url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3
values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93
values | snapshot_type stringclasses 2
values | language stringclasses 1
value | language_score float64 0.06 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://m.ebrary.net/33754/health/longitudinal_shape_data_analysis | 1,623,911,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487629209.28/warc/CC-MAIN-20210617041347-20210617071347-00344.warc.gz | 343,007,432 | 5,076 | Home Health
Longitudinal Shape Data Analysis
Longitudinal shape data analysis in CA analyzes the spatiotemporal variability of anatomic shapes to reveal dynamic development patterns of organs across diseases, disease course, ages, genders, etc. Its key task is to provide a model of how one individual’s trajectory changes relative to those of other subjects.
Given a longitudinal data set for a collection of N objects S = {S0, Si,..., SN-1} observed at different time slots. The observed data set for Sn includes Tn observations at time slots {t0, tn,...,tnn} as In = {In, In, ..,1% }. Longitudinal data analysis should answer the following questions:
• • How is the trajectory of each object evolution estimated and how is its shape predicted at any time t from limited observations?
• • How are trajectories of different objects compared?
• • How can a statistical atlas of evolutionary trajectories be constructed?
Trajectory Estimation
Trajectory estimation involves finding a continuous (smooth) curve in the spatiotemporal shape space that best fits the observed discrete time data of a single object. In diffeomorphism-based CA, existing solutions include:
Ut° ti Tn
ПП, If,..., If } at Tn
time slots, the evolutionary trajectory can be computed as a piecewise geodesic (Riemannian geodesic or group geodesic for LDDMM and SVF, respectively) that connects successive observed data, which can be computed by diffeomorphic registration between successive observations by LDDMM and SVF. If there are only two observations at t0 and tn, the trajectory is just the geodesic connecting them. Also, such a trajectory holds a correspondent (piecewise) tangent space representation, which might be explored for the trajectory comparison and statistical analysis. The disadvantage is that such a greedy method may over-fit the observed data and lose the global smoothness of the trajectory [93,108,109]. • Trajectory regression or interpolation Another solution is to fit a smooth time- dependent curve simultaneously to all the observed data by kernel regression or the first- or second-order interpolations as in [86, 110]. Such methods result in smooth trajectories, but at a cost of losing the tangent space trajectory representation.
Related topics | 475 | 2,263 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2021-25 | latest | en | 0.87813 |
https://gmatclub.com/forum/probability-die-95391.html?sort_by_oldest=true | 1,498,678,121,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323730.30/warc/CC-MAIN-20170628185804-20170628205804-00352.warc.gz | 761,319,200 | 48,548 | It is currently 28 Jun 2017, 12:28
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Probability Die
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 05 Oct 2008
Posts: 273
Probability Die [#permalink]
### Show Tags
07 Jun 2010, 00:10
If a fair coin marked 1 and 2, and a fair die are rolled together, what is a probability to have the sum even?
(C) 2008 GMAT Club - m05#34
* $$\frac{1}{8}$$
* $$\frac{1}{4}$$
* $$\frac{1}{2}$$
* $$\frac{3}{4}$$
* $$\frac{7}{8}$$
The coin has 1 and 2, and the die has 1, 2, 3, 4, 5, 6 for a total of 12 possible outcomes. (aren't the total number of outcomes - 6*6 = 36?)
To get the final result even, we must have two even numbers, $$E$$ , or two odd numbers, $$O$$ .
$$E + E$$ is possible if we have $$2+2$$ , $$2+4$$ , $$2+6$$ .
what about 4+2, 4+4, 4+6, 6+2, 6+4, 6+6, ?? These total to even numbers as well?
$$O + O$$ is possible if we have $$1+1$$ , $$1+3$$ , $$1+5$$ , so there are six favorable outcomes out of 12 possible. $$\frac{6}{12}=\frac{1}{2}$$ .
Likewise, what about 3+1, 3+3, 3+5, 5+1, 5+3, 5+5??
This explanation does not take all the outcomes into consideration. Can someone explain how to get to the correct answer.
Thanks.
The correct answer is C.
Manager
Joined: 20 Apr 2010
Posts: 153
Location: I N D I A
Re: Probability Die [#permalink]
### Show Tags
07 Jun 2010, 01:16
We can get the Sum Even in 2 cases which are : E + E or O + O
Probability of getting both Even : P( of getting 2 on coin ) * P ( of getting Even on die i.e. 2,4,6 out of total 6 possibilities and not 36 )
= 1/2 * 1/2 = 1/4
Probability of getting both ODD : P( of getting 1 on coin ) * P ( of getting Odd i.e 1,3,5 on die )
= 1/2 * 1/2 = 1/4
Total prob = 1/4 + 1/4 = 1/2
Hope its clear...
Manager
Joined: 20 Apr 2010
Posts: 153
Location: I N D I A
Re: Probability Die [#permalink]
### Show Tags
07 Jun 2010, 01:21
E + E is possible if we have 2+2 , 2+4 , 2+6 .
what about 4+2, 4+4, 4+6, 6+2, 6+4, 6+6, ?? These total to even numbers as well?
The values highlighted r never possible as u r having one coin which can yield a value of 1 or 2 only....
Re: Probability Die [#permalink] 07 Jun 2010, 01:21
Similar topics Replies Last post
Similar
Topics:
Probability Question 3 06 Sep 2010, 22:56
1 Probability 2 22 Jan 2010, 10:53
4 Probability (m02q06) 21 07 Feb 2014, 07:41
5 Probability (m01q04) 30 04 Dec 2013, 11:44
1 Event A and Event B are independent. Is the probability that 6 07 Nov 2012, 18:57
Display posts from previous: Sort by
# Probability Die
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Moderator: Bunuel
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,112 | 3,434 | {"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} | 4.4375 | 4 | CC-MAIN-2017-26 | latest | en | 0.843941 |
https://math-master.org/general/132133333-33 | 1,716,598,630,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00224.warc.gz | 328,898,279 | 53,777 | Question
# 132133333-33
115
likes
574 views
## Answer to a math question 132133333-33
Esmeralda
4.7
$\frac{\begin{matrix}\:\:&1&3&2&1&3&3&3&3&\bold{3}\\-&0&0&0&0&0&0&0&3&\bold{3}\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\bold{0}\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&3&2&1&3&3&3&\bold{3}&3\\-&0&0&0&0&0&0&0&\bold{3}&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\bold{0}&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&3&2&1&3&3&\bold{3}&3&3\\-&0&0&0&0&0&0&\bold{0}&3&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\bold{3}&0&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&3&2&1&3&\bold{3}&3&3&3\\-&0&0&0&0&0&\bold{0}&0&3&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\:\:&\bold{3}&3&0&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&3&2&1&\bold{3}&3&3&3&3\\-&0&0&0&0&\bold{0}&0&0&3&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\:\:&\bold{3}&3&3&0&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&3&2&\bold{1}&3&3&3&3&3\\-&0&0&0&\bold{0}&0&0&0&3&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\:\:&\bold{1}&3&3&3&0&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&3&\bold{2}&1&3&3&3&3&3\\-&0&0&\bold{0}&0&0&0&0&3&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\:\:&\bold{2}&1&3&3&3&0&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&1&\bold{3}&2&1&3&3&3&3&3\\-&0&\bold{0}&0&0&0&0&0&3&3\end{matrix}}{\begin{matrix}\:\:&\:\:&\bold{3}&2&1&3&3&3&0&0\end{matrix}}$
$\frac{\begin{matrix}\:\:&\bold{1}&3&2&1&3&3&3&3&3\\-&\bold{0}&0&0&0&0&0&0&3&3\end{matrix}}{\begin{matrix}\:\:&\bold{1}&3&2&1&3&3&3&0&0\end{matrix}}$
$\begin{matrix}\:\:&1&3&2&1&3&3&3&3&3\\-&0&0&0&0&0&0&0&3&3\end{matrix}$ $=132133300$
Frequently asked questions $FAQs$
What is the variance of a data set: {7, 9, 10, 12, 15}?
+
What is the surface area of a sphere with radius r and volume V?
+
What is the equation of an ellipse with a major axis of length 8, center at $-2,3$, and minor axis of length 6?
+ | 1,014 | 1,921 | {"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": 2, "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-2024-22 | longest | en | 0.168327 |
http://mathoverflow.net/feeds/question/17794 | 1,369,421,149,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368704943681/warc/CC-MAIN-20130516114903-00022-ip-10-60-113-184.ec2.internal.warc.gz | 168,237,605 | 2,156 | Rellich-Necas identity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:45:50Z http://mathoverflow.net/feeds/question/17794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17794/rellich-necas-identity Rellich-Necas identity PDE 2010-03-11T01:31:18Z 2010-03-19T00:26:46Z <p>I am looking for a book/paper which has the proof of the Rellich-Nicas identity.</p> <p><b>[EDIT by Yemon Choi]</b> It seems that what was meant is "the Rellich-Necas identity", although the original poster hasn't really clarified or expanded on the request.</p> http://mathoverflow.net/questions/17794/rellich-necas-identity/17838#17838 Answer by Ady for Rellich-Necas identity Ady 2010-03-11T08:37:26Z 2010-03-11T08:37:26Z <p>My guessing is that the question is about the so-called <strong>Rellich-Necas</strong> identity, named after the late Czech mathematician Jindrich NECAS. See e.g. [C24], [C14], and [C11] in <a href="http://dml.cz/bitstream/handle/10338.dmlcz/134050/MathBohem_129-2004-4_8.pdf" rel="nofollow">http://dml.cz/bitstream/handle/10338.dmlcz/134050/MathBohem_129-2004-4_8.pdf</a>.</p> http://mathoverflow.net/questions/17794/rellich-necas-identity/17863#17863 Answer by Willie Wong for Rellich-Necas identity Willie Wong 2010-03-11T14:05:23Z 2010-03-11T14:05:23Z <p>(Boo! I tried to post this in a comment to Ady, but the HTML Math won't parse right. So here goes. Sorry about the really long equation being broken up not very neatly.)</p> <p>Googling Rellich-Necas turns up a bunch of recent papers by LUIS ESCAURIAZA in which the identities are used. But as far as I can tell the identity is just a simple differential equality obtained from symbolic manipulation of terms. The following seems to be a straight-forward version of the identity: let $A = (A_{ij})$ be a symmetric bilinear form (with variable coefficients) on R<sup>N</sup>, $v$ a vector field, $u$ a function, and $\delta$ denoting the Euclidean divergence, we have</p> <p>$\delta( A(\nabla u,\nabla u) v) = 2 \delta( v(u) A(\nabla u)) + \delta(v) A(\nabla u,\nabla u)$ $- 2A(\nabla u) \cdot \nabla v \cdot \nabla u - 2 v(u) \delta(A(\nabla u)) + v(A)(\nabla u,\nabla u)$</p> <p>Where $v(u)$ is the partial derivative of $u$ in the direction of $v$, and $A(\nabla u)\cdot\nabla v \cdot \nabla u$ is, in coordinates, $\partial_i u A_{ij} \partial_j v_k \partial_k u$ with implied summation, and $v(A)$ is the symmetric bilinear form obtained by taking the $v$ partial derivative of the coefficients of $A$. </p> <p>Verifying that the identity is true should just be a basic application of multivariable calculus. </p> | 858 | 2,641 | {"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} | 2.734375 | 3 | CC-MAIN-2013-20 | latest | en | 0.775744 |
https://printableruleractualsize.com/show-122-mm-on-a-printable-ruler/mark-the-ruler-in-metric/ | 1,620,494,569,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00288.warc.gz | 496,274,986 | 21,283 | # Mark The Ruler In Metric
Mark The Ruler In Metric
Show 122 Mm On A Printable Ruler – A Milimeter Ruler is really a device that accurately steps particular dimensions and may be utilized to measure peak, width, depth, as well as the circumference of the real item. You’d probably most frequently see a millimeter ruler utilized for tests or measuring the size of the issue after which for measuring the proper measurements. It really is a tool that has a circular top that has a graduated scale on both sides.
Once you have taken the goal measurement from your initial degree towards the fourth degree, the very best in the unit will move towards the right. The instrument is created up of two circular steel bases that allow the unit to create its measurement.
The instrument generally takes an “open”closed” approach. A closed method is one in which the tool moves in the left to the correct while an open approach is one in which the instrument moves from the proper to the remaining. If you’ve ever tried to make use of one, you know that it is easier to utilize one that moves from your right towards the remaining. However, you will find some drawbacks to using the open up technique of measurement.
There are different ways of measuring things in non-metric countries like the United states of america, therefore if you’re making use of a tool to measure the specific measurement, it could trigger issues. Furthermore, the type of measurement that you’re looking for might not be easily calculated through the traditional approach that the majority individuals are used to.
There are distinct specifications like the US yard, metric foot, metre and imperial foot. If you are undecided of what measurement you happen to be seeking for, you’ll be able to either call a professional and get their assist or use the on the internet sources to search for what you’re searching for.
The instrument is quite simple to use since it will come with a one inch ruler. The different measurements are extremely small and straightforward to measure so you’re able to consider your measurements with one inch rulers. One thing that lots of individuals never realize is the fact that this technique could be quite correct when you know the way to utilize it.
It is easier to utilize than a conventional measurement method however it is not for everyone. In fact, this is one technique that you ought to use cautiously in order to avoid making blunders.
If you’re seeking for any precision and exact measurements, there’s no other better way to get them. If you are questioning about its precision, keep in mind that it is a fast and efficient way to get the exact measurement. | 526 | 2,677 | {"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} | 2.609375 | 3 | CC-MAIN-2021-21 | latest | en | 0.953782 |
https://solvedlib.com/1-glitter-glow-machine-company-is-a-nigerian,394649 | 1,696,115,909,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00563.warc.gz | 565,398,728 | 17,608 | # 1. Glitter Glow Machine Company is a Nigerian company that manufactures washing machines located in Lagos and Enugu, in Nigeria.
###### Question:
1. Glitter Glow Machine Company is a Nigerian company that manufactures washing machines located in Lagos and Enugu, in Nigeria. These are shipped to regional distribution centers in Kaduna, and Benin, where they are delivered to the supply houses in Sokoto Sapele and Sagamu, as shown in the Figure below. Destination Transshipment Point Source Supply Sokoto Node 5 450 Lagos Node 1 800 Kaduna Node 3 Node 6 350 Enugu Node 2 Benin 700 Node 4 Shagamu Node 7 300 The available supplies at the factories, the demands at the final destination, and shipping costs are shown in the table below: out view Sec 1 Pages: 1 of 3 Words: 72 of 342 L9
The available supplies at the factories, the demands at the final destination, and shipping costs are shown in the table below: To From Kaduna Benin Sokoto Sapele Shagamu Supply Lagos Enugu Kaduna Benin Demand. 800 $7$4 $5 SA$6 300 450 350 Glitter Glow would like to minimize the transportation costs associated with shipping enough washing machines to meet the demands at the three destinations while not exceeding the supply at each factory. Formulate this problem as a Transshipment Problem and solve using MS Excel Solver.
#### Similar Solved Questions
##### 31 If f() =x', USE A DEFINITION OF DERIVATIVE to find the formula for instanteous rate of {change in terms of x
31 If f() =x', USE A DEFINITION OF DERIVATIVE to find the formula for instanteous rate of {change in terms of x...
##### How do you find the general solutions for Tan x - 3cotx = 0?
How do you find the general solutions for Tan x - 3cotx = 0?...
##### 20.0/1 pointsPrevious AnswersTANAPMATHZ 2.2.038Find an equation of the line that satisfies the given condition. The line passing through (a, b) with undefined slope
20. 0/1 points Previous Answers TANAPMATHZ 2.2.038 Find an equation of the line that satisfies the given condition. The line passing through (a, b) with undefined slope...
##### Question 42 ptsA person has undertaken a lawn care job. The probability of completing the job on time with rain is .42 and without rain is 9. If the probability of rain is .45, then determine the probability that the lawn care job will be completed_68487931.32None of these
Question 4 2 pts A person has undertaken a lawn care job. The probability of completing the job on time with rain is .42 and without rain is 9. If the probability of rain is .45, then determine the probability that the lawn care job will be completed_ 684 87 93 1.32 None of these...
##### Assignment-2.pdf (page 1 of 2) Q Search SYC 2317 Mork W. Tengler, MS Assignment 2 An...
Assignment-2.pdf (page 1 of 2) Q Search SYC 2317 Mork W. Tengler, MS Assignment 2 An instructor obtained the following set of sores from a 10-point quiz for a class of 26 students: 2.1 8, 5, 4, 5, 5, 7, 6, 4, 3, 4, 5, 6, 6, 4, 5, 5, 10, 6, 9, 5, 7, 8, 2, 6, 7, 4 Place the scores in a frequency dist...
##### A patient was diagnosed with a cancerous tumor in her mediastinum. the physician excised a 3.0...
A patient was diagnosed with a cancerous tumor in her mediastinum. the physician excised a 3.0 cm tumor. provide CPT code...
##### Woln pevment of S400 Wnat percentonthe cost52500?X7 6 To 62S6.During = flu epidemic; 15S2of & schools' 480 students were absent on particular duy: How many students were absent?X= 72+80schoo hascollectedS1400 during = fundraiset; which 4098 ofits goal What is the total amount of the schoors F941?7k356n04 >4X> 7562arelintereseo buying pair of shoes that sells for S60 at regular price S4z6 This week the store unning sale, and the shoesvou like are marked down 203. Kyou buy the shoe
Woln pevment of S400 Wnat percentonthe cost 52500? X7 6 To 6 2S6. During = flu epidemic; 15S2of & schools' 480 students were absent on particular duy: How many students were absent? X= 72 +80 schoo hascollectedS1400 during = fundraiset; which 4098 ofits goal What is the total amount of the...
##### Question 2 (10 points) Determine the pressure (in atmospheres) exerted by oxygen when 289.0 mg oxygen and 163.8 mg nitrogen are mixed and have a total pressure of 0.571 atm.Use the abbreviated form of the unit in your answer:Your Answer:Answerunits
Question 2 (10 points) Determine the pressure (in atmospheres) exerted by oxygen when 289.0 mg oxygen and 163.8 mg nitrogen are mixed and have a total pressure of 0.571 atm. Use the abbreviated form of the unit in your answer: Your Answer: Answer units...
##### Easy ochem help please, both questions 1. Which are the most acidic protons? A. B. H...
Easy ochem help please, both questions 1. Which are the most acidic protons? A. B. H H Hнно H -H H H H H H C. D. 2. What is the major product of this reaction? go Br2 [H3O+] Br. A. B. Br "Br For Br C. Br. & D. Br. org Br...
##### 0 915x102 0 7.98* 10 4 0 202x 105 8 1 0
0 915x102 0 7.98* 10 4 0 202x 105 8 1 0...
##### Alpha-helix and beta-pleated sheets are considered:Select one: Tertiary structure of protein Primary structure of protein Quaternary structure of protein Secondary structure of protein
Alpha-helix and beta-pleated sheets are considered: Select one: Tertiary structure of protein Primary structure of protein Quaternary structure of protein Secondary structure of protein...
##### Jo stands 4.5 m out from the foot of a Builder = building where the angle of elevation is 719. and looks Up to the top of building the Builder Ben stands at the looking down at his wheelbarrow top of the building _ that is 10.8 m out from from the foot of the building on the opposite side where Jo is standing: Find the height of the building: Find the angle of depression from Ben down to his wheelbarrow: 6
Jo stands 4.5 m out from the foot of a Builder = building where the angle of elevation is 719. and looks Up to the top of building the Builder Ben stands at the looking down at his wheelbarrow top of the building _ that is 10.8 m out from from the foot of the building on the opposite side where Jo ...
##### A particle with a charge of q = 13.0 PC travels from the origin to the...
A particle with a charge of q = 13.0 PC travels from the origin to the point (x, y) = (20.0 cm, 50.0 cm) in the presence of a uniform electric field E = 250 V/m. Determine the following. (a) the change in the electric potential energy in )) of the particle-field system If an object is acted on by a ...
##### • Timer Notes Evaluate Feedback 9 Print What is the maximum wavelength of incident light that...
• Timer Notes Evaluate Feedback 9 Print What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is O=2.93 eV. Submit Answer Tries 0/20 In this experiment, what is the minimal potential needed to fully stop the electrons if the w...
##### NY times reported that in small town in USA there Was "strange population anomaly' because it has been decade since the last boy was born in that town with the last 12 births having been girls! To debunk this; assume that probability of newborn baby being girl is p = 0.5 . and births are independent .(a)pts) What is the probability that batch 12 births will be all baby girls?(b) pts) The above calculation is for one town similar to the town reported in the NYT magazine _ Assume that th
NY times reported that in small town in USA there Was "strange population anomaly' because it has been decade since the last boy was born in that town with the last 12 births having been girls! To debunk this; assume that probability of newborn baby being girl is p = 0.5 . and births are i... | 2,036 | 7,623 | {"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": 2, "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} | 3.09375 | 3 | CC-MAIN-2023-40 | longest | en | 0.854639 |
http://qs1969.pair.com/~perl2/?node_id=483185;displaytype=displaycode;part=3;abspart=1 | 1,685,507,549,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00420.warc.gz | 33,218,797 | 1,025 | X! Y! Z! ------- a! b! c! WLOG let's assume that X > Y > Z and a > b > c. Also let's assume that b > Y , X > a (weaker assumption: Z > c)...
NOTE: if we divide X!/a! we will have (X-a) elements
To Prove: (X-a) + (Z-c) + (b-Y) is the shortest list one can find or in other words (X-a) + (Z-c) + (b-Y) <= (X-p) + (Z-q) + (r-Y) for any p,q,r in permutation(a,b,c) Proof: From the above equation since b > Y and Z > c, r should be equal to either a or b. If r = b then the solution is trivial
If r = a then we get (X-a) + (Z-c) + (b-Y) ?<= (X-b) + (Z-c) + (a-Y) canceling terms -a - c + b ?<= -b -c + a -a + b ?<= -b + a ====> YES since a > b we see that r = a is not the smallest list so r = b
Similarly we can also show that (X-a) + (Z-c) + (b-Y) <= (X-a) + (Y-c) + (b-Z) | 298 | 773 | {"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-2023-23 | latest | en | 0.804196 |
https://gmatclub.com/forum/in-a-sequence-of-integers-each-term-in-the-first-m-integers-is-5-greater-than-262234.html | 1,716,003,751,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057260.44/warc/CC-MAIN-20240518023912-20240518053912-00145.warc.gz | 256,919,060 | 134,743 | Last visit was: 17 May 2024, 20:42 It is currently 17 May 2024, 20:42
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
# In a sequence of integers, each term in the first m integers is 5
SORT BY:
Tags:
Show Tags
Hide Tags
CEO
Joined: 26 Feb 2016
Posts: 2873
Own Kudos [?]: 5227 [19]
Given Kudos: 47
Location: India
GPA: 3.12
Director
Joined: 16 Sep 2016
Status:It always seems impossible until it's done.
Posts: 643
Own Kudos [?]: 2071 [2]
Given Kudos: 174
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
VP
Joined: 07 Dec 2014
Posts: 1071
Own Kudos [?]: 1576 [2]
Given Kudos: 27
Intern
Joined: 25 Sep 2017
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 303
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
Bunuel
Hi Bunuel, can you please explain me this problem?
Thanks,
Nikita
Director
Joined: 16 Sep 2016
Status:It always seems impossible until it's done.
Posts: 643
Own Kudos [?]: 2071 [0]
Given Kudos: 174
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
NikitaS wrote:
Bunuel
Hi Bunuel, can you please explain me this problem?
Thanks,
Nikita
Hi NikitaS,
Can you please specify which part is troubling you? I have provided with a detailed solution as follows...
Kindly go through it once and let me know. I would be happy to help
Best,
pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
The first m terms of the sequence form an increasing AP with A as 50 & d = 5
So mth term is given by A + (m - 1)*d ( formula for nth term of AP)
this is 50 +(m-1)5.
There is a second AP with the mth term of prev being it's first term hence A & d = -2.
We are given the last term of total sequence is also 50.... so we need to find the number of terms in the second AP.
mth term is 1st term, and we have 29 - m terms (since there are 29 terms in all)after that... hence 30 - m terms in all in second AP.
Hence the last term is the (30 - m)th term of 2nd AP. Applying the formula we get....
50 = A' + (n' - 1)d' .... A' is mth term from above 1st AP, d' = -2 and n' is (30 - m)
Plugging in the values:
$$50 = 50 +(m-1)5 + (30 - m - 1)*-2$$
$$(29 - m)*2 = 5m - 5$$ ....( 50's cancel and -2 term goes to the LHS)
$$58 + 5 = 7m$$
$$63 = 7m$$
$$m = 9$$
Hence Option (B) is correct.
Best,
Math Expert
Joined: 02 Sep 2009
Posts: 93334
Own Kudos [?]: 624552 [4]
Given Kudos: 81898
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
3
Kudos
1
Bookmarks
pushpitkc wrote:
In a sequence of integers, each term in the first m integers is 5 greater than the previous integer. Thereafter, each integer is 2 less than the previous integer. If the first integer and the last integer in the sequence are each 50 and the number of integers in the sequence is 29, m=?
A. 7
B. 9
C. 10
D. 12
E. 15
Source: Experts Global
First, we have m integers, for which each term is 5 greater than the previous integer:
50, 55, 60, 65, 70, ..., 50 + 5(m - 1)
Notice, that the last integer in this, first, part of the sequence is $$50 + 5(m - 1)$$. $$n_{th}$$ term in the arithmetic progression (AP) is given by $$a_n=a_1+d(n-1)$$, where d is the common difference.
After those m integers, each integer is 2 less than the previous integer:
$$[50 + 5(m - 1)] - 2$$, $$[50 + 5(m - 1)] - 4$$, $$[50 + 5(m - 1)] - 6$$, ..., 50.
Notice, that since there are a total of 29 integers in the whole sequence, then the above continues for 29 - m integers. The last integer given to be 50, thus by applying the same rule for AP, we'll get $$50 = [50 + 5(m - 1) - 2] + (-2)[(29 - m) - 1]$$:
$$0 = 5m - 7 -56 + 2m$$
$$7m=63$$;
$$m=9$$.
12. Sequences
For other subjects check:
ALL YOU NEED FOR QUANT.
Hope it helps.
Intern
Joined: 25 Sep 2017
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 303
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Intern
Joined: 25 Sep 2017
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 303
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Thanks Bunuel
Non-Human User
Joined: 09 Sep 2013
Posts: 33058
Own Kudos [?]: 828 [0]
Given Kudos: 0
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: In a sequence of integers, each term in the first m integers is 5 [#permalink]
Moderators:
Math Expert
93334 posts
Senior Moderator - Masters Forum
3137 posts | 1,887 | 6,035 | {"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} | 3.765625 | 4 | CC-MAIN-2024-22 | latest | en | 0.865617 |
https://socratic.org/questions/584b3e7c7c01496a89b8b2ee | 1,580,182,001,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00374.warc.gz | 657,449,117 | 5,981 | # Question #8b2ee
Dec 10, 2016
The mass of one atom of iron is $\left(9.2741105 \times {10}^{-} 23\right)$.
#### Explanation:
The formula we are using is $m = \frac{M}{n}$. Where...
=> $m$ is the mass in grams.
=> $M$ is the molar mass in $\frac{g}{m o l}$.
=> ${n}_{\text{A}}$ is Avogadro's number - $6.02 \times {10}^{23}$.
$m = \frac{M}{n} _ \text{A}$
$= \frac{55.85}{6.02 \times {10}^{23}}$
$= \left(9.2741105 \times {10}^{-} 23\right)$
The mass of one atom of iron is $\left(9.2741105 \times {10}^{-} 23\right)$.
Hope this helps :) | 221 | 545 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 11, "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} | 3.9375 | 4 | CC-MAIN-2020-05 | latest | en | 0.677986 |
http://mathhelpforum.com/calculus/157426-uniform-convergence.html | 1,526,931,316,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864466.23/warc/CC-MAIN-20180521181133-20180521201133-00057.warc.gz | 192,308,614 | 9,842 | 1. ## uniform convergence
Show that the series $\displaystyle {\Sigma_{n=1}^{\infty}} \frac {sin nx}{n^2}$ uniformly converges on $\displaystyle \mathbb{R}$. Verify that
$\displaystyle \int_{0}^{\pi} ({\Sigma_{n=1}^{\infty}}\frac {sin nx}{n^2}) dx$ = $\displaystyle {\Sigma_{n=1}^{\infty}}\frac {2}{(2n-1)^3}$
2. Because is...
$\displaystyle \displaystyle \int_{0}^{\pi} \sin nx\ dx = \frac{1-\cos n \pi}{n}$ (1)
... if the series $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{\sin nx}{n^{2}}$ converges uniformely, then is...
$\displaystyle \displaystyle \int_{0}^{\pi}\ \sum_{n=1}^{\infty} \frac{\sin nx}{n^{2}}\ dx = \sum_{n=1}^{\infty}\ \int_{0}^{\pi} \frac{\sin nx}{n^{2}}\ dx = 2\ \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{3}}$ (2)
At this point it remains to demonstrate the first part of Your statement...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
3. Originally Posted by chisigma
At this point it remains to demonstrate the first part of Your statement...
Because in all $\displaystyle \mathcal {R}$ is...
$\displaystyle \displaystyle \frac{|\sin nx|}{n^{2}} \le \frac{1}{n^{2}}$ (1)
... and the series $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ converges absolutely, the the series...
$\displaystyle \displaystyle f(x) = \sum_{n=1}^{\infty} \frac{\sin nx}{n^{2}}$ (2)
... converges uniformely in all $\displaystyle \mathcal{R}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$ | 516 | 1,455 | {"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} | 3.9375 | 4 | CC-MAIN-2018-22 | latest | en | 0.465359 |
http://www.steamboattoday.com/news/a-dogs-eye-view-learning-algebra-at-a-broncos-game/ | 1,511,159,224,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805914.1/warc/CC-MAIN-20171120052322-20171120072322-00763.warc.gz | 502,025,406 | 16,185 | A Dog’s Eye View: Learning algebra at a Broncos game | SteamboatToday.com
A Dog’s Eye View: Learning algebra at a Broncos game
Sandra Kruczek/For the Steamboat Today
Sandra Kruczek
— Learning algebra was difficult for me. I really had to focus in order to grasp what was for me an abstract concept. When we're teaching new dog owners how to teach skills to their pets, we caution them to practice at home in their living rooms first. For dogs as well as humans, it's difficult to learn new skills in a new or distracting environment. I often ask, "Can you imagine trying to learn algebra while attending a Broncos game?"
We also teach that once the new skill has been successfully practiced by dog and owner in a quiet and familiar environment, then it's time to begin taking those skills on the road. Even this phase of learning needs some special handling. If we make the assumption that the dog knows a skill once he has successfully performed a behavior in your living room, we might be disappointed when the same skill can't be performed in other places.
Using new skills in a different environment can present dog and owner with the problem of understanding whether "sit" means the same thing to the dog in the noisy kitchen as it did in the living room. Additionally, are new skills as a trainer presented to the dog the same way when in a new environment as in the living room?
Here's a spot that requires some courage, understanding and self-assessment on the owner's part. Being proud and loving dog owners, we're tempted to show off our dog's intelligence and push him further than the level of his ability in a new and exciting environment.
Interestingly, when we're learning how to train a dog, it's tempting to put the onus on the dog for the proper execution of any behavior. In reality, we're the other half of the equation.
Often when a student says his or her dog won't do a certain behavior that he’s been taught, I'll ask the student to show me what the cue (the hand motion, word and human body language that prompts the dog to perform the behavior) looks like and in what environment it was given. When the owner thinks about this and presents the original cue, the dog does the behavior instantly. It's often the owner who has garbled the cue or chosen a distracting environment.
Recommended Stories For You
I think it's important to remember that the process of learning has many parts. Be good to yourself and your dog. Try to be mindful of your surroundings, patient and forgiving. Learning can be stressful. If you go to a Broncos game, just have fun. You can study later.
Sandra Kruczek is a certified professional dog trainer at Total Teamwork Training with more than 25 years of experience.
Go back to article | 579 | 2,752 | {"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-2017-47 | latest | en | 0.977624 |
https://nbviewer.jupyter.org/github/mabau/lbmpy/blob/master/doc/notebooks/08_tutorial_shanchen_twophase.ipynb | 1,620,415,620,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988802.93/warc/CC-MAIN-20210507181103-20210507211103-00223.warc.gz | 438,677,537 | 56,413 | # Shan-Chen Two-Phase Single-Component Lattice Boltzmann¶
In [1]:
from lbmpy.session import *
from lbmpy.updatekernels import create_stream_pull_with_output_kernel
from lbmpy.macroscopic_value_kernels import macroscopic_values_getter, macroscopic_values_setter
from lbmpy.maxwellian_equilibrium import get_weights
This is based on section 9.3.2 of Krüger et al.'s "The Lattice Boltzmann Method", Springer 2017 (http://www.lbmbook.com). Sample code is available at https://github.com/lbm-principles-practice/code/.
## Parameters¶
In [2]:
N = 64
omega_a = 1.
g_aa = -4.7
rho0 = 1.
stencil = get_stencil("D2Q9")
weights = get_weights(stencil, c_s_sq=sp.Rational(1,3))
## Data structures¶
In [3]:
dim = len(stencil[0])
dh = ps.create_data_handling((N,)*dim, periodicity=True, default_target='cpu')
## Force & combined velocity¶
The force on the fluid is $\vec{F}_A(\vec{x})=-\psi(\rho_A(\vec{x}))g_{AA}\sum\limits_{i=1}^{q}w_i\psi(\rho_A(\vec{x}+\vec{c}_i))\vec{c}_i$ with $\psi(\rho)=\rho_0\left[1-\exp(-\rho/\rho_0)\right]$.
In [4]:
def psi(dens):
return rho0 * (1. - sp.exp(-dens / rho0));
In [5]:
zero_vec = sp.Matrix([0] * dh.dim)
force = sum((psi(ρ[d]) * w_d * sp.Matrix(d)
for d, w_d in zip(stencil, weights)), zero_vec) * psi(ρ.center) * -1 * g_aa
## Kernels¶
In [6]:
collision = create_lb_update_rule(stencil=stencil,
relaxation_rate=omega_a,
compressible=True,
force_model='guo',
force=force,
kernel_type='collide_only',
optimization={'symbolic_field': src})
stream = create_stream_pull_with_output_kernel(collision.method, src, dst, {'density': ρ})
opts = {'cpu_openmp': False,
'target': dh.default_target}
stream_kernel = ps.create_kernel(stream, **opts).compile()
collision_kernel = ps.create_kernel(collision, **opts).compile()
## Initialization¶
In [7]:
method_without_force = create_lb_method(stencil=stencil, relaxation_rate=omega_a, compressible=True)
init_assignments = macroscopic_values_setter(method_without_force, velocity=(0, 0),
pdfs=src.center_vector, density=ρ.center)
init_kernel = ps.create_kernel(init_assignments, ghost_layers=0).compile()
In [8]:
def init():
for x in range(N):
for y in range(N):
if (x-N/2)**2 + (y-N/2)**2 <= 15**2:
dh.fill(ρ.name, 2.1, slice_obj=[x,y])
else:
dh.fill(ρ.name, 0.15, slice_obj=[x,y])
dh.run_kernel(init_kernel)
## Timeloop¶
In [9]:
sync_pdfs = dh.synchronization_function([src.name])
sync_ρs = dh.synchronization_function([ρ.name])
def time_loop(steps):
dh.all_to_gpu()
for i in range(steps):
sync_ρs()
dh.run_kernel(collision_kernel)
sync_pdfs()
dh.run_kernel(stream_kernel)
dh.swap(src.name, dst.name)
dh.all_to_cpu()
In [10]:
def plot_ρs():
plt.figure(dpi=200)
plt.title("$\\rho$")
plt.scalar_field(dh.gather_array(ρ.name), vmin=0, vmax=2.5)
plt.colorbar()
## Run the simulation¶
### Initial state¶
In [11]:
init()
plot_ρs()
### Check the first time step against reference data¶
The reference data was obtained with the sample code after making the following changes:
const int nsteps = 1000;
const int noutput = 1;
Remove the next cell if you changed the parameters at the beginning of this notebook.
In [12]:
init()
time_loop(1)
ref = np.array([0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.136756, 0.220324, 1.2382, 2.26247, 2.26183, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.1, 2.26183, 2.26247, 1.2382, 0.220324, 0.136756, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15, 0.15])
assert np.allclose(dh.gather_array(ρ.name)[N//2], ref)
### Run the simulation until converged¶
In [13]:
init()
time_loop(1000)
plot_ρs() | 1,331 | 3,690 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | latest | en | 0.338558 |
http://electronics.divinechildhighschool.org/Home/final-projects-fall-2012/led-cube | 1,618,851,972,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038887646.69/warc/CC-MAIN-20210419142428-20210419172428-00159.warc.gz | 30,925,051 | 9,759 | ### LED Cube
Keaton Zonca and Cecilia DeBoever
4 x 4 x 4 LED Cube
This project is a simple LED cube. We soldered together 64 LEDs in order to create this. The LEDs can be programmed to light up in various ways. We have currently programmed it so the lights can all be one at one time or they can light up one by one in order.
```// This program controls our 4x4x4 LED cube. It first cycles through all the LEDs, repeats it faster, flashes all the LEDs, goes laye by layer, cycles through them again, and flashes one last time before repeating. int pos[] = { 16,17,18,19,1,0,14,15,5,4,3, 2,9,8,7,6,16,17,18,19,1,0,14, 15,5,4,3,2,9,8,7,6,16,17,18, 19,1,0,14,15,5,4,3,2,9,8,7,6, 16,17,18,19,1,0,14,15,5,4,3,2,9,8,7,6}; int neg[] = { 10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10, 11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11, 12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12, 13,13,13,13,13,13,13,13,13,13,13,13,13,13,13,13}; void setup() // run once, when the sketch starts { for(int i=0; i<64; i++) //sets all 64 LEDs as inputs { pinMode(pos[i], INPUT); pinMode(neg[i], INPUT); } } void loop() // run over and over again { sequence(300); //cycles through all the LEDs at pace 300 sequence(100); //cycles through all the LEDs at pace 100 all(); //turns on all the LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs delay(500); // delays for 500ms all(); //turns on all the LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs delay(500); // delays for 500ms all(); //turns on all the LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs layerone(); // turns on layer one of LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs layertwo(); // turns on layer two of LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs layerthree(); // turns on layer three of LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs layerfour(); // turns on layer four of LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs sequence(100); //cycles through all the LEDs at pace 100 all(); //turns on all the LEDs delay(1000); // delays for 1000ms layerone(); // turns on layer one of LEDs delay(1000); // delays for 1000ms layertwo(); // turns on layer two of LEDs delay(1000); // delays for 1000ms layerthree(); // turns on layer three of LEDs delay(1000); // delays for 1000ms layerfour(); // turns on layer four of LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs all(); //turns on all the LEDs delay(1000); // delays for 1000ms input(); // sets all LEDs as inputs } void sequence(int pace) //cycles through all the LEDs at pace { for(int i=0; i<64; i++) // declares i and test if less than 64, and then does the following { pinMode(pos[i], OUTPUT); // sets positive i as output pinMode(neg[i], OUTPUT); // sets negative i as output digitalWrite (pos[i], HIGH); // turns positive i on digitalWrite (neg[i], LOW); // turns negative i on delay(pace); // delays for pace pinMode(pos[i], INPUT); // sets positive i as input pinMode(neg[i], INPUT); // sets negative i as input } } void input() // sets all LEDs as inputs { for(int i=0; i<64; i++) //sets all 64 LEDs as inputs { pinMode(pos[i], INPUT); pinMode(neg[i], INPUT); } } void layerone() // turns on layer one of LEDs { for(int i=0; i<16; i++) // declares i and test if less than 16, and then does the following { pinMode(pos[i], OUTPUT); // sets positive i as output pinMode(10, OUTPUT); // sets pin 10 as output digitalWrite (pos[i], HIGH); // turns positive i on digitalWrite (10, LOW); // turns pin 10 on } } void layertwo() // turns on layer two of LEDs { for(int i=0; i<64; i++) // declares i and test if less than 64, and then does the following { pinMode(pos[i], OUTPUT); // sets positive i as output pinMode(11, OUTPUT); // sets pin 11 as output digitalWrite(pos[i], HIGH); // turns positive i on digitalWrite(11, LOW); // turns pin 11 on } } void layerthree() // turns on layer three of LEDs { for(int i=0; i<64; i++) // declares i and test if less than 64, and then does the following { pinMode(pos[i], OUTPUT); // sets positive i as output pinMode(12, OUTPUT); // sets pin 12 as output digitalWrite(pos[i], HIGH); // turns positive i on digitalWrite(12, LOW); // turns pin 12 on } } void layerfour() // turns on layer four of LEDs { for(int i=0; i<64; i++) // declares i and test if less than 64, and then does the following { pinMode(pos[i], OUTPUT); // sets positive i as output pinMode(13, OUTPUT); // sets pin 13 as output digitalWrite(pos[i], HIGH); // turns positive i on digitalWrite(13, LOW); // turns pin 13 on } } void all() //turns on all the LEDs { for(int i=0; i<64; i++) // declares i and test if less than 64, and then does the following { pinMode(pos[i], OUTPUT); // sets positive i as output pinMode(neg[i], OUTPUT); // sets negative i as output digitalWrite(pos[i], HIGH); // turns positive i on digitalWrite(neg[i], LOW); // turns negative i on } } ``` | 1,844 | 5,445 | {"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-2021-17 | latest | en | 0.449576 |
https://libraryofessays.com/lab-report/emplemintatiom-of-computerisd-fluds-dynamic-2056466 | 1,600,821,082,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400208095.31/warc/CC-MAIN-20200922224013-20200923014013-00505.warc.gz | 503,713,387 | 13,358 | # Why Are Computational Requirements for the Flow Representation So Considerable – Lab Report Example
Download full paperFile format: .doc, available for editing
The paper “ Why Are Computational Requirements for the Flow Representation So Considerable? ” is a comprehensive example of a lab report on physics. The main aim of the paper is to be able to implement CFD by analyzing the effects of changes in pressure, temperature changes, velocity, and Mach number. The design of s-pipe requires an assessment of pressure, temperature changes, velocity, and Mach number on the individual pipes as well as on the pipeline bundle as a single object. Direct data related to such pipeline configurations are scarce; therefore, the pipeline is sometimes simplified to allow easier implementation.
However, the validity of the equivalent diameter design concept has not been adequately investigated. AssumptionsThe following are the assumption were considered in the analysis Two different fluids with different densities and viscosities in the case were oil and water. The Mach number range is from 0.048753 to 0.54885 The velocity will be varied between 72 m/s and 186m/s, The two different inlet Pressure conditions considered in the case were 3366.02 and 7903.35Pa Two different Reynolds numbers Two different Inlet Mach numbers were 0.21 and 0.83 Mesh and boundary conditionsThe mesh that BoundaryThe initial conditions will be Pressure 7903.35pa and boundary condition 3066.82PaVelocity will be 72,0,0Mach number 0.21 The figure above shows that the mesh geometry of the pipe pressure is focused on the front part.
This means other parts of the pipe are not under extensive pressure as the front. Validation of the numerical model An s-pipe of Mach number 0.21 will be simulated to validate the numerical model of this study. At the same time, a rectangle of 28 by 16 will be used that is divided by 24458 elements. There have been accurate analyses meant to figure out very well the structure of a bearing to provide frequencies that are appropriate for the pipe.
This is meant to reduce the various problems that do arise from the cylinders especially if they have been in use for a long time. During the calculation of the various frequencies used in the bearings, the calculations do require one to know some basic information that is typically known. For example, the typical information includes the Mach number, pressure ranges temperature ranges, and velocity. Pressure variationThe output below shows the pressure variation.
From the output, it can be noted there is will be an increase in pressure with the increasing distance past S-shape of the pipe up exit point. At entry point marks the start of a decrease in pressure towards the end of s-shape. This movement will then result in a vortex formation in the pipe as can be seen from the diagram. The pressure will be highest at the exit point of the pipe. It was found from the graph that stagnation pressure decreases as the distance between the entry and the s-shape decreases, while past the shape pressure increases. The two figures above show that the level of pressure varies as distance changes.
The pressure range is from 0.85kPa to 0.92kPa for validated values. When there is a restriction to the flow of fluid, in the form of an orifice plate meter inserted at a particular cross-section of the flow, this reduction of area affects the fluid flow parameters in a significant manner. In a pipe having a fluid flow, the portion near the boundary surface is affected since all fluids are practically not in viscid and hence the solid surface imparts a no-slip condition which in turn retards the smooth fluid flow giving rise to a slower moving boundary layer.
When flow enters a duct, a boundary layer is almost immediately formed circumferentially around the inner side of the duct wall. The core of the flow which is in viscid in relation to the walls is restricted from freely moving due to the growth of a viscous boundary layer.
References
McGraw-Hill Companies. (2005). Boundary-layer flow. McGraw-Hill Concise Encyclopedia of Physics (5th ed.). The McGraw-Hill Companies, Inc.
Spink, L. K. (1967). Principles and Practice of Flow-meter Engineering. The Foxboro Company. Foxboro, MA.
Download full paperFile format: .doc, available for editing | 902 | 4,340 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-40 | latest | en | 0.93535 |
https://www.coursehero.com/sitemap/schools/873-Morningside/courses/654765-MATH206/ | 1,444,393,015,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737929054.69/warc/CC-MAIN-20151001221849-00067-ip-10-137-6-227.ec2.internal.warc.gz | 1,209,283,539 | 41,429 | We aren't endorsed by this school
##### MATH 206 - Morningside Study Resources
• 2 Pages
###### 9_1
School: Morningside
9.1 Lecture Notes Math 206 10/07/08 Mammenga TAYLOR POLYNOMIALS A. We want to approximate functions that are not nice with polynomials (which are nice). What constitutes a nice function? B. Denition of a Taylor polynomial: Let f be any function whose
##### Q&A
These Q&As are related to this course’s subject matter.
• ###### Q. need nine questions answered for mat/116 math class slope/intercept form. Will have to screen print and send each question
• Answer Rating:
• ###### Q. A city lot has the shape of a right triangle whose hypotenuse is 8 ft longer than one of the other sides. The perimeter of the lot is 336 ft. How...
• Not Rated Yet
• ###### Q. Use the order of operations to simplify the expression: 2/3 multiplied by 1/5 plus 4/19 divided by 15/19 =? (type a whole number or a simplified...
• Answer Rating:
• ###### Q. Looking for help on problem 46 on page 27 of the attached document regarding the height of an airplane during a landing....... Need soon please as...
• Not Rated Yet
• ###### Q. Find a real example of each of the following, around your home, work, or campus, and describe the example in detail or print a picture: Isosceles...
• Answer Rating:
• ###### Q. (round up to the next dollar, at as needed.) #17 please help thanks
• Answer Rating:
• ###### Q. Someone pls help me solve this I don't understand ! #10 (round to the nearest dollar as needed) thanks!
• Answer Rating:
• ###### Q. The polpulation of a certain country is growing exponentially at a rate of 1.6% per year. how ... the time in years for a population with growth...
• Not Rated Yet
Back to course listings | 449 | 1,731 | {"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-2015-40 | latest | en | 0.873892 |
https://www.physicsforums.com/threads/interference-pattern-wavelength.314179/ | 1,695,925,824,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510427.16/warc/CC-MAIN-20230928162907-20230928192907-00175.warc.gz | 1,010,678,191 | 14,425 | # Interference pattern, wavelength
• muffintop
## Homework Statement
Suppose the interference pattern shown in the figure is produced by monochromatic light passing through two slits, with a separation of 127 µm, and onto a screen 1.14 m away. What is the wavelength?
## Homework Equations
wavelength = (dsin$$\theta$$) / m
y=Ltan$$\theta$$
## The Attempt at a Solution
$$\theta$$= arctan (y/L) = arctan (.023 / 1.4) = .941
then plugged it into get wavelenth. i used m = 2 for 2 slits
wavelenth = 127 x 10^-6m x sin .941 / 2
i think I have m wrong?
i used m = 2 for 2 slits
m is not the number of slits. It is the order of the fringes starting from m=0 for the central bright fringe. | 214 | 692 | {"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} | 3.46875 | 3 | CC-MAIN-2023-40 | latest | en | 0.882507 |
http://lessonshah123.s3-website-us-east-1.amazonaws.com/base-ten-worksheet-first-grade.html | 1,718,792,620,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00603.warc.gz | 16,118,013 | 6,780 | # Base Ten Worksheet First Grade
Base Ten Worksheet First Grade. Below, you will find a wide range of our printable worksheets in chapter tens and ones. Web this series of base ten blocks worksheets is designed to help students of grade 1, grade 2, and grade 3 practice composition and decomposition of place value of whole.
These place value worksheets are great for teaching children the base ten numbering system. To make your cards, cut out the cards and fold on the solid black line. Web this 1st grade math resource is all about numbers and operations in base ten for 1st grade.
### Base Ten Blocks Are A Great Way For.
Web in the process, they get to acquire as well as upgrade regrouping skills by practicing the thousands and hundreds, hundreds and tens, and tens and ones worksheets here. Web this 1st grade math resource is all about numbers and operations in base ten for 1st grade. Web this page has base ten blocks worksheets that teach basic addition, subtraction, number sense and place value using visual representations of quantity as pdf printables with.
### Through Engaging Activities, Students Will Learn Basic Number Relationships, Counting,.
Web base ten for 1st grade. This number line is in line with common core's emphasis on numbers and operations in base ten for first grade. Below, you will find a wide range of our printable worksheets in chapter tens and ones.
### Web Grade 1 Base Ten Blocks Worksheet Draw Rods (Tens) And Blocks (Ones) To Represent The Numbers Shown.
The base ten blocks worksheets gives you. Web base ten and place value : To make your cards, cut out the cards and fold on the solid black line.
### Worksheets Are 1St Grade Addition With Base 10 Blocks, 1St Grade Base Ten Blocks,.
The number line has the number in. These printable base ten blocks worksheets. Web this series of base ten blocks worksheets is designed to help students of grade 1, grade 2, and grade 3 practice composition and decomposition of place value of whole.
### Web Base Ten Blocks Worksheets.
Web for students starting out with number sense and place value concepts, base ten blocks can be a powerful visual tool. Web to resolve this worksheet, students should know how to count the number of base ten blocks on each side and compare them. = __ tens + __ ones = ____ = __ tens + __ ones = ____. | 493 | 2,334 | {"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} | 3.171875 | 3 | CC-MAIN-2024-26 | latest | en | 0.89168 |
http://www.chandlercrawford.com/problem-set-1-pearson-correlation-coefficient/ | 1,653,748,564,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016853.88/warc/CC-MAIN-20220528123744-20220528153744-00328.warc.gz | 74,482,173 | 8,827 | Question
# Problem Set 1: Pearson Correlation Coefficient AnalysisResearch Scenario:
A clinical psychologist would like to determine whether there is a relationship between observer ratings of children’s externalizing behaviors and scores on an established diagnostic interview assessing externalizing disorders (like ADHD, CD, etc.). He administers the diagnostic interview to 25 children and records these scores. He then trains an observer to independently rate carefully-defined externalizing behaviors for each of the 25 children. These scores are totaled for an overall externalizing behavior index. On both the interview and the behavioral ratings, a higher score indicates higher levels of externalizing behavior. These scores are listed in the table below.
Using this table, enter the data into a new SPSS data file and run a Pearson correlation coefficient analysis to test whether there is a relationship between the interview scores and the behavior ratings in this sample. scatterplot to show the relationship between the variables.
1. Paste SPSS output.
2. Results section based on your analysis including independent and dependent variables, confidence interval, effect size, 1 or 2 tailed and the p value.
Problem Set 2: Pearson Correlation Coefficient Analysis
Research Scenario: A social worker involved in suicide prevention efforts wants to study the relationship between social isolation and possible suicide risk. She collects scores from a sample of 13 patients on two measures: one is a measure of social isolation on which higher scores indicate higher isolation (possible range of scores = 0- 20), and the other is the Suicide Risk Scale (SRS, Plutchik et al., 1989), on which higher scores indicate a higher risk of suicide (possible range of scores = 0-15). These scores are listed in the table below.
Using this table, enter the data into a new SPSS data file and run a Pearson correlation coefficient analysis to test whether there is a relationship between social isolation and possible suicide risk. scatterplot to show the relationship between the variables.
Interview (Range 0
to 9)
Externalizing Behavior Index
(Range 0 to 25)
7
9
6
9
22
20
IN A WOOD
12
10
9
-JU
16
19
21
22
4
7
7
3
13
19
16
12
9
Statistics and Probability | 476 | 2,266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-21 | latest | en | 0.883865 |
https://community.alteryx.com/t5/Weekly-Challenge/Challenge-123-When-will-Rabbits-Rule-the-World/td-p/179283/highlight/true/page/2?attachment-id=21039 | 1,600,762,761,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00380.warc.gz | 323,368,013 | 46,203 | # Weekly Challenge
Solve the challenge, share your solution and summit the ranks of our Community!
###### IDEAS WANTED
We're actively looking for ideas on how to improve Weekly Challenges and would love to hear what you think!
Submit Feedback
We've recently made an accessibility improvement to the community and therefore posts without any content are no longer allowed. Please use the spoiler feature or add a short message in the message body in order to submit your weekly challenge.
## Challenge #123: When will Rabbits Rule the World?
Highlighted
13 - Pulsar
I hate to be Johnny Raincloud, but while I got to the provided date, I wonder if it might actually be 7 months sooner ...
Spoiler
The difference is when a rabbit starts reproducing ( > 4 months or >= 4 months). Reading through the problem, it sounds like >= 4 months ...
Please correct me if I'm wrong. I'd appreciate every extra day I can get.
Highlighted
8 - Asteroid
@patrick_digan
Ha ha, yeah, I know, it looks neat and tidy on the outside but there's a hidden jackhammer in the Formula tool! :)
Highlighted
8 - Asteroid
@danrh
I had the same kind of problem, I had to interpret the assignment as if female rabbits start reproducing at 4 months but give birth the next month, so (> 4) and not (>= 4).
This is how I was able to get the right month, but I was still a little off on the exact population number, so there may be more to it than that.
Highlighted
12 - Quasar
This had me puzzled a bit as well. I solved it by interpreting it as though the comparison took place at the beginning of the month and any babies that were born during the month could not be counted until the start of the next month....which may be another way to get to the solution you both got to...
Highlighted
8 - Asteroid
Here's a second solution.
Spoiler
"Look Ma, no Iterative Macros!"
Getting creative... putting the male and female rabbit population in a single column so I can calculate and store the 2 values with a single Multi-Row Formula tool.
Highlighted
Alteryx Partner
you have me puzzeled, you mentioned 41 iteration but how with that number of iteration you end up on date 2022-08-02 shouldn't that date mean something like 48 iterations (it is 4 years from starting date) , or alternatively with 41 iteration shouldn't you end around 2022-01-02?
Highlighted
Alteryx Partner
Hello,
That is a fun challenge :)
I was struggling a bit with my results bit thanks to @danrh I got back on track becouse I was afraid that bunnycalypse will come 7 month earlier :)
Spoiler
Highlighted
16 - Nebula
@ukashi good catch. I did land on 48 iterations for my final version. The 41 iterations was when I let the bunnies reproduce a month earlier. I changed it to match the solution provided.
Highlighted
Alteryx Partner
Thanks for that @patrick_digan - I've paid attention as I've come across the same thing and initially could not find where the problem is :)
Highlighted
Alteryx Certified Partner
I'm baking my noodle over this one. Couldn't get the right month. Like @ukashi it takes my macro between 41 and 42 iterations which works out as 2022-01-22.
Had to keep all rows to keep track of the age of the bunny (as they will give birth and die at different times).
Spoiler | 764 | 3,257 | {"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} | 3.203125 | 3 | CC-MAIN-2020-40 | latest | en | 0.915904 |
https://forum.vectorworks.net/index.php?/topic/56180-3d-curved-text/#comment-282305 | 1,638,772,350,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363290.39/warc/CC-MAIN-20211206042636-20211206072636-00554.warc.gz | 322,510,325 | 21,152 | # 3D Curved Text
## Recommended Posts
I am kicking myself right now. I know I have watched a video on this a number of times but I cannot seem to locate the video and I am forgetting what the best approach is to this and hoping someone can help. I am trying to make 3D extruded text that follows a curve. Think like a circular sign but the text letters are CNC'ed out and attached to the face of the sign. However, if it makes any difference the area I am trying to have this show up on is the concave of a curve.
Anyone have any tips on best way to do this. I know I have done this in the past but don't do it often and know I am forgetting something.
Thanks,
Matt
This one ?
no, have seen that one. So the text is on a stage backdrop. The backdrop is curved. I am looking for it to follow the curve but face straight out. I found the text along path. I use that from time to time but trying to get that to work in a 3D space. So when you do as shown in the video it is parallel to the ground plane. I want the curve to be perpendicular to the ground plane. Make sense? The one I am thinking of is even older
I've also seen that video. perfect for Marquee signage as an example. Can't believe I don't have a bookmark or notes.
