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https://www.evi.com/q/320_ml_is_how_many_cups | 1,432,640,945,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928831.69/warc/CC-MAIN-20150521113208-00058-ip-10-180-206-219.ec2.internal.warc.gz | 880,091,535 | 5,549 | # 320 ml is how many cups
• 320 milliliters is 1.35 cups.
• tk10publ tk10canl
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• 320cc to cups | 928 | 2,783 | {"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} | 3.03125 | 3 | CC-MAIN-2015-22 | longest | en | 0.899612 |
https://djmarsay.wordpress.com/2015/06/24/instrumental-probabilities/ | 1,500,921,005,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424909.50/warc/CC-MAIN-20170724182233-20170724202233-00436.warc.gz | 658,877,027 | 42,523 | # Instrumental Probabilities
Reflecting on my recent contribution to the economics ejournal special issue on uncertainty (comments invited), I realised that from a purely mathematical point of view, the current mainstream mathematical view, as expressed by Dawid, could be seen as a very much more accessible version of Keynes’. But there is a difference in expression that can be crucial.
In Keynes’ view ‘probability’ is a very general term, so that it always legitimate to ask about the probability of something. The challenge is to determine the probability, and in particular whether it is just a number. In some usages, as in Kolmogorov, the term probability is reserved for those cases where certain axioms hold. In such cases the answer to a request for a probability might be to say that there isn’t one. This seems safe even if it conflicts with the questioner’s presuppositions about the universality of probabilities. In the instrumentalist view of Dawid, however, suggests that probabilistic methods are tools that can always be used. Thus the probability may exist even if it does not have the significance that one might think and, in particular, it is not appropriate to use it for ‘rational decision making’.
I have often come across seemingly sensible people who use ‘sophisticated mathematics’ in strange ways. I think perhaps they take an instrumentalist view of mathematics as a whole, and not just probability theory. This instrumentalist mathematics reminds me of Keynes’ ‘pseudo-mathematics’. But the key difference is that mathematicians, such as Dawid, know that the usage is only instrumentalist and that there are other questions to be asked. The problem is not the instrumentalist view as such, but the dogma (of at last some) that it is heretical to question widely used instruments.
The financial crises of 2007/8 were partly attributed by Lord Turner to the use of ‘sophisticated mathematics’. From Keynes’ perspective it was the use of pseudo-mathematics. My view is that if it is all you have then even pseudo-mathematics can be quite informative, and hence worthwhile. One just has to remember that it is not ‘proper’ mathematics. In Dawid’s terminology the problem seems to be that the instrumental use of mathematics without any obvious concern for its empirical validity. Indeed, since his notion of validity concerns limiting frequencies, one might say that the problem was the use of an instrument that was stunningly inappropriate to the question at issue.
It has long seemed to me that a similar issue arises with many miscarriages of justice, intelligence blunders and significant policy mis-steps. In Keynes’ terms people are relying on a theory that simply does not apply. In Dawid’s terms one can put it blunter: Decision-takers were relying on the fact that something had a very high probability when they ought to have been paying more attention to the evidence in the actual situation, which showed that the probability was – in Dawid’s terms – empirically invalid. It could even be that the thing with a high instrumental probability was very unlikely, all things considered. | 651 | 3,132 | {"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-2017-30 | longest | en | 0.950596 |
http://jbtermpaperektk.rivieramayaexcursions.us/trig-help.html | 1,534,652,262,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221214691.99/warc/CC-MAIN-20180819031801-20180819051801-00647.warc.gz | 217,438,230 | 4,469 | # Trig help
Trigonometry helps us find angles and distances, and is used a lot in science, engineering, video games, and more right-angled triangle the triangle of most interest is the right-angled triangle. Elementary functions part 4, trigonometry lecture 47a, solving problems with inverse trig functions dr ken w smith sam houston state university. Free trigonometric identities - list trigonometric identities by request step-by-step. Trigonometric identities use these fundemental formulas of trigonometry to help solve problems by re-writing expressions in another equivalent form. A in two or more complete sentences, explain how to find the exact value of sec 13pi/6 including quadrant location b in two or more complete sentences, explain how to find the exact value of cot 7pi/4 including quadrant location. Basic and advanced trigonometric equations with problems, examples and solutions. Precalculus review and calculus preview - shows precalculus math in the exact way you'll use it for calculus - also gives a preview to many calculus concepts. Exercise try this paper-based exercise where you can calculate the sine function for all angles from 0° to 360°, and then graph the result it will help you to understand these relatively simple functions.
Free math help tutorial videos on topics including arithmetic, algebra, trigonometry, and calculus. Free tutorials, problems and self tests on trigonometry examples and problems with solutions are included. Trigonometry help resource to help students with trigonometry at all levels. Carbidedepotcom.
Grade 10 trigonometry questions and problems with solutions and answers are presented. Get trigonometry help from chegg now trigonometry guided textbook solutions, expert answers, definitions and more.
Trignometry resources--video tutorials, interactive lessons and free calculators. Click your trigonometry textbook below for homework help our answers explain actual trigonometry textbook homework problems each answer shows how to solve a textbook problem, one step at a time.
## Trig help
At trig, we define marketing as demand management we believe that marketing is a crucial part of your business strategy and it empowers your products and services, strengthens your brand, and authentically connects you with your customers. Online trigonometry video lessons to help students with problems on trigonometric functions and identities, among other important concepts, to improve their math problem solving skills so they can find the solution to their trigonometry. This precalculus review (calculus preview) lesson reviews the unit circle and basic trigonometric (trig) identities and gives great tips on how to remember everything.
Trigonometry homework help and answers popular trigonometry textbooks see all trigonometry textbooks algebra and trigonometry: structure and. Trigonometry, which studies the measure of triangles, takes algebra to the next level its most well-known features include the pythagorean theorem and. Trigonometry is all about angles and right triangles get free trigonometry help from expert online trigonometry tutors of tutorvista join our tutoring, workout trig problems and complete your homework with ease. Trigonometry (from greek trigōnon, triangle and metron, measure) is a branch of mathematics that studies relationships involving lengths and angles of trianglesthe field emerged in the hellenistic world during the 3rd century bc from applications of geometry to astronomical studies. What are similar triangles what are complementary angles how to find all functions from one function what are cofunctions.
Trigonometry & calculus - powered by webmath visit cosmeo for explanations and help with your homework problems. Trigonometry book 1 [] book 1 is pre-calculus trigonometry we assume the student is relatively new to algebra and do algebra step by step many of the pages have closely related free/youtube videos at the khan academy. Finally, there will only be four trig questions on the math test, so even if you aren’t comfortable with trig, it won’t destroy your math score. Your complete trigonometry help that gets you better marks learn with step-by-step video help, instant trigonometry practice and a personal study plan.
Trig help
Rated 4/5 based on 24 review | 862 | 4,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2018-34 | latest | en | 0.87454 |
https://www.coursehero.com/file/6501248/09-09ans/ | 1,519,440,046,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891815034.13/warc/CC-MAIN-20180224013638-20180224033638-00042.warc.gz | 801,250,836 | 71,747 | {[ promptMessage ]}
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09_09ans - STAT 410 Examples for Fall 2011 2.3 1...
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STAT 410 Examples for 09/09/2011 Fall 2011 2.3 Conditional Distributions and Expectations. 1. Consider the following joint probability distribution p ( x , y ) of two random variables X and Y: y x 0 1 2 p X ( x ) 1 0.15 0.10 0 0.25 2 0.25 0.30 0.20 0.75 p Y ( y ) 0.40 0.40 0.20 a) Find the conditional probability distributions p X | Y ( x | y ) = ( ) ( ) y p y x p , Y of X given Y = y , conditional expectation E ( X | Y = y ) of X given Y = y , and E ( E ( X | Y ) ). x p X | Y ( x | 0 ) x p X | Y ( x | 1 ) x p X | Y ( x | 2 ) 1 0.15 / 0.40 = 0.375 1 0.10 / 0.40 = 0.25 1 0.00 / 0.20 = 0.00 2 0.25 / 0.40 = 0.625 2 0.30 / 0.40 = 0.75 2 0.20 / 0.20 = 1.00 E ( X | Y = 0 ) = 1.625 E ( X | Y = 1 ) = 1.75 E ( X | Y = 2 ) = 2.0 Def E ( X | Y = y ) = x x P ( X = x | Y = y ) = x x p X | Y ( x | y ) discrete E ( X | Y = y ) = ( ) - dx y x f x | Y | X continuous Denote by E ( X | Y ) that function of the random variable Y whose value at Y = y is E ( X | Y = y ). Note that E ( X | Y ) is itself a random variable, it depends on the ( random ) value of Y that occurs.
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y E ( X | Y = y ) p Y ( y ) 0 1.625 0.40 0.65 E ( E ( X | Y ) ) = 1.75. 1 1.75 0.40 0.70 2 2.0 0.20 0.40 Recall: E ( X ) = 1.75. E ( a 1 X 1 + a 2 X 2 | Y ) = a 1 E ( X 1 | Y ) + a 2 E ( X 2 | Y ) E [ g ( Y ) | Y ] = g ( Y ) E ( E ( X | Y ) ) = E ( X ) E [ E ( X | Y ) | Y ] = E ( X | Y ) E [ g ( Y ) X | Y ] = g ( Y ) E ( X | Y ) b) Find the conditional probability distributions p Y | X ( y | x ) = ( ) ( ) x p y x p , X of Y given X = x , conditional expectation E ( Y | X = x ) of Y given X = x , and E ( E ( Y | X ) ). y
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09_09ans - STAT 410 Examples for Fall 2011 2.3 1...
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Ask a homework question - tutors are online | 874 | 2,174 | {"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} | 3.59375 | 4 | CC-MAIN-2018-09 | latest | en | 0.724217 |
http://stackoverflow.com/questions/2620068/vba-public-user-defined-function-in-excel?answertab=oldest | 1,394,762,989,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678683421/warc/CC-MAIN-20140313024443-00066-ip-10-183-142-35.ec2.internal.warc.gz | 143,808,310 | 16,590 | VBA Public User Defined Function in Excel
I have created the function below:
Option Explicit
Public Function fyi(x As Double, f As String) As String
Application.Volatile
Dim data As Double
Dim post(5)
post(1) = "Ribu "
post(2) = "Juta "
post(3) = "Milyar "
post(4) = "Trilyun "
post(5) = "Ribu Trilyun "
Dim part As String
Dim text As String
Dim cond As Boolean
Dim i As Integer
If (x < 0) Then
fyi = " "
Exit Function
End If
If (x = 0) Then
fyi = "Nol"
Exit Function
End If
If (x < 2000) Then
cond = True
End If
text = " "
If (x >= 1E+15) Then
fyi = "Nilai Terlalu Besar"
Exit Function
End If
For i = 4 To 1 Step -1
data = Int(x / (10 ^ (3 * i)))
If (data > 0) Then
part = fyis(data, cond)
text = text & part & post(i)
End If
x = x - data * (10 ^ (3 * i))
Next
text = text & fyis(x, False)
fyi = text & f
End Function
Function fyis(ByVal y As Double, ByVal conds As Boolean) As String
Dim datas As Double
Dim posts(2)
posts(1) = "Puluh"
posts(2) = "Ratus"
Dim parts As String
Dim texts As String
'Dim conds As Boolean
Dim j As Integer
Dim value(9)
value(1) = "Se"
value(2) = "Dua "
value(3) = "Tiga "
value(4) = "Empat "
value(5) = "Lima "
value(6) = "Enam "
value(7) = "Tujuh "
value(8) = "Delapan "
value(9) = "Sembilan "
texts = " "
For j = 2 To 1 Step -1
datas = Int(y / 10 ^ j)
If (datas > 0) Then
parts = value(datas)
If (j = 1 And datas = 1) Then
y = y - datas * 10 ^ j
If (y >= 1) Then
posts(j) = "belas"
Else
value(y) = "Se"
End If
texts = texts & value(y) & posts(j)
fyis = texts
Exit Function
Else
texts = texts & parts & posts(j)
End If
End If
y = y - datas * 10 ^ j
Next
If (conds = False) Then
value(1) = "Satu "
End If
texts = texts & value(y)
fyis = texts
End Function
When I return to Excel and type =fyi(500,"USD") in a cell, it returns #name.
Please inform me how to solve.
-
See this related question: http://stackoverflow.com/questions/755812/create-a-custom-worksheet-function-in-excel-vba
In summary:
What you have should work.
Based on the comments to that question, you should place your user-defined function in any module other than ThisWorkbook.
-
Make sure that your function is in a Module, not in the Worksheet.
-
If your UDF is in a workbook other than the workbook your calling from, prefix the udf with the workbook name. E.g.
=PERSONAL.XLS!fyi(500,"USD")
-
Not sure I'd use this method.. but +1 for obscure UDF knowledge Dick! – Mark Nold Apr 13 '10 at 9:47
The best place for functions such as this is in an Addin... To make an addin:
Make a new workbook
hit alt+F11
create a module, call it MyFunctions or something else meaningfull | 860 | 2,595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2014-10 | latest | en | 0.499951 |
http://www.expertsmind.com/library/money-in-savings-account-at-an-annual-rate-51322259.aspx | 1,597,317,426,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738982.70/warc/CC-MAIN-20200813103121-20200813133121-00242.warc.gz | 135,010,044 | 11,320 | ### Money in savings account at an annual rate
Assignment Help Financial Management
##### Reference no: EM131322259
You have just received an endowment and placed this money in a savings account at an annual rate of 19.01 percent. You are going to withdraw the following cash flows for the next five years. End of year 1. \$1406 2. \$9693 3. \$2460 4. \$2265 5. \$846 How much is the endowment that you received.
#### Begin by constructing a time line
A project has an initial cost of \$40,000, expected net cash inflows of \$9,000 per year for 7 years, and a cost of capital of 11%. What is the project's NPV? (Hint: Begin by co
#### Purchase zero coupon bond with a face value
Suppose you purchase a zero coupon bond with a face value of \$1,000, maturing in 22 years, for \$215.75. Zero coupon bonds pay the investor the face value on the maturity date.
#### Trying to determine its cost of debt
Drogo, Inc., is trying to determine its cost of debt. The firm has a debt issue outstanding with 14 years to maturity that is quoted at 104 percent of face value. The issue ma
#### What was flotation cost as percentage of funds raised
The Wiley Oakley Co. has just gone public. Under a firm commitment agreement, the company received \$21.65 for each of the 6.65 million shares sold. The initial offering price
#### Pure expectation hypothesis holds and market expects
Assume the pure expectation hypothesis holds and the market expects that 1-year rate will be 6.1% 4 years from today (4rt1=6.1%). What is the 4-year rate today (rt4)? Show eve
#### Use the compressed adjusted present value model
An unlevered firm has a value of \$900 million. An otherwise identical but levered firm has \$50 million in debt at a 5% interest rate. Its cost of debt is 5% and its unlevered
#### What is its self-supporting growth rate
Maggie's Muffins, Inc., generated \$2,000,000 in sales during 2013, and its year-end total assets were \$1,100,000. Also, at year-end 2013, current liabilities were \$1,000,000,
#### Reporting under international financial reporting standards
From the e-Activity, contrast the impairment of goodwill on the financial statements of the entity reporting under international financial reporting standards (IFRS) that you | 551 | 2,254 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2020-34 | latest | en | 0.953689 |
https://www.unitconverters.net/data-storage/megabit-to-megabyte.htm | 1,719,354,220,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198866422.9/warc/CC-MAIN-20240625202802-20240625232802-00060.warc.gz | 891,743,111 | 3,238 | Home / Data Storage Conversion / Convert Megabit to Megabyte
# Convert Megabit to Megabyte
Please provide values below to convert megabit [Mb] to megabyte [MB], or vice versa.
From: megabit To: megabyte
### Megabit to Megabyte Conversion Table
Megabit [Mb]Megabyte [MB]
0.01 Mb0.00125 MB
0.1 Mb0.0125 MB
1 Mb0.125 MB
2 Mb0.25 MB
3 Mb0.375 MB
5 Mb0.625 MB
10 Mb1.25 MB
20 Mb2.5 MB
50 Mb6.25 MB
100 Mb12.5 MB
1000 Mb125 MB
### How to Convert Megabit to Megabyte
1 Mb = 0.125 MB
1 MB = 8 Mb
Example: convert 15 Mb to MB:
15 Mb = 15 × 0.125 MB = 1.875 MB | 205 | 559 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-26 | latest | en | 0.48499 |
https://www.physicsforums.com/threads/kinetic-force.826404/ | 1,519,184,851,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813322.19/warc/CC-MAIN-20180221024420-20180221044420-00652.warc.gz | 957,771,002 | 16,176 | # Kinetic force
1. Aug 5, 2015
### Dalip Saini
1. The problem statement, all variables and given/known data
A crate on a rough horizontal floor is being moved horizontally by a force applied at a 16 degree angle above the horizontal. This is the smallest force that will move the crate at constant speed. What is the coefficient of kinetic friction?
