Definition of total work energy density
Choose a channel featured in the header of iMechanica:
Like we have the elastic strain energy density for small deformations defined as 0.5* σ :e .
Is the equation PK2:E valid for the total work energy density for elastoplastic regimes ? If not, what would be a valid equation for total energy density ?
How can we decompose total work density into elastic work and plastic work densities for a large deformation case.
Where,
PK2 is the second piola kirchoff stress tensor
E is the Green-Lagrange strain tensor
Thanks,
Prithivi
## Plane_Stress, Plane_Strain and 3D - Simple doubt..
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Hello everybody,
UPDATE : Question can be deemed as closed.. :)
I have a very simple doubt in 3D model simplification. I believe plane stress and plane strain conditions are the two extreme states to simplify a 3D model to 2D case.
## Transfroming applied tractions into statically equivalent nodal force BC
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Hello,
I am trying to produce results for problems with analytical solution to validate my FE code. One of the problmes is the classical plate with a hole configuration with tractions applied at boundaries. However, presently, I can only impose BC in terms of nodal forces and displacements. Therefore I must transform the applied stress in statically equivalent nodal forces.
## Modeling of Localized Inelastic Deformation 2016 (LID 2016)
Dear colleagues,
on behalf of the course organizer, I would like to bring to your attention to the advanced course on Modeling of localized inelastic deformation that will be taught by Milan Jirásek in Prague, Czech Republic on 19-23 September 2016.
## Simpleware at Hartree Summer Schools 2016: Week 2 - Engineering Simulation
Dates: June 27 - July 1, 2016
Venue: The Hartree Centre, STFC Daresbury Laboratory, Warrington, WA4 4AD, UK [Directions]
School Leaders: Lee Margetts (University of Manchester), Anton Shterenlikht (University of Bristol), and Llion Evans (Culham Centre for Fusion Energy)
Registration Fee: £150
Simpleware are contributing to this event.
## Pressure Difference Calculation in ABAQUS
Hi everyone,
I am working on ABAQUS to simulate Pipeline Inspection Gauge (PIG) movement in a pipeline via CEL technique. I am unable to find how to calculate Pressure Difference in the pipeline using Abaqus.
If anyone of you can guide me through it.
Regards,
MD Aftab
## Simplified slip-bearing action by using stop ellement in Abaqus
Hi
can anyone please advise me to define connector ellements with following details:
Using rigid plastic CARTESIAN element and a STOP element in parallel and The connector behaviour must be:
1- rigid up to the slip force limit Rn
2- slip at the bolts up to ±1 mm at constant force Rn
3- rigid bearing hardening
## Simple Physics concepts
Hello......
I have a small doubt. It may feel silly, but please reply..
Suppose I have a pendulum and it is impacting onto a structure. The mass of pendulum is 'm' and the intiial velocity is 'v'. The condition is that the KE at the time of impact should be 'E'. E= 0.5*m*v2 .
Now i change the mass of the pendulum to m1 correspondingly velocity should be changed to v1 so that E1=E.
Now i need to compare the force transmitted to the structure. F = ma.
## Welding analysis with Abaqus
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Hello everyone,
Hello everyone,
## 7th Summer School on Biomechanics of Soft Tissues: Multiscale Modeling, Simulation and Applications
7th Summer School on Biomechanics of Soft Tissues: Multiscale Modeling, Simulation and Applications
Graz University of Technology, Austria
July 4 - 8, 2016
coordinated by:
Gerhard A. Holzapfel, Graz University of Technology, Austria
Ray W. Ogden, University of Glasgow, UK
Choose a channel featured in the header of iMechanica:
Hi
I want to simulate experimental result of a CFRP [45/0/-45/90]s single lap joint in ABAQUS.
I have given geometry, properties & graphical result in the uploaded images.
Yellow color plate in the joint geometry is the E-Glass end tabs to grab both the ends
of the joint in the hydraulic grips. The loading(shear) is carried out at 0.5mm/min.
Thichness of each ply is 0.12mm; so thickness(T) of one CFRP plate is 0.12*8 = 0.96mm.
Thickness of adhesive layer is 0.15mm.
Overlap length(L) is 10mm.
## Help with a LS DYNA compression test model
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Free Tags:
Hi
I amtrying to model a simple compression test om a foam cube with LS DYNA. I have tried a lot of different approaches but there is always something wrong in the model :(. I have managed to get on model up and running, but i am having problems with excessive penetrations! Do anyone have any idea how to fix this?
I have includef the .k file as a text file to this post.
## How to extract angle of twist and shear stress due to torque in Workbench
I set up this model of a shaft with to opposing torque simulating a shaft and pulley system.
Note that the origin (top left) has the rotation X constrained to eliminate rigid body motion and to set a reference of zero twist at that point.
I requested every result that I can imagine, see the Outline here: Ex-3-9-Outline2.png
The torque is shown here: Ex-3-9-TorsionalMoment.png | 1,412 | 5,885 | {"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-2019-09 | latest | en | 0.866726 |
https://hirecalculusexam.com/how-to-schedule-differential-calculus-quiz-taking-services | 1,713,508,884,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817289.27/warc/CC-MAIN-20240419043820-20240419073820-00532.warc.gz | 267,787,169 | 19,019 | # How to schedule Differential Calculus quiz-taking services?
How to schedule Differential Calculus quiz-taking services? http://www.scimedeskim.com/searchforum.php?t=52 http://www.scimedeskim.com/p/SQUUTS_CALMUS Sun, 24 Aug 2018 15:00:17 +0000https://www.scimedeskim.com/?p=52The most advanced topic, which can be applied to any of the topics covered by the K-Edition, is differentiating the mathematics in the “method” area of the CTE. And if mathematics is one area that need to be discussed, it would be good to talk about it in the more general topic. There is a great article, “The ‘Maths of Complex Analysis” by Paul Reubens, by Thomas Hockham and Dan Osterman (see: http//www.schoo-sacla.org/doc/schoo-sacla/pdf/schoo-sacla-2-3.pdf) based on the philosophy of the CTE. However, there are the more fundamental things to notice: Basic question-solving; basic calculus Before explaining an example, it must be noted that “Maths of Complex Analysis” is not necessary to understand the key idea-an important difference between the CTE’s (the theory of “methodcalculus”) and the GED (the concept of theorem “forgetfulness”, “certainty” “proposers gave us”). In the CTE’s it is not necessary to ask some of the basic questions (and avoid confusion with their interpretation), but rather to ask a set of questions to quantify the quality of a proof (e.g. the quality of the general proof that there are $\mathcal{D}$ (of $\mathbb{Z}[x_1,\ldots x_k]$) which cannot be proved without going over the proof the other way) asHow to schedule Differential Calculus quiz-taking services? DETECTIVE CALISTRUE IN NOTES If you’re interested in joining our regular live classes, we’ve got some new and exciting Read More Here to learn. We offer a wide range of classes for you to make sure you’re on the right track, and after having taken these classes this week out on Friday, grab a group meal and come back with some helpful suggestions for how you can best complete your homework assignments. Who is chosen to help Clicks! With homework assignments, you’ll need some practical skills that make getting done easy easy! As you’ll see, you’ll need these skills to be in contact with your students, and those who do homework will be picked for this new opportunity to help them. How to start Check your luck and go with it, because with homework you ask this question? If you came across a group homework in class and could have your knowledge blocked, then you know how to start you trip when you start. | 590 | 2,494 | {"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} | 2.796875 | 3 | CC-MAIN-2024-18 | latest | en | 0.920095 |
http://matlab.izmiran.ru/help/techdoc/creating_plots/annota20.html | 1,726,050,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651383.5/warc/CC-MAIN-20240911084051-20240911114051-00712.warc.gz | 21,395,330 | 2,525 | Graphics
Text Alignment
The `HorizontalAlignment` and the `VerticalAlignment` properties control the placement of the text characters with respect to the specified x-, y-, and z-coordinates. The following diagram illustrates the options for each property and the corresponding placement of the text.
The default alignment is
• `HorizontalAlignment` `= left`
• `VerticalAlignment = middle`
MATLAB does not place the text `String` exactly on the specified `Position`. For example, the previous section showed a plot with a point annotated with text. Zooming in on the plot enables you to see the actual positioning of the text.
The small dot is the point specified by the text `Position` property. The larger dot is the bullet defined as the first character in the text `String` property.
Example -- Aligning Text
Suppose you want to label the minimum and maximum values in a plot with text that is anchored to these points and that displays the actual values. This example uses the plotted data to determine the location of the text and the values to display on the graph. One column from the `peaks` matrix generates the data to plot.
• ```Z = peaks;
h = plot(Z(:,33));
```
The first step is to find the indices of the minimum and maximum values to determine the coordinates needed to position the text at these points (`get`, `find`). Then create the string by concatenating the values with a description of what the values are.
• ```x = get(h,'XData'); % Get the plotted data
y = get(h,'YData');
imin = find(min(y) == y);% Find the index of the min and max
imax = find(max(y) == y);
text(x(imin),y(imin),[' Minimum = ',num2str(y(imin))],...
'VerticalAlignment','middle',...
'HorizontalAlignment','left',...
'FontSize',14)
text(x(imax),y(imax),['Maximum = ',num2str(y(imax))],...
'VerticalAlignment','bottom',...
'HorizontalAlignment','right',...
'FontSize',14)
```
The `text` function positions the string relative to the point specified by the coordinates, in accordance with the settings of the alignment properties. For the minimum value, the string appears to the right of the text position point; for the maximum value the string appears above and to the left of the text position point. The text always remains in the plane of the computer screen, regardless of the view.
Creating Text Annotations with the text or gtext Command Editing Text Objects | 523 | 2,373 | {"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-2024-38 | latest | en | 0.680842 |
http://www.bristol.ac.uk/maths/undergraduate/units1819/levelh6units/random-matrix-theory-math300016/ | 1,566,263,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315174.57/warc/CC-MAIN-20190820003509-20190820025509-00004.warc.gz | 234,883,784 | 7,579 | # Random Matrix Theory
## Unit aims
By the end of the unit you will master some of the most important mathematical techniques used in random matrix theory and have an understanding of how these are relevant in various areas of mathematics, physics, engineering and probability.
## Unit description
Random matrices are often used to study the statistical properties of systems whose detailed mathematical description is either not known or too complicated to allow any kind of successful approach. It is a remarkable fact that predictions made using random matrix theory have turned out to be accurate in a wide range of fields: statistical mechanics, quantum chaos, nuclear physics, number theory, combinatorics, wireless telecommunications and structural dynamics, to name only few examples.
Particular emphasis will be given to computing correlations of eigenvalues of ensembles of unitary and Hermitian matrices. Different ensembles have distinct invariance properties, which in the applications are used to model systems whose physical or mathematical behaviour depends only on their symmetries. In some cases the dimension of the matrices will be treated as a large asymptotic parameter. In addition we will develop several techniques to compute certain types of multiple integrals. There will be general discussion of how this relates to current research in various fields of mathematics and physics. The course will appeal to students in applied and pure mathematics as well as in statistics.
## Relation to other units
The material covered provides a useful background for the level 7 unit Quantum Chaos. Some aspects of this course are related to topics presented in Statistical Mechanics.
## Learning objectives
After completing this unit successfully you should be able to:
• Define and comprehend the notions of spectral statistics for various matrix ensembles.
• Compute typical examples of spectral statistics.
• Recognize and compute with a few common matrix ensembles
## Transferable skills
• Clear, logical thinking.
• Problem solving techniques.
## Syllabus
1. Introduction: Applications of random matrix theory and examples of spectral distributions.
2. Elementary notions of matrix theory.
3. Elementary notions of probability and measure theory.
4. The Poisson process and its spacing distribution.
5. The 2 x 2 Gaussian ensembles and their spacing distributions.
6. The CUE ensemble.
7. Spectral correlation functions for the CUE ensemble.
8. Techniques for calculating spectral statistics
## Reading and References
There is no recommended text but two useful references are:
• Madan Mehta. Random Matrices, Elsevier, 2004
• Peter Forrester. Log-gases and random matrices, Princeton University Press, 2010
Unit code: MATH30016
Level of study: H/6
Credit points: 10
Teaching block (weeks): 2 (13-18)
Lecturer: Dr Nina Snaith
## Pre-requisites
MATH20901 Multivariable Calculus
None
## Methods of teaching
Lectures, homework and exercises. Notes will be made available to the students.
## Methods of Assessment
The pass mark for this unit is 40.
The final mark is calculated as follows:
• 80% from a 1 hour 30 minute exam in May/June
• 20% homework assignments
NOTE: Calculators are NOT allowed in the examination.
For information resit arrangements, please see the re-sit page on the intranet.
Please use these links for further information on relative weighting and marking criteria.
Further exam information can be found on the Maths Intranet. | 707 | 3,497 | {"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} | 3.0625 | 3 | CC-MAIN-2019-35 | latest | en | 0.909536 |
https://www.chiefdelphi.com/t/what-gear-ratio-should-we-use-for-our-off-season-turret/469155 | 1,723,504,964,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641052535.77/warc/CC-MAIN-20240812221559-20240813011559-00327.warc.gz | 531,398,083 | 10,937 | # What gear ratio should we use for our off season turret
our team wants to build a turret for our off-season robot, we would like for the turret to rotate simultaneously with the robot.
in order to achieve, we need to now first which gear ratio to use for the turret’s rotational motor.
our team currently uses an L1 swerve but we are planning to upgrade to an L3 swerve, the turret will use neo 1. 1 and the swerve will use kraken.
please forgive for my lack of English skills, I am not a native speaker
1 Like
There are a few extra bits of information that would need to be known to really make any answer useful to you, a lot of what weight is this and that, and some preferences of things like the gear for your turret ring. There are also probably 1 or 2 other things that aren’t coming to my mind.
So instead of answering your question, I have linked a copy of the Google sheets doc that NoMythic will use to calculate this sort of thing. Not sure where we originally got it as it predates my time on the team, but here it is.
What you should do here is make your own copy of the Google sheets doc, and then enter the info into the single speed drive train sheet for your L1 or L3 Swerve drive train. (The gear ratios for the swerve modules should be on the manufactures’ website.) I would then make a copy of that single speed drive train sheet and fill in the info for your turret as if it is a drive train and play with the gear ratios until you are happy.
To my understanding, this should get you right about what you need.
If you have any questions, I’ll be checking Chief Delphi every now and then tonight.
If your design constraint is to rotate at the same speed of the robot you will also need the size of the frame (wheel is driving on a circumference of the robots rotation circle, so to convert that module speed to an angle you need to know the radius)
I will say trying to do this fast is a good exercise. Since turrets will almost always have wires you will need to quickly slew the turret range of motion to wrap/unwrap, so experience in designing for spinning through 360 degrees of motion quickly is very useful. (This is especially true for full rotation turrets, which is most years, 2020 was the exception with lots of sub 180° designs)
I would advise against using the free speed of the motor as the sole value in your calculations for the module speed and the turrent speed. A lot of these motions that you will be doing are more acceleration dependent than top speed. Look at the motor curves, play with the gearing a little. It would also be a decent idea to keep the moment of inertia low on the turret, this will dramatically reduce the load on the turret motor and give you a bigger window to play with for gearing and controls.
3 Likes
I’d gear for the turret to have about 2x the free rotational speed of the drivetrain, it should be a decent starting point. Generally, you need a lot of torque on turrets because it takes a deceivingly high amount of torque to accelerate the mass of the turret, and for most mechanisms, acceleration matters more than free speed as optimally-geared turrets often aren’t able to reach their free speed for short (under 180 degrees) movements because the motors can only apply so much torque.
It is possible to use a dynamics simulator to find the optimal gear ratio, but there aren’t any good ones out on the internet and you would need to spend a bit of effort making your own and often require knowledge beyond most FRC
teams to successfully write and use.
1 Like
I’ll recommend my AMB Design Calculator as a first step for finding the proper gear ratio. You’ll need to know how fast you want to turn (i.e. faster than your drivetrain can turn), how much force your turret needs in order to turn (a rough estimate will be fine), and the type of motor you want to use.
2 Likes
I want to explain a little more. When your chassis starts rotating and you pass this information to the turret, the turret will be lagged behind and will need to ‘catch up’ to be pointing the right direction. In addition, if you are rotating and start translating (shoot on the move) you will need to turn your turret even faster because you are accounting for the chassis rotation AND the motion of the robot.
So 2x is a great place to start.
I usually come at this sort of question from another angle. Given the motor I’m using and the mass of the turret, just how fast I can I move it? Might as well go as fast as feasible.
If you don’t want to do the math you can also “steal from the best, invent the rest”.
254’s 2020 and 2022 turrets had Falcons powered ~50:1 and ~45:1 reductions. Our 2024 turret had a Kraken powering ~45:1 reduction and its shooter was quite big/heavy because it also included the spindexer. Effectively, these big brushless motors are very strong and overkill for a turret, but it means you don’t need much reduction.
7 Likes
Why didn’t I think to just copy 254 when we were selecting our turret gear ratio this year?
Here’s a snippet of the numerical analysis I did for 118’s 2024 turret. Equations and rough estimates are included for inertia of both the turret head and the motor rotor. I looked at two components to estimate an overall torque: the torque required to accelerate the load/gearbox inertia (τ=Iα), plus a constant drag torque expected from the wire bundle. I’m not sure why the current drops below zero in the third graph but the peak current and time-to-target align about right with what we saw on the built system. If you are going to do your own analysis, be sure to set the endpoint of travel to a velocity of zero, as I forgot to do. This should end up showing that a greater reduction, closer to 254’s 45:1 gives a faster time-to-target with an acceptable current draw.
118 has pretty much halved the gear ratio on our turrets each year we build a new one, and think we found the limit this year. For 2024, we selected a 30:1 Kraken with a spur gear set we could swap to get a little more or less reduction if desired. We probably should’ve swapped to a ratio with a little more torque, closer to what 254 ran, since we ended up running only half duty cycle since the input data from the goal tracking camera was at too low of a frequency to utilize anywhere near the peak of the actuator. 254’s turret was also tuned much better than ours, pretty much fully keeping up with the drivetrain at all times where ours oscillated pretty badly. Our rotating mass must have been greater too with our amp mechanism being carried around by the turret.