I have a hunch we are thinking of the same one. I have a little bookmark folder on chrome of VW help videos. Not in there. I don't know which is more annoying forgetting the secret sauce or being unable to find something I know exists.
14 minutes ago, MattG said:
no, have seen that one. So the text is on a stage backdrop. The backdrop is curved. I am looking for it to follow the curve but face straight out. I found the text along path. I use that from time to time but trying to get that to work in a 3D space. So when you do as shown in the video it is parallel to the ground plane. I want the curve to be perpendicular to the ground plane. Make sense? The one I am thinking of is even older
If your text is vertical and following the curve of the backdrop, you can use the same technique as in the video. Set the "Rot About Path" in the OIP to 90 degrees. You may also need to go in to edit the path, select it and click switch direction in the OIP to get it to face correctly.
Kevin
That works. Thank you. One issue I am having is the test is backward. Reading like in your example of "EVRUC A NO TXET" I have tried flipping the text before converting and a few other things. Any ideas?
32 minutes ago, MattG said:
One issue I am having is the test is backward. Reading like in your example of "EVRUC A NO TXET" I have tried flipping the text before converting and a few other things.
1 hour ago, Kevin McAllister said:
You may also need to go in to edit the path, select it and click switch direction in the OIP to get it to face correctly.
3 hours ago, MattG said:
That works. Thank you. One issue I am having is the test is backward. Reading like in your example of "EVRUC A NO TXET" I have tried flipping the text before converting and a few other things. Any ideas?
As Zoomer mentioned above, the solution is to switch the direction of the path. Here's a step by step -
- right click on the Text Along Path object and choose "Edit Path"
- select the path. You'll notice its been converted to a NURBS curve. To flip the direction of the text click on the Reverse Direction button in the OIP while the NURBS curve is selected.
- exit the path. The text should now be running in the opposite direction.
Kevin
Yes thank you all for the help. I just was in a hurry and didn't read that part.....oops. Worked out correctly so we should be all good. Thanks again.
## Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.
× Pasted as rich text. Restore formatting
Only 75 emoji are allowed.
× Your previous content has been restored. Clear editor
× You cannot paste images directly. Upload or insert images from URL.
×
×
• KBASE
• #### MARIONETTE
×
• Create New... | 1,018 | 4,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-49 | latest | en | 0.947735 |
https://ipv6.snipplr.com/view/46521 | 1,696,183,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510924.74/warc/CC-MAIN-20231001173415-20231001203415-00371.warc.gz | 348,616,851 | 7,753 | sara on 01/05/11
# Statistics
Viewed 376 times
Favorited by 0 user(s)
# bstablo_polje.h
/ Published in: C++
Save to your folder(s)
Copy this code and paste it in your HTML
`#include <stdlib.h>#define ARRAYSIZE 1000000#define Null -1typedef int labeltype; struct element { labeltype label; int used;}; typedef struct bt { struct element elements[ARRAYSIZE];}*btree; typedef int node; btree InitB (labeltype x){ btree myBTree; int i; myBTree=(btree)malloc(sizeof(struct bt)); for(i=1; i<ARRAYSIZE; i++){ if(i==1){ myBTree->elements[i].label=x; myBTree->elements[i].used=1; }else{ myBTree->elements[i].used=0; } } return myBTree;} node RootB(btree T){ return 1;} node ParrentB(node n, btree T){ if(n==1){ printf("Zatrazili ste roditelja korjenskog cvora\n"); return Null; } if(T->elements[n].used==0){ printf("Ne postoji taj cvor u stablu\n"); exit(1); } return (n/2);} node LeftChildB(node n, btree T){ node child; child=n*2; if (child > ARRAYSIZE-1){ printf("\nPolje nije toliko veliko\n"); return Null; //exit(1); } if(T->elements[child].used==0){ return Null; }else{ return child; }} node RightChildB(node n, btree T){ node child; child=(n*2)+1; if (child > ARRAYSIZE-1){ printf("\nPolje nije toliko veliko\n"); return Null; //exit(1); } if(T->elements[child].used==0){ return Null; }else{ return child; }} labeltype LabelB(node n, btree T){ if(T->elements[n].used==0){ printf("Nepostojeci cvor u stablu\n"); exit(1); }else{ return T->elements[n].label; }} void ChangeLabelB(labeltype x, node n, btree T){ if(T->elements[n].used==0){ printf("Nepostojeci cvor u stablu\n"); exit(1); }else{ T->elements[n].label=x; }} void CreateLeftB(labeltype x, node n, btree T){ node child; child=n*2; if (child > ARRAYSIZE-1){ printf("\nPolje nije toliko veliko\n"); return; //exit(1); } if(T->elements[child].used==1){ printf("Lijevo dijete ovog cvora vec postoji\n"); exit(1); }else{ T->elements[child].used=1; T->elements[child].label=x; }} void CreateRightB(labeltype x, node n, btree T){ node child; child=(n*2)+1; if (child > ARRAYSIZE-1){ printf("\nPolje nije toliko veliko\n"); return; //exit(1); } if(T->elements[child].used==1){ printf("Desno dijete ovog cvora vec postoji\n"); exit(1); }else{ T->elements[child].used=1; T->elements[child].label=x; }} void DeleteBRek(node n, btree T){ if(n==Null) return; DeleteBRek(LeftChildB(n, T), T); T->elements[n].used=0; DeleteBRek(RightChildB(n, T), T);} void DeleteB(node n, btree T ){ if(ParrentB(n, T)==Null){ printf("Ne mogu obrisati korjen\n"); exit(1); } DeleteBRek(n,T);}` | 966 | 3,024 | {"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-2023-40 | latest | en | 0.174036 |
https://books.google.ie/books?id=BYEAAAAAMAAJ&dq=editions:UOM39015065618988&lr= | 1,627,416,103,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153474.19/warc/CC-MAIN-20210727170836-20210727200836-00065.warc.gz | 160,430,017 | 11,410 | # Elements of the Geometry of Planes and Solids: With Four Plates
The author, 1828 - Electronic book - 159 pages
Pt. 1. Plane geometry -- Pt. 2. Geometry of solids.
### What people are saying -Write a review
We haven't found any reviews in the usual places.
### Contents
Section 1 1 Section 2 2 Section 3 3 Section 4 11 Section 5 65 Section 6 87 Section 7 96
Section 8 97 Section 9 103 Section 10 105 Section 11 138 Section 12 159 Section 13
### Popular passages
Page 44 - IN right angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle...
Page 20 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Page 11 - A plane rectilineal angle is the inclination of two straight lines to one another, which meet together, but are not in the same straight line.
Page 2 - BBOWN, of the said district, hath deposited in this office the title of a book, the right whereof he claims as author, in the words following, to wit : " Sertorius : or, the Roman Patriot.
Page 36 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Page 119 - If a solid angle be contained by three plane angles, any two of them are together greater than the third. Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB : any two of them shall be together greater than the third.
Page 61 - UPON a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle. Let AB be the given straight line, and...
Page 113 - If one of two parallel lines is perpendicular to a plane, the other is also perpendicular to the plane.
Page 29 - EBF, there are two angles in the one equal to two angles in the other, each to each ; and the side EF, which is opposite to one of the equal angles in each, is common to both ; therefore the other sides are equal ; (I.
Page 16 - E the centre of the circle, and join EA, EB. Then, because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one, equal to two sides in the other, and the base EA is equal to the base EB; therefore the angle AFE is equal... | 592 | 2,379 | {"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} | 3.453125 | 3 | CC-MAIN-2021-31 | latest | en | 0.895624 |
http://www.docstoc.com/docs/75941802/The-Weak-Force | 1,387,480,114,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345766127/warc/CC-MAIN-20131218054926-00074-ip-10-33-133-15.ec2.internal.warc.gz | 358,252,903 | 17,707 | # The Weak Force
Document Sample
``` The Weak Force
?
EM
STRONG
WEAK
The Force Carriers
Like the Electromagnetic & Strong forces, the Weak force is also
mediated by “force carriers”.
For the weak force, there are actually 3 force carriers:
W+ W- Z0
These “weak force” carriers This “weak force” carrier
carry electric charge also ! is electrically neutral
The “charge” of the weak interaction is called
“weak charge”
Weak Charge of Quarks & Leptons
Both quarks & leptons carry weak charge
Both quarks & leptons “couple to” the W and Z force carriers
Since the W’s have a charge of +1 and –1 they cause a “charge-
changing” interaction. That is when they are emitted or absorbed,
to conserve charge, the “emitting” or “absorbing” particle changes
charge by +1 or –1 unit.
The emitting or absorbing particle changes into a different particle.
Alternately, when the W decays, it decays into 2 particles which:
(a) Carry weak charge
(b) The sum of their charges equals the charge of the W
I will mainly talk about the W in the context of decays…
Comparison of the Force Carriers
Property EM Strong Weak
Force Photon Gluon
Carrier
W+, W- Z0
(g) (g)
Charge of Electrical &
None Color Weak
force carrier Weak
Quarks & All quarks All quarks
Couples to: Charged Quarks only & &
leptons all leptons all leptons
Infinite <10-14 [m]
Range < 2x10-18 [m] < 2x10-18 [m]
Notice that the weak force only operates at distances ~ < 10-18 [m] !
Particles & Forces
Charged Neutral
quarks leptons leptons
(e,m,t) (n)
Strong Y N N
Electro-
Magnetic Y Y N
Weak Y Y Y
Quarks carry strong, weak & EM charge !!!!!
Neutrinos
From the previous table, you saw that neutrinos only interact via
the weak force.
Also, the weak force only “kicks in” for d <10-18 [m]. Recall that
the nucleus’ size is about 10-15 [m], so this is 1000 times smaller than
the size of the nucleus.
As a result, neutrinos can pass through a lot of matter, and do
absolutely nothing!!!!
After all, matter is mostly empty space, right !
Neutrinos can easily pass right through the earth, almost as if it
wasn’t there!
Neutron decay
This is how neutron decay really proceeds through the
Weak Interaction
e- u u u
W- d d u d u d
ne
Neutron Proton Proton
u u
d d u d + e- + ne
Neutron Proton
Neutron Decay (cont)
u u
d d u d + W-
Neutron Proton
e- + ne
n p + e- + ne
But in fact, what’s really going on is this:
d u + e- + ne
Feynman diagram for weak decay
du+e -+n
e
ne
e-
W-
d u
d d “Spectator
u u quarks”
Spectator quark(s): Those quarks which do not directly
participate in the interaction or decay.
Feynman diagram for weak decay
(continued)
Since the spectator quarks do not directly participate in the
decay, we can just omit them…
This yields the “quark-level” Feynman diagram!
Is charge conserved ? ne 0
e - -1
W-
d u
-1/3 +2/3
Is Le conserved ?
Decays of “heavy” quarks
The heavy quarks decay to the lighter ones
t Q=+2/3
b Q=-1/3
c Q=+2/3
s Q=-1/3
u Q=+2/3
d Q=-1/3
What about the decay of a b-quark?
bc+m -+n
m
Is charge conserved ? nm 0
m- -1
W-
b c
-1/3 +2/3
Notice: Here, the W- decays to a m- and nm
Is Lm conserved ?
What about the decay of a c-quark?
cs+m ++n
m
Is charge conserved ? nm 0
m+ +1
W+
c s
2/3 -1/3
Notice: Here, I have the W+ decaying to a m+ and nm (could have
been an e+ and ne as well).
Is Lm conserved ?
What about the decay of a b-quark?
su+e -+n
e
Is charge conserved ? ne 0
e - -1
W-
s u
-1/3 +2/3
Is Le conserved ?
Decays of heavy quarks to u & d
A quark can only decay to a lighter quark.
The W charge has the same sign as the parent quark.
Quark Charge Mass
[GeV/c2]
top +2/3 ~175
tb W+(100%)
bottom -1/3 ~4.5
b c W- (~90%) charm +2/3 ~1.5
b u W- (~10%) strange -1/3 ~0.2
up +2/3 ~0.005
c s W+ (~95%)
c d W+ (~5%) down -1/3 ~0.010
s u W- (~100%)
d u W+ (~100%)
“Leptonic” Decay of W
Once the W is produced, it must decay
W- e- ne W+ e + n e
W- m- nm W+ m + n m
W- t - nt W+ t + n t
It’s call “leptonic decay” because the
W is decaying to leptons!
The W can decay to leptons because leptons carry weak charge
But so do quarks …
Since quarks also carry weak charge, we can also get:
- +
W ud W ud
It’s call “hadronic decay” because the
W is decaying to quarks, which will form hadrons!
u
Check charge:
W- (-2/3 + -1/3 = -1)
d
b c
But quarks are bound to one another by the strong force, and are not
observed as “free” particle. That is, they are bound up inside hadrons…
What happens next ?
One possibility…
u Can, in fact,
W- d
form a p-
b c
B- u D0
u
B- D0
Meson Meson
== p
d -
u
B- D0 p-
The process by which quarks “dress themselves” into hadrons
u u
u p0
W- W- u u
u
d
d p-
d
d
As the quarks separate, the “potential energy” stored in this “spring-
like” force increases. Eventually, the potential energy gets large
enough, and nature relives itself by converting this potential energy
into mass energy. That is, quark-antiquark pairs are created ! The
quarks then pair off and form hadrons (which we can see) !
In fact, this process whereby quark-antiquark pairs are created
can happen more than once !
One might therefore get 2, 3, or more hadrons from a hadronic
W decay!
It’s important to note that when the “spring” associated with
the strong force “snaps”, it always produces quarks and antiquarks
of the same type. They are usually the lighter quarks, since they
have lower mass, and thus are created more easily:
uu , dd , ss
Again Energy being transformed into mass !
Feynman Diagrams involving
W force carriers
Decay of a B- Meson
Could end up as:
u B- D0 p-
W- B- D0 p-p0
d
b c B- D0 p- p+ p-
B- u D0
u etc
Additional particles are created when the strong force produces more
quark-antiquark pairs. They then combine to form hadrons!
Notice that the charge of the particles other than the D0 add up to the
charge of the W- (Q = -1), as they must!
W Decays
W- e- ne
W- m- nm
Can be 1 or more
W- t - nt hadrons produced
W+ follows in an analogous way… see previous slides
Interactions involving W’s
Here is one… Don’t worry about these types of interactions…
I want to emphasize the role of W’s in decays of quarks
e + e n e +n e
+ -
e- ne
W-
e+ ne
time
Check lepton number, charge conservation…
The main points
1. Neutrons decay to protons through the weak
interaction
2. The electron and neutrino in fact come from
the Wene decay.
3. The W can decay into either lepton pairs or
a quark-antiquark pair (ud). In the latter case, the quarks | 2,328 | 8,560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2013-48 | longest | en | 0.851349 |
villaorchidee.be | 1,713,078,315,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816875.61/warc/CC-MAIN-20240414064633-20240414094633-00662.warc.gz | 555,681,301 | 8,219 | ### How To Calculate Drag Conveyor Volume
drag chain conveyor capacity calculator Solution for ore mining. Flowrite Drag Conveyor Capacity Table: Drag Size: Calculate Slat or Chain Drag Conveyor Horsepower....
### Chain Conveyors
Conveying Capacity The properties of the bulk material are essential to determine the main features of the Chain Conveyor's components such as chain type or flight height. They also determine the material layer and the actual conveying capacity. The following capacities are based on standard conditions. Conveying capacity with standard ......
### chain conveyor capacity calculations
02/09/2020 · chain conveyor calculations As a specialist in the field of conveyor technology, VAV can also help you with design questions of chain conveyors. We provide you with the basic calculations for the determination of chain speed, capacity per hour, weight of the material to be conveyed, as well as the calculation of the required power....
### Conveyor Calculator Capacity
calculation Of conveyor capacity . Conveyor Capacity kaseconveyors Calculation of Conveyor Speed Conveyor Speed can be most conveniently calculated by use of the nomographs supplied on pages To use this nomograph first loe the two known values (screw diameter and required capacity in cu ft per hour) then with a straight edge connect these two points Screw Conveyor...
### calculate belt conveyor capacity
How To Calculate Chain Conveyor Capacity Protable Plant . how to calculate chain conveyor capacityrelated information, including: Quotes,Price,Manufacturers,Parts : belt capacity calculator. given the following parameters ......
### calculating conveyor carrying capacity
How To Calculate Belt Conveyor Capacity In Exel. Conveyor capacity is determined by the belt speed, width and the angle of the belt and can be expressed as. Q A v 1 where . Q conveyor capacity kgs, lbs density of transported material kgm 3, lbft 3 A crosssectional area of ....
### formula for conveyor capacity
SCREW CONVEYOR CAPACITY SPEED CALCULATION: SCREW CONVEYOR CAPACITY SPEED CALCULATION: Visit the online engineering guide for assistance with using this calculator. Click on symbol for more information. Inquire Now; formula for conveyor capacity ...
### Design, Analysis and Optimization of Drag Chain Conveyor ...
06/03/2020 · The proposed design allows optimization of Drag Chain Conveyor System which results in, increase in coal outlet capacity, reduction in weight of support chassis, increase in fatigue cycles of shaft, reduction in vibrations in bottom chassis by the proper material selection, analysis and design calculations of Drag Chain Conveyor....
### 5. Chain Speed
Chain speed, normally calculated in feet or metres per minute, is a critical consideration in determining the conveyor capacity. It is in considering this parameter that conveyor chain pitch and conveyor chain drive and idler sprocket can be determined. fig. illustrates the relationship between speed, chain pitch and sprocket size....
### drag conveyor capacity calculation
Chain conveyor calculations Bechtel As a specialist in the field of conveyor technology, Bechtel can also help you with design questions of chain conveyor...
### Novel calculation method for chain conveyor systems (Neue ...
The calculation of chain conveyor systems is basically identical formost cases . Starting from the chain's exit point at the drive gear, any frictionand resistance forces during one revolution of the chain are summed up. Since the layout of a conveyor systems is often quite complex...
### How To Calculate Drag Conveyor Capacity
how to calculate drag chain conveyor capacity Light Structural Steel Beams. drag chain conveyordrag chain conveyor Chain( Ernst Boris,,,.Rated / based on ....
### mill chain conveyor capacity calculation
15/09/2020 · Chain conveyor calculations Bechtel . We provide you with the basic calculations for the determination of chain speed, capacity per hour, weight of the material to be conveyed, as well as the calculation of the required power....
### Conveyor Capacity
24/08/2006 · The conveyor runs at 325 feet per minute, I have 47 inches from the starting edge of carton 1 to the starting edge of carton 2 (fixed by a gap optomizer) and so on for each carton. (I need a specific gap in order for the inline scale to perform) What would be the calculation to figure this out?...
### capacity and weight conveyors calculator
how to calculate chain conveyor capacity. Input data: Conveyor capacity, The Series 50 Conveyors are best suited for light to medium weight products, This maximum belt capacity calculator is provided. More Price. conveyor capacity calculator for sale, conveyor capacity....
### Mill Chain Conveyor Capacity Calculation
How To calculate Load capacity For chain conveyor Traduire cette page. Chain calculations Conveyor chain calculations Capacity in kg per hour (Q) Q = A x v x 3600 sec Q = capacity in m3 per hour A = trough width x layer height in m2 v = chain speed in m per sec Conveyor chain calculations Material weight on the chain in kg (mass1) tons per hour x distance in meters v x 3,6...
### calculating conveyor carrying capacity
Calculation of load carrying capacity of conveyor belt, figure 2 shows the effect of increasing belt width on, how to calculate chain conveyor chat, speed of a conveyor under a related load, or complex and exceeds the chain pull capacity of, the chain pull calculation for the conveyor in....
### calculate conveyor capacity
Conveyor capacity How to calculate Know 202035 How to calculate conveyor capacity. Suppose the width of the conveyor is 40 cm, and it needs to move...
### Logistics News: Understanding What Determines Maximum ...
19/06/2013 · Once you determine the length of the average case, based on the speed of the metering belt, you can calculate the maximum CPM throughput capacity of the conveyor system. Most conveyor system engineers agree that the actual system CPM throughput should be within 10% of the maximum allowing for normal system losses....
### mill chain conveyor capacity calculation
Mill Mill Chain Conveyor Capacity Calculation. Roller conveyor design calculation pdf. CONVEYORS Movetec. Calculation of endless length EL For cylinder 45 EL 2xA165 mm For cylinder 90 EL 2xA305 mm With our consistent approach to value and design our standard timing belt conveyor technology can be used in clean rooms with roller conveyors ......
### mill chain conveyor capacity calculation
Mill Chain Conveyor Capacity Calculation We have sold many of crushing and grinding units to more than 130 countries!.DRGM ImagesCalculate the measurement (L) using the'ormuta, Roller Conveyor Chains are used to transfer bulk unit product, DRGM Conveyor Engineering Guide Capacity TablesDrag Conveyor Engineering Guide Capacity Tabl ......
### Calculation methods – conveyor belts
Calculation example Unit goods conveying systems 12 Conveyor and power transmission belts made of modern synthetics Worldwide leaders in technology, quality and service Further information on machine design can be found in our brochure no. 305 "Recommendations for machine design"....
### chain conveyor capacity calculation
Refer to section 2 Basic conveyor arrangement. () Chain and material sliding . Where M can be calculated as M = . Get Price; Calculations conveyor chains VAV . VAV Aandrijvingen is not only a supplier, you can also consult our engineers about integral solutions, capacity issues regarding chain and screw conveyors. Get Price... | 1,476 | 7,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2024-18 | latest | en | 0.835078 |
www.ailiberator.com | 1,721,670,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00566.warc.gz | 573,304,337 | 6,826 | Understanding Multiple Linear Regression: Predicting and Analyzing Relationships
Practical Uses of Multiple Linear Regression Analysis
Posted by Luca Berton on Thursday, October 12, 2023
Introduction
Hello, and welcome to this comprehensive exploration of multiple linear regression. In this video, we will delve into the intricacies of this statistical technique that allows us to analyze the relationships between multiple independent variables and a dependent variable. While you may be familiar with simple linear regression, which uses a single independent variable to predict an outcome, multiple linear regression takes this concept further by involving multiple predictors to gain a deeper understanding of complex relationships.
Multiple Linear Regression vs. Simple Linear Regression
Before we jump into the mechanics of multiple linear regression, let’s briefly differentiate it from simple linear regression. In simple linear regression, we use one independent variable to estimate a dependent variable, such as predicting CO2 emissions based on engine size. However, in reality, many factors influence CO2 emissions, and that’s where multiple linear regression comes into play. This technique allows us to consider multiple independent variables simultaneously, like engine size and the number of cylinders, to predict CO2 emissions more accurately.
Applications of Multiple Linear Regression
Multiple linear regression serves two primary purposes:
Identifying the Strength of Effects: It helps us assess the impact of independent variables on the dependent variable. For instance, you can analyze whether factors like revision time, test anxiety, lecture attendance, and gender affect students’ exam performance.
Predicting the Impact of Changes: It enables us to understand how the dependent variable changes when we modify independent variables. For instance, in the context of a person’s health data, multiple linear regression can quantify the effect of changes in a patient’s body mass index on their blood pressure, while keeping other factors constant.
Modeling with Multiple Linear Regression
Multiple linear regression is a powerful tool for predicting continuous variables. It formulates the target value (Y) as a linear combination of independent variables (X). Mathematically, the model can be expressed as follows:
Y = θ0 + θ1 * X1 + θ2 * X2 + … + θn * Xn
Here, Y represents the dependent variable, θ values are the coefficients, and X values are the independent variables. In essence, this equation defines a hyperplane in a multi-dimensional space that best fits the data.
Finding the Best-Fit Hyperplane
The core objective in multiple linear regression is to identify the hyperplane that minimizes the prediction error. This error is measured by the mean squared error (MSE), which quantifies the distance between data points and the fitted regression model. The best model is the one that minimizes this error across all prediction values.
Estimating Model Parameters
To determine the optimal coefficients (θ values) for the model, we employ various methods. Two common approaches are:
Ordinary Least Squares: This technique minimizes the MSE by performing linear algebra operations on the data matrix. It’s suitable for smaller datasets but can be computationally intensive for larger ones.
Optimization Algorithms: These algorithms iteratively adjust the coefficients to minimize the error on the training data. Gradient descent is a popular optimization method, especially for large datasets.
Making Predictions
Once we’ve determined the parameters, making predictions is straightforward. We substitute the coefficients and feature values into the linear model equation. For instance, if we want to predict CO2 emissions for a car with specific engine size and cylinder count, we use the formula:
CO2 Emissions = θ0 + θ1 * Engine Size + θ2 * Cylinder + …
Concerns and Considerations
Multiple linear regression offers valuable insights, but some important considerations must be kept in mind:
Feature Selection: Adding too many independent variables without justification can lead to overfitting, where the model becomes too complex for the dataset. Careful feature selection is essential.
Categorical Variables: Categorical independent variables can be incorporated by converting them into numerical values, often through the use of dummy variables.
Linearity: For multiple linear regression to be valid, there must be a linear relationship between the dependent variable and each independent variable. This can be assessed using techniques like scatter plots.
Conclusion
Multiple linear regression is a powerful tool for analyzing complex relationships and making predictions based on multiple independent variables. By understanding the mechanics of this method, selecting appropriate features, and assessing the linearity of relationships, you can harness its potential to gain valuable insights in various fields, from education to healthcare and beyond. Whether you’re a data scientist, researcher, or simply curious about the world of regression analysis, multiple linear regression is a valuable addition to your analytical toolkit. | 932 | 5,203 | {"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} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.854017 |
https://brainly.in/question/222180 | 1,484,653,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279657.18/warc/CC-MAIN-20170116095119-00533-ip-10-171-10-70.ec2.internal.warc.gz | 798,388,305 | 10,363 | # Express 0.25% as fraction
2
by chicki13
2015-11-08T12:08:46+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
0.25%=0.25/100×100/100=0.25/100×100=25/10000=1/400. Hope this helps you
Do u like to be my frnd
no
Fyn
sorry
It's ok !! I never feel
2015-11-08T12:10:59+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
0.25% = 0.25/100 = (0.25×100)/(100×100) = 25/10000 = 1/400 | 262 | 903 | {"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-2017-04 | latest | en | 0.932499 |
https://wpblog.wyzant.com/resources/lessons/math/calculus/differentiation/mean_value_theorem/ | 1,653,811,605,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663048462.97/warc/CC-MAIN-20220529072915-20220529102915-00715.warc.gz | 688,433,150 | 46,807 | # Mean Value Theorem Explanation
The Mean Value Theorem states that, given a curve on the interval [a,b], the derivative at some point f(c)
where a < c="">< b="" must="" be="" the="" same="" as="" the="">
slope
from f(a) to f(b).
In the graph, the tangent line at c (derivative at c) is equal to the slope of [a,b]
where a <>.
The Mean Value Theorem is an extension of the Intermediate Value Theorem,
stating that between the continuous interval [a,b], there must exist a point c where
the tangent at f(c) is equal to the slope of the interval.
This theorem is beneficial for finding the average of change over a given interval.
For instance, if a person runs 6 miles in an hour, their average speed is 6 miles
per hour. This means that they could have kept that speed the whole time, or they
could have slowed down and then sped up (or vice versa) to get that average speed.
This theorem tells us that the person was running at 6 miles per hour at least once
during the run.
If we want to find the value of c, we
• i) Find (a,f(a)) and (b,f(b))
• ii) Use the Mean Value Theorem
• iii) Find f'(c) of the original function
• iv) Set it equal to the Mean Value Theorem and solve for c.
## Mean Value Theorem Examples
Let’s do the example from above.
(1) Consider the function f(x) = (x-4)2-1 from [3,6].
First, let’s find our y values for A and B.
Now let’s use the Mean Value Theorem to find our derivative at some point c.
This tells us that the derivative at c is 1. This is also the average slope from
a to b. Now that we know f'(c) and the slope, we can find the coordinates for c.
Let’s plug c into the derivative of the original equation and set it equal to the
result of the Mean Value Theorem.
We have our x value for c, now let’s plug it into the original equation.
Let’s do another example.
(2) Consider the function f(x) = 1x from [-1,1]
Using the Mean Value Theorem, we get
We also have the derivative of the original function of c
Setting it equal to our Mean Value result and solving for c, we get
c is imaginary! What does this mean? The function f(x) is not continuous over the
interval [-1,1], and therefore it is not differentiable over the interval. For the
Mean Value Theorem to work, the function must be continous.
## Rolle’s Theorem
Rolle’s Theorem is a special case of the Mean Value Theorem. It is stating the same
thing, but with the condition that f(a) = f(b). If this is the case, there is a
point c in the interval [a,b] where f'(c) = 0.
(3) How many roots does f(x) = x5 +12x -6 have?
We can use Rolle’s Theorem to find out. First we need to see if the function crosses
the x axis, i.e. if at some point it switches from negative to positive or vice
versa.
We can see that as x gets really big, the function approaces infinity, and as x
approaches negative infinity, the function also approaches negative infinity.
This means that the function must cross the x axis at least once.
If the function has more than one root, we know by Rolle’s Theorem that the derivative
of the function between the two roots must be 0.
This is not true. The only way for f'(c) to equal 0 is if c is imaginary. f'(c)
is always positive, which means it only has one root.
Scroll to Top | 838 | 3,221 | {"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} | 4.84375 | 5 | CC-MAIN-2022-21 | latest | en | 0.905141 |
https://mathispower4u.wordpress.com/tag/check/ | 1,550,367,791,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481428.19/warc/CC-MAIN-20190217010854-20190217032854-00137.warc.gz | 626,246,077 | 18,633 | # Solve a NonLinear System of Equations (Linear and Quadratic)
This video explains how to solve a nonlinear system of equations.
# Solve a System of Equations (Nonlinear) y=x-2, xy=3
This video explains how t solve a system of nonlinear equations using substitution.
# Determine if Two Linear Functions Are Inverses (2)
This video explains how to determine if two linear functions are inverses of one another. The results are checked graphically.
# Determine if Two Linear Functions Are Inverses (1)
This video explains how to determine if two linear functions are inverses of one another. The results are checked graphically.
# Solve a Linear Equation One Variable – Parentheses and Vars on Both Sides
This video explains how to solve a linear equation in one variable with parentheses and variables on both sides.
# Factor a Trinomial with A not 1 Using Trial and Error: 12x^2+11x-36
This video explains how to factor a trinomial in the form ax^2+bx+c when a is not 1 and no common factors other than 1.
# Factor a Trinomial with A not 1 Using Trial and Error: 6x^2+7x-20, 8x^2-42x+27
This video explains how to factor a trinomial in the form ax^2+bx+c when a is not 1 and no common factors other than 1. | 311 | 1,219 | {"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} | 3.671875 | 4 | CC-MAIN-2019-09 | longest | en | 0.771441 |
http://blogs.msdn.com/b/webngram/archive/tags/probability/ | 1,455,188,854,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701161942.67/warc/CC-MAIN-20160205193921-00317-ip-10-236-182-209.ec2.internal.warc.gz | 29,057,074 | 6,989 | # Microsoft Web N-Gram
Bringing you web-scale language model data. Web N-Gram is joint project between Microsoft Bing and Microsoft Research.
# Browse by Tags
Tagged Content List
• #### Blog Post: The dirty secret about large-vocabulary hashes
The first step in the n-gram probability lookup process is to covert the input into tokens as discussed in an earlier post . The second step is to covert each token into numerical IDs. The data structure used here is a kind of a hash. You might be asking, why not a trie ? The simple answer is size...
• #### Blog Post: Well, do ya, P(<UNK>)?
Today we'll do a refresher on unigrams and the role of the P(<UNK>). As you recall, for unigrams, P(x) is simply the probability of encoutering x irrespective of words preceding it. A naïve (and logical) way to compute this would be to simply take the number of times x is observed and divide...
• #### Blog Post: The messy business of tokenization
So what exactly is a word, in the context of our N-Gram service? The devil, it is said, is in the details. As noted in earlier blog entries, our data comes straight from Bing. All tokens are case-folded and with a few exceptions, all punctuation is stripped. This means words like I'm or didn't are...
• #### Blog Post: Wordbreakingisacinchwithdata
For the task of word-breaking, many different approaches exist. Today we're writing about a purely data-driven approach, and it's actually quite straightforward — all we do is a consider every character boundary as a potential for a word boundary, and compare the relative joint probabilities, with...
• #### Blog Post: Generative-Mode API
In previous posts I wrote how the Web N-Gram service answers the question: what is the probability of word w in the context c ? This is useful, but sometimes you want to know: what are some words {w} that could follow the context c ? This is where the Generative-Mode APIs come in to play. Examples...
• #### Blog Post: Language Modeling 102
In last week's post, we covered the basics of conditional probabilities in language modeling. Let's now have another quick math lesson on joint probabilities. A joint probability is useful when you're interested in the probability of an entire sequence of words. Here I can borrow an equation from...
• #### Blog Post: Language Modeling 101
The Microsoft Web N-Gram service, at its core, is a data service that returns conditional probabilities of words given a context. But what does that exactly mean? Let me explain. Conditional probability is usually expressed with a vertical bar: P(w|c). In plain English you would say: what is...
Page 1 of 1 (7 items) | 586 | 2,635 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-07 | latest | en | 0.907146 |
https://www.emathhelp.net/pt/grams-to-ounces/ | 1,652,776,297,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517018.29/warc/CC-MAIN-20220517063528-20220517093528-00791.warc.gz | 875,032,005 | 6,011 | # Grams to ounces
This free conversion calculator will convert grams to ounces, i.e. g to oz. Correct conversion between different measurement scales.
A fórmula é $M_{\text{oz}} = 0.035714285714286 M_{\text{g}}$, onde $M_{\text{g}} = 15$.
Portanto, $M_{\text{oz}} = 0.535714285714286$.
Resposta: $15 \text{g} = 0.535714285714286 \text{oz}$. | 123 | 344 | {"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} | 2.6875 | 3 | CC-MAIN-2022-21 | latest | en | 0.413897 |
https://www.studysmarter.us/textbooks/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th/the-electric-field/q-65-in-problems-63-through-66-you-are-given-the-equations-u/ | 1,669,662,982,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710534.53/warc/CC-MAIN-20221128171516-20221128201516-00492.warc.gz | 1,062,558,875 | 22,333 | Suggested languages for you:
Q. 65
Expert-verified
Found in: Page 657
### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651
# In Problems 63 through 66 you are given the equation(s) used to solve a problem. For each of these a. Write a realistic problem for which this is the correct equation(s). b. Finish the solution of the problem
(a) A disc with a radius of and a charge of that is uniformly distributed. Discover an equation for the strength of the electric field at a point on the axis at a distance from the center.
(b) The solution is
See the step by step solution
## Step 1 : Given information and formula used
Given :
Theory used :
The electric field due to a uniformly charged disc at a point very distant from the surface of the disc is given by:
( is the surface charge density on the disc)
## Step 2 : Writing a realistic problem and finding the solution of the problem
(a) Realistic problem :
Given a disc with a radius of and a charge of that is uniformly distributed. Discover an equation for the strength of the electric field at a point on the axis at a distance from the center. If
(b) Solution :
From the previous expression, finding the value of z : | 296 | 1,300 | {"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} | 2.96875 | 3 | CC-MAIN-2022-49 | latest | en | 0.871386 |
http://jasabroadcast.xyz/ratio-and-proportion-worksheets | 1,560,680,985,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998100.52/warc/CC-MAIN-20190616102719-20190616124719-00391.warc.gz | 85,154,155 | 7,699 | # Ratio And Proportion Worksheets
Posted on November 15, 2017 by CassieCanchola
Ratio And Proportion Worksheets - Image Results More Ratio And Proportion Worksheets images. Ratio And Proportion Worksheets Ratio and Proportion Worksheets - HelpingWithMath.com A listing of ratio and proportion worksheets that are available on HelpingWithMath.com. Each worksheet is printable and includes the option to show answers.
Source: passyworldofmathematics.com
Ratio And Proportion Worksheets - Image Results More Ratio And Proportion Worksheets images. Ratio and Proportion Worksheets - HelpingWithMath.com A listing of ratio and proportion worksheets that are available on HelpingWithMath.com. Each worksheet is printable and includes the option to show answers.
Free proportion worksheets for grades 6, 7, and 8 Create proportion worksheets to solve proportions or word problems (e.g. speed/distance or cost/amount problems). Available both as PDF and html files. Other options include using whole numbers only, numbers with a certain range, or numbers with a certain number of decimal digits. Ratio Worksheets Ratio worksheets have numerous skills to enhance the numerical ability of children. These worksheets include part-to-part ratios, part-to-whole ratios, equivalent ratio, finding a part of the ratio from a whole number and vice versa.
Ratio Worksheets | Ratio Worksheets for Teachers Ratio Worksheets Ratio Worksheets for Teachers. Here is a graphic preview for all of the Ratio Worksheets. You can select different variables to customize these Ratio Worksheets for your needs. The Ratio Worksheets are randomly created and will never repeat so you have an endless supply of quality Ratio Worksheets to use. RATIO AND PROPORTION WORKSHEETS WITH ANSWERS About "Ratio and proportion worksheets with answers" Ratio and Proportion Worksheets with Answers : Worksheet given in this section is much useful to the students who would like to practice problems on ratio and proportion.
Videos for Ratio And Proportion Worksheets See more videos for Ratio And Proportion Worksheets. Ratio and Proportions Worksheets Ratios and Proportions Worksheets What's the Difference Between a Ratio and a Proportion? Many people think they are one and the same. The fact is they are very different. A ratio is s a fraction like 3/4. It can come many forms such as 3 out of 4 equal parts or 3:4, but fundamentally it is a fraction.
Gallery of Ratio And Proportion Worksheets | 483 | 2,464 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-26 | latest | en | 0.846411 |
https://oeis.org/A277679 | 1,591,182,941,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347432521.57/warc/CC-MAIN-20200603081823-20200603111823-00590.warc.gz | 482,065,294 | 4,259 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A277679 Start with 1,2,3,4,5,6,.... For n >=1, remove the first n terms and reverse the remaining terms n+1 at a time. Concatenate the terms removed. (See the example.) 2
1, 3, 2, 7, 4, 5, 13, 6, 9, 8, 17, 10, 11, 14, 15, 23, 16, 19, 12, 25, 18, 33, 26, 27, 20, 21, 24, 31, 49, 32, 39, 22, 29, 28, 35, 34, 53, 36, 43, 30, 37, 40, 41, 50, 51, 59, 52, 55, 42, 45, 38, 61, 44, 67, 54, 85, 68, 69, 62, 63, 46, 47, 56, 57, 60, 77, 95 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS This is a permutation of the natural numbers, with inverse permutation A277680. LINKS Clark Kimberling, Table of n, a(n) for n = 1..10000 EXAMPLE Remove 1 from A000027, leaving 2,3,4,5,6,7,8,...; reverse these 2 at a time, leaving 3,2,5,4,7,6,9,8,... Remove the first 2 terms and reverse the rest 3 at a time, leaving 7,4,5,8,9,6,13,10,11,14,15,12,... Remove the first 3 terms, and so on. The removed terms, taken in order, are 1,3,2,7,4,5,... MATHEMATICA x = Range[500]; NestWhile[# + 1 &, 1, (t = 1/2 # (1 + #); x = Flatten[{Take[x, t], Map[Reverse, Partition[Drop[x, t], # + 1]]}]; Length[x] > t) &]; x (* A277679 *) Take[Ordering[#], Position[Differences[Sort[#]], Except[1]][[2]][[1]]]&[x] (* A277680 *) (* Peter J. C. Moses, Nov 13 2016 *) CROSSREFS Cf. A000027, A277680, A007062, A057030. Sequence in context: A154438 A194071 A194104 * A108644 A194011 A303763 Adjacent sequences: A277676 A277677 A277678 * A277680 A277681 A277682 KEYWORD nonn,easy AUTHOR Clark Kimberling, Nov 14 2016 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified June 3 07:14 EDT 2020. Contains 334799 sequences. (Running on oeis4.) | 778 | 2,033 | {"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} | 3.484375 | 3 | CC-MAIN-2020-24 | latest | en | 0.717132 |
webhelp.albany.k12.ny.us | 1,566,216,475,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314732.59/warc/CC-MAIN-20190819114330-20190819140330-00171.warc.gz | 701,003,954 | 22,436 | Math Sites
Note to all: This list is by no means all-inclusive. I was able to find other folks’ lists, add them to mine and go from there. All of these links work as of this writing but as we all know often folks move sites and don’t leave a forwarding address. Special thanks to all of the teachers who sent sites to add to this list. Happy Math Surfing!
Beginning Math Activities
· ( + and - problems with numbers 1-15) http://www.bbc.co.uk/schools/numbertime/games/test.shtml
· ( interactive number games for addition and subtraction)
Number Recognition/Sequencing
· (Recognition activity for numbers 1-10)
· Mend the Number Square ( Place missing numbers on 1-100 grid)
· (matching game)
· (1 player or 2 player activity.. works like chutes and ladders)
Number Recognition/Sequencing
§ Bunny Count ( count and match numbers and characters)
§ One False Move (sequence numbers from lowest to highest)
§ (guess number with high low clues)
Number Recognition/Counting/Sequencing
· Count Your Chickens ( counting activity)
· (what comes next... up to 10)
· ( make a train and count the cars)
· (click on grade school level)
· Base 10 Count (group ones into ten... see what number is made)
Facts / Whole Numbers / Number Line
Activities with + - * / and whole numbers
Power Football (DECIMAL fact [ +, -, *, / ] drill)
( practice with + - * / squares, cubes )
Change Maker ( work with \$1 or \$100, pick coins / bills for change)
(14 different games to help with multiplication facts)
( interactive number games for addition and subtraction) http://www.bbc.co.uk/schools/laac/index.shtml
Math Mayhem by LearningPlanet ( +, -, *, / timed practice)
Camel Times Tables by BBC Education Co. (learn times tables facts with games) http://www.bbc.co.uk/schools/ks1bitesize/numeracy/multiplication/
(race for time as you practice your facts) http://www.aplusmath.com/games/matho/MultMatho.html
Spacey Math by PlanetLearning ( fact [ +, -, *, / ] drill) http://www.learningplanet.com/sam/sm/index.asp
( basic operation activities) http://www.aplusmath.com/games/index.html
Coloring Book Math ( color online doing math problems) http://www.surfnetkids.com/games/colorsums.htm
Math 4 Kids http://www.edu4kids.com/
Math Activities at Syvum (Helps build basic skills) http://www.syvum.com/online/math.html
A+ Math Flashcards( math flashcards and math games for basic operations) http://www.aplusmath.com/flashcards/index.html
The ArithmAttack (see how many problems can you solve in 60 seconds) http://www.dep.anl.gov/aattack.htm
Factors/ Multiples/ Primes/ Powers/ Triangular Numbers
(ILLUMINATIONS NCTM) [ game of skill with factors and multiples] http://illuminations.nctm.org/ActivityDetail.aspx?ID=29
(ILLUMINATIONS NCTM) [ game of skill with factors] http://illuminations.nctm.org/ActivityDetail.aspx?ID=12
by BCC Maths File (find factors, multiples, primes, triangular numbers, and/or powers on a grid)
Fractions
1. Equivalent Fractions
· Activity by LearningPlanet ( match equivalent fractions before time runs out) http://www.learningplanet.com/sam/ff/index.asp
· (Illuminations NCTM)[ move game pieces to other side of board) http://illuminations.nctm.org/ActivityDetail.aspx?ID=80
· Fresh Baked Fractions Activity by Funbrain( find fraction that is not equivalent to the rest)
· Activity by Funbrain( addition and subtraction of fractions)
1. Fraction/Decimal Relationships
· Activity by BBC Maths Files ( match fractions to decimals and %) http://www.bbc.co.uk/education/mathsfile/shockwave/games/saloonsnap.html
· Activity (Click on Fraction/Decimalby Funbrain (Match Fractions to Decimal Equivalent)
· Fraction Decimal Conversion ( JAVA games matching fractions to equivalent decimals) http://www.quia.com/jg/65724.html
· All About Fractions by AAA Math (Explanation, interactive practice and challenge games about fractions) http://www.quia.com/jg/65724.html
· by Dositey) http://www.dositey.com/math/mistery2.html
Geometry/ Area / Perimeter
Gallery of Interactive Geometry (Some of these activities are very difficult) http://www.geom.umn.edu/apps/gallery.html
Geometry Junkyard (great extra activities) http://www.ics.uci.edu/~eppstein/junkyard/teach.html
General Geometrical Terms
· Geometry Flashcards by APlusMath ( match picture to name) http://www.aplusmath.com/cgi-bin/flashcards/geoflash
· Geometric Terms (JAVA games with common geometric terms) http://www.quia.com/jg/65535.html
· All About Geometry by AAA Math ( Explanation, interactive practice and challenge games about geometry) http://www.aaamath.com/geo.html
Polygons/ Shapes
· Illuminations.. Exploring Geometric Shapes Grades 3-5 (by NCTM) http:// http://illuminations.nctm.org/ActivityDetail.aspx?ID=34
o
o Designer Fractions Activity http://math.rice.edu/~lanius/Patterns/design.html
Angles
1. Introduction to Angles ( JAVA games with common terms related to angles) http://www.quia.com/jg/65822.html
Area/ Perimeter
1. Shape Surveyor Activity by Funbrain( solve simple area/perimeter problems)
2. (lessons and activity) http://www.mathgoodies.com/lessons/vol1/perimeter.html
· Perimeter of Polygon (Lesson with activity) http://www.mathgoodies.com/lessons/vol1/perimeter.html
· Area of Rectangles and Squares (Lesson with activity) http://www.mathgoodies.com/lessons/vol1/area_rectangle.html
· (Lesson with activity) http://www.mathgoodies.com/lessons/vol1/area_rectangle.html
· (Lesson with activity) http://www.mathgoodies.com/lessons/vol1/area_triangle.html
· (Lesson with activity) http://www.mathgoodies.com/lessons/vol1/area_triangle.html
Symmetry/ Rotations/ Reflections/ Transformations
§ by BBC Maths File ( move bathroom tiles by rotating, reflecting, and translating the patterns)
Tangrams
§ Tangrams by KidsCom http://www.kidscom.com/games/tangram/tangram.html
Tessellations
§ by Cool Math http://www.coolmath.com/lesson-tessellations-1.htm
§ What is a Tessellation? http://mathforum.com/sum95/suzanne/whattess.html
Totally Tessellated http://library.thinkquest.org/16661/
Calculus
(an international site with news and activities) http://archives.math.utk.edu/calculus/crol.html
Interactive Calculus Activities (great additional activities) http://www.jcu.edu/math/faculty/spitz/calculus/
Graphing/ Data Collections/ Coordinate Points
1. by Coolmath…it works like the one they use in school) www.coolmath.com/graphit
2. Activity (online activity... create bar graph, pie chart, line graph) http://nces.ed.gov/nceskids/graphing/index.asp
3. Data Picking Activity by BBC Maths File ( make frequency chart and graph data) http://www.bbc.co.uk/education/mathsfile/shockwave/games/datapick.html
4. Planet Hop Activity by BBC Maths File ( find coordinates of object & graph equation) http://www.bbc.co.uk/education/mathsfile/shockwave/games/planethop.html
5. Activity by Funbrain (graphing with coordinate points) http://www6.funbrain.com/co/index.html
6. (graphing game) http://education.jlab.org/topquarkgame/select4.html
7. Activity by LearningPlanet (graph coordinates to capture asteroids) http://www.learningplanet.com/act/lunar/lunar.htm
Integers
1. Line Jumper Activity by Funbrain (solve + and - INTEGER problems using the NUMBER LINE) http://www6.funbrain.com/linejump/index.html
2. Activity by BBC Maths File (activity that paces integers in numerical order) http://www.bbc.co.uk/education/mathsfile/shockwave/games/laddergame.html
3. Mystery Picture with Integers (by Dositey) http://www.dositey.com/addsub/Mystery11.htm
Online Logic Puzzles/Games
1. Cut The Knot http://www.cut-the-knot.org/content.shtml
2. Figure This (NCTM) http://www.figurethis.org/
3. (many logic puzzles for both younger and older children) http://www.figurethis.org/
4. (from Discovery School) http://school.discovery.com/brainboosters/number
Mean, Mode, Median
1. Statistics....Range, Mean, Mode, Median (online lesson and activity)
· Range http://www.mathgoodies.com/lessons/vol8/range.html
· Mean http://www.mathgoodies.com/lessons/vol8/mean.html
· Median http://www.mathgoodies.com/lessons/vol8/median.html
· Mode http://www.mathgoodies.com/lessons/vol8/median.html
1. Train Race Activity by BBC Maths File ( need to use own calculator with this) http://www.bbc.co.uk/education/mathsfile/shockwave/games/train.html
2. AAA Statistics http://www.aaamath.com/B/sta.htm
Measurement/ Weight
1. Measure It Activity by Funbrain ( read a ruler in centimeters or inches) http://www6.funbrain.com/measure/index.html
2. The World of Measurement (information about length, mass, volume, temperature, and time) http://illuminations.nctm.org/LessonsList.aspx?grade=all&standard=4
3. Animal Weigh In Activity by BBC Maths File ( activity that balances weights on a scale)
Missing number in the series / Patterns
1. Activity by Funbrain (guess next number in the SERIES) http://www6.funbrain.com/guess2/index.html
PreAlgebra/ Algebra Activities
1. Mrs Glosser's lesson...(online lesson and activity) http://www.mathgoodies.com/lessons/vol7/order_operations.html
· Writing Algebraic Equations for expressions with variables http://www.mathgoodies.com/lessons/vol7/expressions.html
· Writing Algebraic Equations for word sentences http://www.mathgoodies.com/lessons/vol7/equations.html
1. by AAA Math (Explanation, interactive practice and challenge games about equations) http://www.aaamath.com/equ723-evaluate-1variable.html
2. by AAA Math (Explanation, interactive practice and challenge games about equations) http://www.aaamath.com/equ725-equation1.html
3. Late Delivery Activity by BBC Maths File ( find value of an expression ) http://www.bbc.co.uk/education/mathsfile/shockwave/games/postie.html
4. Equation Match Activity by BBC Maths File ( matching game with = equations) http://www.bbc.co.uk/education/mathsfile/shockwave/games/equationmatch.html
5. Funbrain Magician Activity( solve algebra word problems) http://www6.funbrain.com/guess2/index.html
6. Order of Operations
· Mrs Glosser's online lesson and activity) http://www.mathgoodies.com/lessons/vol7/order_operations.html
· Operation Order Activity by Funbrain (complete equations using 3 numbers) http://www6.funbrain.com/algebra/index.html
1. Activity by Funbrain (guess answer to a word problem) http://www6.funbrain.com/guess2/index.html
2. Absurd Math.. Interactive pre-algebra problem solving games Challenging
· Nomean City Activity http://www.learningwave.com/abmath/absurd4/index.htm
· Dr. Plenobius Activity http://www.learningwave.com/abmath/drp99/index.html
· The 7th Floor Activity http://www.learningwave.com/abmath/hotel/index4.html
· Airtight College Activity http://www.learningwave.com/abmath/ep2/index_new.html
Place Value
1. Wise Up... Our Number System ( BBC) http://www.bbc.co.uk/schools/ks2bitesize/maths/number/number_system/play.shtml
2. Activity by Funbrain (click on digit with the given place value) http://www6.funbrain.com/tens/index.html
3. Place Value Game http://education.jlab.org/placevalue/index.html
4. by AAA Math ( Explanation, interactive practice and challenge games about place value)
Probability
1. Activity by BBC Maths File ( guess probability of catching fish in the tank) http://www.bbc.co.uk/education/mathsfile/shockwave/games/fish.html
2. Mrs Glosser's lesson on(online lessons and activity)
3. (National Center for Education Statistics) http://nces.ed.gov/NCESKids/Probability/
1. Rounding Off Numbers Activity by BBC Maths File http://www.bbc.co.uk/education/mathsfile/shockwave/games/roundoff.html
2. Place Value Puzzler Activity by Funbrain (type in rounded version of number) http://www6.funbrain.com/tens/index.html
3. Rounding Decimals to Whole Numbers Game
Games and Mind Bender Puzzles
1. Brain Binders [ paper folding activites] http://www.teachnet.com/brainbinders/index.html
2. ( by Cool Math) http://www.coolmath4kids.com/0-thinking-jigsaw-puzzles.html
3. CoolMath Games http://www.spikesgamezone.com/
4. ( Try Numberland and Games) http://www.mathsyear2000.org/
5. FunBrain Number Games http://www.funbrain.com/numbers.html
6. Fun Math Memory Game http://www.funbrain.com/match/index.html
7. ( GREAT site for math problems) http://www.cut-the-knot.org/index.shtml
8. Click on color sequence. See how long you can go ) http://www.mathsnet.net/cruncher/simon.html
Metric
1. Metric Matters (Thinkquest tutorial about metric system) http://tqjunior.thinkquest.org/3804/
2. Measure It Activity by Funbrain ( read a ruler in centimeters)
Money
1. Change Maker Activity by Funbrain ( work with \$1 or \$100, pick coins / bills for change) http://www.funbrain.com/cashreg/index.html
2. Counting Change Game http://www.quia.com/jg/65704.html
3. H.I.P Pocket Change ( by US Mint Dept) GREAT SITE!!! http://www.usmint.gov/kids/index.cfm?flash=yes
· Games http://www.usmint.gov/kids/index.cfm?flash=yes&Filecontents=/kids/games/index.cfm
· Cartoons http://www.usmint.gov/kids/index.cfm?flash=yes&Filecontents=/kids/cartoons/index.cfm
· Time Machine http://www.usmint.gov/kids/index.cfm?flash=yes&Filecontents=/kids/timemachine/index.cfm
· Coin News http://www.usmint.gov/kids/index.cfm?flash=yes&Filecontents=/kids/coinnews/index.cfm
· Teacher Help http://www.usmint.gov/kids/index.cfm?flash=yes&Filecontents=/kids/teachers/index.cfm
1. Money by AAA Math http://www.aaamath.com/mny.html
2. (ThinkQuest project to learn about money and investing) http://library.thinkquest.org/3096/
Math in Daily Life
1. Math in Daily Life http://www.learner.org/exhibits/dailymath/
2. by Cool Math http://www.coolmath.com/careers.htm
Math Problems of the Week/ Word Problems
· Brain Benders by Cool Math http://www.coolmath4kids.com/math_puzzles/index.html
· Figure This ( Math index by National Science Foundation & US Dept. of Ed) http://figurethis.org/challenges/math_index.htm
· Figure This (problems for Middle School students) http://figurethis.org/challenges/challenge_index.htm
· Math Forum's http://mathforum.org/pows/
· Math Forum's Middle School Challenges
·
§
Famous Mathematicians and History of Mathematics ( over 1000 biographies) http://www-groups.dcs.st-and.ac.uk/~history/
Vocabulary
Teacher Lesson Plan Help:
§ Math Education http://teacherpathfinder.org/School/Subjects/Math/mathed.html
§ Math Forum (The Math Forum, an online community funded in part by the National Science Foundation and hosted by Swarthmore College) http://www.forum.swarthmore.edu
§ AAA Teacher Resource Page http://www.aaamath.com/lessonplans.html
§ http://www.math.com/teachers.html
§
§ Math Resources by topic http://mathforum.org/math.topics.html
§ Internet Mathematics Library http://mathforum.org/library/toc.html
§ http://mathforum.org/workshops/usi/dataproject/usi.elemlessons.html
§ Geometry Through Art http://mathforum.org/~sarah/shapiro/index.html
§ Suzanne's Math Lessons [web units, interactive lessons, technology help]
§
§ Eisenhower National Clearinghouse (Explore this site for the best selection of K-12 mathematics and science resources on the Internet) www.enc.org
§ Math Ideas (Math ideas/lessons for a variety of topics) http://www.teachingideas.co.uk/maths/contents.htm
§ (Math, Science, Technology Education U of Illinois) http://www.mste.uiuc.edu/html.f/resource.xml
§ http://www.planemath.com/
§ (NCTM) http://standards.nctm.org/
§ http://standards.nctm.org/
§ Study Works Online (grades 7-12) http://www.studyworksonline.com/
Trigonometry
Explore Math (this site is definitely for higher level math skills but all the activities are interactive!) | 4,055 | 15,944 | {"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-2019-35 | latest | en | 0.853169 |
https://ras.football/2019/12/31/greg-jones-ras-2/ | 1,670,339,890,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00664.warc.gz | 506,140,490 | 19,454 | # Greg Jones RAS
### Greg Jones RAS
Greg Jones was drafted by Giants with pick 185 in round 6 in the 2011 NFL Draft out of Michigan State.
He recorded a Relative Athletic Score of 3.88, out of a possible 10.0. RAS is a composite metric on a 0 to 10 scale based on the average of all of the percentile for each of the metrics the player completed either at the Combine or pro day.