• A :
.11
• B :
.27
• C :
.96
• D :
It depends on the size of the force.
• E :
.29
2. Relevant equations
friction force = (coefficient of kinetic friction) (normal force)
3. The attempt at a solution
I am really confused on how to solve this problem without the magnitude of force applied or the weight of the crate.
2. Aug 5, 2015
### tommyxu3
You could decompose the force and analyze if the information is enough for you to solve the problem. Maybe you should have a graph and try to list their relation first and conclude?
3. Aug 6, 2015
### haruspex
I recommend trusting the question. Put in unknowns (m, F, ...) as necessary and expect them to disappear later in the algebra.
Often you can predict that they will. In the present case, if you double the mass then quite clearly you can correspondingly double the force. If the crate moves with one set of values it will move with the other set. The value of g can't matter either. | 322 | 1,306 | {"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} | 3.375 | 3 | CC-MAIN-2018-09 | latest | en | 0.90507 |
https://www.cnblogs.com/EdSheeran/p/9924561.html | 1,606,295,060,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181482.18/warc/CC-MAIN-20201125071137-20201125101137-00555.warc.gz | 613,335,032 | 7,424 | # 平时二十四测
维护联通性是边联通，不是点联通
#include<bits/stdc++.h>
using namespace std;
const int M = 1e6 + 5;
int deg[M], loop, fa[M], l[M];
int find(int x){return fa[x] == x ? x : fa[x] = find(fa[x]);}
int main(){
freopen("tour.in","r",stdin);
freopen("tour.out","w",stdout);
int n, m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) fa[i]=i;
for(int i=1;i<=m;i++){
int v;
scanf("%d%d",&l[i],&v);
fa[find(l[i])]=find(v);
if(l[i]==v)loop++;
else deg[l[i]]++,deg[v]++;
}
int lst=find(l[1]);
for(int i=2;i<=m;i++)
if(find(l[i]) != lst) return !puts("0");
long long ans=1LL*loop*(loop-1)/2 + 1LL*(m-loop)*loop;
for(int i=1;i<=n;i++){
ans+=1LL*deg[i]*(deg[i]-1)/2;
}
printf("%lld\n",ans);
}
View Code
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e7;
int a[105], tot, n;
ll yz[M], k;
bool vis[M];
void chai(int x){
for(int d = x; d > 0; ){
int fm = (x + d - 1) / d;
int nw = (x + fm - 1) / fm;
if(fm < 1e6 && !vis[fm]){
yz[++tot] = fm;
vis[fm]=1;
}
else yz[++tot] = fm;
d = nw - 1;
}
}
bool check(ll now){
ll ans = 0;
for(int i = 1; i <= n; i++)
ans = ans + 1LL*(a[i] + now - 1)/now * now - a[i];
return ans <= k;
}
int main(){
freopen("cut.in","r",stdin);
freopen("cut.out","w",stdout);
scanf("%d%lld", &n, &k);
for(int i = 1; i <= n; i++){
scanf("%d", &a[i]);
chai(a[i]);
k += a[i];
}
ll ans = -1;
yz[++tot] = 1e12;
sort(yz + 1, yz + 1 + tot);
tot = unique(yz + 1, yz + 1 + tot) - yz - 1;
for(int i = 1; i <= tot; i++){
ll cur = 0;
for(int j = 1; j <= n; j++)
cur = cur + 1LL*(a[j] + yz[i] - 1)/yz[i];
ll mx = k/cur;
if(mx>=yz[i]) ans = max(ans, mx);
}
printf("%lld\n", ans);
}
View Code
#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll dp[305][305], mod;
inline void up(ll &a, ll b){
a += b; if(a >= mod) a -= mod;
}
int main(){
freopen("tree.in","r",stdin);
freopen("tree.out","w",stdout);
int n;
cin>>n>>mod;
dp[1][0] = dp[1][1] = 1;
for(int i=2;i<=n;i++){
int lim=n-i+2;
for(int l=0;l<=lim;l++)
for(int r=0;r+l-1<=lim;r++){
ll num=dp[i-1][l]*dp[i-1][r]%mod;
up(dp[i][l+r], num);
up(dp[i][l+r+1], num);
up(dp[i][l+r], num*(l+r)*2%mod);
if(l+r){
up(dp[i][l+r-1], num*l*r*2%mod);
up(dp[i][l+r-1], num*(l*(l-1)+r*(r-1))%mod);
}
}
}
printf("%lld\n", dp[n][1]%mod);
}
View Code
posted @ 2018-11-07 18:25 Ed_Sheeran 阅读(80) 评论(0编辑 收藏 | 965 | 2,275 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2020-50 | latest | en | 0.205403 |
https://stackoverflow.com/questions/5243983/what-is-the-most-efficient-way-to-implement-a-convolution-filter-within-a-pixel | 1,627,357,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152168.38/warc/CC-MAIN-20210727010203-20210727040203-00547.warc.gz | 544,582,215 | 42,285 | # What is the most efficient way to implement a convolution filter within a pixel shader?
Implementing convolution in a pixel shader is somewhat costly as to the very high number of texture fetches.
A direct way of implementing a convolution filter is to make N x N lookups per fragment using two for cycles per fragment. A simple calculation says that a 1024x1024 image blurred with a 4x4 Gaussian kernel would need `1024 x 1024 x 4 x 4 = 16M` lookups.
1. Can one use some optimization that would need less lookups? I am not interested in kernel-specific optimizations like the ones for the Gaussian (or are they kernel specific?)
2. Can one at least make these lookups faster by somehow exploiting the locality of the pixels one would work with?
Thanks!
Gaussian kernels are separable, which means you can do a horizontal pass first, then a vertical pass (or the other way around). That turns O(N^2) into O(2N). That works for all separable filters, not just for blur (not all filters are separable, but many are, and some are "as good as").
Or,in the particular case of a blur filter (Gauss or not), which are all kind of "weighted sums", you can take advantage of texture interpolation, which may be faster for small kernel sizes (but definitively not for large kernel sizes).
EDIT: image for the "linear interpolation" method
EDIT (as requested by Jerry Coffin) to summarize the comments:
In the "texture filter" method, linear interpolation will produce a weighted sum of adjacent texels according to the inverse distance from the sample location to the texel center. This is done by the texturing hardware, for free. That way, 16 pixels can be summed in 4 fetches. Texture filtering can be exploited in addition to separating the kernel.
In the example image, on the top left, your sample (the circle) hits the center of a texel. What you get is the same as "nearest" filtering, you get that texel's value. On the top right, you are in the middle between two texels, what you get is the 50/50 average between them (pictured by the lighter shader of blue). On the bottom right, you sample in between 4 texels, but somewhat closer to the top left one. That gives you a weighted average of all 4, but with the weight biased towards the top left one (darkest shade of blue).
The following suggestions are courtesy of datenwolf (see below):
"Another methods I'd like suggest is operating in fourier space, where convolution turns into a simple product of fourier transformed signal and fourier transformed kernel. Although the fourier transform on the GPU itself is quite tedious to implement, at least using OpenGL shaders. But it's quite easy done in OpenCL. Actually I implement such things using OpenCL, now, a lot of image processing in my 3D engine happens in OpenCL.
OpenCL has been specifically designed for running on GPUs. A Fast Fourier Transform is actually the piece of example code on Wikipedia's OpenCL article: en.wikipedia.org/wiki/OpenCL and yes the performance gain is tremendous. A FFT executes with at most O(n log n), the reverse the same. The filter kernel fourier representation can be precomputed. The way is FFT -> multiply with kernel -> IFFT, which boils down to O(n + 2n log n) operations. Take note the the actual convolution is just O(n) there.
In the case of a separable, finite convolution like a gaussian blur the separation solution will outperform the fourier method. But in case of generalized, possible non-separable kernels the fourier methods is probably the fastest method available. OpenCL integrates nicely with OpenGL, e.g. you can use OpenGL buffers (textures and vertex) for both input and ouput of OpenCL programs."
• To elaborate the "texture filter" method, you can get all 16 pixels in a 4x4 block summed/averaged with only 4 samples placed in between the texels. For box blur, this is entirely trivial, just place them half-way. For a Gaussian, it's a bit more complicated, the taps need to be placed a bit closer to the center so these pixels have more weight (though I don't have the exact distance at hand now). All in all, that would be 4 instead of 8 fetches for separating the kernel, and 16 for the non-optimized version. – Damon Mar 9 '11 at 10:09
• Linear interpolation helps insofar as instead of sampling the texture so exactly exactly one texel is hit, you can place your sample points ("taps") so that they are somewhere in between. The interpolation hardware will return a weighted average (weighted by the inverse distance to the adjacent texel centers) at no extra cost. I will try to come up with a drawing, that's maybe easier. – Damon Mar 9 '11 at 10:40
• @Albus Dumbledore: The gaussian profile is a continous, differentiable distribution. For a discrete filtering process this profile must be turned into a finite vector of discrete elements. One may call it samples, but usually those refer to the samples of the filter. So on the kernel it's taps on the filter profile. – datenwolf Mar 9 '11 at 10:43
• Another methods I'd like suggest is operating in fourier space, where convolution turns into a simple product of fourier transformed signal and fourier transformed kernel. Although the fourier transform on the GPU itself is quite tedious to implement, at least using OpenGL shaders. But it's quite easy done in OpenCL. Actually I implement such things using OpenCL, now, a lot of image processing in my 3D engine happens in OpenCL. – datenwolf Mar 9 '11 at 10:46
• Added an image that hopefully helps understand. On the top left, your sample (the circle) hits the center of a texel. What you get is the same as "nearest" filtering, you get that texel's value. On the top right, you are in the middle between two texels, what you get is the 50/50 average between them (pictured by the lighter shader of blue). On the bottom right, you sample in between 4 texels, but somewhat closer to the top left one. That gives you a weighted average of all 4, but with the weight biased towards the top left one (darkest shade of blue). – Damon Mar 9 '11 at 10:50
More than being separable, Gaussian filters are also computable in O(1) :
There are recursive computations like the Deriche one :
http://hal.inria.fr/docs/00/07/47/78/PDF/RR-1893.pdf
• Recursive Gaussian sounds cool, though I am not sure what you mean by O(1) in this situation. Thanks! – Albus Dumbledore Mar 9 '11 at 11:03
• O(1) in this case means the number of samples you take does not depend on the size of the kernel. It does not necessarily mean that it will be faster or that there will be any fewer, it just says that the number of samples does not depend on the kernel size. Importance sampling does a similar job, but it is only an approximation. Getting O(1) is not hard if "approximately" is good enough. You could for example settle for a fixed "budget" of 10 samples. Then you use some scattered sample pattern and multiply each value with the correct gaussian weight according to its distance. – Damon Mar 9 '11 at 11:34
• Summed area tables are another example of doing a blur (box, in this case, not gaussian) in O(1). The downside is that you have to generate the SAT in the first place. Whether it is "win" or "lose" overall depends on your situation. – Damon Mar 9 '11 at 11:44
• Thanks dm.skt! Nicely explained. By the way, I forgot about SAT. :-D Good of you to remind me... – Albus Dumbledore Mar 9 '11 at 12:19
Rotoglup's answer to my question here my be worth reading; in particular, this blog post about Gaussian blur really helped me understand the concept of separable filters.
One more approach is approximating Gaussian curve with step-wise function: https://arxiv.org/pdf/1107.4958.pdf (I guess piece-wise linear functions can be also used of course). | 1,808 | 7,735 | {"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-2021-31 | latest | en | 0.925879 |
https://www.mrexcel.com/board/threads/formula-returning-n-a-error.590499/ | 1,695,752,150,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510219.5/warc/CC-MAIN-20230926175325-20230926205325-00116.warc.gz | 1,009,419,944 | 17,554 | # Formula returning #N/A error
#### jeepnfl
##### New Member
I have one spreadsheet that pulls data from another spreadsheet using this formula:
=SUMPRODUCT(--('[FY12 Main Log macro.xlsm]Sheet1'!\$A\$2:\$A\$3000>="01-Nov-11"+0),--('[FY12 Main Log macro.xlsm]Sheet1'!\$A\$2:\$A\$3000<"01-Dec-11"+0),--('[FY12 Main Log macro.xlsm]Sheet1'!\$C\$2:\$C\$3000=B30), '[FY12 Main Log macro.xlsm]Sheet1'!\$O\$2:\$O\$3000)
The result from this formula is #N/A and I wonder if it has something to do with the fact that the range "Sheet1'!\$C\$2:\$C\$3000" that this sheet references also uses a formula for it's result.
When I use this formula and the range "Sheet1'!\$C\$2:\$C\$3000" is just a value (not a formula) the result is the correct calculation.
I'm not even really sure how to ask this question. Please let me know if you need any further explanation.
Any words of wisdom?
### Excel Facts
How to calculate loan payments in Excel?
Use the PMT function: =PMT(5%/12,60,-25000) is for a \$25,000 loan, 5% annual interest, 60 month loan.
You'll get #N/A error if there are #N/A errors in the sum range - if so then either eliminate those errors at source by amending the formulas in that range or perhaps try using this "array formula"
=SUM(IF('[FY12 Main Log macro.xlsm]Sheet1'!\$A\$2:\$A\$3000>="01-Nov-11"+0,IF('[FY12 Main Log macro.xlsm]Sheet1'!\$A\$2:\$A\$3000<"01-Dec-11"+0,IF('[FY12 Main Log macro.xlsm]Sheet1'!\$C\$2:\$C\$3000=B30,IF(ISNUMBER('[FY12 Main Log macro.xlsm]Sheet1'!\$O\$2:\$O\$3000),'[FY12 Main Log macro.xlsm]Sheet1'!\$O\$2:\$O\$3000)))))
confirmed with CTRL+SHIFT+ENTER
Wow, that was too easy! I had some #N/A's in the sum range but didn't realize that was a problem, I just assumed those would result in a "false". I fixed those and my other spreadsheet sudenly populated with correct data!
Thank you Mr. Houdini, you helped me escape great frustration!
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Go back | 896 | 2,947 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-40 | latest | en | 0.865943 |
https://www.goshen.edu/physix/310/gco/09.3.php | 1,429,375,637,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1429246635639.11/warc/CC-MAIN-20150417045715-00255-ip-10-235-10-82.ec2.internal.warc.gz | 984,505,686 | 5,039 | # Chemistry and Phases
...in which we relentlessly max out the Gibbs-Duhem equation.
$$S\, dT-V\, dP+\sum_{j=1}^m n_j\, d\mu_j=0.$$
## Phase equilibrium
Imagine two (sub)systems--one of ice $A$ and one of water $B$ enclosed in an adiabatic, fixed size container. Particles and heat can move back and forth between the two phases, subject to these constraints: $$n_A+n_B = n = {\rm constant}\ \Rightarrow dn_A = -dn_B$$ $$V_A+V_B = V= {\rm constant}$$ $$U_A+U_B = U= {\rm constant}$$
At equilibrium, $S=S_A+S_B$ should be a maximum, therefore: $$dS = dS_A+dS_B=0.$$
We'll make the entropy a bit easier to deal with by assuming the more boring picture below, where all of the ice is together, and all of the water is together, and particles and heat move back and forth freely across the moveable phase boundary.
For you to answer, and then compare with someone else...
1.) Why, in a sentence or two, is the total internal energy constant?
2.) Solve this equation below for $dS$, $$dU = T\,dS-P\,dV+\mu\,dn.$$
3.) Use the result above to write out the equilibrium condition $dS=(dS_A+dS_B)=0$ in terms of quantities for each sub-system.
4.) Now, using differential relationships that are consequences of the constraints $n_A+n_B=$constant, $V_A+V_B=$constant, and $U_A+U_B=$constant etc, re-write the equilibrium condition into this form... (that is, get rid of the differentials with $B$ subscripts...) $$$\ \$dU_A+$\ \$dV_A+$\ \$dn_A = 0.$$
The expression above works out to $$$\frac{1}{T_A}-\frac{1}{T_B}$dU_A+$\frac{P_A}{T_A}-\frac{P_B}{T_B}$dV_A+$\frac{\mu_A}{T_A}-\frac{\mu_B}{T_B}$dn_A = 0.$$
5.) Can you show these three consequences?
$T_A = T_B\$ (thermal equilibrium)
$P_A=P_B\$ (mechanical equilibrium)
$\mu_A=\mu_B$ (diffusive equilibrium)
• This last condition, that the chemical potentials are equal at equilibrium, is worth remembering!
• Isn't $\mu$ a constant for a certain substance?
• NO: You may be thinking of $\Delta G_f^o$. But recall that in general $\mu=\mu(T,P)$
## The Gibbs phase rule
For a closed system with one component (e.g. an ideal gas of a pure substance), which has reached equilibrium, specifying 2 thermodynamic parameters are enough to uniquely specify the state.