@Torrance I really only share ours because for once, 118 had less reduction on a mechanism than 254. With all the linear mechanisms y’all build, I’m certain 254 has the highest ratio of blue banners to gear teeth of any team.
11 Likes
There’s an easy solution with cascasde control in which you feed both a position and velocity target to the robot where the velocity target is added to the result of a position controller.
How is this this spreadsheet formulated? An “efficiency” value is not a particularly accurate way to represent the total losses in the system and I’ve had better luck accounting for them by deriving models from the fundamental dynamics and accounting for losses as dry and viscous friction, resistance, and the torque constant.
I came up with a very rough estimate for the inertias of the motor rotor and a theoretical turret “head”. This was before any part of the shooter itself was designed. I then put a spring scale on a prototype wire bundle on a testbed to find a constant torque load. I multiplied my measured force by something like 3 or 5. All very rough estimates for load.
I built a discrete solution using small time steps starting from what torque the motor experiences at each point in time, which has the two contributions I had estimated, one that is constant in time and another that varies with the speed of the motor, requiring some more data.
So I went to the 8V curve data .csv from motors.vex.com for a Falcon 500. Again, very rough estimation because we planned on using Krakens if they showed up on time, although I could not find similar data for them anyway. That table gives a torque that the motor can supply for the given speed it is currently at. The torque found from that table goes towards the the drag torque (coming from friction, so expected to be essentially constant), with any leftovers accelerating the system’s inertias. From Newton’s second law in a rotary reference frame, this gives an acceleration, which is applied over a discrete time step to give the speed of the motor for the next point, which then gives the torque the motor can exert based on the table from VEX, and so on. I iterated smaller time steps until the solution appeared to converge.
I did not apply any mechanical efficiency factor. There are certainly more accurate and precise ways to estimate performance but they’re only as good as the known variables, which at the time were very few. Build season is fast. I’m still learning the valuable skill of doing less math and more doing.
3 Likes
Can you elaborate on this? I’m not up to speed on cascaded Pid control.
If you just request the position while trying to track a moving target, you will end up with significant lag because that is equivalent to requesting the position with 0 velocity when you really want to be traveling at the input velocity when tracking a moving target, which results in the D term in standard PID
to try to bring the system to 0 velocity which is not what you want. With cascade control, you can simply feed the position controller output + the target velocity into your velocity controller which will fix the lag caused by moving inputs.
1 Like
Got ya. So a position Pid, with target position as the input, added to a velocity pid with the target velocity as the input. Is that right?
Yea that’s will get you tracking moving target, because you have the velocity target. You can also really easily add feedforwards and real-time motion profiling for improved performance. | 2,179 | 10,077 | {"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} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.949273 |
https://scipy.github.io/devdocs/reference/generated/scipy.special.k0e.html | 1,696,388,428,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00643.warc.gz | 533,863,753 | 6,742 | # scipy.special.k0e#
scipy.special.k0e(x, out=None) = <ufunc 'k0e'>#
Exponentially scaled modified Bessel function K of order 0
Defined as:
```k0e(x) = exp(x) * k0(x).
```
Parameters:
xarray_like
Argument (float)
outndarray, optional
Optional output array for the function values
Returns:
Kscalar or ndarray
Value of the exponentially scaled modified Bessel function K of order 0 at x.
`kv`
Modified Bessel function of the second kind of any order
`k0`
Modified Bessel function of the second kind
Notes
The range is partitioned into the two intervals [0, 2] and (2, infinity). Chebyshev polynomial expansions are employed in each interval.
This function is a wrapper for the Cephes [1] routine `k0e`. `k0e` is useful for large arguments: for these, `k0` easily underflows.
References
[1]
Cephes Mathematical Functions Library, http://www.netlib.org/cephes/
Examples
In the following example `k0` returns 0 whereas `k0e` still returns a useful finite number:
```>>> from scipy.special import k0, k0e
>>> k0(1000.), k0e(1000)
(0., 0.03962832160075422)
```
Calculate the function at several points by providing a NumPy array or list for x:
```>>> import numpy as np
>>> k0e(np.array([0.5, 2., 3.]))
array([1.52410939, 0.84156822, 0.6977616 ])
```
Plot the function from 0 to 10.
```>>> import matplotlib.pyplot as plt
>>> fig, ax = plt.subplots()
>>> x = np.linspace(0., 10., 1000)
>>> y = k0e(x)
>>> ax.plot(x, y)
>>> plt.show()
``` | 432 | 1,456 | {"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} | 3.171875 | 3 | CC-MAIN-2023-40 | latest | en | 0.59418 |
http://wiki.fool.com/How_to_Configure_Equity_to_Assets_Ratio | 1,440,996,395,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644065534.46/warc/CC-MAIN-20150827025425-00306-ip-10-171-96-226.ec2.internal.warc.gz | 266,583,454 | 14,291 | What is Foolsaurus?
It's a glossary of investing terms edited and maintained by our analysts, writers and YOU, our Foolish community. Get Started Now!
# How to Configure Equity to Assets Ratio
Original post by Christopher Carter of Demand Media
A company’s equity to asset ratio exists as a measure of solvency that indicates the company’s ability to pay debts if the business has to immediately terminate, as explained by the Michigan State University website. Also, the equity to asset ratio illustrates the amount of a company’s assets that are financed by the owner’s investment in the business. To configure equity to assets, divide stockholders’ equity or net worth by a company’s total assets.
## Contents
### Step 1
Calculate a company’s total assets. Add all current assets that the company will convert to cash within one year such as accounts receivable, prepaid rent and cash. Tally the company’s total long-term assets that the company will convert to cash in over one year, like notes receivable, land and equipment. Add total current assets with total long-term assets to determine a company’s total assets. For example, if a company has total current assets of \$100,000 and total long-term assets of \$150,000, the company has total assets of \$250,000
### Step 2
Compute stockholders’ equity. A company can subtract total liabilities from total assets to determine stockholders’ equity. For instance, a company with \$250,000 in total assets and \$150,000 in total liabilities has stockholders’ equity of \$100,000.
### Step 3
Divide stockholders’ equity by total assets. Assume a company has stockholders’ equity of \$100,000 and total assets of \$250,000. In this scenario, a company’s equity to asset ratio is .40, or 40 percent.
```
inline.find('script').remove();
jQuery('#article p').eq(1).after(inline);
});
``` | 406 | 1,849 | {"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} | 3.296875 | 3 | CC-MAIN-2015-35 | longest | en | 0.903384 |
https://www.univerkov.com/the-bases-of-the-rectangular-trapezoid-are-6-cm-and-8-cm-and-the-height-is-5-cm/ | 1,696,366,934,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511220.71/warc/CC-MAIN-20231003192425-20231003222425-00278.warc.gz | 1,142,322,111 | 7,536 | # The bases of the rectangular trapezoid are 6 cm and 8 cm, and the height is 5 cm.
The bases of the rectangular trapezoid are 6 cm and 8 cm, and the height is 5 cm. Find the length of the segment connecting the midpoints of the bases.
Let us draw the height MH from the middle of the base of the BC to the base of AD.
In the formed rectangle BMHA, side AH = BM = BC / 2 = 6/2 = 3 cm.
Then the segment НК = АD – КD – АН = 8 – 4 – 3 = 1cm.
Then from the triangle MHK, by the Pythagorean theorem, we determine the length of the hypotenuse MK.
MK ^ 2 = MH ^ 2 + HK ^ 2 = 5 ^ 2 + 1 ^ 2 = 26.
MK = √26 cm.
Answer: The length of the segment connecting the midpoints of the bases is 26 cm.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities. | 278 | 1,001 | {"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} | 4.15625 | 4 | CC-MAIN-2023-40 | longest | en | 0.905034 |
http://media-1.web.britannica.com/eb-diffs/738/492738-18233-62854.html | 1,405,160,423,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776432874.14/warc/CC-MAIN-20140707234032-00056-ip-10-180-212-248.ec2.internal.warc.gz | 71,621,685 | 3,136 | reaction rate,the speed , or velocity, at which a chemical reaction proceeds, . It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed per in a unit of time or the amount concentration of a reactant used per unit that is consumed in a unit of time. Thus, for the reaction of two compounds X and Y that form a product Z, the equation is X + YZ, and the reaction rate may be given by the rate of increase of the concentration of Z or by the rate of decrease of the concentration of X or Y. Mathematically, the reaction rate is given by dCZ/dt, -dCX/dt, or -dCY/dt, in which C represents the concentration (e.g., moles per litre) of the species denoted by the subscript, and the symbol d/dt is the mathematical expression for the rate of change of some quantity with respect to time (the derivative with respect to time).Chemical reactions proceed at Alternatively, it may be defined in terms of the amounts of the reactants consumed or products formed in a unit of time. For example, suppose that the balanced chemical equation for a reaction is of the form
A + 3B ⟶ 2Z.
The rate could be expressed in the following alternative ways:d[Z]/dt, –d[A]/dt, –d[B]/dt, dz/dt, −da/dt, −db/dtwhere t is the time, [A], [B], and [Z] are the concentrations of the substances, and a, b, and z are their amounts. Note that these six expressions are all different from one another but are simply related. Chemical reactions proceed at vastly different speeds depending on the nature of the reacting substances and , the type of chemical transformation, the temperature, and other factors. In general, reactions in which atoms or ions (electrically charged particles) combine or separate occur very rapidly, while those in which covalent bonds are formed or (bonds in which atoms share electrons) are broken are much slower. For a given set of reactantsreaction, the speed of the reaction will vary with the temperature or pressure imposed on the reacting system , the pressure, and the amounts of reactants used. Ordinarily the reaction will gradually present. Reactions usually slow down as time goes on because of the depletion of the reactants become depleted. In some cases the addition of a substance that is not itself a reactant, called a catalyst, accelerates a reaction that normally takes place at a very low rate. The reaction- rate constant, or the specific rate constant, is the proportionality constant in the equation that describes expresses the relationship between the rate of a chemical reaction and the concentrations of the reacting substances. If r represents reaction rate, k is the symbol customarily used for the reaction-rate constant, and f1.5PT(C1PT) is an expression for the concentrations of the reactants, then the equation for these values is r = kf1.5PT(C1PT). If the reaction rate, or velocity, is visualized as being determined by two factors, one representing the amount of molecules present and the other the type and the condition of those molecules, then the rate constant is a quantity that represents the latter. The prediction, measurement , and interpretation of reaction rates are subjects of reactions constitute the branch of chemistry known as chemical kinetics. | 702 | 3,251 | {"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} | 2.875 | 3 | CC-MAIN-2014-23 | longest | en | 0.939943 |
https://everything.explained.today/Epsilon_number/ | 1,696,465,728,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511424.48/warc/CC-MAIN-20231004220037-20231005010037-00820.warc.gz | 263,401,028 | 8,155 | Epsilon number explained
In mathematics, the epsilon numbers are a collection of transfinite numbers whose defining property is that they are fixed points of an exponential map. Consequently, they are not reachable from 0 via a finite series of applications of the chosen exponential map and of "weaker" operations like addition and multiplication. The original epsilon numbers were introduced by Georg Cantor in the context of ordinal arithmetic; they are the ordinal numbers ε that satisfy the equation
\varepsilon=\omega\varepsilon,
in which ω is the smallest infinite ordinal.
The least such ordinal is ε0 (pronounced epsilon nought or epsilon zero), which can be viewed as the "limit" obtained by transfinite recursion from a sequence of smaller limit ordinals:
\varepsilon0=
⋅ ⋅ ⋅
\omega
\omega
\omega
=\sup\{\omega,\omega\omega,
\omega\omega \omega
,
\omega\omega \omega
\omega
,...\},
where is the supremum function, which is equivalent to set union in the case of the von Neumann representation of ordinals.
Larger ordinal fixed points of the exponential map are indexed by ordinal subscripts, resulting in
\varepsilon1,\varepsilon2,\ldots,\varepsilon\omega,\varepsilon\omega+1,\ldots,
\varepsilon \varepsilon0
,\ldots,
\varepsilon \varepsilon1
,\ldots,
\varepsilon
\varepsilon
\varepsilon
⋅ ⋅ ⋅
,\ldots
.[1] The ordinal ε0 is still countable, as is any epsilon number whose index is countable (there exist uncountable ordinals, and uncountable epsilon numbers whose index is an uncountable ordinal).
The smallest epsilon number ε0 appears in many induction proofs, because for many purposes, transfinite induction is only required up to ε0 (as in Gentzen's consistency proof and the proof of Goodstein's theorem). Its use by Gentzen to prove the consistency of Peano arithmetic, along with Gödel's second incompleteness theorem, show that Peano arithmetic cannot prove the well-foundedness of this ordering (it is in fact the least ordinal with this property, and as such, in proof-theoretic ordinal analysis, is used as a measure of the strength of the theory of Peano arithmetic).
Many larger epsilon numbers can be defined using the Veblen function.
A more general class of epsilon numbers has been identified by John Horton Conway and Donald Knuth in the surreal number system, consisting of all surreals that are fixed points of the base ω exponential map x → ωx.
defined gamma numbers (see additively indecomposable ordinal) to be numbers γ>0 such that α+γ=γ whenever α<γ, and delta numbers (see multiplicatively indecomposable ordinals) to be numbers δ>1 such that αδ=δ whenever 0<α<δ, and epsilon numbers to be numbers ε>2 such that αε=ε whenever 1<α<ε. His gamma numbers are those of the form ωβ, and his delta numbers are those of the form ωωβ.
Ordinal ε numbers
The standard definition of ordinal exponentiation with base α is:
\alpha0=1,
\alpha\beta=\alpha\beta-1\alpha,
when
\beta
has an immediate predecessor
\beta-1
.
\alpha\beta=\sup\lbrace\alpha\delta\mid0<\delta<\beta\rbrace
, whenever
\beta
is a limit ordinal.
\beta\mapsto\alpha\beta
is a normal function, so it has arbitrarily large fixed points by the fixed-point lemma for normal functions. When
\alpha=\omega
, these fixed points are precisely the ordinal epsilon numbers.
\varepsilon0=\sup\lbrace1,\omega,\omega\omega,
\omega\omega \omega
,
\omega\omega \omega
\omega
,\ldots\rbrace,
\varepsilon\beta=\sup\lbrace{\varepsilon\beta-1+1},
\varepsilon\beta-1+1 \omega
,
\varepsilon\beta-1+1 \omega
\omega
,
\varepsilon\beta-1+1 \omega
\omega
\omega
,\ldots\rbrace,
when
\beta
has an immediate predecessor
\beta-1
.
\varepsilon\beta=\sup\lbrace\varepsilon\delta\mid\delta<\beta\rbrace
, whenever
\beta
is a limit ordinal.
Because
\varepsilon0+1 \omega
=
\varepsilon0 \omega
\omega1=\varepsilon0\omega,
\varepsilon0+1 \omega
\omega
=
(\varepsilon0 ⋅ \omega) \omega
=
\varepsilon0 {(\omega
)}\omega=
\omega \varepsilon 0
,
\varepsilon0+1 \omega
\omega
\omega
=
{\varepsilon0 \omega
\omega}=
{\varepsilon0 \omega
1+\omega
} = \omega^ = ^ = ^ \,,a different sequence with the same supremum,
\varepsilon1
, is obtained by starting from 0 and exponentiating with base ε0 instead:
\varepsilon1=\sup\{1,\varepsilon0,
\varepsilon0 {\varepsilon 0}
,
{\varepsilon0 {\varepsilon 0}
\varepsilon0
}, \ldots\},
Generally, the epsilon number
\varepsilon\beta
indexed by any ordinal that has an immediate predecessor
\beta-1
can be constructed similarly.
\varepsilon\beta=\sup\{1,\varepsilon\beta-1,
\varepsilon\beta-1 \varepsilon \beta-1
,
\varepsilon\beta-1 \varepsilon \beta-1
\varepsilon
\beta-1
,...\}
In particular, whether or not the index β is a limit ordinal,
\varepsilon\beta
is a fixed point not only of base ω exponentiation but also of base δ exponentiation for all ordinals
1<\delta<\varepsilon\beta
.
Since the epsilon numbers are an unbounded subclass of the ordinal numbers, they are enumerated using the ordinal numbers themselves. For any ordinal number
\beta
,
\varepsilon\beta
is the least epsilon number (fixed point of the exponential map) not already in the set
\{\varepsilon\delta\mid\delta<\beta\}
. It might appear that this is the non-constructive equivalent of the constructive definition using iterated exponentiation; but the two definitions are equally non-constructive at steps indexed by limit ordinals, which represent transfinite recursion of a higher order than taking the supremum of an exponential series.
The following facts about epsilon numbers are straightforward to prove:
• Although it is quite a large number,
\varepsilon0
is still countable, being a countable union of countable ordinals; in fact,
\varepsilon\beta
is countable if and only if
\beta
is countable.
• The union (or supremum) of any nonempty set of epsilon numbers is an epsilon number; so for instance $\varepsilon_\omega = \sup\$ is an epsilon number. Thus, the mapping
\beta\mapsto\varepsilon\beta
is a normal function.
Representation of ε0 by rooted trees
\varepsilon=\omega\varepsilon
, which means that the Cantor normal form is not very useful for epsilon numbers. The ordinals less than ε0, however, can be usefully described by their Cantor normal forms, which leads to a representation of ε0 as the ordered set of all finite rooted trees, as follows. Any ordinal
\alpha<\varepsilon0
has Cantor normal form
\beta1 \alpha=\omega
\beta2 +\omega
\betak + … +\omega
where k is a natural number and
\beta1,\ldots,\betak
are ordinals with
\alpha>\beta1\geq\geq\betak
, uniquely determined by
\alpha
. Each of the ordinals
\beta1,\ldots,\betak
in turn has a similar Cantor normal form. We obtain the finite rooted tree representing α by joining the roots of the trees representing
\beta1,\ldots,\betak
to a new root. (This has the consequence that the number 0 is represented by a single root while the number
1=\omega0
is represented by a tree containing a root and a single leaf.) An order on the set of finite rooted trees is defined recursively: we first order the subtrees joined to the root in decreasing order, and then use lexicographic order on these ordered sequences of subtrees. In this way the set of all finite rooted trees becomes a well-ordered set which is order-isomorphic to ε0.