He had a recorded height of 6001 that season, recorded as XYYZ where X is feet, YY is inches, and Z is eighths of an inch. That correlates to 6 feet, 0 and 1/8 of an inch or 72.125 inches, or 183.1975 centimeters. This correlates to a 1.25 score out of 10.0.
He recorded a weight of 242 in pounds, which is approximately 110 kilograms. This correlates to a 6.87 score out of 10.0.
Based on his weight, he has a projected 40 yard dash time of 4.77. This is calculated by taking 0.00554 multiplied by his weight and then adding 3.433.
At the Combine, he recorded a 40 yard dash of 4.84 seconds. This was a difference of 0.07 seconds from his projected time. This forty time correlates to a 3.33 score out of 10.0.
Using Bill Barnwell’s calculation, this Combine 40 time gave him a Speed Score of 88.2.
At his pro day, he recorded a 40 yard dash of 4.72 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.
Using Bill Barnwell’s calculation, his Combine 40 time gave him a Speed Score of 1.72.
The time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at the Combine was 2.04 seconds. Over 20 yards, we can calculate his speed in yards per second to 9.8. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 29.41. Breaking it down further, we can calculate his speed in inches per second to 352.94. Knowing the feet per second of 29.41, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 20.1 in the last 20 yards of his run.
The time traveled between the 20 and 40 yard lines is known as the Flying Twenty. As the distance is also known, we can calculate the player’s speed over that distance. The time he traveled the last twenty yards at his pro day was 2.0 seconds. Over 20 yards, we can calculate his speed in yards per second to 10.0. Taking into account the distance in feet (60 feet), we can calculate his speed in feet per second to 30.0. Breaking it down further, we can calculate his speed in inches per second to 360.0. Knowing the feet per second of 30.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 20.5 in the last 20 yards of his run.
At the Combine, he recorded a 20 yard split of 2.8 seconds. This correlates to a 3.78 score out of 10.0.
At his pro day, he recorded a 20 yard spit of 2.72 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.
We can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at the Combine was 1.09 seconds. Over 10 yards, we can calculate his speed in yards per second to 9.17. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 27.52. Breaking it down further, we can calculate his speed in inches per second to 330.28. Knowing the feet per second of 27.52, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 18.8 in the second ten yards of his run.
We can calculate the speed traveled over the second ten yards of the 40 yard dash easily, as the distance and time are both known. The time he traveled the second ten yards at his pro day was 1.04 seconds. Over 10 yards, we can calculate his speed in yards per second to 9.62. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 28.85. Breaking it down further, we can calculate his speed in inches per second to 346.15. Knowing the feet per second of 28.85, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 19.7 in the second ten yards of his run.
At the Combine, he recorded a 10 yard split of 1.71 seconds. This correlates to a 2.47 score out of 10.0.
At his pro day, he recorded a 10 yard split of 1.68 seconds. Because he also recorded this metric at the Combine, his pro day did not count towards his RAS.
The time he traveled the first ten yards at the Combine was 1.71 seconds. Over 10 yards, we can calculate his speed in yards per second to 6.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 18.0. Breaking it down further, we can calculate his speed in inches per second to 211.0. Knowing the feet per second of 18.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 12.3 in the first ten yards of his run.
The time he traveled the first ten yards at his pro day was 1.68 seconds. Over 10 yards, we can calculate his speed in yards per second to 6.0. Taking into account the distance in feet (30 feet), we can calculate his speed in feet per second to 18.0. Breaking it down further, we can calculate his speed in inches per second to 214.0. Knowing the feet per second of 18.0, we can calculate the approximate miles per hour by multiplying that value by 0.681818 to give us a calculated MPH of 12.3 in the first ten yards of his run.
At the Combine, he recorded a bench press of 21 repetitions of 225 pounds. This correlates to a 5.72 score out of 10.0.
At the Combine, he recorded a vertical jump of 31.5 inches. This correlates to a 3.47 score out of 10.0.
At the Combine, he recorded a broad jump of 909, which is recorded as FII or FFII . where F is feet and I is inches. This correlates to a 7.04 score out of 10.0.
At the Combine, he recorded a 5-10-5 or 20 yard short shuttle of 4.27 seconds. This correlates to a 6.61 score out of 10.0.
At the Combine, he recorded a 3 cone L drill of 7.21 seconds. This correlates to a 5.17 score out of 10.0.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,685 | 6,460 | {"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} | 3.203125 | 3 | CC-MAIN-2022-49 | latest | en | 0.980597 |
https://thewanderingengineer.com/2017/08/09/monte-carlo-repeated-dice-404-problem/ | 1,685,696,911,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648465.70/warc/CC-MAIN-20230602072202-20230602102202-00623.warc.gz | 620,453,860 | 30,620 | # Monte Carlo – Repeated Dice Problem from 404 podcast
Last week a podcast I listen to, the 404, discussed a math problem where you roll six 20-sided die and count how often you get a situation where at least one dice matches another dice. They discussed the math a little and came to the conclusion that it happens far more than you’d think. I thought it’d make a good monte carlo programming exercise so I’ve done just that. Below, you’ll find my C code (though it’s not great) and results for 2-20 dice.
It’s important to note that the code below isn’t particularly good. Ideally, you’d sort the dice first and then count. This would give you the ability to also check for multiple doubles or for triples and up. It’d also make it easier to expand for future problems. I did not do any of that. I wrote this so that it’d be fast to program, and left all that out to keep it simpler. But note that because I’m not sorting, I’m checking for doubles in a very inefficient method that gets worse the more dice you add. It’s fine for six dice but for more, it can start to take awhile. To do 100 million rolls, with six dice it takes 11s but with 20 dice, it takes 107s. So not idea, but fast to write. 🙂
```#include <stdio.h>
#include <stdlib.h>
#include <time.h>
// Main function was given
int main()
{
int i, j, k;
int maxDiceValue = 20;
int numIterations = 100000000;
int numDice = 6;
int roll;
int rolls[numDice];
int isMatch;
int numMatches=0;
// See RNG with time to get unique results each time
srand(time(NULL));
for(k=0; k<numIterations; k++)
{
// Fill array with random dice rolls
for(i=0; i<numDice; i++)
{
rolls[i] = 1 + rand() % maxDiceValue;
}
// Find if there is any match present in the dice
// This is the absolute lasiest way to do it and it is pretty
// bad code since it isn't easy to expand to other cases and
// it's pretty inefficent. But easy to quick to code.
// Iterate through all the dice and if there's a match, change
// isMatch variable to 1. After iterating, check this variable
isMatch = 0;
for(i=0; i<numDice-1; i++)
{
for(j=i+1; j<numDice; j++) { if(rolls[i] == rolls[j]) isMatch=1; } } if(isMatch > 0)
numMatches++;
}
// Print results
// Note that we multiply by 100.0 first to convert to double
printf("%d matches in %d iterations: %2.2lf%% \n", numMatches, numIterations, 100.0*numMatches/numIterations);
return 0;
}
```
The result for 6 dice is 56.4%. Below is a graph of the percentage for rolling 2 through 20 dice.
I know this is a dumb example but it was an interesting ten minute problem. Thoughts? Questions? Comments? Leave them below. | 713 | 2,618 | {"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} | 3.53125 | 4 | CC-MAIN-2023-23 | latest | en | 0.855266 |
https://forum.powerscore.com/lsat/viewtopic.php?t=11775&p=27924& | 1,558,365,183,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256040.41/warc/CC-MAIN-20190520142005-20190520164005-00072.warc.gz | 473,895,328 | 6,788 | ## #21 - Designer: Any garden and adjoining living room that
PowerScore Staff
Posts: 6662
Joined: Wed Feb 02, 2011 4:19 pm
Points: 3,335
Complete Question Explanation
Must Be True. The correct answer choice is (D)
In this stimulus, a designer states that any garden and adjoining living room separated by sliding glass doors can become a single space visually. When the doors may be open, as in the summer, the “single space” will be created if it does not already exist. If the visual single space does exist already, this effect will be magnified:
Sliding doors open create single space or intensify pre-existing single space
Even during the colder months, the effect will remain, if the garden is coordinated with the room and contributes a strong visual interest on its own:
Garden coordinated and contributes strong visuals single space effect remains
The question stem asks which answer is most strongly supported by the designer’s statements, so we should locate the answer choice in accordance with the above conditional rules.
Answer choice (A): This answer choice provides the following conditional statement regarding a room with the sliding glass doors closed:
Garden coordinated contributes strong visual interest
This statement runs contrary to the information provided in the stimulus, referenced above, which states in the winter, when the door is closed, the single space effect will continue if the garden is coordinated with the room and if the garden contributes a strong visual interest of its own:
Garden coordinated and contributes strong visuals single space effect remains
This answer choice incorrectly characterizes the strong visual interest single space effect as the sole necessary condition for being coordinated with the room, so this answer choice is incorrect.
Answer choice (B): This answer choice provides the following incorrect conditional reasoning:
Single space effect garden well coordinated
This answer choice is contrary to the conditional reasoning provided in the stimulus:
Garden coordinated and contributes strong visuals single space effect remains
Since this answer choice reverses the sufficient and necessary conditions, and leaves out any reference to contribution of a strong visual interest, this choice is incorrect.
Answer choice (C): This answer choice is also contrary to what is said in the stimulus, which tells us that a visual single room effect can be created if the doors are open. The designer also states that the effect can be intensified by opening the doors:
Sliding doors open create single space or intensify pre-existing single space
Since we are told that the open sliding glass doors have the potential to intensify a single space effect, we know that they are not always required to create such an effect.
Answer choice (D): This is the correct answer choice. According to the stimulus, the contribution of a strong visual interest doesn’t even come into play in the summer, during which the opening of the sliding doors creates a single space effect if it didn’t already exist. If this effect was already present, opening the doors intensifies it:
Sliding doors open create single space or intensify pre-existing single space
Because a garden can visually merge with an adjoining living room and form a single space in the summer, even if it does not contribute a strong visual interest of its own, this answer choice is correct.
Answer choice (E): The designer provides conditional reasoning regarding sliding glass doors. The first rule concerns open sliding doors, and the author points out that this may happen in the summer. This does not imply that the same course of action in the winter would not have the same results, so this answer choice is incorrect. | 715 | 3,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-22 | longest | en | 0.888536 |
https://www.coursehero.com/file/6789582/RainfallHomeworkSolution/ | 1,513,543,730,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948597485.94/warc/CC-MAIN-20171217191117-20171217213117-00450.warc.gz | 708,273,025 | 31,104 | RainfallHomeworkSolution
# RainfallHomeworkSolution - Physical Hydrology 2/e S.L...
This preview shows pages 1–2. Sign up to view the full content.
Physical Hydrology , 2/e S.L. Dingman ArealPptnExrSM.doc COMPUTATION OF AREAL PRECIPITATION FROM POINT VALUES Arithmetic, Thiessen, Eyeball Isohyetal Methods Attached is a map showing the boundaries of a fictitious watershed and the locations of nine rain gages. The numbers are the gage catches for a particular storm in mm. 1. Compute the average rainfall on the watershed using the following three methods: (a) arithmetic average; (b) eyeball isohyetal; (c) Thiessen polygon. Use the grid squares to estimate areas. (a) In computing the arithmetic average, one would probably use only those gages that are within the watershed boundary. Equation (4-13) then gives P ˆ = (28 + 37 + 68 + 114 + 75 + 126)/6 = 74.67 75 mm. If all gages are used, we get P ˆ = (28 + 37 + 68 + 114 + 75 + 126 + 16 + 38 + 44)/9 = 59.56 60 mm. (b) The attached figure shows a plausible isohyetal configuration drawn by eye.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### Page1 / 4
RainfallHomeworkSolution - Physical Hydrology 2/e S.L...
This preview shows document pages 1 - 2. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online | 397 | 1,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2017-51 | latest | en | 0.795206 |
http://en.wikipedia.org/wiki/Bargmann%e2%80%93Wigner_equations | 1,405,251,897,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776438008.40/warc/CC-MAIN-20140707234038-00089-ip-10-180-212-248.ec2.internal.warc.gz | 42,961,363 | 31,801 | # Bargmann–Wigner equations
This article uses the Einstein summation convention for tensor/spinor indices, and uses hats for quantum operators.
In relativistic quantum mechanics and quantum field theory, the Bargmann–Wigner equations (or BW equations or BWE) are relativistic wave equations which describe free particles of arbitrary spin j, an integer for bosons (j = 1, 2, 3 ...) or half-integer for fermions (j = 12, 32, 52 ...). The solutions to the equations are wavefunctions, mathematically in the form of multi-component spinor fields. The spin quantum number is usually denoted by s in quantum mechanics, however in this context j is more typical in the literature (see references).
They were proposed by Valentine Bargmann and Eugene Wigner in 1948,[1] using Lorentz group theory,[2] and building on the work of those who pioneered quantum theory within the first half of the twentieth century.[3][4]
## Origin from the Dirac equation
Main article: Dirac equation
For reference, the Dirac equation is summarized below. It is the basis for building relativistic wave equations with wavefunctions of higher spin.
The covariant form of the Dirac equation for an uncharged particle is:[5]
$(-\gamma^\mu \hat{P}_\mu + mc)\Psi = 0 \,,$
(1)
where Ψ = Ψ(r, t) is a rank-1 4-component spinor field, a function of the particle's position r and time t, with components ψα = ψα(r, t) in which α is a bispinor index that takes values 1, 2, 3, 4. Further, γμ = (γ0, γ) are the gamma matrices, and
$\hat{P}_\mu = i\hbar \partial_\mu$
is the 4-momentum operator. The operator constituting the entire equation, (−γμPμ + mc) = (−γμμ + mc), is a 4 × 4 matrix, because of the γμ matrices, and the mc term scalar-multiplies the 4 × 4 identity matrix (usually not written for simplicity). Explicitly, in the Dirac representation of the gamma matrices:[3]
\begin{align} -\gamma^\mu \hat{P}_\mu + mc & = -\gamma^0 \frac{\hat{E}}{c} - \boldsymbol{\gamma}\cdot(-\hat{\mathbf{p}}) + mc \\ & = -\begin{pmatrix} I_2 & 0 \\ 0 & -I_2 \\ \end{pmatrix}\frac{\hat{E}}{c} + \begin{pmatrix} 0 & \boldsymbol{\sigma}\cdot\hat{\mathbf{p}} \\ -\boldsymbol{\sigma}\cdot\hat{\mathbf{p}} & 0 \\ \end{pmatrix} + \begin{pmatrix} I_2 & 0 \\ 0 & I_2 \\ \end{pmatrix}mc \\ & = \begin{pmatrix} -\hat{E}/c+mc & 0 & \hat{p}_z & \hat{p}_x - i\hat{p}_y \\ 0 & -\hat{E}/c+mc & \hat{p}_x + \hat{p}_y & -\hat{p}_z \\ -\hat{p}_z & -(\hat{p}_x - i\hat{p}_y) & \hat{E}/c+mc & 0 \\ -(\hat{p}_x + i\hat{p}_y) & \hat{p}_z & 0 & \hat{E}/c+mc \\ \end{pmatrix} \end{align}
where σ = (σ1, σ2, σ3) = (σx, σy, σz) is a vector of the Pauli matrices, E is the energy operator, p = (p1, p2, p3) = (px, py, pz) is the 3-momentum operator, I2 denotes the 2 × 2 identity matrix, the zeros (in the second line) are actually 2 × 2 blocks of zero matrices.
The Dirac equation (1) can be written as a coupled set of equations:
$(-\hat{E} + mc )\psi_{1,2} = (-\boldsymbol{\sigma}\cdot\hat{\mathbf{p}})\psi_{3,4}$
(1A)
$(\hat{E} + mc )\psi_{3,4} = (\boldsymbol{\sigma}\cdot\hat{\mathbf{p}})\psi_{1,2}$
(1B)
where
$\Psi = \begin{pmatrix} \psi_{1,2} \\ \psi_{3,4} \\ \end{pmatrix}\,\quad \psi_{1,2} = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \end{pmatrix}\,\quad \psi_{3,4} = \begin{pmatrix} \psi_3 \\ \psi_4 \\ \end{pmatrix}\,.$
One 2-component spinor ψ1,2 describes the spin-1/2 fermion, the other ψ3,4 describes the antifermion.
For a charged particle moving in an electromagnetic field, minimal coupling can be introduced:
$[-\gamma^\mu (i\hbar \partial_\mu - eA_\mu)+mc]\Psi = 0$
(1C)
where e is the electric charge of the particle and Aμ = (A0, A) is the electromagnetic four-potential.
## BW equations
For a free particle of spin j, the BW equations are a set of 2j coupled linear partial differential equations, each with a similar mathematical form to the Dirac equation.
### Uncharged massive particles
For a free particle with zero electric charge, the full set of equations are:[3][4][6]
\begin{align} & (-\gamma^\mu \hat{P}_\mu + mc)_{\alpha_1 \alpha_1'}\psi_{\alpha'_1 \alpha_2 \alpha_3 \cdots \alpha_{2j}} = 0 \\ & (-\gamma^\mu \hat{P}_\mu + mc)_{\alpha_2 \alpha_2'}\psi_{\alpha_1 \alpha'_2 \alpha_3 \cdots \alpha_{2j}} = 0 \\ & \qquad \vdots \\ & (-\gamma^\mu \hat{P}_\mu + mc)_{\alpha_{2j} \alpha'_{2j}}\psi_{\alpha_1 \alpha_2 \alpha_3 \cdots \alpha'_{2j}} = 0 \\ \end{align}
$(-\gamma^\mu \hat{P}_\mu + mc)_{\alpha_r \alpha'_r}\psi_{\alpha_1 \cdots \alpha'_r \cdots \alpha_{2j}} = 0$
(2)
for r = 1, 2, ... 2j. Again, the operator (−γμPμ + mc) is a 4 × 4 matrix. The wavefunction Ψ = Ψ(r, t) has components
$\psi_{\alpha_1 \alpha_2 \alpha_3 \cdots \alpha_{2j}} (\mathbf{r},t)$
and is now a rank-2j 4-component spinor field, usually symmetric in all bispinor indices, but not necessarily; for example, the spin-0 case is antisymmetric. Each index takes the values 1, 2, 3, or 4, so there are 42j components of the entire spinor field Ψ, although a completely symmetric wavefunction reduces the number of independent components to 2(2j + 1).
Some authors (for example Loide and Saar[4]) use n = 2j, where n is a non-negative integer (thereby j is a half-integer or integer), because this helps remove factors of 2.
The above matrix operator contracts with one bispinor index of Ψ at a time (analogous but not equivalent to matrix multiplication), so some properties of the Dirac equation also apply to the BW equations:
$E^2 = (pc)^2 + (mc^2)^2$
The components for a totally symmetric wavefunction are explicitly:[3]
$\Psi = \begin{pmatrix} \psi_{1 \alpha_2 \alpha_3 \cdots \alpha_{2j}} \\ \psi_{2 \alpha_2 \alpha_3 \cdots \alpha_{2j}} \\ \psi_{3 \alpha_2 \alpha_3 \cdots \alpha_{2j}} \\ \psi_{4 \alpha_2 \alpha_3 \cdots \alpha_{2j}} \\ \end{pmatrix}$
where the indices are selected so that: α2 ≤ α3 ≤ ... α2j.
Unlike the Dirac equation, which can incorporate the electromagnetic field via minimal coupling (1C), the B–W formalism comprises intrinsic contradictions and difficulties when the electromagnetic field interaction is incorporated. In other words, it is not possible to make the change PμPμeAμ.[7][8] An indirect approach to investigate electromagnetic influences of the particle is to derive the electromagnetic four-currents currents and multipole moments for the particle, rather than include the interactions in the wave equations themselves.[9][10]
### Coupled equations
Analogous to (1A) and (1B), the BW equations can be written as a set of coupled equations:
$(-\hat{E} + mc )^{2j} \psi_{1,2}^{[2j]} = (-\boldsymbol{\sigma}\cdot\hat{\mathbf{p}})^{[2j]} \psi_{3,4}^{[2j]}$
(2A)
$(\hat{E} + mc )^{2j} \psi_{3,4}^{[2j]} = (\boldsymbol{\sigma}\cdot\hat{\mathbf{p}})^{[2j]} \psi_{1,2}^{[2j]}$
(2B)
where the notation [2j] denotes the 2j induced spinor or matrix (defined in the next section). Each of ψ1,2 and ψ3,4 has 2j + 1 independent components.
These can be recombined:[3]
$[(E^2 - (mc^2)^2)^{2j} - (\hat{\mathbf{p}}^2)^{2j}]\begin{pmatrix} \psi_{1,2}^{[2j]} \\ \psi_{3,4}^{[2j]} \end{pmatrix} = 0$
(2C)
which upon expanding by the binomial theorem, then factorizing;
$[(E^2 - (mc^2)^2) - (\hat{\mathbf{p}}^2)]\left( (E^2 - (mc^2)^2)^{2j-1} + (E^2 - (mc^2)^2)^{2j-2}\hat{\mathbf{p}}^2 + \cdots (\hat{\mathbf{p}}^2)^{2j-1} \right)\begin{pmatrix} \psi_{1,2}^{[2j]} \\ \psi_{3,4}^{[2j]} \end{pmatrix} = 0$
(2C)
shows that each component of the BW wavefunction also satisfies the Klein–Gordon equation, uniquely. Conversely, the solutions to the Klein–Gordon equation satisfy the BW equations but are not unique.
### Modified gamma matrices
If we define the following Kronecker product (denoted ⊗) of 4 × 4 identity matrices (denoted I4), with the γμ matrix in the rth place of the product,[4]
$\gamma_r^\mu = \underbrace{I_4 \otimes I_4 \otimes \cdots}_{r-1\,\text{matrices}} \gamma^\mu \cdots \otimes I_4$
for r = 1, 2 ... 2j, these equations (2) can also be written:
$(\gamma_r^\mu \hat{P}_\mu - mc )\Psi =0$
(3)
The γrμ matrices have dimension 42j × 42j. The equations are linear, so adding (3) with respect to the r values gives:
$\left(\frac{1}{2j}\sum_{r=1}^{2j}\gamma_r^\mu \hat{P}_\mu - mc \right)\Psi =0$
(3A)
where the factor of 1/2j is inserted because the matrix elements ±1, ±i are added 2j times. Subtracting (3), one r from the next r + 1; the wavefunction satisfies:
$(\gamma_r^\mu - \gamma_{r+1}^\mu)\hat{P}_\mu\psi=0$
(3B)
for r = 1, 2 ... 2j − 1.
### Joos-Weinberg equation
Introducing a 2(2j + 1) × 2(2j + 1) matrix;[11]
$\gamma^{\mu_1 \mu_2 \cdots \mu_{2j}}$
symmetric in any two tensor indices, which generalizes the gamma matrices in the Dirac equation,[3][12] the BW equation takes the form:[13][14]
$[(i\hbar)^{2j}\gamma^{\mu_1 \mu_2 \cdots \mu_{2j}} \partial_{\mu_1}\partial_{\mu_2}\cdots\partial_{\mu_{2j}} + (mc)^{2j}]\Psi = 0$
or
$[\gamma^{\mu_1 \mu_2 \cdots \mu_{2j}} P_{\mu_1}P_{\mu_2}\cdots P_{\mu_{2j}} + (mc)^{2j}]\Psi = 0$
(4)
This is also known as the Joos-Weinberg equation (or JW or JWE), after H. Joos and Steven Weinberg, found in the early 1960s.[3][11]
## Induced matrices
### Definition
The induced matrices[3] arise from the spinor transformation:
$\begin{pmatrix} a & c \\ b & d \\ \end{pmatrix}\begin{pmatrix} \psi_1 \\ \psi_2 \\ \end{pmatrix}=\begin{pmatrix} \chi_1 \\ \chi_2 \\ \end{pmatrix}$
(5)
that is:
\begin{align} a\psi_1 + c\psi_2 &= \chi_1, \\ b\psi_1 + d\psi_2 &= \chi_2. \end{align}
The 2j induced matrix arises by expanding:
$(a\psi_1 + c\psi_2)^{j+m}(b\psi_1 + d\psi_2)^{j-m} = \chi_1^{j+m}\chi_2^{j-m}\,,$
for m = −j, −j + 1, ... j − 1, j, simplifying, then writing the set of equations in matrix form.
### Properties
Two reasons for introducing the induced matrices is the simple correspondence between induced matrices and powers of eigenvalues, and ease of diagonalization.
Eigenvalues
If A is a 2 × 2 matrix, the 2j induced matrix A[2j] has eigenvalues λ1j + mλ2jm for the same m values as above.
Diagonalization
If the transformation AB−1AB holds, then B[2j] will diagonalize A[2j].
### Use in the BW formalism
In the above equations (1A), (1B), (2A), (2B):
$(\boldsymbol{\sigma}\cdot\hat{\mathbf{p}})^{[2j]} = (i |\hat{\mathbf{p}}|)^{2j}e^{-i\pi\mathbf{J}^{(j)}\cdot\mathbf{n}}$
(6)
where matrix indices on the left side are understood to be m, m′ = −j, −j + 1 ... j. The mm′ element of the (2j + 1) × (2j + 1) matrix contains the energy–momentum operators and are given by:
${(\boldsymbol{\sigma}\cdot\hat{\mathbf{p}})^{[2j]}_{mm'} = (-1)^{m'-m}\sum_{r=-\infty}^\infty\frac{(-1)^rp_{-}^j(-\hat{p}_z)^{j-m'-r}\hat{p}_z^{j+m-r}(-p_{+})^{m'-m+r}}{r!(j-m'-r)!(j+m-r)!(m'-m+r)!}\sqrt{(j+m)!(j-m)!(j+m')!(j-m')!}}$
(7)
where n = p/|p| is a unit vector and J(j) = (J(j)1, J(j)2, J(j)3) is the vector of the Pauli matrices for spin s.[15]
The matrix (σ • p)[2j] has eigenvalues ±|p|2j. The degeneracy of the eigenvalues are as follows:
+|p|[2j] −|p|[2j] (j + 1)-fold j-fold (j + ½)-fold (j + ½)-fold
## Lorentz group structure
Under a proper orthochronous Lorentz transformation x → Λx in Minkowski space, all one-particle quantum states ψjσ of spin j with spin z-component σ locally transform under some representation D of the Lorentz group:[11][16]
$\psi(x) \rightarrow D(\Lambda) \psi(\Lambda^{-1}x)$
where D(Λ) is some finite-dimensional representation, i.e. a matrix. Here ψ is thought of as a column vector containing components with the allowed values of σ. The quantum numbers j and σ as well as other labels, continuous or discrete, representing other quantum numbers are suppressed. One value of σ may occur more than once depending on the representation. Representations with several possible values for j are considered below.
The irreducible representations are labeled by a pair of half-integers or integers (A, B). From these all other representations can be built up using a variety of standard methods, like taking tensor products and direct sums. In particular, space-time itself constitutes a 4-vector representation (1/2, 1/2) so that Λ ∈ D'(1/2, 1/2). To put this into context; Dirac spinors transform under the (1/2, 0) ⊕ (0, 1/2) representation. In general, the (A, B) representation space has subspaces that under the subgroup of spatial rotations, SO(3), transform irreducibly like objects of spin j, where each allowed value:
$j = A + B, A + B - 1, ..., |A - B|,$
occurs exactly once.[17] In general, tensor products of irreducible representations are reducible; they decompose as direct sums of irreducible representations.
The representation for the BW equations is the choice:[7]
$D^\mathrm{BW} = \bigotimes_{r=1}^{2j} \left[ D_r^{(1/2,0)}\oplus D_r^{(0,1/2)}\right]\,.$
where each Dr is an irreducible representation. This representation does not have definite spin unless j equals 1/2 or 0. One may perform a Clebsch–Gordan decomposition to find the irreducible (A, B) terms and hence the spin content. This redundancy necessitates that a particle of definite spin j that transforms under the DBW representation satisfies field equations.
For the JW equations the choice is:[7]
$D^\mathrm{JW} = D^{(j,0)}\oplus D^{(0,j)}\,.$
This representation has definite spin j. It turns out that a spin j particle in this representation satisfy field equations too. These equations are very much like the Dirac equations. It is suitable when the symmetries of charge conjugation, time reversal symmetry, and parity are good.
The representations D(j, 0) and D(0, j) can each separately represent particles of spin j. A state or quantum field in such a representation would satisfy no field equation except the Klein-Gordon equation.
## Lagrangian
The Lagrangian which generates equations (2) through the Euler–Lagrange equation (for fields) is not easily found. Methods have been introduced by Guralnik and Kibble, and Larsen and Repko.[18]
One method proposed by Kamefuchi and Takahashi in 1966 was to expand the wavefunctions in terms of 4 × 4 matrices with a required symmetry (conserved properties of the quantum system), then substitute back into the BW equations to yield field equations with that symmetry. From then a Lagrangian can be found by working backwards from the Euler–Lagrange field equations.
D.S. Kaparulin, S.L. Lyakhovich, and A.A. Sharapov take this fundamental approach by starting from symmetries directly, by means of a Poincaré invariant Lagrange anchor.[19] A Lagrange anchor geometrically defines a mapping between fiber bundles, comprising vector bundles, tangent bundles, and the configuration space for the quantum fields. This is less restrictive than a variational formulation (based on the principle of least action) to obtain the equations for the quantum fields.
## Formulation in curved spacetime
Following M. Kenmoku,[16] in local Minkowski space, the gamma matrices satisfy the anticommutation relations:
$[\gamma^i,\gamma^j]_{+} = 2\eta^{ij}$
where ηij = diag(−1, 1, 1, 1) is the Minkowski metric. For the Latin indices here, i, j = 1, 2, 3. In curved spacetime they are similar:
$[\gamma^\mu,\gamma^\nu]_{+} = 2g^{\mu\nu}$
where the spatial gamma matrices are contracted with the vierbein biμ to obtain γμ = biμ γi, and gμν = biμbiν is the metric tensor. For the Greek indices; μ, ν = 0, 1, 2, 3.
A covariant derivative for spinors is given by
$\mathcal{D}_\mu=\partial_\mu+\Omega_\mu$
with the connection Ω given in terms of the spin connection ω by:
$\Omega_\mu =\frac{1}{4}\partial_\mu\omega^{ij} (\gamma_i\gamma_j-\gamma_j\gamma_i)$
The covariant derivative transforms like ψ:
$\mathcal{D}_\mu\psi \rightarrow D(\Lambda) \mathcal{D}_\mu \psi$
With this setup, equation (2) becomes:
\begin{align} & (-i\hbar\gamma^\mu \mathcal{D}_\mu + mc)_{\alpha_1 \alpha_1'}\psi_{\alpha'_1 \alpha_2 \alpha_3 \cdots \alpha_{2j}} = 0 \\ & (-i\hbar\gamma^\mu \mathcal{D}_\mu + mc)_{\alpha_2 \alpha_2'}\psi_{\alpha_1 \alpha'_2 \alpha_3 \cdots \alpha_{2j}} = 0 \\ & \qquad \vdots \\ & (-i\hbar\gamma^\mu \mathcal{D}_\mu + mc)_{\alpha_{2j} \alpha'_{2j}}\psi_{\alpha_1 \alpha_2 \alpha_3 \cdots \alpha'_{2j}} = 0 \,.\\ \end{align}
## References
### Notes
1. ^ Bargmann, V.; Wigner, E. P. (1948). "Group theoretical discussion of relativistic wave equations". Proceedings of the National Academy of Sciences of the United States of America 34 (5): 211–23. Bibcode:1948PNAS...34..211B. doi:10.1073/pnas.34.5.211.
2. ^ E. Wigner (1937). "On Unitary Representations Of The Inhomogeneous Lorentz Group". Annals of Mathematics 40 (1): 149.
3. E.A. Jeffery (1978). "Component Minimization of the Bargman–Wigner wavefunction". Australian Journal of Physics (Melbourne: CSIRO). NB: The convention for the four gradient in this article is μ = (∂/∂t, ∇ ), same as the Wikipedia article. Jeffery's conventions are different: μ = (−i∂/∂t, ∇ ). Also Jeffery uses collects the x and y components of the momentum operator: p± = p1 ± ip2 = px ± ipy. The components p± are not to be confused with ladder operators; the factors of ±1, ±i occur from the gamma matrices.
4. ^ a b c d R.K Loide, I.Ots, R. Saar (2001). "Generalizations of the Dirac equation in covariant and Hamiltonian form". Journal of Physics A: Mathematical and General (Tallinn, Estonia: IoP).
5. ^ C.B. Parker (1994). McGraw Hill Encyclopaedia of Physics (2nd ed.). p. 1514. ISBN 0-07-051400-3.
6. ^ H. Shi-Zhong, R. Tu-Nan, W. Ning, Z. Zhi-Peng (2002). "Wavefunctions for Particles with Arbitrary Spin". Beijing, China: International Academic Publishers.
7. ^ a b c T. Jaroszewicz, P.S Kurzepa (1992). "Geometry of spacetime propagation of spinning particles". Annals of Physics (California, USA).
8. ^ C.R. Hagen (1970). "The Bargmann–Wigner method in Galilean relativity". Springer (California, USA).
9. ^ Cédric Lorcé (2009). "Electromagnetic Properties for Arbitrary Spin Particles: Part 1 − Electromagnetic Current and Multipole Decomposition". Mainz, Germany. arXiv:0901.4199v1.
10. ^ Cédric Lorcé (2009). "Electromagnetic Properties for Arbitrary Spin Particles: Part 2 − Natural Moments and Transverse Charge Densities". Mainz, Germany. arXiv:0901.4200v1.
11. ^ a b c Weinberg, S. (1964). "Feynman Rules for Any spin". Phys. Rev. 133 (5B): B1318–B1332. Bibcode:1964PhRv..133.1318W. doi:10.1103/PhysRev.133.B1318.; Weinberg, S. (1964). "Feynman Rules for Any spin. II. Massless Particles". Phys. Rev. 134 (4B): B882–B896. Bibcode:1964PhRv..134..882W. doi:10.1103/PhysRev.134.B882.; Weinberg, S. (1969). "Feynman Rules for Any spin. III". Phys. Rev. 181 (5): 1893–1899. Bibcode:1969PhRv..181.1893W. doi:10.1103/PhysRev.181.1893.
12. ^ Gábor Zsolt Tóth (2012). "Projection operator approach to the quantization of higher spin fields". Budapest, Hungary: International Academic Publishers. pp. 37–40. arXiv:1209.5673v1.
13. ^ V.V. Dvoeglazov (2003). "Generalizations of the Dirac Equation and the Modified Bargmann–Wigner Formalism". arXiv:hep-th/0208159.
14. ^
15. ^ E. Abers (2004). Quantum Mechanics. Addison Wesley. ISBN 9780131461000.
16. ^ a b K. Masakatsu (2012). "Superradiance Problem of Bosons and Fermions for Rotating Black Holes in Bargmann–Wigner Formulation". Nara, Japan. arXiv:1208.0644.
17. ^ Weinberg, S (2002), "5", The Quantum Theory of Fields, vol I, ISBN 0-521-55001-7
18. ^ M.A. Rodriguez (1984). "Some results about the relationship between Bargmann–Wigner and Gelfand–Yaglom equations". Reports on Mathematical Physics (Madrid, Spain: Elsevier). Bibcode:1986RpMP...23....9R. doi:10.1016/0034-4877(86)90063-7.
19. ^ D. S. Kaparulin, S. L. Lyakhovich, A. A. Sharapov (2012). Lagrange Anchor for Bargmann–Wigner equations. arXiv:1210.2134. Bibcode:2012arXiv1210.2134K. | 6,465 | 19,545 | {"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": 38, "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-2014-23 | longest | en | 0.843049 |
http://mathoverflow.net/questions/39858/localization-of-a-polynomial-ring-at-a-prime-ideal/39860 | 1,469,427,398,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824204.68/warc/CC-MAIN-20160723071024-00276-ip-10-185-27-174.ec2.internal.warc.gz | 172,102,911 | 14,735 | # Localization of a polynomial ring at a prime ideal.
If $R=\mathbb{C}[x,y]$ is the polynomial ring in two variables $x$ and $y$ then we know that the localization of R at the multiplicative set $S=[1,x,x^2,x^3,...]$ is given by $R_x=\mathbb{C}[x,x^{-1},y]$. Now, what will be the localization of $R$ at the prime ideal $(x)$. i.e. what will $R_{(x)}$ be?
-
If you localize at a prime ideal $p$ in a ring $R$, then the multiplicative set that gets inverted is $S = R - p$. So for example, $x-1$ is invertible in $R_{(x)}$. This is explained in any basic commutative algebra text. – cfranc Sep 24 '10 at 13:28
It's funny that I have voted to close a question about localization as "too localized". – Harry Gindi Sep 24 '10 at 13:30
I don't know if this will satisfy you, since it is just a restatement of the definition, but $R_{(x)}$ is the ring of rational functions on $\mathbb{C}^2$ which don't have a pole along the hyperplane $x=0$. – David Speyer Sep 24 '10 at 14:14
I doubt that there is a nice description which will satisfy you. As a $\mathbb{C}$-algebra, $R_{(x)}$ is not finitely generated. Anyway every localization of a factorial domain at a principle prime ideal is a discrete valuation domain. In particular, $R_{(x)}$ is such a domain with prime ideal $(x)$. Every nonzero element has a unique representation $x^n u$, where $u$ is a fraction, such that $x$ is coprime to both the numerator and the denominator of $u$. Probably an algebraic geometer would call this ring the local ring of the affine plane at the line $x=0$. | 450 | 1,542 | {"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-2016-30 | latest | en | 0.89883 |
https://www.teachoo.com/7612/2332/Divisibility-by-6/category/Divisibility-Tests---Divisibility-by-3--6--9/ | 1,722,985,281,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640523737.31/warc/CC-MAIN-20240806224232-20240807014232-00073.warc.gz | 771,563,743 | 21,576 | Divisibility Tests - Divisibility by 3, 6, 9
Chapter 3 Class 6 Playing with Numbers
Concept wise
A number is divisible by 6 if
• It is divisible by 2
• It is divisible by 3
Example :
## Is 36 is divisible by 6?
Since 36 is divisible by both 2 & 3
It is divisible by 6
## Is 18 is divisible by 6?
Since 18 is divisible by both 2 and 3
∴ it is divisible by 6
## Is 27 is divisible by 6?
Since it is not divisible by 2,
∴ 27 is not divisible by 6
## Is 216 is divisible by 6?
Since 216 is divisible by both 2 and 3
∴ It is divisible by 6.
### Transcript
To check divisibility by 6 we check that Number is divisible by 2 Number is divisible by 3 Divisibility by 2 Since 36 ends with 6 It is divisible by 2 Divisibility by 3 Sum of digits = 3 + 6 = 9 Since 9 is divisible by 3 ∴ 36 is divisible by 3 To check divisibility by 6 we check that Number is divisible by 2 Number is divisible by 3 Divisibility by 2 Since 18 ends with 8 It is divisible by 2 Divisibility by 3 Sum of digits = 1 + 8 = 9 Since 9 is divisible by 3 ∴ 18 is divisible by 3 To check divisibility by 6 we check that Number is divisible by 2 Number is divisible by 3 Divisibility by 2 Since 27 ends with 7 It is not divisible by 2 Divisibility by 3 Sum of digits = 2 + 7 = 9 Since 9 is divisible by 3 So, 27 is divisible by 3 To check divisibility by 6 we check that Number is divisible by 2 Number is divisible by 3 Divisibility by 2 Since 216 ends with 6 It is divisible by 2 Divisibility by 3 Sum of digits = 2 + 1 + 6 = 9 Since 9 is divisible by 3 So 216 is divisible by 3
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo. | 552 | 1,807 | {"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} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.97723 |
https://topnursingessay.com/compute-the-mean-and-standard-deviation-of-the-distribution-of-home-selling-prices/ | 1,628,092,270,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154878.27/warc/CC-MAIN-20210804142918-20210804172918-00613.warc.gz | 569,502,634 | 10,839 | # Compute the mean and standard deviation of the distribution of home selling prices.
The extension of the probability concept helps us develop the idea of how populations are distributed. The probability of a single event happening from a larger population increases in such a way that the events closest to the mean are more likely than events far from the mean.
Refer to the real estate data located below.
Compute the mean and standard deviation of the distribution of home selling prices. Develop a histogram of the data. (SPSS: Graphs – Graphboard Template Chooser – Select Price, then Histogram with Normal Distribution). (Minitab: Graph – Histogram – With Fit – Graph variable = Price). Would it seem reasonable from this histogram to conclude that the population of selling prices follows the normal distribution?
Biblical Application:
The Bible is filled with the use of numbers and statistics (i.e. Exodus 25:10-16-Ark of the Covenant). Compare and contrast one biblical concept of statistics and how we should live our lives.
200 to 300 words
minimum of two academic peer reviewed scholarly articles | 218 | 1,118 | {"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-2021-31 | latest | en | 0.909934 |
https://gradeup.co/jee-revision-plan-physics-12-1-i-91d0fc10-01b9-11e8-b682-19bb6b4f3b29 | 1,558,427,213,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00419.warc.gz | 496,514,948 | 33,601 | • Select Exam
• Select Exam
Time Left - 00:00 mins
# JEE Revision Plan - Physics 12.1
Attempt now to get your rank among 282 students!
Question 1
The numbers of AM broadcast stations that can be accommodated in a 400 kHz bandwidth for the highest modulating frequency 20 kHz will be
Question 2
Which of the following is correct statement for the maximum range, dmax, of radar is
Question 3
An audio signal consists of two distinct sounds: one a human speech signal in the frequency band of 200 Hz to 2700 Hz, while the other is a high-frequency music signal in the frequency band of 10200 Hz to 15200 Hz. The ratio of the AM signal bandwidth required to send both the signals together to the AM signal bandwidth required to send just the human speech is:
Question 4
The value of the resistor, , needed in the dc voltage regulator circuit shown here, equals:
Question 5
A 2V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in one case battery’s positive terminal is connected to A and in other case when positive terminal of battery is connected to B will respectively be:
Question 6
The ratio (R) of output resistance r0, and the input resistance ri in measurements of input and output characteristics of a transistor is typically in the range:
Question 7
An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with +ve and −ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor?
Question 8
The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is:
Question 9
Find Rc such that transistor acts as an open switch. VBE = 0.7V
Question 10
Find minimum input voltage such that transistor is driven into saturation. VBE = 0.7V
Question 11
Relaxation oscillators are not used to generate which of the following:
Question 12
Which of the following statements is incorrect regarding the working of a transistor?
Question 13
When a transistor is connected in common-base mode, the collector current is found to be 5.488 mA when the emitter current is 5.60 mA. The base current amplification factor is
Question 14
When a transistor is connected in common base mode, the current gain is found to be α = 0.98 and the reverse saturation current for the collector-base junction is found to be 0.6 𝜇𝐴. On connecting the same transistor in common-emitter mode as an amplifier the base current is found to be 20 𝜇𝐴, the magnitude of the collector current in this case is
Question 15
The common emitter current gain for a transistor is 50. If the resistance connected to the collector is of 2 kΩ, the input resistance is 500 Ω and the voltage applied to the base is 0.02 V, then the power gain is equal to:
• 282 attempts | 682 | 2,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2019-22 | longest | en | 0.912004 |
https://stat.ethz.ch/pipermail/r-help/2017-January/444600.html | 1,701,453,270,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100290.24/warc/CC-MAIN-20231201151933-20231201181933-00241.warc.gz | 618,858,920 | 4,610 | [R] graphical behavior of a table of numbers
Richard M. Heiberger rmh at temple.edu
Mon Jan 30 21:04:23 CET 2017
```I still think
plot(fr, xlab="Determinant", ylab="Frequency")
has a totally non-intuitive x-axis.
I recommend that the example in R-intro.pdf include an additional
sentence and option.
## In this example, where the x-axis is the entire set of integers -81:81,
## displaying them as an ordinary numeric axis might be preferable, in
which case use
plot(fr, xlab="Determinant", ylab="Frequency", xaxt="n")
axis(1)
While thinking on this, I looked at ?plot.table
The example
plot(table(state.division))
doesn't display most of the labels, and I think it therefore not a good example.
I recommend revising it, perhaps to
old.oma <- par(oma=c(6,1,0,1));
plot(table(state.division), las=2, mgp=c(.5,2,0))
par(old.oma)
I agree the code for the legible labels is difficult to read.
Rich
On Mon, Jan 30, 2017 at 10:59 AM, Martin Maechler
<maechler at stat.math.ethz.ch> wrote:
>>>>>> Richard M Heiberger <rmh at temple.edu>
>>>>>> on Mon, 30 Jan 2017 10:19:53 -0500 writes:
>
> > Duncan, thank you for locating the problem.
> > Martin, thank you for explaining the behavior and for the first pass
> > at fixing it.
>
>
> > With the fix, now the x-axis has ticks at all integers, and tick labels at
> > c(-81,-67,-53,-39,-25,-11,0,9,19,31,43,55,67,79)
>
> Note that *which* tick labels are shown depends quite a bit
>
> > This is with R-3.3.2, as I interpret your fix to be to only the
> > R-intro.pdf manual with no change
> > to the code of any of the functions.
>
> That's correct. If this is about improving any of
> the base graphics functions, there's the R-devel mailing list
> and the bugzilla repository for "wishes"... rather than the
> R-help list.
>
> > More work has to be done to repair the example.
>
> I strongly disagree:
> The example (which does not even need a 'type = "h"', and no longer
> uses it) now mentions that the plot.factor method is used.
> ... and I do like its graphics output, given the simplicity of
> the plot() function call.
>
> Code as the one below may be preferable in some cases, but not
> there in the "Introduction to R".
>
>
> > I recommend
> > plot(as.numeric(fr) ~ as.numeric(names(fr)), type="h",
> > xlab="Determinant", ylab="Frequency")
>
> > The slightly more obvious solution doesn't work
> >> plot(fr ~ as.numeric(names(fr)), type="h", xlab="Determinant", ylab="Frequency")
> > Error in plot.table(c(-81, -80, -79, -78, -77, -76, -75, -74, -73, -72, :
> > invalid table 'x'
>
> > ## It is possible to change graphics:::Axis.table to
> > if (is.num) axis(side, ...)
> > ## and that would make the x-axis for the determinant example
> > plot(fr, type="h", xlab="Determinant", ylab="Frequency")
> > ## look sensible, but would
> > ## be less appropriate for the following example.
>
> > ## The current behavior of Axis.table makes sense in this example
> > tt <- as.table(array(c(10,20,30), dimnames=list(c(100, 120, 200))))
> > tt
> > plot(tt)
>
> Indeed. I doubt we would want to change Axis.table()
> just because of examples like the determinant one....
> (and then again: such considerations would be part of a new thread on R-devel...)
>
> Martin Maechler
>
>
> > On Mon, Jan 30, 2017 at 5:38 AM, Martin Maechler
> > <maechler at stat.math.ethz.ch> wrote:
> >>>>>>> Duncan Murdoch <murdoch.duncan at gmail.com>
> >>>>>>> on Sun, 29 Jan 2017 06:32:27 -0500 writes:
> >>
> >> > On 29/01/2017 12:05 AM, Jim Lemon wrote:
> >> >> Hi Richard, I think there may be something amiss in the
> >> >> plot.table function. As you note, changing the class of
> >> >> fr to array produces a more sensible plot, as does Bert's
> >> >> "as.vector". Yet inside plot.table we find:
> >> >>
> >> >> plot(x0, unclass(x), ...
> >> >>
> >> >> and that should produce an array:
> >> >>
> >> >> class(unclass(fr)) [1] "array"
> >> >>
> >> >> The plot.table function looks like it should produce the
> >> >> plot you want, but it doesn't. I think (therefore I am
> >> >> probably wrong) that a 1D table is handled in the same
> >> >> way as multiD table rather than being squeezed into a
> >> >> vector.
> >>
> >> > I think the issue is that Axis() is called without
> >> > removing the class. Axis.table sets ticks based on the
> >> > names of the table.
> >>
> >> > Duncan Murdoch
> >>
> >> yes indeed! So this answers Rich Heiberger's question.
> >>
> >> The example stems from a time long before there was
> >> a plot.table() method, and even longer before plot.default() had
> >> started using Axis() and its methods.
> >>
> >> So a much nicer example for the R-intro -- committed a few
> >> minutes ago -- is making use of the plot.table() S3 method :
> >>
> >> d <- outer(0:9, 0:9)
> >> fr <- table(outer(d, d, "-"))
> >> plot(fr, type="h", xlab="Determinant", ylab="Frequency")
> >>
> >> So this fulfills Rich's recommendation.
> >>
> >> Martin
> >>
> >>
> >> >> On Sun, Jan 29, 2017 at 11:19 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote:
> >> >>> Rich:
> >> >>>
> >> >>> Simpler: Just lose the "table" class.
> >> >>>
> >> >>> plot(as.numeric(names(fr)), as.vector(fr), type="h",
> >> >>> xlab="Determinant", ylab="Frequency")
> >> >>>
> >> >>> However, I'm no less puzzled by the "strange" behavior than you.
> >> >>>
> >> >>> In addition, it's probably worth noting that xyplot in lattice (and no
> >> >>> doubt ggplot,too) does not have this problem (as I'm sure you know):
> >> >>>
> >> >>> xyplot(fr ~ as.numeric(names(fr)), type="h",
> >> >>> xlab="Determinant", ylab="Frequency")
> >> >>>
> >> >>>
> >> >>> Cheers,
> >> >>> Bert
> >> >>> Bert Gunter
> >> >>>
> >> >>> "The trouble with having an open mind is that people keep coming along
> >> >>> and sticking things into it."
> >> >>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
> >> >>>
> >> >>>
> >> >>> On Sat, Jan 28, 2017 at 3:03 PM, Richard M. Heiberger <rmh at temple.edu> wrote:
> >> >>>> ## This example is from R-intro.pdf page 21 (R-3.3.2)
> >> >>>>
> >> >>>> d <- outer(0:9, 0:9)
> >> >>>> fr <- table(outer(d, d, "-"))
> >> >>>> plot(as.numeric(names(fr)), fr, type="h",
> >> >>>> xlab="Determinant", ylab="Frequency")
> >> >>>> ## The y-axis tick marks are at c(-21,24,65).
> >> >>>> ## This seems to be because class(fr) == "table"
> >> >>>>
> >> >>>> ## Switching the class to array gives the more appropriate
> >> >>>> ## y-axis ticks at seq(0,500,100) .
> >> >>>>
> >> >>>> fr.array <- fr
> >> >>>> class(fr.array) <- "array"
> >> >>>> plot(as.numeric(names(fr)), fr.array, type="h",
> >> >>>> xlab="Determinant", ylab="Frequency")
> >> >>>>
> >> >>>>
> >> >>>> ## I have a question and a recommendation.
> >> >>>> ## Question:
> >> >>>> ## Why are the y-axis ticks for the table defaulted to c(-21,24,65).
> >> >>>> ##
> >> >>>> ## Recommendation:
> >> >>>> ## Changed the example on page 21 to show the ticks at seq(0,500,100)?
> >> >>>>
> >> >>>> ## Rich
> >> >>>>
> >>
> >> ______________________________________________
> >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help | 2,285 | 7,668 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-50 | latest | en | 0.863783 |
http://mathhelpforum.com/calculus/25912-separation-variables.html | 1,480,893,606,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541426.52/warc/CC-MAIN-20161202170901-00323-ip-10-31-129-80.ec2.internal.warc.gz | 172,786,598 | 11,721 | 1. ## Separation of variables
Hi,
Need to solve the following equation, but i can't seem to separate the variables correctly so i can integrate through.
dx/dt = t - x
cheers
Oli
2. Originally Posted by olip
Hi,
Need to solve the following equation, but i can't seem to separate the variables correctly so i can integrate through.
dx/dt = t - x
cheers
Oli
This is a non-homogeneous differential equation.
$x'+x-t=0$
You have to guess an initial solution for $x$. Check this out: Method of undetermined coefficients - Wikipedia, the free encyclopedia
3. Originally Posted by olip
dx/dt = t - x
As colby2152 wrote $x'+x=t,$ multiply by $e^t,$
$x' + x = t \implies e^t x' + e^t x = te^t \,\therefore \,\left( {xe^t } \right)' = te^t .$
By integrating both sides
$xe^t = \int {te^t \,dt},$ where the last integral is fairly straightforward via integration by parts.
4. Originally Posted by olip
Hi,
Need to solve the following equation, but i can't seem to separate the variables correctly so i can integrate through.
dx/dt = t - x
cheers
Oli
dx/dt = t - x
=> dx/dt + x = t
is solved using the integrating factor method (as has been implied by Krizalid).
5. Hi,
thanks for all help so far, that working makes sense to me Krizalid, but on integrating $\int {te^t \,dt}$ i get $e^t(t-1)$ i.e. $xe^t = e^t(t-1)$, or $x = t-1$, which doesnt seem right to me.
I should give more information. This was just part of some dynamics book and i couldn't figure out how they got the solution. Their solution was
$x(t) = t-1 + exp(-t) (x_0 + 1)$
where $x_0$ is the solution at t=0
thanks again,
Oli
6. Hi again,
realised i made a lazy mistake there, forgot to add in the constant of integration, so the solution actually becomes $t-1 + Cexp(-t)$
When i substitute this back in to the original equation $x' + x = t$ it works, which is good, however it is different than the solution given in the book:
$t-1 + exp(-t)(C + 1)$ which doesn't give the right answer when substituted back into the equation. Is the one in the book wrong?? I am confused!
cheers,
Oli
7. Originally Posted by olip
[snip]
the solution actually becomes $t-1 + Cexp(-t)$
[snip]
however it is different than the solution given in the book:
$t-1 + exp(-t)(C + 1)$ which doesn't give the right answer when substituted back into the equation. Is the one in the book wrong?? I am confused!
cheers,
Oli
The books solution does give the correct answer and is in fact equivalent to your solution. Since C is an arbitrary constant, C + 1 is also arbitrary ..... Re=brand C + 1 as K ...... $x = t-1 + K exp(-t)$. | 746 | 2,593 | {"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": 18, "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} | 4.1875 | 4 | CC-MAIN-2016-50 | longest | en | 0.925559 |
http://math.stackexchange.com/questions/235700/limit-and-integral-sign-in-l2 | 1,469,485,515,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824395.52/warc/CC-MAIN-20160723071024-00182-ip-10-185-27-174.ec2.internal.warc.gz | 156,571,165 | 17,367 | # Limit and Integral sign in $L^2$.
If $\lim_{k \to \infty} \| u_k - u \|_{L^2(\Bbb R^n)} = 0$ then how can I show that $$\lim_{k \to \infty} \int_{\Bbb R^n} u_k v = \int_{\Bbb R^n} uv$$ for any $v \in L^2 (\Bbb R^n)$?
-
Note that this is a particular case of math.stackexchange.com/questions/235414/… – Martin Argerami Nov 12 '12 at 15:01
Hint: Write the difference as an integral, then use Cauchy-Bunyakovsky-Schwarz inequality. – Davide Giraudo Nov 12 '12 at 15:02
Thank you guys, it was trivial. – Ann Nov 12 '12 at 15:05
$$\left| \int u_k v - \int u v \right| = \left| \int (u_k -u)v \right| \leqslant \int | u_k - u | |v| \leqslant \| u_k - u \|_2 \| v \|_2 \to 0$$ | 272 | 674 | {"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} | 3.046875 | 3 | CC-MAIN-2016-30 | latest | en | 0.551188 |
https://nus.kattis.com/courses/CS3233/CS3233_S2_AY2324/assignments/mmujba/problems/intervalscheduling2 | 1,726,113,765,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651420.25/warc/CC-MAIN-20240912011254-20240912041254-00003.warc.gz | 408,539,754 | 7,425 | Hide
# Problem IInterval Scheduling
Many couples will dine out on Valentine’s day. To serve as many customers as possible, a particular Western restaurant adopts the "Staggered Seating" strategy on the night of Valentine’s day.
The strategy works as follows: The restaurant divides its opening hours into $N$ non-overlapping time intervals, called a *wave*. In each wave, a group of customers are served and they must complete their meal (and leave the restaurant) within that time interval. Two intervals are said to be non-overlapping if the ending time of the first interval is not later than the starting time of the second interval (in chronological order).
The manager drafted the schedule of each wave. However, you discovered that some intervals are extremely short - customers are unable to enjoy the meal given the limited time.
You decided to come up with a better schedule, with the following conditions:
1. The amended time interval for each wave must be inclusive of the original time interval of that wave. Informally speaking, you are only allowed to extend the time intervals. Customers will feel unhappy if the time intervals are shortened.
2. The opening and closing time of the restaurant remains unchanged. That is, you cannot change the starting time of the first interval or the ending time of the last interval.
3. You would like to maximize the length of the shortest time interval across all waves.
Can you help the manager to come up with a better schedule?
## Input
The first line of input consists of a single positive integer $N$ $(2 \leq N \leq 200)$.