The equilibrium state.
For an open single-component system, two parameters are no longer enough to completely specify the state of the system.
If we write $$dG = -S\,dT+V\,dP+\mu\,dn$$ There are three differentials that can vary independently, so we'd need to specify 3 quantities, $T$, $P$, and $n$ for example, to uniquely specify the state.
### Multiple components
Now, consider that we have a closed system, at constant temperature and constant pressure: $$[dG]_{T,P} = \sum_{j=1}^m \mu_j dn_j.$$
Let's say that the system consists of...
• $k$ different components: e.g. CO${}_2$, H${}_2$O, and O${}_2$, numbered as $i=1..k$, with $k=3$.
• $\pi$ different phases for each component: e.g. solid, gas, and liquid, numbered as $\gamma=1..\pi$, with $\pi=3$.
There are $m=k\pi$ different kinds of things in the system. To uniquely specify the state, we'd need the standard 2 thermodynamic parameters plus perhaps $k\pi$ *more* numbers, which could be the the number of moles in each phase and each component: $$2+\pi k.$$
But it turns out the number of quantities needed to uniquely specify the state is not quite so high:
### Phase constraints and degrees of freedom
In order to group the $n_j$ by component and phase, write: $n_i^{\gamma}$ = the number of moles in component $i$, and in phase $\gamma$, where $\gamma$ is not an exponent, but just a superscript.
For one particular phase, the mole fraction (percentage) of each component in that phase can be written as... $$x_i^{\gamma}=\frac{n_i^{\gamma}}{\sum_i n_i^{\gamma}}.$$
Such that $$\sum_{i=1}^{k} x_i^{\gamma} = 1.$$
For example, we could freely choose that 30% of the gas phase is CO${}_2$, and 50% of the gas phase is H${}_2$O. But now we have enough information to know that the O${}_2$ mole fraction must be 20%.
There is one of these phase constraint equations for each phase--$\pi$ of them in all. So, the total number of numbers we need to completely specify the state of our multi-component system is not $2+\pi k$, but rather just: $$2+\pi k -\pi = 2+\pi(k-1).$$
Diffusive equilibrium: For the different phases of one component $i$ to all be in equilibrium with each other, the criteria is that the chemical potentials must all be equal: $$\mu_i^{\alpha} = \mu_i^{\beta}=..=\mu_i^{\pi}.$$
This is a total of $\pi-1$ separate equations for each of the $k$ phases. There are a total of $$k(\pi-1)$$ constraint equations related to the condition for diffusive equilibrium.
So, the total number of degrees of freedom $f$, (also called the "variance") is:
$$f=2+\pi(k-1) -k(\pi-1) =2 + k -\pi,$$
where $k$ is the number of components, and $\pi$ is the number of phases. This is the Gibbs phase rule.
#### For example:
An ideal gas of a single type of molecule?
$k=$___?, $\pi=$___?, $\Rightarrow f=$___? What *are* those two free parameters you can specify?
$k=1$; $\pi=1$ $\Rightarrow f=2$.
A single component with a single phase has 2 degrees of freedom. $n$ is a constant for a closed system. So, we could arbitrarily pick $P$ and $T$.
Three phases of water (liquid, gas, vapor) all present in equilibrium with each other.
$k=$___?, $\pi=$___?, $\Rightarrow f=$___? What does that mean?
$k=1$; $\pi=3$ $\Rightarrow f=0$.
This is the case for the triple point of water. If three phases are to coexist, this must happen at fixed $P$ and $T$.
Liquid water and water vapor co-exist in equilibrium.
$k=$___?, $\pi=$___?, $\Rightarrow f=$___? What does that mean?
$k=1$; $\pi=2$ $\Rightarrow f=1$.
This is the case for liquid-vapor coexistence. We can chose $P$, but then $T$ is determined (or vice versa).
A mixture of three chemically distinct substances, carbon-dioxide, oxygen, and nitrogen, but all in the same (gas) phase.
$k=$___?, $\pi=$___?, $\Rightarrow f=$___? What could those parameters be?
$k=3$; $\pi=1$\Rightarrow f=4$. The 4 quantitities we specify could be$P$,$T$, and the partial pressures (concentrations) of two of the gases. ### Chemical reactions Consider $$2{\rm H}_2+{\rm O}_2 \rightarrow 2{\rm H}_2{\rm O}.$$ The stoichiometric coefficients for such a reaction are integers which represent the molecular ratios, with the convention that they are negative for initial reactants, and positive for products. E.g. $$\nu_{{\rm H}_2}=-2 ; \nu_{{\rm O}_2}=-1 ; \nu_{{\rm H}_2{\rm O}}=+2.$$ The ratios of the changes in the number of kilomoles of each species is, neatly: $$dn_{{\rm H}_2}:dn_{{\rm O}_2}:dn_{{\rm H}_2{\rm O}} = -2:-1:+2 = \nu_{{\rm H}_2}:\nu_{{\rm O}_2}:\nu_{{\rm H}_2{\rm O}}.$$ At equilibrium,$dG=0$. In particular, for constant$T$, and$P$, we have $$[dG]_{T,P} =0= \sum_{j=1}^m\mu_j\, dn_j.$$ Since the$dn_j$are proportional to the$\nu_j\$, we have... $$\sum_{j=1}^m\mu_j\, \nu_j = 0.$$
Work out what relationship this implies for the chemical potentials of the three species discussed above.
This is a useful internal consistency check for chemical potentials. | 2,080 | 7,115 | {"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": 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} | 3.4375 | 3 | CC-MAIN-2015-18 | latest | en | 0.782501 |
https://www.scirra.com/forum/solved-what-is-the-8-direction-moving-angle-equation_t149602 | 1,527,192,073,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866772.91/warc/CC-MAIN-20180524190036-20180524210036-00246.warc.gz | 828,038,011 | 10,664 | # [SOLVED] What is the 8 Direction Moving Angle equation?
Get help using Construct 2
### » Thu Jul 09, 2015 12:17 am
Hi C2 community
I'm trying to calculate the angle of an object moved without 8 Direction behavior in relation to another object, but I'm not succeeding.
I tried the formula floor(angle(Self.X,Self.Y,object.X,object.Y)), but didn't work for the end result I want.
Last edited by lukezero on Thu Jul 09, 2015 1:34 pm, edited 2 times in total.
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### » Thu Jul 09, 2015 12:22 am
Just for the record, I'm wondering that to implement the walking animation of a character not controlled by the player in a isometric 8 directions game.
I'm currently using int(((IA.calc_angle+360) %360)/45), where the instance variable calc_angle is floor(angle(Self.X,Self.Y,object.X,object.Y)).
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### » Thu Jul 09, 2015 3:39 am
@R0J0hound, sorry to bother you again, but I tried your formule above, and unfortunately it didn't work for a character just moved by the Move command (with some instance variables).
"walk" & int(((Object.8Direction.MovingAngle+360+22.5) %360)/45)
As said object doesn't use the 8 Direction behavior, you know the best way to apply their walking animations with the angle changes?
The calculations of my instance variables are correct for the character walk animations that move only in diagonal isometric perspective?
Last edited by lukezero on Thu Jul 09, 2015 4:22 am, edited 2 times in total.
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### » Thu Jul 09, 2015 3:46 am
I don't quite understand the first bit. Do you want the direction the enemy is moving or the direction from the enemy to the player?
if you're using a behavior they often have a angleofmotion expression or x and y velocity expressions from which you can calculate the angle of motion with angle(0,0,veloctyx,velocityY).
As for calculating a value of 0 to 7 for animation.
This converts a from the range (-180,180) to (0,360):
(a+360)%360
This rounds the angle to the nearest 45degree angle 0 to 7. The 22.5 is half of 45 and is needed to round correctly.
Int((a+22.5)/45)
Edit:
If you are using the move at angle action to move the object then you should keep track of the direction you moved. Another way is to save the objects position to variables oldx and oldy before you move and calculate the angle of motion with angle(oldx,oldy,sprite.x,sprite.y) after you move.
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### » Thu Jul 09, 2015 3:54 am
To be precise, @R0J0hound, the uncontrolled character follows a trail of fixed pre-established points by command Move with angle and distance calculations basically based on the distance between both. So I used the equation floor(angle(Self.X,Self.Y,object.X,object.Y)).
I'm using the 0-7 calculation, but after some tests, I came to the conclusion that the problem is the lack of connection between the movement and the angle of the object that don't use any behavior.
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### » Thu Jul 09, 2015 12:42 pm
lukezero wrote:To be precise, @R0J0hound, the uncontrolled character follows a trail of fixed pre-established points by command Move with angle and distance calculations basically based on the distance between both. So I used the equation floor(angle(Self.X,Self.Y,object.X,object.Y)).
I'm using the 0-7 calculation, but after some tests, I came to the conclusion that the problem is the lack of connection between the movement and the angle of the object that don't use any behavior.
Are you specifying which object you mean in the expression "floor(angle(Self.X,Self.Y,object.X,object.Y))"
I'm guessing there are more than one..
I told my dentist I had trouble with my teeth and asked her to fix it without looking in my mouth..
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### » Thu Jul 09, 2015 1:10 pm
Hi @LittleStain! ^^
The object is a trail Dot that the avatar follows every time.
I managed to solve this problem this morning after the @R0J0hound tip. It was enough after put the calc_angle in the detection event of the next point to be found.
Thanks for the attention and cooperation of always!
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https://www.hackmath.net/en/example/2199?tag_id=99 | 1,558,966,933,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232262600.90/warc/CC-MAIN-20190527125825-20190527151825-00280.warc.gz | 794,645,084 | 6,849 | # Rectangle - parallelogram
It is given a rectangle that is circumscribed by a circle with a radius of 5 cm. The short side of the rectangle measures 6 cm. Calculate the perimeter of a parallelogram ABCD whose vertices are the midpoints of the sides of the rectangle.
Result
x = 20 cm
#### Solution:
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#### To solve this example are needed these knowledge from mathematics:
Pythagorean theorem is the base for the right triangle calculator.
## Next similar examples:
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How many different divisors has number ?? | 821 | 3,233 | {"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} | 3.546875 | 4 | CC-MAIN-2019-22 | latest | en | 0.942132 |
http://mathhelpforum.com/math-challenge-problems/9389-riddle.html | 1,529,434,477,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863109.60/warc/CC-MAIN-20180619173519-20180619193519-00376.warc.gz | 197,280,805 | 10,313 | 1. ## riddle
We need to know how to do this without having to open and close this many doors. There must be a way.
We don't need the answer, but a method of finding it would be wonderful.
On graduation day, 200 seniors line up outside the school. As they enter the school, they pass the school lockers, aptly numbered 1 to 200. The first student opens all of the lockers. The second student closes every other locker beginning with the second locker. The third student changes the status of every third locker beginning with the third one (if opened, the student closes it; if closed, the student opens it). The fourth student changes the status of every fourth locker. The fifth student changes the status of every fifth locker, and so on. Which lockers remain open after all 200 students have entered the school?
Use any method you want to solve the problem, but I MUST see all work and be given a short explanation of how you solved the problem.
4 extra credit points for the correct answer(s) AND a complete explanation of how the problem was solved.
Do you recognize a pattern? What is it? 2 extra credit points
4 extra credit points for being able to explain WHY the pattern exists.
2. Originally Posted by namanta
We need to know how to do this without having to open and close this many doors. There must be a way.
We don't need the answer, but a method of finding it would be wonderful.
On graduation day, 200 seniors line up outside the school. As they enter the school, they pass the school lockers, aptly numbered 1 to 200. The first student opens all of the lockers. The second student closes every other locker beginning with the second locker. The third student changes the status of every third locker beginning with the third one (if opened, the student closes it; if closed, the student opens it). The fourth student changes the status of every fourth locker. The fifth student changes the status of every fifth locker, and so on. Which lockers remain open after all 200 students have entered the school?
Use any method you want to solve the problem, but I MUST see all work and be given a short explanation of how you solved the problem.
4 extra credit points for the correct answer(s) AND a complete explanation of how the problem was solved.
Do you recognize a pattern? What is it? 2 extra credit points
4 extra credit points for being able to explain WHY the pattern exists.
I solved this problem hier.
The general formula I give,
$\displaystyle [ \sqrt{n} ]$.
Where $\displaystyle [ \, \, \, \, ]$.
Denote the greatest integer function.
Meaning the integer part of the number.
Thus, $\displaystyle [ \sqrt{200} ] = 14$
3. Here is a link to a solution of this very famous problem.
Locker problem
4. ## Thanks
I appreciate the help. Can do it now. | 623 | 2,786 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-26 | latest | en | 0.95282 |
https://www.wyzant.com/resources/blogs/383896/neat_mental_math_tricks_some_finger_math_and_the_reasoning_behind_them | 1,606,713,154,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141205147.57/warc/CC-MAIN-20201130035203-20201130065203-00337.warc.gz | 896,654,424 | 13,206 | # Neat Mental Math Tricks + Some Finger Math and the reasoning behind them
Multiplying by 4:
(1) Double the multiplicand you want to multiply 4 by
(2) Double it one more time
e.g. 8 * 4 = 32
8 * 2 = 16
16 * 2 = 32
Why does this work?
4 can be broken up into 2 * 2
8 * 4 = 32
8 * (2 * 2) = 32
Thanks to the associative property of multiplication, we can multiply factors in whatever grouping or order we chose and still get the same answer. We start by multiplying multiplicand we want to multiply 4 by 2 because this computation is easy for most people to do in their heads.
(8 * 2) * 2
(16) * 2
We then multiply our product by the remaining multiplicand, which is 2.
16 * 2 = 32
Multiplying by 10:
Stick a zero behind whatever number you wish to multiply 10 by
988 * 10
= 9880
Why does this work?
Consider what we’re doing in terms of place value. When multiplying a number by 10, we’re essentially moving everything up a place value.
10 ones = 10
10 tens = 100
10 hundreds = 1,000
10 thousands = 10,000
etc.
Going back to our original example of 988 x 10 we see that:
8 ones x 10 becomes 8 tens or 80
8 tens x 10 becomes 8 hundreds or 800
9 hundreds x 10 becomes 9 thousands or 9,000
Let’s take a look at the process again, this time using long multiplication:
988
x 10
80 We’re multiplying each digit first by zero ones, then by one ten.
800 That combination keeps getting pushed over as you multiply by larger and larger
+9000 places so the original digits are simply moved over one place.
9880
You probably knew those already. Let’s move on to some lesser known ones:
Multiplying by 5:
(1) Multiply the factor you want to multiply 5 by by 10
(2) Divide that product by 2
66 * 5
66 * 10 = 660
660/2 = 330
Why does this work?
That 10 trick was pretty easy, right? Well, the five trick is basically a modified version of that.
We multiply by 10 because it’s pretty easy to do in our heads. Then we divide by 2 because 5 is half of 10 so our final product must be half of our product from step one.
Multiplying by 9 (only works for 9x1 - 9x10):
(1) Spread your fingers out in front of you (doesn’t matter whether palms are facing up or down)
(2) Going from left to right, fold the finger down that you wish to multiply 9 by (if you wish to multiply 9 * 3 then fold down your middle finger on your left hand)
(3) Everything to the left of your folded down finger represents the tens digit and everything to the right represents the ones digit
Why does this work?
Notice that the sum of all the digits of numbers divisible by 9 is equal to or divisible by 9
3 * 9 = 27
2 + 7 = 9
8 * 9 = 72
7 + 2 = 9
This is true of all numbers divisible by 9.
5418/9 = 602
5 + 4 = 9
1 + 8 = 9
999963/9 = 111107
9 = 9
9 = 9
9 = 9
9 = 9
6 + 3 = 9
But how does this work? We need to rethink how we think about numbers:
(1) Take the number that you’re interested in seeing if it’s divisible by 9 and split it up into its places.
5418 becomes
5(1000) + 4(100) + 1(10) + 8(1)
(2) If n represents a place, rewrite what’s in the parentheses to ((n - 1) + 1)
5(999 + 1) + 4(99 + 1) + 1(9 + 1) + 8
How are we allowed to do this? Well, you’ll see that if you add everything up and then multiply, you’ll still get 5418. We’re simply expressing the number differently.
(3) Distribute
5 + (5 * 999) + 4 + (4 * 99) + 1 + (1 * 9) + 8
(4) We know that everything within the parenthesis is divisible by 9 because we are multiplying what’s inside by a multiple of nine. So that just leaves that string on the end, which added together also has to be divisible by 9. Notice anything interesting about them? Yep, they are our original digits!
(5 * 999) + (4 * 99) + (1 * 9) + 8 + 5 + 4 + 1
8 + 5 + 4 + 1 = 18
1 + 8 = 9
Multiplying by 11:
(1) Erase all numbers in between the ones digit and the largest place value of the multiplicand you wish to multiply 11 by (e.g. 54321 becomes 5 _ _ _ 1). Skip to step two if you’re multiplying a two digit number.