This representation is related to the proof of the hydra theorem, which represents decreasing sequences of ordinals as a graph-theoretic game.
Veblen hierarchy
See main article: Veblen function. The fixed points of the "epsilon mapping"
x\mapsto\varepsilonx
form a normal function, whose fixed points form a normal function; this is known as the Veblen hierarchy (the Veblen functions with base φ0(α) = ωα). In the notation of the Veblen hierarchy, the epsilon mapping is φ1, and its fixed points are enumerated by φ2.
Continuing in this vein, one can define maps φα for progressively larger ordinals α (including, by this rarefied form of transfinite recursion, limit ordinals), with progressively larger least fixed points φα+1(0). The least ordinal not reachable from 0 by this procedure—i. e., the least ordinal α for which φα(0)=α, or equivalently the first fixed point of the map
\alpha\mapsto\varphi\alpha(0)
—is the Feferman–Schütte ordinal Γ0. In a set theory where such an ordinal can be proved to exist, one has a map Γ that enumerates the fixed points Γ0, Γ1, Γ2, ... of
\alpha\mapsto\varphi\alpha(0)
; these are all still epsilon numbers, as they lie in the image of φβ for every β ≤ Γ0, including of the map φ1 that enumerates epsilon numbers.
Surreal ε numbers
n\mapsto\omegan
; this mapping generalises naturally to include all surreal numbers in its domain, which in turn provides a natural generalisation of the Cantor normal form for surreal numbers.
It is natural to consider any fixed point of this expanded map to be an epsilon number, whether or not it happens to be strictly an ordinal number. Some examples of non-ordinal epsilon numbers are
\varepsilon-1=\{0,1,\omega,\omega\omega,\ldots\mid\varepsilon0-1,
\varepsilon0-1 \omega
,\ldots\}
and
\varepsilon1/2=\{\varepsilon0+1,
\varepsilon0+1 \omega
,\ldots\mid\varepsilon1-1,
\varepsilon1-1 \omega
,\ldots\}.
There is a natural way to define
\varepsilonn
for every surreal number n, and the map remains order-preserving. Conway goes on to define a broader class of "irreducible" surreal numbers that includes the epsilon numbers as a particularly-interesting subclass. | 2,503 | 9,512 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 3.53125 | 4 | CC-MAIN-2023-40 | latest | en | 0.907983 |
https://it.mathworks.com/matlabcentral/answers/730968-writing-a-closed-while-loop?s_tid=prof_contriblnk | 1,708,941,018,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474659.73/warc/CC-MAIN-20240226094435-20240226124435-00733.warc.gz | 331,457,043 | 27,048 | # Writing a closed while loop
1 visualizzazione (ultimi 30 giorni)
David Scidmore il 30 Gen 2021
Risposto: Walter Roberson il 30 Gen 2021
I'm trying to write part of a code that continues a function for the time set of 10.
So something like
while t =< 10
delta_R = -10;
Phi2 = beta*(1-((omega_n/omega_d)*exp(-sigma*t).*cos((omega_d*t)-phi)));
Phi2dot = diff(Phi2);
I have values for the terms, just don't know how to write a closed while loop (or closed loop) and I only get things about transfer loops which I'm not looking for.
##### 5 CommentiMostra 3 commenti meno recentiNascondi 3 commenti meno recenti
dpb il 30 Gen 2021
See the examples at
doc while
The first one would seem to match your need with an incrementing operation instead of decrementing.
David Scidmore il 30 Gen 2021
@Walter Roberson, it would be t increments between 0 to 10. So it would be a function using t as time. So I guess that would be a simulated 10 seconds.
Accedi per commentare.
### Risposte (1)
Walter Roberson il 30 Gen 2021
t = 0;
while t < 10
some action
t = t + appropriate increment
end
However if the increment is constant then a lot of the time it makes more sense to write a for loop
tvals = linspace(0,10,75); %use appropriate number of divisions
numt = length(tvals) ;
results = zeros(1 numt) ń
for tidx = 1:numt
t = tvals(tidx) ;
someaaction
results(tidx) = value;
end
plot(tvals,results)
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
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Translated by | 496 | 1,733 | {"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} | 2.875 | 3 | CC-MAIN-2024-10 | latest | en | 0.713983 |
https://gamedev.stackexchange.com/questions/158222/2d-top-down-cant-follow-the-path-accurately/172178 | 1,579,770,705,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250609478.50/warc/CC-MAIN-20200123071220-20200123100220-00150.warc.gz | 446,128,946 | 33,298 | # 2D Top Down: Can't follow the path accurately
In my game, there are many agents. Agents request a path, then after attaining a path to "goal" they follow the path. However player and many other things in the world can impact their position by pushing them or pulling them.
When agents get pushed and they happen to advance forward closer to "goal", I want them to smartly follow the previously given path.
I have pictures below to enhance your understanding of my situation.
I COULD recalculate path every a few seconds or when agent gets out of path BUT I have DOZENS of these little agents. I don't want to do path finding for DOZENS of agents again, again, again, and again whenever they walk out of the path.
• Do the agents know when they've been acted on by external forces? Are said forces instantaneous or applied over time? Given the options between the green & orange solutions, how concerned are you trying to minimize the distance to re-establish the path versus minimizing the time to the goal? – Pikalek May 3 '18 at 21:14
• Are your agents all moving toward one or a small set of destinations (like mobs in a tower defense all seeking the closest target building)? If so, you may have simpler options available like flow fields. – DMGregory May 3 '18 at 22:59
• Nice pictures. :) – HolyBlackCat Jun 4 '18 at 15:42
EDIT2: A series of heuristics is likely in order using the below. If within distance that node scores some points. Then if it's angular difference to the next node is close to 0, it scores (a lot) more points. Then if none of that works out, recalculate. If it does work out, simply move to the highest scoring node.
First, you should probably maintain a rough angle between the agent and it's goal, that way you can approximate the direction it's supposed to be going.
Second, you might have an impacted agent perform a distance calc on the next nodes in the path. If it's within distance then just go to the nearest node (+1) of that path and follow it. If it's outside of distance then recalculate. The nearest node (+1) (maybe recalculate just to this node) will look fairly natural I think.
This should save you most of your A* calculations and give a natural looking move.
The way you generally do this is to run the path-finding algorithm again every so often.
For lots of units, you can speed things up using hierarchical pathfinding and/or swarm pathfinding and/or using predefined paths between waypoints and merely pathing to the nearest waypoint.
There are many possibilities to do it:
1) Repath
2) Use path segments directions - and check if at goal like follows:
If (dot(playerToGoalDirection,SegmentDirection) < 0) Go to the next path point update path segment.
You can add some tracing to sieve out waypoints that you dont want to move to after being hit.
For example trace to the next point if success trace to the next-next point if success remove the next point and path to next-next point
Remark: PlayerToGoalDirection is the direction to the current waypoint
You can see the second in action here: | 687 | 3,068 | {"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} | 3.15625 | 3 | CC-MAIN-2020-05 | latest | en | 0.952525 |
http://en.allexperts.com/q/Advanced-Math-1363/2015/1/coordinate-geometry-variation-1.htm | 1,480,725,925,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540798.71/warc/CC-MAIN-20161202170900-00369-ip-10-31-129-80.ec2.internal.warc.gz | 90,609,413 | 7,888 | You are here:
Question
Coordinate Geometry Va
QUESTION: Dear Prof Scott
As per the attached image, Will you consider this Coordinate Geometry
Variation as correct or incorrect ?.
The Four Quadrants Positions are changed along with the positive and negative numbers on the x and y axis.
Thanks & Regards,
Prashant S Akerkar
ANSWER: Yes, that looks correct, for it just the normal coordinate system rotated by 180.
---------- FOLLOW-UP ----------
QUESTION: Dear Prof Scott
Thanks.
The equations, curves plotted using
these coordinate geometry variation would
be interesting?.
Thanks & Regards,
Prashant s Akerkar
From what I can tell, it looks like with that variation on plotting the curve would be to rotate it by 180 degrees. I'm not sure if that's really of any interest since the piece of paper they're plotted on just needs to be turned over. Doing so would make both look the same.
Volunteer
#### Scott A Wilson
##### Expertise
I can answer any question in general math, arithetic, discret math, algebra, box problems, geometry, filling a tank with water, trigonometry, pre-calculus, linear algebra, complex mathematics, probability, statistics, and most of anything else that relates to math. I can also say that I broke 5 minutes for a mile, which is over 12 mph, but is that relevant?
##### Experience
Experience in the area; I have tutored people in the above areas of mathematics for over two years in AllExperts.com. I have tutored people here and there in mathematics since before I received a BS degree back in 1984. In just two more years, I received an MS degree as well, but more on that later. I tutored at OSU in the math center for all six years I was there. Most students offering assistance were juniors, seniors, or graduate students. I was allowed to tutor as a freshman. I tutored at Mathnasium for well over a year. I worked at The Boeing Company for over 5 years. I received an MS degreee in Mathematics from Oregon State Univeristy. The classes I took were over 100 hours of upper division credits in mathematical courses such as calculus, statistics, probabilty, linear algrebra, powers, linear regression, matrices, and more. I graduated with honors in both my BS and MS degrees. Past/Present Clients: College Students at Oregon State University, various math people since college, over 7,500 people on the PC from the US and rest the world.
Publications
My master's paper was published in the OSU journal. The subject of it was Numerical Analysis used in shock waves and rarefaction fans. It dealt with discontinuities that arose over time. They were solved using the Leap Frog method. That method was used and improvements of it were shown. The improvements were by Enquist-Osher, Godunov, and Lax-Wendroff.
Education/Credentials
Master of Science at OSU with high honors in mathematics. Bachelor of Science at OSU with high honors in mathematical sciences. This degree involved mathematics, statistics, and computer science. I also took sophmore level physics and chemistry while I was attending college. On the side I took raquetball, but that's still not relevant.
Awards and Honors
I earned high honors in both my BS degree and MS degree from Oregon State. I was in near the top in most of my classes. In several classes in mathematics, I was first. In a class of over 100 students, I was always one of the first ones to complete the test. I graduated with well over 50 credits in upper division mathematics.
Past/Present Clients
My clients have been students at OSU, people who live nearby, friends with math questions, and several people every day on the PC. I would guess that you are probably going to be one more. | 815 | 3,677 | {"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 | 0.965541 |
https://marketingwithanoy.com/what-is-the-ideal-gas-law-marketingwithanoy/ | 1,679,432,637,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943746.73/warc/CC-MAIN-20230321193811-20230321223811-00610.warc.gz | 436,311,657 | 20,819 | # What is the ideal gas law? | MarketingwithAnoy
It may seem like a big volume, but it’s not. It’s almost half a liter, so it’s half a bottle of soda.
Moles and particles
These moles are not the furry creatures that make holes in the ground. The name comes from molecules (which are apparently too long to write).
Here is an example to help you understand the idea of a mole. Suppose you run an electric current through water. A water molecule is made up of one oxygen atom and two hydrogen atoms. (It is h2O.) This electric current breaks up the water molecule and you get hydrogen gas (H2) and oxygen gas (O2).
This is actually a pretty simple experiment. Check it out here:
Since water has twice as many hydrogen atoms as oxygen, you get twice as many hydrogen molecules. We can see this if we collect the gases from that water: we know the ratio of the molecules, but we do not know the number. That is why we use moles. It’s basically just a way of counting the innumerable.
Don’t worry, there’s actually a way to find the number of particles in a mole – but you’ll need Avogadro’s numbers for that. If you have a liter of air at room temperature and normal pressure (we call it atmospheric pressure), then there will be about 0.04 mol. (That would be nine the ideal gas law.) Using Avogadro’s numbers, we get 2.4 x 1022 particles. You can not count so high. Nobody can. But it is N, the number of particles, in the second version of the ideal gas law.
Constants
Just a quick note: You almost always need some kind of constant for an equation with variables representing different things. Just look at the right side of the ideal gas law, where we have pressure multiplied by volume. The units for this left side would be newton-meters, which is the same as a joule, the unit of energy.
On the right side is the number of moles and the temperature in Kelvin – the two times clearly not to give units of joules. But you shall have the same units on both sides of the equation, otherwise it would be like comparing apples and oranges. This is where constant R comes to the rescue. It has units of joules / (mol × Kelvin), so mol × Kelvin cancels and you just get joules. Bom: Now both sides have the same devices.
Now let’s look at some examples of the ideal gas law using a regular rubber balloon.
To inflate a balloon
What happens when you blow up a balloon? You are obviously adding air into the system. When you do this, the balloon becomes larger so that its volume increases.
What about the temperature and the pressure inside? Let’s just assume they are constant.
I will include arrows next to the variables that change. An up arrow means a rise and a down arrow means a fall. | 617 | 2,703 | {"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} | 3.875 | 4 | CC-MAIN-2023-14 | latest | en | 0.929703 |
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## #1 2009-08-12 16:26:19
macy
Guest
can any expert here tell me how to work out the last digit of the result when six multiply with eight to the power 2009?
Thanks
## #2 2009-08-12 20:44:31
juriguen
Member
Registered: 2009-07-05
Posts: 59
Hi!
First, lets see after how many powers does the last digit repeat:
OK, so after every 4 cycles, starting at some number N, eight to the power of N + 4k repeats its last digit.
For instance, 8^5 = 8^{1 + 4} has the same last digit as 8^1, both equal to 8.
So, for 2009
Hope it helps!
Jose
PS: there are other numbers which seem more strange, since they have 0 remainder when dividing by 4, but then N is also 4.
For example:
And, indeed, 8^20 = 1152921504606846976
Last edited by juriguen (2009-08-12 20:50:25)
Make everything as simple as possible, but not simpler. -- Albert Einstein
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## #3 2009-08-13 00:54:42
Guest
Hi macy
We can use the language of congruences as follows:
8^2=64=4(mod 10) multiply both sides withn 8^2 we get
8^4=64*4(mod 10)=256(mod 10 ) = 6(mod 10) by squaring both sides we get
8^8=36(mod 10)= 6(mod 10) so in general we have
8^(4k)=6(mod 10) where k is a positive integer so
8^(4k+1)= 8*6(mod 10)=48(mod 10)= 8 (mod 10 )
Now for 2009 we have 2009=4*502+1 therefore
8^2009= 8^(4*502+1)= 8 (mod 10 ) multiplying both sides by 6 we get
6*(8^2009)= 6*8 (mod 10 )=48 (mod 10 ) =8 (mod 10 )
therefore the last digit is the remainder when dividing by 10 and so the remainder is 8.
Best Regards
## #4 2009-08-13 01:23:28
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Hi Macy;
Lets forget about the 6 for a moment.
Now do the inner parentheses.
8^8 ends in a 6. So 8^8 * 8^8 * 8^8 * 8^8 *8^8 * 8^8 also ends in a 6 . Now times by the last 8 =48 so 8^49 ends in 8.
Now do the outer power:
We already know that any succession of 8^8 * 8^8 * 8^8... ends in 6. The last 8 * 6 =48 which ends in 8.
So
Which is the math way of saying that 8^2009 ends in 8.
Now don't forget the 6 which is just 6*8 =48 the last digit is again an 8.
Last edited by bobbym (2009-08-13 01:33:28)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #5 2009-08-13 18:42:14
Macy
Guest
Geezzz... you guys make me impressed by:
1. Fast response
big Hug to all
## #6 2009-08-13 19:24:44
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
Hi Macy;
Thanks from all 3 of us.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #7 2009-08-13 19:41:01
juriguen
Member
Registered: 2009-07-05
Posts: 59
Yes, thanks Macy!
I like the other approaches You can always learn many new things in this forum, even if you got the correct answer!
Jose
Make everything as simple as possible, but not simpler. -- Albert Einstein
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## #8 2009-08-13 20:04:39
bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606
B. M. wrote:
Everyone here has knowledge, so they are all my friends.
Last edited by bobbym (2009-08-13 20:05:08)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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## #9 2009-08-27 09:54:01
Fruityloop
Member
Registered: 2009-05-18
Posts: 136
Nothing like beating a dead horse, but here we go...
So the last digit is either 3 or 8, since all of the factors are even the last digit must be an 8.
Last edited by Fruityloop (2009-08-27 09:56:37)
The eclipses from Algol come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.
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https://math.stackexchange.com/questions/404941/i-cannot-understand-how-this-matrix-works-or-how-it-is-defined | 1,575,859,556,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540517156.63/warc/CC-MAIN-20191209013904-20191209041904-00439.warc.gz | 458,047,006 | 31,417 | # I cannot understand how this matrix works or how it is defined
I'm currently reading Ranking a Stream of News and have trouble on page 100 (don't be afraid, the math starts at 99). I cannot understand a matrix they define.
In this article, the authors are providing a mathematical representation -as a graph- of a stream of news consisting of articles and sources. This paragraph should, if I'm understanding correctly, be about how to determine how important edges are. An edge should be more important, for example, when it links an article to a authoritative source or when it links two articles with high similarity to each other.
In the paragraph before, they introduced $G_\omega$ as a stream of news, generally a graph consisting of a set $N$ with articles $n_1, ...$ and a set $S$ with sources $s_1, ...$ (both nodes) and edges sets $E_1$ (source - article) and $E_2$ (article - article).
Let A be the (weighted) adjacency matrix associated with $G_\omega$. We can attribute an identifier to the nodes in $G_\omega$ so that any source precedes the pieces of news. We define the matrix $$A = \begin{bmatrix}O & B\\B^T & \Sigma\end{bmatrix},$$ where $B$ refers to edges from sources to news articles, and $b_{ij} = 1$ if the source $s_i$ emitted article $n_j$ and is the similarity matrix. Assuming one can learn similarity of sources, the matrix $A$ can be modified in the upper-left corner incorporating a submatrix taking into account a source-source information.