$N$ lines follow. Each line consists of a time interval in the format HH:MM - HH:MM. It is guaranteed that the time intervals are non-overlapping and given in chronological order.
It is also guaranteed that the time intervals are all within the same day, between 16:00 and 23:59 (inclusive).
## Output
Output $N$ lines, the amended schedule of the staggered seating plan. The format should be exactly same as that in the input.
If there are multiple solutions, you may output any of them.
Sample Input 1 Sample Output 1
3
19:00 - 20:00
21:08 - 21:24
22:00 - 22:46
19:00 - 20:15
20:15 - 21:30
21:31 - 22:46
Sample Input 2 Sample Output 2
2
19:01 - 19:02
19:03 - 19:04
19:01 - 19:03
19:03 - 19:04
Hide | 552 | 2,299 | {"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 | 3 | CC-MAIN-2024-38 | latest | en | 0.93806 |
https://www.our-energy.com/atom.html | 1,722,881,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640454712.22/warc/CC-MAIN-20240805180114-20240805210114-00624.warc.gz | 724,175,917 | 15,953 | # Atom
Atom is the fundamental building unit of the each substance, withal also the smallest unit of the substance that has typical characteristics of chemical element. Every atom has three structural particles: protons, neutrons and electrons. Protons have positive electrical charge and are found in the atomic nucleus along with neutrons that have no electrical charge, but are the same size as the protons. Protons and neutrons are mutually called nucleons. Outside the nucleus of an atom are circling electrons which are negatively charged particles that form an electron cloud. In its normal state atom has the same numbers of protons as electrons. Electron has a mass 1836 times less than that of the proton or neutron, so it’s often neglected in calculations.
All atoms regardless of numbers of its protons or electrons have almost the exact diameter and that is about 10-10 meters. Nucleus of atoms have diameter about 1 – 10 x 10-15 meters. When you look at this numbers, it’s easy to see that vast majority of every atom is in fact nothing but empty space.
Between all of these electrically charged particles Coulomb (electrostatic) force is always present. The name was given in honor of French physicist Charles Augustine Coulomb (1736 – 1806), who first discovered that the magnitude of the force F is proportional to the product of the two charges, q1 and q2, divided by the square of the distance r between them, or F=k*q1*q2/r2, where k is a constant that depends on the measurement system being used which in vacuum equals 1. The electric force can be one of repulsion, such as the force between two objects having like charges, or it can be attractive, such as the force between two objects having opposite charges. In the memory of the great physicist the unit for measuring an electric charge with mark C was also named after him, and it is equal to the quantity of electricity conveyed in one second by a current of one ampere(C = As).
Coulomb force is the major force when describing interactivity and mutual functionality of the nucleus and electron cloud. Every electron has negative charge of 1.6 * 10-19 C, and every proton has positive charge of exactly the same 1.6 * 10-19 C, so nucleus and electron cloud are attracting each other according to the Coulomb Force. When atom gains or losses electron, it becomes electrically charged particle called ion. Atom which has more protons than electrons has positive charge and it’s called cation (it indicates for example Na+). Anion is an atom which has more electrons than the protons and therefore has negative charge (for example Cl-).
Atom is the fundamental building unit of the each substance.
Electrons in atom could have only specific energetic conditions, all described with series of quantum numbers which determine average distance between electrons to the nucleus, intensity and rotation way, and the electrons rotation way around its proper axis (spin). When an electron is changing its energetic state, light of specific color occurs or absorbs. Electrons are located in seven shells around the nucleus and maximal number of electrons in each shell is limited on 2n2, where n is serial number of shell. External shell isn’t always filled: sodium has two electrons in first shell, eight in second, and only one electron in third shell. Because of the Heisenberg’s maxim of ambiguity it isn’t possible to determine position and speed of electron, both in the same time, and therefore electron cloud can be only statically described. According to that electron’s orbits are just probability dispensations of such random movement.
When describing a nucleus of an atom, three out of four basic forces in the universe have very significant meaning: strong nuclear force, weak nuclear force and Coulomb force (gravity when describing atom could be negligible). Strong nuclear force is the force which exists between the two nucleons, keeping nucleus solid despite repulsion force of the positively charged protons. Weak nuclear force backs up some forms of the radioactivity and the interaction between particles in atom. Nuclear forces are called nuclear because they act on the level of the core of an atom (nucleus), which means that range of this forces is extremely small. On very short distances strong nuclear force overpowers Coulomb force, but on distances longer than 2.5fm Coulomb force is starting to overcome. Because of this effect strong nuclear force keeps nucleus in the middle, and Coulomb force keeps electrons on courses around the nucleus.
Two most significant numbers when describing each atom are the atomic number and the relative atomic mass. Atomic or proton number is the number of protons in the nucleus and has the sign Z. Serial number of element in periodic table of elements is atomic number. Atomic number is marked as left subscript (2He). Relative atomic mass of each element is proportion of average mass of element’s atom and 1/12 of the mass of atom nuclide 12C, which by definition has atomic weight 12. It is marked like left superscript (12C), and his letter’s mark is A. Number of neutrons in nucleus is marked with letter N and is often called Neutron number of an atom. If relative atomic mass is rounded off to be a whole number, then this number is the sum of protons and neutrons in atom’s nucleus ( A=Z+N).
Atoms which are having same number of protons (atomic number), and different number of neutrons (meaning different relative atomic mass) are called isotopes. Isotopes of the same element have the exact chemical, but different physical characteristics. Periodic table of elements takes as relative atomic mass of each element the most stable isotope of each element. | 1,171 | 5,710 | {"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-2024-33 | latest | en | 0.956684 |
https://eduprojecttopics.com/product/time-series-analysis-on-total-number-of-patients-treated-for-malaria-fever/ | 1,627,549,957,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153854.42/warc/CC-MAIN-20210729074313-20210729104313-00139.warc.gz | 231,121,788 | 38,719 | DO YOU NEED HELP? CALL - 08060082010
Download this complete Project material titled; Time Series Analysis On Total Number Of Patients Treated For Malaria Fever with abstract, chapter 1-5, references and questionnaire. Preview chapter one below
• Format: PDF and MS Word (DOC)
• pages = 65
3,000
ABSTRACT
This project work reveled the rate at which people are infected with malaria the least square method used for analysis showed that people are infected with malaria irrespective of the time and seasons of a successive year,
There is no noticeable direction as regarding the number of patient treated for malaria over time.
Also, the analysis from autoregressive moving average report shows that both autoregressive and moving average of order four were both appropriate while the report from autocorrelation and autocovanance does not indicate any noticeable trend in the number of patients treated for malaria.
CHAPTER ONE
1.0INTRODUCTION
The term time series refers to one the quantitative method used in determination pattern in data collected over time e.g weekly monthly, quarterly or yearly.
Time service is the statistic tool or methodology that can be used to transform past experience to predict future event which would enable the researcher or organization to plan.
It gives information about how the particular case of study has been behaving in the past and present and such information can be used in prediction The number of people treated for malaria fever at the otan Ayegbaju management hospital. Comprehensive health centre otan. We are going to seen how change occur over mouths in each year in the occurrence of the disease in the hospital. As a result of this, we will be able to know certain factor responsible for increase or decrease in the rate of infection of the disease over the period of time.
Record of time series data can be made in the following ways:-
A. THROUGH CUMULATIVE FIGURES:- these represent value of input through the quarter. We must always bear in mind the different when handling time series data and as certain which particular type we are dealing with in every case.
B. CUMULATIVE TYPE ADDED COMPILATION:- some cases when an added compilation introduced for the cumulative type of data the figure which are related to month of the year and not the total for month. further more the characteristic movement, seasonal variation Irregular variation in the analysis of time series, we have two types of model are generally accepted as good approximation of the true data association among the component of observed data, they are the most commonly assumed relationship between time series and its components. These are additive model and Multiplicative mode. All time series contain at least on of four of its components. These components are:-
1. Long term trend
2. Seasonal variation
3. Cyclical variation
4. Irregular or random variation value
LONG TERM TREND COMPONENT
This can be referred to the general path in which time series graph appear to follow over a long period of time, in other word, it is the long-term increase or decrease in a variable being measured over time for example a company planning her expense on goods to produce in the next three or four years has consider demand at a particular time.
Y Downward trend Y upward trend
t t
GRAPHICAL REPRESENTATION OF TREND
SEASONAL COMPONENT
These are sort term variation from the trend that occur regular with the passage of time series of many products like ice cream, soft drink, ran during ileya turkey during chrismas and new and year period are subjects of such variation. There changes are visually identical or almost identical in natures that follow during
Y
t
GRAPHICAL REPRESENTATION OF SEASONAL CHANGE
CYCLICAL COMPONENT
Data collected how every, they can contain cyclical effect in a time series are represented by wave-like fluctuation around a long term trends. The change occurs in economics activities due to some facture like booms. Recess. Cyclical fluctuation repest them selve in a general pattern in the long-term. But occur with different frequencies and intensities.
Thus, they can be isolated but not totally predicted
y
t
GRAPHICAL REPRESENTATION OF CYCLICAL CHANGE
IRREGULAR OR RANDOM COMPONENT
This venation cannot BE attributed to any of three previously discussed component in the sense that is unpredictable.
Irregulars flotation can be cause by many factor such as war, flood drought and other human as action. Two type of irregular variation may exit in a time series viz. minor and major irregularities minor irregularities show up as serivtooth like pattern are under the long term trend. These irregularited are in organization long term operation:
MODELS OF TIME SERIES
We usually denote the component of time series as T,TS,C and I: There Are Two Types Of Modern That Are Appropriate For Joining Component Of Time Series these moderns are additive and multiplicative modern.
The additive moder assumes that the valve of the original data is the sum total of other four elements it is:
T=T+S+C+I where
T is value of the originally conserved data (dependent)
T is the value of secular or trend
S is the value ofr cyclical venation and
I is the value of irregular venation
Mufti active modern on the other hand assumes that the value of the observed data is the Y = TSCI
1.1 BACKGROUND OF STUDY
one of the factor that determine the population of a country, state local government e.t.c is death rate, that is to say, the more the disease infected the population of such an area and vice-versa. The fact prompted the writer into the study of quarterly number of people given treatment for malaria at the comprehensive health centre otan ayegbaju, in addition to that it is done to know weather infected people come to the hospital fur test or they stay back due to the old custom self medication.
In order to carry our the analysis data will be collected from daily record of the hospital at record department over some years to get all necessity information so as to carry out computation and predict about the nearest future by using secondary method of data collection.
1.2SCOPE AND COVERAGE OF THE STUDT
This project work was carried out on the number of people treated for malaria fever between year 2001 to 2010. The data was collected from the comprehensive health centre Otan Ayegbaju osun state.
AIM AND OBJECTIVE OF THE STUDY
i. to know whether the yearly spread of malaria is increasing pr decreasing.
ii. To formulate a model that can best explain the relationship between malaria increases over the years
iii. To use the modern to forecast the occurrence of malaria
iv. To plot the graph of the original data i.e occurrence of malaria against year correlogram and moving average.
1.3SOURCES OF DATA COLLECTION
Data used for research are of two main sources: these source are primary and secondary data.
Primary data are fresh data which are collected for the task at hand. An example of such is census registration for cards.
Secondary data on the other hand are data dreaded in existence. They are originally collected for some purpose other than research current problem. They can be collected from school, hospital organization, government agencies, newspaper, monthly or annual report e.t.c thus a secondary data is used in this project.
1.4 LIMITATION OF THE STUDY
As we know that a researcher is bound to face certain problem. various problem were encountered before during and after data collection
The problem are:-
Poor storage, which make the transfer of data difficult Data were not properly recorded and in some cases, we have missing value.
Also data for some period are missing from the record office: those were available were poorly recorded this establishing one of the major.
Disadvantage of a secondary data usage data
collection :- before the data was released to me, I had to present a cover letter from the HOD and my student identify card and also promised that the data would be used for statistical purpose only.
Do you need help? Talk to us right now: (+234) 08060082010, 08107932631 (Call/WhatsApp). Email: edustoreng@gmail.com.
Disclaimer: This PDF Material Content is Developed by the copyright owner to Serve as a RESEARCH GUIDE for Students to Conduct Academic Research.
You are allowed to use the original PDF Research Material Guide you will receive in the following ways:
1. As a source for additional understanding of the project topic.
2. As a source for ideas for you own academic research work (if properly referenced).
3. For PROPER paraphrasing ( see your school definition of plagiarism and acceptable paraphrase).
4. Direct citing ( if referenced properly). | 1,821 | 8,944 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-31 | latest | en | 0.904341 |
https://www.wordunscrambler.net/unscramble-akomecr | 1,637,972,425,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00202.warc.gz | 1,205,134,028 | 14,897 | # Unscramble AKOMECR
AKOMECR unscrambles into 142 different words! We have all of them and the meanings below! Enter any word and we will UNSCRAMBLE IT!
### 2 letter words made by unscrambling AKOMECR
#### Unscrambled 17 2 Letter Words
Above are the words made by unscrambling AKOMECR (ACEKMOR). To further help you, here are a few lists related to/with the letters AKOMECR
### The Value of AKOMECR In Word Scramble Games
The letters AKOMECR are worth 15 points in Scrabble
The letters AKOMECR are worth 17 in points Words With Friends
• A = 1 points in WWF & 1 points in Scrabble
• K = 5 points in WWF & 5 points in Scrabble
• O = 1 points in WWF & 1 points in Scrabble
• M = 4 points in WWF & 3 points in Scrabble
• E = 1 points in WWF & 1 points in Scrabble
• C = 4 points in WWF & 3 points in Scrabble
• R = 1 points in WWF & 1 points in Scrabble
## What Does AKOMECR Mean... If you Unscramble it?
### Possible Definitions of AKOMECR
If we unscramble these letters, AKOMECR, it and makes several words. Here is one of the definitions for a word that uses all the unscrambled letters:
### comaker
• Sorry. I don't have the meaning of this word.
## Permutations of AKOMECR
According to our other word scramble maker, AKOMECR can be scrambled in many ways. The different ways a word can be scrambled is called "permutations" of the word.
#### Definition of Permutation
a way, especially one of several possible variations, in which a set or number of things can be ordered or arranged.
How is this helpful? Well, it shows you the letters akomecr scrambled in different ways That way you will recognize the set of letters more easily. It will help you the next time AKOMECR comes up in a word scramble game.
We stopped it at 50, but there are so many ways to scramble AKOMECR!
### Scramble Words
Unscramble these letters to make words...
scrambled using word scrambler... | 532 | 1,892 | {"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-2021-49 | latest | en | 0.89301 |
https://kr.mathworks.com/matlabcentral/cody/problems/1121-make-a-logical-diamond-using-gallery-function/solutions/1997744 | 1,582,173,348,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144637.88/warc/CC-MAIN-20200220035657-20200220065657-00480.warc.gz | 457,527,312 | 15,998 | Cody
# Problem 1121. Make a logical diamond using GALLERY function
Solution 1997744
Submitted on 31 Oct 2019 by rolf harkes
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('your_fcn_name.m') assert(~isempty(strfind(filetext, 'gallery'))) assert( isempty(strfind(filetext, '% gallery')))
filetext = 'function y = your_fcn_name(n) y=(gallery('fiedler',n)==floor(n/2)|fliplr(gallery('fiedler',n)==floor(n/2))) end %This code written by profile_id 9801364 '
2 Pass
n = 5; expected = [0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 0]; y_correct = logical(expected) assert(isequal(your_fcn_name(n),y_correct))
y_correct = 5×5 logical array 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 0 y = 5×5 logical array 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 0
3 Pass
n = 9; expected = [ 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0] y_correct = logical(expected) assert(isequal(your_fcn_name(n),y_correct))
expected = 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 y_correct = 9×9 logical array 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 y = 9×9 logical array 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 | 1,072 | 1,722 | {"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-2020-10 | latest | en | 0.599513 |
https://robotics.stackexchange.com/questions/24089/question-about-getting-global-coordinates-of-lidar-point-cloud-from-relative-in | 1,723,526,782,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00305.warc.gz | 381,517,241 | 41,764 | # Question about getting global coordinates of lidar point cloud from relative in Webots
I need to do custom mapping of surroundings with lidar using mobile robot in Webots. What I use for that:
• GPS for getting robot position.
• Compass for getting direction robot.
• Lidar for getting info about surroundings.
I did translation and rotation of relative points from lidar, which worked well when robot is on flat surface.
But no matter how much I tried I can't figure out how to get accurate global coordinates from point cloud relative points, when robot is even a bit tilted.
My guess is that it suppose to use spatial transformation matrices, but I not sure how to use Webots Compass values in rotation matrix.
Maybe someone can show code example or explain the math behind it or there is a method that I missed in Webots?
• You can also use a InertialUnit node and retrieve the ground truth quaternion for the orientation of the robot. Then, you may want to compare the results you get with the InertialUnit to the one you get with the Compass. This sample robot controller may be helpful to understand how to compare them. Commented Nov 30, 2022 at 8:47
• @OlivierMichel, didn't know such device existed before. So in the end InertialUnit.getQuaternion() is what I was looking for. Commented Dec 1, 2022 at 0:56
Basic Example of solution on Python:
from scipy.spatial.transform import Rotation as Rotation
RobotPoint = gps.getValues()
STR = Rotation.from_quat(InertialUnit.getQuaternion())
for RelativeCloudPoint in lidar.getPointCloud():
Point2 = STR.apply(RelativeCloudPoint)
GlobalCloudPoint = RelativeCloudPoint + RobotPoint
Using InternalUnit to get Quaternion for spartial rotation matrix. Then apply it to relative coordinates. After that add to it real robot coordinates from GPS. In the end you will get global coordinates of points you need.
I guess you’re using GPS to get the initial transformation, matching lidar points with ICP and re-calculating the pose update.
• P’1 = Pose_gps * P1
• Match P2 with P’1 using ICP
• Pose = P2 * pinv(P1)
I would use a Kalman filter with the pose in state vector and use the motion model in the state machine to constraint the motion.
Don’t forget to use quaternions and avoid gimble lock ;) | 512 | 2,263 | {"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} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.859107 |
http://au.metamath.org/mpegif/infxp.html | 1,508,535,848,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824357.3/warc/CC-MAIN-20171020211313-20171020231313-00662.warc.gz | 32,349,032 | 8,880 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > infxp Unicode version
Theorem infxp 7841
Description: Absorption law for multiplication with an infinite cardinal. Equivalent to Proposition 10.41 of [TakeutiZaring] p. 95. (Contributed by NM, 28-Sep-2004.) (Revised by Mario Carneiro, 29-Apr-2015.)
Assertion
Ref Expression
infxp
Proof of Theorem infxp
StepHypRef Expression
1 sdomdom 6889 . . 3
2 infxpabs 7838 . . . . . 6
3 infunabs 7833 . . . . . . . . 9
433expa 1151 . . . . . . . 8
54adantrl 696 . . . . . . 7
6 ensym 6910 . . . . . . 7
75, 6syl 15 . . . . . 6
8 entr 6913 . . . . . 6
92, 7, 8syl2anc 642 . . . . 5
109expr 598 . . . 4
121, 11syl5 28 . 2
13 domtri2 7622 . . . 4
15 xpcomeng 6954 . . . . . . 7
1615ad2ant2r 727 . . . . . 6
1716adantr 451 . . . . 5
18 simplrl 736 . . . . . . 7
19 simplr 731 . . . . . . . 8
20 domtr 6914 . . . . . . . 8
2119, 20sylan 457 . . . . . . 7
22 infn0 7119 . . . . . . . . 9
2322ad2antlr 707 . . . . . . . 8
2423adantr 451 . . . . . . 7
25 simpr 447 . . . . . . 7
26 infxpabs 7838 . . . . . . 7
2718, 21, 24, 25, 26syl22anc 1183 . . . . . 6
28 uncom 3319 . . . . . . . 8
29 infunabs 7833 . . . . . . . . 9
3018, 21, 25, 29syl3anc 1182 . . . . . . . 8
3128, 30syl5eqbr 4056 . . . . . . 7
32 ensym 6910 . . . . . . 7
3331, 32syl 15 . . . . . 6
34 entr 6913 . . . . . 6
3527, 33, 34syl2anc 642 . . . . 5
36 entr 6913 . . . . 5
3717, 35, 36syl2anc 642 . . . 4
3837ex 423 . . 3
3914, 38sylbird 226 . 2
4012, 39pm2.61d 150 1
Colors of variables: wff set class Syntax hints: wn 3 wi 4 wb 176 wa 358 wcel 1684 wne 2446 cun 3150 c0 3455 class class class wbr 4023 com 4656 cxp 4687 cdm 4689 cen 6860 cdom 6861 csdm 6862 ccrd 7568 This theorem is referenced by: alephmul 8200 infxpg 25095 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-3 7 ax-mp 8 ax-gen 1533 ax-5 1544 ax-17 1603 ax-9 1635 ax-8 1643 ax-13 1686 ax-14 1688 ax-6 1703 ax-7 1708 ax-11 1715 ax-12 1866 ax-ext 2264 ax-rep 4131 ax-sep 4141 ax-nul 4149 ax-pow 4188 ax-pr 4214 ax-un 4512 ax-inf2 7342 This theorem depends on definitions: df-bi 177 df-or 359 df-an 360 df-3or 935 df-3an 936 df-tru 1310 df-ex 1529 df-nf 1532 df-sb 1630 df-eu 2147 df-mo 2148 df-clab 2270 df-cleq 2276 df-clel 2279 df-nfc 2408 df-ne 2448 df-ral 2548 df-rex 2549 df-reu 2550 df-rmo 2551 df-rab 2552 df-v 2790 df-sbc 2992 df-csb 3082 df-dif 3155 df-un 3157 df-in 3159 df-ss 3166 df-pss 3168 df-nul 3456 df-if 3566 df-pw 3627 df-sn 3646 df-pr 3647 df-tp 3648 df-op 3649 df-uni 3828 df-int 3863 df-iun 3907 df-br 4024 df-opab 4078 df-mpt 4079 df-tr 4114 df-eprel 4305 df-id 4309 df-po 4314 df-so 4315 df-fr 4352 df-se 4353 df-we 4354 df-ord 4395 df-on 4396 df-lim 4397 df-suc 4398 df-om 4657 df-xp 4695 df-rel 4696 df-cnv 4697 df-co 4698 df-dm 4699 df-rn 4700 df-res 4701 df-ima 4702 df-iota 5219 df-fun 5257 df-fn 5258 df-f 5259 df-f1 5260 df-fo 5261 df-f1o 5262 df-fv 5263 df-isom 5264 df-ov 5861 df-oprab 5862 df-mpt2 5863 df-1st 6122 df-2nd 6123 df-riota 6304 df-recs 6388 df-rdg 6423 df-1o 6479 df-2o 6480 df-oadd 6483 df-er 6660 df-en 6864 df-dom 6865 df-sdom 6866 df-fin 6867 df-oi 7225 df-card 7572 df-cda 7794
Copyright terms: Public domain W3C validator | 1,793 | 3,351 | {"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-2017-43 | longest | en | 0.100721 |
http://mathforum.org/kb/message.jspa?messageID=6743564 | 1,524,237,012,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125938462.12/warc/CC-MAIN-20180420135859-20180420155859-00065.warc.gz | 206,612,595 | 13,509 | Search All of the Math Forum:
Views expressed in these public forums are not endorsed by NCTM or The Math Forum.
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic: Answer to Dik T. Winter
Replies: 441 Last Post: Feb 5, 2013 6:25 AM
Messages: [ Previous | Next ]
mueckenh@rz.fh-augsburg.de Posts: 18,076 Registered: 1/29/05
Re: Answer to Dik T. Winter
Posted: Jun 8, 2009 11:32 AM
On 8 Jun., 00:15, Virgil <virg...@nowhere.com> wrote:
> In a maximal infinite binary tree a path can have only one "end" and
> all paths have the same end: the root node.
That is the beginning. If you dislike the name "end", then speak of X
= "path without its first n nodes".
There are infinitely many nodes mapped on every X, namely those which
belong to that X. The X exhaust the complete binary tree, because
there is no node that remains unmapped. The X of every real number you
wish, including 1/3, 1/sqrt(2), 1/pi and so on, is among all paths
constructed.
Regards, WM
Date Subject Author
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Dik T. Winter
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Virgil
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 Virgil
5/27/09 Virgil
5/28/09 mueckenh@rz.fh-augsburg.de
5/28/09 Virgil
5/28/09 Dik T. Winter
5/28/09 G. Frege
5/28/09 Jesse F. Hughes
5/29/09 Dik T. Winter
5/29/09 G. Frege
5/29/09 Dik T. Winter
5/30/09 G. Frege
5/30/09 Jesse F. Hughes
6/2/09 Dik T. Winter
5/30/09 Jesse F. Hughes
6/1/09 mueckenh@rz.fh-augsburg.de
6/1/09 Jesse F. Hughes
6/1/09 Virgil
6/2/09 george
6/2/09 Denis Feldmann
6/2/09 Dik T. Winter
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 Guest
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/14/09 Owen Jacobson
6/14/09 Virgil
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/13/09 mueckenh@rz.fh-augsburg.de
6/13/09 Virgil
6/13/09 William Hughes
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 William Hughes
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 William Hughes
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 William Hughes
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 William Hughes
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 William Hughes
6/15/09 mueckenh@rz.fh-augsburg.de
6/15/09 Virgil
6/15/09 William Hughes
6/15/09 mueckenh@rz.fh-augsburg.de
6/15/09 Virgil
6/15/09 William Hughes
6/15/09 mueckenh@rz.fh-augsburg.de
6/15/09 Virgil
6/15/09 mueckenh@rz.fh-augsburg.de
6/15/09 Virgil
6/15/09 William Hughes
6/15/09 mueckenh@rz.fh-augsburg.de
6/16/09 Virgil
6/16/09 mueckenh@rz.fh-augsburg.de
6/16/09 Virgil
2/5/13
6/16/09 William Hughes
6/16/09 mueckenh@rz.fh-augsburg.de
6/16/09 Virgil
6/16/09 William Hughes
6/16/09 mueckenh@rz.fh-augsburg.de
6/16/09 Virgil
6/16/09 mueckenh@rz.fh-augsburg.de
6/16/09 Virgil
6/16/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
7/17/09 scriber77@yahoo.com
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 William Hughes
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Virgil
6/17/09 mueckenh@rz.fh-augsburg.de
6/17/09 Owen Jacobson
6/17/09 Virgil
6/17/09 William Hughes
6/18/09 mueckenh@rz.fh-augsburg.de
6/18/09 Virgil
6/18/09 William Hughes
6/18/09 mueckenh@rz.fh-augsburg.de
6/18/09 Virgil
6/18/09 William Hughes
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 William Hughes
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 William Hughes
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 William Hughes
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 William Hughes
6/20/09 mueckenh@rz.fh-augsburg.de
6/20/09 Virgil
6/20/09 William Hughes
6/20/09 mueckenh@rz.fh-augsburg.de
6/20/09 Virgil
6/20/09 William Hughes
6/20/09 mueckenh@rz.fh-augsburg.de
6/20/09 Virgil
6/20/09 William Hughes
6/20/09 george
6/20/09 george
6/20/09 Virgil
6/20/09 george
6/2/09 george
6/2/09 Jesse F. Hughes
6/3/09 george
6/3/09 mueckenh@rz.fh-augsburg.de
6/3/09 george
6/3/09 Virgil
6/3/09 Guest
6/3/09 Marshall
6/3/09 Jack Markan
6/3/09 Dik T. Winter
6/3/09 Guest
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
6/8/09 mueckenh@rz.fh-augsburg.de
6/8/09 Virgil
6/9/09 mueckenh@rz.fh-augsburg.de
6/9/09 Virgil
6/9/09 William Hughes
6/9/09 Virgil
6/10/09 mueckenh@rz.fh-augsburg.de
6/10/09 Virgil
6/10/09 mueckenh@rz.fh-augsburg.de
6/10/09 Virgil
6/10/09 mueckenh@rz.fh-augsburg.de
6/10/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Rainer Rosenthal
6/11/09 Virgil
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 Dik T. Winter
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 Dik T. Winter
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/19/09 Dik T. Winter
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/19/09 Virgil
6/22/09 Dik T. Winter
6/25/09 mueckenh@rz.fh-augsburg.de
6/25/09 Virgil
6/30/09 Dik T. Winter
6/30/09 mueckenh@rz.fh-augsburg.de
7/1/09 Dik T. Winter
7/1/09 Guest
7/2/09 mueckenh@rz.fh-augsburg.de
7/2/09 Virgil
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Rainer Rosenthal
6/12/09 Virgil
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Guest
6/12/09 Rainer Rosenthal
6/12/09 Virgil
6/3/09 george
6/3/09 Dik T. Winter
5/28/09 William Hughes
5/29/09 mueckenh@rz.fh-augsburg.de
5/29/09 Virgil
5/30/09 mueckenh@rz.fh-augsburg.de
5/30/09 Virgil
5/31/09 mueckenh@rz.fh-augsburg.de
5/31/09 Virgil
6/2/09 Dik T. Winter
6/2/09 Virgil
6/3/09 mueckenh@rz.fh-augsburg.de
6/3/09 Virgil
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
6/3/09 mueckenh@rz.fh-augsburg.de
6/3/09 Virgil
6/3/09 Dik T. Winter
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
6/11/09 Dik T. Winter
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 Dik T. Winter
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/19/09 Dik T. Winter
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/22/09 Dik T. Winter
6/25/09 mueckenh@rz.fh-augsburg.de
6/25/09 Virgil
6/30/09 Dik T. Winter
6/30/09 mueckenh@rz.fh-augsburg.de
6/30/09 MeAmI.org
6/30/09 Virgil
7/1/09 Dik T. Winter
7/2/09 G. Frege
7/2/09 Dik T. Winter
7/2/09 herb z
7/2/09 Dik T. Winter
7/2/09 herb z
7/2/09 G. Frege
7/3/09 Aatu Koskensilta
7/3/09 Daryl McCullough
7/2/09 G. Frege
7/2/09 G. Frege
7/2/09 G. Frege
7/2/09 Jack Markan
7/2/09 G. Frege
7/2/09 Jack Markan
7/2/09 G. Frege
7/2/09 Jack Markan
7/3/09 Dik T. Winter
7/3/09 G. Frege
7/3/09 Dik T. Winter
7/3/09 G. Frege
7/4/09 Aatu Koskensilta
7/6/09 Dik T. Winter
7/6/09 G. Frege
7/6/09 Dik T. Winter
7/6/09 Jesse F. Hughes
7/3/09 mueckenh@rz.fh-augsburg.de
7/3/09 Virgil
7/2/09 G. Frege
7/2/09 Virgil
7/2/09 mueckenh@rz.fh-augsburg.de
7/2/09 ross.finlayson@gmail.com
7/2/09 mueckenh@rz.fh-augsburg.de
7/2/09 Jack Markan
7/2/09 Jack Markan
7/2/09 Virgil
7/2/09 ross.finlayson@gmail.com
7/2/09 Jack Markan
7/2/09 ross.finlayson@gmail.com
7/6/09 Jack Markan
7/6/09 ross.finlayson@gmail.com
7/6/09 Jack Markan
7/6/09 ross.finlayson@gmail.com
7/6/09 Jack Markan
6/8/09 mueckenh@rz.fh-augsburg.de
6/8/09 Virgil
6/9/09 Jack Markan
6/9/09 Jack Markan
6/9/09 Virgil
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 William Hughes
6/11/09 Virgil
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/12/09 herb z
6/12/09 Virgil
6/12/09 Owen Jacobson
6/12/09 Virgil
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/11/09 Virgil
6/11/09 Dik T. Winter
6/11/09 William Hughes
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 YBM
6/11/09 Virgil
6/11/09 Dik T. Winter
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/12/09 mueckenh@rz.fh-augsburg.de
6/12/09 Virgil
6/19/09 Dik T. Winter
6/19/09 mueckenh@rz.fh-augsburg.de
6/19/09 Virgil
6/22/09 Dik T. Winter
6/25/09 mueckenh@rz.fh-augsburg.de
6/25/09 Virgil
6/30/09 Dik T. Winter
6/30/09 Virgil
6/30/09 Guest
6/30/09 mueckenh@rz.fh-augsburg.de
7/1/09 Dik T. Winter
6/11/09 mueckenh@rz.fh-augsburg.de
6/11/09 Virgil
6/3/09 mueckenh@rz.fh-augsburg.de
6/3/09 Virgil
6/3/09 Dik T. Winter
6/7/09 mueckenh@rz.fh-augsburg.de
6/7/09 Virgil
6/11/09 Dik T. Winter
6/3/09 george
5/29/09 mueckenh@rz.fh-augsburg.de
5/29/09 Virgil
5/30/09 mueckenh@rz.fh-augsburg.de
5/30/09 Virgil
5/31/09 mueckenh@rz.fh-augsburg.de
5/31/09 Virgil
5/31/09 Guest
5/27/09 Virgil
5/28/09 mueckenh@rz.fh-augsburg.de
5/28/09 Virgil
5/30/09 mueckenh@rz.fh-augsburg.de
5/30/09 Virgil
5/30/09 Virgil
5/30/09 Virgil
5/30/09 Virgil
5/27/09 mueckenh@rz.fh-augsburg.de
5/27/09 David C. Ullrich
5/27/09 Virgil
5/27/09 Guest
5/27/09 mueckenh@rz.fh-augsburg.de
5/28/09 mueckenh@rz.fh-augsburg.de
5/28/09 Virgil
5/29/09 Peter Webb
5/29/09 Virgil
5/29/09 mueckenh@rz.fh-augsburg.de
5/29/09 Virgil
5/29/09 Peter Webb
5/30/09 mueckenh@rz.fh-augsburg.de
5/30/09 Virgil
5/30/09 Peter Webb
5/31/09 mueckenh@rz.fh-augsburg.de
5/31/09 Virgil
5/31/09 mueckenh@rz.fh-augsburg.de
5/31/09 Peter Webb
5/31/09 Virgil
6/1/09 mueckenh@rz.fh-augsburg.de
6/1/09 David Bernier
6/1/09 Jack Markan
6/1/09 Virgil
6/2/09 george
6/1/09 mueckenh@rz.fh-augsburg.de
6/1/09 Virgil
6/2/09 george
6/2/09 george
6/2/09 george
6/3/09 george
6/12/09 Guest
2/5/13
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/14/09 mueckenh@rz.fh-augsburg.de
6/14/09 Virgil
6/15/09 Owen Jacobson
6/15/09 Virgil
6/15/09 mueckenh@rz.fh-augsburg.de
6/15/09 Owen Jacobson
6/15/09 Virgil
6/15/09 Virgil
6/16/09 MeAmI.org
6/16/09 mueckenh@rz.fh-augsburg.de
6/16/09 Virgil
6/18/09 mueckenh@rz.fh-augsburg.de
6/18/09 Virgil
6/20/09 Guest | 5,136 | 10,992 | {"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-2018-17 | longest | en | 0.622268 |
https://worldctx.com/ecology/how-do-you-do-an-ecological-footprint.html | 1,653,518,699,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662594414.79/warc/CC-MAIN-20220525213545-20220526003545-00734.warc.gz | 705,783,245 | 19,102 | # How do you do an ecological footprint?
Contents
The Ecological Footprint of a person is calculated by adding up all of people’s demands that compete for biologically productive space, such as cropland to grow potatoes or cotton, or forest to produce timber or to sequester carbon dioxide emissions.
## What is Ecological Footprint example?
The Ecological Footprint tracks the use of productive surface areas. Typically these areas are: cropland, grazing land, fishing grounds, built-up land, forest area, and carbon demand on land. … If a region’s biocapacity exceeds its Ecological Footprint, it has a biocapacity reserve.
## What is a good Ecological Footprint?
The world-average ecological footprint in 2013 was 2.8 global hectares per person. The average per country ranges from over 10 to under 1 global hectares per person. There is also a high variation within countries, based on individual lifestyle and economic possibilities.
## What is ecological footprint in simple words?
The simplest way to define ecological footprint would be to call it the impact of human activities measured in terms of the area of biologically productive land and water required to produce the goods consumed and to assimilate the wastes generated.
## How do you use ecological footprint in a sentence?
I believe we really must reduce our ecological footprint. We also need to consider the whole ecological footprint of the industry, to see whether such cuts are, in the end, justified. Our ecological footprint, in other words the environmental effect of our consumption, is much greater than what the earth can cope with.
## What is your ecological footprint?
The Ecological Footprint measures the amount of biologically productive land and sea area an individual, a region, all of humanity, or a human activity that compete for biologically productive space.
## How does ecological footprint affect the environment?
If everyone observed his or her ecological footprint, there will be less environmental problems today. Problems like carbon emissions, lack of fresh air, increased desertification, global warming and increased environmental pollution would be reduced.
## How do you calculate your ecological footprint?
Ecological Footprint Calculation. EF = ΣTi/Yw x EQFi, where Ti is the annual amount of tons of each product i that are consumed in the nation, Yw is the yearly world-average yield for producing each product i, and EQFi is the equivalence factor for each product i.
## What is ecological footprint and why is it important?
This is what the Ecological Footprint does: It measures the biologically productive area needed to provide for everything that people demand from nature: fruits and vegetables, meat, fish, wood, cotton and other fibres, as well as absorption of carbon dioxide from fossil fuel burning and space for buildings and roads.
## Why do we need to know about ecological footprint?
What we eat, how much we travel and which products we use are factors in determining how much we consume as humans. Ecological footprints are the measure of that consumption. … In order to preserve our remaining resources, it’s crucial that we reduce our consumption.
IMPORTANT: Are Barbie boxes recyclable?
## How are ecological footprint and carbon footprint related?
An ecological footprint, as explained earlier compares the total resources people consume with the land and water area that is needed to replace those resources. A carbon footprint also deals with resource usage but focuses strictly on the greenhouse gases released due to burning of fossil fuels.
## What factors affect your ecological footprint?
Resource consumption such as electricity, oil or water higher a person’s ecological footprint. Therefore, electricity consumption, oil consumption and water consumption are all factors that contribute to ecological footprint size.
## How do you use ecological in a sentence?
They have so far failed to come to grips with the ecological problems.
1. Each animal has its ecological niche.
2. They were warned of the ecological catastrophe to come.
3. Large dams have harmed Siberia’s delicate ecological balance.
4. The Black Sea is facing ecological catastrophe as a result of pollution.
## What does ecology deal with?
Ecology is the study of organisms and how they interact with the environment around them. An ecologist studies the relationship between living things and their habitats. | 852 | 4,444 | {"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-2022-21 | latest | en | 0.924171 |
https://www.univerkov.com/on-the-coordinate-line-points-a-and-a1-are-symmetric-about-point-p-1-if-1-a-3/ | 1,660,655,409,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572304.13/warc/CC-MAIN-20220816120802-20220816150802-00020.warc.gz | 881,162,668 | 6,699 | # On the coordinate line, points A and A1 are symmetric about point P (1) If: 1) A (3),
On the coordinate line, points A and A1 are symmetric about point P (1) If: 1) A (3), 2. A (-2), 3. A (-4), 4. A (5) Find the coordinate of point A1.
If points A and A1 are symmetrical about point P, then the distance between points A and P is equal to the distance between points A1 and P. That is, the length of the segment AP is equal to the length of the segment A1P.
If you need to find the coordinate of the center of symmetry, for example, for points O (-9) and K (5):
It is necessary to find the length of the segment between the symmetric points, the length of the segment is equal to the modulus of the difference in the coordinates of the points, that is, | -9 – 5 | = | -14 | = 14;
divide the length of the segment in half, 14: 2 = 7;
set aside from the point located to the left (from the point O (-9)), to the right 7 unit segments, (-9) + 7 = (-2);
or set aside from the point located to the right (from the point K (5)), to the left 7 unit segments, 5 – 7 = -2;
the coordinate of the center of symmetry has coordinate (-2).
Algorithm for finding a symmetric point
If the coordinates of one of the symmetric points and the center of symmetry are known, then you need to find the length of the segment between the point and the center of symmetry and postpone the same segment to the other side of the center of symmetry.
For example: given point B (8) and center of symmetry O (2). Let’s find the length of the segment IN: | 8 – 2 | = | 6 | = 6. Point B is located to the right of point O, which means that we put off to the left of point O 6 unit segments: 2 – 6 = -4. The coordinate of the point symmetrical to point B is (-4).
Find the coordinate of point A1
By condition: coordinate of the center of symmetry P (1).
1) A (3).
Let’s find the length of the segment PA: | 3 – 1 | = 2. Point A lies to the right of point P, therefore, to the left of point P, we postpone 2 unit segments: 1 – 2 = -1.
The coordinate of point A1 is (-1).
2) A (-2).
We find the length of PA: | -2 – 1 | = | -3 | = 3. Point A lies to the left of point P, so we move to the right of point P by 3 unit segments: 1 + 3 = 4.
The coordinate of point A1 is (4).
3) A (-4).
Let’s find the length of the segment PA: | -4 – 1 | = | -5 | = 5. Point A lies to the left of point P, so we move to the right of point P by 5 unit segments: 1 + 5 = 6.
The coordinate of point A1 is (6).
4) A (5).
Let’s find the length of the segment PA: | 5 – 1 | = 4. Point A lies to the right of point P, therefore, to the left of point P, we postpone 4 unit segments: 1 – 4 = -3.
The coordinate of point A1 is (-3).
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. | 840 | 2,998 | {"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} | 4.53125 | 5 | CC-MAIN-2022-33 | latest | en | 0.896849 |
https://s127041.gridserver.com/.tmb/journal/6nmqq7b.php?e5b311=dead-rising-2-cheats-pc | 1,632,050,773,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00653.warc.gz | 548,571,977 | 14,444 | # dead rising 2 cheats pc
dead rising 2 cheats pc
October 28, 2020
Finding in-vestable strategies entails the construction of optimal portfolios. # In my portfolio, I show how the popular Fama-MacBeth (1973) procedure is constructed in R. # The procedure is used to estimate risk premia and determine the validity of asset pricing models. I first create a dataset containing ``rankdates'', which are the date identifiers for the rolling regression. Rolling regression viewer. Using R: Fama MacBeth Regression - Portfolio formation and Stock return ranking. The sum βs are the sum of the slopes from a regression of the post‐ranking monthly returns on the current and prior month's VW NYSE returns. As common, i test the null hypothesis, if the average \$λ_t\$ is statistically different from zero. How do I properly run a Fama-Macbeth regression with a cross-sectional invariant variable? I have looked at the Fama-MacBeth two step regression because it essentially regresses y on m, n and o to determine how exposed it is to each one in the first step, while in the second step y is regressed on the found factor exposures derived from m, n and o, so that y … The Fama French 3-factor model is an asset pricing model that expands on the capital asset pricing model by adding size risk and value risk factors to the market risk factors. Two Stage Fama-Macbeth Factor Premium Estimation The two stage Fama-Macbeth regression estimates the premium rewarded to a particular risk factor exposure by the market. Get βp. Systematic Investor Toolbox, (includes the Three Factor Rolling Regression Viewer by forum member mas) Spreadsheet. Risk, Return, and Equilibrium: Empirical Tests Eugene F. Fama and James D. MacBeth University of Chicago This paper tests the relationship between average return and risk for To do this, I first get the first and last date for each permno, and then … I have uploaded the portfolios pr1, p2.. pr6 + SMB, RF, MKT_RF, MKT and HML to eviews from excel spreadsheet. Stat/Transfer is a cute tool to switch the data types. R_it = beta_i * F_t. Ask Question Asked 4 years, 9 months ago. The purpose of the regression test is to observe whether the five-factor model captures average returns on the variables and to see which variables are positively or negatively correlated to each other and additionally identifying the size of the regression slopes and how all these factors are related to and affect average returns of stocks values. Hurn (QUT) Applied Financial Econometrics using Stata 11 / 40. The CAPM is prolific, but doesn’t appear to work! 3. Fama-MacBeth Standard Errors. Fama-MacBeth (FM) (1973) represents a landmark contribution toward the empirical validation or refusal of the basic implications of the Capital Asset Pricing Model. Forum: FanChart: 2016/04/27: Creates a Bank of England style fan chart using forecast mode, uncertainty and skewness data. Fama-MacBeth (FM) cross-sectional regressions (see Fama and French, 2008, for a recent review). We find that the convergence of the OLS two-pass estimator depends critically on the time series sample size … )For example, in the figures below I’ve plotted the Fama-French 25 (portfolios ranked on size and book-to-market) against beta.. Fama and Macbeth (1973) regression(by Dr. Jeff Wongchoti)Fama and Macbeth regression is “a special type of regression methodology (very)widely used in financial research to handle panel data” (data series with both crosssectional (e.g. So I think it will be easier for me to run the time series regressions when the data is in time-series set and then reformate the data into a panel set for the second step regression in the Fama-MacBeth two-pass regressions. It includes Fama-MacBeth regressions, fixed effects, and bootstrapped standard errors, etc. Stata is easy to use but it is a little painful to save the outputs. In this paper, we provide the asymptotic theory for the widely used Fama and MacBeth (1973) two-pass regression in the usual case of a large number of assets. Downloadable (with restrictions)! Fama-MacBeth regression are cross sectional, as mentioned above and are predictive in nature. FAVARSF* The analysis is based on asset returns and factor returns published on Professor Kenneth French's data library. A side effect which is based on the market capitalization of a company is SMB. Many of the documented patterns are highly significant and seem almost certainly to be real, i.e., they are unlikely to be due to random chance or data-snooping biases. Second, our analysis is from an econometric, rather than from an investments, perspective. Do pass 1 for each stock. second stage cross-sectional regression leads to less efficient estimates of risk premia. I have a panel of 53 firms (id) with 18 years of daily observations of: Return (ret), Five Fama and French (2015) factors (rmrf, smb, hml, rmw, cma) - Calculate monthly returns for each from 1930-1934 (60 months) for the 20 portfolios. panel.xlsx I've computed the returns Ri for every asset in every month from my testing period (2007-2017), in rolling with estimation period 2004-2006, the same with beta. I've done a kind of panel in Excel, at my teacher's suggestion, but I'm really not sure how I should upload this to EViews and then apply the Fama-MacBeth add-in on this data. Rolling Your Own: Three Factor Analysis William Bernstein EF (Winter 2001) - an excellent tutorial on how to do this in Excel. \$\begingroup\$ Just a typical Fama/MacBeth regression on a test of the Fama-French-3-factor model. In this paper, we provide the asymptotic theory for the widely used Fama and MacBeth (1973) two-pass risk premia estimates in the usual case of a large number of assets. Stata does not contain a routine for estimating the coefficients and standard errors by Fama-MacBeth (that I know of), but I have written an ado file which you can download. Regression is a statistical measurement that attempts to determine the strength of the relationship between one dependent variable (usually denoted by Y) and a … (Note: see here for our epic post on the history of factor investing. - Rank securities by βand form into portfolios 1-20. # Google shows that the original paper has currently over 9000 citations (Mar 2015), making the methodology one of the most I am very new to R (I used Stata before). Active 4 years, 9 months ago. Get β. (Note that this line is important: etdata = data.set_index(['firm','year']), else Python won't know the correct dimensions to run F&McB on.) regression z it = i + if t + u it: Comparing the model and the expectation of the time-series regression, it follows that all the regression intercepts i should be zero. 2. • Example: Fama-MacBeth (1973) Data: 1926-1968 NYSE stocks Rm= Returns on the NYSE Index - Start with 1926-1929 (48 months). Despite its simplicity, the Fama-MacBeth method suffers from the well-known errors-in-variables (EIV) problem: That is, because estimated betas are used in place of true betas in the second stage cross-sectional regression, the second-stage regression estimates in the Fama-MacBeth method do not have the usual OLS or GLS properties. In the first figure, I plot the average excess return to the FF 25 against the average excess return one would expect, given beta. The independent variables in the Fama‐MacBeth regressions are defined for each firm at the end of December of each year t − 1. Stated practically, if you have a theory about what particular factors drive Perform Fama-French three-factor model regression analysis for one or more ETFs or mutual funds, or alternatively use the capital asset pricing model (CAPM) or Carhart four-factor model regression analysis. In my actual study, the risk factors are unobserved and extracted from Kalman filter process. Hey I have download the fama-macbeth add-in but I still have trouble to do the test for 6 Portfolios Formed on Size and Book-to-Market (2 x 3) from Europe. First, import the library readxl to read Microsoft Excel files, it can be any kind of format, as long R can read To add some detail to /u/Gymrat777's explanation, suppose that your asset returns are R_it and your factors are F_t.. First, run the following time-series regression for each stock i:. Fama-Macbeth approach is an innovative two-stage approach meant to minimize within-portfolio variance while capturing the across-portfolio characteristics... Their 1974 paper is not a landmark in terms of econometric modelling, but the approach is nice. A relevant portion of the available financial literature, see for example the remarkable work by Roll (1977), devoted its attention to the issue of determining the mean-variance Do pass 1 for portfolios. I got it to work in one go. See this site and run the lines of code for OLS below: "Here the difference is presented using the canonical Grunfeld data on investment." Data is from Kenenth R. French website. Fama-Macbeth: 2013/04/18: Performs Fama-MacBeth regression on a set of portfolio or asset returns and factors and returns summary results including the output of a simple cross-sectional average regression. In other words the regression intercepts are equal to the pricing errors. This yields an estimated betahat_i for each stock.. Second, for each time period t, run a cross-sectional regression:. Viewed 1k times 3. Its factor’s coefficient is calculated via linear regression, and it can have negative and positive values. A rankdate of 31Dec2001, for instance, uses data from 31Jan2000 to 31Dec2001, inclusive. The Fama-MacBeth Approach • Fama and MacBeth (1973) used the two stage approach to testing the CAPM outlined above, but using a time series of cross-sections • Instead of running a single time-series regression for each stock and then a single cross-sectional one, … In the next example, use this command to calculate the height based on the age of the child. A linear regression can be calculated in R with the command lm. Finally, our setting assumes Fama and French regressions, specifically in 1993 paper, are time-series, i.e., they develop portfolios and risk factors, then the time-series returns of each portfolio are … The direct output in Excel by double portfolio sorting SAS macro [ Newey-West (1987) t-stat in bracket] Sample Output 3 The direct output in Excel by Fama-MacBeth regression code [Newey-West (1987) t … Apologize for attaching screenshot instead of proper tables — still have to learn how to do it. Sometimes it is convenient to handle raw data in SAS and then perform statistical analysis in Stata. Questions: 1. Again, the logic behind the Fama-French model is that higher returns come from small-cap companies, rather than large-cap companies. The ado file fm.ado runs a cross-sectional regression for each year in the data set. Average \$ λ_t \$ is statistically different from zero calculated via linear regression, and bootstrapped standard errors,..: Creates a Bank of England style fan chart using forecast mode, uncertainty and skewness data a Bank England. A side effect which is based on asset returns and factor returns published on Professor Kenneth 's. Instance, uses data from 31Jan2000 to 31Dec2001, inclusive estimated betahat_i for stock... Factor Rolling regression Viewer by forum member mas ) Spreadsheet the logic behind the Fama-French model that. Cross-Sectional regression: data types as common, i test the null hypothesis if!, 9 months ago to do it Creates a Bank of England style fan chart using mode... The regression intercepts are equal to the pricing errors 31Dec2001, inclusive have. To 31Dec2001, inclusive Bank of England style fan chart using forecast,. Capitalization of a company is SMB − 1 year t − 1 regression, and it can negative. Uses data from 31Jan2000 to 31Dec2001, inclusive the 20 portfolios that higher returns come from small-cap,! To use but it is convenient to handle raw data in SAS and then perform statistical analysis in.. Cross sectional, as mentioned above and are predictive in nature a cute tool to switch data. Proper tables — still have to learn how to do it, run a cross-sectional regression: regression. ’ s coefficient is calculated via linear regression, and bootstrapped standard errors, etc Note! Switch the data types returns come from small-cap companies, rather than large-cap companies predictive nature! Is easy to use but it is a little painful to save the outputs to work defined for firm! Is easy to use but it is a little painful to save the outputs rankdate! Question Asked 4 years, 9 months ago to work portfolios 1-20 is statistically different from zero form... Three factor Rolling regression using forecast mode, uncertainty and skewness data in other words regression... Mas ) Spreadsheet λ_t \$ is statistically different from zero our analysis is based asset. In Stata Professor Kenneth French 's data library from an econometric, rather than companies. Fama/Macbeth regression on a test of the child cute tool to switch the data types Applied Financial using..., uses data from 31Jan2000 to 31Dec2001, inclusive to Calculate the height based on the market of! Date identifiers for the 20 portfolios 20 portfolios variables in the data set is based on returns. And skewness data t − 1, 2008, for instance, uses data from 31Jan2000 31Dec2001. Of factor investing of factor investing βand form into portfolios 1-20: Fama regression. Used Stata before ) December of each year t − 1 Econometrics using Stata 11 /.. Systematic Investor Toolbox, ( includes the Three factor Rolling regression Fama-French model is that higher returns from... A fama-macbeth regression are cross sectional, as mentioned above and are predictive in nature ’ s coefficient is via. S coefficient is calculated via linear regression, and bootstrapped standard errors, etc how to do.! Can have negative and positive values a test of the Fama-French-3-factor model the age the! Stata 11 / 40 i first create a dataset containing `` rankdates '', which are date! Stock.. Second, our analysis is based on the age of the Fama-French-3-factor model of the model... Each stock.. Second, for each stock.. Second, our is... To the pricing errors above and are predictive in nature ( includes the Three Rolling. I am very new to R ( i used Stata before ) Toolbox, includes. I properly run a fama-macbeth regression are cross sectional, as mentioned above and are predictive in.! Skewness data fama-macbeth ( FM ) cross-sectional regressions ( see Fama and French,,. Securities by βand form into portfolios 1-20 style fan chart using forecast,... Regression Viewer by forum member mas ) Spreadsheet ( see Fama and French, 2008, for time! Is calculated via linear regression, and it can have negative and positive values using R Fama. The 20 portfolios Creates a Bank of England style fan chart using mode... And then perform statistical analysis in Stata year t − 1 fama-macbeth regressions, fixed effects and., inclusive QUT ) Applied Financial Econometrics using Stata 11 / 40,... Investments, perspective to learn how to do it the Rolling regression by. Of the child year in the next example, use this command to Calculate the height based on the of..., for a recent review ) portfolios 1-20 betahat_i for each year in Fama‐MacBeth! With a cross-sectional regression for each time period t, run a cross-sectional regression for each from (... Professor Kenneth French 's data library am very new to R ( i used Stata before ) of... Three factor Rolling regression next example, use this command to Calculate the height based on the history factor! Intercepts are equal to the pricing errors regression on a test of the child for each year in data! File fm.ado runs a cross-sectional regression for each stock.. Second, our analysis is an. And stock return ranking other words the regression intercepts are equal to the pricing errors tables — still to! The logic behind the Fama-French model is that higher returns come from small-cap,. Return ranking negative and positive values on Professor Kenneth French 's data library Fama-French model is that returns... Standard errors, etc bootstrapped standard errors, etc t, run a fama-macbeth are... Defined for each year t − 1 than large-cap companies investments, perspective:! From zero regressions ( see Fama and French, 2008, for instance, uses data 31Jan2000. Then perform statistical analysis in Stata to switch the data types the Three factor regression! The outputs used Stata before ) proper tables — still have to learn how to do.! Bank of England style fan chart using forecast mode, uncertainty and skewness data to. Have negative and positive values year in the Fama‐MacBeth regressions are defined for each t!: Fama MacBeth regression - Portfolio formation and stock return ranking fama-macbeth regression excel ``... Epic post on the age of the child into portfolios 1-20 to 31Dec2001, inclusive history of investing! Note: see here for our epic post on the market capitalization of company... Calculate monthly returns for each time period t, run a fama-macbeth regression with a cross-sectional:! From 31Jan2000 to 31Dec2001, for a recent review ) fama-macbeth regression are sectional! On the age of the Fama-French-3-factor model defined for each firm at the end of December of each year the!, 9 months ago a dataset containing `` rankdates '', which are the date for. Is from an investments, perspective the Fama-French model is that higher returns come from small-cap companies, than... Identifiers for the 20 portfolios and then perform statistical analysis in Stata attaching... Each time period t, run a fama-macbeth regression with a cross-sectional regression each! Defined for each year t − 1 econometric, rather than large-cap companies predictive in nature first a. An econometric, rather than from an econometric, rather than large-cap.... Appear to work fama-macbeth ( FM ) cross-sectional regressions ( see Fama and French, 2008 for! French 's data library in the Fama‐MacBeth regressions are defined for each firm at the end of December of year... Properly run a cross-sectional invariant variable the 20 portfolios year in the data set epic on. Regression - Portfolio formation and stock return ranking variables in the next example, use this command to Calculate height! Econometric, rather than large-cap companies the market capitalization of a company is SMB on. Construction of optimal portfolios very new to R ( i used Stata before ) Rolling regression by... To R ( i used Stata before ): FanChart: 2016/04/27: Creates a Bank England. Then perform statistical analysis in Stata ) for the 20 portfolios a company is SMB how to do it forum... \Begingroup \$ Just a typical Fama/MacBeth regression on a test of the child are defined for each 1930-1934! Cross sectional, as mentioned above and are predictive in nature to handle raw in. French 's data library from 31Jan2000 to 31Dec2001, for each firm at the end December... Is statistically different from zero the construction of optimal portfolios for a recent review ) t, run a regression! Ado file fm.ado runs a cross-sectional invariant variable the construction of optimal.... Pricing errors the independent variables in the next example, use this command to Calculate height... Common, i test the null hypothesis, if the average \$ λ_t \$ is different! Proper tables — still have to learn how to do it is based the... R ( i used Stata before ) create a dataset containing `` rankdates '' which... A fama-macbeth regression are cross sectional, as mentioned above and are predictive in nature 4 years 9... Above and are predictive in nature a fama-macbeth regression with a cross-sectional variable. Words the regression intercepts are equal to the pricing errors run a fama-macbeth regression with a cross-sectional invariant?... Rankdates '', which are the date identifiers for the Rolling regression and standard... Pricing errors fama-macbeth regressions, fixed effects, and it can have and! Each year t − 1 do i properly run a cross-sectional regression: above and are predictive in nature Kenneth..... Second, our analysis is based on the age of the child regression are sectional... Company is SMB in SAS fama-macbeth regression excel then perform statistical analysis in Stata screenshot instead of tables... | 4,522 | 20,235 | {"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} | 2.859375 | 3 | CC-MAIN-2021-39 | latest | en | 0.857103 |
https://www.wyzant.com/resources/answers/845601/water-is-draining-from-an-inverted-conical-tank-that-has-a-height-of-12-met | 1,623,704,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00550.warc.gz | 966,757,242 | 13,631 | Akira I.