(2) Add an additional space in between the ones and the largest place
(5 _ _ _ _1)
(3) Add together the ones digit and the tens digit from the multiplicand that you wish to multiply 11 by and insert the sum into the right-most blank place.
54321 2 + 1 = 3 5 _ _ _ 31
(4) Working from right to left, move one place over so now you’re working with the tens and hundreds digits. Add those two together and insert the sum into the right-most empty spot.
54321 3 + 2 = 5 5 _ _ 531
(5) Keep moving through the places until you run out of numbers.
54321 4 + 3 = 7 5 _ 7531
54321 5 + 4 = 9 597531
And there you have it!
What happens when the sum in step two is greater than 9 - i.e. how to carry the 1?
Simple! You carry the one over to the next place.
157 * 11
1_7
1_ _ 7
1 _ 27 because 7 + 5 = 12
1727 you’d think it’d be 6 because 1 + 5 = 6 HOWEVER, even though we usually don’t mess with the largest place, we still have that one leftover from the previous step. Thus, the hundreds place becomes 7 and the 1 gets bumped to the thousands place in this case.
How does this work?
This “trick” is a lot more straightforward than you’d think. When you do the long multiplication, you see that you’re simply adding up adjacent digits
54321
x 11
11
220
3300
44000
+550000
597531
\$40p/h
Sierra S.
Math and Comp Sci Tutor w experience w the Singapore method | 1,621 | 5,464 | {"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} | 4.625 | 5 | CC-MAIN-2020-50 | latest | en | 0.923461 |
https://jerrylwalls.com/brain-teaser-test-your-iq-and-find-the-missing-number-in-30-seconds-max/ | 1,696,246,666,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510994.61/warc/CC-MAIN-20231002100910-20231002130910-00873.warc.gz | 361,844,211 | 64,648 | Home » Brain teaser: Test your IQ and find the missing number in 30 seconds max!
# Brain teaser: Test your IQ and find the missing number in 30 seconds max!
Feeling brave? Put your math skills to the test with this brain-teasing challenge! See if you can find the hidden number in 30 seconds or less!
Are you ready to test your brainpower? Today’s challenge is to find the missing number in this logical sequence. The statement of the mathematical problem is summarized in the picture below.
Logic and thinking tests like this one can be a fun way to practice your problem-solving skills and really get your brain working. So take a few moments to think it through and see if you can figure out the missing number.
## Understanding the medium level challenge of identifying the correct number
You may come across a challenge that requires you to identify the number that should be in the first circle instead of the question mark.
This challenge is of a medium level and it requires you to find the calculation logic of the circles to perform the operation that will guide you to the correct answer. It is important to understand how each of the circles are related and how they interact with each other in order to solve the challenge.
Read also: Visual brain teaser: Find the odd one among clouds in less than 15 seconds!
It is necessary to train our logical minds in order to find the missing number in this logical sequence. Such training helps to strengthen the brain’s ability to think logically, identify patterns, and draw conclusions.
Training your brain to think logically helps you become better at solving problems, making decisions and understanding complex ideas. It can also help you better understand relationships between abstract concepts, which can be extremely useful in many areas of life.
Overall, training your brain strengthens your mind and helps you think more clearly and effectively.
Have you found the solution? You did it! Let’s check on the next page if your solution worked!
## Logical puzzle: Can you crack the code in 20 seconds or less and prove your intelligence?
Écrit par : Marley Poole
I fell into the Web pot at a young age and I now have a thirst for knowledge. Very curious, I don't hesitate to document myself on all subjects. I hope my articles are interesting and useful and that you will have a good time reading them. | 475 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2023-40 | longest | en | 0.939862 |
https://webwork.maa.org/mediawiki_new/index.php?title=RecursivelyDefinedFunctions&diff=prev&oldid=6869 | 1,643,216,554,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00007.warc.gz | 648,565,105 | 7,449 | # Difference between revisions of "RecursivelyDefinedFunctions"
## Recursively Defined Functions (Sequences)
This PG code shows how to check student answers that are recursively defined functions.
PG problem file Explanation
```DOCUMENT();
"PGstandard.pl",
"MathObjects.pl",
"parserFunction.pl",
);
TEXT(beginproblem());
```
Initialization: We will be defining a new named function and adding it to the context, and the easiest way to do this is using `parserFunction.pl`. There is a more basic way to add functions to the context, which is explained in example 2 at AddingFunctions
```Context("Numeric")->variables->are(n=>"Real");
parserFunction(f => "sin(pi^n)+e");
\$fn = Formula("3 f(n-1) + 2");
```
Setup: We define a new named function `f` as something the student is unlikely to guess. The named function `f` is, in some sense, just a placeholder since the student will enter expressions involving `f(n-1)`, WeBWorK will interpret it internally as `sin(pi^(n-1))+e`, and the only thing the student sees is `f(n-1)`. If the recursion has an closed-form solution (e.g., the Fibonacci numbers are given by f(n) = (a^n - (1-a)^n)/sqrt(5) where a = (1+sqrt(5))/2) and you want to allows students to enter the closed-form solution, it would be good to define f using that explicit solution in case the student tries to answer the question by writing out the explicit solution (a^n - (1-a)^n)/sqrt(5) instead of using the shorthand f(n).
```Context()->texStrings;
BEGIN_TEXT
The current value \( f(n) \) is three
times the previous value, plus two. Find
a recursive definition for \( f(n) \).
Enter \( f_{n-1} \) as \( f(n-1) \).
\$BR
\( f(n) \) = \{ ans_rule(20) \}
END_TEXT
Context()->normalStrings;
```
Main Text: The problem text section of the file is as we'd expect. We should tell students to use function notation rather than subscript notation so that they aren't confused about syntax.
```\$showPartialCorrectAnswers=1;
ANS( \$fn->cmp() );
ENDDOCUMENT();
``` | 503 | 1,982 | {"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-2022-05 | latest | en | 0.816577 |
https://sis4teachers.org/2021/05/math-mights-teachers-guide-episodes-401-402/?utm_source=rss&utm_medium=rss&utm_campaign=math-mights-teachers-guide-episodes-401-402 | 1,624,268,473,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488269939.53/warc/CC-MAIN-20210621085922-20210621115922-00629.warc.gz | 488,879,379 | 26,489 | Select Page
# Math Mights Teacher’s Guide: Episodes 401-402
May 13, 2021
Thanks so much for joining us this week for our teacher’s guide to Math Might shows 401 and 402.
You might wonder where 319 and 320 went…you didn’t miss them! The Math Mights Show has eighteen shows per quarter, per grade level. This week is really just a continuation from last week, but the numbering system just signifies we’ve moved into the 4th quarter.
# Teacher's Guide
Episodes 401-402
400 Series Focus: Numeracy/Number Talks
We revisit number talks as our warm-up in the 400 series, but there are plenty of twists! Kindergarten is practicing conservation to 20 in multiple modalities, 1st grade is experience their first actual number talks after doing numeracy talks so far this year, 2nd grade will be meeting a NEW Math Might friend with a brand new, “magical” strategy, and 3rd graders are putting a fraction twist on the traditional number talk!
### Kindergarten
“I Can” statement: I can compose numbers 11 through 19 using ten, ones, and some more ones. / I can show numbers with 10-frames and dots or counters.
Extension Activity: Deck o’ Dots Teen Match-Up / Teen Bingo
For Kindergarten, in show 401, we’re going to bring back numeracy talks! Previously, in the late 200s shows, we were doing numeracy talks with conservation to 10, but this time, Dotson is going to help us with conservation to 20. As we did before, we’re going to have the red carpet and flash the double-10 frame for students. They’ll “take a picture” of what they saw and tell us how many they see. Our friends Nora and Layla are going to tell us two different ways that they knew that the 10-frames in this example were 13. The big idea, especially this time of year, is to help kids to see numbers in different ways or in multiple modalities.
The “I Can” Statement is: I can compose numbers 11 through 19 using ten, ones, and some more ones.
We started off playing a really fun game called Deck o’ Dot Teen Match-Up. Students are going to flip over two Deck o’ Dots cards. One of them will have the quantity of 10 and the other one will be any of the numbers one through nine. As we flip over a card, we see that we have one full 10-frame and one with just four, so we can tell the total is going to be 14. Of course, we can’t do 10s and 1s without our Math Might friend, Value Pak! We want kids to really see the relationship between the number and its value. When they see that total 10-frame, it’s worth 10. The four is worth 4. Instead of just writing a “1” and a “4”, Value Pak wants to make sure students know that the amounts on their bellies actually show 10 and 4, which make “14” when you put it together. We use a really great recording sheet here where students are able to color in the 10-frame, and then complete the sentence 10 + _____ = 14.
As we continue, we have a Two out of Three game, also played with the Deck o’ Dots. Dotson wants students to select which two cards make the target number, for example 13. We might have six in a 10-frame, 10 in a 10-frame, and three in a 10-frame. The sentence stem says “___ is a group of 10 and ____ ones.” In this example, we know that 13 is a group of 10 and three ones, so students would select which cards support that sentence. We do several different examples that reinforce the idea of teen numbers.
The fun game that students get to play as their independent activity is the game that we played in the show called Deck o’ Dot Teen Match-Up.
As we move on to episode 402, we’re again doing a numeracy talk. This time, instead of doing it with the double 10-frame, we want kids to be able to switch modalities and see the quantity in a linear way. Are students in your Kindergarten classroom just memorizing the 10-frame? Well, make sure you get a Counting Buddy Senior for your Kindergarteners at this time of the year so they can see quantities up to 20 in a linear way! The Counting Buddy Senior has 10 of one color beads and 10 of another color. We flash a quantity, and Nora and Layla once again give their feedback for how they figured out the number for the total that they saw.
The “I Can” statement is: I can show numbers with 10-frames and dots or counters. We give students counters that fill up a complete 10 frame with 10 in it, and we want them to create a special teen number. The first number we want them to build is 11, so of course we call again on Value Pak to help us! When students look at the digits 1 and 1, what is the actual value? As we know, it’s one 10 and one 1. We build a variety of different numbers this way where students are figuring out how many counters are needed to make a total.
We then play a really fun game called Teen Bingo. The bingo board is filled with quantities shown in double 10-frames and on a rekenrek, which again is another modality in which students should be able to see 20. Then, just like a bingo game, we pull a card, and students have to find the number on their double 10-frame or rekenrek to try to get three in a row. Students have to be really careful here, and teachers might need to offer a scaffold to students by helping them actually see the number built on a 10-frame built and build it themselves on the rekenrek to help them transfer their understanding of teen numbers.
“I Can” statement: I can sort, describe, and create solid shapes. / I can sort flat shapes and create a data display to represent our sort.
Extension Activity: Sorting Solid Shapes / Sorting Flat Shapes
In first grade, episode 401, we’re going to be looking at shapes in this series. We start off the show, just like we do in Kindergarten, however this show marks the first time in the Math Might Show that we’re doing an actual number talk with first graders. Typically we do numeracy talks the first half of the year, maybe even throughout January, then switch to number talks as you’re working on conservation to 20, 40, maybe then to 100 and beyond. For number talks in first grade, we want to remember to pose a problem with operations students are familiar with. This time of year, it might be compensation, which is also known as “doubles plus one” or “doubles minus one,” which is the strategy that Abracus helps us with.
In a number talk at this stage, students might also be familiar with being able to make a 10 with D.C., or they might be able to add 10s and 10s, and ones and ones. Eventually students might even be able to do a subtraction problem in a first grade number talk, where they’re doing something like 12 – 7 using Springling.
In this number talk, we’re going to feature Abracus for the first time on the show! He’s asking students to solve 7 + 6. As we’re solving this, it’s great to use a visual and try not to use the terms “doubles plus one,” “doubles minus one,” “doubles plus two,” “doubles minus two” to name the strategy, because to most first graders, that sounds like four different strategies. However, they’re all just using compensation.
As we solve this problem, 7 + 6, it’s nice to build these two addends in a double 10-frame, with seven in red on the top, and six in yellow on the bottom. This helps kids see a quantity they already know, like maybe 6 + 6, and then they can add one more to make 13. Other students might say, I see seven and I can zap that six with Abracus’ wand to see it as 7 + 7, and then minus one
Compensation is a really great strategy for first graders to know. Obviously, it’s helpful if students understand their doubles facts in order to apply this strategy, but by creating problems with concrete tools to help them visualize what’s happening, students can be successful.
Our “I Can” statement is: I can sort, describe, and create solid shapes. We offer students four different pictures and ask them which one doesn’t belong. Some of the shapes are flat shapes and some of the shapes are 3D shapes. We then start talking about how to sort solid shapes. You might sort the shapes by ones that are flat versus round, ones that roll or don’t roll. Maybe straight sides or not straight sides? Does it have squares or not have squares? Tall or short? We get the kids to sort the shapes in different ways and they can even guess how someone else sorted the shapes by looking at their attributes.
Then, we look at a bridge that’s built out of blocks, and we want to see if students can see what shapes that particular bridge is made up of. It’s made up of cubes and triangles and rectangle blocks. The idea is for students to look at a geo block and create a new geo block shape with it. It’s pretty fun to do an activity that provides exploratory ways for students to visualize and picture what they’re doing.
In the extension activity, they’re going to be sorting solid shapes by the attributes. Providing kids with the language to describe how shapes are created is really helpful for being successful in this standard.
In show 402 for first grade, we continue with a number talk. Just as we did in the previous session, we are still focusing on Abracus. We’re hoping that, in the second show, students become more independent with being able to answer a problem like 7 + 8. We build the problem again on the double 10-frame for students to observe and solve.
The “I Can” Statement is: I can sort flat shapes and create a data display to represent our sort. Again, we offer four images and ask students which one doesn’t belong. This time, the majority of the shapes are 3D and only one of them is flat. But as we know by now, students can figure a reason based on one attribute as to why each shape may or may not belong.
We then do a sort with flat shapes. We take a bunch of different flat shapes and see if we can sort them into triangles, squares or rectangles. You could also sort by different categories, like color.
Then, we try to take the idea of shapes and apply it to data collection. We take three handfuls of pattern blocks and see if we can determine the data. We find that we have nine triangles, four trapezoids, and seven squares. How can we use this data that we’ve collected on shapes to answer questions such as How many triangles and trapezoids are there in all? We might even ask How many more triangles are there than squares
For the independent activity, students are going to be doing something similar to what we did in the show, which is sorting flat shapes.
“I Can” statement: I can compare numbers and add or subtract. / I can add and subtract 10s and 100s.
Extension Activity: Solving with Springling / Add and Subtract 10s and 100s
For second grade show 401, students are going to be doing a number talk, like we’ve done in the past. This time, we also use Abracus with second graders. Now we’ve talked about Abracus as the “doubles plus one” or “doubles minus one” strategy. He likes to zap a number to change it temporarily, holds that change in his wand, and then zaps it back when he’s done solving. The example that we have here is 25 + 26. Some students might think of this problem like quarters – 25 plus 25 is 50. So, we’re temporarily changing, or compensating, the number 26 to make it 25 by taking away one. You know 25 + 25 = 50 quite quickly, but, don’t forget, you have to zap it back!
The “I Can” Statement is: I can compare numbers and add or subtract. I think the theme of this show makes a lot of sense at this time of the year for second graders. A lot of second graders have lots of different strategies in their math tool belts to figure things out by now, and they often just stick to the one that is their favorite if they’re only required to solve problems one way. Instead, we want students to start to look at problem solving analytically.
When we say “compare numbers” we don’t necessarily mean decide if they’re greater than or less than, but we want to have students look critically at two numbers and see what strategy will be most appropriate. If we are subtracting 81 – 79, should we use T-Pops? Or would Springling be more appropriate? In this show, Tyler and Elena work on solving problems two different ways. Tyler uses T-Pops, and Elena solves with Springling.
In Mathville, there are two different vehicles you might see going around – a pokey little car with a windsock hanging from his antenna on the back of the car with a hat that is usually just putzing along, and a jet plane that you have to watch out for because it zooms around really quickly. Both kinds of transportation will get you there, but the jet plane is clearly more efficient. We don’t want kids to feel like they need to rush through math, but what we’re really talking about is being able to determine which strategy is most efficient, based on the problem we’re looking at.
When we look closely at this problem, 81 – 79, we notice that using counting up with Springling makes a whole lot more sense because the two numbers are really close together. Springling is the jet plane strategy. Using T-Pops for that problem would get you the right answer, but it will take much longer to get there. We give a few other examples, such as 680 minus 673, and students have to decide if it is more efficient to use Springling or T-Pops.
For the extension activity, we’re going to drive home the idea of Springling for students with numbers that are very close to each other.
As we move into 402, we’re doing another talk with our friend Abracus. This time, it’s 58 + 22. It’s kind of interesting when you see a problem like this, because you could add 2 and subtract two from the respective addends, and it would kind of equal each other out. If I added two to 58, it’s going to be 60. If I took away two from 22, it’s going to be 20. Then, I’m left with a pretty easy problem to answer 60 + 20.