I'm not able to understand this paragraph. This may partially be a lack of English skills which makes it hard to read for me, but I got problems with the math as well. Here are my questions:
• What is $A$? What is its concrete meaning?
• What is $O$? With "$A$ can be modified in the upper-left corner", do they mean they will change $O$?
• What "data type" is $B$? I'd think this is a set, "edges from sources to news articles", thus $E_1$. However, they're calculating $B^T$ - what does that mean, if $B$ is a set of edges?
• What is $T$? I don't see it anywhere else. Is it some constant like $i$ that I don't know of?
• What is $\Sigma$? I know the sum function, but don't see how it's relevant here. I only see that in the paragraph before this one they define $\sigma_{ij}$ as the continous similarity between article $n_i$ and article $n_j$. Would $\Sigma$ be the set of all $\sigma$, just like $N$ is the set of all $n$ (in this article)?
Nota Bene: I am not very familiar with matrices or set theory. Actually, I'm on the level of a just-finished high school student. I'd like some layman explanation, please.
• 1. This could go with a more descriptive title, but I couldn't think up one due to my lack of expertise - feel free to edit. 2. If it's easier to discuss this in chat, that's okay, just give a shout :) – user63495 May 28 '13 at 17:39
• $A$ is a block matrix which can be seen on wikipedia. $B^T$ denotes the transpose of the matrix $B$ (exchange the rows and columns) which appears to be an indicator matrix of sorts which tells us which sources put out which articles. $\Sigma$ is another matrix which probably has some special property like diagonal or symmetric. They are all matrices though. – Jemmy May 28 '13 at 17:48
I discussed this on the Electrical Engineering chat (permalink) and ThePhoton explained it to me.
$O$ is the zero matrix. This matrix contains only zeros.
$B$ is the matrix where in one direction (horizontal) the sources $s \in S$ are listed and in the other direction (vertical) the articles $n \in N$. This matrix shows which source produced an article.
$B^T$ is the transpose of $B$ (wrapped over the diagonal).
$\Sigma$ is the similarity matrix. It tells you how much the articles are like each other.
Now the meaning of
Assuming one can learn similarity of sources, the matrix $A$ can be modified in the upper-left corner incorporating a submatrix taking into account a source-source information.
We saw that $\Sigma$ was the similarity matrix. It shows similarity between articles. We could also calculate similarity of sources. 'By default', this is disabled: in the upper left corner of $A$, a zero matrix is put. Now assuming you can calculate similarity of sources, you could replace $O$ with $\Sigma_s$, which would be a matrix determining similarity of sources.
Now we can see that $A$ is a matrix containing both the relations between sources and articles (which articles were produced by which sources), and the relations between different articles (how similar they are), and, if $O$ is replaced, the relations between different sources (how similar they are). | 1,118 | 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": 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} | 3.703125 | 4 | CC-MAIN-2019-51 | latest | en | 0.922037 |
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Economics Economics Forum - Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance
January 19th, 2012, 09:03 PM #2 Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,556 Thanks: 101 Math Focus: The calculus Re: Very difficult question (need help) Every six months $1800 is deposited and interest compounded, so after 35 years, or 70 compounding periods, the account has a value V in dollars given by: $V=1800\sum_{k=1}^{70}(1.05)^k\approx1112318.89$ Now subtract the$10,000.00 going to the university and we are left with: $1102318.89 To find the first year's withdrawal W, we set: $1.05^{18}V-W\sum_{k=9}^{18}1.05^k=0$ $W=\frac{1.05^{18}V}{\sum_{k=9}^{18}1.05^k}\approx\ frac{2652861.8308853353}{19.5124395879424565}\appr ox135957.47$ Close, but no cigar. I post my incorrect response so perhaps someone who knows how to actually solve this can show me where I went wrong. January 20th, 2012, 01:27 AM #3 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,073 Thanks: 32 Re: Very difficult question (need help) Quote: Originally Posted by MarkFL Every six months$1800 is deposited and interest compounded, so after 35 years, or 70 compounding periods, the account has a value V in dollars given by: $V=1800\sum_{k=1}^{70}(1.05)^k\approx1112318.89$
You have to obtain a MONTHLY equivalent rate (since deposits are monthly):
(1 + r)^12 = 1.05^2
Then use 300 (not 1800) and 420 (not 70).
January 20th, 2012, 09:06 AM #4 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,073 Thanks: 32 Re: Very difficult question (need help) Code: Time Dep/-Wd Interest Balance 0 .00 1 300.00 .00 300.00 2 300.00 2.45 602.45 3 300.00 4.92 907.37 ..... 419 300.00 8,680.68 1,072,157.76 420 300.00 8,754.01 1,081,211.77 1 -132,841.67 948,370.10 (same time as 420th deposit) 2 -139,483.75 97,207.94 906,094.29 ..... 9 -196,267.65 36,468.57 195,991.85 10 -206,081.03 20,089.18 10,000.00 I make the 1st withdrawal $132,841.67 (instead of 132,691.91), as shown above. I am interpreting your "unclear problem" this way (make sure you tell your teacher to quit making up stories that are not clear math-wise!) : 420 monthly deposits of$300 are made in an account earning 10% compounded semi-annually; the deposits are made at month-end. On the same day as the last deposit is made, $W is withdrawn; withdrawals continue annually for the next 9 years, increasing by 5% annually. The rate remains at 10% compounded semi-annually (note that this was not stated in problem). "EDIT" After the last withdrawal,$10,000.00 remains in the account. Calculate W If instead I started the monthly deposits 1 month sooner (beginning of month), it would make the situation worse, since the account would be higher after 420 months. Anyway, 132,891.67 - 132,691.91 = 199.76: such a difference is really negligible when you consider a span of 45 years. If the "scenario" is different from what I assumed, let me know.
January 20th, 2012, 01:31 PM #5 Global Moderator Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 11,556 Thanks: 101 Math Focus: The calculus Re: Very difficult question (need help) To obtain the equivalent monthly rate: $$$1+\frac{r}{12}$$^12=$$1+\frac{0.10}{2}$$^2$ $$$1+\frac{r}{12}$$=$$1.05$$^{\frac{1}{6}}$ $V=300\sum_{k=1}^{420}$$1.05$$^{\frac{k}{6}}\approx 1090039.70$ To find the first year's withdrawal W, we set: $1.05^{18}V-W\sum_{k=9}^{18}1.05^k=10000$ $W=\frac{1.05^{18}V-10000}{\sum_{k=9}^{18}1.05^k}\approx\frac{2613310. 5045594033}{19.5124395879424565}\approx133930.49$
January 20th, 2012, 02:26 PM #6
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Re: Very difficult question (need help)
Quote:
Originally Posted by MarkFL $W=\frac{1.05^{18}V-10000}{\sum_{k=9}^{18}1.05^k}\approx\frac{2613310. 5045594033}{19.5124395879424565}\approx133930.49$
Agree, IF 1st $300 deposit is at START of month. Using actual year, last (420th)$300 deposit is made Dec.1(year 35).
Interest is earned/credited on Dec.31(year 35).
The 1st withdrawal is theoretically on same date...even if Jan.1(year 36).
So really we're at the 44 year point when last withdrawal is made.
Death is 1 year later: so I think we have to "re-visit" this, and calculate
to end up with ~9,070.29 after the last withdrawal:
so at death (1 year later or end of 45th year), we end up with
9070.20 * 1.1025 = 10,000.00
What you say to that, Mark? Am I clear?
January 20th, 2012, 02:41 PM #7 Joined: Jan 2012 Posts: 2 Thanks: 0 Re: Very difficult question (need help) The answer is what is stated in the problem, I have asked the professor. I am pulling my hair out because of this problem. There is some trick with that 5% inflation rate. I am getting an answer very similar to you Dennis, but apparently, that is the wrong answer.
January 20th, 2012, 07:17 PM #8
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Re: Very difficult question (need help)
Quote:
Originally Posted by xtifer The answer is what is stated in the problem, I have asked the professor. I am pulling my hair out because of this problem. There is some trick with that 5% inflation rate. I am getting an answer very similar to you Dennis, but apparently, that is the wrong answer.
Did you ask Mr Prof about the final $300 deposit and the first annual withdrawal: are they together, or a month apart, or something else? And did you ask about the$10,000 donation: was it made 1 year after the last annual withdrawal?
Problem states: "You also plan to bequeath, upon your death at 75 years of age, ....";
but the last withdrawal is at 74 years old (beginning of year).
Ask him/her if this is the intended cash flow (starting with year 1 for simplicity):
Jan.1/01 300.00 (deposit#1)
Feb,1/01 300.00 (deposit#2)
...
Nov.1/35 300.00 (deposit#419)
Dec.1/35 300.00 (deposit#420)
Jan.1/36 - W (wd#1)
Jan.1/37 - (1.05)W (wd#2)
Jan.1/38 - (1.05^2)W (wd#3)
...
Jan.1/45 - (1.05^9)W (wd#10)
Jan.1/46 -10,000.00 .....leaves account at exactly ZERO
Surely they'll tell you if that's the correct interpretation of their confusing "wording"!
January 21st, 2012, 03:00 AM #9
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Re: Very difficult question (need help)
I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that.
So I will merely toss in my thoughts on what's been posted and you two can grab it from there.
Quote:
Originally Posted by Denis You have to obtain a MONTHLY equivalent rate (since deposits are monthly): (1 + r)^12 = 1.05^2 Then use 300 (not 1800) and 420 (not 70).
I did this an entirely different way, just on my own, and came up with something about $600 off Mark's second shot. Because it's not compounding, I just figured out what six$300 payments would come out to at the end of six months, starting at the beginning of the month. (The problem doesn't say beginning or end, which is a serious flaw in the problem, but since the retirement is at the beginning, I'm going to just go with that throughout. I haven't yet re-worked it for end of the month.)
$300 a month, not compounding but earning, at the end of six months is$1853.24. (i.e. $300 for 6 months,$300 for 5 months, etc.) So that's how much there is at the end of the first six months. That amount will now just continue to compound at 5% per six months. The next six months the same thing is going to happen. So there's another 1853.24 plus the amount compounded on the first 1853.24. This seems to me is the equivalent of making a 1853.24 payment at the end of every six months for 70 periods at 5%. No?
That gave me 1,090,681.80.
Quote:
The rate remains at 10% compounded annually (note that this was not stated in problem).
Why annually when it was semi-annual the rest of the time? Again, the problem lacks this information, but I also see no reason it would suddenly stop compounding. But I see even less reason it would suddenly change to annual.
Quote:
And did you ask about the $10,000 donation: was it made 1 year after the last annual withdrawal? Problem states: "You also plan to bequeath, upon your death at 75 years of age, ...."; but the last withdrawal is at 74 years old (beginning of year). I think this is clear. Last withdrawal is at the beginning of age 74, and the$10,000 is one year later. I don't see the two as being connected or any reason to interpret it differently.
Quote:
Originally Posted by MarkFL Now subtract the $10,000.00 going to the university and we are left with:$1102318.89
I totally disagree with this one. Assuming that we continue the semi-annual compounding, that $10,000 would come from compounded money. We're not pulling out$10,000 at age 65 and just letting it sit doing nothing. It can be separated out, but we need the present value of it, which would be 3768.89. And I think it's easier just to separate that out.
But once that's subtracted off, I have no clue how to do the rest of the retirement part of it.
January 21st, 2012, 06:04 AM #10
Math Team
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Re: Very difficult question (need help)
Quote:
Originally Posted by Erimess
I
Quote:
The rate remains at 10% compounded annually (note that this was not stated in problem).
Why annually when it was semi-annual the rest of the time? Again, the problem lacks this information, but I also see no reason it would suddenly stop compounding. But I see even less reason it would suddenly change to annual.
Thanks...typo (changed it); but at least I did say "remains"
January 21st, 2012, 06:20 AM #11
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Re: Very difficult question (need help)
Quote:
Originally Posted by Erimess Because it's not compounding, I just figured out what six $300 payments would come out to at the end of six months, starting at the beginning of the month. (The problem doesn't say beginning or end, which is a serious flaw in the problem, but since the retirement is at the beginning, I'm going to just go with that throughout. I haven't yet re-worked it for end of the month.)$300 a month, not compounding but earning, at the end of six months is $1853.24. (i.e.$300 for 6 months, $300 for 5 months, etc.) So that's how much there is at the end of the first six months. That amount will now just continue to compound at 5% per six months. The next six months the same thing is going to happen. So there's another 1853.24 plus the amount compounded on the first 1853.24. This seems to me is the equivalent of making a 1853.24 payment at the end of every six months for 70 periods at 5%. No? That gave me 1,090,681.80. Yes, ok to do it that way...I get 1852.14425...to end with 1,090,039.69877... I think you used .10/12 as rate: slightly too high: rate compounds semi-annually, not monthly. .10/12 = .008333.... ; should be .008164... It is USUAL to make the interest rate factor match the frequency of payments. January 21st, 2012, 11:49 AM #12 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 3,073 Thanks: 32 Re: Very difficult question (need help) Quote: Originally Posted by Erimess I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that. Code: j = 10% i = 12% Year Deposit Interest Balance 0 .00 .00 .00 1 1000.00 .00 1000.00 2 1100.00 120.00 2220.00 3 1210.00 266.40 3696.40 Li'l "picture",$1000 deposit increasing by 10% annually, rate 12% annually.
F = Future value (?)
D = initial deposit (1000)
j = deposit increase (.10)
i = interest rate (.12)
n = number of years (3)
F = D*[(1 + i)^n - (1 + j)^n] / (i - j)
F = 1000(1.12^3 - 1.10^3) / (.12 - .10) = 3696.40
Basically all that simple...google will give ya lotsa sites, like:
http://www.financeformulas.net/Growing- ... Value.html
January 21st, 2012, 08:33 PM #13
Joined: Apr 2011
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Re: Very difficult question (need help)
Quote:
Originally Posted by Denis Yes, ok to do it that way...I get 1852.14425...to end with 1,090,039.69877... I think you used .10/12 as rate: slightly too high: rate compounds semi-annually, not monthly. .10/12 = .008333.... ; should be .008164... It is USUAL to make the interest rate factor match the frequency of payments.
I didn't miss the compounding. Looks like you're doing this as though the 10% were a yield and I did it as though it's a rate.
January 21st, 2012, 08:49 PM #14
Joined: Apr 2011
From: USA
Posts: 782
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Re: Very difficult question (need help)
Quote:
Originally Posted by Denis
Quote:
Originally Posted by Erimess I do not know how to work this out using the math you guys are, and I don't know how to work out the last part of it. That is, I don't know how to increase a payment by 5% a year, and at the same time still compound the 10% on the remainder. I only have my "built-in" equations and I don't know any that do that.
Code:
j = 10% i = 12%
Year Deposit Interest Balance
0 .00 .00 .00
1 1000.00 .00 1000.00
2 1100.00 120.00 2220.00
3 1210.00 266.40 3696.40
Li'l "picture", \$1000 deposit increasing by 10% annually, rate 12% annually.
F = Future value (?)
D = initial deposit (1000)
j = deposit increase (.10)
i = interest rate (.12)
n = number of years (3)
F = D*[(1 + i)^n - (1 + j)^n] / (i - j)
F = 1000(1.12^3 - 1.10^3) / (.12 - .10) = 3696.40
Basically all that simple...google will give ya lotsa sites, like:
http://www.financeformulas.net/Growing- ... Value.html
This isn't the same thing. I could've done that in Excel easily enough. (Though it never hurts to add an equation to the repertoire.)
This is a present value relative to the withdrawals, payments growing at 5% annually, but earning (we assume) 10% semi-annually. That equation is growing to a future value, not shrinking from a present value.
January 22nd, 2012, 07:00 AM #15
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Re: Very difficult question (need help)
Quote:
Originally Posted by Erimess
Quote:
Originally Posted by Denis Yes, ok to do it that way...I get 1852.14425...to end with 1,090,039.69877... I think you used .10/12 as rate: slightly too high: rate compounds semi-annually, not monthly. .10/12 = .008333.... ; should be .008164... It is USUAL to make the interest rate factor match the frequency of payments.
I didn't miss the compounding. Looks like you're doing this as though the 10% were a yield and I did it as though it's a rate.
Huh? I simply can't follow that...
This was the rate per the problem: "in an account that earns an average of 10% compounded semi-annually".
10% cpd semi-annually means 10.25% annually ; 1.05^2 - 1 = .1025
So the problem statement could as well be: ....10.25% compounded annually (same thing).
It is standard practice to convert the effective ANNUAL rate such that it matches the payments frequency;
on this case, to 12 periods annually; so: (1 + i)^12 = 1.1025 ; i = .0081648...
So the "semiannual equivalent deposit" is ~1852.14, not ~1853.24.
Another example:If given rate is 12% annual cpd. quarterly and payment frequency is 6 (every 2 months),
then equivalent rate becomes: (1 + i)^6 = (1 + .12/4)^4 ; i = .019901...slightly lower than 2%, as expected.
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=SignalProcessing/GenerateFiniteImpulseResponseFilterTaps | 1,462,485,494,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461861623301.66/warc/CC-MAIN-20160428164023-00055-ip-10-239-7-51.ec2.internal.warc.gz | 643,663,408 | 28,766 | SignalProcessing - Maple Help
Home : Support : Online Help : Science and Engineering : Signal Processing : Filtering : SignalProcessing/GenerateFiniteImpulseResponseFilterTaps
SignalProcessing
GenerateFiniteImpulseResponseFilterTaps
generate a taps array for a finite impulse response filter
Calling Sequence GenerateFiniteImpulseResponseFilterTaps( n, freq )
Parameters
n - posint, the size of the resulting taps Array freq - realcons or [realcons,realcons]}, the frequency for a low-pass or high-pass filter, or a pair [lowfreq, highfreq] of frequencies for a band-pass or band-stop filter. Frequencies f must satisfy 0 < f and f < 1/2.