# Water is draining from an inverted conical tank that has a height of 12 metres and diameter of 8 metres.
a) As the water is draining out, the depth of the water, h, in the conical tank is decreasing at a rate of
(h – 12) m/minute. At what rate is the volume of the water in the conical tank changing when the depth of
the water, h, is 3 metres? (Need to use volume of a cone)
b) The conical tank is draining into a cylindrical tank that has a base with an area of 400(pi)(m2).At what rate is the depth of the water in the cylinder increasing when the depth of the water in the conical tank, h, has reached 3 metres? (Need to use volume of a cylinder) | 176 | 666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2021-25 | latest | en | 0.946378 |
https://www.coursehero.com/file/8728374/Chapter-11-t-test-for-two-related-samples/ | 1,529,702,473,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00313.warc.gz | 796,019,081 | 127,466 | Chapter 11- t test for two related samples
# Chapter 11- t test for two related samples - Chapter 11 The...
This preview shows pages 1–2. Sign up to view the full content.
Chapter 11: The t Test for Two Related Samples Terminology Matched-Subjects Design: Individuals in one sampled are matched to individuals in another sample. This is done so that the two individuals are matched to be as close to equal with respect to a specific variable. Related-Samples Design: Two research designs that are statistically equivalent and share a common name. Difference Scores: The difference between treatments for each individual Estimated Standard Error for M D : The “D” subscripts are used to emphasize that the variables are dealing with difference scores as opposed to X values. Repeated-Measures t Statistic: Refer to chapter 10 Individual Differences: Differences that are personal to the test subject that can skew results. This makes the repeated measures design more useful. Order Effects: Changes in scores that are caused by participation in earlier treatment. Wilcoxon Test: An alternative analysis to test if on the r tests has ben violated Formulas and Symbols Difference Scores: ࠵?
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
{[ snackBarMessage ]}
### What students are saying
• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.
Kiran Temple University Fox School of Business ‘17, Course Hero Intern
• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.
Dana University of Pennsylvania ‘17, Course Hero Intern
• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.
Jill Tulane University ‘16, Course Hero Intern | 482 | 2,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2018-26 | latest | en | 0.930505 |
http://hi.gher.space/forum/viewtopic.php?f=32&t=1468&start=30 | 1,670,241,237,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711016.32/warc/CC-MAIN-20221205100449-20221205130449-00310.warc.gz | 22,471,563 | 17,234 | ## Johnsonian Polytopes
Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.
### Re: Johnsonian Polytopes
I've created a wiki page for this: CRF polychora discovery project
I've also coined the term "extratruncate" to refer to the CD string xoo...oox, i.e. the 3D cantellate and 4D runcinate. See the Truncation page for etymology.
Keiji
Posts: 1969
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Johnsonian Polytopes
quickfur wrote:I just thought of another polychoron with square antiprisms: this one has 4 square antiprisms, so it's non-trivial. Basically, take 4 of these antiprisms and join them together by their square faces, then fold them up to form a 4-membered ring in 4D. If I'm not wrong, you should be able to fill in the gaps with tetrahedra and square pyramids. I think all edge lengths should be equal, so this should be a CRF polychoron too. I'm still trying to figure out how many tetrahedra/pyramids are needed to close up the shape.
It looks like alternated 8,4-duoprism. You'll need 24 tetrahedra to fill gaps, but I'm not sure that they will be regular. To make uniform antiprism you have to take high 8-prisms as cells, so instead of cubes you'll get non-regular square prisms. And resulting tetrahedra will have different edge lengths But may be there is more that one way to fill gaps? I doubt it - we have only one convex hull of vertices.
Attached is the file with model of "ortho" for my viewer...
Attachments
ortho.txt
"ortho" model
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
Looks like there is complete family of skew 3,n-duoprisms: you make combine two n-antiprisms with 1 n-prism, n tetrahedra and n square pyramids and get CRF body. Contructions based on 4,n rings give nothing new: if we take 2 prisms and 2 antiprsms, we get antiprismatic prism that is uniform, and for 4 antiprisms I guess that resulting body should be either uniform or non-CRF. But we don't have such uniform polychora in the list, so there is no new CRF in that side.
But what is skew 3,3? It has one triangular prism, 3 square pyramids attached to its sides, 3 tetrahedra between them and the rest is filled by two octahedra. 6 cells, 9 vertices. Does it really exist?
Last edited by Mrrl on Tue Nov 22, 2011 12:30 am, edited 1 time in total.
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
Mrrl wrote:Coordinates of "ortho" are (±(1+sqrt(2)),±1,0,0), (±1,±(1+sqrt(2)),0,0), (±1,±1,1,±1). For "gyro" it's more difficult, because height of 4-antiprism is a difficult value itself
OK, here's a render of the ortho variant:
I freely reoriented in order to get the best view. Because this is a relatively simple polytope, I turned off visibility clipping, so you should be able to locate all the cells here. The 4 tetrahedra, 4 triangular prisms, and cube should be immediately obvious. There is actually an octagonal face cutting through the middle of this thing, which isn't easy to see (maybe i'll re-render this with a slightly different color for the octagon); the two resulting halves then are the two square cupola.
OK, here's a slightly more enhanced render in an attempt to show the octagon:
It turns out that the octagon covers the entire 2D area of the 3D->2D projection, so changing the color of the octagon didn't help make it stand out. So instead, I colored the cube red so that you can see where the octagon intersects with it. And I made the cyan for the other ridges more opaque so that its presence is more easily seen.
Last edited by quickfur on Tue Nov 22, 2011 5:33 pm, edited 1 time in total.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
Thanks! To begin with it didn't look like the same shape, and I was about to moan about you missing out edges, but then I realized it did have all the facets after all. It just shows how much cleaner things look when rendered properly rather than just sketched in Inkscape
Keiji
Posts: 1969
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Johnsonian Polytopes
Mrrl wrote:
quickfur wrote:I just thought of another polychoron with square antiprisms: this one has 4 square antiprisms, so it's non-trivial. Basically, take 4 of these antiprisms and join them together by their square faces, then fold them up to form a 4-membered ring in 4D. If I'm not wrong, you should be able to fill in the gaps with tetrahedra and square pyramids. I think all edge lengths should be equal, so this should be a CRF polychoron too. I'm still trying to figure out how many tetrahedra/pyramids are needed to close up the shape.
It looks like alternated 8,4-duoprism. You'll need 24 tetrahedra to fill gaps, but I'm not sure that they will be regular. To make uniform antiprism you have to take high 8-prisms as cells, so instead of cubes you'll get non-regular square prisms. And resulting tetrahedra will have different edge lengths But may be there is more that one way to fill gaps? I doubt it - we have only one convex hull of vertices.
I thought it was the alternated 8,4-duoprism too, but then I'm no longer sure about that. It may be that the ring of square antiprisms doesn't make 90° internal angles, so the antiprisms form two pairs with different angles between them. So the coordinates would be different from the alternated 8,4-duoprism. There's still a chance it may be CRF. I'm still trying to work out the details.
Mrrl wrote:Looks like there is complete family of skew 3,n-duoprisms: you make combine two n-antiprisms with 1 n-prism, n tetrahedra and n square pyramids and get CRF body. Contructions based on 4,n rings give nothing new: if we take 2 prisms and 2 antiprsms, we get antiprismatic prism that is uniform, and for 4 antiprisms I guess that resulting body should be either uniform or non-CRF. But we don't have such uniform polychora in the list, so there is no new CRF in that side.
I think square antiprisms may be a special case, though, because the ring closes up fast enough that it may be possible to fill in the remaining gaps with johnson polyhedra. With larger rings, the chances of this becomes more and more unlikely because the resulting edge topology becomes too complicated and may not be able to fit regular polygons anymore.
But what is skew 3,3? It has one triangular prism, 3 square pyramids attached to its sides, 3 tetrahedra between them and the rest is filled by two octahedra. 6 cells, 9 vertices. Does it really exist?
I think so! Your cell list seems correct, and from what I can tell, it should be possible to make all edge lengths equal. But there's a danger that the square pyramids may be trapezoidal pyramids though. So this is not 100% confirmed yet.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
quickfur wrote:
Mrrl wrote:[...]
But what is skew 3,3? It has one triangular prism, 3 square pyramids attached to its sides, 3 tetrahedra between them and the rest is filled by two octahedra. 6 cells, 9 vertices. Does it really exist?
I think so! Your cell list seems correct, and from what I can tell, it should be possible to make all edge lengths equal. But there's a danger that the square pyramids may be trapezoidal pyramids though. So this is not 100% confirmed yet.
OK, a quick (informal) check with projection diagrams (buahaha) confirms that the square pyramids are CRF. So there are 2 octahedra, 1 triangular prism, 3 tetrahedra and 3 square pyramids.
And the construction is general; the shapes of the lateral cells are independent of the degree of the degree of the antiprisms and prisms (only the number is different). So there is such a family: for all n>=3, the corresponding member of this family has a 3-membered ring made with two n-gonal antiprisms, 1 n-gonal prism, n square pyramids, and n tetrahedra. In other words, this is an infinite family of CRF polychora. Would it be the first non-uniform infinite family discovered? Way to go, Andrey!
The existence of this family can be confirmed mathematically by noting that if we attach the two antiprisms together first, then the height of the prism controls the width of two parallel edges in the square pyramids, but since this height is free to vary, it can always be set to the edge length. So you always get equal edge lengths, and so the prism has square faces. So now you have this ring of faces on the pyramids: square, triangle, triangle, and so the other two triangles must also be equilateral. Likewise, the tetrahedra must be regular because all edge lengths are equal. So the existence of this family is proven.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
You should to note also that height of antiprism is more than half of prism's height, so the triangle really exists
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
quickfur wrote:I thought it was the alternated 8,4-duoprism too, but then I'm no longer sure about that. It may be that the ring of square antiprisms doesn't make 90° internal angles, so the antiprisms form two pairs with different angles between them. So the coordinates would be different from the alternated 8,4-duoprism. There's still a chance it may be CRF. I'm still trying to work out the details.
What if you take two n,3 skew duoprisms and glue the together by the prismatic cells? Is there a chance to get convex body? Also we may try to glue them by antiprismatic cells (and get two new bodies).
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
Mrrl wrote:You should to note also that height of antiprism is more than half of prism's height, so the triangle really exists
You're right, otherwise the antiprisms have to be stretched and the polychoron will no longer be CRF.
Well, if we fix edge length to L, then an n-gonal prism always has height L. As n approaches infinity, the band of triangles of an n-gonal antiprism approaches the horizontal tiling of equilateral triangles, whose height is the height of an equilateral triangle, L*sqrt(3)/2. Since sqrt(3)/2 > 1/2, the height of an n-gonal antiprism is never shorter than half the height of the corresponding n-gonal prism, so we're safe: all members of this family exist.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
Small comment to the discussion in Wiki: you can build a CRF pyramid over CRF polyhedron only if it is inscribed in the sphere - otherwise legths of triangles may be different. The same is about bipyramids: vertices of the base lay on the intersection of two hypespheres that is 2D sphere. So the base polyhedron should be inscribed in the sphere again. And the radius of sphere should be less than edge length.
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
Coordinates of "gyro":
octagon: (±(1+sqrt(2)),±1,0,0), (±1,±(1+sqrt(2)),0,0)
first square: (±1,±1,S,H)
second square: (±sqrt(2),0,S,-H), (0,±sqrt(2),S,-H)
here H=sqrt(sqrt(1/2)), S=sqrt(2-sqrt(1/2)).
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
Mrrl wrote:Coordinates of "gyro":
octagon: (±(1+sqrt(2)),±1,0,0), (±1,±(1+sqrt(2)),0,0)
first square: (±1,±1,S,H)
second square: (±sqrt(2),0,S,-H), (0,±sqrt(2),S,-H)
here H=sqrt(sqrt(1/2)), S=sqrt(2-sqrt(1/2)).
Are you sure your H value is correct? I generated the convex hull for these coordinates but got inconsistent edge lengths (2 and 3.46).
On another note, the ring of 4 square antiprisms is not a CRF polychoron. Just confirmed that it has tetrahedra with different edge lengths.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
i'm not sure in anything. Do you know what are the points that gave you this distance? And 3.46 is actually 2*sqrt(3)=3.4641016?
Mrrl
Trionian
Posts: 165
Joined: Sun May 29, 2011 7:37 am
### Re: Johnsonian Polytopes
Actually, nevermind. Your coordinates are correct. I don't know how I ended up with the wrong coordinates; I think I probably copy-and-pasted the wrong value from the calculator I used to evaluate H and S, so it screwed up the result. It's working now. Sorry for the false alarm!
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
Alright, after fixing the wrong coordinates that I don't know how it got into my polytope definition, I verified that all edge lengths are equal, and here's a nice projection of the result:
Again, I turned off visibility clipping because this is a relatively simple polytope. There is an octagonal face cutting through the middle of the image, like the previous ortho version, where the two cupola meet. So anyway. Have at it.
Side note: it's not so easy to tell from the 3D viewpoint, but the projection image is quite narrow; the radius of the octagon is significantly larger than the height of the square antiprism. So this particular projection image is almost like a lens shape.
Thanks, Mrrl, for the coordinates!
Last edited by quickfur on Tue Nov 22, 2011 5:34 pm, edited 1 time in total.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
BTW, I just got an inspiration about naming Keiji's polychora. I shall name them the cubic ortho-square-cupola tacotope and the square antiprismic gyro-square-cupola tacotope.
You see, a taco is basically a bunch of meat and veggies sandwiched by a circular tortilla skin folded in half. So the 4D analogue of a taco is a bunch of meat and veggies sanwiched by a spherical skin folded in half. The two cupolae, resemble halves of a 4D taco skin (if you unfold them back into 3D, the approximate a sphere), and they are joined at their octagonal bases so they are basically 4D taco shells. The cube and the square antiprism are the "meat" of the taco, and the triangular prisms, tetrahedra, and square pyramids are the veggies, and all of these are sandwiched between the two halves of the taco shell. So these polychora are tacotopes.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
And we can generalize this name to Mrrl's family of antiprismic CRF polychora too: they are just tacotopes with exotic-shaped skins, but they are essentially two halves of a shell with stuff sandwiched in between. So that family of tacotopes would be called the n-prismic n-antiprism tacotopes.
The first modifier is an adjectival form, because the meat of the taco is what gives it its name, for example, a taco containing fish is called a fish taco, a taco containing beef is called a beef taco, so the polyhedron of the 3-membered ring that isn't part of the skin gets to be the "meat" in adjectival form. The second modifier is in nominative form because it describes what kind of shell the taco has. For example, a taco with a hard shell is called a hard taco, so a taco made of two n-antiprisms is called an n-antiprism tacotope.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
Seems that diminished icositetrachora ARE part of augmented tesseracts, EXCEPT:
Icositetrachoron has 24 vertices, while tesseract only has 8 cells. Augmented tesseracts correspond to icositetrachora with vertices cut off from a specific 8-vertex subset. But it should be possible to cut off two or more vertices from DIFFERENT subsets, if they are far enough from each other - and those shapes wouldn't correspond to any augmented tesseract.
I think that the search for Johnsonian polychora needs to be two-fold:
1. Search for the primitive polychora, i.e. those that can't be diminished.
2. Put them together in various convex ways (bearing in mind that, unlike 3D case, here it's possible for two cells to be cohyperplanar if it means they will create a larger cell.
I'd suggest starting by enumerating the possible configurations of vertices and edges.
Let's look at possible dihedral angles first (all numbers in degrees). I assume that heptagonal, nonagonal and 11+-gonal prisms and antiprisms can't form anything more than duoprisms and antiprismatic prisms (although... some of those might be possible to augment?), so I omit those. Angles were taken from Great Stella.
So far it contains dihedral angles up to gyroelongated square dipyramid, feel free to extend
31,7175 - 4-10 angle of pentagonal cupola
37.3774 - 3-5 angle of pentagonal pyramid
3-10 angle of pentagonal cupola
Relations: 138.19 - 100.812, 127.377 - 90, half of 74.7547
45 - 4-8 angle of square cupola
Relations: half of 90, third of 135, 135 - 90
54.7356 - 3-4 angle of square pyramid
4-6 angle of triangular cupola
3-8 angle of square cupola
Relations: half of 109.471; 125.264 - 70.5288, 144.736 - 90, 158.572 - 103.836
60 - 4-4 angle of triangular prism, 4-4 angle of elongated triangular pyramid, 4-4 angle of elongated triangular dipyramid
Relations: Complementary to 120, half of 120
63.4349 - 5-10 angle of pentagonal rotunda
Relations: 142.623 - 79.1877
70.5288 - the dihedral angle of tetrahedron, 3-3 angle of elongated triangular pyramid, axial angle of triangular dipyramid, 3-3 angle of elongated triangular dipyramid
6-6 angle of truncated tetrahedron
3-6 angle of triangular cupola
Relations: Complementary to 109.471; 125.264 - 54.7356, 160.529 - 90, half of 141.058
74.7547 - equatorial angle of pentagonal dipyramid
Relations: double of 37.3774
79.1877 - 3-10 angle of pentagonal rotunda
Relations: 142.623 - 63.4349
90 - the dihedral angle of cube, 4-4 angle of elongated square pyramid, 4-4 angle of elongated square dipyramid
8-8 angle of truncated cube
3-4 angle of triangular prism, base 3-4 angle of elongated triangular pyramid
4-5 angle of pentagonal prism, 4-5 angle of elongated pentagonal pyramid
4-6 angle of hexagonal prism
4-8 angle of octagonal prism
4-10 angle of decagonal prism
Relations: double of 45, 135 - 45, 144.736 - 54.7356, 160.529 - 70.5288, 127.337 - 37.3774
95.2466 - 3-10 angle of decagonal antiprism
96.5945 - 3-8 angle of octagonal antiprism
98.8994 - 3-6 angle of hexagonal antiprism
100.812 - 3-5 angle of pentagonal antiprism, 3-5 angle of gyroelongated pentagonal pyramid
Relations: 138.19 - 37.3774
103.846 - 3-4 angle of square antiprism, 3-4 angle of gyroelongated square pyramid
Relations: 158.572 - 54.7356
108 - 4-4 angle of pentagonal prism, 4-4 angle of elongated pentagonal pyramid
109.471 - dihedral angle of octahedron, 3-3 angle of square pyramid, 3-3 angle of elongated square pyramid, pyramid 3-3 angle of gyroelongated square pyramid, 3-3 angle of elongated square dipyramid, pyramidal angle of gyroelongated square dipyramid
3-6 angle of truncated tetrahedron
6-6 angle of truncated octahedron
Relations: Complementary to 70.5288, double of 54.7356
116.565 - dihedral angle of dodecahedron, 10-10 angle of truncated dodecahedron
120 - 4-4 angle of hexagonal prism
Relations: Complementary to 60, double of 60
125.264 - dihedral angle of cuboctahedron, 3-4 angle of triangular cupola
4-6 angle of truncated octahedron
3-8 angle of truncated cube
6-8 angle of truncated cuboctahedron
Relations: 54.7356 + 70.5288
127.377 - 3-4 angle of elongated pentagonal pyramid
Relations: 37.3774 + 90
127.552 - 3-3 angle of square antiprism, equatorial 3-3 angle of gyroelongated square pyramid, antiprismatic angle of gyroelongated square dipyramid
135 - 4-4 angle of rhombicuboctahedron, 4-4 angle of octagonal prism, 4-4 angle of square cupola
4-8 angle of truncated cuboctahedron
Relations: triple of 45, 45 + 90
138.19 - dihedral angle of icosahedron, 3-3 angle of pentagonal antiprism, 3-3 angle of pentagonal pyramid, 3-3 angle of elongated pentagonal pyramid, 3-3 angle of gyroelongated pentagonal pyramid, axial angle of pentagonal dipyramid
6-6 angle of truncated icosahedron
Relations: 37.3774 + 100.812
141,058 - equatorial angle of triangular dipyramid
Relations: double of 70,5288
142.623 - dihedral angle of icosidodecahedron, 3-5 angle of pentagonal rotunda
5-6 angle of truncated icosahedron
3-10 angle of truncated dodecahedron
6-10 angle of truncated icosidodecahedron
Relations: 63.4349 + 79.1877
142.983 - 3-4 angle of snub cube
144 - 4-4 angle of decagonal prism
144.736 - 3-4 angle of rhombicuboctahedron, 3-4 angle of square cupola, 3-4 angle of elongated square pyramid, 3-4 angle of elongated square dipyramid
4-6 angle of truncated cuboctahedron
Relations: 54.7356 + 90
145.222 - 3-3 angle of hexagonal antiprism
148.283 - 4-5 angle of rhombicosidodecahedron, 4-5 angle of pentagonal cupola
4-10 angle of truncated icosidodecahedron
152.93 - 3-5 angle of snub dodecahedron
153.235 - 3-3 angle of snub cube
153.962 - 3-3 angle of octagonal antiprism
158.572 - join 3-3 angle of gyroelongated square pyramid, join 3-3 angle of gyroelongated square dipyramid
Relations: 54.7356 + 103.836
159.095 - 3-4 angle of rhombicosidodecahedron, 3-4 angle of pentagonal cupola
4-6 angle of truncated icosidodecahedron
159.187 - 3-3 angle of decagonal antiprism
160.523 - side 3-4 angle of elongated triangular cupola
Relations: +
160.529 - apex 3-4 angle of elongated triangular pyramid, 3-4 angle of elongated triangular dipyramid
Relations: 70.5288 + 90
164.172 - 3-3 angle of snub dodecahedron
Vertex data (and these are complete):
3-3-3: tetrahedron, elongated triangular pyramid, triangular dipyramid, elongated triangular dipyramid, augmented tridiminished icosahedron
3-3-4: square pyramid
3-3-5: pentagonal pyramid
3-4-4: triangular prism, elongated triangular pyramid, gyrobifastigium, augmented triangular prism
3-4-6: triangular cupola
3-4-8: square cupola
3-4-10: pentagonal cupola
3-5-5: metabidiminished icosahedron, tridiminished icosahedron, augmented tridiminished icosahedron, bilunabirotunda
3-5-10: pentagonal rotunda
3-6-6: truncated tetrahedron, augmented truncated tetrahedron
3-8-8: truncated cube, augmented truncated cube, biaugmented truncated cube
3-10-10: truncated dodecahedron, augmented truncated dodecahedron, parabiaugmented truncated dodecahedron, metabiaugmented truncated dodecahedron, triaugmented truncated dodecahedron
4-4-4: cube, elongated square pyramid
4-4-5: pentagonal prism, elongated pentagonal pyramid, augmented pentagonal prism, biaugmented pentagonal prism
4-4-6: hexagonal prism, elongated triangular cupola, augmented hexagonal prism, parabiaugmented hexagonal prism, metabiaugmented hexagonal prism
4-4-8: octagonal prism, elongated square cupola
4-4-10: decagonal prism, elongated pentagonal cupola, elongated pentagonal rotunda
4-5-10: diminished rhombicosidodecahedron, paragyrate diminished rhombicosidodecahedron, metagyrate diminished rhombicosidodecahedron, bigyrate diminished rhombicosidodecahedron, parabidiminished rhombicosidodecahedron, metabidiminished rhombicosidodecahedron, gyrate bidiminished rhombicosidodecahedron, tridiminished rhombicosidodecahedron
4-6-6: truncated octahedron
4-6-8: truncated cuboctahedron
4-6-10: truncated icosidodecahedron
5-5-5: dodecahedron, augmented dodecahedron, parabiaugmented dodecahedron, metabiaugmented dodecahedron, triaugmented dodecahedron
5-6-6: truncated icosahedron
3-3-3-3: octahedron, square pyramid, elongated square pyramid, gyroelongated square pyramid, triangular dipyramid, pentagonal dipyramid, elongated square dipyramid, gyroelongated square dipyramid, augmented triangular prism, biaugmented triangular prism, triaugmented triangular prism, augmented pentagonal prism, biaugmented pentagonal prism, augmented hexagonal prism, parabiaugmented hexagonal prism, metabiaugmented hexagonal prism, triaugmented hexagonal prism, snub disphenoid, augmented sphenocorona, sphenomegacorona
3-3-3-4: square antiprism, gyroelongated square pyramid, augmented triangular prism, biaugmented triangular prism, sphenocorona, augmented sphenocorona
3-3-3-5: pentagonal antiprism, gyroelongated pentagonal pyramid, metabidiminished icosahedron, tridiminished icosahedron, augmented tridiminished icosahedron, triangular hebesphenorotunda
3-3-3-6: hexagonal antiprism, gyroelongated triangular cupola
3-3-3-8: octagonal antiprism, gyroelongated square cupola
3-3-3-10: decagonal antiprism, gyroelongated pentagonal cupola, gyroelongated pentagonal rotunda
3-3-4-4: elongated triangular pyramid, elongated square pyramid, elongated pentagonal pyramid, elongated triangular dipyramid, elongated square dipyramid, elongated pentagonal dipyramid, triangular orthobicupola, square orthobicupola, pentagonal orthobicupola, sphenocorona, sphenomegacorona, hebesphenomegacorona, disphenocingulum
3-3-4-5: pentagonal gyrocupolarotunda, augmented pentagonal prism, biaugmented pentagonal prism
3-3-4-6: augmented hexagonal prism, parabiaugmented hexagonal prism, metabiaugmented hexagonal prism, triaugmented hexagonal prism, triangular hebesphenorotunda
3-3-5-5: pentagonal orthobirotunda, augmented dodecahedron, parabiaugmented dodecahedron, metabiaugmented dodecahedron, triaugmented dodecahedron, augmented tridiminished icosahedron
3-4-3-4: cuboctahedron, triangular cupola, elongated triangular cupola, gyroelongated triangular cupola, gyrobifastigium, triangular orthobicupola, square gyrobicupola, pentagonal gyrobicupola, elongated triangular orthobicupola, elongated triangular gyrobicupola, gyroelongated triangular bicupola, augmented truncated tetrahedron
3-4-3-5: pentagonal orthocupolarotunda, bilunabirotunda, triangular hebesphenorotunda
3-4-3-6: augmented truncated tetrahedron
3-4-3-8: augmented truncated cube, biaugmented truncated cube
3-4-3-10: augmented truncated dodecahedron, parabiaugmented truncated dodecahedron, metabiaugmented truncated dodecahedron, triaugmented truncated dodecahedron
3-4-4-4: rhombicuboctahedron, square cupola, elongated triangular cupola, elongated square cupola, elongated pentagonal cupola, gyroelongated square cupola, square orthobicupola, square gyrobicupola, elongated triangular orthobicupola, elongated triangular gyrobicupola, elongated square gyrobicupola, elongated pentagonal orthobicupola, elongated pentagonal gyrobicupola, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda, gyroelongated square bicopula, augmented truncated cube, biaugmented truncated cube
3-4-4-5: elongated pentagonal rotunda, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda, elongated pentagonal orthobirotunda, elongated pentagonal gyrobirotunda, gyrate rhombicosidodecahedron, parabigyrate rhombicosidodecahedron, metabigyrate rhombicosidodecahedron, trigyrate rhombicosidodecahedron, paragyrate diminished rhombicosidodecahedron, metagyrate diminished rhombicosidodecahedron, bigyrate diminished rhombicosidodecahedron, gyrate bidiminished rhombicosidodecahedron
3-4-5-4: rhombicosidodecahedron, pentagonal cupola, elongated pentagonal cupola, gyroelongated pentagonal cupola, pentagonal orthobicupola, pentagonal gyrobicupola, pentagonal orthocupolarotunda, pentagonal gyrocupolarotunda, elongated pentagonal orthobicupola, elongated pentagonal gyrobicupola, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda, gyroelongated pentagonal bicupola, gyroelongated pentagonal cupolarotunda, augmented truncated dodecahedron, parabiaugmented truncated dodecahedron, metabiaugmented truncated dodecahedron, triaugmented truncated dodecahedron, gyrate rhombicosidodecahedron, parabigyrate rhombicosidodecahedron, metabigyrate rhombicosidodecahedron, trigyrate rhombicosidodecahedron, diminished rhombicosidodecahedron, paragyrate diminished rhombicosidodecahedron, metagyrate diminished rhombicosidodecahedron, bigyrate diminished rhombicosidodecahedron, parabidiminished rhombicosidodecahedron, metabidiminished rhombicosidodecahedron, gyrate bidiminished rhombicosidodecahedron, tridiminished rhombicosidodecahedron
3-5-3-5: icosidodecahedron, pentagonal rotunda, elongated pentagonal rotunda, gyroelongated pentagonal rotunda, pentagonal orthocupolarotunda, pentagonal gyrocupolarotunda, pentagonal orthobirotunda, elongated pentagonal orthocupolarotunda, elongated pentagonal gyrocupolarotunda, elongated pentagonal orthobirotunda, elongated pentagonal gyrobirotunda, gyroelongated pentagonal cupolarotunda, gyroelongated pentagonal birotunda, bilunabirotunda, triangular hebesphenorotunda
3-3-3-3-3: icosahedron, pentagonal pyramid, elongated pentagonal pyramid, gyroelongated square pyramid, gyroelongated pentagonal pyramid, pentagonal dipyramid, elongated pentagonal dipyramid, gyroelongated square dipyramid, biaugmented triangular prism, triaugmented triangular prism, augmented dodecahedron, parabiaugmented dodecahedron, metabiaugmented dodecahedron, triaugmented dodecahedron, metabidiminished icosahedron, snub disphenoid, snub square antiprism, sphenocorona, augmented sphenocorona, sphenomegacorona, hebesphenomegacorona, disphenocingulum
3-3-3-3-4: snub cube, gyroelongated triangular cupola, gyroelongated square cupola, gyroelongated pentagonal cupola, gyroelongated triangular bicupola, gyroelongated square bicopula, gyroelongated pentagonal bicupola, gyroelongated pentagonal cupolarotunda, snub square antiprism, augmented sphenocorona, sphenomegacorona, hebesphenomegacorona, disphenocingulum
3-3-3-3-5: snub dodecahedron, gyroelongated pentagonal rotunda, gyroelongated pentagonal cupolarotunda, gyroelongated pentagonal birotunda
Marek14
Pentonian
Posts: 1183
Joined: Sat Jul 16, 2005 6:40 pm
### Re: Johnsonian Polytopes
quickfur wrote:tacotopes
I don't like this at all.
Firstly, the names are a freakin' mouthful, no pun intended. Secondly, they're full of redundancy. "cubic ortho" and "antiprismatic gyro"? Each word in those terms implies the other.
You're not recognising the way I wanted to classify these shapes. I thought I made it clear that they are not formed by an operation adding one dimension to a 3D shape. They are formed by adding two dimensions to a 2D shape, just like the duoprisms. There is something called the hexagonal-pentagonal duoprism for example. Your system would probably call this a "hexagonal prism ortho-pentagonal prism tacotope", which is just silly.
The names for my series of 9 shapes need to be like this: (triangular|square|pentagonal) (ortho|gyro|magna)something. Even "square orthoringcupola", the logical name I'd devised and then rejected from it sounding bad, sounds better than your taco naming. The important parts of the name are the n-gon from which it is formed and the particular sub-operation being used to take that n-gon into 4D. In fact, I'll just go ahead and adjust my original name slightly, to "square orthobicupolic ring", which I have no problems with.
Similarly, for Mrrl's infinite family. The important thing is that from an n-gon a ring of two antiprisms and a prism are assembled to form the resulting duoprism-like shape. In the same vein as my ring polychora, these should be named "n-gonal biantiprismatic ring". Notice I didn't even include the prism in the name, it is implied by the fact that it is a ring.
It's also worth pointing out that you shouldn't be naming these tacotopes based on how the projection happens to look a bit like a sandwich. The shapes look nothing like sandwiches in 4D.
Keiji
Posts: 1969
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Johnsonian Polytopes
Keiji wrote:[...]
It's also worth pointing out that you shouldn't be naming these tacotopes based on how the projection happens to look a bit like a sandwich. The shapes look nothing like sandwiches in 4D.
No, I didn't base the name on the projection, but on the 4D shape itself. It consists of two facets joined at the wide bases, with a facet of a different kind connecting their two narrow tops, forming a triangular configuration with two equal members and one unequal member. This is the basic form of the shape; the rest of the cells are added by the convex hull, and the added cells basically lie around the one facet of a different kind and between the two facets of the same kind.
And yes, the name was a bit silly but I was trying to capture this A-A-B triangular ring structure that forms the basis of these shapes. I wasn't trying to add a dimension to a 3D shape; I was trying to describe the 4D construction process using an analogy with how a widely-known object in 3D is similarly constructed: take a "shell" consisting of two A's joined at their wide bases, and fold them up in 4D and stick a B at the narrow end to make a 3-membered ring, then fill in the remaining gaps with other miscellaneous pieces.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
Alright, I was bored this morning so I decided to calculate the coordinates for the triangular cupola version of Keiji's CRF polychora. So here's the ortho variant:
Enjoy!
EDIT: and the coordinates are:
Hexagon:
(±sqrt(3), ±1, 0, 0)
(0, ±2, 0, 0)
Triangles:
(-1/sqrt(3), ±1, ±1, sqrt(5/3))
(2/sqrt(3), 0, ±1, sqrt(5/3))
Last edited by quickfur on Tue Nov 22, 2011 5:36 pm, edited 1 time in total.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
You should really get these images on the wiki sometime so they can get decent pages.
Keiji
Posts: 1969
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Johnsonian Polytopes
Keiji wrote:You should really get these images on the wiki sometime so they can get decent pages.
Alright, I was going to work on the coordinates for the gyro 3-prism cupola tritope (still trying to come up with a satisfactory name for these things), but I think i should start uploading all these renders to the wiki instead, 'cos i'm starting to accumulate too many temporary renders on my server. I think from now on I should just upload them to the wiki; it's a bad idea to have forum posts depending on external images in the first place.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
quickfur wrote:[...](still trying to come up with a satisfactory name for these things)[...]
Actually, I like your latest terminology on the wiki (bicupolic rings and biantiprismic rings). I'll stick to that from now.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
OK, here's the triangular gyrobicupolic ring:
Again I turned off visibility clipping because this polytope is relatively simple. The cells are easy to pick out: 6 square pyramids, 1 octahedron, and two triangular cupola.
The coordinates I'm using are:
Hexagon:
(±sqrt(3), ±1, 0, 0)
(0, ±2, 0, 0)
Triangle 1:
(-1/sqrt(3), ±1, sqrt(2/3), sqrt(2))
(2/sqrt(3), 0, sqrt(2/3), sqrt(2))
Triangle 2:
(1/sqrt(3), ±1, -sqrt(2/3), sqrt(2))
(-2/sqrt(3), 0, -sqrt(2/3), sqrt(2))
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
I see you just uploaded this:
Could you render it with a cubical envelope, i.e. the rotated square face in the middle? I think that would nicer, since the projection would be more symmetrical and easier to interpret.
Keiji
Posts: 1969
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Johnsonian Polytopes
I was going to do multiple renders of this one, because none of the usual views are very easy to interpret. At the very least, I would highlight some of the cells in order to show the structure better.
But how do I rename that file? "test.png" is a totally stupid name.
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
### Re: Johnsonian Polytopes
In Kawachan, files don't have names, they only have hashes. There's no way I or anyone else would have known it was called "test.png" if you hadn't just said so.
Keiji
Posts: 1969
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England
### Re: Johnsonian Polytopes
That's what I thought, I guess I got thrown off by the post-upload page that says "successfully uploaded test.png", so i thought it was actually going to use that lousy name. Sighh....
quickfur
Pentonian
Posts: 2841
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North
PreviousNext | 10,668 | 36,796 | {"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} | 2.75 | 3 | CC-MAIN-2022-49 | latest | en | 0.925659 |
https://getallcalculator.com/how-many-days-in-a-year/ | 1,558,891,094,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232259327.59/warc/CC-MAIN-20190526165427-20190526191427-00450.warc.gz | 484,558,659 | 8,695 | # how many days in a year
So here we are, one thing we are going to talk about is, a day, a year and then how these are two things are linked.
A day, being more specific exactly is the time duration In twenty-four hours. (How come?) Let see.
Generally, the smallest unit of time is 1 second, sixty of which makes a “minute.” Sixty of these “minutes” makes an “hour” and again 24 of these “hours” makes a “day.” That’s a little logical definition if a day. People have a different definition of a day, according to their perspective. Also, clear day of differ definition, day or simple day is different. Let us see different definitions of a “day.”
A Day is a time spent in between two consecutive midnight hour or reckoned from midnight. Also defines that week, months and ultimately a year. Due to earth’s rotation on its axis. It literary also means for a particular past time, or an era (in literature).
## how many days in a year
Technically, a day is unit of time, in which earth completes one rotation on its axis, concerning the sun. A public day is usually, 86,400 seconds long. Typically, the definition varies from people using it in different contexts. People sleep during the night and wake up in “day,” so another definition would be, time spent between two consecutive nights. The word “day” may also refer to a particular “day” of a week; it can be any day, Tuesday, Monday or Saturday. But a “day to years
A scientific day or stellar day Is termed as the time taken by earth to complete one full rotation on its axis. Although the period being, four minutes less than twenty-four hours. (23hours, 56 min, 4 seconds). Day time, on an average, is the time when light reaches the ground and leaves the field. So it comes out to be, 7 minutes longer than 12 hours.
#### Let us see how the year is defined,
It is the orbital period of earth that earth takes to revolve around the sun. Due to tilt in earth’s axis, we see different seasons and other factors including vegetation, soil fertility, and weather.
A calendar year is equal to the approximate orbital time and number of days counted in Georgian calendar or modern calendar of 365 days. And in a leap year, (366 days). For the current schedule, the exact number if days is equal to 365.245 days. This sword “year” is also used for different terms like a lunar year, seasonal year, business year, academic year, etc. A year too, means the orbital time for other planets too, like Martian year, Venusian year. The term may also be used to define the certain long interval of time about any incident or happening.
Different types of years include the sidereal year, anomalistic year, tropical year, draconic year, lunar year, vague year, heliacal year, Sothic year Julian year, etc. A mean tropical year is defined as the period for mean ecliptic longitude for the sun to add 360 degrees.
A business year and academic year may be the same for a country, depends upon the education system and time lime followed. But is a period of 12 months (usually from April to march-next year)
But for now, let’s stick to our basic definition of a year.
For better understanding, an essential follow-up I saw like following,
1 year = 365 days 1 day= 24 hours
1 year = 12 months 1 day = 1440 minutes
1 year = 52 weeks 1 day= 86,400 seconds
1 year = 3, 15, 36, 000 seconds 1 day = 1/365 year. (1/366 in leap year) | 803 | 3,384 | {"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-2019-22 | longest | en | 0.962753 |
http://www.dbforums.com/showthread.php?1622709-Calculate-Data-between-two-dates | 1,513,286,179,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948550986.47/warc/CC-MAIN-20171214202730-20171214222730-00191.warc.gz | 333,138,042 | 17,826 | # Thread: Calculate Data between two dates
1. Registered User
Join Date
Sep 2007
Posts
2
## Unanswered: Calculate Data between two dates
A newbie and I have learned a lot by reading posts but cannot find this one. I have a database with several yes/no checkboxes ie, referredtoRn,Followup,phoneCall, etc. I can calculate how many yes boxes are in each field - I researched and found that answer. However I want a report of how many in each field for between this date and this date such as a monthly report. Thanks in advance.
2. Registered User
Join Date
Oct 2002
Location
Posts
697
Originally Posted by oscarpoo
A newbie and I have learned a lot by reading posts but cannot find this one. I have a database with several yes/no checkboxes ie, referredtoRn,Followup,phoneCall, etc. I can calculate how many yes boxes are in each field - I researched and found that answer. However I want a report of how many in each field for between this date and this date such as a monthly report. Thanks in advance.
Something like this would work as a subquery: (you could paste it into a field in the query design grid)
SELECT COUNT(*) FROM someTable WHERE referredtoRn = true AND someDateField BETWEEN fromDate AND toDate
If the BETWEEN operator doesn't work, try fromDate <= someDateFIeld AND someDateField <= toDate instead.
3. Registered User
Join Date
Sep 2007
Posts
2
Ok I have tried and can not figure it out. I am using a query Clinic Reports Query that refers to a table (Clinic Info) I have a DateModified field
In each place of the Yes/No such as
First grid - Count of ReferredToRN: Sum(IIf([ReferredToRN],1,0))
Second Grid - Count of FollowUp: Sum(IIf([FollowUp],1,0))
etc
with total as expression. it works great. but as soon as I put in under Datemodified it gives me one line for each yes instead of one line for each with a total I think I am supposed to place this
SELECT COUNT(*) FROM Clinic Info WHERE referredtoRn = true AND DateMotified BETWEEN Start Date AND End Date
in an empty grid but it did not work. It keeps saying the syntax of the subquery in the expression is incorrect. Check the subquery syntax and enclose the subquery in parenthesis.
Am I on the right track? Missing something or just totally confused at this point?
Join Date
Nov 2004
Location
out on a limb
Posts
13,692
you can count all your values in one pass
however Id suggest using sum, and only including true / false, or Boolean columns.
the SQL is somehting like...
select abs(sum(<mybooleancolumn01>)) as <myalias01>, abs(sum(<mybooleancolumn02>)) as <myalias02>,....
abs(sum(<mybooleancolumnxx>)) as <myaliasxx>
from <mytablename>
where <maydatecolumn> between <mysepecifiedstartdate> and mysepecifiedenddate);
..this takes advantage of the fact that Access stores true as -1, and false as zero, it should go without saying that this will only work for as long as JET represents true as -1 and false as zero
so summing the value of the column gives the (negative) number of occurances of that column, the abs converts its to positive
HTH
Last edited by healdem; 09-24-07 at 05:30.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 807 | 3,229 | {"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-2017-51 | latest | en | 0.953165 |
http://forum.arduino.cc/index.php?topic=94973.msg713783 | 1,475,237,585,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662166.99/warc/CC-MAIN-20160924173742-00234-ip-10-143-35-109.ec2.internal.warc.gz | 93,228,785 | 18,044 | Go Down
### Topic: Why is there such a large difference in the current capacity of these vregs? (Read 4504 times)previous topic - next topic
#### scswift
These are both 3.3v regs, they have the same package, they appear to have nearly identical specs, yet the first one is rated for 250mA and \$1.50, and the second is rated for 150mA, and \$0.50:
The first one seems to have a slightly lower dropout. 350mV @ 125 degrees, versus 400mV @ 125 degrees for the second.
But the second one has a 150 degrees max junction temperature, while the first has only a 125 degrees max junction temperature, and I know this is used in the power dissipation calculations.
They both can handle up to 16v, and both list their PEAK output current as 350mA.
So what's the difference? Is it the dropout voltage? If so, where does that factor into any power dissipation calculations? I looked up how to calculate the power dissipation, and found this equation:
PD = IOUT * (VIN - VOUT) + VIN*IQ
where VIN = input voltage; VOUT = output voltage; IOUT = output current, A; and IQ = quiescent current, A
I assume quiescent current is also known as ground current as that seems to match up with the minumum quiescent current listed in the absolute maximum ratings and there's no other quiescent current listed. In which case for both vregs, it's 850uA typical at 150mA output. Which means as far as the power dissipation equation goes, the two regulators are identical.
So why is the current rating so different for these parts?
#### johnwasser
Maybe they just justify the price difference by degrading the specs on the lower-cost option.
Send Bitcoin tips to: 1G2qoGwMRXx8az71DVP1E81jShxtbSh5Hp
#### James C4S
It's a result of TI acquiring National. Two different companies, two different products, two different manufacturing processes, two different designs, for the essentially the same end product.
Capacitor Expert By Day, Enginerd by night. || Personal Blog: www.baldengineer.com || Electronics Tutorials for Beginners: www.addohms.com
#### scswift
So you're saying they're actually identical?
I took another look at the spec sheet and I noticed they had some graphs labeled "short circuit current". If I'm not mistaken those show a worst case scenario of the reg trying to put out as much current as it can at 6v and 16v. I think the graph also shows the thermal regulation kicking in shortly after the start of the test.
If that's the case, then for the first reg @ 6v, it peaks at 450mA, and drops to 300mA over the course of 2 seconds, then takes a nose dive, leveling out again just under 250mA.
For the second reg, it peaks at 350mA, drops to 250mA after half a second, then over the next 1.5 seconds slowly drops to 225mA.
That would seem to indicate the first part is slightly better. It can provide an average of 325mA for 2 seconds before thermal regulation kicks in, whereas the second part really can't provide more than an average of 250mA over that same period.
But it also seems to indicate the parts are largely the same, if all you're interested in is their long term current handling ability, and that the second part can handle quite a bit more than 150mA if you've got the input voltage low enough. (Seems at 16V they both crap out around 100mA.)
On the other hand, I just noticed the graphs for the two regulators aren't measuring exactly the same stuff. The first is measuring 2.8v out, and the second 3.3v out. I'm not sure how the difference would affect the graphs. It could either make the two regs more equal or more different. But still, with the second reg putting out 3.3v it's serving up quite a bit more current than I expected.
#### johnwasser
So you're saying they're actually identical?
No, he said they are different because they were designed and made by different companies before the companies merged.
Send Bitcoin tips to: 1G2qoGwMRXx8az71DVP1E81jShxtbSh5Hp
#### scswift
Yes, but he also said they were essentially the same. When I said "identical" I meant in specs, not internals.
#### jwatte
They are not identical in specs, because they have difference current capacity, and different cost. One was designed for lower cost, lower current capacity, and the other for higher cost, higher current capacity. They may look the same on the outside, but they are not on the inside!
Also, the fact that they both can go to 16V is probably more because of use cases (car batteries go to 13.8V, say.)
Btw: 400 mA drop-out compared to 250 mA of drop-out means 60% higher power dissipation at a minimum. Coupled that with a potentially different way to couple the hot parts to the outside, and you get parts that perform differently.
#### zoomkat
Yes, but he also said they were essentially the same. When I said "identical" I meant in specs, not internals.
Perhaps you can get somebody to help you interpet the data sheets.
Why I like my 2005 Rio Yellow Honda S2000 https://www.youtube.com/watch?v=pWjMvrkUqX0
#### scswift
Quote
Btw: 400 mA drop-out compared to 250 mA of drop-out means 60% higher power dissipation at a minimum.
That's what I would have thought myself, but none of the power dissipation equations I have seen in the datasheets or online appear to include drop out in the power dissipation calculations. And since the Iq for both parts is the same, the amount of power that needs to be dissipated for both is equal.
I think I found the difference between the two regulators though.
It's not that the two packages are mounted differently. The landing pad sizes are virtually identical.
The difference can be seen on page 5 note 3 in the first datasheet, and in the Absolute Maximum ratings on page 4 of the second datasheet. The junction-to-ambient thermal resistance: ?j-a
For the first part, ?j-a = 220°C/W
For the second, ?j-a = 206°C/W
And the equation for calculating the amount of power that can be dissipated at a specific ambient temperature is:
p(max) = ( Tj(max) - Ta ) / ?j-a
So, the power which can be dissipated is the maximum junction temperature, minus the ambient temperature, divided by the junction-to-ambient thermal resistance.
This power calculation isn't new to me, but yesterday when I was tired I could not for the life of me find those thermal impedance ratings and I saw something in one of the datasheets indicating what a typical value would be on FR4 pcb, and I thought that it must be the same for both parts because the two packages are the same size and neither has any kind of heat sinking.
This leaves me with a little problem though. A lower thermal impedance is better. And doing the calculations for am ambient temperature of 25 degrees, I get:
(125-25) / 220 = .455
(150 - 25) / 206 = .607
Which means the first part, the one which is 3x as expensive and has a higher current rating, can actually dissipate less power than the second part.
But that's for the same package. The first part comes in a second type of package with a heat sink on the bottom. And that package has a thermal resistance of 282 C/W on a 2 layer board with no heat sink (worse than the first two) and 62 C/W on a 4-layer board with 6 thermal vias.
Plugging that into my equation I get:
(125-25) / 62 = 1.61
Which is 2.65x better than the 50 cent regulator. So by all rights it ought to be rated for 400mA if the cheaper one is 150mA, but I don't know what criteria they used to decide to rate it... ie what amient temperature, what current, what input and output voltages. So that could explain the difference.
Anyway after all this, assuming I don't go with the third option, the part with the heat sink on the bottom:
http://search.digikey.com/us/en/products/LP2992ILD-3.3%2FNOPB/LP2992ILD-3.3CT-ND/567557
It would seem that the 50 cent part would be the better of the two choices.
Now I just have to figure out how much current it can really carry by doing the rest of those power calculatons.
#### scswift
Just did some calculations for the 50 cent 3.3v regulator:
Assuming a max PD of .607:
A = .607 / (Vin - Vout)
Vout = 3.3v, so:
.607 / (5 - 3.3) = .357 (Piggybacking off my 5A 5v regulator)
.607 / (8.4 - 3.3) = .119 (7.2v NiMh -> 8.4v at full charge)
.606 / (12 - 3.3) = .070 (12v battery - probably more like 13v+ but whatever)
.606 / (16 - 3.3) = .048 (16v worst case)
So based on this information, I think I should be okay using the same 50 cent 3.3v regulator I used for my SD card. Using a 7.2v rechargable or 6v alkaline will be what I reccomend, and if someone absolutely must run at 12v, they can plug the expansion board with the regulator into one of my servo ports which supply a regulated 5v instead of Vcc, which will put some more stress on the 5v 5A regulator but take a lot off the 3v.
#### scswift
Decided to do the calcs for my 5v reg as well:
http://www.micrel.com/_PDF/mic29150.pdf
Junction temp max - Tj(max) = 125°C
Ambient temp - Ta = 25°C
Thermal resistance - ?j-a = 2°C/W
Pd(max) = (Tj(max) - Ta) / ?j-a
Pd(max) = (125°C - Ta) / 2°C/W
Pd(max) = 25W @ 75°C
Pd(max) = 50W @ 25°C
I'm not sure which ambient temp to use, the regulator docs mention 75°C but 167°F seems excessive to me. Still, I'll go with 75 for my calculations.
This regulator gives a slightly different calculation for power dissipation which takes into account the ground current:
Pd = Iout (1.01Vin - Vout)
So if Vout = 5v, and Pd = 25W per my above calculation, and I rearrange the equation like so:
Iout = Pd / (1.01Vin - Vout)
Then, @ 75°C ambient and:
6v in, Iout(max) = 23.5A
8.4v in, Iout = 7.2A
12V in, Iout = 3.5A
16V in, Iout = 2.24A
#### oric_dan
Quote
A = .607 / (Vin - Vout)
Vout = 3.3v, so:
.607 / (5 - 3.3) = .357 (Piggybacking off my 5A 5v regulator)
.607 / (8.4 - 3.3) = .119 (7.2v NiMh -> 8.4v at full charge)
.606 / (12 - 3.3) = .070 (12v battery - probably more like 13v+ but whatever)
.606 / (16 - 3.3) = .048 (16v worst case)
Let me toss in a few ideas here. These last calculations are the most pertinent. They show
you are quite limited in how much current the v.regs can "realistically" supply, without melting
your pcb. Assuming the NiMH cells are the most likely to be used, you've got ~120 mA max.
HOWEVER, and it's a BIG however, you're allowing 150degC temp on the v.reg. How hot is that?
100degC is the temp of boiling water, and how long could you keep your hand in boiling water
for? 0.5 sec, that's it. IOW, you're really pushing it here.
Quote
PD = IOUT * (VIN - VOUT) + VIN*IQ
where VIN = input voltage; VOUT = output voltage; IOUT = output current, A; and IQ = quiescent current, A
I assume quiescent current is also known as ground current as that seems to match up with the minumum quiescent current listed in the absolute maximum ratings and there's no other quiescent current listed. In which case for both vregs, it's 850uA typical at 150mA output. Which means as far as the power dissipation equation goes, the two regulators are identical.
Secondly, you're initial ideas here were offbase. The PD eqn is simply how you calculate PD from
Ohm's Law and circuit theory, and has nothing to do with the characteristics of the v.regs. The 1st
term is the important one, and has nothing to do with what's "inside" the box, only with the
input-output characteristics. The 2nd term is basically insignificant, since IQ is > 150X smaller than
IOUT.
Thirdly, for my part [and I've designed a lot of pcbs], I wouldn't even think about using such a tiny
smt part at all, unless I was sure PD would always be maybe < 50% of your 0.6W value. Then the v.reg
would not overheat so much, and it would still be a 75degC = 167degF on the part. Still pretty
darn hot.
Fourthly, I don't even like the SOT223 parts [160degC/W], let alone the tiny ones you're using. I
go for the NCP1117 DPAKs [67 degC/W].
Fifthly, I think the answer to your initial question ["difference in current capacity"] mainly has to
do with the physical characteristics of the PNP output transistor in the parts, and not with the
size of the package, etc. Eg, the smaller the collector area of the transistor, the lower the
max current.
#### scswift
Thanks for the info Dan. :-)
Quote
Let me toss in a few ideas here. These last calculations are the most pertinent. They show
you are quite limited in how much current the v.regs can "realistically" supply, without melting
your pcb. Assuming the NiMH cells are the most likely to be used, you've got ~120 mA max.
HOWEVER, and it's a BIG however, you're allowing 150degC temp on the v.reg. How hot is that?
100degC is the temp of boiling water, and how long could you keep your hand in boiling water
for? 0.5 sec, that's it. IOW, you're really pushing it here.
Well, I was calculating absolute maximums, not necessarily the average current the regs would be putting out.
In my particular setup for example, I would like to power a 3v vibration motor or a few lasers. A massive 10G vibration motor or a green laser might need 200mA. A smaller vibration motor or three red lasers might need 150mA. And a small vibration motor or a single laser might need onl 60mA. But in any case, they would only be on for brief periods of time. A few seconds at most, usually.
Quote
Thirdly, for my part [and I've designed a lot of pcbs], I wouldn't even think about using such a tiny
smt part at all, unless I was sure PD would always be maybe < 50% of your 0.6W value. Then the v.reg
would not overheat so much, and it would still be a 75degC = 167degF on the part. Still pretty
darn hot.
Well, my plan as of this moment is to forget trying to run the 3.3v off the battery because even a NiMh doesn't leave me with much amperage to work with before the 3.3v reg would overheat. So the plan now is to run both 3.3v regs on my board off the 5v regulator. That will allow them to put put almost 350mA, which is 150mA more than what I ever expect the reg will need to put out, and even if does get to 200mA, it will only be for brief periods as I mentioned earlier.
Oh, and in case you're wondering why I have two 3.3v regs instead of just one bigger one, it has to do with the fact that the board is modular. The 3v3 reg that would drive the motor/lasers would be on its own board, and the other one drives an SD card. It would be nice if I could just use one big 3v3 reg on the main board, but I don't really have the space for it, and it would make the way everything connects to the main board a lot more complicated for people. Trying to keep things simple. The board is already studded with pinouts:
http://shawnswift.com/arduino/layout2.png
Quote
Fourthly, I don't even like the SOT223 parts [160degC/W], let alone the tiny ones you're using. I
go for the NCP1117 DPAKs [67 degC/W].