The “I Can” Statement is: I can add and subtract 10s and 100s. Of course, we have to have our friend Value Pak here! Most of you have seen Value Pak with their red and white, but this time you’re going to see that Value Pak has a new member – orange (hundreds)! We’re going to start with a number, 297. We roll a number cube, and add that many hundreds to complete the equation – 297 + (whatever you roll as hundreds) = ____. As we use place value strips, we want kids to understand that they’re adding in the 10s, or they’re adding in the 100s and how that can help. We do the same thing with the idea with subtraction. Starting with a number like 982, and this time we want the dice this time to represent 10s, so students have to roll and complete the equation.
The idea that there are different ways to look at numbers and figure out how many there are all together is one we come back to often. In this show, for example, Mia has two 100s, two 10s, and three ones, and someone else has two 100s. How could we figure out their value all together? We want to really make sure kids understand place value!
This transfers into students being able to write an equation by looking at place value blocks. They work on this objective by using a combination of place value blocks and even place values strips.
For their independent activity, students are going to add and subtract 10s and 100s. It’s a great way for students to spin and quickly figure out how they can add those together without feeling like they have to write out a whole algorithm.
“I Can” statement: I can measure length in halves or quarters of an inch. / I can measure length using a ruler marked with halves and quarters of an inch.
Extension Activity: Measure to the Nearest Half or Quarter Inch / Measure to find Equivalent Lengths
To begin episode 401, we’re going to be doing a number talk with a topic that the third graders just learned about – fractions! We’re going to be doing an area model fraction number talk so we can see exactly what they remember. Students are presented with a piece of paper divided into fourths, two are yellow and the other two are blue. Of course, we want to keep this really open ended so we just ask the students What fraction of space is occupied by each color?
I really enjoy doing these fraction talks because they are so open-ended. Students oftentimes will give me a right answer, like, 2/4 are yellow, or they might say 4/8 are blue because they can see those, but sometimes they don’t understand that both of those equal half. They don’t see that yellow is half or 2/4, but it could also be 4/8. This kind of activity really creates wonderful conversation and inquiry-based learning in the classroom.
The “I Can” Statement is: I can measure length in halves or quarters of an inch. We start by brainstorming what students already know about inches. Some students might remember that inches are used to measure length. Some might remember that there are different tick marks on yardsticks and rulers, or even tape measures. Others remember that inches are shorter than feet, but they’re also longer than centimeters.
We want to incorporate the idea of fractions with measurement, so we start with a paperclip on a ruler. All of our rulers are enlarged on the show so that students can see how we’re measuring from endpoint to endpoint. Students can look at the object, see if it’s halfway between three and four and see how that would measure three and a half. We look at a pencil and different objects in this way.
We also bring in the idea of what happens if I measure something and it is past an inch, but not quite to one and a half inches? Well, we know that would be a quarter of an inch. We talk about how to label that and how it would look on a ruler. We especially look at these marks of 1/4, 2/4, 3/4, and 4/4, and show that 2/4 actually equals a half.
The extension activity is to measure to the nearest 1/2 or 1/4 inch. It gets kids to really look at exactly where objects are lying on a ruler, and helps them understand the parts of fractions that we’ve covered in previous shows as it applies to something in the real world.
In episode 402, we’re going to be doing another fraction number talk. If you’re interested in learning more about this type of number talk, click the link in the episode guide for Love for Math. I love how they set up fraction number talks! In this number talk, the fractions are occupied again by different pieces – we have half as orange, a fourth is yellow, and two eighths are blue. Again, those fractions can be named different things, so you’re getting kids to be able to add to their knowledge of equivalent fractions by telling us what color is occupying each space.
Our “I Can” Statement is: I can measure length using a ruler marked with halves, and a quarter of an inch. We look at two different rulers that now go beyond just zero and one. Our rulers go all the way from zero inches up to nine inches, and we see different tick marks in between. We talk about what we notice, and what we wonder. As always, it’s great to throw out this question as a way to catch kids’ attention and really get them the gist of what we’re going to be talking about.
In this show, we do a lot of measurement with worms! One worm measures four and a half inches. Jayda says the worm is four and a half long, but Kiran says the worm is four and two fourths inches long. Who is correct? Obviously with us numbering, or labeling, the fraction tick marks in between the whole numbers, you could have something as four and a half or it could be four and two fourths, if you were to mark each tick mark by fourths. We measure a variety of different worms, discussing the different ways that you could talk about how you could name that measurement.
We then look at finding the lengths and equivalent lengths of scissors, a stapler and a hole punch. We want to get students to be able to rename the length of objects using their knowledge of fractions.
The extension activity is to measure with a ruler to find equivalent lengths. They have different objects that they’ll also be measuring in the extension activity to apply what they’ve learned in the show.
Wow, we have all kinds of things to offer this week with all the different topics we’re doing – from adding and subtracting strategies, to shapes, to measuring, and even our teen numbers!
Thanks so much for joining us. I can’t wait to hear how you enjoy the Math Might shows this week!
## Celebrating 112 Shows! Math Might Reflections
Wow, it's hard to believe that we have recorded, produced, edited, and sent off to be hosted on the PBS Michigan Learning Channel, our 112th Math Might Show. As we tie a bow on this season, and on this school year, I thought it would be fun to look back at how this...
## Math Might Teacher’s Guide: Episodes 403-404
I can't believe that we've made it to the end of our Math Might shows for this second semester, shows 403 and 404! They both have some really great material for you to check out!Episodes 403-404400 Series Focus: Numeracy/Number Talks We’re going to continue the number...
## Math Mights Teacher’s Guide: Episodes 317-318
Thanks for joining us for our blog this week for our Math Mights recap on shows 317 and 318! This week’s shows will be the last of the 300 level shows. As you know from the 200 level shows, we stopped at 217 and 218, giving you eighteen shows per quarter. And so, next... | 4,890 | 21,182 | {"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-2021-25 | latest | en | 0.937089 |
www.fasshotel.ch | 1,669,612,344,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710473.38/warc/CC-MAIN-20221128034307-20221128064307-00414.warc.gz | 820,127,484 | 340,132 | # Why is variance important?
Most of the time in statistics, you’ll want to find the sample variance, not the population variance. Because statistics is usually all about making inferences from samples, not populations. If you had all of the data from a population, there would be no need for statistics at all! That said, there really is very little difference between the formula for the population variance and the formula for the sample variance. You’d just need to insert your data into the columns instead of your population data. If you prefer to plug the numbers straight into the formula, just make sure you use the population mean and not the sample mean(). In addition, the most common sample variance formula uses n-1 in the denominator instead of n.
• Each has its own R2 and, the larger the R2, the more important the source.Once you find an important source of variation, turn your attention to creating business advantage.
• The widely used types of variances that are analyzed by management are given above.
• We’ll walk you through the budget vs actual variance analysis formula in excel.
• And could this variance be related to other way in which thing vary ?
• Variance refers to the expected deviation between values in a specific data set.
• It is essential to be aware of when to include information on the spread or variance of data in reports to the interprofessional team.
In probability theory and statistics, the variance of a random variable is a measure of its statistical dispersion, indicating how far from the expected value its values typically are. The variance of a real-valued random variable is its second central moment, and it also happens to be its second cumulant. The variance of a random variable is the square of its standard deviation. Variance analysis, also described as analysis of variance or ANOVA, involves assessing the difference between two figures. It is a tool applied to financial and operational data that aims to identify and determine the cause of the variance.
## Raw Material Price Variance
He helps companies and people, including start-ups, multinationals, executives, and leaders at all levels, chart their courses to data-driven futures. He places special emphasis on quality, analytics, and organizational capabilities. Importantly, this manager’s job was much easier starting at week 24. Her process performed better, and three-quarters of the variation was https://online-accounting.net/ removed, making it easier to predict a brighter future. Estimation of the average correlation coefficient for stratified bivariate data. This expression can be used to calculate the variance in situations where the CDF, but not the density, can be conveniently expressed. A square with sides equal to the difference of each value from the mean is formed for each value.
Intuitively, computing the variance by dividing by instead of underestimates the population variance. This is because we are using the sample mean as an estimate of the unknown population mean , and the raw counts of repeated elements in the sample instead of the unknown true probabilities. Note that the term in the denominator above contrasts with the equation for , which has in the denominator.
## Step 1: Gather Data
Variance, in the context of Machine Learning, is a type of error that occurs due to a model’s sensitivity to small fluctuations in the training set. High variance would cause an algorithm to model the noise in the training set. Variance refers to an algorithm’s sensitivity to small changes in the training set. High variance is a result of the algorithm fitting to random noise in the training set.
• Depending on your service line and business goals, you will choose what variance analysis makes the most sense to track to ensure you are maximising efficiency and minimising costs.
• What is the difference between cost variance and schedule… Schedule variance shows the deviation in time consumed and the estimated time.
• Analysts take ample time to prepare for these presentations as management often asks questions that they need to be ready to answer.
• Variance is a mathematical expression of how data points are spread across a data set.
• The version of this I’ve always found helpful is thinking of them in terms of accuracy and precision and picturing arrows around a target with a bullseye.
One of the ways that I feel that variance is important is that analyzing what causes variances is key to doing inference. For example, when you do the ANOVA decomposition in linear regression, you are breaking down SST, the total variance, into SSE and SSR . So here, doing this analysis of variance gives you an indication of what each predictor is contributing to the response, which is the essense of statistical inference. A variance is the difference between actual and budgeted income and expenditure. As you dive into the numbers, it’s important to understand the sources of variation. For instance, everyone knows that some full-grown adults are taller than others, and it is easy enough to observe that men, on average, are taller than women.
## Distribution of the sample variance
The manager can now safely predict that unless they take active steps to change it, the process will perform within these limits for the foreseeable future. Some new deformation formulas about variance and covariance. Proceedings of 4th International Conference on Modelling, Identification and Control. The generalized variance can be shown to be related to the multidimensional scatter of points around their mean.
However, a variance is indicated in larger units such as meters squared while the standard deviation is expressed in original units such as meters. Most of the time, however, variance analysis catches operational inefficiencies. Operational anomalies are common in every business environment. By identifying these, companies can uncover any problematic areas within their process and correct any errors. Similarly, variance analysis allows companies to consider material variances only.
## Population Variance
The variance is often incorporated into a reference range provided with each lab result. For example, a resting heart rate of 65 beats per minute is generally not concerning. Although the mean resting heart rate might be in the 70s or 80s, the corresponding reference range is 60 to 100 beats per minute. Since 65 falls within this reference range, it does not fall far enough from the mean to be of concern. Arranging the squares into a rectangle with one side equal to the number of values, n, results in the other side being the distribution’s variance, σ2. In statistics, the variance is used to determine how well the mean represents an entire set of data. The formula shows that the variance of X (Var) is equal to the average of the square of X minus the square of its mean.
In the process analysts might work with various department leaders to understand what occurred to lead to a variance. The value of the first quartile is 50; the value of the second quartile (equal to the average of the 21st- and 22nd-smallest values) is 51.5; and the value of the third quartile is 54. A box plot is often used to plot some of the summarizing statistics of a data set.
The whole point is the real world is not deterministic, isn’t perfect, it isn’t what traditional math assume it to be. In Statistic/probability world there are variance from the expected value. You expect Stephen Curry to why is variance important make those 3 points often but he will miss once and awhile. If you take all Stephen Curry 3 points attempts it’ll converge to the true probability value of Stephen Curry hitting that 3 points (probably close to 90%).
Resource allocation based on real-time route planning & route optimization. Instead of re-entering data into QuickBooks for estimating and invoicing, mHelpDesk makes it easy to reuse the data you’ve collected out in the field. MHelpDesk supports different sync versions for QuickBooks Desktop and QuickBooks Online that determine what data you can import and export. Free 7-Day Test Drive with sample data for your Service Manager and one field technician via our Mobile App . Suggestions FormAnd the software will tell you the best time slots for your appointment.
• This software automatically updates inventory as line items get added to work orders.
• Sometimes there is a long lapse between windows; generally it is several seconds but at time it can be 15 to 20 seconds.
• Your access to this site was blocked by Wordfence, a security provider, who protects sites from malicious activity.
• Office staff includes any other relevant notes in the record and then posts the information to QuickBooks.
• Then as they close a work order, they can find and add parts to the invoice.
• Learn how your businesses can use FieldCircle to achieve more efficient, transparent, and profitable service operations.
Also, your preventive maintenance work orders are auto-created and easy to schedule right alongside your current work orders. Using mobile payments, your technicians take payment on-site and get paid instantly. There’s no waiting, no double entries and no smudged credit card numbers. Increase your cash flow and reduce the effort it takes to get paid. With Intuit Field Serive https://quickbooks-payroll.org/ management, you can trade in your clipboards for cell phones, tablets, and PCs. Both solutions come with a mobile app, but the sales representative told me most Smart Service 365 clients prefer to use their phone or tablet’s web browser. By clicking on Suggestions at the bottom of the schedule form, you can view the most efficient schedule times for each appointment.
## Third Party Integrations
ReachOut is a field service management suite to streamline field processes with customizable mobile-based forms and workflow. My Service Depot developed its Smart Service software to serve as a one-stop solution for managing service operations in the field.
Most of the comments indicated that the software did wonders for their business operation. Several people noted the end to problems with penmanship on service orders and receipts. Others mentioned the highly effective scheduling process, including the route optimization function. Being able to optimize service routes by geography is a great way to instantly save money on gas. By reducing travel time, it also allows technicians the ability to see more customers each day. From the auto-assignment of work orders to field service technicians to instant route optimization, field service scheduling software simplifies your workforce logistics.
## Customer and Equipment Service History
At the beginning of each day, your techs will find their schedules waiting for them on their mobile devices. Smart Service field service software adds scheduling and dispatching to QuickBooks to create a single, seamless software system. Smart Service is an excellent solution to manage field service companies digitally. The software’s simple and intuitive design is completely customizable, and its features have great depth.
However, the application offers several functions that could assist with the coordination of site visits when they do occur. The link to QuickBooks could be especially helpful to campaign staff as a way to track donations in the field. The donor could receive an email receipt at the same time campaign headquarters received a record of the funds.
## Onboarding New Customers
Others have noted that the system seems slow and clunky to them while others note that integration with QuickBooks is only one-way, not a true synchronizing. Some have found it to be plagued by too many bugs, and are unhappy that it is based on Microsoft Access, which seems woefully out of date. Smart Service is a QuickBooks scheduling software that lets you add scheduling, dispatch, and more to QuickBooks. This field service software helps turn your QuickBooks into a single, seamless software and supports QuickBooks Pro, Premier, Enterprise and Online. Let’s all get on the same page by looking at what field service scheduling software is. That means immediate information on all your work orders and technician schedules as well as fast rescheduling.
• Analyze the data and information with advanced analytics to make strategic decisions, growth plans and evaluating the current scenario.
• This includes scheduling and routing jobs in a manner that maximizes technician utilization and service delivery quality.
• This tool lets your office team know exactly what work your field teams have completed by providing live updates on service delivery.
• The most often-mentioned advantage is how easy it is to use Smart Service, which also makes it easy to learn for newbies.
• However, many US customers would love the cultural and language benefits that come from receiving technical support from within the States.
Here are some of the advantages of using Smart Service with QuickBooks. Smart Service functions as a smart service quickbooks direct QuickBooks add-on, so all the work you do in the software updates QuickBooks in real time.
Social workers for example also set appointments and make site visits. Although many of them generally make their own daily schedules, it may be helpful for supervisors to have access to employee calendars as part of Smart Service. Food and safety inspectors are other categories of government workers that could use this application. Even political campaigns could likely use this software to coordinate outreach and fundraising activities. The vast majority of small and medium-sized businesses report that moving to some kind of digital tool helped their operations . Now the question is finding the right software to help your business.
### Search results – iTWire
Search results.
Posted: Tue, 11 Oct 2022 03:40:37 GMT [source]
Process forms are downloaded and emailed, and photocopies are taken with a phone. Synchroteam cloud based Field Service Management solution optimize costs, dispatch, scheduling and reporting. Better yet, you can automatically send your invoices to customers and collect payment via online portals. What’s more, is that it’s a great way to connect your in-office team and field crews. This tool lets your office team know exactly what work your field teams have completed by providing live updates on service delivery. Cloud hosting ensures that users have sufficient power for operating seamlessly at all times. The bill can be sent to customers at the tap of a button.
## Field Service Management QuickBooks Integration
Sometimes there is a long lapse between windows; generally it is several seconds but at time it can be 15 to 20 seconds. Having to update the customer list almost everyday; it tedious and with over 10K clients can take a long time. If there is an issue on weekends, holidays or after business hours then you have to wait and that is the only con I can think of.
The QuickBooks Self-Employed platform was designed with a very specific audience in mind. If you need to file a Schedule C document with Form 1040 during tax season, this is your accounting solution.