Options
• filtertype : string or name, the type of filter
• window : string or name, the window type
• normalize : truefalse, whether to normalize
• container : Array, predefined Array for holding the result
Description
• The GenerateFiniteImpulseResponseFilterTaps( n, freq ) command generates a taps array for a finite impulse response filter. The resulting array may be used with the command FiniteImpulseResponseFilter to effect filtering operations on a sample.
• For a low-pass or high-pass filter, a single frequency freq is required. It must be strictly positive and less than $\frac{1}{2}$.
• For a band-pass or band-stop filter, a pair [lowfreq, highfreq] of low and high frequencies are required. Each frequency must be positive and less than $\frac{1}{2}$; they must additionally satisfy $\mathrm{lowfreq}<\mathrm{highfreq}$.
• Specify the type of filter with the filtertype option. It can take any of the values "lowpass", "highpass", "bandpass" and "bandstop". The option values can be given as a name or as a string.
• The type of window to use is indicated by the window option. It can take any of the following values: "Bartlett", "Blackman", "Hamming" and "Hann". The option values can be given as a name or as a string.
• The normalize option specifies whether to normalize the filter coefficients produced, and can be either of the values true or false. The default is normalize = true.
• If the container=C option is provided, then the results are put into C and C is returned. With this option, no additional memory is allocated to store the result. The container must be an Array of size $n$ having datatype float[8].
• The SignalProcessing[GenerateFiniteImpulseResponseFilterTaps] command is thread-safe as of Maple 17.
Examples
> $\mathrm{with}\left(\mathrm{SignalProcessing}\right):$
> $\mathrm{GenerateFiniteImpulseResponseFilterTaps}\left(8,\frac{1}{3}\right)$
$\left[\begin{array}{cccccccc}{0.}& {-}{0.0357142857250822}& {4.31857549578176}{}{{10}}^{{-11}}& {0.535714285681896}& {0.535714285681896}& {4.31857549578176}{}{{10}}^{{-11}}& {-}{0.0357142857250822}& {0.}\end{array}\right]$ (1)
> $\mathrm{GenerateFiniteImpulseResponseFilterTaps}\left(8,\frac{1}{3},':-\mathrm{normalize}'=\mathrm{true}\right)$
$\left[\begin{array}{cccccccc}{0.}& {-}{0.0357142857250822}& {4.31857549578176}{}{{10}}^{{-11}}& {0.535714285681896}& {0.535714285681896}& {4.31857549578176}{}{{10}}^{{-11}}& {-}{0.0357142857250822}& {0.}\end{array}\right]$ (2)
> $C≔\mathrm{Array}\left(1..8,'\mathrm{datatype}'={'\mathrm{float}'}_{8},'\mathrm{order}'='\mathrm{C_order}'\right):$
> $\mathrm{GenerateFiniteImpulseResponseFilterTaps}\left(8,\frac{1}{3},':-\mathrm{container}'=C\right)$
$\left[\begin{array}{cccccccc}{0.}& {-}{0.0357142857250822}& {4.31857549578176}{}{{10}}^{{-11}}& {0.535714285681896}& {0.535714285681896}& {4.31857549578176}{}{{10}}^{{-11}}& {-}{0.0357142857250822}& {0.}\end{array}\right]$ (3)
> $C$
$\left[\begin{array}{cccccccc}{0.}& {-}{0.0357142857250822}& {4.31857549578176}{}{{10}}^{{-11}}& {0.535714285681896}& {0.535714285681896}& {4.31857549578176}{}{{10}}^{{-11}}& {-}{0.0357142857250822}& {0.}\end{array}\right]$ (4)
> $\mathrm{GenerateFiniteImpulseResponseFilterTaps}\left(8,\frac{1}{3},':-\mathrm{filtertype}'="highpass"\right)$
$\left[\begin{array}{cccccccc}{0.}& {0.0220588235076414}& {-}{0.147058823517644}& {0.330882352974715}& {-}{0.330882352974715}& {0.147058823517644}& {-}{0.0220588235076414}& {-}{0.}\end{array}\right]$ (5)
> $\mathrm{GenerateFiniteImpulseResponseFilterTaps}\left(8,\left[0.2,0.4\right],':-\mathrm{filtertype}'="bandstop",':-\mathrm{window}'="Hann"\right)$
$\left[\begin{array}{cccccccc}{-}{0.}& {0.0250125469199284}& {0.0729369300004544}& {0.402050523079617}& {0.402050523079617}& {0.0729369300004544}& {0.0250125469199284}& {-}{0.}\end{array}\right]$ (6)
>
Compatibility
• The SignalProcessing[GenerateFiniteImpulseResponseFilterTaps] command was introduced in Maple 17.
• For more information on Maple 17 changes, see Updates in Maple 17. | 1,471 | 4,690 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 19, "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-2016-18 | longest | en | 0.618918 |
https://git.ist.ac.at/alois.schloegl/sonets/blame/master/run_ei_balanced.cpp | 1,571,061,892,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986653247.25/warc/CC-MAIN-20191014124230-20191014151730-00068.warc.gz | 489,417,592 | 16,465 | run_ei_balanced.cpp 8.63 KB
Alois SCHLOEGL committed Sep 17, 2015 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 ``````#include #include #include #include #include #include #include "secorder_rec_2p.hpp" #include "calc_stats_2p.hpp" using namespace std; int main(int argc, char *argv[]) { // optional flags int deterministic_seed = 0; // if nonzero, use for random num gen seed int N_nodes[2] = {2000,500}; double p[2][2] = {{0.01, 0.01},{0.01,0.01}}; double alpha_recip[2][2] = {{0.0, 0.0},{99,0.0}}; // form of large N covariance matrix centered around population x // conv_x00 conv_x01 chain_0x0 chain_1x0 // conv_x11 chain_0x1 chain_1x1 // div_00x div_01x // div_11x /////////////////////////////////////////////// // correlations centered around population zero // diagonal entries double alpha_conv_hom_00 = 0.5; // zeros onto zero (1,1) double alpha_conv_hom_01 = 0.5; // ones onto zero (2,2) double alpha_div_hom_00 = 0.5; // zero onto zeros (3,3) double alpha_div_hom_10 = 0.5; // zero onto ones (4,4) // cross entries for divergence and convergence // constrained by the diagonal entries double cc_conv_0_01 = 0.5; // mix onto zero (2,1) double cc_div_01_0 = 0.0; // zero onto mix (4,3) // remaining four cross entries are chains through population 0 // constrained by diagonal entries double cc_chain_000 = 0.8; // (3,1) double cc_chain_100 = 0.0; // (4,1) double cc_chain_001 = 0.8; // (3,2) double cc_chain_101 = 0.0; // (4,2) /////////////////////////////////////////////// // correlations centered around population one // diagonal entries double alpha_conv_hom_10 = 0.5; // zeros onto one (1,1) double alpha_conv_hom_11 = 0.5; // ones onto one (2,2) double alpha_div_hom_01 = 0.5; // one onto zeros (3,3) double alpha_div_hom_11 = 0.5; // one onto ones (4,4) // cross entries for divergence and convergence // constrained by the diagonal entries double cc_conv_1_01 = 0.5; // mix onto one (2,1) double cc_div_01_1 = 0.0; // one onto mix (4,3) // remaining four cross entries are chains through population 1 // constrained by diagonal entries double cc_chain_010 = 0.0; // (3,1) double cc_chain_111 = 0.0; // (4,1) double cc_chain_110 = 0.0; // (3,2) double cc_chain_011 = 0.0; // (4,2) // double alpha_conv[2][2][2] = {{{0.8,0.1},{99,0.8}},{{0.8,0.1},{99,0.8}}}; // double alpha_div[2][2][2] = {{{0.8,0.8},{0.1,0.1}},{{99,99},{0.8,0.8}}}; // //double alpha_chain[2][2][2] = {{{0.5,0.2},{0.3,0.4}},{{0.2,0.1},{0.4,0.5}}}; // double alpha_chain[2][2][2] = {{{0.0,0.0},{0.0,0.0}},{{0.0,0.0},{0.5,0.0}}}; double alpha_conv_hom[2][2] = {{alpha_conv_hom_00,alpha_conv_hom_01},{alpha_conv_hom_10,alpha_conv_hom_11}}; double cc_conv_mixed[2][2][2] = {{{99,cc_conv_0_01},{99,99}},{{99,cc_conv_1_01},{99,99}}}; double alpha_div_hom[2][2] = {{alpha_div_hom_00,alpha_div_hom_01},{alpha_div_hom_10,alpha_div_hom_11}}; double cc_div_mixed[2][2][2] = {{{99,99},{cc_div_01_0,cc_div_01_1}},{{99,99},{99,99}}}; double cc_chain[2][2][2] = {{{cc_chain_000,cc_chain_001},{cc_chain_010,cc_chain_011}},{{cc_chain_100,cc_chain_101},{cc_chain_110,cc_chain_111}}}; //{{{000,001},{010,011}},{{100,101},{110,111}}} gsl_rng *r = gsl_rng_alloc(gsl_rng_mt19937); // set the seed // if deterministic_seed, use that for seed if(deterministic_seed) gsl_rng_set(r,deterministic_seed); // else use time in seconds for the seed else gsl_rng_set(r, time(NULL)); // matrix files will be stored in data directory // with filename determined by the six input parameters mkdir("data",0755); // make data directory, if it doesn't exist char FNbase[200]; sprintf(FNbase,"_%i_%i_%1.3f_%1.3f_%1.3f_%1.3f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f_%1.2f", N_nodes[0], N_nodes[1], p[0][0],p[0][1],p[1][0],p[1][1], alpha_recip[0][0], alpha_recip[0][1], alpha_recip[1][1], alpha_conv_hom[0][0], alpha_conv_hom[0][1], alpha_conv_hom[1][0], alpha_conv_hom[1][1], cc_conv_mixed[0][0][1], cc_conv_mixed[1][0][1], alpha_div_hom[0][0], alpha_div_hom[0][1], alpha_div_hom[1][0], alpha_div_hom[1][1], cc_div_mixed[0][1][0], cc_div_mixed[0][1][1], cc_chain[0][0][0], cc_chain[0][0][1], cc_chain[0][1][0], cc_chain[0][1][1], cc_chain[1][0][0], cc_chain[1][0][1], cc_chain[1][1][0], cc_chain[1][1][1]); // generate second order matrix W gsl_matrix *W; W = secorder_rec_2p(N_nodes,p, alpha_recip, alpha_conv_hom, cc_conv_mixed, alpha_div_hom, cc_div_mixed, cc_chain, r); // if failed to generate a matrix, write error and quit if(!W) { cerr << "Failed to generate the matrix\n"; return -1; } char FN[200]; FILE *fhnd; //////////////////////////////////////////////////////////// // Calculate the covariance structure of the matrix // This should approximately agree with the input alphas //////////////////////////////////////////////////////////// cout << "Testing statistics of matrix..."; cout.flush(); double phat[2][2]; double alphahat_recip[2][2]; double alphahat_conv[2][2][2]; double alphahat_div[2][2][2]; double alphahat_chain[2][2][2]; calc_phat_alphahat(W, N_nodes, phat, alphahat_recip, alphahat_conv, alphahat_div, alphahat_chain); cout << "done\n"; strcpy(FN, "data/stats"); strcat(FN, FNbase); strcat(FN, ".dat"); fhnd = fopen(FN, "w"); if(fhnd==NULL) { cerr << "Couldn't open outfile file " << FN << "\n"; exit(-1); } fprintf(fhnd, "%i %i ", N_nodes[1], N_nodes[2]); cout << "Actual statistics of matrix:\n"; cout << "phat = "; for(int i=0; i<2; i++) for(int j=0; j<2; j++) { cout << phat[i][j] << " "; fprintf(fhnd, "%e ", phat[i][j]); } cout << "\n"; cout << "alphahat_recip = "; for(int i=0; i<2; i++) for(int j=i; j<2; j++) { cout << alphahat_recip[i][j] << " "; fprintf(fhnd, "%e ", alphahat_recip[i][j]); } cout << "\n"; cout << "alphahat_conv = "; for(int i=0; i<2; i++) for(int j=0; j<2; j++) for(int k=j; k<2; k++) { cout << alphahat_conv[i][j][k] << " "; fprintf(fhnd, "%e ", alphahat_conv[i][j][k]); } cout << "\n"; cout << "alphahat_div = "; for(int i=0; i<2; i++) for(int j=i; j<2; j++) for(int k=0; k<2; k++) { cout << alphahat_div[i][j][k] << " "; fprintf(fhnd, "%e ", alphahat_div[i][j][k]); } cout << "\n"; cout << "alphahat_chain = "; for(int i=0; i<2; i++) for(int j=0; j<2; j++) for(int k=0; k<2; k++) { cout << alphahat_chain[i][j][k] << " "; fprintf(fhnd, "%e ", alphahat_chain[i][j][k]); } cout << "\n"; cout.flush(); fclose(fhnd); // balance excitation and inhibition int balance_average=1; if(balance_average) { // balance so expected value of inputs is zero double ee_tot = p[0][0]*N_nodes[0]; double ei_tot = p[0][1]*N_nodes[1]; double ie_tot = p[1][0]*N_nodes[0]; double ii_tot = p[1][1]*N_nodes[1]; // calculate magnitudes to make the expected values balance //double Wee_mag = 1.0; double Wei_mag = -ee_tot/ei_tot; //double Wie_mag = 1.0; double Wii_mag = -ie_tot/ii_tot; gsl_matrix_view Wei = gsl_matrix_submatrix (W, 0, N_nodes[0], N_nodes[0], N_nodes[1]); gsl_matrix_scale(&Wei.matrix, Wei_mag); gsl_matrix_view Wii = gsl_matrix_submatrix (W, N_nodes[0], N_nodes[0], N_nodes[1], N_nodes[1]); gsl_matrix_scale(&Wii.matrix, Wii_mag); } else { // balance so each row sum is exactly zero for(int i=0; i | 3,081 | 8,085 | {"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} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.176482 |
https://community.esri.com/t5/arcgis-survey123-blog/using-formulas-in-survey123/bc-p/898177 | 1,611,721,940,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704820894.84/warc/CC-MAIN-20210127024104-20210127054104-00711.warc.gz | 289,771,893 | 66,575 | # Using Formulas in Survey123
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08-23-2015 06:03 PM
Regular Contributor
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Formulas supported in Survey123 include the following operators:
SymbolFunctionExample
+
2 + 2
-
Subtraction
2 - 2
*
Multiplication
2 * 2
div
Division
2 div 2
=
Equal
.=10
!=
Not equal
.!=10
<
Less than
.<10
<=
Less than or equal to
.<=10
>
Greater than
.>10
>=
Greater than or equal to
.>=10
or
or
.=10 or .=20
and
and
.>10 and .<20
Formulas can be used in Survey123 when building your forms in the following columns:
Constraint
Adding a constraint to a survey question will restrict the accepted inputs for a response. This could include a specific range of numbers, combinations of letters and numbers or general pattern matching. In your spreadsheet, the constraint expression is entered into the constraint field and helpful text is entered into the constraint_message field. In the constraint expressions, the input for the question is always represented by a full stop.
For example, the following formula can be used to restrict the input of an integer field to positive numbers only:
.>=0
Regular expressions can also be used for pattern matching to constrain valid responses. This regular expression can constrain the input of a string field to a well-known email address format:
regex(., '[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}')
Relevant
A question, or a sets of questions, can be hidden and revealed based on previous answers using relevant expressions. These expressions are entered into the relevant column and the answers to previous questions are always referred to as follows: \${field_name}. You can apply a relevant expression to a single question, or group questions together and set the relevant expression for the entire group. For example, this expression will reveal a question if the answer to the previous question is true:
\${previous_question} = 'true'
This example will hide questions if the answer to the previous question was greater than 100:
\${previous_question} <= 100
This example combines multiple operators and questions:
\${previous_question} + \${other_previous_question} <= 100
When using mathematical operators, be aware that sometimes you may need to cast values into numbers. In the previous examples we assumed that previous_question and other_previous_question where integers or decimals... but what if the question types were strings? Then you cast them as follows:
int(\${previous_question}) + int(\${other_previous_question}) <= 100
Calculations
Calculations are performed in the calculation field of a question. Calculations are often associated with calculate type of questions, but can also be applied to integer, text etc type of questions. The outcome of the calculation can be used to populate relevant or constraint expressions by referring to the field name of the calculate question. They can be used to hold values that do not need to be displayed on the form, but are included in the feature service.
For example, you can create a question of type calculate and name it calc, then insert the following expression int its calculation column:
\${question_1} + \${question_2} + \${question_3}
And then use the result to set the relevance for the next question:
\${calc} <= 100
Tags (1)
by
New Contributor
Is there a way to assign text answers with values? And then in turn calculate them at the end?
New Contributor
I had the same question but managed to figure it out in the end.. after a bit of fiddling it turned out be rather simple.
Let's say you have a multiple-choice question ('select_one' type) with 5 possible answers, and you want to score the answer on a scale of 1 to 5.
In the choices worksheet, first create a list with the possible text responses, let's call the list_name 'Q1_choices'. In the label column you'll write the text for each possible choice (so that's 5 lines of text in total, with the Q1_choices name repeated in the list_name column for every line). All you have to do now is to write 1,2,3,4 or 5 in the 'name' column as appropriate (one per line) which will be the score for the response.
In the survey sheet, you'll have as type 'select_one Q1_choices'. For the name column let's use 'Q1_response'. For label just type the text of your question.
Now to work out and show the score add another line in the survey sheet (before the 'end group' line). For the type column use 'integer', let's call the name column 'Q1_score', and in the label column you can write something like 'Your score is:'. Finally, and crucially, in the calculation column insert '\${Q1_response}' (without quotation marks).
When you test the questionnaire in survey123 you should now see the score displayed. As you click through the different responses in the form you should see the score changing between 1 and 5 in real time.
This approach can be repeated for multiple questions, so you could have lists called Q2_choices, Q3_choices, etc all with 1,2,3,4 or 5 in the name column. The duplication of identical 'name' options between list_names doesn't seem to matter, the responses are all uniquely tied to their particular question.