That looks like a nice part, but the problem is, that regulator doesn't have an enable pin, and the plan is to switch the motor and laser using the enable pin on the regulator. Keeps things simple. I've calculated the regulator should be able to switch at the PWM frequency of the Arduino (around 500hz) but if not, it's no big deal, I can do software PWM more slowly or not at all. It's just a vibration motor. Can't do PWM on the laser anyway; the driver boards usually don't like being switched faster than 10-100hz from what I've read.
Now usually, someone would use a mosfet for switching, but like I said, I don't really need to switch fast, I think the reg can handle PWM speeds, it saves me a component, and I don't have to worry about finding a mosfet which can handle the same current the regulator can. I had that problem with the first 5A reg I chose, because it didn't have an enable pin, and I need to be able to turn the power off, and I didn't want to have to use a 5A switch, but a mosfet that could handle 5A was the same size as the regulator. In the end I decided to just choose a larger regulator with an enable pin, and I probably got more current capacity as a result to boot.
I'd be interested in your thoughts on this plan though.
#### scswift
On another note, I'm looking for a suitable flyback diode for the aforementioned vibration motors.
This motor here is probably the worst case scenario:
https://catalog.precisionmicrodrives.com/order-parts/product/320-100-20mm-vibration-motor-25mm-type
So that would be run at 3.3v around 170mA, but I'd like to assume 200mA just to be on the safe side.
I found this diode on Digikey:
And I found this thread on flyback diodes:
http://www.electro-tech-online.com/electronic-projects-design-ideas-reviews/85310-when-flyback-diode-necessary.html
Retrolefty there states the diode should be 4x the voltage rating of the coil, and 40v is well over 12v. And it's got a 5.5A surge current rating which is over 25x as much as I think I need.
#### takao21106
the cheaper one's maybe just will make "pop" one day if you work them too much towards the margin.
It happened to me, using very tiny SMD diodes for a dc/dc converter.
Fine at 1.2 volts, at 5 volts it only survived for some minutes.
Then I changed to a big, way overdimensionated diode (2A rating, max. current maybe 100 mA).
I would not bother too much about it.
I also have PIC related web domain.
Go Up
Please enter a valid email to subscribe | 4,631 | 17,198 | {"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-2016-40 | longest | en | 0.918239 |
https://physics.stackexchange.com/questions/596918/why-it-is-necessary-that-phase-of-incident-reflected-and-refracted-wave-must-eq | 1,718,400,563,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861575.66/warc/CC-MAIN-20240614204739-20240614234739-00220.warc.gz | 420,371,262 | 40,239 | # Why it is necessary that phase of incident, reflected and refracted wave must equal at the interface of two medium?
1.Why it is necessary that phase of incident, reflected and refracted wave must equal at the interface of two medium to satisfy the boundary conditions at the interface? 2. According to boundary conditions fields left to the interface must join fields right to interface why we just add amplitudes of fields E(I) + E (R) = E(T) here E(I), E (R) and E(T) are amplitudes of incident,reflected and transmitted wave
• It's hard to answer without knowing your level. Are you familiar the way to derive conditions on fields at boundaries using Maxwell's equations? Commented Nov 28, 2020 at 20:39
• You may want to look into the Fresnel equations. Commented Nov 28, 2020 at 21:12
• Prof.Andrew Steane my particular doubt is if according to boundary conditions fields left to the interface must join fields right to interface why we just add amplitudes of fields E(I) + E (R) = E(T) here E(I), E (R) and E(T) are amplitudes of incident,reflected and transmitted wave Commented Nov 29, 2020 at 4:54
It's not necessary that the phase should be equal, as in the case of $$v_2, the reflected and incident wave is out of phase.
What is important that the field must satisfy the electrodynamic boundary conditions : $$(1) \ \ \epsilon_1E_1^\perp=\epsilon_2E^\perp_2, \ \ \ \ \ (2) \ \ \ \ E^\parallel_1=E_2^\parallel$$ $$(3) \ \ \ B_1^\perp=B_2^\perp, \ \ \ \ \ \ (4) \ \ \ \ \ \frac{1}{\mu_1}B^\parallel_1=\frac{1}{\mu_2}B^\parallel_2$$ These equations relate the electric and magnetic fields just to the left and just to the right of the interface between two linear media.
The rest follows from here.
It's not necessary to use Amplitude, You can, if want to, use the fields. Suppose like as in Griffith, the fields are : $$\mathbf{E}_I(z,t)=E_{0I}e^{i(k_1z-\omega t)}\hat{x}$$ $$\mathbf{E}_R(z,t)=E_{0R}e^{i(-k_1z-\omega t)}\hat{x}$$ $$\mathbf{E}_T(z,t)=E_{0T}e^{i(k_1z-\omega t)}\hat{x}$$
The interface is at $$z=0$$ so $$\mathbf{E}_I(0,t)+\mathbf{E}_R(0,t)=\mathbf{E}_T(0,t)$$
$$\Rightarrow E_{0I}e^{i(k_1\cdot 0-\omega t)}+E_{0R}e^{i(-k_1\cdot 0-\omega t)}=E_{0T}e^{i(k_1\cdot 0-\omega t)}$$ $$E_{0I}+E_{0R}=E_{0T}$$ That's the same thing as to add amplitudes.
• As I said, The reflection point occurs at a point in the reflection plane. You can choose it to be origin so that $(x,y,z)=(0,0,0)$. Commented Nov 29, 2020 at 5:39 | 789 | 2,443 | {"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": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-26 | latest | en | 0.855558 |
http://googology.wikia.com/wiki/Graham%27s_number | 1,550,673,581,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247495001.53/warc/CC-MAIN-20190220125916-20190220151916-00616.warc.gz | 120,770,917 | 37,907 | FANDOM
10,405 Pages
For other uses, see Graham's number (disambiguation).
Graham's number G is a famous large number, defined by Ronald Graham.[1]
Using up-arrow notation, it is defined as the 64th term of the following function:
\begin{eqnarray*} G_0 &=& 4 \\ G_1 &=& 3 \uparrow\uparrow\uparrow\uparrow 3 \\ G_2 &=& 3 \underbrace{\uparrow\uparrow\cdots\uparrow\uparrow}_{G_1 \text{ arrows}} 3 \\ G_{k + 1} &=& 3 \underbrace{\uparrow\uparrow\uparrow\cdots\uparrow\uparrow\uparrow}_{G_k \text{ arrows}} 3 \\ G_{64} &=& \text{Graham's number} \end{eqnarray*}
Graham's number is commonly celebrated as the largest number ever used in a serious mathematical proof, although much larger numbers have since claimed this title (such as TREE(3) and SCG(13)). The smallest Bowersism exceeding Graham's number is corporal, and the smallest Saibianism exceeding Graham's number is graatagold.
History
Graham's number arose out of the following unsolved problem in Ramsey theory:
Let N* be the smallest dimension n of a hypercube such that if the lines joining all pairs of corners are two-colored for any nN*, a complete graph K4 of one color with coplanar vertices will be forced. Find N*.
To understand what this problem asks, first consider a hypercube of any number of dimensions (1 dimension would be a line, 2 would be a square, 3 would be a cube, 4 would be a tesseract (4-dimensional cube), etc.), and call that number of dimensions N. Then, imagine connecting all possible vertex pairs with lines, and coloring each of those red or blue - one such way you can color all of the vertex pairs of the 3-dimensional cube is shown to the right. What is the smallest number of dimensions N such that all possible colorings would have a monochromatic complete graph of four coplanar vertices (that is a set of four points that are connected in all possible ways, with all lines being the same color)?
Graham published a paper in 1971 proving that the answer exists, providing the upper bound $$F^{7}(12)$$, where $$F(n) = 2 \uparrow^{n} 3$$ in arrow notation.[2] Sbiis Saibian calls this number "Little Graham". Martin Gardner, when discovering the number's size, found it difficult to explain, and he devised a larger, easier-to-explain number which Graham proved in an unpublished 1977 paper. Martin Gardner wrote about the number in Scientific American and it even made it to Guiness World Records in 1980 as the largest number used in a mathematical proof, although a few years later the title was removed from Guinness World Records.
In 2013, the upper bound was further reduced to N' = 2↑↑2↑↑2↑↑9 using the Hales–Jewett theorem.[3] As of 2014, the best known lower bound for N* is 13, shown by Jerome Barkley in 2008.[4]
Comparison
Since g0 is 4 and not 3, Graham's number cannot be expressed efficiently in chained arrow notation $$g_{64} \approx 3 → 3 → 64 → 2$$ or BEAF $$\{3,65,1,2\} < g_{64} < \{3, 66, 1, 2\}$$. Using Jonathan Bowers' G functions it is exactly G644 in base 3. It can be also exactly expressed in the Graham Array Notation as $$[3,3,4,64]$$.
Tim Chow proved that Graham's number is much larger than the Moser.[5] The proof hinges on the fact that, using Steinhaus-Moser Notation, n in a (k + 2)-gon is less than $$n\underbrace{\uparrow\uparrow\ldots\uparrow\uparrow}_{2k-1}n$$. He sent the proof to Susan Stepney on July 7, 1998.[6] Coincidentally, Stepney was sent a similar proof by Todd Cesere several days later.
It has been proven that Graham's number is much less than $$\Sigma(64)$$,[7] and later a better upper bound $$\Sigma(17)$$ was proven.
In the fast-growing hierarchy, Graham's number can be shown to be less than $$f_{\omega+1}(64)$$.
Approximations
Notation Approximation
Fast-growing hierarchy $$f_{\omega+1}(64)$$
Chained arrow notation $$3 \rightarrow 3\rightarrow 64\rightarrow 2$$
Hardy hierarchy $$H_{\omega^{\omega+1}}(64)$$
BEAF $$\{3,65,1,2\}$$
Extended Hyper-E Notation E[3]3##4#64
The Q-supersystem $$Q_{1,64}(3)$$
Hyperfactorial array notation $$64![2]$$
Strong array notation $$s(3,64,2,2)$$
X-Sequence Hyper-Exponential Notation $$3\{X+1\}65$$
Graham array notation $$[3,3,4,64]$$ (exact)
Slow-growing hierarchy $$g_{\Gamma_0}(65)$$
Calculating last digits
The final digits of Graham's number can be computed by taking advantage of the convergence of last digits, because Graham's number is a power tower of threes. Here is a simple algorithm to obtain the last $$x$$ digits $$N(x)$$ of Graham's number:
• $$N(0) = 3$$
• $$N(x) = 3^{N(x-1)} \text{ mod } 10^x$$
For example:
• $$N(1) = 3^{N(0)} \equiv 3^3 \equiv 27 \equiv 7 \pmod{10}$$, so the last digit is 7.
• $$N(2) = 3^{N(1)} \equiv 3^7 \equiv 2187 \equiv 87 \pmod{100}$$, so the last two digits are 87.
• $$N(3) = 3^{N(2)} \equiv 3^{87} \equiv 323,257,909,929,174,534,292,273,980,721,360,271,853,387 \equiv 387 \pmod{1,000}$$, so the last three digits are 387.
• etc.
This naive method is not very efficient, since number of digits in the leftmost expression grows exponentially. We can use right-to left binary method instead:
• Convert the exponent into binary form. E.g. $$87_{10} = 1010111_2$$
• If last digit of exponent is 1, then multiply base to result and square base.
• Otherwise, just square base.
Using this, it can be shown that last 20 digits of Graham's number are: $$...04,575,627,262,464,195,387$$.[8]
While it is impossible to calculate the leading digit of Graham's number in base 10, the leading digit must be 1 in base 2 (because all positive integers except 0 have this property), 1 in base 3 (because it is a power of 3), and 3 in base 9 (because it is an odd-numbered power of 3). Nor it is possible to calculate the leading digit of Graham's number in any other base, unless if it is a power of 3 (such as 27), or at least comparable to Graham's number (such as Graham's number - 64). | 1,721 | 5,840 | {"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} | 3.828125 | 4 | CC-MAIN-2019-09 | latest | en | 0.859232 |
http://www.pinktruth.com/2019/05/15/mary-kay-destination-red-cruise/ | 1,618,369,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038076454.41/warc/CC-MAIN-20210414004149-20210414034149-00190.warc.gz | 161,294,307 | 26,554 | # Mary Kay Destination Red Cruise
Let’s talk about this cruise that Mary Kay is offering to consultants in 2019. It’s a cruise to the Bahamas October 7 to 11, and when you qualify, you get to bring someone along. The cruise is on Royal Caribbean’s Navigator of the Seas, leaving out of Miami. The current cost if you just book yourself a room directly with Royal Caribbean is \$491 per person (double occupancy) for the least expensive room (an interior). Mary Kay is also covering base airfare (no baggage or other fees, though). Consultants can take a “cash option” of \$1,500 if they don’t want the cruise.To get this incentive, you have to purchase \$3,600 of inventory in the first 6 months of 2019. You also have to recruit 6 new consultants who order at least \$600 each in the same time period. (During January, qualified recruits counted as double, so you only needed to recruit 3 in January to meet this one.) Three of your recruits must also become “star team builders” for the first time during this period, which means at the end of a month, she has 3 active recruits. How much inventory is being sold to get this incentive:
• Winning consultant \$3,600
• 6 recruits x \$600 = \$3,600
• 3 star team builders x 3 active x \$225 = \$2,025
• TOTAL = \$9,225
Those total inventory orders of \$9,225 are the minimum the company will receive as a result of this promotion. There will likely be other orders placed under the winning consultant as various recruits of hers go active and inactive, trying to achieve the goal.
So Mary Kay Inc. benefits to the tune of \$9,225. They don’t actually pay \$1,000 for the rooms. They get a discount because they’re buying in bulk. But even if they pay something on the order of \$750 for a room, that’s only about 8% of the total they see coming in (at a minimum) from all this activity. MK can’t lose with this promotion.
The fine print:
• You can’t bring along a consultant or director in MK. So if you don’t have a spouse or significant other, don’t try to bring the friend or aunt that you signed up to round out your numbers.
• You pay for your drinks and your excursions. Mary Kay is covering the base cost of your room only.
All in all, it’s a better prize than they usually offer to consultants. We’re used to dollar store trinkets and dirty diamonds (for the “big” prizes at Seminar.) I can’t help but think promotions like this are evil genius. There are all sorts of consultants who will try and fail. The company gets all the money from the inventory orders, but doesn’t have to provide a prize. This is a well-oiled machine. I bet they can predict with 90% accuracy exactly how much of a bump in production they will get from this and how many prizes they will actually have to give out.
Another tidbit: Rumor had it that around the end of April, only 45 people had qualified for the cruise. That’s out of something like 500,000 consultants in the U.S. That’s pretty sad.
1. Enorth
Airfare to/from Miami *is* included.
MK will not pay airline baggage fees.
MK will not pay for transportation from your home to the airport.
You’ll need a passport, so factor in time and money if you don’t have one.
1. TRACY
I stand corrected and I’ll fix it. Do you have any docs with the details?
1. Enorth
I don’t. Just info gleaned from videos and flyers.
Also:
— If you earn the cruise, you can choose the cash option instead: \$1,500
— The sales force member will receive a 1099 for the entire amount of travel expenses for the sales force and the guest.
2. rdr
About the passport, MS Navigator of the Seas is registered in the Bahamas, not the US. Anyone know how Bahamas financial law treats MLMs?
3. Lazy Gardens
When traveling by AIR from the Bahamas, all US citizens must provide a valid US passport to enter or re-enter the states. Citizens traveling TO the Bahamas on short trips for tourist or business purposes do not require a visa.
So add the cost of the passport to the cruise, if you want to leave the ship.
PASSPORT VALIDITY: Must be valid at time of entry.
BLANK PASSPORT PAGES: Two pages are required for entry stamp.
TOURIST VISA REQUIRED: No.
For the car program MK offers a cash option that’s about half of the car lease supposed value. If the cash option for the cruise is \$1.500 I wonder what those cruise 1099s are going to say.
Oh yes, October is at the peak of the hurricane season. Matthew, Maria, Irma and Michael all hit in late September or October. Enjoy your trip, ladies.
1. Deflated Pink Bubble
I live in South Florida so rhe first thing that came to mind was HURRICANE SEASON! 🤣🤣
2. Wendy Duke Lambright
You are just so rude Good lord
1. Char
Scamming ladies for thousands of dollars in a pyramid scheme to profit is so far beyond “rude” that you are in no position to judge. Plus, MLM Radar stating facts about hurricane season isn’t rude. Not wishing a ship of con artists a good time perpetuating their scheme is appropriate. Of course, no one here wishes physical harm on anyone; but financial harm on shysters is another story. A bit of discomfort would be nice too!
2. BestDecision
So, you haven’t even finished your Mini Cooper, yet I bet you say you make an executive income. I was a Director, too, and the numbers don’t lie. \$75,000 over 6 months is FAR from “executive”. Please tell me you haven’t uttered those words to your unit or to any prospective recruits. Please tell me you’re honest and admit you gross about \$3,000/month in commissions and keep way less than that.
Honesty is the ONLY policy accepted on here.
3. parsonsgreen
Lol also how did taking goodie bags to jury duty work out for you Wendy?
1. Megan
Oh my gosh, please tell me this is a real thing LOL I thought I was surprised at nurses and hospitals! All these warm chatting opportunities make it seem so pathetic and desperate. If this is considered ‘working hard at my business’, I’d rather not be in this “business” at all.
Please don’t take this as an insult; I see how it’s so easy to get indoctrinated. But was there ever embarrassment in this? Or are you so deep in it that nothing ever seems inappropriate? I have seen some youtube “scripts”, and I could never morally do things that include outright lying to random people.
4. BestDecision
Wendy, you haven’t responded. \$36,000/year gross commissions aren’t “executive income”, are they??
3. When the 1099 includes \$1500 for “winning” the cruise, it’s costing you approx. \$500 in taxes. That’s NOT FREE.
1. TRACY
Precisely why it’s much better to take the cash option. You still pay the \$500 in taxes, but you have \$1,000 left in your pocket when all is said and done.
The cash option is \$1,500, but MK’s cash option has historically been half of the “prize worth” amount. So I’m thinking the 1099 for the women who actually take the trip will say something more like \$3,000.
Remember, the trip is for you and guest, but the cash option is just for you. Also, MK always makes a profit on prizes
Of course MK won’t come right out and say that. You’ll find out next January when that 1099 arrives. Surprise!
1. Deflated Pink Bubble
This trip was supposed to be later this week. My comment about hurricane season was considered rude by some drive by troll but alas, a hurricane has hit and pretty much leveled the Bahamas. I imagine Mary Kay will be paying out the cash equivalent to whomever qualified.
1. TRACY
No, it’s a month from now. Royal Caribbean will still have the cruise, they’ll simply go to different ports.
1. Deflated Pink Bubble
Whoops! It is a month away!
2. enorth
If you earn the trip, you select whether you want the trip or the cash.
Once you select the trip, your travel arrangements are made and you cannot change your mind to the cash option.
— “Cancellations after registration will result in the forfeiture of the cash option.”
1. TRACY
I saw that. Makes sense, because once you choose the trip, MK will confirm the reservation with RCL to lock in the room and the rate.
1. enorth
And also make airline reservations for you and your guest.
4. Lazy Gardens
“When the 1099 includes \$1500 for “winning” the cruise, it’s costing you approx. \$500 in taxes.”
So you spend \$3600 in product ordfers AND then have to PAY \$500 more in taxes?
1. BestDecision
Yes! Taxes on all our prizes. I hated it!
1. Lexi
Doesn’t it work like that with all cash prizes as they count as income?
At least with the cash you have something real.
When you take the prize instead, Mary Kay sticks you with an IRS Form 1099 that claims your income included the value of your prize. It makes a nice tax deduction for Mary Kay Corporate, because they subtract the total of all those 1099s as business expenses.
But as for you, surprise! You don’t get a statement detail showing that the 1099 is supported by anything except Mary Kay’s own inflated estimate of what the prize is supposedly worth. But you get stuck paying taxes on that dollar amount.
If you don’t think that 1099 figure is inflated for a Mary Kays own benefit, just ask anyone who ever tried to sell a Mary Kay prize ring. You’re told it’s “valued at over \$800!” You pay tax on \$800. Then you find out it’s really just dirty spit diamond scraps and junk gold worth maybe \$100.
2. BestDecision
Yes, but these are tangible items that are taxed and even subtracted from inventory refunds. Any item we received was counted as income on our tax form.
5. OnelessSD
I feel sorry for ALL the Non-MK people on this boat…. but on the flip side… they will be over exposed to the crazy that is all MK people.
I remember going to Hawaii for my 10th anniversary. My husband and I were so excited to go on vacation for the 1st time without the kids, etc… looking forward to a real relaxing time. Unfortunately- we had a MLM trip on our flight… Party Lite Candles…. SO MANY OBNOXIOUS WOMEN! Sadly… it reminded me of MK (which I was heavily still working)… but I never put 2 & 2 together.. as that is how the outside world viewed me when I was with my MK friends.
God help those non-MK people who paid good money to go on a relaxing vacation.. and they get stuck with all that. Ugh.
1. Char
I wonder if I could get a cheap deal with RC if I mention that I know MK nut jobs will be on the ship? LOL
6. didpink
That sounds like a lot of product to line your shelves as well as your recruits. You’re better off just booking a cruise and enjoying BEFORE hurricane season.
7. Cindylu
I have been on many better vacations with my husband and adult children. In the end in MK, I figured out that those trips, cars, jewelry etc. were all low end bribes. (With a catch). It’s like the MK cult indoctrinates us into chasing that pot of gold. Like a gambling addiction, we spend time, money and energy for that NSD prize. Sadly this NSD lottery is a long shot. Worst is the amount of debt most MK women get themselves into. By the time we pay for the kit, the samples, the gas, the clothes, the fake training, the Open Houses along with decor, coffee etc., the conferences, Seminar and inventory etc. our meager earnings dwindle and debt ensues.
8. Megan
Surely this cruise aren’t all full of desperate scammers? I feel awful if regular people just wanted to go on a nice vacation only to be harassed the entire time.
1. Megan
*isn’t.
But considering how homophones are so difficult for these whenever you post the pro-MK emails, it makes me sad how much time is wasted instead of going to college, if that suits you, or just a normal job. Actual professionals wouldn’t put up with that. I don’t why the grammar in particular bothers me so much. Very middle-school-ish. | 2,753 | 11,621 | {"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-2021-17 | latest | en | 0.950944 |
https://studysoup.com/tsg/981032/mathematical-statistics-with-applications-7-edition-chapter-5-problem-5-141 | 1,597,369,497,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439739134.49/warc/CC-MAIN-20200814011517-20200814041517-00572.warc.gz | 504,586,042 | 11,329 | ×
×
# Let Y1 have an exponential distribution with mean and the conditional density of Y2
ISBN: 9780495110811 47
## Solution for problem 5.141 Chapter 5
Mathematical Statistics with Applications | 7th Edition
• Textbook Solutions
• 2901 Step-by-step solutions solved by professors and subject experts
• Get 24/7 help from StudySoup virtual teaching assistants
Mathematical Statistics with Applications | 7th Edition
4 5 1 277 Reviews
10
4
Problem 5.141
Let Y1 have an exponential distribution with mean and the conditional density of Y2 given Y1 = y1 be f (y2 | y1) = 1/y1, 0 y2 y1, 0, elsewhere. Find E(Y2) and V(Y2), the unconditional mean and variance of Y2.
Step-by-Step Solution:
Step 1 of 3
STAT 110: Notes for Week of 10/25/16 Chapter 21 o This section will focus on interpreting data and inferential statistics (making inferences about a population based on data from a population sample). o The sample proportion probably is not equal to the parameter. Sample proportion: p = (# of successes/n) n = number...
Step 2 of 3
Step 3 of 3
##### ISBN: 9780495110811
Mathematical Statistics with Applications was written by and is associated to the ISBN: 9780495110811. The full step-by-step solution to problem: 5.141 from chapter: 5 was answered by , our top Statistics solution expert on 07/18/17, 08:07AM. This textbook survival guide was created for the textbook: Mathematical Statistics with Applications , edition: 7. Since the solution to 5.141 from 5 chapter was answered, more than 209 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 32 chapters, and 3350 solutions. The answer to “Let Y1 have an exponential distribution with mean and the conditional density of Y2 given Y1 = y1 be f (y2 | y1) = 1/y1, 0 y2 y1, 0, elsewhere. Find E(Y2) and V(Y2), the unconditional mean and variance of Y2.” is broken down into a number of easy to follow steps, and 41 words.
#### Related chapters
Unlock Textbook Solution | 539 | 2,044 | {"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-2020-34 | latest | en | 0.880156 |
https://toph.co/p/nothing-is-absolute-everything-is-relative | 1,642,972,608,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00015.warc.gz | 615,015,531 | 13,047 | # Practice on Toph
Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.
# Nothing Is Absolute, Everything Is Relative!
By shuvon · Limits 1.5s, 512 MB
You are given an integer array, $A[ ]$ of size $N$, and an integer $M$. Then you have to perform a total of $M$ operations on the array. The operations can be of $2$ types. They are:
Operation type 1:
$1$ $L$ $R$ $X$
Add $X$ to each element of the sub-array $[L,R]$. For example, if the current array is $A[ ] = [1, 3, 8, 5, 10, 6]$, after performing the operation $1$ $2$ $5$ $3$ the updated array will be $A[ ] = [1, 6, 11, 8, 13, 6]$.
Operation type 2:
$2$ $L$ $R$
Find the summation of absolute differences between the adjacent elements of the sub-array $[L,R]$. More formally, find the value of $|A_L - A_{L+1}| + |A_{L+1} - A_{L+2}| + \ldots + |A_{R-1} - A_R|$. Here, $|x|$ means absolute (non-negative) value of $x$.
## Input
The first line contains two space-separated integers $N$ and $M$, the size of the array and the number of operations, respectively. The second line contains $N$ space-separated integers denoting the array. The $i^{th}$ integer represents the $i^{th}$ element, $A[i]$ of the array. After that $M$ lines follow. The $j^{th}$ line represents the $j^{th}$ operation.
Constraints:
$2 \leq N,M \leq 5\times10^5$
$1 \leq L \leq R \leq N$
$-10^7 \leq A[i],X \leq 10^7$
In case of operation type $2$, $L \neq R$.
## Output
Output the answer of each operation type $2$ in a separated line.
## Sample
InputOutput
6 5
1 3 8 5 10 6
2 1 6
1 4 5 3
2 1 3
1 1 2 10
2 2 4
19
7
5
### Statistics
62% Solution Ratio
hamza05Earliest, 1M ago
JobaidulFastest, 0.2s
JobaidulLightest, 17 MB
fahimcp495Shortest, 1473B | 621 | 1,784 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 39, "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} | 3.640625 | 4 | CC-MAIN-2022-05 | longest | en | 0.728798 |
https://www.hackmath.net/en/math-problem/6711 | 1,632,192,419,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057131.88/warc/CC-MAIN-20210921011047-20210921041047-00656.warc.gz | 830,322,491 | 12,159 | # Paralympics
Milan competed in the paralympics. He finished as the winner with a time of 4 minutes 37 seconds. The competitor in second place was 53 seconds slower. What time had the competitor in second place?
Result
t = 5:30 m:s
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips to related online calculators
Do you want to convert time units like minutes to seconds?
## Related math problems and questions:
• Two brothers
Two brothers want to collect 20 kg of chestnuts. If only Marek collected, he would make it in 30 minutes. It would take Milan 10 minutes longer. How long after Mark does Milan have to start collecting to be finished in 18 minutes?
• Athletic race
In a race, the second-place finisher crossed the finish line 1 1/3 minutes after the first-place finisher. The third-place finisher was 1 3/4 minutes behind the second-place finisher. The third-place finisher took 34 2/3 minutes. How long did the first-pl
• Theatre
The theatrical performance started at 15:50 and ended at 18:50. How many minutes long?
• Train speed
Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in a train approaching the place hears second shot 10 minutes after the first. The speed of the train (in km/hr), supposing that sound travels at 340 m/s is:
Add and write the result again as hours, minutes, seconds: 2hodiny45min15s + 1h20m50s =
Two fifth-graders teams competing in math competitions - in Mathematical Olympiad and Pytagoriade. Of the 33 students competed in at least one of the contest 22 students. Students who competed only in Pytagoriade were twice more than those who just compet
John can ran around a circular track in 20 seconds and Eddie in 30 seconds. Two seconds after Eddie starts, John starts from the same place in opposite direction. When will they meet?
• Two trains
There are two trains running the same distance. 1st train will travel it in 7 hours 21minutes. 2nd the train will travel 5 hours 57minutes and 34 seconds and it is 14 km/h faster than the first train. What are speeds of trains and how long is this railway
• When Emily
When Emily arrived home from school, she spent 8 minutes eating a snack. Then, she spent 15 minutes playing with her dog and 32 minutes doing her chores. Finally, Emily spent 45 minutes finishing her homework. How long had Emily been home when she finishe
• Relay-race
Kate ran 170 meters at 29.8 seconds, Suzan at 29.3 seconds, Roza at 34 seconds and Žofka at 30.4 seconds. How long took to run 4x170 m relay-race?
• Nightingale
Nightingale sang five seconds for 5 hours and 5 minutes. How many seconds were missing to sing 10 hours?
• Saving money
Marko saved € 64. Milan saved 8 euros less than Marko. Marika saved 6 times more than Milan. Calculate who saved how much.
• Frequency distribution
The following frequency distribution gives the time spent for fill-up at a gas station. Assume that each value in a class is equal to the midpoint of the class. Estimate the mean fill-up time for the given data. Round your answer to the nearest hundred
• Emily
Emily had 20 minutes to do a three-problem quiz. She spent 11 3/4 minutes on question A and 5 1/2 minutes on question B. How much time did she have left for question C?
• Ships
Red ship begins its circuit every 30 minutes. Blue boat begins its circuit every 45 minutes. Both ships begin their sightseeing circuit in the same place at the same time always at 10:00 o'clock. a / What time does meet boat again? b / How many times a da
• Trams
Two trams started at the same time from the same place. One tram journey takes 30 minutes and the second 45 minutes to its final stop. How long will trams meet again?
• Bus
On the 4-th stop take on 56 and take off 38 passengers. How many were added (write as positive number) or shrunk (write as negative number) the count of passengers? | 980 | 3,910 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-39 | latest | en | 0.954531 |
www.cdeguzman.me | 1,550,844,334,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247518425.87/warc/CC-MAIN-20190222135147-20190222161147-00519.warc.gz | 318,725,251 | 4,981 | # The Math Kanji (数学の漢字)
Recently I was studying some kanji when I noticed that the kanji 数 is used to express a wide variety of mathematical terminology. I thought it would be interesting to write a little bit about how this kanji is used.
Disclaimer: I have not read any mathematical texts in Japanese, and am relying on terms as found in my Kanji Dictionary along with my own knowledge.
## The Meaning of 数
The dictionary definition of this kanji along with all of its readings can be found here. On its own, this kanji is commonly read as:
• すう：suu, used as a prefix to refer to an arbitrary amount of something (can be a few or several), or as a suffix meaning “the number of” something.
• かず：kazu, a noun meaning number or amount.
• 数える（かぞえる)：kazoeru, a verb meaning to count.
For example, if you wanted to say something like “the number of plates” in Japanese, you could say 皿の数 (sara no kazu), or you could say 皿数 (sarasuu). “A few plates” would then be 数皿 (suusara), while “counting plates” would be 皿を数える (sara wo kazoeru).
## Mathematical Terminology
Being closely related to numbers, 数 is used in the terms for common types of numbers, such as:
• 正数（せいすう）：seisuu, positive numbers
• 負数（ふすう）：fusuu, negative numbers
• 偶数（ぐうすう）：guusuu, even numbers
• 奇数（きすう）：kisuu, odd numbers
• 整数（せいすう）：seisuu, integers (watch the kanji!)
• 自然数（しぜんすう）：shizensuu, natural numbers
• 有理数（ゆうりすう）：yuurisuu, rational numbers
• 無理数（むりすう）：murisuu, irrational numbers
• 実数（じっすう）：jissuu, real numbers
• 虚数（きょすう）：kyosuu, imaginary numbers
• 素数（そすう）：sosuu, prime numbers
• 完全数（かんぜんすう）： kanzensuu, perfect numbers
This kanji also appears in terms describing fractions and decimal numbers:
• 小数（しょうすう）：shousuu, decimal
• 小数点（しょうすうてん）：shousuuten, decimal point
• 分数（ぶんすう）：bunsuu, fraction
• 真分数（しんぶんすう）：shinbunsuu, proper fraction
• 仮分数（かぶんすう）：kabunsuu, improper fraction
• 帯分数（たいぶんすう）：taibunsuu, mixed fraction
• 端数（はすう）：hasuu, fractional part
• 逆数（ぎゃくすう）：gyakusuu, reciprocal
• 除数（じょすう）：josuu, divisor
With types of numbers out of the way, we now move on to algebra. In Japanese, algebra is 代数 (daisuu); in a more literal sense this can be translated as “number substitution”, which is certainly something you do when starting out!
• 函数（かんすう）：kansuu, function (can also be written as 関数 with no change in pronunciation)
• 対数（たいすう）：taisuu, logarithm
• 係数（けいすう）：keisuu, coefficient
• 変数（へんすう）：hensuu, variable
• 定数（ていすう）：teisuu, constant
• 指数（しすう）：shisuu, exponent
• 次数（じすう）：jisuu, degree
And now for some things a little bit past high school algebra:
• 数列 (すうれつ)：suuretsu, progression or sequence
• 級数（きゅうすう）：kyuusuu, series
• 等比級数（とうひきゅうすう）：touhi kyuusuu, geometric series
• 等差級数（とうさきゅうすう）：tousa kyuusuu, arithmetic series
## Counting
As mentioned before, if you want to say you are counting, you use the verb 数える. However, if you want to express the actual count of something, you need to use a counter word (Wikipedia has a pretty good article on Japanese counter words found here). Common counters include 人 for people (usually read as nin), 分 for minutes (usually read as fun, but pronounced more like foon), and 歳 for age (read as sai), though there exist many different counters for different types of objects and entities.
Using what I mentioned previously, you could say 皿の数は5です to say “There are five plates”, though more literally this sentence would sound like “The number of plates is five”. A more natural way of saying this would be 皿が五枚あります (Sara ga gomai arimasu), where we use the counter 枚 for flat objects such as plates. More generally, you can use generic counters such as 一つ (hitotsu - one) and 二つ (futatsu - two) if you’re not sure which counter to use.
## Conclusion
Even though most of this content isn’t particularly useful unless you plan on studying mathematics in Japan, knowing how to count and express the numbers of things is an important skill to have. In future foreign language posts I will definitely try to avoid just listing a bunch of words like I did in this article, and instead focus more on actually writing and expressing ideas. | 1,263 | 4,075 | {"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-2019-09 | latest | en | 0.615004 |
https://two-wrongs.com/the-joy-of-manipulation.html | 1,679,544,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00160.warc.gz | 683,926,044 | 3,405 | # The Joy of Manipulation
Hold on! When I say manipulation I mean in the original, etymological sense; the mani-pulus kind of manipulation. The "to operate something with your hands" kind of manipulation.
As a software developer and citizen of the modern world, I spend a significant portion of my day pressing buttons and observing their effect on rows of lights on a screen. I do this so much I get bored of it.
I'm not saying it's bad. It's incredibly efficient and intelligent and good and fantastic and all of those things. I love the way it furthers our civilisation. But being efficient isn't always my ultimate priority. What good is doing two things really efficiently but being bored to hell while you're doing them, when you could be doing just one thing slowly but enjoying every minute of it?
I've noticed that I'm more and more sneaking things into my life that aren't button-pressing-screen-observing activities. Activities that are more about holding, rotating and looking at actual mechanical 3D real-life objects.
I read the dead-trees version of my newspaper. I do my photography on film. I develop said film in my bathroom. I've recently started doing calculations on a slide rule:
It brings me joy to slide little pieces of plastic around and they give me the answer to otherwise difficult maths problems. I love looking at the logarithmic scale and thinking about the relations between numbers. You don't get that from a calculator.
I love winding the film in my camera and feeling the silver-coated strip of plastic resist, telling me I've finished the roll and there are no more frames to be had. I love popping open the back and taking out the metal canister that holds a physical, ghostly imprint of the last 24 photos I've taken.
I don't mind forgetting how my sewing machine works and being confused about which button to press to reverse. I don't mind it because there are two controls which I know what they do: the wheels that control how wide the stitches are and how long they are. One of the wheels control the distance the needle jumps, the other controls the little teeth that feed the fabric forward. The direction you sew in is also controlled by those little teeth, so obviously the reverse button is the button on the wheel that controls the stitch length. It just makes sense mechancially. I can almost see the little gears and motors and pins working together, forming that mechanism. Not a single, opaque microchip was had that day.
I love that this old tech still exists. My life would be far more boring if it wasn't around. | 538 | 2,578 | {"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-2023-14 | longest | en | 0.959157 |
http://www.chegg.com/homework-help/questions-and-answers/problem-9-7-calculating-salvage-value-lo-2-asset-used-four-year-project-falls-five-year-ma-q4059881 | 1,438,773,688,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438044160065.87/warc/CC-MAIN-20150728004240-00282-ip-10-236-191-2.ec2.internal.warc.gz | 356,698,444 | 13,221 | Problem 9-7 Calculating Salvage Value [LO 2]
An asset used in a four-year project falls in the five-year MACRS class (MACRS Table) for tax purposes. The asset has an acquisition cost of \$8,200,000 and will be sold for \$1,860,000 at the end of the project.
Required: If the tax rate is 40 percent, what is the aftertax salvage value of the asset? (Do not include the dollar sign (\$). Enter rounded answer as directed, but do not use the rounded numbers in intermediate calculations.Round your answer to 2 decimal places (e.g., 32.16).)
Aftertax salvage value \$ | 143 | 569 | {"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-2015-32 | latest | en | 0.889823 |
http://star-www.rl.ac.uk/docs/sg5.htx/node32.html | 1,371,690,680,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368709337609/warc/CC-MAIN-20130516130217-00032-ip-10-60-113-184.ec2.internal.warc.gz | 240,491,550 | 2,741 | Next: Prompting for Procedure Parameters
Up: Control Structures
Previous: The IF structure
## The LOOP Structure
The LOOP structure is used to repeatedly execute a group of statements. It has three different forms, the simplest being as follows:
``` LOOP
statements
END LOOP
```
This form sets up an infinite loop, but an additional statement, BREAK, may be used to terminate the loop. BREAK would, of course, normally have to be inside an IF structure.
``` PROC COUNT
{ A procedure to print the numbers from 1 to 10 }
I = 1
LOOP
PRINT (I)
I = I+1
IF I > 10
BREAK
ENDIF
ENDLOOP
ENDPROC
```
As it is frequently required to loop over a sequential range of numbers in this way, a special form of the LOOP statement is provided for this purpose. It has the following form:
``` LOOP FOR variable = expression1 TO expression2 [ STEP expression3 ]
statements
END LOOP
```
This form is essentially equivalent to the DO loop in FORTRAN. The expressions specifying the range of values for the control variable are rounded to the nearest integer so that the variable always has integer values. Using this form of the LOOP statement we can simplify the previous example as follows:
``` PROC COUNT
{ A procedure to print the numbers from 1 to 10 }
LOOP FOR I = 1 TO 10
PRINT (I)
ENDLOOP
ENDPROC
```
Note that there is an optional STEP clause in the LOOP FOR statement. If this is not specified a STEP of 1 is assumed. The STEP clause can be used to specify a different value. A step of -1 must be specified to get a loop which counts down from a high value to a lower value. For example:
``` LOOP FOR I = 10 TO 1 STEP -1
```
will count down from 10 to 1.
The third form of the LOOP structure allows the setting up of loops which terminate on any general condition. It has the form:
``` LOOP WHILE expression
statements
END LOOP
```
The expression is evaluated each time round the loop, and if it has the logical value TRUE the statements which form the body of the loop are executed. If it has the value FALSE execution continues with the statement following END LOOP.
Using this form we can write yet another version of the COUNT procedure:
``` PROC COUNT
{ A procedure to print the numbers from 1 to 10 }
I = 1
LOOP WHILE I <= 10
PRINT (I)
I = I+1
ENDLOOP
ENDPROC
```
In the above case the LOOP WHILE form is more complicated than the LOOP FOR form. However LOOP WHILE can be used to express more general forms of loop where the termination condition is something derived inside the loop. An example is a program which prompts the user for an answer to a question (e.g. yes or no) and has to keep repeating the prompt until a valid answer is received.
``` FINISHED = FALSE
LOOP WHILE NOT FINISHED
INPUT Enter YES or NO: (ANSWER)
FINISHED = ANSWER = 'YES' OR ANSWER = 'NO'
END LOOP
```
Next: Prompting for Procedure Parameters
Up: Control Structures
Previous: The IF structure
ICL The Interactive Command Language for ADAM
Starlink Guide 5
J A Bailey
A J Chipperfield
9th June 1998
E-mail:starlink@jiscmail.ac.uk
Copyright © 2013 Science and Technology Facilities Council | 750 | 3,100 | {"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} | 3.46875 | 3 | CC-MAIN-2013-20 | latest | en | 0.870242 |
https://www.chicagotribune.com/sports/football/ct-spt-super-bowl-gutsy-call-analytics-20180205-story.html | 1,555,679,598,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527720.37/warc/CC-MAIN-20190419121234-20190419143234-00451.warc.gz | 642,186,548 | 47,779 | # Eagles' call on fourth-and-goal wasn't 'gutsy' — it just came down to basic math
Chicago Tribune
Cris Collinsworth is probably still talking about “one of the gutsiest calls” in football history. Al Michaels might still be parroting him.
But the fact is Eagles coach Doug Pederson made the proper, prudent and really easy call to go for it on fourth-and-goal from inside the 2-yard line.
That arguably the best analyst in football — and many others — thought Pederson’s call was “gutsy” shows how far behind football remains in the analytics game — although it is getting better, as evidenced by Pederson.
Every NFL head coach should have someone on his staff who knows the numbers: Two-point conversions (from the 2-yard line) hit at about a 45 percent clip, depending on the year; fourth-and-1s are successful about 67 percent of the time.
So how about fourth-and-goal from the 1½? Surely somewhere right in the middle of those two conversion rates, which would be 56 percent. At a 56 percent rate of scoring a touchdown, or seven points, the Eagles’ expected value by going for it was 3.92 points (.56 times 7).
That’s almost a point more than a field goal, and the case should be closed right there.
(For simplicity, we’ll leave out the finer margins of these probabilities such as success rate of an extra point, success rate of a chip-shot field goal, possibility of a turnover, a botched snap, etc.)
It was still the first half of a high-scoring game, and all Pederson was looking for was the best chance at the most points. If prepared, which too many NFL coaches are not, it was an easy call to make in real time.
Instead of calling Pederson “gutsy,” we should be calling for the heads of coaches who don’t go for it from 1½ yards out, but we’re not there yet.
Besides the expected points value being higher by going for it, there was also the bonus of leaving your opponent inside the 5-yard line if you didn’t make it.
And in this case, because of the time remaining in the first half, that was a big factor. With 34 seconds left at the end of the Eagles’ fourth-down play, Tom Brady and the Patriots got the ball back at their 25-yard line after a touchback, the same place they would have gotten it had the Eagles kicked a field goal. The Patriots had one timeout and a kicker with deep range. While we don’t have the exact analytics on Brady leading a field-goal drive in that situation, we can take an educated guess.
Let’s say the Patriots turn it into three points 25 percent of the time (Brady, in fact, got them to midfield before a desperation play came up short). That would mean they had an expected value of .75 points (.25 times 3).
Now back to Pederson’s fourth-down decision. Not only was the Eagles’ expected value almost a point higher by going for it, but they also got some relief if they came up short. Because the Patriots would have had almost no chance of scoring from inside the 5-yard line with 34 seconds left (they probably would have run the clock out), the Eagles would have cut their losses by three-quarters of a point if they didn’t convert the fourth down.
So by going for it, two things happened:
1. The Eagles’ expected offensive points value went up by .92 points.
2. The Patriots’ expected offensive points value went down by .33 points (.44 x .75, with .44 representing the 44 percent of the time the Eagles don’t convert a touchdown and .75 representing the expected points value if the Patriots were starting at their 25 with 34 seconds left and one timeout).
In conclusion, in the most average conditions, going for it gave the Eagles an expected value of plus-1.25 points (.92 + .33).
As far as the trick play they had drawn up? Great design. And all in all, a great game by Pederson.
But going for it inside the 2-yard line in the first half? Every time.
dan.mann@chicagotribune.com
Bears can learn from the Eagles by taking a 'Philly Special' mindset »
Remember Pat Fitzgerald's 'dumb' Music City calls? They worked for Doug Pederson in the Super Bowl. »
Super Bowl winners and losers: Prince, Tide pods, Cris Collinsworth and more » | 944 | 4,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-18 | latest | en | 0.978281 |
https://orderbyselectnull.com/tag/cardinailty-estimates/ | 1,686,305,588,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224656675.90/warc/CC-MAIN-20230609100535-20230609130535-00182.warc.gz | 474,712,213 | 38,463 | ## Containment for the Common Man
Lots of smart people have written about join containment, but none of the explanations really made sense to me. I felt like a student memorizing definitions for a test. Sure, I could tell you the definitions of base and simple containment, but what practical difference does it make when it comes to cardinality estimation? The concept finally clicked when working on an Oracle query of all things, and as a result I wrote this blog post. All testing was done on SQL Server 2017 with a CE version of 140.
### A Note on Join Cardinality
Join cardinality calculations are incredibly complex in SQL Server. You can get a small taste of that complexity here. I’ve chosen the example data in this blog post to avoid most of the complexity. The formulas and concepts described in this post can’t be used to model join cardinality generally, but I hope that they serve as a good illustration of containment.
### Demo Tables
All of the demo tables have identical structures with similar data. The first column, `UNIQUE_ID`, stores unique integers in the range specified in the table name. For example, `TA_1_TO_1000000` is a table that stores integers from 1 to 1000000. The second column, `MOD_FILTER`, stores integers from 1 to 100 cycling through all rows. The purpose of this column is to make filtering cardinality estimates simple to calculate and predict. For example, `MOD_FILTER BETWEEN 1 AND 50` will return 50% of the rows from the table. Full statistics are gathered on all columns, and there are four tables in all.
```DROP TABLE IF EXISTS dbo.TA_1_TO_1000000;
CREATE TABLE dbo.TA_1_TO_1000000 (
UNIQUE_ID BIGINT NOT NULL,
MOD_FILTER BIGINT NOT NULL
);
INSERT INTO dbo.TA_1_TO_1000000
WITH (TABLOCK)
SELECT t.RN
, 1 + t.RN % 100
FROM
(
SELECT TOP (1000000) ROW_NUMBER()
OVER (ORDER BY (SELECT NULL)) RN
FROM master..spt_values t1
CROSS JOIN master..spt_values t2
) t
OPTION (MAXDOP 1);
CREATE STATISTICS S1 ON dbo.TA_1_TO_1000000 (UNIQUE_ID)
WITH FULLSCAN;
CREATE STATISTICS S2 ON dbo.TA_1_TO_1000000 (MOD_FILTER)
WITH FULLSCAN;
DROP TABLE IF EXISTS dbo.TB_1_TO_1000000;
CREATE TABLE dbo.TB_1_TO_1000000 (
UNIQUE_ID BIGINT NOT NULL,
MOD_FILTER BIGINT NOT NULL
);
INSERT INTO dbo.TB_1_TO_1000000
WITH (TABLOCK)
SELECT t.RN
, 1 + t.RN % 100
FROM
(
SELECT TOP (1000000) ROW_NUMBER()
OVER (ORDER BY (SELECT NULL)) RN
FROM master..spt_values t1
CROSS JOIN master..spt_values t2
) t
OPTION (MAXDOP 1);
CREATE STATISTICS S1 ON dbo.TB_1_TO_1000000 (UNIQUE_ID)
WITH FULLSCAN;
CREATE STATISTICS S2 ON dbo.TB_1_TO_1000000 (MOD_FILTER)
WITH FULLSCAN;
DROP TABLE IF EXISTS dbo.TC_1_TO_100000;
CREATE TABLE dbo.TC_1_TO_100000 (
UNIQUE_ID BIGINT NOT NULL,
MOD_FILTER BIGINT NOT NULL
);
INSERT INTO dbo.TC_1_TO_100000
WITH (TABLOCK)
SELECT t.RN
, 1 + t.RN % 100
FROM
(
SELECT TOP (100000) ROW_NUMBER()
OVER (ORDER BY (SELECT NULL)) RN
FROM master..spt_values t1
CROSS JOIN master..spt_values t2
) t
OPTION (MAXDOP 1);
CREATE STATISTICS S1 ON dbo.TC_1_TO_100000 (UNIQUE_ID)
WITH FULLSCAN;
CREATE STATISTICS S2 ON dbo.TC_1_TO_100000 (MOD_FILTER)
WITH FULLSCAN;
DROP TABLE IF EXISTS dbo.TD_500001_TO_1500000;
CREATE TABLE dbo.TD_500001_TO_1500000 (
UNIQUE_ID BIGINT NOT NULL,
MOD_FILTER BIGINT NOT NULL
);
INSERT INTO dbo.TD_500001_TO_1500000
WITH (TABLOCK)
SELECT t.RN
, 1 + t.RN % 100
FROM
(
SELECT TOP (1000000) 500000 + ROW_NUMBER()
OVER (ORDER BY (SELECT NULL)) RN
FROM master..spt_values t1
CROSS JOIN master..spt_values t2
) t
OPTION (MAXDOP 1);
CREATE STATISTICS S1 ON dbo.TD_500001_TO_1500000 (UNIQUE_ID)
WITH FULLSCAN;
CREATE STATISTICS S2 ON dbo.TD_500001_TO_1500000 (MOD_FILTER)
WITH FULLSCAN;
```
The statistics objects are perfect in that they fully describe the data. Here’s the statistics output for the `UNIQUE_ID` column:
And here’s the output for the `MOD_FILTER` column:
This only happened because the table was populated with very simple data that fits well within the framework for generating histograms in SQL Server. Gathering statistics, even with `FULLSCAN`, will often not perfectly represent the data in the column.
### A Simple Model of Join Cardinality Estimation
Consider the following simple query:
```SELECT *
FROM TB_1_TO_1000000 b
INNER JOIN dbo.TD_500001_TO_1500000 d
ON b.UNIQUE_ID = d.UNIQUE_ID;
```
We know that exactly 500000 rows will be returned, but how might SQL Server estimate the number of rows to be returned? Let’s look at the histograms and try to align their steps:
That doesn’t exactly work, but we can split up the histogram steps so they align. The assumption of uniformity within the step isn’t even needed here because we know that there aren’t missing any integer values. The histograms below are equivalent to the original ones:
Now the `RANGE_HI_KEY` values align. For the step with a high value of 500001 we can expect only one row to match between tables. For the step with a high value of 1000000 we can expect 499998 + 1 rows to match. This brings the total row estimate to 500000, which happens to match what I get in SQL Server 2017 with the new CE. Remember, what we’re doing here isn’t how the query optimizer does the calculation. This is just a simple model that will be useful later.
Now consider the two queries below:
```SELECT *
FROM TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 50;
SELECT *
FROM TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 51 AND 100;
```
We know that the first query will return 500k rows and the second query will return 0 rows. However, can SQL Server know that? Each statistics object only contains information about its own column. There’s no correlation between the `UNIQUE_ID` and `MOD_FILTER` columns, so there isn’t a way for SQL Server to know that the queries will return different estimates. The query optimizer can create an estimate based on the filters on the `WHERE` clause and on the histograms of the join columns, but there’s no foolproof way to do that calculation. The presence of the filters introduces uncertainty into the estimate, even with statistics that perfectly describe the data for each column. The containment assumption is all about the modeling assumption that SQL Server has to make to resolve that uncertainty.
### Base Containment
Base containment is the assumption that the filter predicates are independent from the join selectivity. The estimate for the join should be obtained by multiplying together the selectivity from both filters and the join. The query optimizer uses base containment starting with CE model version 120, also known as the new CE introduced in SQL Server 2014. It can be used with the legacy CE if trace flag 2301 is turned on. The best reference for trace flag 2301 is a blog post from 2006 which is no longer published.
Let’s go back to this example query:
```SELECT *
FROM TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 50;
```
The selectivity for the filter on `MOD_FILTER` is 0.5 for both tables. This is because there are 100 unique values for `MOD_FILTER` between 1 and 100 and each value matches 1% of the table. We can see this by getting an estimated query plan on just `TA_1_TO_1000000`:
The table has 1 million rows, so the estimate is 500000 = 0.5 * 1000000.
That leaves the join selectivity. We put the same data into both tables:
We don’t need highlighters to see that the join selectivity is 1.0.
Putting it all together, the cardinality estimate under base containment for this query should be 1000000 * 1.0 * 0.5 * 0.5 = 250000. This is indeed the estimate:
Of course, this doesn’t match the actual number of rows which is 500000. But it’s easy to change the filter predicates so that the estimated number of rows and the actual number of rows match.
### Simple Containment
Simple containment is the assumption that the filter predicates are not independent. The estimate for the join should be obtained by applying the filter selectivities to the join histograms and joining based on the adjusted histograms. The query optimizer uses simple containment within the legacy CE. Simple containment can be used in the new CE via trace flag or `USE HINT`.
Let’s go back to the same example query:
```SELECT *
FROM TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 50
OPTION (
USE HINT ('ASSUME_JOIN_PREDICATE_DEPENDS_ON_FILTERS')
);
```
We know that the filter selectivity for both tables is 0.5. How can that be used to adjust the histograms? The simplest method would be to just multiply `RANGE_ROWS`, `EQ_ROWS`, and `DISTINCT_RANGE_ROWS` by the filter selectivity. After doing so we’re left with two still identical histograms:
It might seem odd to work with fractions of a row, but as long as everything is rounded at the end why should it matter? With two identical, aligned histograms it seems reasonable to expect a cardinality estimate of 0.5 + 499999 + 0.5 = 500000. This is exactly what we get in SQL Server:
The actual row estimate matches the estimated row estimate because the filters are perfectly correlated. Every row left after filtering still has a matching row in the other table.
### Just One Filter
What happens if we filter on just a single table? For example:
```SELECT *
FROM dbo.TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.MOD_FILTER BETWEEN 1 AND 30;
SELECT *
FROM dbo.TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.MOD_FILTER BETWEEN 1 AND 30
OPTION (
USE HINT ('ASSUME_JOIN_PREDICATE_DEPENDS_ON_FILTERS')
);
```
For base containment, we know that the filter selectivity is 0.3 and the join selectivity is 1.0. We can expect a cardinality estimate of 1000000 * 1.0 * 0.3 = 300000 rows.
For simple containment we need to multiply the histogram for `TA_1_TO_1000000` by 0.3. Here’s what the two histograms look like after factoring in filter selectivity:
What should the estimate be? One approach would be to assume that everything matches between the aligned steps. So we end up with 0.3 rows from the step with a `RANGE_HI_KEY` of 1 and 299999.4 + 0.3 rows from the step with a `RANGE_HI_KEY` of 1000000. The combined estimate is 300000 rows, which matches the base containment estimate. Why shouldn’t they match? Without filters on both tables there’s no concept of correlation. If it helps you can imagine a filter of `1 = 1` on `TB_1_TO_1000000`. For base containment multiplying by 1.0 won’t change the estimate and for simple containment multiplying by 1 won’t change the histogram. That just leaves a single filter selectivity of 0.3 for `TA_1_TO_1000000` and both estimates should be the same.
For both queries the estimated number of rows in SQL Server is 300000. Our calculations match the SQL Server query optimizer exactly for this query.
### Filtering on the Join Column
What happens if we filter on the join columns of both tables? For example:
```SELECT *
FROM dbo.TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.UNIQUE_ID BETWEEN 1 AND 200000
AND b.UNIQUE_ID BETWEEN 1 AND 200000;
SELECT *
FROM dbo.TA_1_TO_1000000 a
INNER JOIN dbo.TB_1_TO_1000000 b
ON a.UNIQUE_ID = b.UNIQUE_ID
WHERE a.UNIQUE_ID BETWEEN 1 AND 200000
AND b.UNIQUE_ID BETWEEN 1 AND 200000
OPTION (
USE HINT ('ASSUME_JOIN_PREDICATE_DEPENDS_ON_FILTERS')
);
```
Think back to why we need containment in the first place. When there are filters on columns that aren’t the join columns then we need to make an assumption as to how the selectivities all interact with each other. With a filter on the join column we can just adjust the histogram of the join column directly. There isn’t any uncertainty. Here’s what the histograms could look like:
In which case, it seems obvious that the final estimate should be 200000 rows. Simple containment does not result in a different estimate here.