• I haven’t used all its features yet but I’m having a lot of issues with setting rules for expense transactions; namely they don’t work at all!
• First of all, QuickBooks Self-Employed includes built-in mileage tracking—a crucial feature if you plan to include business travel as a tax write-off.
• The tax bundle plans allow you to file your state and local tax returns and also your quarterly estimated taxes.
• ZipBooks cashes out to your bank in as little as two business days.
• Data access is subject to cellular/internet provider network availability and occasional downtime due to system and server maintenance and events beyond your control.
• See for yourself why Bonsai Tax is hands down the best tax software for self-employed folks.
Here you can create invoices to send to clients directly from QuickBooks Self-Employed. The Quarterly Taxes tab tells you what your recommended payments are for each quarter, as well as the amount you have to pay for each of those quarters. Once you get successfully logged in through the 30-day free trial, you can then experience all of the benefits for yourself. Once you reach the end of your 30-day trial, you will receive an email to ask you to continue your subscription. This gives you some peace of mind and lets you try their product without any risk. You can also customize reports so you can get a better insight on specific business needs. You can run reports on what matters the most to your business.
## QuickBooks Self-Employed Customer Service
The thing I am most fond of with QuickBooks Self-Employed version is the ease of use. The software offers a user the functions that cover all necessary withholding categories that are required on a state and federal level. It inputs the correct taxes, insurance, fees etc. required by law within your area. And if you are not within regulation it notifies you and offers the user direction into making sure you are financially in accord to the law. Because it is not full-featured accounting software and is designed for freelancers, it doesn’t have advanced features that are difficult to master. QuickBooks Self-Employed is cloud-based software that is easy to navigate, and the new UI makes the software even more of a joy to use.
• It also connects to Shopify and automatically adds sales tax to your invoices, two crucial features for sole proprietors who sell products online.
• However, I understand how important accounting is, thus why I put such a focus on it in my business.
• If you are self-employed, or you are considered to be an Independent Contractor you can get a 30-day free trial of QuickBooks for the self-employed.
• From creating proposals that clients can’t say no to, to sealing the deal with a professional contract – Bonsai will revolutionize the way you do business as a freelancer.
• There are many other disadvantages to Quickbooks Self-Employed, which makes it important to know about tools that overcome these cons pretty well.
Software Advice’s FrontRunners report ranks top products based based on user reviews, which helps businesses find the right software. Tim worked as a tax professional for BKD, LLP before returning to school and receiving his Ph.D. from Penn State. He then taught tax and accounting to undergraduate and graduate students as an assistant professor at both the University of Nebraska-Omaha and Mississippi State University. Tim is a Certified QuickBooks Time Pro, QuickBooks ProAdvisor for both the Online and Desktop products, as well as a CPA with 25 years of experience.
## How to Get a QuickBooks Self Employed Free Trial With a Virtual Credit Card
It caters to business owners with simple business structures. This means no employees and/or contractors, and very few customers that need to be invoiced.
You can use QuickBooks for your home or personal expenses, such as managing groceries and other home bills. If you’re looking for live bookkeeping support, QuickBooks Self-Employed is a more affordable option over Wave. Very simple time tracking is a new feature in Intuit QuickBooks Self-Employed. You can record business trip mileage automatically or manually and let Intuit QuickBooks Self-Employed calculate your tax deduction. With QuickBooks Accountant, you also get access to free training and valuable product support.. This comes with free access to the QuickBooks ProAdvisor program..
## Mighty Taxes
There’s no phone support for QBSE, only email and the guy tells me to do the same thing over and over again, to no avail. So for now while it may be convenient for keeping track of my direct expenses it’s very limited due to the problems I am having. The tech support is just terrible by today’s standards.
When evaluating offers, please review the financial institution’s Terms and Conditions. If you find discrepancies with your credit score or information from your credit report, please contact TransUnion® directly. It is extremely useful that this software integrates banking, car/fuel usage in the internal structure. The fact that income/expense entires can be custom-automated makes for an easy and intuitive workflow.
## QuickBooks Self-Employed Live Tax Bundle for \$35 Per Month
To make your Schedule C tax return easy, the software will transfer relevant information to TurboTax if you opt for this feature. 100% accurate calculations qbo login and your maximum refund, guaranteed. Massage Book This all-in-one practice management tool will help you simplify and grow your practice.
• What’s more, you don’t have to rely on a hiring a lawyer to explain all that legal jargon anymore.
• These include a study guide, practice test, and exam voucher.
• If you are a sole proprietor who files a federal tax return with TurboTax, the cost to use their service starts at \$119.99.
• There’s no confusing accounting jargon or unnecessary features crowding your dashboard, and setting up your account takes very little time.
There’s no confusing accounting jargon or unnecessary features crowding your dashboard, and setting up your account takes very little time. There’s also built-in support if you need extra help setting up and navigating the software. However, QuickBooks Self-Employed does have its drawbacks.
Many people have trouble understanding https://www.bookstime.com/ because they mistake it for an escrow account, so it’s important to know the difference. Earnest money is a deposit made to a seller, often in real estate transactions, that shows the buyer’s good faith in a transaction. Your real estate agent will oversee this entire escrow process, so don’t be too concerned if you don’t understand every detail. At this point, monthly escrow payments for the following year are adjusted up or down based on whether there was a shortage or surplus in the account for the current year’s payment. Mortgage-holders are obligated to send you an annual statement regarding the activity of your escrow account, which may also be referred to as a mortgage impound account. With traditional mortgages, your experience with escrow usually ends at this point. If you are buying a house with a Federal Housing Administration loan, however, your dealings with escrow accounts continue in a different way, for different reasons.
The buyer, seller, and lender are all parties to the creation of the document. Real estate brokerages can also establish escrow accounts to hold deposits for rental applications. Escrow accounts exist to protect all parties in the case of a real estate deal, so money does not get exchanged before a contract is executed. The 2017 version of the FAR/BAR contract contains provisions in the event of a buyer or seller default. If the buyer defaults, the seller may elect to recover and retain the deposit as agreed upon liquidated damages, and buyer and seller shall be relieved from all further obligations under the contract.
## An escrow account is necessary to seal the deal.
In the process of buying a home, it is used to protect both buyers and sellers. This ensures the seller that the buyer has the money to go through with the purchase. It also protects home buyers from getting scammed and losing large sums of cash. It is essentially the middle man who makes sure no property or money is transferred before both parties to the transaction get the benefit of their margin – their money or their property. If the parties used the 2017 FAR/BAR contract, the buyer and seller will have 10 days after the date demands are made for the deposit in order to resolve the dispute. If there are still unresolved issues after the 10 days, the buyer and seller must go to mediation, and if mediation does not resolve the issues, an action may be filed in court. Pursuant to the 2017 contract prevailing party in a court action shall be entitled to recover from the non-prevailing party attorney’s fees and costs incurred.
### What does in escrow mean?
When you hear the phrase in escrow, it means that all items placed in the escrow account (e.g., earnest money, property deed, loan funds) are held with an escrow agent until all conditions of the escrow arrangement have been met. The conditions usually involve receiving an appraisal, title search and approved financing.While the earnest money is in escrow, neither you nor the seller can touch it. Once conditions are met, the earnest money will likely be applied toward the purchase price or your down payment on the home.
The best way to avoid falling out of escrow is to prevent it from happening altogether. Prior to making an offer, the buyer should have a reasonable budget in mind and be confident they will qualify for the loan. On the other end, the seller should be transparent about any damage or potential problems with the property.
## How Do Escrow Accounts Work in NYC Real Estate?
Typically, the buyer will instruct the escrow officer to release funds only when all conditions have been met, title insurance has been issued and the seller’s deed has been signed. Escrow is not complete until all the terms have been fully satisfied and all the parties have signed the appropriate documentation. There is no defined length of time for how long a house must stay in escrow, although usually, a purchase contract does specify a closing date. How long it actually takes to close depends on factors such as financing details, an appraisal, a title search, inspections, and more. A house is no longer in escrow once all closing documents have been signed and the title has been transferred to the new owner. At different stages of a home purchase, the use of escrow accounts (sometimes called “impound accounts”) has benefits for the homebuyer and if the home is financed, the mortgage lender. An escrow agent facilitates the closing of a home sale and disperses all the funds to the appropriate parties.
• In this article, we’ll break down the definition of escrow and share how it affects buyers and sellers during the closing process.
• Delegating the responsibility of paying for taxes and insurance to the mortgage lender also adds predictability to monthly payments.
• This compensation may impact how, where and in what order products appear.
• Accounts are typically managed by licensed real estate professionals in their state.
• Paying a predictable amount each month makes it easier to budget and you don’t have to worry about tracking the due dates for your taxes and insurance policies.
In a real estate transaction, a trusted third party is hired to hold all documents and funds for both buyer and seller. Homeowners with a mortgage on their home may set up an escrow account at the request of their lender. Escrow is when a neutral third party holds on to funds during a transaction. In real estate, it’s used as a way to protect both the buyer and seller during the home purchasing process. After you purchase a home, your lender will establish an escrow account to pay for your taxes and insurance. After closing, your mortgage servicer takes a portion of your monthly mortgage payment and holds it in the escrow account until your tax and insurance payments are due. The second type of escrow account, aka a mortgage escrow account, comes into play once you’ve actually purchased a home.
## How long do you pay escrow?
Then, at the end of the year, you get a bill from your bank for hundreds or thousands of dollars. This is unpleasant news under any circumstances, particularly when it’s unexpected. But depending on how the sales process goes – or doesn’t go – the deposit might go to the seller or the lender, or be returned to the escrow real estate buyer. In order to ensure that the money is handled properly, the buyer deposits it into an escrow account that’s set up by both parties. There will need to be a home inspection, and you’ll need to get final approval for your mortgage. Depending on the results of the inspection, some repairs may be necessary.
Most buyers would not be comfortable allowing the seller to hold their funds, and, in fact, this may not be allowed through real estate law. It’s the buyer’s or seller’s agent who generally recommends which escrow agency to use. It’s a procedure where a neutral, third-party entity—typically an attorney or title company representative— holds a buyer’s deposit and important documents related to a home sale. The calculator shows you your monthly payments, including principal, interest, property taxes, and insurance (collectively referred to as P.I.T.I.). Just like an escrow agent, escrow officers are neutral third parties that account for everything when dealing with two independent parties.
The escrow agency, guided by the escrow instructions, will take care of the money and documents accordingly. Placing funds and important paperwork in escrow acts as a safeguard, to maintain integrity in the homebuying process. Depending your state’s laws, your escrow contact might be considered either an attorney or an officer.
### Housing market first-timer? 15+ real estate terms you should know, from FICO to escrow – STL.News
Housing market first-timer? 15+ real estate terms you should know, from FICO to escrow.
Posted: Wed, 16 Nov 2022 14:15:52 GMT [source]
When your tax bills and insurance premiums are due, your mortgage servicer will make sure those bills are paid on time, every time. Your servicer will even cover bills for you if your escrow account is short on funds. Title companies are not your only option if you need to hold earnest money in escrow for a for sale by owner real estate transaction. You may also contact a real estate lawyer to get help in this situation. A lawyer may be able to place earnest money into a trust account until the sale of the property is completed. Such accounts are used to assure mortgage lenders that property tax and homeowners insurance payments will be made on time.
Disbursing the funds and closing the escrow account is one of the last steps. Home buying process, the escrow account holds specific funds (i.e., earnest money and any prepaid taxes or other items) until the real estate purchase is complete. In California, for instance, homeowners who make mortgage or property tax payments through an escrow account are entitled to the interest earned on that money. Not all states have these types of rules, and it may depend on the bank involved. Even if there is no requirement to use an escrow account in the home purchase context, using such an account may provide additional protection to all parties involved in the transaction.
Assetsare items of value the firm owns or controls, acquired at a measurable cost, which the firm uses for earning revenues. Balance Sheet Assets, therefore, represent the book value of everything the firm has to work with to bring income. Note especially that the first equation shows clearly that the firm’s assets are partly owned by owners and partly owned by creditors .
However, the amount credited to the partner’s capital account is only equivalent to their profit sharing ratio. Business owners may think of owner’s equity as an asset, but it’s not shown as an asset on the balance sheet of the company. Because technically owner’s equity is an asset of the business owner—not the business itself. Stockholders‘ equity is the remaining amount of assets available to shareholders after paying liabilities. A final type of private equity is a Private Investment in a Public Company .
## Definition of Owner’s Equity
The withdrawals are considered capital gains, and the owner must pay capital gains tax depending on the amount withdrawn. Another way of lowering owner’s equity is by taking a loan to purchase an asset for the business, which is recorded as a liability on the balance sheet. The liabilities represent the amount owed by the owner to lenders, creditors, investors, and other individuals or institutions who contributed to the purchase of the asset.
• Equity is defined as the owner’s interest in the company assets.
• Free AccessFinancial Modeling ProUse the financial model to help everyone understand exactly where your cost and benefit figures come from.
• The amount of treasury stock is deducted from a company’s total equity.
• And, you can compare your owner’s equity from one period to another to determine whether you are gaining or losing value.
• Successful branding is why the Armani name signals style, exclusiveness, desirability.
Owner’s equity and retained earnings are largely synonymous in many circumstances, but there are key differences in exactly how they’re calculated. Many small businesses with just a few owners will prefer to use owner’s equity. Retained earnings statement of stockholders equity are more useful for analyzing the financial strength of a corporation. Owners of limited liability companies also have capital accounts and owner’s equity. The owners take money out of the business as a draw from their capital accounts.
## Example Detailed Balance Sheet
Owner’s equity is equal to a company’s total assets minus its total liabilities. It represents the potential capital available to use for a sole proprietorship. It is also the capital left if all the liabilities are deducted from the assets. Owner’s equity in a business can decrease over time as well, depending on the owner’s actions. Withdrawals are considered capital gains, which are subjected to a capital gains tax. Additionally, owner’s equity can be reduced by taking out loans to purchase assets. Therefore, they reduce the value of the business’s assets when calculating equity.
On the other hand, market capitalization is the total market value of a company’s outstanding shares. Apple’s current market cap is about \$2.2 trillion, so investors clearly think Apple’s business is worth many times more than the equity shareholders have in the company. Naturally, having insight into a business’s performance will be important to existing and potential lenders and investors.
## Examples of Owners Equity
Additionally, higher business profits and decreased expenses can increase owner’s equity. To further increase that worth, business expenses can be decreased. Treasury Stock → Share buybacks are used by companies seeking to compensate shareholders.
Due to the cost principle the amount of owner’s equity should not be considered to be the fair market value of the business. The CFS is, therefore, more comprehensive with regard to understanding the financial health of a company, but does not offer the same type of transparency into any specific line item. Each of the components that impact the equity account is listed in the top row, with the corresponding change listed below.
He Owners equity concept applies to companies in business, but it is similar to the notion in personal finance, where a homeowner speaks of „equity“ in a home property. In that case, Equity represents the initial down payment on the property plus the part of the mortgage loan principal that has been „paid off.“ These funds are profits the company earns and uses to grow equity. The other primary use for earnings that a company may choose is to distribute them directly to shareholders as dividends.
Equity can be found on a company’s balance sheet and is one of the most common pieces of data employed by analysts to assess a company’s financial health. Equity is also what a shareholder owns in a corporation, entitling him or her to part of that entity’s profits and a measure of control . Business owners should be aware of the impact of their decisions on owner’s equity. For example, it is possible to have a negative amount as owner’s equity if an owner has withdrawn a higher amount than they have invested. Profit margin is a measure of a business’s profit relative to its revenue. Learn about the types of profit margin and the formulas to calculate each. Jean Murray, MBA, Ph.D., is an experienced business writer and teacher who has been writing for The Balance on U.S. business law and taxes since 2008.
Subtract total liabilities from total assets to arrive at shareholder equity. Equity is important because it represents the value of an investor’s stake in a company, represented by the proportion of its shares. Owning stock in a company gives shareholders the potential for capital gains and dividends.
### What are examples of owner’s equity?
Owner's equity is the amount that belongs to the business owners as shown on the capital side of the balance sheet, and the examples include common stock, preferred stock, and retained earnings. Accumulated profits, general reserves, other reserves, etc.
This form of depreciation can be hard to predict with absolute certainty. For example, at the beginning of the year, the asset has a remaining life of 8 years. The following year, the asset has a remaining life of 7 years, etc. If, for example, you buy a car for \$10,000, but a year later it is only worth \$8,000 due to wear and tear, then the depreciation on the car is \$2,000.
• This is the simplest and most straightforward method of depreciation.
• The tricky bit of this equation is the sum of the years’ digits piece.
• For the decrease in value of a currency, see Currency depreciation.
• An asset’s basis includes the cost of buying the asset, transporting it, and setting it up.
• An asset is an item of value that you expect will provide future benefit.