Good luck, hope this makes sense!
Occasional Contributor II
I had a question regarding using the regex commands. I have a survey where we are using a work order id to start the survey but we want to control what the user types in by using a constraint as well as have a default value set on the same field so it is always pre-populated with the letter "T" at the beginning. So, I have the default value set to the letter "T" and have the following formula written in the constraint :
regex(.,'[0-9]{10}')
The goal here was to allow the user to type in whatever numbers necessary after the letter "T" but they would be limited to 10 characters long including this letter "T". Every little alteration with these regex commands seems to publish fine with no errors but then doesn't work the way I need it to when testing it on the survey itself. Is there an issue with the default value being pre-populated already? I need the result to look like something like this "T101819654" but produce a message when the string is less or more than 10 characters long.
I also ran across the types that ESRI supports on the xls forms themselves and I see this:
re{ n, m} Matches at least n and at most m occurrences of preceding expression.
So, if I need to use the "re" where do I insert it to make my equation post successfully? Do I need anything additional to my equation above to make it work? It seems so simple and even had the same type of regex command working in an online tester but then realized that ESRI uses some slightly different syntax and now I am stuck. Any help with fixing my expression would be great!
Thanks, Jason Sphar
by
Occasional Contributor II
Hi Jason,
I think the regular expression you need would look like this:
regex(.,'T+[0-9]{9,9}')
{m,n} matches the preceding element at least m and not more than n times, so it should be {9,9} in your case as you need exactly 9 numbers after the letter 'T'. Mind you, in practice the form only seems to use the value for m and completely ignores n. It throws an error if you have less than 9 numbers after T, but lets you enter more than 9 as well. (Same issue with any combination, tried {3,4} and so on... )
It's the same with the example in the article for email addresses. Based on the expression regex(., '[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}') , it should check that you have a proper email address ending @domain.xx, @domain.xxx or @domain.xxxx. However, it accepts anything after @ if you have at least 4 characters, like @mmmm which should throw an error. (Using Survey connect 1.5.35)
If this is not a bug, then the documentation needs some updating...
Zoltan
by
Occasional Contributor II
I had another think about it and you may be better off with using an input mask for that field instead of a regular expression. With that you can control the number of characters, whether they are letters or numbers etc. Have a look at "Input masks" in here. I hope it helps.
Zoltan
New Contributor III
Hi,
I have a select_one (cascade select) question that i use to hide or not subsequent question and it works fine but i have an issue because i have another question that i use to filter the same subsequent question like previous question that i described. I need to hide some questions using one or two conditions. it is possible?
Esri Esteemed Contributor
Hi Pablo,
It is possible to use and/or/not in the relevant statement to generate more complex functions. Given what you describe above, let's do a scenario around disaster preparation. Sometimes extra assistance is needed to evacuate; you want a question asking if extra assistance is needed when the household contains more than 8 people or there are people with significant mobility limitations. The relevant statement for the third question would be:
\${numPeople} > 8 or \${mobilityLimitations} = 'yes'
New Contributor II
Is it possible to "calculate" if an image question has been filled? ie has the photo been collected in the survey. I want to do this so I can use that as a condition for displaying a subsequent question.
Esri Frequent Contributor
Hi Terry. Information in this thread may be useful:https://community.esri.com/message/662807-re-make-a-new-question-dependent-on-an-image-capture
Essentially, the expression string-length(\${imageQuestion})>0 evaluates to true only if a photo has been selected or taken. You can use the expression in the relevant field.
New Contributor II
Great info Hannah. I just wanted to let you know the links for constraint, relevant, and calculations do not work. Thanks!
#survey123
Esri Esteemed Contributor
Hi Sean,
I've updated the links- we revised our documentation system a couple of months ago.
Occasional Contributor III
@Ismael Chivite, @James Tedrick
How do I go about populating a value based on a previous select_one answer? Here's my set up:
• Q1: Infrastructure type (select one) to equal 'residential meter'
• Q2: Enclosure (select one) populate with the answer 'meter box'
Thanks
Esri Esteemed Contributor
Hi Andrew,
Calculations with select_one questions is coming with the 2.7 release of Survey123, which will be released in the near future. You can test out your functions by grabbing an beta version in the Esri Early Adopter Community .
New Contributor II
I have created surveys for street sign inspections and have collected about 7k sign poles with 13000 related signs (with groups and repeats). I am now working on another survey to collect the reflectivity of those signs. The current form auto-populates most of the fields based upon the sign inspections collected previously and through the use of external choices and "pull data". Is it possible to calculate values from those choices using a range? What I am hoping to accomplish is populate a field with a condition based upon the reflectivity reading. For example, a "white" "regulatory" sign would be considered in "fair" condition if the reflectivity reading fell between the range of 250 -500 and a "brown" "guide" sign would be in fair condition if the reading was between 25 - 40. So the condition is determined by the results of the following fields 1) SignType 2) SignColor 3) Reflectivity (integer). I am just trying to figure out how to incorporate ranges for different sign types and color into the equation to determine condition based upon the entered integer (reflectivity). Hope that makes sense! Any thoughts?
Esri Esteemed Contributor
Hi,
This should be possible. I would probably find it easiest to set up a table of sign types and conditions with minimum values per condition level, import those with pulldata, and then use a series of nested if statements to find the appropriate level
table structure:
signtypemin_poormin_fairmin_good
white_regulatory10250501
brown_guide52541
The if statement would be something like:
if(\${reflectivity_measurement} < \${min_fair}, 'poor', if(\${reflectivity_measurement} < \${min_good}), 'fair', good))
New Contributor II
Thanks for the quick response! So... here is my external table (SignReflect2.csv):
color min_poor max_poor min_fair max_fair min_good max_good Blue 0 19 20 39 40 60 Brown 0 24 25 44 45 55 FY 0 149 150 374 375 450 FYG 0 249 250 474 475 650 Green 0 69 70 114 115 125 Red 0 49 50 99 100 160 White 0 249 250 499 500 750
How would I nest the following expression using SignColor and Reflectivity(integer) entries from Survey123 form?
if(\${Reflectivity} < \${min_fair}, 'poor'), if(\${Reflectivity} < \${min_good}, 'fair'), if(\${Reflectivity} > \${min_good}, 'good')
When using the following pulldata parameters:
pulldata ('SignReflect2','min_fair','color',{SignColor})
pulldata ('SignReflect2','min_good','color',{SignColor})
or am I not approaching this right? I appreciate any feedback you can provide as I may be over thinking this.
Esri Esteemed Contributor
Hi,
That looks pretty close to correct, it looks like it's a matter of getting the nesting correct. A nested set of if statements will look correct if all (or nearly all, if more complex functions/evaluations are present) of the closing parenthesis are all bunched together - think of each if statement being an individual matryoshka doll:
if(\${Reflectivity} < \${min_fair}, 'poor', if(\${Reflectivity} < \${min_good}, 'fair', 'good'))
You also don't need the last evaluation since there is no other category beyond 'good'. Sentence-wise this can read as:
"If the reflectivity is less than the min_fair value, mark as poor; otherwise, look to see if reflectivity is less than the min_good value; if it is less than, mark as fair; otherwise, mark as good."
New Contributor II
Thanks again. One last question. The min_fair, min_poor, etc is determined by the color. The color answer from a previous question should be used to determine the min and max values for condition in the table above. see below for description of statements:
if color is Blue and Reflectivity is less the min_fair then condition is poor
if color is Blue and Reflectivity is less then min_good and greater than max_poor then condition is fair
if color is Blue and Reflectivity is greater then min_good then condition is good
if color is Brown and Reflectivity is less the min_fair then condition is poor
if color is Brown and Reflectivity is less then min_good and greater than max_poor then condition is fair
if color is Brown and Reflectivity is greater then min_good then condition is good
etc.
Is it possible to use the pulldata function for multiple fields (min_poor, max_poor, min_fair, max_fair, etc.) using the external table above and then use the above "if" statement based upon the color (which is determined by a previous question in the form)? I have not been able to get this to work but am sure I am missing something in the syntax.
Esri Esteemed Contributor
Yes, this should be possible. The loading of values based on the color would be the job of the pulldata function. One caveat is that values read from pulldata are treated as text by default, so you would need to convert to a number. This can be done by setting the bind::type of the pulldata calculate questions to int or decimal. Here's a quick mock up (blog comments can't add attachments):
New Contributor II
Worked like a charm!! Thanks a bunch!
New Contributor
I followed the same steps on a survey that I am working on, but I tried to do it with the select_multiple question type. My final calculation would only show up if a single answer was selected, not if I picked more than one. Do you know of any workarounds for this issue?
New Contributor III
Can or/and statements be used in constraint statements too?
I have the survey questions 'Select the most abundant', 'Select the 2nd most abundant', and 'Select the 3rd most abundant'. I'd like to constrain the response to the 2nd and 3rd most abundant questions so that users cannot choose a response they selected for the previous question.
.!=\${mostabundant} seems to prevent the 2nd most abundant response from being the same as the most abundant answer.
.!=\${mostabundant} or .!=\${2ndabundant} however does not achieve the desired response; the 3rd most abundant can be the same as either the most or 2nd most abundant answer.
Occasional Contributor II
Is there a way to do this without using the name field? What if I need the name to populate the result in the backend? Can another field be added that can hold a value for an answer and then be able to add it up at the end?
Thanks
Occasional Contributor
If you're using a select_one question, then in most cases (the most common exception being when using a cascading select) the label of the choice will be used in the results.
Esri Frequent Contributor
Hi Santokh,
Which pulldata() calculation is not working as there are 2 shown. For the first one I do not see the PR column in your CSV file screenshot, which should be the return column?
pulldata(<csvfile>, <returnColumn>, <lookupColumn>, <lookupValue>)
Are you able to provide a copy of your xlsx file and csv file so we can take a closer look.
Phil.
New Contributor II
Hi Phil,
I got it working.
Thanks for help.
Santokh
New Contributor II
Hi,
Is it possible to secure the feature layer for Survey123, so that data uploaded to ArcGIS cannot be changed? Any help will be appreciated.
Thanks,
Santokh
Esri Esteemed Contributor
Hi Santokh,
When you share a Survey123 form, you have the option to share it for viewing; this creates a read-only endpoint that can be used in applications. See Share survey results—Survey123 for ArcGIS | ArcGIS for more information. Additionally, you can restrict querying on the feature layer or feature layer view being used for submission (if the form is not being used to edit data) - see Manage hosted feature layers—ArcGIS Help | ArcGIS for more information.
Occasional Contributor III
Hi James,
The 'Manage Hosted feature layers' link was useful, especially when talking about 'Further control of edits for layers'. I'm wondering if making a small group as described in that section is my only answer to this question, posted in the ArcGIS group and maybe similar to what Santokh Randhawa was asking....
New Contributor II
Hi James, Thanks for reply. I would like to explain my question in a better way. When you are viewing your submitted data from Survey123 in ArcGIS , you can double tap on any field and you can change its value. Is there way to lock that edit option?
Santokh
Esri Esteemed Contributor
Hi Santokh,
Which product are you viewing the survey data in? The table view of the Survey123's Data tab does not allow editing. That being said, other tools (like ArcGIS 's Map Viewer table view) can let you edit the data; you should 'Disable editing' in the layer's properties if this is what you are referring to.
New Contributor II
Hi James,
I have inserted the image of data table. I did disable editing the layer in the settings. But still I was able to edit all data fields by double clicking on it.
Santokh
Esri Esteemed Contributor
Hi Santokh,
This is an issue with regard to ArcGIS 's data view. As the item owner, it may be expected that you would have edit access to the data- I would ask the ArcGIS team for clarity on this.
New Contributor II
Hi James,
I thought about that and then checked it using the login of one of the group members but I was still able to edit the submitted data in ArcGIS data view.
Another question I have is that is there a way to lock editing of sent surveys from Survey123 app. As of now I can click on "sent surveys" on the app and reopen it, edit and send again which will edit the previously submitted data. Is there a way to lock it?
Thanks,
Santokh
Esri Esteemed Contributor
In terms of 'locking' from the Survey123 app, you can disable the Sent folder
New Contributor II
Thanks James. That was helpful. How about my other question on blocking editing of ArcGIS online data? did you find something on that.
Santokh
New Contributor II
Hi, I have a similar situation, but I have three conditions to choose from. Is it possible to have more than two conditions in an "if" statement? For more information please see my question: How to calculate a field based on multiple conditions?
Esri Frequent Contributor
Hi Michael,
Yes, you can have more than two conditions in an if statement if you use nested if statements, after each condition add a new if statement. An example can be seen here:
Hope this helps.
Phil. | 4,817 | 20,674 | {"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-2021-04 | latest | en | 0.867597 |
http://www.helpteaching.com/questions/122063/solve-the-following-equation-2407-106 | 1,511,127,433,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805809.59/warc/CC-MAIN-20171119210640-20171119230640-00291.warc.gz | 423,576,152 | 5,128 | ##### Question Info
This question is public and is used in 88 tests or worksheets.
Type: Multiple-Choice
Category: Decimals
Standards: 5.NBT.B.7
Score: 1
Author: TiffTwns
View all questions by TiffTwns.
# Decimals Question
View this question.
Add this question to a group or test by clicking the appropriate button below.
## Grade 5 Decimals CCSS: 5.NBT.B.7
Solve the following equation: $24.07 - 1.06 =$
1. 13.4
2. 1.467
3. 22.01
4. 23.01
You need to have at least 5 reputation to vote a question down. Learn How To Earn Badges. | 172 | 537 | {"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} | 2.625 | 3 | CC-MAIN-2017-47 | latest | en | 0.861518 |
https://zh.khanacademy.org/math/algebra/algebra-functions/average-rate-of-change-word-problems/a/average-rate-of-change-review | 1,723,728,872,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641291968.96/warc/CC-MAIN-20240815110654-20240815140654-00582.warc.gz | 805,671,542 | 132,102 | If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
# 平均变化率复习
## 何为平均变化率？
$\frac{f\left(b\right)-f\left(a\right)}{b-a}$
## 求平均变化率
### 例题 1：从图中求平均变化率
$\begin{array}{rl}\text{平均变化率}& =\frac{f\left(9\right)-f\left(0\right)}{9-0}\\ \\ & =\frac{3-\left(-7\right)}{9}\\ \\ & =\frac{10}{9}\end{array}$
### 例题 2: 求方程式的平均变化率
$g\left(1\right)={1}^{3}-9\cdot 1=-8$
$g\left(6\right)={6}^{3}-9\cdot 6=162$
$\begin{array}{rl}\text{平均变化率}& =\frac{g\left(6\right)-g\left(1\right)}{6-1}\\ \\ & =\frac{162-\left(-8\right)}{5}\\ \\ & =34\end{array}$
$g$ 在区间 $-8\le x\le -2$ 上的平均变化率是多少? | 311 | 638 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "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": 6, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2024-33 | latest | en | 0.246226 |
http://openstudy.com/updates/52ca2394e4b04f95cb84ece0 | 1,444,407,588,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737933027.70/warc/CC-MAIN-20151001221853-00043-ip-10-137-6-227.ec2.internal.warc.gz | 234,218,272 | 17,592 | ## desalexus one year ago How can you determine whether a matrix product AB is defined?
1. Loser66
2. Mertsj
You're doing fine. Give an example. That will help.
3. desalexus
its HER and like this A is 2×3 and B is 3×2, so AB is (2×3)(3×2).
4. RolyPoly
$A_{m\times n}B_{n\times p}$? The number of columns of A = number of rows of B
5. RolyPoly
Dimension of AB would be m×p
6. desalexus
Thanks
7. Mertsj
It depends on the dimensions:
8. mathmale
The example that Desalexus presents is illustrative: A is 2×3 and B is 3×2, so AB is (2×3)(3×2). Note how Matrix A has 2 rows and 3 columns, and B 3 rows and 2 columns. We can indeed multiply matrices A and B (in that order).
9. Mertsj
|dw:1388979444164:dw|
10. Mertsj
|dw:1388979475429:dw|
11. RolyPoly
|dw:1388979474597:dw|
12. mathmale
Yes, Mertsj, and you can make that statement more precise by stating that "the number of columns of the first matrix must equal the number of rows of the second. RolyPoly is correct in his drawing (immediately above), whereas I found that I was wrong at the onset (and have deleted my incorrect statement accordingly)!
13. mathmale
I'd rather find my own mistakes than have someone else find them for me! :)
14. Mertsj
Sometimes it is easier for the asker to see an example than to decipher a bunch or words but it is good to have both so the asker can choose. | 414 | 1,370 | {"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} | 3.328125 | 3 | CC-MAIN-2015-40 | longest | en | 0.903801 |
https://www.airmilescalculator.com/distance/the-to-lec/ | 1,606,781,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00560.warc.gz | 588,371,952 | 27,652 | Distance between Teresina (THE) and Lençóis (LEC)
Flight distance from Teresina to Lençóis (Teresina Airport – Lençóis Airport) is 521 miles / 838 kilometers / 453 nautical miles. Estimated flight time is 1 hour 29 minutes.
Driving distance from Teresina (THE) to Lençóis (LEC) is 695 miles / 1118 kilometers and travel time by car is about 14 hours 42 minutes.
Map of flight path and driving directions from Teresina to Lençóis.
Shortest flight path between Teresina Airport (THE) and Lençóis Airport (LEC).
How far is Lençóis from Teresina?
There are several ways to calculate distances between Teresina and Lençóis. Here are two common methods:
Vincenty's formula (applied above)
• 520.925 miles
• 838.348 kilometers
• 452.672 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 523.580 miles
• 842.621 kilometers
• 454.979 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Teresina Airport
City: Teresina
Country: Brazil
IATA Code: THE
ICAO Code: SBTE
Coordinates: 5°3′35″S, 42°49′24″W
B Lençóis Airport
City: Lençóis
Country: Brazil
IATA Code: LEC
ICAO Code: SBLE
Coordinates: 12°28′56″S, 41°16′37″W
Time difference and current local times
There is no time difference between Teresina and Lençóis.