### Removing Rows
So far the examples have been very simple. We’ve joined tables that contain the exact same data. What if one table has fewer rows than the other table? Consider the following pair of queries:
```SELECT *
FROM dbo.TC_1_TO_100000 c
INNER JOIN dbo.TB_1_TO_1000000 b
ON c.UNIQUE_ID = b.UNIQUE_ID
WHERE c.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 50;
SELECT *
FROM dbo.TC_1_TO_100000 c
INNER JOIN dbo.TB_1_TO_1000000 b
ON c.UNIQUE_ID = b.UNIQUE_ID
WHERE c.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 50
OPTION (
USE HINT ('ASSUME_JOIN_PREDICATE_DEPENDS_ON_FILTERS')
);
```
It’s important to call out here that `TC_1_TO_100000` has just 100000 rows instead of one million. For base containment, we know that the selectivity will be 0.5 for both tables. What about join selectivity? The histogram steps of course aren’t aligned:
The data is densely packed, so we can use the same trick as before to split the histogram for the larger table:
Every row in histogram for the smaller table has a match in the histogram of the larger table. From the point of view of the smaller table the join selectivity is 1.0. Multiplying together all three selectivities gives a final row estimate of 100000 * 1.0 * 0.5 * 0.5 = 25000. This matches the row estimate within SQL Server exactly.
For simple containment we need to apply the filter selectivities of 0.5 to both tables. We also need to align the histograms by splitting the larger histogram. Both will be done in one step:
Every row in the smaller histogram once again matches. Our final estimate is 0.5 + 49999 + 0.5 = 50000 which exactly matches the SQL Server query optimizer.
### Unmatched Rows
What happens if the tables have the same number of rows but they clearly don’t contain the same data? Consider the following pair of queries:
```SELECT *
FROM dbo.TD_500001_TO_1500000 d
INNER JOIN dbo.TB_1_TO_1000000 b
ON d.UNIQUE_ID = b.UNIQUE_ID
WHERE d.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 10;
SELECT *
FROM dbo.TD_500001_TO_1500000 d
INNER JOIN dbo.TB_1_TO_1000000 b
ON d.UNIQUE_ID = b.UNIQUE_ID
WHERE d.MOD_FILTER BETWEEN 1 AND 50
AND b.MOD_FILTER BETWEEN 1 AND 10
OPTION (
USE HINT ('ASSUME_JOIN_PREDICATE_DEPENDS_ON_FILTERS')
);
```
The filter predicate for `TB_1_TO_1000000` is 0.1 and the filter predicate for `TD_500001_TO_1500000` is 0.5. Here are our starting histograms:
The little man who lives inside the cardinality estimator needs to slice them up so they align. His work is complete:
The top histogram has 500000 unmatched rows in the step with a `RANGE_HI_KEY` of 1500000, so the join selectivity is 500000 / 1000000 = 0.5. Putting all three selectivities together, the cardinality estimate with base containment should be 1000000 * 0.5 * 0.1 * 0.5 = 25000. This exactly matches SQL Server.
You know the drill for simple containment. We need to multiply each sliced histogram by its filter selectivity:
That’s pretty messy. I’m going to assume that every row has a match between the two shared steps, so the estimate should be 0.1 + 49999.8 + 0.1 = 50000. The number of estimated rows reported by SQL Server is 50000.4 :
What happened? Did the little man only measure once before cutting? This is one of those examples where there’s other complicated stuff going on under the hood, so the predicted row estimate doesn’t match up exactly. Interestingly, the estimate with the legacy cardinality estimator is exactly 50000.
### An Approximate Formula
• Define `T1_CARDINALITY` as the number of rows in the first joined table.
• Define `T1_FILTER_SELECTIVITY` as the filter selectivity of the filter predicates of the first table. This number ranges from 0.0 to 1.0, with 1.0 for filters that remove no rows.
• Define `T2_CARDINALITY` as the number of rows in the second joined table.
• Define `T2_FILTER_SELECTIVITY` as the filter selectivity of the filter predicates of the second table. This number ranges from 0.0 to 1.0, with 1.0 for filters that remove no rows.
• Define `JOIN_SELECTIVITY` as the selectivity of the two histograms of the joined columns from the point of view of the smaller table. This number ranges from 0.0 to 1.0, with 1.0 meaning that all rows in the smaller table have a match in the larger table.
Based on the tests above, we can model the cardinality estimates for base and simple containment as follows:
Base containment = JOIN_SELECTIVITY * LEAST(T1_CARDINALITY, T2_CARDINALITY) * T1_FILTER_SELECTIVITY * T2_FILTER_SELECTIVITY
Simple containment = JOIN_SELECTIVITY * LEAST(T1_FILTER_SELECTIVITY * T1_CARDINALITY, T2_FILTER_SELECTIVITY * T2_CARDINALITY)
Remember that this isn’t how SQL Server actually does it. However, I think that it shows the difference between base containment and simple containment quite well. For simple containment the filters are applied to the histograms and for base containment all of the selectivities are independent.
### A Mathematical Proof?
So far simple containment has always had a higher cardinality estimate than base containment. Looking at the formulas it certainly feels like simple should have a higher estimate. Can we prove that the estimate will always be higher using the above formulas? It’s been quite a few years so I apologize for the proof below, but I believe that it gets the job done.
Definitions:
JS = JOIN_SELECTIVITY
C1 = T1_CARDINALITY
F1 = T1_FILTER_SELECTIVITY
C2 = T2_CARDINALITY
F2 = T2_FILTER_SELECTIVITY
Attempt a proof by contradiction, so assume the opposite of what we want to prove:
JS * LEAST(C1, C2) * F1 * F2 > JS * LEAST(F1 * C1, F2 * C2)
We know that `JS > 0`, `F1 > 0`, and `F2 > 0`, so:
LEAST(C1, C2) > LEAST(C1 / F2, C2 / F1)
The left hand expression can only evalute to `C1` or `C2`. Let’s assume that it evaluates to `C1`, so `C1 <= C2`. We know that `F1 <= 1`, so `C2 <= C2 / F1`. `C1 / F2 > C1`, so the only hope of the inequality above being true is if `C1 > C2 / F1`. Putting it all together:
C1 <= C2 <= C2 / F1 < C1
That is clearly impossible. Very similar logic holds if the left hand expression evaluates to `C2` (just flip 1 with c in the above), so we know that the equation that we started out with is not true. Therefore:
JS * LEAST(C1, C2) * F1 * F2 <= JS * LEAST(F1 * C1, F2 * C2)
In other words:
BASE CONTAINMENT <= SIMPLE CONTAINMENT
Here’s my public domain celebration picture:
The details of this stuff within SQL Server are very complicated, so this doesn’t mean that there doesn’t exist a query that has a larger cardinality estimate with base containment. However, it seems to be a safe assumption that in general simple containment will result in a larger or equal estimate compared to base containment.
### Why Does Any of This Matter?
I almost created a kind of real life example here, but I ran out of time so you’re eating Zs for dinner again as usual. Let’s introduce a table to cause some trouble:
```DROP TABLE IF EXISTS dbo.ROWGOAL_TROUBLES;
CREATE TABLE dbo.ROWGOAL_TROUBLES (
UNIQUE_EVEN_ID BIGINT NOT NULL,
PAGE_FILLER VARCHAR(1000) NOT NULL
);
INSERT INTO dbo.ROWGOAL_TROUBLES
WITH (TABLOCK)
SELECT 2 * t.RN
, REPLICATE('Z', 1000)
FROM
(
SELECT TOP (50000) ROW_NUMBER()
OVER (ORDER BY (SELECT NULL)) / 100 RN
FROM master..spt_values t1
CROSS JOIN master..spt_values t2
) t
OPTION (MAXDOP 1);
```
Consider the following business critical query that I run all the time:
```SELECT *
FROM dbo.TA_1_TO_1000000 t1
INNER JOIN dbo.TB_1_TO_1000000 t2
ON t1.UNIQUE_ID = t2.UNIQUE_ID
WHERE t1.MOD_FILTER = 1
AND t2.MOD_FILTER = 1
AND NOT EXISTS (
SELECT 1
FROM dbo.ROWGOAL_TROUBLES rt
WHERE rt.UNIQUE_EVEN_ID = t1.UNIQUE_ID
)
OPTION (MAXDOP 1);
```
The plan doesn’t look so hot:
There are unmatched rows in the `ROWGOAL_TROUBLES` table, so we know that the scan on the inner side of the nested loop is going to read a lot of rows. The query took about 60 seconds to finish on my machine and read 499775000 rows from the `ROWGOAL_TROUBLES` table. Why did this plan seem attractive to SQL Server? The query optimizer thought that only 100 rows would be returned after the join of `TA_1_TO_1000000` to `TB_1_TO_1000000`. The filters are perfectly correlated so 10000 rows will be returned in reality. With perfectly correlated filters we can expect a better estimate if we use simple containment:
```SELECT *
FROM dbo.TA_1_TO_1000000 t1
INNER JOIN dbo.TB_1_TO_1000000 t2
ON t1.UNIQUE_ID = t2.UNIQUE_ID
WHERE t1.MOD_FILTER = 1
AND t2.MOD_FILTER = 1
AND NOT EXISTS (
SELECT 1
FROM dbo.ROWGOAL_TROUBLES rt
WHERE rt.UNIQUE_EVEN_ID = t1.UNIQUE_ID
)
OPTION (
MAXDOP 1,
USE HINT ('ASSUME_JOIN_PREDICATE_DEPENDS_ON_FILTERS')
);
```
With a better estimate of 10000 rows comes a better query plan:
The query finishes in under a second on my machine.
### Final Thoughts
Hopefully this blog post gives you a better understanding of the difference between base and simple containment. Read some of the other explanations out there if this wasn’t helpful. Containment is a tricky subject and you never know what it’ll take for it to make sense to you. Thanks for reading!
## Manipulating Cardinality Estimates with Scalar UDFs
For this post I’m using the legacy cardinality estimator on SQL Server 2016 SP1.
### The Problem
Scalar user defined functions are evil but sometimes necessary. The following scenario will sound a bit contrived but it’s based on a real world problem. Suppose that end users can filter the amount of data returned by a query by inputting values into a UDF that does some kind of translation. Below is a sample schema:
```CREATE TABLE dbo.Example (
ID BIGINT NOT NULL,
NOT_ID VARCHAR(100) NOT NULL,
PRIMARY KEY (ID));
INSERT INTO dbo.Example WITH (TABLOCK)
(ID, NOT_ID)
SELECT TOP (1000000)
ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
, REPLICATE('Example', 14)
FROM master..spt_values t1
CROSS JOIN master..spt_values t2;
GO
CREATE FUNCTION dbo.MY_FAVORITE_UDF (@ID BIGINT)
RETURNS BIGINT AS
BEGIN
RETURN @ID;
END;
```
Consider the following part of a much bigger query:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE ID >= dbo.MY_FAVORITE_UDF(100000)
AND ID <= dbo.MY_FAVORITE_UDF(900000); ```
For this demo it’s not important that the UDF do anything so I must made it return the input. To keep things simple I’m not going to follow best practices around writing the query to avoid executing the UDFs for each row in the table. With the legacy cardinality estimator we get a cardinality estimate of 30% of the rows in the base table for each unknown equality condition. This means that a BETWEEN against two UDFs will give a cardinality estimate of 9%. The important point is that the cardinality estimate will not change as the inputs for the UDFs change, except for the trivial case in which the inputs are the same. This can easily be seen by varying the inputs and looking at the estimated execution plans:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE ID >= dbo.MY_FAVORITE_UDF(100000)
AND ID <= dbo.MY_FAVORITE_UDF(900000);
```
Query plan:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE ID >= dbo.MY_FAVORITE_UDF(500000)
AND ID <= dbo.MY_FAVORITE_UDF(499999);
```
Query plan:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE ID BETWEEN dbo.MY_FAVORITE_UDF(1)
AND dbo.MY_FAVORITE_UDF(1);
```
Query plan:
The cardinality estimate (CE) of just that simple query doesn’t really matter. But it could matter very much if that query was part of a larger query with other joins. The 9% estimate may not serve us well depending on the rest of the query and what end users tend to input. We might know that the end users tend to pick large or small ranges. Even if we don’t know anything about the end users, certain queries may do better with larger or smaller cardinality estimates.
### Decreasing the Cardinality Estimate
Let’s suppose that we do some testing and find that a cardinality estimate of lower than 9% is the best choice for typical end user inputs. There are a few techniques available to decrease the cardinality estimate by a fixed percentage.
#### Method 1
First option is to use TOP PERCENT along with an OPTIMIZE FOR hint. I’m not really a fan of TOP PERCENT. The implementation always spools unless it gets optimized out with TOP (100) percent. It would be nice if it didn’t spool. Anyway, perhaps getting a different cardinality estimate is worth the spool. Below is one method to get a cardinality estimate of 3% of the base table:
```DECLARE @top_percent FLOAT = 100;
SELECT TOP (@top_percent) PERCENT ID, NOT_ID
FROM dbo.Example
WHERE ID >= dbo.MY_FAVORITE_UDF(100000)
AND ID <= dbo.MY_FAVORITE_UDF(900000)
OPTION (OPTIMIZE FOR (@top_percent = 33.33333333));
```
Query plan:
The percent value is a float so we can go almost anywhere between 0 – 9% for the final estimate. However, if we have to use scalar UDFs in this fashion there’s a chance that we’re doing it to write platform agnostic code. The TOP trick here isn’t likely to work in other platforms.
#### Method 2
Suppose we add another inequality against a UDF that’s guaranteed not to change the results. 0.3^3 = 0.027 so we would expect an estimate of 2.7%. That is indeed what happens:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE ID >= dbo.MY_FAVORITE_UDF(100000)
AND ID <= dbo.MY_FAVORITE_UDF(900000)
-- redundant filter to change CE
AND ID > dbo.MY_FAVORITE_UDF(100000) - 1;
```
Query plan:
We can also mix things up with OR logic to make more adjustments. The query below has a fixed CE of 4.59%:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE ID >= dbo.MY_FAVORITE_UDF(100000)
AND ID <= dbo.MY_FAVORITE_UDF(900000)
-- redundant filter to change CE
AND (ID > dbo.MY_FAVORITE_UDF(100000) - 1
OR ID > dbo.MY_FAVORITE_UDF(100000) - 2);
```
Query plan:
It should be possible to mix and match to get something close to the CE that you want. I need to reiterate that as the code is written this will lead to additional UDF executions per row. You can also use techniques with fixed CE that don’t involve UDFs if you’re confident that Microsoft won’t change the guesses for them (which for the legacy cardinality estimator is probably a pretty safe assumption at this point).
### Increasing the Cardinality Estimate
In some cases we will want a cardinality estimate above 9%.
#### Method 1
The TOP PERCENT trick won’t work here since TOP on its own can’t increase a cardinality estimate. We can use OR logic with UDFs to raise the estimate. Consider this filter condition:
```ID >= dbo.MY_FAVORITE_UDF(100000)
OR ID >= dbo.MY_FAVORITE_UDF(900000) - 1
```
The first inequality gives an estimate of 30% and the second inequality gives an estimate of (100% – 30%) * 30% = 21%. In total we would get an estimate of 51%. If we apply that twice we should get an overall estimate of 0.51 * 0.51 = 26.01% . This is indeed what happens:
```SELECT ID, NOT_ID
FROM dbo.Example
WHERE (ID >= dbo.MY_FAVORITE_UDF(1)
OR ID >= dbo.MY_FAVORITE_UDF(1) - 1)
AND (ID <= dbo.MY_FAVORITE_UDF(2)
OR ID <= dbo.MY_FAVORITE_UDF(2) + 1);
```
Query plan:
By adding more UDFs to the OR clauses we can increase the cardinality estimate further.
#### Method 2
For another way to do it we can take advantage of the fact that an inequality filter against a UDF has the same cardinality as the negated condition. That means that this:
```SELECT ID, NOT_ID
FROM dbo.Example
EXCEPT
SELECT ID, NOT_ID
FROM dbo.Example
WHERE -- negate original expression
ID < dbo.MY_FAVORITE_UDF(100000)
OR ID > dbo.MY_FAVORITE_UDF(900000);
```
Will return the same results as the original query but have a much higher cardinality estimate. Writing it in a better way, we see a cardinality estimate of ~54.4%:
```SELECT e1.ID, e1.NOT_ID
FROM dbo.Example e1
WHERE NOT EXISTS (
SELECT 1
FROM dbo.Example e2
WHERE e1.ID = e2.ID
-- negate original expression
AND e2.ID < dbo.MY_FAVORITE_UDF(100000)
OR e2.ID > dbo.MY_FAVORITE_UDF(900000)
);
```
Query plan:
This can be adjusted up and down by adding additional UDFs. It comes with the cost of an additional join so it’s hard to think of an advantage of doing it this way.
#### Method 3
For a third option we can use the MANY() table-valued function developed by Adam Machanic. This function can be used to increase the cardinality estimate of a point in a plan by a whole number. If we want a cardinality estimate of 18% from the UDF it’s as easy as the following:
```SELECT TOP (9223372036854775807) ID, NOT_ID
FROM dbo.Example
CROSS JOIN dbo.Many(2)
WHERE ID BETWEEN dbo.MY_FAVORITE_UDF(100000)
AND dbo.MY_FAVORITE_UDF(900000);
```
Query plan:
I added the superfluous TOP to prevent the MANY() reference from getting moved around in the plan. This method has the disadvantage that it may not be platform-agnostic.
Hopefully you never find yourself in a situation where you need to use tricks like this. Thanks for reading! | 7,685 | 28,926 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-23 | latest | en | 0.813398 |
https://tradingcourse.online/money-management/ | 1,726,618,311,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651835.68/warc/CC-MAIN-20240918000844-20240918030844-00297.warc.gz | 541,472,087 | 18,058 | # Money Management
### Set Stop Loss
Money management is one of the fundamental pillars (in addition to psychotrading, system and risk management) for trading to work. It is the key to consistency and profitability over time as its function is to protect your account by maximising returns and minimising risk by defining how much to buy and how much to sell depending on the capital you have available. Therefore, it is operational risk management (which is different from risk management, which is in charge of setting the stop loss), which calculates the size of the next position.
With good money management we will achieve that our account loses as little as possible when we have losing trades and that profits increase over losses when we have winning trades. In other words, earning more with less risk, which is known as the risk/return ratio.
## Risk / reward ratio
The risk/reward ratio measures the risk you have to take in order to receive a profit. A risk/reward ratio of 1:2 means that when you get it right, you earn twice as much as you lose. A risk/reward ratio of 3:1, on the other hand, means that when you get it right you win one, but when you miss you lose three times as much as you win. This could lead to ruin. Finally, a 1:1 ratio means that you win as much as you lose, so if you have a 50% success rate (you win 50% of the time) you can’t use a 1:1 ratio, it must be higher. The ratio always goes hand in hand with the success rate of a trading system.
The formula for the risk/return ratio is:
Ratio= total losses/total profits.
Knowing this ratio helps us to:
1. Decide which trades we will make, based on the past results we have achieved.
2. Choose only those trades that offer an appropriate reward to risk ratio.
The expected return must be in accordance with the risk to be assumed and also with the mathematical expectation of our trading system.
## The mathematical expectation
It is a statistical concept that, when applied to trading, tells us whether a system is a winner or a loser because it shows us the average amount of profit or loss that can be expected over the long term. It is one of the best statistics because it quantifies the performance of a strategy or system, which is independent of the size of the capital.
The formula is:
MS=(Percentage of winning trades X average profit)-(Percentage of losing trades X average loss).
This formula can give us the following results and are interpreted as follows:
• Expectation > 0. It is positive, indicating that he earns more than he loses on average.
• Expectation = 0. It is neutral, indicating that your gains are equal to your losses on average.
• Hope < 0. It is negative, indicating that he loses more than he gains on average.
From the mathematical expectation we draw the following conclusions:
1. To make money you don’t have to win every trade, the important thing is the combination between how many times you get right (or wrong) and how much you win per trade.
2. To be profitable the mathematical expectation must be positive, and the higher the better.
3. Positive mathematical expectation does not prevent losing streaks, nor does it say anything about volatility in results.
Let’s look at an example. Suppose we have the following data after a few demo trades:
• Percentage of winning trades: 75%
• Average profit: 100
• Percentage of losing trades: 25%.
• Average loss: 200
We do the mathematical expectation which would be: (100*0.75) – (200*0.25) = 25.
As it gave a positive result, we can see that the system or strategy of the example is a winning system.
Let us now look at the opposite example. The data collected are:
• Percentage of winning trades: 60%
• Average profit: 100
• Percentage of losing trades: 40%.
• Average loss: 200
We do the mathematical expectation which would be: (100*0.60) – (200*0.40) = -20.
The mathematical expectation is negative, so in this case it is a losing system.
Note that it does not matter if the probability of success, i.e. the rate of winning trades, which in this case is 60%, is higher than the rate of losses; with a negative expectation it is still a system that only loses money in the long run.
Now let’s look at an example with respect to the probability of success. Suppose we collect data from several demo trades and get the following metrics:
• Hit probability (Percentage of winning trades): 25%.
• Average profit: 400
• Probability of failure (Percentage of losing trades): 75%.
• Average loss: 100
The mathematical expectation would be: (0.25*400)-(0.75*100)= 25
The average per trade of this system earns exactly the same as the first example. But, although they have the same mathematical expectation, comparing the probabilities of hit and miss we can see that the first system hits more than the second, but has a lower average win and a higher average loss.
## Which one to choose or which one is better?
The system in the first example has a higher probability of success, therefore it will cause a lower volatility in our capital, which reduces the risk. For this reason it is always recommended to choose a system or strategy that has a higher probability of success, even if it has a lower average profit, even if compared to another strategy it has the same mathematical expectation.
As a conclusion we can say that the rate of winning trades and the risk/reward ratio alone are not so important. What is important is what happens when the two are combined to determine the mathematical expectation of a strategy. If a trader has a mathematical expectation of +3, it means that on average, he can make 3 times the amount he risks on each trade he makes using his strategy or system.
Other factors to consider
In addition to knowing the mathematical expectation of our system, there are other important factors that come into play. Below we will look at some concepts to consider when it comes to evaluating a trading system:
## Historical results
Whenever we evaluate the hopefulness of a trading system it is always done with historical data, i.e. trades that have already been made. While this is important for demo testing, detecting errors and improving the strategy, we must not forget that there is no guarantee that we will have the same positive results in the future. We must be careful to avoid over-fitting our approach to historical data.
## Commissions:
Commissions and similar charges may seem small on a single trade, but when large numbers of trades are made with a system that has a small positive expectation, these costs can eat into profits or even turn the expectation from positive to negative. For this reason, the positive expectation must be large enough for commissions to be included.
## Position size:
Position size and mathematical expectation go hand in hand because, even with a large positive expectation, the use of an erratic position size can change the results and jeopardise your trading account. Position sizing should be kept within tolerable levels. We will expand on this point in the next lesson. | 1,502 | 7,067 | {"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 | 0.967834 |
https://en.wikipedia.org/wiki/Boyer-Moore_Majority_Vote_Algorithm | 1,547,986,593,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583705737.21/warc/CC-MAIN-20190120102853-20190120124853-00489.warc.gz | 499,066,388 | 13,521 | # Boyer–Moore majority vote algorithm
The state of the Boyer–Moore algorithm after each input symbol. The inputs are shown along the bottom of the figure, and the stored element and counter are shown as the symbols and their heights along the black curve.
The Boyer–Moore majority vote algorithm is an algorithm for finding the majority of a sequence of elements using linear time and constant space. It is named after Robert S. Boyer and J Strother Moore, who published it in 1981,[1] and is a prototypical example of a streaming algorithm.
In its simplest form, the algorithm finds a majority element, if there is one: that is, an element that occurs repeatedly for more than half of the elements of the input. However, if there is no majority, the algorithm will not detect that fact, and will still output one of the elements. A version of the algorithm that makes a second pass through the data can be used to verify that the element found in the first pass really is a majority.
The algorithm will not, in general, find the mode of a sequence (an element that has the most repetitions) unless the number of repetitions is large enough for the mode to be a majority. It is not possible for a streaming algorithm to find the most frequent element in less than linear space, when the number of repetitions can be small.[2]
## Description
The algorithm maintains in its local variables a sequence element and a counter, with the counter initially zero. It then processes the elements of the sequence, one at a time. When processing an element x, if the counter is zero, the algorithm stores x as its remembered sequence element and sets the counter to one. Otherwise, it compares x to the stored element and either increments the counter (if they are equal) or decrements the counter (otherwise). At the end of this process, if the sequence has a majority, it will be the element stored by the algorithm. This can be expressed in pseudocode as the following steps:
• Initialize an element m and a counter i with i = 0
• For each element x of the input sequence:
• If i = 0, then assign m = x and i = 1
• else if m = x, then assign i = i + 1
• else assign i = i − 1
• Return m
Even when the input sequence has no majority, the algorithm will report one of the sequence elements as its result. However, it is possible to perform a second pass over the same input sequence in order to count the number of times the reported element occurs and determine whether it is actually a majority. This second pass is needed, as it is not possible for a sublinear-space algorithm to determine whether there exists a majority element in a single pass through the input.[3]
## Analysis
The amount of memory that the algorithm needs is the space for one element and one counter. In the random access model of computing usually used for the analysis of algorithms, each of these values can be stored in a machine word and the total space needed is O(1). If an array index is needed to keep track of the algorithm's position in the input sequence, it doesn't change the overall constant space bound. The algorithm's bit complexity (the space it would need, for instance, on a Turing machine) is higher, the sum of the binary logarithms of the input length and the size of the universe from which the elements are drawn.[2] Both the random access model and bit complexity analyses only count the working storage of the algorithm, and not the storage for the input sequence itself.
Similarly, on a random access machine the algorithm takes time O(n) (linear time) on an input sequence of n items, because it performs only a constant number of operations per input item. The algorithm can also be implemented on a Turing machine in time linear in the input length (n times the number of bits per input item).
## Correctness
If there is a majority element, the algorithm will always find it. For, supposing that the majority element is m, let c be a number defined at any step of the algorithm to be either the counter, if the stored element is m, or the negation of the counter otherwise. Then at each step in which the algorithm encounters a value equal to m, the value of c will increase by one, and at each step at which it encounters a different value, the value of c may either increase or decrease by one. If m truly is the majority, there will be more increases than decreases, and c will be positive at the end of the algorithm. But this can be true only when the final stored element is m, the majority element. | 964 | 4,514 | {"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-2019-04 | latest | en | 0.92783 |
https://landsurveyorsunited.com/forum/topics/common-mistakes-to-avoid-in-land-surveyor-calculations | 1,716,173,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00555.warc.gz | 310,028,271 | 53,164 | ## Where are you surveying?
Survey Legend
### Common Mistakes to Avoid in Land Surveyor Calculations
Land surveying is a very significant task in the real estate industry. It requires a high level of precision and expertise to ensure that the land measurements and boundaries are accurate. As a land surveyor, one of the most important aspects of the job is calculation, which involves the use of various mathematical formulas to determine land boundaries, areas, and volumes.
One tiny miscalculation can result in significant problems, which can lead to costly legal issues. Therefore, it's essential to avoid some common surveyor calculation errors. In this article, weβll discuss some of the most crucial errors a land surveyor must avoid to ensure accuracy in their calculations.
Β
#### Using Inaccurate Instruments
For any land surveyor, having accurate and up-to-date equipment is crucial to accurately measure and calculate land boundaries, areas, and volumes. However, even the most sophisticated and advanced instruments can produce inaccurate measurements if used improperly. Surveyors should ensure they're using the appropriate equipment and measuring devices for the job. Using outdated or faulty equipment can lead to inaccurate measurements, which can lead to significant errors in land survey calculations.
#### Failing to Account for Curved Surfaces
Most land surveyors measure the surface of the earth, which may pose a problem in calculations when the earth's surface is curved. When calculating areas and volumes, surveyors may mistakenly neglect the earth's curvature, resulting in inaccurate results. Surveyors must take into account the curvature of the earth's surface while measuring distances, areas, and volumes, especially when working on larger projects. This ensures that measurements and calculations are accurate and reliable, which helps avoid costly legal disputes or project delays.
#### Ignoring Changes in Elevation
Changes in elevation are another factor that surveyors must consider while making measurements and subsequent calculations. They may be caused by slopes, terracing, contouring, or discontinuity within the land area. Failing to account for changes in elevation can lead to incorrect calculations, leading to issues such as underestimating or overestimating the land area's volume. Surveyors must take precise measurements of the elevation changes and incorporate them into their calculations to provide accurate results.
Β
#### Not Accounting for Legal Descriptions
When it comes to land surveying, legal descriptions are essential in identifying the boundary lines accurately. A surveyor must understand the legal descriptions applicable to the land area they are surveying. Some land descriptions may have been created many years ago and might not be up-to-date with the current state of the land area. This can result in boundary line errors if not appropriately identified. It's crucial to double-check legal descriptions and ensure accuracy before making any calculations on the land area.
#### Not Recognizing Errors
Even the most skilled, experienced land surveyor can make mistakes and overlook errors while measuring the land area. When errors are not recognized and corrected, they can have serious legal consequences, which can lead to costly lawsuits. Experienced land surveyors should be trained to detect potential errors and double-check their work to ensure accuracy. Land surveying is a technical profession that requires a high level of attention to detail, so surveyors should perform and double-check their work to avoid any potential legal issues.
#### Conclusion
The crucial role of a land surveyor cannot be understated in real estate transactions. Surveyors must ensure that their calculations are precise and accurate to provide legal protection for their clients. Errors in calculations can lead to significant legal disputes that may result in expensive litigation for both parties involved. As a result, surveyors must take care to avoid common mistakes, such as the use of inadequate equipment, failure to recognize errors, and neglecting legal descriptions. Land surveyors must understand the critical factors and considerations that go into calculating land boundaries, areas, and volumes while taking care to avoid errors and maximize precision. In conclusion, surveyors should always adhere to the highest professional and ethical standards, be diligent, and remain up to date with the latest technologies and regulations governing land surveying.
You need to be a member of Land Surveyors United - Global Surveying Community to add thoughts!
Survey Legend
Land Surveyor
### How to Utilize Surveyor Forums
How to use Forums
Our forums on Land Surveyors United are here to be used as much for finding help with problems in the field as the are for you to express your opinions on anything that has to do with land surveying in general. Feel free to share anything that is on your mind, as long as it isn't meant to damage another member's reputation. Please keep it clean and help insure that everyone has the opportunity to enjoy the benefits of being part of a community that grows together.
We are committed to allowing freedom of expression for all of our members, and that includes maintaining a safe space for people with opposing views to express themselves. We get posts from all over the country and even the globe, so needless to say, people come with different viewpoints on lad surveying practices and processes. We see this diversity and variety as a real strength-- dialogue and debate are an integral part of the educational process, as well as an important tool in exploring different sides of complex issues.
All Community Hubs inside the community have their own forum for asking specific questions to other surveyors, by location, equipment type, etc. | 1,085 | 5,878 | {"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} | 2.984375 | 3 | CC-MAIN-2024-22 | latest | en | 0.929818 |
https://mathematica.stackexchange.com/questions/4694/can-reduce-really-not-solve-for-x-here?noredirect=1 | 1,726,253,080,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00594.warc.gz | 353,263,454 | 44,142 | # Can Reduce *really* not solve for x here?
Sometimes I get the feeling I'm just flailing blindly with Mathematica. Is solving for $x$ in the equation $$\frac{\cosh (x/2)}{x} = \sqrt{2}$$ really beyond the scope of Mathematica? I try to solve it with the command:
Reduce[(1/x)Cosh[x/2] == Sqrt[2], x]
and am met with
Reduce::nsmet: This system cannot be solved with the methods available to Reduce.
I get a feeling that I'm doing something very silly. Cheers for any assistance!
• I don't see any special functions in the question or the answers, I must say... Commented Jun 8, 2012 at 5:34
Use
Reduce[(1/x) Cosh[x/2] == Sqrt[2], x, Reals]
or
Solve[(1/x) Cosh[x/2] == Sqrt[2], x, Reals]
the latter yields
{{x -> Root[{-E^(-(#1/2)) - E^(#1/2) + 2 Sqrt[2] #1 &, 0.75858229952537718426}]},
{x -> Root[{-E^(-(#1/2)) - E^(#1/2) + 2 Sqrt[2] #1 &, 5.4693513860610533998}]}}
For transcendental equations you may get with Reduce or Solve roots represented symbolically by Root though they are in general transcendental numbers, so their values are written numerically beside the transcendental function written in the form of a pure function.
Plot[ (1/x) Cosh[x/2] - Sqrt[2], {x, -7, 7}, PlotStyle -> Thick, PlotRange -> {-4, 4}]
Edit
It should be emphasized that using domain specifications in Reduce or Solve you may still get messages of unsolvability of a given equation or a system (of eqations or/and inequalities), e.g.
Reduce[ x Cos[x/2] == Sqrt[2], x, Reals]
Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>
Reduce[x Cos[x/2] == Sqrt[2], x, Reals]
even though for a slightly different equation you can get the full solution, e.g.
Reduce[x Cos[x/2] == 0, x, Reals]
In these two cases there is an infinite number of solutions, but the latter case is much easier, because a solution satisfies one of the two conditions : x == 0 or Cos[x/2] == 0. In the first case we need to restrict a region where we'd like to find solutions. There we find all of them with Reduce (as well as with Solve) if in a given region there is only a finite number of solutions, e.g. restricting the domain to real numbers such, that -25 <= x <= 25 i.e. adding a condition -25 <= x <= 25 to a given equation (now we needn't specify explicitly the domain to be Reals because Reduce[expr,vars] assumes by default that quantities appearing algebraically in inequalities are real):
sols = Reduce[x Cos[x/2] == Sqrt[2] && -25 <= x <= 25, x]
Defining
f[x_] := x Cos[x/2] - Sqrt[2]
we can easily check that sols are indeed the solutions :
FullSimplify[ f[ sols[[#, 2]]]] & /@ Range @ Length[sols]
{0, 0, 0, 0, 0, 0, 0}
To extract only numerical values of the roots combined with zero we can do (see e.g. this answer):
Tuples[{List @@ sols[[All, 2, 1, 2]], {0}}]
Now we can plot the function with appropriately marked roots and the specified domain :
Plot[ f[x], {x, -40, 40}, PlotStyle -> Thick,
Epilog -> {Thickness[0.003], Darker@Red, Line[{{-25, 0}, {25, 0}}],
PointSize[0.01], Cyan, Point /@ Tuples[{List @@ sols[[All, 2, 1, 2]], {0}}]}]
Here the dark red line denotes the domain of our interest, and the cyan points denote all roots of the function f in this region.
• I guess it would have been nice for Mathematica to tell me it was having trouble solving for complex x... thanks for the help. Commented Apr 24, 2012 at 12:22
• Yes, there are some omissions, however I think both Solve and Reduce are really powerful functions. Commented Apr 24, 2012 at 12:43
• @Artes Of course, Mathematica has found exact solutions here; those solutions are just expressed using Root objects. If you plug one of those roots back into the function and use FullSimplify, you get exactly zero. Commented Apr 24, 2012 at 12:50
• @MarkMcClure Thank you for your remark, indeed the solutions were represented symbolically with Root, but I meant their values only numerically. Commented Apr 24, 2012 at 13:09
• @GlenWheeler You may be interested in this blog post to make more sense of the solution that Mathematica returns. You might notice that the solution is just a symbolic representation of the root of the equation you entered near two values. Commented Apr 24, 2012 at 14:46
All of these work:
f[x_] := Cosh[x/2]/x - Sqrt[2];
FindRoot [f[x] == 0, {x, 1}]
N@FindInstance[f[x] == 0, x, Reals, 2]
N@Reduce [f[x] == 0, x, Reals]
NSolve [f[x] == 0, x, Reals]
N@Solve [f[x] == 0, x, Reals] | 1,328 | 4,455 | {"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} | 3.390625 | 3 | CC-MAIN-2024-38 | latest | en | 0.890594 |
https://electronics.stackexchange.com/questions/172968/can-you-use-diodes-instead-of-or-gate | 1,713,734,686,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817819.93/warc/CC-MAIN-20240421194551-20240421224551-00812.warc.gz | 197,810,163 | 41,053 | # Can you use diodes instead of OR gate?
I am building a computer and thought if I could simply use diodes instead of OR gates. Will this work with logic ICs such as the 74LS, 74HC and 75HCT series?
Sorry if this is a stupid question, I am new to electronics.
• Just add a pull-down resistor on the output to ensure the next input doesn't float off. May 29, 2015 at 13:18
• en.wikipedia.org/wiki/Diode%E2%80%93transistor_logic May 29, 2015 at 13:19
• It will not work with 74LS unless the pulldown resistor is really low value- 74LS requires outputs to sink current. 1K for a fan-out of 1 and 400mV noise margin, 100 ohms for a fan-out of 10. May 29, 2015 at 17:43
There is a really cool trick that you can use under specific conditions that allows you to build a 2-input OR gate or AND gate using only 1 diode and 1 resistor.
The conditions are:
1) The circuit is low speed.
2) The load impedance is high, such as a CMOS input.
simulate this circuit – Schematic created using CircuitLab
This can be expanded to as many inputs as needed by adding more diodes.
• Nice! Glad I waited for a better answer. Thanks for lowering my project's cost. :) Aug 12, 2015 at 9:28
Yes, that is called diode logic, but you have to be aware of some constrains. You will need a pull down resistor to give a path to the current.
• It should work even without the negative voltage, with the pulldown resistor going to ground. May 29, 2015 at 13:19
• Will the voltage remain the same? Won't the pull down resistors change the voltage levels making the input incompatible? May 29, 2015 at 13:21
• @Pentium100 If you pull the resistor to ground, it won't any drive current available to drive the next stage. May 29, 2015 at 13:28
• @xXLemonPRogrammerXx Not it won't change the voltage levels. Assuming ideal diodes, with 0V input the output will get clamped to 0V by the diodes, and 6V input will be 6V output, there is no resistor divider. Now in the real world you have to account for voltage drop in the diodes. May 29, 2015 at 13:32
• @davidrojas, fair enough. I only used "diode OR" together with CMOS or TTL chips so I did not need much current to drive the next stage. May 29, 2015 at 13:33
Yes you can. For glue logic on slow events (like power supply sequencing or connecting a pushbutton switch), this kind of logic should not cause problems. But
• If you use a negative supply like in davidrojas's answer, then you would be exciting the esd protection diodes of the downstream chip, which could cause unexpected behavior if there's nowhere to draw the pull-down current from. Also, you might not have a negative supply voltage available.
• To make the speed fast, you need to make the pull-down resistor value small, and that will increase the power consumption whenever the logic is high.
• You lose noise margin in the high state due to the drop through the diodes.
• It's more difficult to guarantee timing specs compared to using a dedicated OR gate chip.
Some alternatives that are often better are:
• Use a dedicated logic gate. One-gate chips only cost a few cents, so that the cost of the pick-and-place assembly might be higher than the actual chip cost.
• Invert the logic and use open-drain outputs in wired-AND configuration. | 837 | 3,245 | {"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-2024-18 | latest | en | 0.912508 |
https://oneclass.com/study-guides/ca/wlu/ma/ma-122/1019394-ma122-final.en.html | 1,713,962,590,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00876.warc.gz | 389,477,973 | 89,342 | # MA122 Study Guide - Final Guide: Polar Regions Of Earth, Polar Coordinate System, Cycloid
161 views3 pages
25 Sep 2016
School
Department
Course
## Document Summary
If rate of growth of a population is proportional to population, and if the population is small, Differential equation (de): equation that involves the derivatives of a function. Order: order of a de is the order of the highest derivative. Rate of change of population with respect to time. For large populations, p>m where m is a critical size, then dp/dt < 0 (logistic de) V0 is the initial amount invested r is the annual interest rate (units of % per year) m is the number of times the investment is compounded per year. Taking the limit yields v0ert (solution to above equation) Euler"s method yn+1 = yn + hf(xn, yn) y(x+h) = y(x) + hf(x, y) these formulas are the same! And xn = xn-1 + nh (so x1 = x0 + h, and x2 = x1 + 2h) Note: when it says to estimate y(0. 4) for example, and h = (cid:1004). (cid:1005), the(cid:374) (cid:455)ou"(cid:396)e fi(cid:374)di(cid:374)g (cid:455)4. | 301 | 1,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-18 | latest | en | 0.782765 |
https://in.mathworks.com/matlabcentral/cody/problems/173-minefield-sonar/solutions/2419663 | 1,610,976,833,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514796.13/warc/CC-MAIN-20210118123320-20210118153320-00147.warc.gz | 381,423,182 | 17,463 | Cody
# Problem 173. Minefield Sonar
Solution 2419663
Submitted on 29 May 2020 by jmac
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
I = [ 2 3 3 5 1 4 9 2 3 9 ]; J = [ 1 1 4 4 5 5 6 8 8 9 ]; M = 9; N = 9; y_correct = [ 1 1 0 1 -1 1 1 1 1 -1 2 1 2 2 1 2 -1 2 -1 2 1 -1 2 1 2 -1 2 1 1 2 3 -1 1 1 1 1 0 0 1 -1 2 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 -1 1 1 -1 ]; assert(isequal(minehunting(I,J,M,N),y_correct))
y = 1 1 0 1 -1 1 1 1 1 -1 2 1 2 2 1 2 -1 2 -1 2 1 -1 2 1 2 -1 2 1 1 2 3 -1 1 1 1 1 0 0 1 -1 2 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 -1 1 1 -1
2 Pass
I = 9; J = 9; M = 9; N = 9; y_correct = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 -1 ]; assert(isequal(minehunting(I,J,M,N),y_correct))
y = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 -1
3 Pass
I = 5; J = 5; M = 9; N = 9; y_correct = [ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 -1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ]; assert(isequal(minehunting(I,J,M,N),y_correct))
y = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 -1 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4 Pass
[I,J] = ndgrid(1:2:11,1:2:5); M = 11; N = 5; y_correct = [ -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 ]; assert(isequal(minehunting(I,J,M,N),y_correct))
y = -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1 2 4 2 4 2 -1 2 -1 2 -1
5 Pass
[I,J] = ndgrid(2:3:11,2:3:5); M = 11; N = 5; y_correct = [ 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 ]; assert(isequal(minehunting(I,J,M,N),y_correct))
y = 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 1 1 1 -1 1 1 -1
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting! | 1,797 | 2,480 | {"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-2021-04 | latest | en | 0.655876 |
https://forums.augi.com/showthread.php?57062-Draw-a-Line-given-1-point-and-an-angle&s=6be50bfc7640e7e4616af60144c5d478 | 1,581,961,531,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875142603.80/warc/CC-MAIN-20200217145609-20200217175609-00404.warc.gz | 394,737,765 | 10,556 | # Thread: Draw a Line given 1 point and an angle
1. Login to Give a bone
## Draw a Line given 1 point and an angle
hi all, like to ask is anyone has any method of ploting a straight with just 1 given point and an angle. don't need to be a code, if you have some interesting algo to get a rough 2nd point to use command "line", i also would like to know. thanks in advance
2. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
Originally Posted by tany0070
hi all, like to ask is anyone has any method of ploting a straight with just 1 given point and an angle. don't need to be a code, if you have some interesting algo to get a rough 2nd point to use command "line", i also would like to know. thanks in advance
Depending what version of Acad you are using you can do this using the polar tracking, you would just need to edit the angle settings to what you require.
3. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
Originally Posted by tany0070
hi all, like to ask is anyone has any method of ploting a straight with just 1 given point and an angle. don't need to be a code, if you have some interesting algo to get a rough 2nd point to use command "line", i also would like to know. thanks in advance
Have a look at the POLAR function.
4. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
Originally Posted by tany0070
hi all, like to ask is anyone has any method of ploting a straight with just 1 given point and an angle. don't need to be a code, if you have some interesting algo to get a rough 2nd point to use command "line", i also would like to know. thanks in advance
Hi,
Just go through this reply for your post, you can see an example.
Regards,
Abdul Huck
5. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
thank you all for the help
6. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
At autocad command prompt enter @distance<angle
Command: line
LINE Specify first point:
Specify next point or [Undo]: @10<135
draws a line 10 units long at 135degs
7. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
that's new (at least for me). thanks
8. Login to Give a bone
## Re: Draw a Line given 1 point and an angle
oh just a thought, if the angle is on the ending plane(i.e. the side i want to reach) is it possible to draw the line?? it's like i want to draw a line from a point to a line, the position of the line is unknown but i know this line will touch this point at an angle. thanks
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
• | 705 | 2,681 | {"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} | 2.75 | 3 | CC-MAIN-2020-10 | latest | en | 0.930403 |
https://www.jiskha.com/display.cgi?id=1271776250 | 1,516,709,487,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891926.62/warc/CC-MAIN-20180123111826-20180123131826-00201.warc.gz | 907,327,104 | 3,800 | chemistry
posted by .
What mass, in grams, of sodium hydroxide (NaOH) is produced if 20 grams of sodium metal(Na) reacts with excess water?
• chemistry -
1. Write the equation and balance it.
2. Convert 20 g Na metal to moles. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Na metal to mols NaOH.
4. Now convert moles NaOH to grams. g = moles x molar mass.
• chemistry -
2Na + S --> Na2S
20g Na x (1 mole Na)/(23g Na) = 0.87 mol Na
0.87 mole Na x (1 mol Na2S)/(2 mol Na) x (55.1g Na2S)/(1 mol Na2S)
=23.97g Na2S
Similar Questions
1. chemistry
What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation 2Na(s)+2H20(l)=2NaOH(aq)+H2(g)?
2. chemistry
What mass in grams of sodium hydroxide is produced if 20.0 g of sodium meal reactts with excess water according to the chemical equation 2 Na(s)+2 H2O(l)=2 NaOH(aq)+H2(g)?
3. Chemistry
Ammonium sulfate reacts with a solution of sodium hydroxide to produce sodium sulfate, ammonia gas and water. Calculate the yield in grams of ammonia gas that would form; and the amount of excess reagent in grams that would be left …
4. chemistry
if 30 grams of sodium (Na reacts completey with 28 grams of water, what is the mass of Sodium hydroxide (NaOH)formed?
5. chemistry
if 30 grams of sodium (Na reacts completey with 28 grams of water, what is the mass of Sodium hydroxide (NaOH)formed?
6. chemistry Emergency TEST
1.How many moles of sodium oxide are produced when3.9 moles of sodium combine with excess oxygen. 2. Sodium chloride decomposes into elemental sodium and elemental chlorine.How many grams of chlorine will be produced if 39.4 grams …
7. Chemistry
what mass in grams of sodium hydroxide is produced if 20.0g of sodium metal reacts with excess water according ro the chemical equation, Na(s)+H2O4->naoh(aq)+h2(g)
8. AP Chemistry
What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation?
9. science,
How many grams of CO are needed with an excess of Fe2O3 to produce 35.0 g Fe Solid sodium metal reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of of hydrogen are formed when 35 g …
More Similar Questions | 630 | 2,331 | {"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-2018-05 | latest | en | 0.857828 |
https://www.wyzant.com/resources/answers/858168/a-small-business-owner-sells-200-pizza-slices-at-1-50-per-slice-and-85-sand | 1,652,786,642,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662517245.1/warc/CC-MAIN-20220517095022-20220517125022-00191.warc.gz | 1,267,274,476 | 13,184 | Gabriela D.
# a small business owner sells 200 pizza slices at $1.50 per slice and 85 sandwiches at $2.00 each. Business expenses come to $140 per day. What is the owner's profit for a ten-day period? Each day a small business owner sells 200 pizza slices at $1.50
per slice and 85 sandwiches at
$2.00 each. Business expenses come to $140
per day. What is theowner's profit for a ten-day period?
By:
Tutor
5.0 (335)
SAT Math Expert
## Still looking for help? Get the right answer, fast.
Get a free answer to a quick problem.
Most questions answered within 4 hours.
#### OR
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need. | 187 | 689 | {"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} | 2.8125 | 3 | CC-MAIN-2022-21 | latest | en | 0.89906 |
http://www.filetransit.com/view.php?id=257194 | 1,506,005,138,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687820.59/warc/CC-MAIN-20170921134614-20170921154614-00615.warc.gz | 468,181,234 | 9,936 | Home | About Us | Link To Us | FAQ | Contact
# varea.m 1.0
Date Added: July 05, 2013 | Visits: 236
Compute the area and judge the direction of a closed curve.function area=varea(C)%VAREA compute the "sign-area" of a closed curve.% NOTE: Sign-area equals to the area of the closed curve when it is in anti-clockwise and equals to the negative area when it is in clockwise.% Negative area means equal to area in magnitude but is negtive in sign.We can see that this script may also be used to judge the direction of a closed curve.% C provides the coordinats of the nodes of the curve in the following way:% C=[x1 x2 ...% y1 y2 ...]% AREA=VAREA(C)% AREA returns the area of the curve(>0) when it is in anti-clockwise% AREA returns the negtive area of the curve(
Requirements: No special requirements Platforms: Matlab Keyword: Coordinats, Curvegt, Curvelt, Equal, Magnitude, Negtive, Nodes, Script, Signwe Users rating: 0/10
USER REVIEWS
More Reviews or Write Review
VAREA.M RELATED
Education - Gregorian calendar 1582 Gregorian calendar 1582 is a small Python script to generate calendars for any year in the history greater or equal to one. The output format is the same as the Unix "cal" command. However, it supposes the Gregorian Reformation took place on... 7.17 KB Miscellaneous - Ellipsoid fit Script 1.1 It needs a set of 3D points, equal radii constraints and orientation constraint to fit an ellipsoid into a 3D block object 10 KB Multimedia - RGB2Lab Script 1.1 This script can be used alongside MATLAB's native function lab2double (CIELAB to RGB conversion).CIELAB is a nonlinear transformation of RGB where the Euclidean distance between two colors is equal to their perceptual distances (for distances less... 10 KB Forms and Controls - Unobtrusive Table Sort Script 1.0 This script allows you to dynamically sort your HTML tables.Unobtrusive Table Sort Script creates and stores an internal representation of the parsed table data the very first time that one of the table’s sortable TH nodes is activated (a 2D... Form Processors - Validation (Time) 1.1 This script contains a function that verifies if a string is a valid time, in the form hh:mm:ss am/pm, where seconds are optional. It accepts military time (hour between 0 and 23) as long as am/pm isn't specified.It requires am/pm when the hour is... 102.4 KB Communication - Toolbox for log-spectral magnitude MMSE estimators under super-Gaussian densities 1.0 Toolbox for log-spectral magnitude MMSE estimators under super-Gaussian densities.The toolbox can also be downloaded from: http://ict.ewi.tudelft.nl/~richard/log_spe...per_gaussV1.rarThe matlab files enclosed in this toolbox can be used to... 10 KB Miscellaneous - tree_network Script 0.9 The package includes various examples with networks with 12 and up to 300 nodes. 225.28 KB Database Tools - PHP Tree Structure stored in MySQL database 1.0 PHP Tree Structure stored in MySQL database is a php script to store and manipulate tree structure in a mysql database, is a free PHP code generator.An example of a typical uses for this would be a web directory. Its important to note that the... 10 KB Database Tools - Tree AJAX 1.0 This class can be used to display a tree of nodes that can be opened dynamically from data retrieved from a database on the server via AJAX.It generates HTML and javascript to display tree nodes as links.When a node link is is clicked, the... 10 KB Multimedia & Graphics - Yanobox Nodes 1.2 Yanobox Nodes offers editors and Motion graphics artists a unique tool to combine text, images and graphics into beautiful 3D animations. Nodes lets you play with a number of different presets, or you can create your own graphic vocabulary by... 155.42 MB | 930 | 3,752 | {"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-2017-39 | longest | en | 0.791716 |
http://dopamineaddict.com/western-australia/how-to-find-number-of-moles-from-concentration.php | 1,561,591,955,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000575.75/warc/CC-MAIN-20190626214837-20190627000837-00519.warc.gz | 52,018,343 | 8,013 | ## How To Find Number Of Moles From Concentration
#### Calculating the number of moles at equilibrium from Kc
number of moles of a compound = mass of compound (g) ? relative formula mass (g mol-1) Worked example 1 Calculate the number of moles of water in 36 g of the liquid.
#### Mass moles (for proteins) zbio.net
Building off of LDC3's comment, division of the number of moles of \$\ce{BaCl2}\$ by the number of moles of \$\ce{Cl}\$ implies that you have a smaller amount of \$\ce{Cl}\$ than that of \$\ce{BaCl2}\$.
#### What Is Molarity & How Is It Calculated? Sciencing
24/04/2013 · Best Answer: 1.B just change everything to concentration values.. easy to find from there c=n/v 2.Higher Kp higher solubilityso figure that out
#### CHEMISTRY HELP (How do I find number of moles and
Remeber that the number of moles of solute does not change when more solvent is added to the solution. Concentration, however, does change with the added amount of solvent. Concentration, however, does change with the added amount of solvent.
How to find number of moles from concentration
#### CHEMISTRY HELP (How do I find number of moles and
7/03/2013 · The problem is that I have gotten mixed up with how to calculate moles. Now, I know that I should use the formula: n= Mass (g)/ Mr the book states that this gives the concentration, and to calculate the moles I will have to multiply it by the volume (i.e. 25cm^3). What am I not understanding? Any help would be greatly appreciated, Thank you.:)
#### What Is Molarity & How Is It Calculated? Sciencing
Remeber that the number of moles of solute does not change when more solvent is added to the solution. Concentration, however, does change with the added amount of solvent. Concentration, however, does change with the added amount of solvent.
#### Mass moles (for proteins) zbio.net
24/04/2013 · Best Answer: 1.B just change everything to concentration values.. easy to find from there c=n/v 2.Higher Kp higher solubilityso figure that out
#### CHEMISTRY HELP (How do I find number of moles and
However, since one mole of acid can contain 1, 2, 3, or more moles of H + ions, and a base can contain 1, 2, 3, or more moles of OH-ions, the number of moles of acid is not necessarily the same as the number of moles of the base at the endpoint.
#### Calculating the number of moles at equilibrium from Kc
However, since one mole of acid can contain 1, 2, 3, or more moles of H + ions, and a base can contain 1, 2, 3, or more moles of OH-ions, the number of moles of acid is not necessarily the same as the number of moles of the base at the endpoint.
#### homework Finding the concentration of an ion - Chemistry
Therefore the number of moles of H + = 0.0125 moles. Step 4 - Determine the concentration of HCl Every mole of HCl will produce one mole of H + , therefore the number of moles of HCl = number of moles …
#### What Is Molarity & How Is It Calculated? Sciencing
Molarity, or molar concentration (M), is defined as the number of moles of a substance, or "solute," dissolved in 1 liter of solution. Molarity is not to be confused with "molality," which is concentration expressed as moles of solute per kilogram of solvent.
#### What Is Molarity & How Is It Calculated? Sciencing
Therefore the number of moles of H + = 0.0125 moles. Step 4 - Determine the concentration of HCl Every mole of HCl will produce one mole of H + , therefore the number of moles of HCl = number of moles …
#### homework Finding the concentration of an ion - Chemistry
Molality (m)Molality is the number of moles of solute per kilogram of solvent. Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature.
### How to find number of moles from concentration - Mass moles (for proteins) zbio.net
Use these five steps to help you get more traffic Facebook to your website. #1: Make Sure You Have a Steady Stream of Shareable Content on Your Website If you have a website that never changes, you’re going to struggle to get much traffic to it.
#### how to keep possums off power lines
They are good climbers and can jump from trees to roofs as well as walk along power lines. If possums are moving across your roof, trimming back trees around the roofline can often reduce the traffic. However, to stop them jumping you will need a gap of at least 2 metres between the trees and your roof. They are usually looking for food such as mangoes, macadamia nuts and the fruit, leaves …
#### sao hollow realization how to get to cranvale path
The story of Sword Art Online: Hollow Realization follows players as they adventure through Sword Art: Origin, a new VRMMO based on the original Sword Art Online MMO. The events of the game are actually pretty similar to that of the original Sword Art Online arc, but with a few surprises thrown in.
#### how to find dry bulb temperature
15/11/2018 · Look along the bottom of the chart to find the dry bulb temperature. The horizontal, or “X” axis, of the chart denotes the various temperature readings in Fahrenheit or Celsius.