To calculate depreciation using this method, first, the salvage value of an asset is subtracted from the cost of the asset, and this total is divided by the useful life of the asset. Examples include oil & gas, automobiles, real estate, metals & mining. Titus, the plant supervisor, determined the technical feasibility test of the bottling machine. Titus believes it will last for 5 years with a salvage value of \$8000. Find out the depreciated expense for each year using the straight-line method. Accumulated DepreciationThe accumulated depreciation of an asset is the amount of cumulative depreciation charged on the asset from its purchase date until the reporting date. It is a contra-account, the difference between the asset’s purchase price and its carrying value on the balance sheet.
## Importance Of Depreciation
Depreciation is considered a non-cash charge because it doesn’t represent an actual cash outflow. The entire cash outlay might be paid initially when an asset is purchased, but the expense is recorded incrementally for financial reporting purposes. That’s because assets provide a benefit to the company over a lengthy period of time.
They take the amount you’ve written off using the accelerated depreciation method, compare it to the straight-line method, and treat the difference as taxable income. In other words, it may increase your tax bill in the year of sale. Usually business owners using accelerated methods will set up a depreciation schedule — a table that shows the depreciation expense for each year of the asset’s life — so they only have to do the calculations once. That is, a business does not write a check to „depreciation.“ Instead, the business records or recognizes the cost of the asset over time on the income statement. The annual depreciation using the straight-line method is calculated by dividing the depreciable amount by the total number of years. The depreciation formula is used to calculate the depreciation of a business’s fixed assets. Units Of Production MethodUnit of production depreciation is an activity method to ascertain asset value through its usage.
At the time of the purchase, the equipment vendor guaranteed that the bulldozer would function properly for a span of five to seven years. Record depreciation for assets still in use toward the end of the fiscal year. The following section will review the different types of depreciation that are used. It’s a good idea to consult with your accountant before you decide which fees to lump in with the cost of your property. If you can determine what you paid for the land versus what you paid for the building, you can simply depreciate the building portion of your purchase price. As a reminder, it’s a \$10,000 asset, with a \$500 salvage value, the recovery period is 10 years, and you can expect to get 100,000 hours of use out of it. Since hours can count as units, let’s stick with the bouncy castle example.
• The carrying value of an asset after all depreciation has been taken is referred to as its salvage value.
• The purchase price minus accumulated depreciation is your book value of the asset.
• Every business can take advantage of depreciation by deducting the expense of using up a portion of the value of an asset from taxable income.
• Company A purchased an asset for \$8,000 with an expected useful life of seven years.
• Let’s assume that if a company buys a piece of equipment for \$50,000, it may expense its entire cost in year one or write the asset’s value off over the course of its 10-year useful life.
Each method is used for different types of businesses and types of assets. Depreciation is the process of deducting the cost of a business asset over a long period of time, rather than over the course of one year. In this article, we define depreciation, review the types of depreciation, explain how they are calculated and show you how to record it in financial documents.
## Synonyms & Antonyms For Depreciate
By taking the time to understand important financial aspects that should be incorporated in your company, you can produce clever financial decisions and give your company the best chance to succeed. When you buy property, many fees get lumped into the purchase price. You can expense some of these costs in the year you buy the property, while others have to be included in the value of property and depreciated. Since the asset is depreciated over 10 years, its straight-line depreciation rate is 10%.
Total cost of the asset (called „asset basis“) – salvage value/number of years of the asset’s useful life. Depreciation is defined as a reduction in the value of an asset that occurs over time as the asset gets older or as wear and tear occurs, or the decline of one currency in relation to others. The fall in value of capital, such as machinery, due to wear and tear, old age, obsolescence, or a fall in the market price. Such a decrease as allowed in computing the value of property for tax purposes. This section will provide a step-by-step guide on how to record depreciation on your company’s financial documents. Without Section 1250, strategic house-flippers could buy property, quickly write off a portion of it, and then sell it for a profit without giving the IRS their fair share.
## Is Depreciation A Fixed Cost?
In other words, it is the reduction in the value of an asset that occurs over time due to usage, wear and tear, or obsolescence. The four main depreciation methods mentioned above are explained in detail below. The main advantage of the units of production depreciation method is that it gives you a highly accurate picture of your depreciation cost based on actual numbers, depending on your tracking method. Its main disadvantage, though, is that it is difficult to apply to many real-life situations, as it is not always easy to estimate how many units an asset can produce before it reaches the end of its useful life. This method, also called declining balance depreciation, allows you to write off more of an asset’s value right after you purchase it and less as time goes by. To compensate for this loss of value caused by depreciation, businesses may write off some of the depreciation of long-lasting assets as an expense every year until the item reaches the end of its useful life.
• Also, it is seen as a business expense despite being a non-cash expense.
• The following section will review the different types of depreciation that are used.
• Depreciation allows businesses to write off through taxes costs incurred through the operation of assets and is typically arrived at using the straight-line depreciation method.
• The following year, the asset has a remaining life of 7 years, etc.
Straight-line depreciation is the most common depreciation method to use. It requires that a company spreads out an asset’s depreciation expenses evenly across each year that an asset is estimated to work and function properly. Under this method, the more units your business produces , the higher your depreciation expense will be. Thus, depreciation expense is a variable cost when using the units of production https://accountingcoaching.online/ method. Let’s say that, according to the manufacturer, the bouncy castle can be used a total of 100,000 hours before its useful life is over. To get the depreciation cost of each hour, we divide the book value over the units of production expected from the asset. If you use it with the wrong type of asset, you can easily overstate or understate your net income in a given accounting period.
The depreciation rate is used in both the declining balance and double-declining balance calculations. There are many types of depreciation, including straight-line and various forms of accelerated depreciation. She is an expert in personal finance and taxes, and earned her Master of Science in Accounting at University of Central Florida.
## T Century Insurance Review 2022: Car & Home
Calculate the amount of depreciation for each year and the closing value of the asset at the end of each year. The business should charge \$1,000 to its income statement each year and reduce the value of the asset by \$1,000 per year as well. With this method of depreciation, the value of the asset is reduced uniformly over its Depreciation Definition useful life. This expense will then show up on the business’s income statement. Asset Lifetime – The number of periods an asset will be depreciated for . Generally, if you’re depreciating property you placed in service before 1987, you must use the Accelerated Cost Recovery System or the same method you used in the past.
Under such a convention, all property of a particular type is considered to have been acquired at the midpoint of the acquisition period. One half of a full period’s depreciation is allowed in the acquisition period .
Instead, the company only has to expense \$4,000 against net income. The company expenses another \$4,000 next year and another \$4,000 the year after that, and so on until the asset reaches its \$10,000 salvage value in 10 years. As stated earlier, carrying value is the net of the asset account and the accumulated depreciation. The salvage value is the carrying value that remains on the balance sheet after which all depreciation is accounted for until the asset is disposed of or sold. The carrying value of an asset on the balance sheet is its historical cost minus all accumulated depreciation. For example, cars start depreciating as soon as they are driven off the lot after a sale. The reason why cars and other items lose value over time is because wear and tear reduces the quality of the item, and oftentimes, newer models are produced.
• You can also use the declining balance method, which allows a business to take either 200% or 150% depreciation in each year.
• Its salvage value is \$500, and the asset has a useful life of 10 years.
• When you invest in certain types of property for your business, even though you may use cash to make the purchase, you typically can’t immediately deduct the entire cost as expense against revenue.
• Following are examples where the depreciated amount is calculated using different methods.
• With the double-declining-balance method, the depreciation factor is 2x that of the straight-line expense method.
Bonus depreciation on purchases of new, qualified business assets is 100% if acquired and placed into service after September 27, 2017, and before January 1, 2023. Businesses can write off the full cost of depreciable property such as machinery, equipment, computers, appliances, and furniture. Sum of the years’ digits depreciation is similar to the double-declining method in that it is also an accelerated depreciation calculation. Instead of decreasing the book value, SYD calculates a weighted percentage based on the remaining useful life of the asset. To calculate depreciation, you need to know the item’s useful life or total possible output, its cost (including taxes, shipping, and preparation/setup expenses), and its residual value. For the unit-of-production method, you will also need to know how many units were produced for the production run. So, if you use an accelerated depreciation method, then sell the property at a profit, the IRS makes an adjustment.
The main drawback of SYD, though, is that it is markedly more complex to calculate than the other methods. Depreciation allows businesses to write off through taxes costs incurred through the operation of assets and is typically arrived at using the straight-line depreciation method. Learn the definition of the process of depreciation and the formula used to calculate it through examples.
United States rules require a mid-quarter convention for per property if more than 40% of the acquisitions for the year are in the final quarter. Depreciation is often what people talk about when they refer to accounting depreciation. This is the process of allocating an asset’s cost over the course of its useful life in order to align its expenses with revenue generation. For example, an asset with a useful life of five years would have a reciprocal value of 1/5, or 20%. Double the rate, or 40%, is applied to the asset’s current book value for depreciation. Although the rate remains constant, the dollar value will decrease over time because the rate is multiplied by a smaller depreciable base for each period. The diminishing balance method of depreciation, or as it is also known, the reducing balance method, calculates depreciation as a percentage of the diminishing value of an asset.
Thus, the depreciable value diminishes every year, and so does the depreciated expense. The value of the assets gets depleted due to constant use for business purposes. Companies depreciate to account for this value throughout the useful life of that asset. It is a fixed cost for the companies, and the amount depreciated can be used to purchase new machinery after the old one turns into a scrap. Also, it is seen as a business expense despite being a non-cash expense. In the sum-of-the-years digits depreciation method, the remaining life of an asset is divided by the sum of the years and then multiplied by the depreciating base to determine the depreciation expense.
A payroll journal is a specific type of accounting journal that includes data regarding the human resources part of the company. The final step is making all payments with the IRS EFTPS and other third parties, such as insurance companies, 401 vendors and state agencies.
• Select the QuickBooks vendor that you wish to assign an equivalent CenterPoint vendor to.
• If you receive an Application Requirements screen, click Next to install the necessary requirements.
• Your company is not paying the Federal or State tax agencies directly.
• Instead of combining them, enter each individual paycheck on separate lines.
• (Otherwise, the client will not be included in the Client drop-down list in the QuickBooks Online Export dialog).
• With the help of this software, you can import, export, as well as erase lists and transactions from the Company files.
Input the current date and assign a number to the entry. To use the service, you have to open both the software QuickBooks and Dancing Numbers on your system. To import the data, you have to update the Dancing Numbers file and then map the fields and import it. Multiple items can be selected by holding the Shift and Ctrl key while simultaneously clicking each item. In the Profit Center Type box, select the profit center type that the class should be assigned to. Repeat the process of assigning an account category to all accounts you want to import from QuickBooks to CenterPoint.
## Step 2: Enter the payroll paychecks into QuickBooks Online
A journal entry to a payroll clearing account is a journal entry that summarizes the total expenses that are included in all net payroll checks. In other words, this is an entry that helps you determine exactly how much you are paying out in payroll in a given period. The „clearing“ in a payroll clearing account means that you are using this accounting method to „clear“ revenues and expenses and reconcile any potential errors you may have in determining payroll.
• This screen allows you to keep the accounts, employees, vendors that are to be paid payroll liabilities, and QuickBooks classes that are used in payroll transactions in sync.
• We created some default accounts for you, so all you need to do is choose your Bank Account.
• The key to doing journal entries is to ensure that the total amount debited and credited is the same so that the general ledger will remain balanced.
• If you use a third-party payroll service like Paychex or ADP, the payroll company sends you reports.
In QuickBooks, a QuickBooks – Application Certificate screen will display. Verify that QuickBooks Integration is selected and then click OK. On the Setup Type screen, select Custom, and then click Next to continue.
## Plano based Computer/IT Training Company
Get more detail connect to QuickBooks Payroll Customer Service Number. However, creating .iif files from adapting data from another program is technically complex.
Select the pay run and click Detail to see a detail error log and list of attempts for that pay run. Pay runs can be reposted whether or not they were originally successful. If a pay run failed, it will not be successful until the issue causing failure is corrected. Once mapping is removed, items can be re-imported to a new item or remapped to a different existing item if needed. In the Account Category box, select the account category that the account should be assigned to.
## How Does a Payroll Clearing Account Work?
The next step is to gather and enter any payroll expenses your business has in your accounting records. Payroll expenses can include any payment made by your business during an accounting https://quickbooks-payroll.org/ period, such as wages, salaries, or benefits. You must boost your expenditure account because payroll costs are paid. Your debits must increase in tandem with your expenses.
## Features of Dancing Numbers for QuickBooks Desktop
Information created in Quickbooks can be imported into CenterPoint, or information created in both can be mapped. New information cannot be created in QuickBooks by using this process. This tool will also allow items previously mapped to be unmapped.
This check may be paid through the corporate accounts payable bank account, rather than its payroll account, so you may need to make this entry through the accounts payable system. If you are recording it directly into the general ledger or the payroll journal, then use the same line items already payroll journal quickbooks noted for the primary payroll journal entry. The salary or other remuneration that employers give to their employees is documented in a payroll journal entry. Payroll journal entries are used by an accountant to record these transactions in the company’s general ledger and to track payroll costs.
CS Professional Suite Integrated software and services for tax and accounting professionals. However, even business owners need to be aware of certain aspects of Payroll processing to ensure payroll activities are being completed authentically and accurately. Also, these entries should be reflected in General Ledger in a systematized way. Instead of combining them, enter each individual paycheck on separate lines. Best of all, you can easily import your QuickBooks Desktop data into QuickBooks Online.
### How to Make Payroll Journal Entries: A Small Business Guide – The Motley Fool
How to Make Payroll Journal Entries: A Small Business Guide.
Posted: Wed, 18 May 2022 07:00:00 GMT [source]
On the other hand, mature businesses can put this money toward building reserves that’ll protect company value if managers aren’t able to secure capital from elsewhere. In the case of the Unearned Revenue, the account is supposed to be settled in exchange for goods and services, whereas in the case of Accounts Payable, the liability is settled with Cash. Hence in this regard, the revenue has been collected but has not been ‘earned’, in the sense that the company is yet to provide goods and services against this particular amount. Finally, if a dividend was paid out, the balance is transferred from the dividends account to retained earnings. Janet Berry-Johnson is a CPA with 10 years of experience in public accounting and writes about income taxes and small business accounting.
Then, another \$200,000 worth of revenues was seen in 2017, as well as \$400,000 in 2018. If the temporary account was not closed, the total revenues seen would be \$900,000. It is shown as the part of owner’s equity in the liability side of the balance sheet of the company. Income summary account– Step three is to square off the income summary. The amount of the income summary is expenses and revenue transferred to the capital account. Revenue AccountRevenue accounts are those that report the business’s income and thus have credit balances.
## What are the four closing entries in order?
You forget to close the temporary account at the end of 2018, so the balance of \$50,000 carries over into 2019. Now you know a bit about permanent and temporary accounts. Let’s move on to learn about how to record closing those temporary accounts. Deferred revenue is classified as either a current liability or a long-term liability. This classification depends on how long it will take the company to earn the revenue. If services will be performed, or goods shipped, within one year, the deferred revenue is a current liability. If services will be performed, or goods shipped, over a period of more than one year, the deferred revenue is a long-term liability.
Harold Averkamp has worked as a university accounting instructor, accountant, and consultant for more than 25 years. He is the sole author of all the materials on AccountingCoach.com. Read more about the author. Liability accounts such as Accounts Payable, Notes Payable, Accrued Liabilities, Deferred Income Taxes, etc. Speaking of studying; @studydotcom is a great website and app that students can utilize. It has changed my life for the better and I never run out of resources to assist me as I take very challenging courses at times. Have you ever thought about getting a tattoo?
## Which is not a temporary account accounting?
The type of service provider depends on what they offer, so you might hire an accountant if you is service revenue a permanent account need tax advice or take your car to get fixed at a mechanic’s shop if something breaks down.
In the event of a loss for the period, the income summary account needs to be credited and retained earnings reduced through a debit. Temporary accounts are not carried onto the next accounting period.
## Are unearned revenues a credit?
Owner’s drawings account is not a permanent account. This is because the drawing account of the owner is closed to the owner’s capital account at the end of each year, and the next year it begins with the balance of zero. On the other hand, owner’s retained earnings, owner’s capital account are all permanent accounts. Retained earnings is the cumulative money left after all expenses are done. It is not closed at the end of the accounting period.
The amount of net income for the period. In preparing closing entries a. Each revenue account will be credited.
## Related Posts
A corporation’s temporary accounts are closed to the retained earnings account. The temporary accounts of a sole proprietorship are closed to the owner’s capital account. Closing an account means exactly what it says.
### Is service revenue a current liability?
No, service revenue is not a current asset for accounting purposes. A current asset is any asset that will provide an economic value for or within one year. Service revenue refers to revenue a company earns from performing a service.