-03
-03
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 102 kg (224 pounds).
Frequent Flyer Miles Calculator
Teresina (THE) → Lençóis (LEC).
Distance:
521
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
521
Round trip? | 498 | 1,784 | {"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-2020-50 | latest | en | 0.794137 |
https://www.freemathhelp.com/forum/printthread.php?t=109285&pp=10&page=1 | 1,545,133,908,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829140.81/warc/CC-MAIN-20181218102019-20181218124019-00457.warc.gz | 888,998,628 | 4,308 | # maybe a stupid question?
• 12-06-2017, 12:00 PM
derekc
maybe a stupid question?
my kid is finishing up his first semester in a college level algebra class. he currently has an 86% but his is 'white knuckling it'. he told me he is nervous about the semester test because they have gone over so much, he feels that he's not retaining it. he says he doesn't know which formula to use when approaching i guess a word problem...?
my question is this: is there a clue in any given problem as to which formula must be used?
i cannot tell you what formulas they covered in class but is it like: "when you see this, you must use the quadratic formula or the binomial theorem...etc"
thank you for the help
by the way i am not a math person. but i know math is straightforward and doesn't change. he is kind of like me, we both like geometry but algebra is like putting a needle in the eye:(
• 12-06-2017, 02:34 PM
tkhunny
Rule of Thumb: Concepts are more important than formula memorization - almost always.
Rule of Word Problem: If you don't know what something is, give it a name. Then you can talk about it.
Rule of Understanding: Why is the student not here to represent himself? We help with homework.
• 12-06-2017, 03:33 PM
JeffM
In word problems, you must FIRST translate into mathematical symbols the relevant information. Until that translation is done, you cannot possibly know what techniques or formulas apply.
• 12-07-2017, 04:46 AM
mmm4444bot
Quote:
Originally Posted by derekc
… my question is this: is there a clue in any given problem as to which formula must be used?
Sometimes. Sometimes not.
For example, if a word problem involves moving objects, and it asks for a distance, or an amount of time, or a rate of travel, then the formula Distance=Rate·Time will probably be useful.
With some exercises, the student must create a formula (i.e., an equation) because we don't have formulas (to memorize) for covering every possible scenario.
For example, a taxi meter has a drop charge of \$2.50 and adds \$0.35 per tenth mile, with \$36 per hour added for waiting time. A customer pays the driver \$17.49 total. How many miles was the trip, if the taxi waited a total of seven minutes at traffic lights and the customer gave the driver a 20% tip? We don't memorize formulas for this type of exercise; it's easier to just work it out, in steps.
Word problems in algebra almost always require writing an equation, followed by solving it. To write an equation, one must think about the given information, and determine two quantities that are equal. Writing these quantities as algebraic expressions comes from first assigning symbols to represent unknowns (especially the unknown for which the problem asks).
In my opinion, when math students lack confidence going into an exam, it's often because they have not practiced enough outside of class. With more practice comes more exposure to different types of word problems. With more exposure comes greater pattern recognition and recall.
At the University of Washington, the math department recommends studying five hours outside of class for every one hour of class attended, to be successful. Most math courses are three hours per week, in class. That means 15 hours per week studying outside of class.
I'm glad that you're taking an active interest in your child's education. With a lot of practice, things do get easier. Encourage your child to keep trying; they will know when they feel confident.
"There are no shortcuts to any place worth going." ~ Helen Keller :cool:
PS: We're also here to help, with specific exercises. If your child gets stuck, have them post the complete exercise statement (in a new thread) and show whatever work they can.
• 12-07-2017, 09:40 AM
derekc
thank you
thank you for the replies. i will direct him to this site to see what you have written and to engage with his own questions under his own username.
TKHUNNY, i will get him him here. i partially wanted to know myself because i am a pattern seeker. i wanted to see if there was a pattern to make this less daunting for us. i get hopelessly lost in algebra and i am trying not to let my confusion and frustration rub off on him.
you guys are heroes!
derek | 965 | 4,222 | {"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} | 2.6875 | 3 | CC-MAIN-2018-51 | latest | en | 0.960852 |
https://se.mathworks.com/matlabcentral/cody/problems/1070-next-higher-power-of-b/solutions/803579 | 1,610,725,824,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495901.0/warc/CC-MAIN-20210115134101-20210115164101-00053.warc.gz | 638,956,203 | 17,662 | Cody
# Problem 1070. Next Higher Power of B
Solution 803579
Submitted on 9 Jan 2016 by William
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
%% assert(nextpowb(2,126) == 7)
2 Pass
%% assert(nextpowb(3,6560) == 8)
3 Pass
%% assert(nextpowb(4,262141) == 9)
4 Pass
%% assert(nextpowb(5,21) == 2)
5 Pass
%% assert(nextpowb(6,1294) == 4)
6 Pass
%% assert(nextpowb(7,5) == 1)
7 Pass
%% assert(nextpowb(8,134217726) == 9)
8 Pass
%% assert(nextpowb(9,4782966) == 7)
9 Pass
%% assert(nextpowb(10,99993) == 5)
10 Pass
%% assert(nextpowb(11,1771559) == 6)
11 Pass
%% assert(nextpowb(12,429981693) == 8)
12 Pass
%% assert(nextpowb(13,2194) == 3)
13 Pass
%% assert(nextpowb(14,537814) == 5)
14 Pass
%% assert(nextpowb(15,2562890613) == 8)
15 Pass
%% assert(nextpowb(16,249) == 2)
16 Pass
%% assert(nextpowb(17,2015993900438) == 10)
17 Pass
%% assert(nextpowb(18,3570467226613) == 10)
18 Pass
%% assert(nextpowb(19,6131066257790) == 10)
19 Pass
%% assert(nextpowb(20,3199997) == 5)
20 Pass
%% assert(nextpowb(21,85766100) == 6)
21 Pass
%% assert(nextpowb(22,467) == 2)
22 Pass
%% assert(nextpowb(23,519) == 2)
23 Pass
%% assert(nextpowb(24,2641807540202) == 9)
24 Pass
%% assert(nextpowb(25,95367431640600) == 10)
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 562 | 1,521 | {"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} | 3.21875 | 3 | CC-MAIN-2021-04 | latest | en | 0.466479 |
http://mathhelpforum.com/algebra/145355-arithmetic-progression.html | 1,524,224,174,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937440.13/warc/CC-MAIN-20180420100911-20180420120911-00529.warc.gz | 198,653,229 | 10,593 | 1. ## arithmetic progression
how do i find common difference if the first term is 3 and the seventh term is twice the third term. bit of help.
2. Originally Posted by finnkeers
how do i find common difference if the first term is 3 and the seventh term is twice the third term. bit of help.
Hi finnkeers,
Use $\displaystyle a_n=a_1+(n-1)d$
Let $\displaystyle x=a_3$
$\displaystyle a_3=a_1+(n-1)d$
$\displaystyle x=3+(3-1)d$
$\displaystyle x=3+2d$
$\displaystyle {\color{red}d=\frac{x-3}{2}}$
Do a similar calculation using:
Let $\displaystyle a_7=2x$
$\displaystyle a_7=3+(7-1)d$
$\displaystyle 2x=3+6d$
$\displaystyle {\color{red}d=\frac{2x-3}{6}}$
Now set the two d's equal to each other and solve.
3. sorry but now i have two variables so how do i solve for d by getting rid of x
4. Originally Posted by finnkeers
sorry but now i have two variables so how do i solve for d by getting rid of x
Well, let's see.
$\displaystyle d=\frac{x-3}{2}$ and $\displaystyle d=\frac{2x-3}{6}$
Setting them equal to each other, we have
$\displaystyle \frac{x-3}{2}=\frac{2x-3}{6}$
$\displaystyle x=6$
If x = 6, then substituting into either equation with d, we get
$\displaystyle d = \frac{6-3}{2} =\boxed{\frac{3}{2}}$
$\displaystyle d = \frac{2(6)-2}{6}=\frac{9}{6}=\boxed{\frac{3}{2}}$
And there you have it!
5. first term (u1) = 3
u1 + 6d = 7th term But 7th term = 2(u1+2d)
set equations equal to one another:
3 + 6d = 2(3+2d)
3+6d = 6+4d
6d-4d = 6-3
2d= 3
Therefore, d=3/2
6. thanks everyone. much appreciated. | 541 | 1,533 | {"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} | 4.5 | 4 | CC-MAIN-2018-17 | latest | en | 0.803307 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-6-rational-expressions-and-equations-review-exercises-chapter-6-page-437/3 | 1,575,736,249,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540500637.40/warc/CC-MAIN-20191207160050-20191207184050-00163.warc.gz | 732,095,940 | 13,244 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
When $t=3$, the numerator of the given expression evaluates to zero. However, this just means the overall expression evaluates to zero since the denominator does not evaluate to zero. Thus, the expression is not undefined. | 59 | 301 | {"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} | 3.15625 | 3 | CC-MAIN-2019-51 | longest | en | 0.895797 |
https://helpful.knobs-dials.com/index.php/Data_modeling,_restructuring,_and_massaging | 1,708,983,482,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474663.47/warc/CC-MAIN-20240226194006-20240226224006-00867.warc.gz | 291,407,534 | 9,748 | # Data modeling, restructuring, and massaging
This is more for overview of my own than for teaching or exercise. Arithmetic · 'elementary mathematics' and similar concepts Set theory, Category theory Geometry and its relatives · Topology Elementary algebra - Linear algebra - Abstract algebra Calculus and analysis Logic Semi-sorted : Information theory · Number theory · Decision theory, game theory · Recreational mathematics · Dynamical systems · Unsorted or hard to sort Math on data: Statistics as a field some introduction · areas of statistics types of data · on random variables, distributions Virtues and shortcomings of... on sampling · probability glossary · references, unsorted Footnotes on various analyses Other data analysis, data summarization, learning Data modeling, restructuring, and massaging Statistical modeling · Classification, clustering, decisions, and fuzzy coding · dimensionality reduction · Optimization theory, control theory · State observers, state estimation Connectionism, neural nets · Evolutionary computing More applied: Formal grammars - regular expressions, CFGs, formal language Signal analysis, modeling, processing Image processing notes
# NLP data massage
## Bag of words / bag of features
The bag-of-words model (more broadly bag-of-features model) use the collection of words in a context, unordered, in a multiset, a.k.a. bag.
In other words, we summarize a document (or part of it) it by appearance or count of words, and ignore things like adjacency and order - so any grammar.
#### In text processing
In introductions to Naive Bayes as used for spam filtering, its naivety essentially is this assumption that feature order does not matter.
Though real-world naive bayes spam filtering would take more complex features than single words (and may re-introduce adjacenct via n-grams or such), examples often use 1-grams for simplicity - which basically is bag of words, exc.
Other types of classifiers also make this assumption, or make it easy to do so.
#### Bag of features
While the idea is best known from text, hence bag-of-words, you can argue for bag of features, applying it to anything you can count, and may be useful even when considered independently.
For example, you may follow up object detection in an image with logic like "if this photo contains a person, and a dog, and grass" because each task may be easy enough individually, and the combination tends to narrow down what kind of photo it is.
In practice, the bag-of-features often refers to models that recognize parts of a whole object (e.g. "we detected a bunch of edges of road signs" might be easier and more robust than detecting it fully), and used in a number image tasks, such as feature extraction, object/image recognition, image search, (more flexible) near-duplicate detection, and such.
The idea that you can describe an image by the collection of small things we recognize in it, and that combined presence is typically already a strong indicator (particularly when you add some hypothesis testing). Exact placement can be useful, but often easily secondary.
### N-gram notes
N-grams are contiguous sequence of length n.
They are most often seen in computational linguistics.
Applied to sequences of characters it can be useful e.g. in language identification, but the more common application is to words.
As n-grams models only include dependency information when those relations are expressed through direct proximity, they are poor language models, but useful to things working off probabilities of combinations of words, for example for statistical parsing, collocation analysis, text classification, sliding window methods (e.g. sliding window POS tagger), (statistical) machine translation, and more
For example, for the already-tokenized input This is a sentence . the 2-grams would be:
This is
is a
a sentence
sentence .
...though depending on how special you do or do not want to treat the edges, people might fake some empty tokens at the edge, or some special start/end tokens.
#### Skip-grams
This article/section is a stub — probably a pile of half-sorted notes and is probably a first version, is not well-checked, so may have incorrect bits. (Feel free to ignore, or tell me)
An extension of n-grams where components need not be consecutive (though typically stay ordered).
A k-skip-n-gram is a length-n sequence where the components occur at distance at most k from each other.
They may be used to ignore stopwords, but perhaps more often they are intended to help reduce data sparsity, under a few assumptions.
They can help discover patterns on a larger scale, simply because skipping makes you look further for the same n. (also useful for things like fuzzy hashing).
Skip-grams apparently come from speech analysis, processing phonemes.
In word-level analysis their purpose is a little different. You could say that we acknowledge the sparsity problem, and decide to get more out of the data we have (focusing on context) rather than trying to smooth.
Actually, if you go looking, skip-grams are now often equated with a fairly specific analysis. | 1,054 | 5,142 | {"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-2024-10 | longest | en | 0.842146 |
https://roman-numerals.info/1995 | 1,709,481,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476396.49/warc/CC-MAIN-20240303142747-20240303172747-00736.warc.gz | 478,414,978 | 4,011 | Roman Numerals
Roman Numerals: 1995 = MCMXCV
Convert Roman Numerals
Arabic numerals:
Roman numerals:
Arabicnumerals 0 1 M C X I 2 MM CC XX II 3 MMM CCC XXX III 4 CD XL IV 5 D L V 6 DC LX VI 7 DCC LXX VII 8 DCCC LXXX VIII 9 CM XC IX
The converter lets you go from arabic to roman numerals and vice versa. Simply type in the number you would like to convert in the field you would like to convert from, and the number in the other format will appear in the other field. Due to the limitations of the roman number system you can only convert numbers from 1 to 3999.
To easily convert between roman and arabic numerals you can use the table above. The key is to handle one arabic digit at a time, and translate it to the right roman number, where zeroes become empty. Go ahead and use the converter and observe how the table shows the solution in realtime!
Current date and time in Roman Numerals
2024-03-03 16:59:40 MMXXIV-III-III XVI:LIX:XL
Here is the current date and time written in roman numerals. Since the roman number system doesn't have a zero, the hour, minute, and second component of the timestamps sometimes become empty.
The year 1995
The year 1995 began on a Sunday and was not a leap year. Here you can read more about what happened in the year 1995.
The number 1995
The number 1995 is divisble by 3, 5, 7, 15, 19, 21, 35, 57, 95, 105, 133, 285, 399 and 665 and can be prime factorized into 3×5×7×19.
1995 as a binary number: 11111001011
1995 as an octal number: 3713
1995 as a hexadecimal number: 7CB
Numbers close to 1995
Below are the numbers 1992 through 1998, which are close to 1995. The right column shows how each roman numeral adds up to the total.
1992 = MCMXCII = 1000 + 1000 − 100 + 100 − 10 + 1 + 1 1993 = MCMXCIII = 1000 + 1000 − 100 + 100 − 10 + 1 + 1 + 1 1994 = MCMXCIV = 1000 + 1000 − 100 + 100 − 10 + 5 − 1 1995 = MCMXCV = 1000 + 1000 − 100 + 100 − 10 + 5 1996 = MCMXCVI = 1000 + 1000 − 100 + 100 − 10 + 5 + 1 1997 = MCMXCVII = 1000 + 1000 − 100 + 100 − 10 + 5 + 1 + 1 1998 = MCMXCVIII = 1000 + 1000 − 100 + 100 − 10 + 5 + 1 + 1 + 1 | 712 | 2,082 | {"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} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.791653 |
https://staging4.aicorespot.io/dual-annealing-optimization-with-python/ | 1,695,349,763,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00346.warc.gz | 598,128,220 | 37,327 | >Business >Dual Annealing Optimization with Python
### Dual Annealing Optimization with Python
Dual Annealing is a stochastic global optimization algorithm.
It is an implementation of the generalized simulated annealing algorithm, an extension of simulated annealing. Also, it is coupled with a local search algorithm that is automatically carried out at the end of the simulated annealing procedure.
This combo of efficient global and local search procedures furnishes a potent algorithm for challenging nonlinear issues.
In this guide, you will find out about the dual annealing global optimization algorithm.
After going through this guide, you will be aware of:
• Dual annealing optimization being a global optimization that is an altered version of simulated annealing that also leverages a local search algorithm.
• How to leverage the dual annealing optimization algorithm API within Python.
• Instances of leveraging dual annealing to find solutions to global optimization problems with several optima.
Tutorial Summarization
The tutorial is subdivided into three portions, which are:
1] What is dual annealing
2] Dual annealing API
3] Dual annealing instance
What is Dual Annealing
Dual annealing is a global optimization algorithm.
As such, it is developed for objective functions that possess a nonlinear response surface. It is a stochastic optimization algorithm, implying that it leverages randomness in the search procedure and every run of the search might identify a differing solution.
Dual Annealing is based on the Simulated Annealing optimization algorithm.
Simulated annealing is a variant of stochastic hill climbing where a candidate solution is altered in an arbitrary way and the altered solutions are accepted to substitute the present candidate solution probabilistically. This implies that it is doable for worse solutions to substitute the present candidate solution. The probability of this variant of replacement is high at the start of the search and reduces with every iteration, managed by the “temperature” hyperparameter.
Dual annealing is an implementation of the classical simulated annealing (CSA) algorithm. It is based on the generalized simulated annealing (GSA) algorithm detailed in the 1997 paper “Generalized Simulated Annealing Algorithm and its Application to the Thomson Model.”
It brings together the annealing schedule (rate at which the temperature reduces over algorithm iterations) from “fast simulated annealing” (FSA) and the probabilistic acceptance of an alternate statistical process “Tsallis statistics” named after the author.
Experimental outcomes identify that this generalized simulated annealing algorithm seems to feature better performance in a better fashion than the classical or the quick versions of the algorithms to which it was contrasted.