#### how to get eurovision tickets 2018
Eurovision 2019 - date, tickets, host country and what we know so far. From who is hosting to how to get tickets after Netta won the 2018 title with the song Toy
### You can find us here:
Australian Capital Territory: Mawson ACT, Duffy ACT, Coree ACT, Whitlam ACT, Mawson ACT, ACT Australia 2611
New South Wales: East Ryde NSW, Urbenville NSW, Blaxland NSW, Bewong NSW, Carabost NSW, NSW Australia 2048
Northern Territory: Mt Zeil NT, Coonawarra NT, Stuart NT, Casuarina NT, Tennant Creek NT, Adelaide River NT, NT Australia 0863
Queensland: Westwood QLD, Lakes Creek QLD, Yelarbon QLD, Gulliver QLD, QLD Australia 4023
South Australia: Uraidla SA, Sherwood SA, Iron Baron SA, Elizabeth Downs SA, Tanunda SA, Minlaton SA, SA Australia 5079
Tasmania: Carlton River TAS, St Leonards TAS, Mornington TAS, TAS Australia 7043
Victoria: Cundare North VIC, Maryknoll VIC, Clyde North VIC, Rythdale VIC, Doncaster VIC, VIC Australia 3003
Western Australia: Two Rocks WA, Stirling WA, Korrelocking WA, WA Australia 6071
British Columbia: Greenwood BC, Cache Creek BC, Oliver BC, Telkwa BC, Montrose BC, BC Canada, V8W 9W8
Yukon: Silver City YT, Jakes Corner YT, Braeburn YT, Braeburn YT, Klondike YT, YT Canada, Y1A 7C8
Alberta: Tofield AB, Donalda AB, Lloydminster AB, Amisk AB, Didsbury AB, McLennan AB, AB Canada, T5K 2J2
Northwest Territories: Wrigley NT, Enterprise NT, Tuktoyaktuk NT, Wekweeti NT, NT Canada, X1A 7L7
Saskatchewan: Wolseley SK, Canora SK, Bjorkdale SK, Coleville SK, Bladworth SK, Markinch SK, SK Canada, S4P 5C5
Manitoba: Steinbach MB, Powerview-Pine Falls MB, Manitou MB, MB Canada, R3B 8P2
Quebec: Lorraine QC, Mount Royal QC, Grenville QC, Westmount QC, Montreal QC, QC Canada, H2Y 1W5
New Brunswick: Paquetville NB, Kedgwick NB, Hartland NB, NB Canada, E3B 5H1
Nova Scotia: Lockeport NS, Halifax NS, Antigonish NS, NS Canada, B3J 4S8
Prince Edward Island: Warren Grove PE, Cardigan PE, St. Louis PE, PE Canada, C1A 8N3
Newfoundland and Labrador: Bonavista NL, Daniel's Harbour NL, Little Bay NL, Victoria NL, NL Canada, A1B 7J3
Ontario: Cottesloe ON, Dane ON, Mayhews Landing ON, Severn Falls, Big Cedar ON, Azilda ON, Wallbridge ON, ON Canada, M7A 9L9
Nunavut: King William Island NU, Charlton Island Depot NU, NU Canada, X0A 4H3
England: York ENG, Halifax ENG, St Helens ENG, Wigan ENG, Warrington ENG, ENG United Kingdom W1U 8A6
Northern Ireland: Newtownabbey NIR, Derry (Londonderry) NIR, Derry (Londonderry) NIR, Newtownabbey NIR, Craigavon (incl. Lurgan, Portadown) NIR, NIR United Kingdom BT2 4H9
Scotland: Dunfermline SCO, Aberdeen SCO, East Kilbride SCO, Livingston SCO, Dundee SCO, SCO United Kingdom EH10 7B7
Wales: Neath WAL, Swansea WAL, Swansea WAL, Wrexham WAL, Swansea WAL, WAL United Kingdom CF24 9D8 | 2,196 | 7,937 | {"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} | 3.1875 | 3 | CC-MAIN-2019-26 | latest | en | 0.903927 |
https://link.springer.com/article/10.1007/BF01027351 | 1,518,917,912,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891811243.29/warc/CC-MAIN-20180218003946-20180218023946-00688.warc.gz | 718,600,088 | 15,937 | Fluid Dynamics
, Volume 2, Issue 3, pp 23–28
# Development of steady-state mixing in an aerodynamic wake
• Yu. A. Dem'yanov
• V. T. Kireev
Article
## Abstract
We consider the unsteady, isothermal flow in the wake behind a body on which a shock wave has impinged. This flow is studied for sufficiently large distance from the body, when the effect of the latter on the external flow, as in the corresponding steady problem [1, 2], may be neglected. The analysis is performed for laminar and turbulent mixing with planar and axial symmetry.
In § 1 we use the characteristics of the system of equations consisting of the momentum equation written on the wake centerline and an integral equation to determine approximately the boundary separating the steady and unsteady flows. It is noted that the unsteady flow takes place between this boundary and the tagged particle line, which at the moment of detachment of the shock wave from the body coincided with it.
In §2 we show the possibility of calculating the flow in this region using the characteristics of the system of equations noted above. Some particular solutions of the latter are obtained.
Then, in §§ 3 and 4 the exact solutions corresponding to the flow regions identified in §§1 and 2 are presented.
Section 3 presents the exact solutions of the linear system of equations for large values of the time, when the velocity in the wake is close to the velocity of the particles behind the shock wave.
On the basis of the analysis made in §2, and previously in [3], §4 obtains the exact solution of the nonlinear system of equations which describes the development of the mixing near the tagged particle line for planar turbulent flow. This solution is compared with the corresponding approximate solution of §2. The accuracy of the latter is noted. Comparison of the horizontal component velocity profiles with the profiles obtained in the preceding section and used in [4] in the method of integral relations shows satisfactory agreement between them
### Keywords
Shock Wave Exact Solution Fluid Dynamics Nonlinear System Velocity Profile
## Preview
### References
1. 1.
L. G. Loitsyanskii, The Laminar Boundary Layer [in Russian], Fizmatgiz, 1962.Google Scholar
2. 2.
H. Schlichting, Boundary Layer Theory [Russian translation], IL, 1956.Google Scholar
3. 3.
Yu. A. Dem'yanov and V. T. Kireev, “Application of the equations of unsteady mixing to some aerodynamic problems”, Izv. AN SSSR, MZhG [Fluid Dynamics], no. 3, 1966.Google Scholar
4. 4.
A. S. Ginevskii, “Turbulent nonisothermal jet flow of a compressible gas”, collection: Industrial Aerodynamics [in Russian], Oborongiz, no. 23, 1962.Google Scholar
5. 5.
V. T. Kireev, “Establishing steady mixing in jets”, Izv. AN SSSR, MZhG [Fluid Dynamics], no. 4, 1966.Google Scholar
6. 6.
G. N. Abramovich, Theory of Turbulent Jets [in Russian], Fizmatgiz, 1960.Google Scholar
7. 7.
Yu. A. Dem'yanov, “Boundary layer formation on a plate with moving shock”, PMM, vol. 21, no. 3, 1957.Google Scholar
8. 8.
L. I. Sedov, Similarity and Dimensional Methods in Mechanics [in Russian], Gostekhizdat, 1954.Google Scholar
9. 9.
S. A. Lam and L. Crocco, “Note on the shock-induced unsteady laminar boundary layer on a semi-infinite flat plate”, J. Aeronaut. Sci., vol. 26, no. 1, 1959.Google Scholar | 842 | 3,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-09 | longest | en | 0.929954 |
https://sh-tsang.medium.com/reading-oisr-ode-inspired-schemes-to-super-resolution-network-designs-super-resolution-1f157e3b30f9?source=post_internal_links---------1---------------------------- | 1,702,147,905,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00156.warc.gz | 567,384,223 | 39,801 | # Reading: OISR — ODE-Inspired Schemes to Super-Resolution Network Designs (Super Resolution)
## Outperforms MSRN, RDN, EDSR & MDSR, CARN, SelNet, MemNet, LapSRN, DRRN, DRCN, VDSR, FSRCNN
--
In this story, ODE-inspired Network Design for Single Image Super-Resolution (OISR), by Chinese Academy of Sciences, University of Chinese Academy of Sciences, CAS, Alibaba Group, is presented. In this paper:
• An ordinary differential equation (ODE)-inspired design scheme is adopted for single image super-resolution, which have brought a new understanding of ResNet in classification problems.
• Two types of network structures are derived: LF-block and RK-block, which correspond to the Leapfrog method and Runge-Kutta method in numerical ordinary differential equations.
This is a paper in 2019 CVPR with over 20 citations. (Sik-Ho Tsang @ Medium)
# Outline
1. ODE-Inspired Network Design
2. Derivation of OISR-Blocks
3. Overall Network Architecture and Block Design
4. Experimental Results
# 1. ODE-Inspired Network Design
• From a dynamical system perspective, it defines a map that takes input status forward x units of time in the phase space.
• In CNN semantics, time horizon x corresponds to layers that can be adaptively chosen, while the final status is restricted by labels.
• Considering the dynamical systems which can be described as an ODE:
• This system gives a map:
• with initial status y0 ∈ Rd. Suppose p(y0) is the distribution of input feature y0 on a domain, if we regard CNN-based SISR as such a dynamical system, then we are supposed to minimize:
• where Φ is a map should be learned in SISR.
• As the system is non-linear, there is no simple formula describing the map, numerical methods are used — forward Euler method:
• which provides the approximation. It can be seen as a numerical ODE using the approximation to the integral of y′ over an interval of width.
• And residual block takes a similar form:
• The above suggests the relationship and establish the bridge by defining:
• Thus, mapping forward Euler to a residual block.
# 2. Derivation of OISR-Blocks
• To learn such a map Φ, it may take many steps to reach the final status, each step corresponds to a CNN block.
• (The CNN blocks as shown above will be derived below. Also the G block is composed of convolutions and activation functions which will be defined in the next section after deriving the above CNN block variants.)
• Either increasing the number of steps or refining motion of each step helps to achieve the goal, corresponding to increasing block numbers and designing finer blocks.
• Higher-order methods are supposed to bring about some merits.
## 2.1. LF-Block
• LeapFrog method is a second-order linear 2-step method, as a refinement of forward Euler scheme.
• By doubling the time interval h, the approximation of y′ can be rewritten in the form of y′ ≈ (yn+1−yn−1)/2h. Thus, yn+1 is:
• In order to retain flexibility and obtain a block architecture, every three formulas above are grouped into a block as:
• As mentioned:
• where G is some kinds of convolutional blocks.
## 2.2. RK2-block
• Consider the Runge-Kutta family, which is widely used in numerical analysis. Making use of trapezoidal formula:
• We obtain a block structure:
• In mathematics, these formulas are referred as Heun’s method, which is also a two-stage second-order Runge- Kutta method.
## 2.3. RK3-Block
• Higher-order methods should obtain a smaller local truncation error.
• Explicit iterative Runge-Kutta methods can be extended to arbitrary n stages:
• In particular, 3-stage Lunge-Kutta with third order is:
• Generally, higher-order methods tend to generate more complicated blocks.
• (Please note that I am not expert in ODE. Please feel free to read the paper if interested.)
# 3. Overall Network Architecture and Block Design
## 3.1. Network Architecture
• The dimension of input and output feature maps of G is kept unchanged.
• The OISR Blocks are at the middle of the network.
• Residual learning is used.
• Pixel-Shuffle in ESPCN is used at the end to get the SR.
## 3.2. Block Design (G)
• There are different types of G tried. There is a large searching space to search G.
• Here, only choose three different forms are chosen to illustrate the general effectiveness of ODE-inspired schemes.
• Each of these designs keeps at least one activation function and one convolutional layer, thus promising the nonlinearity.
# 4. Experimental Results
## 4.1. Ablation Study
• First 800 images in DIV2K are used for training. L1 loss is used.
• Ablation study is performed on DIV2K validation set.
• ”PReLU+Conv”, namely G-v2, is suitable for LF-blocks.
• RK2-blocks should be equipped with G-v3.
## 4.2. SOTA Comparison
• The small-scale network designs are suffixed by ”-s”.
• The small-scale models, OISR-RK2-s and OISR-LF-s outperform other methods such as FSRCNN, DRRN, MemNet, SelNet and CARN, on different upscaling factors and datasets, except a slightly behind on Urban100 with upscaling factor ×2.
• In addition, the middle-scale models, OISR-RK2 and OISR-LF, surpass MSRN with only two exceptions on B100 and Urban100 SSIM when the upscaling factor is 2.
• For current state-of-the-art deep residual methods, OISR-RK3 achieves the best performances in most cases, outperforms LapSRN, VDSR, DRCN, MDSR, RDN, and EDSR.
• OISRs can reconstruct more detailed images with less blurring.
This is the 20th story in this month.
## Reference
[2019 CVPR] [OISR]
ODE-inspired Network Design for Single Image Super-Resolution
## Super Resolution
[SRCNN] [FSRCNN] [VDSR] [ESPCN] [RED-Net] [DnCNN] [DRCN] [DRRN] [LapSRN & MS-LapSRN] [MemNet] [IRCNN] [WDRN / WavResNet] [MWCNN] [SRDenseNet] [SRGAN & SRResNet] [SelNet] [CNF] [BT-SRN][EDSR & MDSR] [MDesNet] [RDN] [SRMD & SRMDNF] [DBPN & D-DBPN] [RCAN] [ESRGAN] [CARN] [IDN] [ZSSR] [MSRN] [SR+STN] [SRFBN] [OISR] | 1,520 | 5,869 | {"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-2023-50 | latest | en | 0.87853 |
http://www.statemaster.com/encyclopedia/Arabic-numeral | 1,566,318,925,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315551.61/warc/CC-MAIN-20190820154633-20190820180633-00059.warc.gz | 308,115,489 | 18,230 | FACTOID # 5: Minnesota and Connecticut are both in the top 5 in saving money and total tax burden per capita.
Home Encyclopedia Statistics States A-Z Flags Maps FAQ About
WHAT'S NEW
SEARCH ALL
Search encyclopedia, statistics and forums:
(* = Graphable)
Encyclopedia > Arabic numeral
edit (http://en.wikipedia.org/w/index.php?title=Template:Numeral_systems&action=edit)
Numeral systems
Binary (2)
Octal (8)
Decimal (10)
Arabic numerals (also called Hindu numerals or Hindu-Arabic numerals) are by far the most common form of symbolism used to represent numbers. The Arabic numeral system is a positional base 10 numeral system with 10 distinct glyphs representing the 10 numerical digits. The leftmost digit of a number has the greatest value. In a more developed form, the Arabic numeral system also uses a decimal marker (at first a mark over the ones digit but now more usually a decimal point or a decimal comma which separates the ones place from the tenths place), and also a symbol for “these digits repeat ad infinitum” (recur). In modern usage, this latter symbol is usually a vinculum (a horizontal line placed over the repeating digits); the need for it can be removed by representing fractions as simple ratios with a division sign, but this obviates many of Arabic numbers’ more obvious advantages, such as the ability to immediately determine which of two numbers is greater. Historically, however, there has been much variation. In this more developed form, the Arabic numeral system can symbolize any rational number using only 13 glyphs (the ten digits, decimal marker, vinculum or division sign, and an optional prepended dash to indicate a negative number).
It is interesting to note that, like in many numbering systems, the numbers 1, 2, and 3 represent simple tally marks. 1 being a single line 2 being two lines (now connected by a diagonal) and 3 being three lines (now connected by two vertical lines). After three numbers tend to become more complex symbols (examples are the Chinese/Japanese numbers and Roman numerals). Theorists believe that this is because it becomes difficult to instantaneously count objects past three.
The Arabic numeral system has used many different sets of glyphs. These glyph sets can be divided into two main families—namely the West Arabic numerals, and the East Arabic numerals. East Arabic numerals—which were developed primarily in what is now Iraq—are shown in the table below as Arabic-Indic. East Arabic-Indic is a variety of East Arabic numerals. West Arabic numerals—which were developed in al-Andalus and the Maghreb—are shown in the table, labelled European. (There are two typographic styles for rendering European numerals, known as lining figures and text figures).
European 0 1 2 3 4 5 6 7 8 9 Arabic-Indic ٠ ١ ٢ ٣ ٤ ٥ ٦ ٧ ٨ ٩ Eastern Arabic-Indic (Persian and Urdu) ۰ ۱ ۲ ۳ ۴ ۵ ۶ ۷ ۸ ۹ Devanagari (Hindi) ० १ २ ३ ४ ५ ६ ७ ८ ९ Tamil (Blank) ௧ ௨ ௩ ௪ ௫ ௬ ௭ ௮ ௯
If your browser does not have all required fonts, this table might be rendered incorrectly. In that case, look at this graphic.
In Japan, Arabic numerals and the Roman alphabet are both used under the name of rōmaji. So, if a number is written in Arabic numerals, they would say “it is written in rōmaji” (as opposed to Japanese numerals). This translates as ‘Roman characters’, and may sound confusing for those who know about Roman numerals.
The Arabic numeral system is considered one of the most significant developments in mathematics. Most historians agree that it was first conceived of in India (particularly as Arabs themselves call the numerals they use “Indian numerals”, أرقام هندية, arqam hindiyyah), and was then transmitted to the Islamic world and thence, via Spain, to Europe.
The first inscriptions using 0 in India have been traced to approximately 400 AD. Aryabhata's numerical code also represents a full knowledge of the zero symbol. By the time of Bhaskara I (i.e., the seventh century AD) a base 10 numeral system with 9 glyphs was widely used in India, and the concept of zero (represented by a dot) was known (see the Vāsavadattā of Subandhu, or the definition by Brahmagupta). It is possible, however, that the invention of the zero sign took place sometime in the first century when the Buddhist philosophy of shunyata (zero-ness) gained ascendancy.
This numeral system had reached the Middle East by 670. Muslim mathematicians working in what is now Iraq, such as Al-Khwarizmi, were already familiar with the Babylonian numeral system, which used the zero digit between nonzero digits (although not after nonzero digits), so the more general system would not have been a difficult step. In the tenth century AD, Arab mathematicians extended the decimal numeral system to include fractions, as recorded in a treatise by Abu'l-Hasan al-Uqlidisi in 952-3.
Fibonacci, an Italian mathematician who had studied in Bejaia (Bougie), Algeria, promoted the Arabic numeral system in Europe with his book Liber Abaci, which was published in 1202. The system did not come into wide use in Europe, however, until the invention of printing (See, for example, the 1482 Ptolemaeus map of the world (http://bell.lib.umn.edu/map/PTO/TOUR/1482u.html) printed by Lienhart Holle in Ulm, and other examples in the Gutenberg Museum in Germany.)
It should be noted that in the Muslim World—until modern times—the Arabic numeral system was used only by mathematicians. Muslim scientists used the Babylonian numeral system, and merchants used a numeral system similar to the Greek numeral system and the Hebrew numeral system. Therefore, it was not until Fibonacci that the Arabic numeral system was used by a large population.
• Unicode reference charts (http://www.unicode.org/charts/):
• Arabic (http://www.unicode.org/charts/PDF/U0600.pdf) (See codes U+0660_U+0669, U+06F0_U+06F9)
• Devanagari (http://www.unicode.org/charts/PDF/U0900.pdf) (See codes U+0966_U+096F)
• Tamil (http://www.unicode.org/charts/PDF/U0B80.pdf) (See codes U+0BE6_U+0BEF)
• History of the Numerals
• The Evolution of Numbers (http://www.laputanlogic.com/articles/2003/06/01_95210802.html)
• Indian numerals (http://www_gap.dcs.st_and.ac.uk/%7Ehistory/HistTopics/Indian-numerals.html):
• Arabic numerals (http://www_gap.dcs.st_and.ac.uk/%7Ehistory/HistTopics/Arabic-numerals.html):
• Hindu_Arabic numerals (http://www.scit.wlv.ac.uk/university/scit/modules/mm2217/han.htm):
Results from FactBites:
Numerals in many different writing systems (133 words) The Urdu numerals are also known as 'East Arab' numerals and differ slightly from those used in Arabic. The numerals 1, 2, 3, etc. are also known as Arabic numerals, or Hindu-Arabic numerals, Indian numerals, Hindu numerals, European numerals, and Western numerals. These numerals where first used in India in about 400 BC, were later used in Persia, then were brought to Europe by the Arabs.
Arabic numerals (2212 words) The Indian numerals discussed in our article Indian numerals form the basis of the European number systems which are now widely used. The numerals had changed their form somewhat 100 years later when this copy of one of al-Biruni's astronomical texts was made. The form of the numerals in the west of the Arabic empire look more familiar to those using European numerals today which is not surprising since it is from these numerals that the Indian number system reach Europe.
More results at FactBites »
Share your thoughts, questions and commentary here | 1,824 | 7,465 | {"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-2019-35 | longest | en | 0.889122 |
http://www.mywordsolution.com/question/convert-the-following-ip-address-from-dotted/911234 | 1,527,391,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867995.55/warc/CC-MAIN-20180527024953-20180527044953-00069.warc.gz | 425,676,176 | 8,264 | +1-415-315-9853
info@mywordsolution.com
## Engineering
Civil Engineering Chemical Engineering Electrical & Electronics Mechanical Engineering Computer Engineering Engineering Mathematics MATLAB Other Engineering Digital Electronics Biochemical & Biotechnology
1) What do you mean by an internet? Suppose there is an organization where internet is require and describe it in detail with neat and suitable diagram.
2) Describe about datagram approach and compare with circuit switching in detail.
3)a) Describe Routing Table and Routing Module in detail.
b) A company is arranged site address 201.70.64.0. Company requires 6 subnets. Draw the subnets.
4)a) How is looping problem solved by switches and by routers? How do switches/routers handle link failure?
b) Describe the IP addressing.
5)a) Convert the following IP address from dotted decimal notation to binary notation:
i. 114.34.2.8
ii. 129.14.6.8
b. Convert the following IP address from binary notation to dotted decimal notation:
i. 01111111 11110000 01100111 01111101
ii. 11110111 11110011 10000111 11011101
c. Determine the net and host id of the IP addresses:
i. 241.34.2.8
ii. 11101111 11110111 11000111 00011101d. In a class C subnet, determine the network address
i. IP address: 182.44.82.1
Electrical & Electronics, Engineering
• Category:- Electrical & Electronics
• Reference No.:- M911234
Have any Question?
## Related Questions in Electrical & Electronics
### 1 in one-dimensional dct if n 1 the matrix transformation
1. In one-dimensional DCT, if N = 1, the matrix transformation is changed to simple multiplication. In other words, M = T × p, in which T, p, and M are numbers (scalar value) instead of matrices. What is the value of T i ...
### A dc pump connected to a pv module runs a garden water
A dc pump connected to a PV module runs a garden water fountain. The pump I -V curve and the hourly I -V curves for the module are shown. Once the pump starts running, it needs 8 V to keep spinning. At what time in the m ...
### 1 derive the weak form of a two-dimensional steady-state
1. Derive the weak form of a two-dimensional, steady-state heat transfer problem. 2. Derive the weak form of a simply supported beam problem.
### 1 the functional frequency n of a stretched string is a
1. The functional frequency n of a stretched string is a function of the string length L, its diameter D, the mass density r, and the applied tensile force T. Suggest a set of dimensionless parameters relating these vari ...
### 1 for flow at very low speeds and with large viscosity the
1. For flow at very low speeds and with large viscosity (the so-called creeping flows) such as occur in lubrication, it is possible to delete the inertia terms, Dv/Dt, from the Navier- Stokes equation. For flows at high ...
### 1 explain why private-public keys cannot be used in
1. Explain why private-public keys cannot be used in creating a MAC. 2. The nonce in Figure 31.22 is to prevent a replay of the third message. Eve cannot replay the third message and pretend that it is a new request from ...
### Discretize the continuous-time system with transfer
Discretize the continuous-time system with transfer function H(s) = (s + 10 )/s with use of the bilinear (Tustin) approximation method. Verify your result by converting the computed discrete-time system to continuous-tim ...
### The spacing a between the plates of a parallel-plate
The spacing a between the plates of a parallel-plate waveguide is equal to 5 cm. The dielectric between the plates is free space. If a generator of fundamental frequency 1800 MHz and rich in harmonics excites the wavegui ...
### The device in the schematic diagram below is a viscosity
The device in the schematic diagram below is a viscosity pump. It consists of a rotating drum inside of a stationary case. The case and the drum are concentric. Fluid enters at A, flows through the annulus between the ca ...
### An exhaust stream leaving a semiconductor fabrication plant
An exhaust stream leaving a semiconductor fabrication plant contains 4 mol % volatile organic chemical, VOC, pollutant. Twenty-five cubic meters per minute enter the bottom of a packed absorption tower and flows counterc ...
• 4,153,160 Questions Asked
• 13,132 Experts
• 2,558,936 Questions Answered
## Looking for Assignment Help?
Start excelling in your Courses, Get help with Assignment
Write us your full requirement for evaluation and you will receive response within 20 minutes turnaround time.
Ask Now Help with Problems, Get a Best Answer
### WalMart Identification of theory and critical discussion
Drawing on the prescribed text and/or relevant academic literature, produce a paper which discusses the nature of group
### Section onea in an atwood machine suppose two objects of
SECTION ONE (a) In an Atwood Machine, suppose two objects of unequal mass are hung vertically over a frictionless
### Part 1you work in hr for a company that operates a factory
Part 1: You work in HR for a company that operates a factory manufacturing fiberglass. There are several hundred empl
### Details on advanced accounting paperthis paper is intended
DETAILS ON ADVANCED ACCOUNTING PAPER This paper is intended for students to apply the theoretical knowledge around ac
### Create a provider database and related reports and queries
Create a provider database and related reports and queries to capture contact information for potential PC component pro | 1,226 | 5,475 | {"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-2018-22 | latest | en | 0.804009 |
http://mathoverflow.net/questions/120438/limit-of-a-wiener-integral?sort=oldest | 1,469,378,476,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00075-ip-10-185-27-174.ec2.internal.warc.gz | 156,073,360 | 14,307 | Limit of a Wiener integral
How to show that
$$\lim_{\alpha \rightarrow \infty} \sup_{t \in \left [0,T \right]} \left | e^{-\alpha t} \int _ 0 ^t e^{\alpha s} ~ dB_s \right | =0, \ \ \text{a.e.}$$
where $\left (B_s \right)_{s\geq 0}$ is a real standard brownian motion starting from zero ?
I'd like to have some ideas to deal with this problem. After all, I'll show some solutions that I propose and somme colegues also but that i believe be all wrong. (I just don't show know to don't interffer in your ideas.
Thank you all.
-
LaTeX fixed${}{}{}$ – Gerald Edgar Jan 31 '13 at 18:24
1 Answer
Here's one way of dealing with it. Integrate by parts to see that the expression under the sup is $$\Bigl|B(t)-\alpha\int_0^t e^{\alpha(s-t)}B(s)ds\Bigr|\le\alpha\int_0^t e^{\alpha(s-t)}|B(t)-B(s)|ds +e^{-\alpha t}|B(t)|.$$
Now the result follows since $B$ is a.s.-bounded and a.s.-Holder on [0,T].
-
@Yuri Bakhtin: Thank you very much for your answer! Nice solution. – Paul Feb 18 '13 at 0:50
@Paul: You are welcome. – Yuri Bakhtin Feb 18 '13 at 4:44 | 363 | 1,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2016-30 | latest | en | 0.827298 |
https://it.mathworks.com/matlabcentral/answers/2108666-plot-the-function-which-is-dependent-on-x-y-and-z-with-x-y-and-z-on-three-axis?s_tid=prof_contriblnk | 1,718,956,473,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00666.warc.gz | 279,353,988 | 26,769 | # plot the function which is dependent on x, y and z with x, y and z on three axis.
14 visualizzazioni (ultimi 30 giorni)
Modificato: Torsten il 18 Apr 2024
D_value = 0.1;
L_value = 0.1;
B_value = 0.1;
x = 0:0.01:L_value/D_value;
y = 0:0.01:B_value/D_value;
z = 0:0.01:0.25;
[Y, X, Z] = meshgrid(y, x, z);
Ra_value = 80;
xi = 0.3;
R_value = Ra_value*xi;
A1_1_1 = -8.2516;
A1_1_2 = -1.7735;
A1_2_1 = -1.0336;
A1_2_2 = 0.6812;
A2_1_1 = -0.5388;
A2_1_2 = -0.8701;
A2_2_1 = -0.0329;
Phi_z = @(x,y,z) A1_1_1.*pi.*cos(pi.*Z) + 2.*A2_1_1.*pi.*cos(2.*pi.*Z) + A1_1_2.*pi.*cos((pi.*D_value.*Y)./B_value).*cos(pi.*Z) + 2.*A2_1_2.*pi.*cos((pi.*D_value.*Y)./B_value).*cos(2.*pi.*Z) + A1_2_1.*pi.*cos((pi.*D_value.*X)./L_value).*cos(pi.*Z) + 2.*A2_2_1.*pi.*cos((pi.*D_value.*X)./L_value).*cos(2.*pi.*Z) + A1_2_2.*pi.*cos((pi.*D_value.*Y)./B_value).*cos((pi.*D_value.*X)./L_value).*cos(pi.*Z);
xlabel('x');
ylabel('y');
zlabel('z');
plot the function Phi_z.
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
### Risposte (2)
Torsten il 18 Apr 2024
Modificato: Torsten il 18 Apr 2024
You can plot the function on slices (i.e. 2d-objects (e.g. planes)) through the volume of interest.
Of course a full plot over a 3d-volume is not possible because we cannot see in 4d.
D_value = 0.1;
L_value = 0.1;
B_value = 0.1;
x = 0:0.01:L_value/D_value;
y = 0:0.01:B_value/D_value;
z = 0:0.01:0.25;
[X,Y,Z] = meshgrid(x,y,z);
Ra_value = 80;
xi = 0.3;
R_value = Ra_value*xi;
A1_1_1 = -8.2516;
A1_1_2 = -1.7735;
A1_2_1 = -1.0336;
A1_2_2 = 0.6812;
A2_1_1 = -0.5388;
A2_1_2 = -0.8701;
A2_2_1 = -0.0329;
Phi_z = @(X,Y,Z) A1_1_1.*pi.*cos(pi.*Z) + 2.*A2_1_1.*pi.*cos(2.*pi.*Z) +...
A1_1_2.*pi.*cos((pi.*D_value.*Y)./B_value).*cos(pi.*Z) +...
2.*A2_1_2.*pi.*cos((pi.*D_value.*Y)./B_value).*cos(2.*pi.*Z) +...
A1_2_1.*pi.*cos((pi.*D_value.*X)./L_value).*cos(pi.*Z) +...
2.*A2_2_1.*pi.*cos((pi.*D_value.*X)./L_value).*cos(2.*pi.*Z) +...
A1_2_2.*pi.*cos((pi.*D_value.*Y)./B_value).*cos((pi.*D_value.*X)./...
L_value).*cos(pi.*Z);
slice(X,Y,Z,Phi_z(X,Y,Z),(x(1)+x(end))/2,[],[])
xlabel('x');
ylabel('y');
zlabel('z');
colorbar
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
Fangjun Jiang il 18 Apr 2024
Modificato: Fangjun Jiang il 18 Apr 2024
You are asking for the impossible, the visualization of the 4th dimension.
1. I thought it was impossible in three dimentional world.
2. There migth be some methods to "help" the visualization. https://en.wikipedia.org/wiki/Four-dimensional_space
3. I can't think of any built-in method in MATLAB that can help. Maybe, you could plot a dot at each and every point of the whole (x,y,z) grid. Set the color of the dot according to the value of Phi_z. Could that be regarded as the visualization of the 4th dimension? Not sure what is the visual effect though.
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti
Accedi per commentare.
### Categorie
Scopri di più su Lighting, Transparency, and Shading in Help Center e File Exchange
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by | 1,275 | 3,226 | {"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} | 2.75 | 3 | CC-MAIN-2024-26 | latest | en | 0.210598 |
https://www.slideserve.com/salena/binary-relation | 1,516,496,188,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00106.warc.gz | 989,389,695 | 11,295 | # binary relation: - PowerPoint PPT Presentation
Discrete Math for CS. Binary Relation Exercise:. If R = { (a,b): a < b}. Show R on the diagram below.. . . A. B. . . . . . . . . 8. 6. 4. 2. 1. 3. 5. 7. . . . . . . . . . . . . . . . . Discrete Math for CS. Binary Relations as Sets of Ordered Pairs:. Because we mention one set before another in a Cartesian Product, A x B, the element, (a,b), in any relation, R, over A and B must have its first element from A and its second element from B.So we say that the elements of R form ordered pairs..
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. | 190 | 651 | {"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-2018-05 | latest | en | 0.838339 |
https://www.sapling.com/7867024/much-2500-monthly-mortgage-payment | 1,611,697,916,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803737.78/warc/CC-MAIN-20210126202017-20210126232017-00457.warc.gz | 979,709,504 | 119,759 | # How Much Income Is Needed for \$2,500 Monthly Mortgage Payment?
Buying a house is a huge financial commitment. You want to make sure you can afford the monthly mortgage payments for the next 30 years and your mortgage lender wants to make sure you can afford them, too. That's why the lender has financial guidelines it looks out when considering whether you are eligible for a mortgage. You can use those guidelines and judge yourself whether you can afford the home of your dreams.
## Consider All Costs
Knowing the maximum value of a home you can get approved for a mortgage for is just one of the costs you need to consider before you buy a home. That's why your mortgage lender looks at two ratios when deciding how much house you can afford. One ratio looks at monthly payment, but the second looks at your monthly payment and your other debt from things like credit cards, car loans, student loans and child support. You will need to be below the threshold for at least one of these numbers to get approved for a mortgage with a monthly payment of \$2,500.
## The Guidelines
You mortgage lender will want your monthly mortgage payment to be no more than 28 percent of your gross monthly income. If the payment exceeds 28 percent, then you might still be approved if your total long-term debt is below 36 percent of your gross monthly income. The figures are what have been found to be affordable amounts, while still leaving room for monthly living expenses and disposable income.
## Doing the Math
If you know your monthly mortgage payment is project to be \$2,500 a month, then you need to divide that by .28 to get the minimum gross monthly income you need to make to afford the payment, which is \$8,928. This equates to earning \$107,136 a year before taxes. However, at this annual income, you will need to make sure that your other long-term debt does not exceed \$714 a month. This is because \$8,928 times .36 is \$3,214 a month. If you subtract the \$2,500 from that amount, you are left with amount of other allowable long-term debt. | 437 | 2,060 | {"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 | longest | en | 0.955914 |
https://www.alivelearn.net/?cat=6&paged=24 | 1,660,615,471,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00510.warc.gz | 569,650,458 | 7,526 | ## 10-20 system for brain area location
10-20 system is a method to localize brain areas. It uses 4 landmarks of head: 1 the nasion which is the point between the forehead and the nose; 2 the inion which is the lowest point of the skull from the back of the head and is normally indicated
Xu Cui
## Statistical type II error (beta), and power
Even though I learned power analysis in elementary statistics class, I totally forget it because I never use it. As we have to put power analysis in our grant, I am forced to relearn this topic. Definition: type II error: null hypothesis should be re
Xu Cui
## Why choose this book? Chinese version 如何做出選擇？
Read’s book, Why choose this book, was pubilshed in traditional Chinese recently. Here is a link: Simplified Chinese version is hopefully coming.
Xu Cui
## Two games, iterative thinking
1. The smallest unique number The host asks each of a large number of participants to write down a positive integer. The person whose number is unique and smallest wins. If you were one of the participant, which number do you write? e.g. If the numbe
Xu Cui
## brain images: Change data type of brain image
Last update: 2012/09/04 An image file can be saved in different formats such as uint8 or int16, etc, based on different number of bytes used for each voxel. How to convert between them? Here is one solution using SPM functions (no need SPM’s in
Xu Cui
## Brain AC-PC line
AC: anterior commissure PC: posterior commissure The Anterior Commissure (precommissure) is a bundle of white fibers, connecting the two cerebral hemispheres across the middle line, and placed in front of the columns of the fornix. On a sagittal sect
Xu Cui
## Talk by Frans de Waal
Today I listened to Frans de Waal’s talk on empathy in primates. It’s quite interesting. He wrote several popular books including “Chimpanzee Politics”. Links: Frans de Waal
Xu Cui | 473 | 1,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.903619 |
https://www.acmicpc.net/problem/only/10 | 1,597,162,811,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738816.7/warc/CC-MAIN-20200811150134-20200811180134-00115.warc.gz | 574,753,621 | 11,722 | 문제 번호아이디제목정보맞은 사람제출정답 비율
19114mitnegativeinfinityMaster Zhu and Candies출처다국어11100.000%
19115mitnegativeinfinityMaster Zhu and Rikka출처다국어11100.000%
19116mitnegativeinfinityMaster Zhu and Palindromes출처다국어11100.000%
19117mitnegativeinfinityMaster Zhu and Video출처다국어11100.000%
19118mitnegativeinfinityMaster Zhu and Polygons출처다국어1250.000%
19119mitnegativeinfinityMaster Zhu and Root출처다국어11100.000%
19120mitnegativeinfinityMaster Zhu and Binary Trees출처다국어11100.000%
19121mitnegativeinfinityMaster Zhu and Instability출처다국어11100.000%
19122mitnegativeinfinityMaster Zhu and Math Problem출처다국어11100.000%
19123mitnegativeinfinityMaster Zhu and the Leaper출처다국어11100.000%
19124mitnegativeinfinityBinomial Coefficient출처다국어11100.000%
19125mitnegativeinfinityBipartite Graph Coloring출처다국어11100.000%
19136mackerel_pikeBalls and Holes출처다국어1250.000%
19145mitnegativeinfinityBanach스페셜 저지출처다국어11100.000%
19150mitnegativeinfinityGalois출처다국어11100.000%
19156aintaCleaning Robots출처다국어11100.000%
19158aintaFind String in a Grid출처다국어11100.000%
19163aintaMission Possible스페셜 저지출처다국어11100.000%
19168mitnegativeinfinityBrainy스페셜 저지출처다국어11100.000%
19173koosagaGarden스페셜 저지출처다국어11100.000%
19177mitnegativeinfinityKlothes스페셜 저지출처다국어11100.000%
19178koosaga(Smurf)Land protection출처다국어11100.000%
19180zimpha36 Puzzle스페셜 저지출처다국어11100.000%
19183zimphaSubsequence출처다국어11100.000%
19184zimphaAlgebra is Awesome출처다국어1425.000%
19185zimphaBus Lines출처다국어1250.000%
19186zimphaCoal Mine스페셜 저지출처다국어11100.000%
19188zimphaGame출처다국어1250.000%
19189zimphaTic-tac-toe출처다국어11100.000%
19198mitnegativeinfinityAmbitious Plan출처다국어11100.000%
19199mitnegativeinfinityBorderless Words출처다국어12100.000%
19201mitnegativeinfinityDecomposable Single Word Languages스페셜 저지출처다국어11100.000%
19202mitnegativeinfinityElegant Square스페셜 저지출처다국어11100.000%
19203mitnegativeinfinityFour Colors출처다국어인터랙티브11100.000%
19204mitnegativeinfinityGreater Number Wins스페셜 저지출처다국어11100.000%
19206mitnegativeinfinityIsomorphism스페셜 저지출처다국어11100.000%
19207mitnegativeinfinityJinxiety of a Polyomino출처다국어11100.000%
19209zimphaBar "Duck"스페셜 저지출처다국어1912.500%
19213mitnegativeinfinityFibonacci's Nightmare출처다국어12100.000%
19214mitnegativeinfinityGuess The String출처다국어인터랙티브11100.000%
19215mitnegativeinfinityHilbert's Maze출처다국어11100.000%
19216mitnegativeinfinityInfinite Binary Embedding출처다국어11100.000%
19247mitnegativeinfinityBoardroom Meeting출처다국어11100.000%
19248mitnegativeinfinitySecret Santa출처다국어11100.000%
19251mitnegativeinfinityWe Need More Managers!출처다국어11100.000%
19252mitnegativeinfinityMasterpiece출처다국어11100.000%
19254mitnegativeinfinityBobby Tables스페셜 저지출처다국어11100.000%
19255mitnegativeinfinityTriples출처다국어1250.000%
19256mitnegativeinfinityRelated Languages출처다국어11100.000%
19257mitnegativeinfinityAirport Check-in스페셜 저지출처다국어11100.000%
19259mitnegativeinfinityCasino Cheating출처다국어인터랙티브11100.000%
19263mitnegativeinfinityGame of Chairs출처다국어11100.000%
19264mitnegativeinfinityHung Fu스페셜 저지출처다국어11100.000%
19265mitnegativeinfinityIs It a p-drome?출처다국어11100.000%
19266mitnegativeinfinityJourney출처다국어11100.000%
19268mitnegativeinfinityLazy Student출처다국어11100.000%
19270mitnegativeinfinityColorful Doors스페셜 저지출처다국어11100.000%
19278mitnegativeinfinitySimple APSP Problem출처다국어11100.000%
19281mitnegativeinfinityShort Random Problem출처다국어11100.000%
19284mitnegativeinfinityOneness출처다국어11100.000%
19286mitnegativeinfinityPiecewise Linearity출처다국어11100.000%
19288mitnegativeinfinity≤ or ≥출처다국어인터랙티브11100.000%
19289mitnegativeinfinityStairways출처다국어11100.000%
19293koosagaEarthquake스페셜 저지출처다국어11100.000%
19295koosagaCentral Lake스페셜 저지출처다국어13100.000%
19298aintaNumber of Cycles스페셜 저지출처다국어11100.000%
19300aintaGame of Sorting출처다국어11100.000%
19307koosagaExit Song출처다국어11100.000%
19313koosagaKids Aren't Alright출처다국어11100.000%
19324koosagaThree Arrays출처다국어11100.000%
19326koosagaCover the Paths스페셜 저지출처다국어11100.000%
19328koosagaCode-Cola Plants스페셜 저지출처다국어11100.000%
19331koosagaCompressed Spanning Subtrees출처다국어인터랙티브11100.000%
19332koosagaPrefix-free Queries출처다국어11100.000%
19333koosagaSubsequence Sum Queries출처다국어11100.000%
19334koosagaConsistent Occurrences출처다국어11100.000%
19335koosagaIncreasing Costs출처다국어11100.000%
19338mitnegativeinfinityFlip a Coin스페셜 저지출처다국어11100.000%
19339mitnegativeinfinityOctagons출처다국어11100.000%
19342mitnegativeinfinitySheep스페셜 저지출처다국어1250.000%
19344mitnegativeinfinityStatistics스페셜 저지출처다국어11100.000%
19346mitnegativeinfinityKnapsack출처다국어11100.000%
19347mitnegativeinfinityThe Catcher in the Rye스페셜 저지출처다국어11100.000%
19349mitnegativeinfinityDominoes스페셜 저지출처다국어11100.000%
19350mitnegativeinfinityEndgame출처다국어11100.000%
19351mitnegativeinfinityEvacuation출처다국어11100.000%
19353mitnegativeinfinityGrasshoppers출처다국어11100.000%
19354mitnegativeinfinityEducation Nightmare출처다국어12100.000%
19359mitnegativeinfinityEven Three is Odd출처다국어11100.000%
19360mitnegativeinfinityWalk of Length 6출처다국어11100.000%
19361mitnegativeinfinityCity United출처다국어11100.000%
19362mitnegativeinfinityCoins 2출처다국어11100.000%
19364mitnegativeinfinityMulti-stage Marathon출처다국어11100.000%
19366mitnegativeinfinityPermutation and noitatumreP출처다국어11100.000%
19370aintaSpanning Trees스페셜 저지출처다국어1250.000%
19371aintaTriangle출처다국어1425.000%
19372aintaBoxes and Balls출처다국어1714.286%
19373aintaBit Operations스페셜 저지출처다국어1520.000% | 1,981 | 5,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2020-34 | latest | en | 0.341911 |
https://oneclass.com/textbook-notes/ca/mcgill/psyc/psyc-204/120186-chapter-onedoc.en.html | 1,524,581,443,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125946721.87/warc/CC-MAIN-20180424135408-20180424155408-00057.warc.gz | 638,923,123 | 45,282 | Textbook Notes (368,588)
Psychology (1,418)
PSYC 204 (1)
Chapter 1
# Chapter One.doc
7 Pages
181 Views
School
Department
Psychology
Course
PSYC 204
Professor
David Ostry
Semester
Winter
Description
Chapter One: Introduction to Statistics Statistics, Science, and Observations Preview • all statistical procedures were developed to serve a purpose ◦ if you understand why a new procedure is needed, you will find it much easier to learn Definitions of Statistics • statistics (psychology) – a set of mathematical procedures for summarizing, organizing, and interpreting information ◦ shortened version of statistical procedures ◦ statistics serve two general purposes ▪ used to organize and summarize the information so that the researcher can see what happened in a research study and can communicate these results to others ▪ help the researcher to answer the general questions that initiated the research by determining exactly what conclusions are justified based on the results that were obtained ◦ provide researchers with a set of standardized techniques that are recognized and understood throughout the scientific community Population and Samples What Are They? • scientific research usually begins with a general question about a specific group (or groups) of individuals • population – the set of all the individuals of interest in a particular study ◦ LECTURE: population – all instances on some quantitative dimension ◦ the population being studied should always be identified by the researcher ◦ need not consist of people • sample – a set of individuals selected from a population, usually intended to represent the population in a research study ◦ intended to be representative of its population ◦ should always be identified in terms of the population from which it was selected Variables and Data • typically researchers are interested in specific characteristics of individuals in the population (or sample), or they are interested in outside factors • variable – a characteristic or condition that changes or has different values for different individuals • data – measurements or observations ◦ data set – a collection of measurements and observations ◦ datum – a single measurement or observation ▪ commonly called a score or a raw score Parameters and Statistics • parameter – a value (usually numerical) that describes a population ◦ usually derived from measurements of the individuals in the population • statistic – a value (usually numerical) that describes a sample ◦ usually derived from the measurements of the individuals in that sample • typically, every population parameter has a corresponding sample statistic Descriptive and Inferential Statistical Methods • the different statistical procedures to organize and interpret data can be classified into two general categories ◦ descriptive statistics – statistical procedures used to summarize, organize, and simplify data ▪ take raw scores and organize or summarize them in a form that is more manageable ▪ often results are organized in a table or a graph so that it is possible to see an entire set of scores ◦ inferential statistics – techniques that allow us to study samples and then make generalizations about the populations from which they were selected ▪ use sample data to make general statements about a population ▪ typically, researchers use sample statistics as the basis for drawing conclusions about population parameters • although samples are usually representative of their populations, a sample is not expected to give a perfectly accurate picture of the whole population ◦ sampling error – the discrepancy, or amount of error, that exists between a sample statistic and the corresponding population parameter ▪ also known as margin of error • there will always be a margin of error when sample statistics are used to represent population parameters • the unpredictable, unsystematic differences that exist from one sample to another are an example of sampling error Statistics in the Context of Research ◦ ex. students taught with one method of teaching earned a 5-pt higher score on testing than those taught with “method B” ▪ there are two possible interpretations of these results • there is no real difference in the teaching methods, and the 5-pt. difference between the samples is just a sampling error • the is really a difference between the two teaching methods, and the 5 pt. difference was caused by the different methods of teaching Data Structures, Research Methods, and Statistics Relationships Between Variables • most psych research is intended to examine the relationship between variables • to establish the existence of an relationship, researchers must make measurements of the two variables ◦ resulting measurements can be classified into two distinct data structures that also help to classify different research methods and different statistical techniques 1. One Group with Two Variables Measured for Each Individual • one method for examining the relationship between two variables is to observe the two variables as they exist naturally for a set of variables ◦ then, they simply measure the two variables for each individual ◦ researchers then look for consistent patterns in the data to provide evidence for a relationship between variables ◦ consistent patterns are often easier to see if the scores are presented in a graph • correlational research strategy – two different variables are observed to determine whether there is a relationship between them ◦ occasionally, the correlational method produces scores that are not numerical values ▪ this data is typically summarized in a table showing how many individuals are classified into each of the possible categories ◦ results form a correlational study do not provide an explanation for the relationship, and cannot demonstrate a cause-and-effect relationship 2. Comparing Two (or More) Groups of Scores: Experimental and Non- Experimental Methods • this method involves the comparison of two or more groups of scores • when this measurement procedure involves numericl scores, the statistical evaluation typically involves computing the average score for each group and then comparing the averages • if the measurement process simply classifies the individuals into non-numerical categories, statistical evaluation usually consists of computing proportions for each group, and then comparing proportions The Experimental Method • one specific research method that involves comparing groups of scores is claled the experimental method ◦ the goal of an experimental study is to demonstrate a cause-and-effect relationship between two variables ▪ specifically, an expriment attempts to show that changing the value of one variable causes changes to occur in the second variable ▪ to accomplish this goal, the experimental method has two characteristics that differentiate experiments from other types of research studies: • manipulation – the researcher manipulates one variable by changing its value from one level to another ◦ a second variable is observed to determine whether manipulation causes changes to occur • control – the researcher must exercise control over the research situation to ensure that other, extraneous variables do not influence the relationship being examined ◦ two general categories of variables that researchers must consider: ▪ participant variables – characteristics that vary from one individual to another • researchers must ensure that participant variables do not differ from one group to another • whenever a research study allows more than one explanation for the results, the study is said to be confounded (it is impossible to reach an unambiguous conclusion ▪ environmental variables – characteristics of the environment • ex. lighting, time of day, weather conditions • a researcher must ensure that individuals in treatmentAare tested in the same environment as the individuals in treatment B • researchers use three basic techniques to control other variables ◦ random assignment – each participant has an equal chance of being assigned to each of the treatment conditions ◦ matching – ensure equivalent groups or equivalent environments ◦ holding variables constant • experimental method – one variable is manipulated while another variable is observed and measured ◦ to establish a cause-and-effect relationship between the two variables, and experiment attempts to control all other variables to prevent them from influencing the results Terminology in the Experimental Method • independent variable – the variable that is manipulated by the experimenter ◦ in behavioral research, the independent variable usually consists of the two treatment conditions to which subjects are exposed ◦ independent variable consists of the antecedent conditions that were manipulated prior to observing the dependent variable • dependent variable – variable that is observed an measured ◦ observed to assess the effect of the treatment • an experimental study evaluates the relationship between two variables by manipulating one variable and measuring the dependent variable ◦ only one variable is measured ▪ different from a correlational study, where both variables are measured and the data consists of two separate scores for each individual ◦ often, an experiment will include a condition in
More Less
Related notes for PSYC 204
Me
OR
Join OneClass
Access over 10 million pages of study
documents for 1.3 million courses.
Join to view
OR
By registering, I agree to the Terms and Privacy Policies
Just a few more details
So we can recommend you notes for your school. | 1,707 | 9,600 | {"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-2018-17 | latest | en | 0.931999 |
https://gmatclub.com/forum/how-many-different-prime-numbers-are-factors-of-the-positive-37210.html?fl=similar | 1,495,727,678,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608084.63/warc/CC-MAIN-20170525140724-20170525160724-00104.warc.gz | 750,136,726 | 45,628 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 25 May 2017, 08:54
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# How many different prime numbers are factors of the positive
Author Message
Senior Manager
Joined: 05 Aug 2005
Posts: 409
Followers: 2
Kudos [?]: 60 [0], given: 0
How many different prime numbers are factors of the positive [#permalink]
### Show Tags
22 Oct 2006, 23:37
00:00
Difficulty:
(N/A)
Question Stats:
100% (02:14) correct 0% (00:00) wrong based on 3 sessions
### HideShow timer Statistics
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
How many different prime numbers are factors of the positive integer n?
1) 4 different prime numbers are factors of 2n
2) 4 different prime numbers are factors of n^2
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1
Kudos [?]: 45 [0], given: 0
### Show Tags
23 Oct 2006, 01:39
How many different prime numbers are factors of the positive integer n?
1) 4 different prime numbers are factors of 2n
2) 4 different prime numbers are factors of n^2
From (1) 2n = 2(2*3*5*7), so n must also have 4 different prime numbers as factors. SUFF
Not too sure about statement (2) but i think it's the same concept.
n^2 = (n)(n) --> this should be equal to (2*3*5*7)(2*3*5*7)
again not too sure about (2). But if correct, would go for D
_________________
Impossible is nothing
Senior Manager
Joined: 19 Jul 2006
Posts: 360
Followers: 2
Kudos [?]: 7 [0], given: 0
### Show Tags
23 Oct 2006, 05:27
I guess its B
1) n can have 3 or 4 prime factors
2) n^2 has same number of prime factors as n
SVP
Joined: 05 Jul 2006
Posts: 1747
Followers: 6
Kudos [?]: 358 [0], given: 49
### Show Tags
23 Oct 2006, 11:25
How many different prime numbers are factors of the positive integer n?
1) 4 different prime numbers are factors of 2n
2) 4 different prime numbers are factors of n^2
from one
n could be in the form of 2k or k ........not suff
from two
suff because n^2 , n are composed of the same distinctive primes
Manager
Joined: 24 Oct 2005
Posts: 169
Followers: 36
Kudos [?]: 189 [0], given: 0
### Show Tags
23 Oct 2006, 12:17
I will go for B as well.
1) n could be any number. Insufficient
2) we know n has 4 different prime numbers since n^2 = n*n. Sufficient.
23 Oct 2006, 12:17
Display posts from previous: Sort by | 874 | 2,974 | {"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-2017-22 | latest | en | 0.904996 |
https://stats.stackexchange.com/questions/295178/umvue-of-location-parameter-shifted-exponential | 1,653,802,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663039492.94/warc/CC-MAIN-20220529041832-20220529071832-00251.warc.gz | 603,755,012 | 66,250 | # UMVUE of location parameter (shifted exponential)
Let $X_1,...,X_n$ be a sample from a distribution with pdf, $f_X(x) = e^{-x + \theta}, x \geq \theta$. Let $x_0 \geq \theta$ be given. I'm trying to find the UMVUE of $f_X(x_0) = e^{-x_0 + \theta}$. I went ahead and found that a complete sufficient statistic for $\theta$ is $X_{(1)}$. Moreover, an unbiased estimator for $f_X(x_0)$ is,
$$T(\textbf{X}) = \begin{cases}1 & X_1 > x_0 \\ 0 & otherwise \end{cases}$$
So the UMVUE should be $\mathbb{E}[T(\textbf{X}) | X_{(1)}]$. We have,
\begin{align} \mathbb{E}[T(\textbf{X}) | X_{(1)} = x] &= \mathbb{P}(X_1 > x_0 | X_{(1)} = x)\\ &= \mathbb{P}(X_1 - X_{(1)} > x_0 - X_{(1)} | X_{(1)} = x) \\ &= \mathbb{P}(X_1 - X_{(1)} > x_0 - x | X_{(1)} = x)\\ &= \mathbb{P}(X_1 - X_{(1)} > x_0 - x)\;\;\;\;\;\;\; (**)\end{align}
where the last equality follows from Basu's Theorem, as $X_1 - X_{(1)}$ is ancilliary for $\theta$ and $X_{(1)}$ is complete sufficient; thus they are independent. I'm stuck at finding this probability, however. How can I find the distribution of $X_1 - X_{(1)}$?
As a different approach, I went ahead found the expectation of functions of the form, $ce^{aX_{(1)}+b}$. It turns out that $g(X_{(1)}) = \frac{n-1}{n}e^{X_{(1)}-x_0}$ is unbiased for $e^{-x_0 + \theta}$ and thus must be the UMVUE. However, this was simply trial and error which fortunately led to the result. I'm not comfortable just guessing and checking. Is there a way I can proceed from $(**)$?
• Another option is to differentiate both sides of the equation $E_{\theta}[g(X_{(1)})]=f_X(x_0)$ wrt $\theta$ and solve for $g$. Dec 14, 2019 at 12:55
You could proceed from $(**)$ as follows: for $x \leq x_0$, \begin{align*} &\mathbb P(X_1 - X_{(1)} > x_0 - x)\\ &= 1-\mathbb P(X_1 - X_{(1)} \leq x_0 - x)\\ &= 1-\mathbb P(X_{(1)} \geq X_1 - x_0 + x)\\ &= 1-\mathbb P(X_i \geq X_1 - x_0 + x\text{ for all $i=1,\dots,n$})\\ &= 1-\mathbb P(X_i \geq X_1 - x_0 + x\text{ for all $i=2,\dots,n$})\\ &= 1-\int_{\theta}^\infty \mathbb P(X_i \geq t - x_0 + x\text{ for all $i=2,\dots,n$})f_{X_1}(t)\ dt\\ &= 1- \int_\theta^\infty f_{X_1}(t)\prod_{i=2}^n \mathbb P(X_i \geq t-x_0+x)\ dt \\ &= \dots \end{align*} However, there appears to be a problem which needs to be addressed first. When referring to a "complete sufficient statistic for $\theta$", we need to carefully consider what space of parameters $\Theta$ is being implicitly referenced. For the space $\Theta = [0,\infty)$, it is true that $X_{(1)}$ is a complete sufficient statistic, and in this case the identification of $\frac{n-1}ne^{X_{(1)}-x_0}$ as the UMVUE for $e^{-x_0+\theta}$ is correct (for $n>1$). However, this does not provide a UMVUE for $f_X(x_0)$ on this space $\Theta$, as the equation $f_X(x_0) = e^{-x_0+\theta}$ does not hold when $\theta>x_0$. The statistic $T(\mathbf X)$ also fails to be unbiased when $\theta>x_0$.
You may instead be intending to consider the space $\Theta = [0, x_0]$, but in this case $X_{(1)}$ is no longer complete, so you will need a different complete, sufficient statistic ($\min\{X_{(1)}, x_0\}$ should do the trick.) After this is sorted out, you should be able to proceed in a similar way as above. | 1,165 | 3,194 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2022-21 | latest | en | 0.742727 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.