GSA not just converges quicker than FSA and CSA, but also has the capability to escape from a local minimum more easily than FSA and CSA.
On top of these modifications of simulated annealing, a local search algorithm can be applied to the solution identified by the simulated annealing search.
This is desired as global search algorithms are typically good at situating the basin (area in the search space) for the optimal solution but are typically poor at identifying the most optimal solution in the basin. While local search algorithms excel at identifying the optima of a basin.
Coupling a local search with the simulated annealing procedure makes sure the search obtains the most out of the candidate solution that is located.
Now that we are acquainted with the dual annealing algorithm from a high level, let’s delve into the API for dual annealing in Python.
Dual Annealing API
The dual annealing global optimization algorithm is available in Python through the dual_annealing() SciPy function.
The function gets the name of the objective function and the bounds of every input variable as minimum arguments for the search.
[Control]
1 2 3 … # perform the dual annealing search result = dual_annealing(objective, bounds)
There are a plethora of extra hyperparameters for the search that possess default values, even though you can set them up to customize the search.
The “maxiter” argument mentions the cumulative number of iterations of the algorithm (not the cumulative number of function evaluations) and defaults to 1,000 iterations. The “maxfun” can be mentioned if desired to restrict the cumulative number of function evaluations and defaults to 10 million.
The initial temperature of the search is mentioned by the “initial_temp” argument, which defaults to 5,230. The annealing procedure will begin again after the temperature attains a value equivalent or less than (initial_temp * restart_temp_ratio). The ratio defaults to a very small number 2e-05 (i.e., 0.00002), so the default trigger for re-annealing is a temp of (5230 * 0.00002 or 0.1046.
The algorithm also furnishes control over hyperparameters particular to the generalized simulated annealing on which it has its basis. This consists of how far jumps can be made during the search through the “visit” argument, which defaults to 2.62. (values less than 3 are preferred), and the “accept” argument that controls the likelihood of accepting new solutions, which defaults to -5.
The minimize() function is called for the local search with default hyperparameters. The local search can be setup by furnishing a dictionary of hyperparameter names and values to the “local_search_options” argument.
The local search aspect of the search can be disabled by setting the “no_local_search” argument to True.
The outcome of the search is an OptimizeResult object where attributes can be accessed like a dictionary. The success (or not) of the search can be accessed through the “success” or ‘message’ key.
The cumulative number of function evaluations can be accessed through “nfev” and the optimal input identified for the search is accessible through the ‘x’ key.
Now that we are acquainted with the dual annealing API in Python, let’s look at a few worked instances.
Dual Annealing Instance
In this portion of the blog, we will look at an instance of leveraging the dual annealing algorithm on a multi-modal objective function.
The Ackley function is an instance of a multimodal objective function that has a singular global optima and several local optima in which a local search might get stuck.
As such, a global optimization strategy is needed. It is a two-dimensional objective function that possess a global optima at [0,0] which evaluates to 0.0.
The instance below implements the Ackley and develops a 3D surface plot demonstrating the global optima and several local optima.
[Control]
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 # ackley multimodal function from numpy import arange from numpy import exp from numpy import sqrt from numpy import cos from numpy import e from numpy import pi from numpy import meshgrid from matplotlib import pyplot from mpl_toolkits.mplot3d import Axes3D # objective function def objective(x, y): return -20.0 * exp(-0.2 * sqrt(0.5 * (x**2 + y**2))) – exp(0.5 * (cos(2 * pi * x) + cos(2 * pi * y))) + e + 20 # define range for input r_min, r_max = -5.0, 5.0 # sample input range uniformly at 0.1 increments xaxis = arange(r_min, r_max, 0.1) yaxis = arange(r_min, r_max, 0.1) # create a mesh from the axis x, y = meshgrid(xaxis, yaxis) # compute targets results = objective(x, y) # create a surface plot with the jet color scheme figure = pyplot.figure() axis = figure.gca(projection=’3d’) axis.plot_surface(x, y, results, cmap=’jet’) # show the plot pyplot.show()
Running the instance creates the surface plot of the Ackley function displaying the massive number of local optima.
We can go about applying the dual annealing algorithm to the Ackley objective function.
To start with, we can define the bounds of the search space as the limits of the function in every dimension.
123 …# define the bounds on the searchbounds = [[r_min, r_max], [r_min, r_max]]
We can then go about applying the search by mentioning the name of the objective function and the bounds of the search.
In this scenario, we will leverage the default hyperparameters.
123 …# perform the simulated annealing searchresult = dual_annealing(objective, bounds)
After the search is finished, it will report the status of the search and the number of iterations carried out, in addition to the outcome identified with its evaluation.
12345678 …# summarize the resultprint(‘Status : %s’ % result[‘message’])print(‘Total Evaluations: %d’ % result[‘nfev’])# evaluate solutionsolution = result[‘x’]evaluation = objective(solution)print(‘Solution: f(%s) = %.5f’ % (solution, evaluation))
Connecting these together, the full example, of applying dual annealing to the Ackley objective function is detailed here.
123456789101112131415161718192021222324252627 # dual annealing global optimization for the ackley multimodal objective functionfrom scipy.optimize import dual_annealingfrom numpy.random import randfrom numpy import expfrom numpy import sqrtfrom numpy import cosfrom numpy import efrom numpy import pi # objective functiondef objective(v):x, y = vreturn -20.0 * exp(-0.2 * sqrt(0.5 * (x**2 + y**2))) – exp(0.5 * (cos(2 * pi * x) + cos(2 * pi * y))) + e + 20 # define range for inputr_min, r_max = -5.0, 5.0# define the bounds on the searchbounds = [[r_min, r_max], [r_min, r_max]]# perform the dual annealing searchresult = dual_annealing(objective, bounds)# summarize the resultprint(‘Status : %s’ % result[‘message’])print(‘Total Evaluations: %d’ % result[‘nfev’])# evaluate solutionsolution = result[‘x’]evaluation = objective(solution)print(‘Solution: f(%s) = %.5f’ % (solution, evaluation))
Running the instance executes the optimization, then reports the results.
Your outcomes may demonstrate variance provided the stochastic nature of the algorithm or evaluation procedure, or variations in numerical accuracy. Take up running the instance a few times and contrast the average outcome.
In this scenario, we can observe that the algorithm situated the optima with inputs very near to zero and an objective function evaluation that is virtually zero.
We can observe that a total of 4,136 function evaluations were carried out.
123 Status : [‘Maximum number of iteration reached’]Total Evaluations: 4136Solution: f([-2.26474440e-09 -8.28465933e-09]) = 0.00000
This section furnishes more resources on the subject if you are seeking to delve deeper.
Papers
Generalized Simulated Annealing Algorithm and its Application on the Thomson Model, 1997
Generalized Simulated Annealing, in 1996.
Generalized Simulated Annealing for Global Optimization: TheGenSA Package, 2013
APIs
scipy.optimize.dual_annealing API
scipy.optimize.OptimizeResult API
Articles
Global optimization, Wikipedia
Simulated annealing, Wikipedia
Ackley Function, Wikipedia
Conclusion
In this guide, you found out about the dual annealing global optimization algorithm.
Particularly, you learned:
• Dual annealing optimization is a global optimization that is an altered variant of simulated annealing that also leverages a local search algorithm.
• How to leverage the dual annealing optimization algorithm API within Python.
• Instances of leveraging dual annealing to find solutions to global optimization problems with several optima. | 2,640 | 11,438 | {"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-2023-40 | longest | en | 0.890651 |
https://loialan.wordpress.com/2015/06/18/4-season/ | 1,500,582,896,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423486.26/warc/CC-MAIN-20170720201925-20170720221925-00154.warc.gz | 681,256,920 | 46,367 | # 4 season
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To exclude the and from you, you have to understand the proclaw to each and every unit before and after it dies off which also means the more you do it the less you will have to point into each and every position of time, to conclude the manuscript of ambition, and to de-straight the percentages of value to stole together and to lighten the path this also means the more and the less you will consider your position to 24for35eightyeight and 26 by forming into a deformation. Which also exaggerate into a section of 26for35and to conclude detention which is “OUT OF CONTROL” – By this you have to profound each and every unit to simplify the issues of matters and to profound each and every network once you ambush yourself inner-self without knowing the awareness of time. Which also means the more you tender the meat the less power you will have, if you do this you can’t reward yourself into a position of time to exclude a degradation.
To episode 26andfor25of31for29andeach way it goes it will fell the pain of sorrow but if you do this you can’t mark up your language to and from a unit to section 39 before and after it does not die off you know why? In this realtion it will build a certain amount of transactions before and after it conclude the material to distinguish yourself before and after it conclude the inner PEACE!
## 3 thoughts on “4 season”
1. In this rotations you have to understand the needed once they comply with maturity only once as each units will rectify itself into a pathshion which also exaggerate into a certain amount of income. Which also means the m ore or the less you have you will rejoin yourself into a path of finder which also means the more and the less you have you will NOT degree c6by2 1 which also means t12and d4 in reversity of time to consider the percentages of value and to convert into a position of 29for36 by and given to an element of NATURE!
To do this you can’t comply with your mark up language to defense yourself into the element of design this also means the array of 296andfor728 out of control which also means the less you do it the more you will get there right? – Simple but not really in this point of investment you have to defra a certain amount of income before and after it may dies off which also exclude the and from you, which is out of control by this you have to know instead of understanding the fundamental awareness of each and every network into section 55by44, and to from each unify the element of design to conclude the animission. By each section of 24for35 an to exclude the degradation of 5to5 by 238andforeachone which is NOT a return to and from each way you go you have to comply within your heart before and after a path way to find a clue – Next time everyone 😉
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• Season 4, winter, around a and each maths which also mean the more the less you will have to conclude a material which will reverse out of order and to conclude each and every way you go. Which also means the less the more the more the lesson you have psychologically (later). To profound a clue you have to understand the needed once you comply with maturity you can’t learn the maths which also means the more you have the less power you will assume to invest into the stock marker before and after it CANNOT turn into a section 66by44 which means NOT you but in rotation to time to sister your brother and to find a clue of YOU!
To conclude this you have to understand the needed once you comply with yourself you will NOT exclude the degree of SISTER before it dies off which also means the more you have the less you will get to DIE! – Which is NOT a conclusion it is a form of each way it goes and then therefore it becomes your ambition, which also degrade into your pollution of time to rewind yourself into pathology.
To export the sentence you have to fine a claw to learn and to point into each position of 24for38by39 which is section 2 2 for 9 4 and 5 7 which sector itself into a certain amount of income to infra red itself into each section which also exaggerate into point of INVESTMENT into section 5 ONLY 😉
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2. This was from my OLD CAM!
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https://byjusexamprep.com/gate-2024-subject-name-foundation-quiz-65-i-18dfb140-ee79-11ed-8124-ac866922202e | 1,708,488,653,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473370.18/warc/CC-MAIN-20240221034447-20240221064447-00039.warc.gz | 162,001,207 | 48,186 | Time Left - 15:00 mins
# GATE 2024 Subject Name Foundation Quiz 65
Attempt now to get your rank among 63 students!
Question 1
The current gain of a bipolar transistor drops at high frequencies because of
Question 2
To get higher cut-off frequency in a BJT, base sheet resistance should be.
Question 3
For CE BJT amplifier, high frequency response is limited by
Question 4
In a BJT circuit, capacitor CBE will result in:
Question 5
A common emitter BJT amplifier is operating at high frequency, it π- circuit model as shown below-
If the transistor current Ic= 1.2mA, Cm = 2pf & hfe= 20 at 40 MHz
Then find out the frequency at which gain of amplifier is unity. (VT = 25mV & η = 1)
• 63 attempts | 192 | 709 | {"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} | 2.578125 | 3 | CC-MAIN-2024-10 | latest | en | 0.878465 |
https://math.stackexchange.com/questions/4164048/is-a-continuous-function-necessarily-the-pointwise-limit-of-a-sequence-of-unifor | 1,716,085,107,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00886.warc.gz | 346,230,048 | 35,227 | # Is a continuous function necessarily the pointwise limit of a sequence of uniformly continuous functions?
The pointwise limit of a sequence of uniformly continuous functions needn't be continuous. I've been wondering about the converse:
Is a continuous function $$f$$ necessarily the pointwise limit of some sequence of uniformly continuous functions $$(f_n)_{n\in\mathbb{N}}$$?
I'm having trouble establishing or disproving this.
If the domain of $$f$$ is $$\Bbb R$$, then the answer is yes. The proof is quite easy: let $$f_n: \Bbb R \to \Bbb R$$ defined by $$f_n(x)= \begin{cases} f(-n) & x<-n \\ f(x) & -n \le x \le n \\ f(n) & x>n \end{cases}$$ these are uniformly continuous and they pointwise converge to $$f$$.
• Why are the $f_n$ uniformly continuous (in $-n\leq x\leq n$)? We only know that $f$ is continuous. Jun 5, 2021 at 15:01
• That's because on a compact set every continuous function is uniformly continuous. If you extend as a constant function outside the boundary of $[-n,n]$ it remains uniformly continuous. Jun 5, 2021 at 15:07 | 289 | 1,055 | {"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": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-22 | latest | en | 0.902933 |
https://www.careerride.com/mcq-tag-wise.aspx?Key=Properties%20of%20Fluids&Id=16 | 1,679,994,596,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948817.15/warc/CC-MAIN-20230328073515-20230328103515-00771.warc.gz | 786,448,167 | 8,892 | # Properties of Fluids - Mechanical Engineering (MCQ) questions and answers
1) According to Archimede's principle, if a body is immersed partially or fully in a fluid then the buoyancy force is _______ the weight of fluid displaced by the body.
a. equal to
b. less than
c. more than
d. unpredictable
Answer Explanation ANSWER: equal to Explanation: No explanation is available for this question!
2) The specific weight of the fluid depends upon
a. gravitational acceleration
b. mass density of the fluid
c. both a. and b.
d. none of the above
Answer Explanation ANSWER: both a. and b. Explanation: No explanation is available for this question!
3) One litre of a certain fluid weighs 8N. What is its specific volume?
a. 2.03 x 10– 3 m3/kg
b. 20.3 x 10– 3 m3/kg
c. 12.3 x 10– 3 m3/kg
d. 1.23 x 10– 3 m3/kg
Answer Explanation ANSWER: 1.23 x 10– 3 m3/kg Explanation: No explanation is available for this question!
4) What is the correct formula for absolute pressure?
a. Pabs = Patm – Pgauge
b. Pabs = Pvacuum – Patm
c. Pabs = Pvacuum + Patm
d. Pabs = Patm+ Pgauge
Answer Explanation ANSWER: Pabs = Patm+ Pgauge Explanation: No explanation is available for this question!
5) When is a liquid said to be not in a boiling or vaporized state?
a. If the pressure on liquid is equal to its vapour pressure
b. If the pressure on liquid is less than its vapour pressure
c. If the pressure on liquid is more than its vapour pressure
d. Unpredictable
Answer Explanation ANSWER: If the pressure on liquid is more than its vapour pressure Explanation: No explanation is available for this question!
6) Bulk modulus is the ratio of
a. shear stress to volumetric strain
b. volumetric strain to shear stress
c. compressive stress to volumetric strain
d. volumetric strain to compressive stress
Answer Explanation ANSWER: compressive stress to volumetric strain Explanation: No explanation is available for this question!
7) When the angle between surface tension with the liquid (θ) is greater than 90o, the liquid becomes
a. flat
b. concave upward
c. convex upward
d. unpredictable
Answer Explanation ANSWER: convex upward Explanation: No explanation is available for this question!
8) The fluid will rise in capillary when the capillary is placed in fluid, if
a. the adhesion force between molecules of fluid and tube is less than the cohesion between liquid molecules
b. the adhesion force between molecules of fluid and tube is more than the cohesion between liquid molecules
c. the adhesion force between molecules of fluid and tube is equal to the cohesion between liquid molecules
d. cannot say
Answer Explanation ANSWER: the adhesion force between molecules of fluid and tube is more than the cohesion between liquid molecules Explanation: No explanation is available for this question!
9) The below diagram is a graph of change in shear stress with respect to velocity gradient in a fluid. What is a type of the fluid?
a. Newtonian fluid
b. Non-Newtonian fluid
c. Ideal fluid
d. Dilatent fluid
Answer Explanation ANSWER: Non-Newtonian fluid Explanation: No explanation is available for this question!
10) What is an ideal fluid?
a. A fluid which has no viscosity
b. A fluid which is incompressible
c. A fluid which has no surface tension
d. All of the above
Answer Explanation ANSWER: All of the above Explanation: No explanation is available for this question!
1 2 | 861 | 3,418 | {"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} | 2.78125 | 3 | CC-MAIN-2023-14 | latest | en | 0.761384 |
https://www.zentralblatt-math.org/matheduc/en/?id=2755&type=txt | 1,560,827,211,000,000,000 | text/plain | crawl-data/CC-MAIN-2019-26/segments/1560627998605.33/warc/CC-MAIN-20190618023245-20190618045245-00089.warc.gz | 950,760,565 | 1,256 | id: 06492944 dt: j an: 2016e.00657 au: Murty, Gurajada Suryanarayana ti: A note on theorem of unequal pair of lunes. so: J. Indian Acad. Math. 37, No. 1, 13-18 (2015). py: 2015 pu: Indian Academy of Mathematics, Indore, Madhya Pradesh, India la: EN cc: G40 G30 ut: area; lune; mathematical modeling ci: li: ab: The paper proves the following result: Consider a circle and a right triangle \$ABC\$ such that \$AB\$ is a diameter of the circle. Draw semicircles with diameters \$AC\$ and \$BC\$. Each of this semicircles, together with the original circle determines a lune. The sum of the areas of these lunes equals the area of the triangle \$ABC\$. The result is elementary and not very surprising, Nevertheless it can be a nice excercise for secondary school students and it is suitable to be worked out with GeoGebra. rv: Antonio M. Oller (Zaragoza) | 242 | 852 | {"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} | 3.140625 | 3 | CC-MAIN-2019-26 | latest | en | 0.769336 |
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