Business owners love Patriot’s accounting software. Businesses typically list their accounts using a chart of accounts, or COA. Your COA allows you to easily organize your different accounts and track down financial or transaction information. https://business-accounting.net/ Free access to premium services like Tuneln, Mubi and more. Daniel Liberto is a journalist with over 10 years of experience working with publications such as the Financial Times, The Independent, and Investors Chronicle.
In order to close the owner’s drawing account, the a. Income summary account should be debited.
• There are courses online for just about anything, and you can charge whatever pricing you feel is right for your instruction.
• Our events are great thanks to amazing people like you.
• I used to host estate sales online and offline, and it can earn you a nice amount of money if you have customers with a full house of stuff.
• Consequently, you should make sure that you reserve the first four hours of your work day, for the most important work.
• Whatever small amount of time you do have matters, because the truth is “you’re never going to have an uninterrupted 38-hour block of time to knock out a project,” he says.
• Harry at The Ride Share Guy has a lot of great resources about getting started, including a course about starting to RideShare for Profit.
Get your images on something physical and sell them as hangable and wearable art. Getting quality prints of your photos is easy and cheap these days, so there’s a real market if you’ve got the right image. Side hustles require an investment, but with a full-time job on the go, you might be able to leverage some of this income to get it started. Whether it’s for financial reasons or through a lack of fulfilment, many South Africans embrace a hybrid model that extracts the best of both worlds, called side hustles. If executed correctly, a side hustle retains the security and comfort of a day job but allows for the exploration of a passion or a money-making venture outside of its confines. And for some, it’s a less intimidating way to transition from the confines of the corporate world to the untapped possibilities of an entrepreneurial one. Creating extra income with a side hustle is a great way to pay off debt, build your emergency fund, or invest for retirement.
## Hack your wellbeing, productivity and goals
The only downside to a part-time job is that you need to invest your time for money.. The manufacturer not only carries the inventory but also ships it directly to the customer for you.
• These can be ongoing side hustles, like a weekly or bi-weekly service, or one time deals .
• Because you’re hands on with your products and suppliers, you’re able to keep a close eye on the quality of your products.
• If you have time on the weekends, and are okay lifting heavy boxes and furniture, you could join up with a moving service and get paid to help people move.
• At my peak, I was making over \$2,000 per month buying items and reselling them on eBay.
• In fact, 83% of people who find their jobs through a current contact do so through people they see only occasionally, if at all.
• It’s hard to hustle in something you don’t believe in, so don’t be afraid of self-promoting and showing your belief in yourself and your startup.
Sometimes, the best advice is to try something else or to shelf your idea for another time so that you don’t end up doing something you regret. Starting a side hustle with a conflict of interest could lead to costly lessons learned.
## Does the idea align with your passions and interests?
Since the rise of the technological era, it’s never been easier to connect with others at the touch of a button. Despite the advantages this provides, an increase in texting and socializing on social media may result in less substantial relationships, leaving users feeling isolated and disconnected. Snuggling services offer a new way to fill this void and respond to a growing need without judgment. Although not every hobby should be burdened by the need for profitability, most of us have student loans and bills to pay. By our definition of a good side hustle, we are looking to create some kind of return on time invested.
### My side hustle earns me £1,000 a month and it doesn’t even feel like work… – The Sun
My side hustle earns me £1,000 a month and it doesn’t even feel like work….
Posted: Mon, 03 Oct 2022 08:08:29 GMT [source]
The Handshake app allows you access to the wholesale marketplace where you can choose from thousands of products that you can purchase and sell in your store. Developing and building relationships with those who are brilliantly doing what you want to do, and then finding ways add value is highly significant on How To Hustle and win. Therefore, it’s important to create a list of specific goals, immediate next steps, and deadlines. A clear understanding of WHY you’re hustling so hard will give you a reason to keep your head down and focus on the difficult days, instead of giving up. Anything can be a hustle, if you’ve got the talent for it. If you’re good at basketball, you can build a basketball hustle around playing guys down at the park. If you’re really good at video games, a video game hustle might be in the cards.
## Online Freelancing
Starting a side hustle is a way to teach yourself valuable skills and help yourself grow as a professional and as an entrepreneur. In our free 40-minute video workshop, we’ll get you from product idea to setting up an online store to getting your first print-on-demand sale. Print-on-demand businesses are fun, low-risk ventures if you’ve got a passion for design and are looking to dip your feet into the entrepreneurial waters. As with grocery delivery, food https://wave-accounting.net/ delivery has become increasingly popular over the past couple of years. Food delivery is an easy-to-start side hustle, with a ton of apps like UberEats, DoorDash, Grubhub, Postmates, SkipTheDishes , and Menulog where you can sign up to become a driver. These days there are several options for those wanting to drive for a rideshare company. Sage Business Cloud Accounting software can easily track income and expenses from anywhere and on any device.
### Your Side Hustle Success Begins With Finding The Money – Forbes
Your Side Hustle Success Begins With Finding The Money.
Posted: Sun, 18 Sep 2022 07:00:00 GMT [source]
Prepare to burn the midnight oil in order to see your startup be successful. Be ready for haters and don’t let them slow you down. One characteristic of hustlers is their ability to brush off criticism and rejection without getting embarrassed or upset.
## How to Make Money Online: 28 Real Ways to Earn Money Online
Entrepreneurial and working life can be taxing, chaotic and at times, unplannable. And really, you don’t need a huge, detailed plan of everything . However, if you’re so focused on moving forward at any cost, you risk sacrificing key elements of achieving goals and frankly, wasting your time. In his defense, the sentiment of that section is in general, to work super hard at reaching your goals and achieving your dreams. We all should find the drive to work at something that brings us happiness and fulfillment.
• Taking consistent and intentional, action steps towards your goals is a key strategy to building momentum and success in your business or career.
• If you can successfully master this skill , you will have the world at your fingertips.
• Websites like Branded Surveys, Swagbucks, and Survey Junkie have a ton of surveys you can get paid to take.
• Nearly everyone could stand to benefit from a little extra income, but sometimes saving extra money and having a full-time job aren’t enough.
• But, try as you might, sometimes you just won’t want to work.
• Trust and the ability to let go and delegate are very important for any leader – you can’t, and shouldn’t, do it all yourself.
Sometimes one person can be your FTE 100%, or FTE 1.0, which is another way to call it. An FTE calculation for all employees in a company needs to be rounded down to the nearest whole number (usually, 1.0 FTE or greater). Using % Availability, will take the incoming number and convert it to hours using the Resource’s Calendar when entering data into a record tied directly to a specific resource in Assignments and Staffing. Inpatients are medical, surgical, maternity, specialty, and intensive-care unit patients whose length of stay exceeds 23 hours.
• If you are such an employer, then you are eligible to apply for a tax credit in the amount of 50% of employer-paid health care premiums.
• However, bear in mind that this only applies to positions that work 30 hours per week when assigned with a 0.75 FTE position, i.e. the IRS minimum for a full-time definition.
• Divide the total hours worked by the number of full-time hours for the given time period to find the FTE.
• It also allows employers to examine total hours worked across the business rather than looking only at how many full-time and part-time employees are on staff.
• From there, if your full-time hours are 40 per week, divide the number by 2,080.
• Alternative measures of FTE that incorporate these additional assumptions can place the number of hours for one FTE as low as 1,680 hours per year.
She has experience working as an accountant in public accounting firms, nonprofits, and educational institutions, and has also honed her communication skills via an MA in English, writing jobs, and as a teacher. She received her CPA from the Accountancy Board of Ohio in 1994 and has a BS in Business Administration/Accounting. Holiday hours and other paid leave are already accounted for as part of the hours worked so you don’t need to make any special calculations. If your business has a 32- or 35-hour workweek, then multiply the number of employees by that amount instead of 40.
## FTE Calculation Examples
Each employee who worked less than 40 hours per week on average during a specific calculation period counts as 0.5 FTE and is viewed as a part-time employee. Each employee who worked more than 40 hours per week on average during a specific calculation period counts as 1.0 FTE and is viewed as a full-time employee. Once you have a list of employees, you’ll need to determine how many hours each employee works each week.
This calculator helps understand the future penalties that employers are faced with pertaining to the health care coverage that is supplied to their employees. Employers Resource encourages you as an employer to contact an HR Specialist for any further explanation. full time equivalent ‡– Enter total number of hours all part time employees work each month. This is the sum of the hours of each employee that works less than 30 hours a week. This is the number of employees you have working for you every month that work 30 or more hours a week.
## Using an online calculator
Project management can be a hassle when using both full-time and part-time employees. For that reason, companies may need to calculate the full-time equivalent. It allows companies to determine the number of full-time employees they employ.
Besides ALEs needing to know their FTE status to comply with the employer mandate or not, understanding how many FTEs you have at your organization is good for your company’s metrics. Being able to track your employees’ workloads and output between your full-time employees and FTEs gives you more insight into how efficiently your employees are working. Below, Equations 1 and 2 indicate the formulas to calculate the FTE. If you have an estimate for how long a work project will take, divide the estimated hours by the hours in a workweek. The result is the number of full-time equivalent employees you’ll need to complete the project. To get FTE for the year, divide the total annual hours worked by 2080, which assumes a 40-hour workweek for 52 weeks of the year.
## Company
No employee can be greater than 1.0 FTE, as overtime is not counted for this calculation. Small businesses are eligible for small-employer health care tax credit — they apply to employers who employ less than 50 full-time equivalent employees. They also use it to calculate labor costs for the expected https://www.bookstime.com/ project workload and subsequently define the amount of funds needed to carry out the project to its end. In our example, the company gives all its workers the same four weeks off, but employees may have different benefits packages in some companies, which may change the FTE calculations.
• You can seamlessly collaborate with remote and in-person colleagues in a professional and stylish office space.
• The FTE computation is based on the total number of credits carried by all students enrolled in classes at a specific level, also divided by the number of credits in a full-time load.
• Therefore, companies must start by defining what full-time is.
• According to the IRS and the ACA, part-time employees work fewer than 30 hours a week on average.
• The Paycheck Protection Program is designed to prevent job losses during the pandemic by giving loans to small businesses to cover payroll expenses and other costs.
• 💡To track employee work hours and decide whether your employees are full-time equivalents, try Clockify, our free time tracker for teams.
• The Full-Time Equivalent is the number of hours a business considers a full-time employee to work.
The Paycheck Protection Program is designed to prevent job losses during the pandemic by giving loans to small businesses to cover payroll expenses and other costs. A portion of this loan can be forgiven based on the number of FTE workers an employer has on its payroll. Back to our sample company ABC, employee Gina works 0.25 times the amount of a full-time employee.
## How to Calculate FTE
Next, she uses this figure to estimate the company’s health (awesome, of course!) by comparing it to the amount Toggl Track needs to operate properly. By examining revenues instead of profits , you can judge profitability on an employee’s actual work, not your other costs (ad costs, etc.). Of course, employee profitability factors heavily into hiring decisions. Remember, to average the number of employees you employ across the year, especially if you run a seasonal business.
If you are using FTE calculations to determine future hiring needs, it makes sense to adjust the number of hours to fit your business. A full-time equivalent employee is a worker with an FTE of 1.0, which will usually mean they’re employed full-time at your company. A full-time employee is considered to work at least 30 hours per week on average. Businesses may find it difficult to determine the salary of part-time workers, especially when they work only a few days a week. Using the FTE calculations, a company can clearly see how much the employee works in comparison to a full-time employee. You’ll also need to know the actual number of weekly work hours to complete the FTE calculation. Actual hours worked per week is the exact number of hours an employee worked.
Learn financial statement modeling, DCF, M&A, LBO, Comps and Excel shortcuts. How to track your team’s time off Here’s how you can easily track your team’s time off in Clockify…. Business owners who want to understand whether their business is eligible to apply for a Paycheck Protection Program. A 100% FTE is the same as a 1.0 FTE — it may point to one person working a full-time schedule or several people fulfilling the duties of 1 full-time position.
### Appleton’s 2023 budget would raise city property taxes by 6.6%, in part due to heavy debt load – Post-Crescent
Appleton’s 2023 budget would raise city property taxes by 6.6%, in part due to heavy debt load.
Posted: Wed, 12 Oct 2022 10:01:30 GMT [source]
This is similar to how certain laws only apply to a full-time employee, but not a part-time employee. Let’s say that you need to calculate the total amount of hours, time, and money required for your business to run successfully.. FTE is the answer, full time equivalent will help you do this.
The Employee Retention Credit is another measure designed to incentivize employers to keep hold of staff during the COVID-19 pandemic. It comes in the form of a tax credit worth half of employee wages over the course of a covered period. The Full-Time Equivalent is the number of hours a business considers a full-time employee to work.
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https://preprod.motherjones.com/kevin-drum/2013/07/apparently-undergraduate-finance-students-believe-esp/ | 1,659,891,048,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00129.warc.gz | 454,403,371 | 44,039 | # Apparently Undergraduate Finance Students Believe in ESP
Via Andrew Sullivan, Chris Dillow points us toward a weird experiment in a fourth-year finance class at the Autonomous University of Barcelona. One student flipped a coin five times and a group of 20 students predicted heads or tails for each toss. Then a second group of students was told that there would be another set of five coin tosses. The best and worst guessers from the first group would try their luck again, and everyone in the second group would earn money for each coin flip that the worst guesser got right.
But there’s more! If students in the second group were willing to pay for the privilege, they could instead earn money for each coin flip that the best guesser got right. There’s no reason to do this, of course, since guessing is just guessing. Nonetheless, 82 percent of the students paid to switch to the better guesser.
But there’s even more! Just to add some baroque complexity to the whole thing, students in the second group had to say beforehand how much they’d pay to switch from the worst guesser to the best guesser, and they had to provide a different payment depending on how good the guessers were. For example, maybe you’ll pay one euro to switch to a guesser who outpredicted the worst guesser by two flips, but you’ll pay two euros to switch to a guesser who outpredicted the worst guesser by three flips. Then the best and worst guessers were introduced, along with their guessing records, and the professors employed yet another baroque method to decide which students were allowed to switch. Apparently they did this solely to get a matrix of prices from the students. Here’s what they got:
Sure enough, students were willing to pay more depending how much better the top guesser was compared to the worst guesser. So what does this all mean? Here are some guesses:
1. Fourth-year finance students in Barcelona have no clue about the independence of coin tossing events.
2. FYF students like to play mind games with their professors.
3. FYF students believe in telekinesis.
4. FYF students believe in supernatural forecasting abilities. (N.B. This would explain much Wall Street behavior.)
5. FYF students in this class didn’t really understand the experiment.
6. FYF students are skeptical sorts and figured there was a trick involved in all this.
7. The sample size was too small to be meaningful.
8. Something else about the methodology invalidates the results.
### We've never been very good at being conservative.
And usually, that serves us well in doing the ambitious, hard-hitting journalism that you turn to Mother Jones for. But it also means we can't afford to come up short when it comes to scratching together the funds it takes to keep our team firing on all cylinders, and the truth is, we finished our budgeting cycle on June 30 about \$100,000 short of our online goal.
This is no time to come up short. It's time to fight like hell, as our namesake would tell us to do, for a democracy where minority rule cannot impose an extreme agenda, where facts matter, and where accountability has a chance at the polls and in the press. If you value our reporting and you can right now, please help us dig out of the \$100,000 hole we're starting our new budgeting cycle in with an always-needed and always-appreciated donation today.
### We've never been very good at being conservative.
And usually, that serves us well in doing the ambitious, hard-hitting journalism that you turn to Mother Jones for. But it also means we can't afford to come up short when it comes to scratching together the funds it takes to keep our team firing on all cylinders, and the truth is, we finished our budgeting cycle on June 30 about \$100,000 short of our online goal.
This is no time to come up short. It's time to fight like hell, as our namesake would tell us to do, for a democracy where minority rule cannot impose an extreme agenda, where facts matter, and where accountability has a chance at the polls and in the press. If you value our reporting and you can right now, please help us dig out of the \$100,000 hole we're starting our new budgeting cycle in with an always-needed and always-appreciated donation today. | 904 | 4,234 | {"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-2022-33 | longest | en | 0.972555 |
https://www.freemathhelp.com/forum/threads/109797-How-do-I-apply-multivariate-regression-to-my-dataset-to-formulate-an-equation?p=423266 | 1,550,374,511,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481612.36/warc/CC-MAIN-20190217031053-20190217053053-00577.warc.gz | 842,985,698 | 12,688 | # Thread: How do I apply multivariate regression to my dataset to formulate an equation?
1. ## How do I apply multivariate regression to my dataset to formulate an equation?
I have the following data points (for example)
x, y, z, a, b
Lets say those are each a column in a database (or Excel) and I have 1000 rows of data (so 5000 total values). I need to create an equation that solves for "b". How do I go about doing this? Excel will do it based on a collection of just x and y values, but I don t know if Excel goes higher than two variables? Does it? If so, how do I do that? DO NOT ASSUME THAT THIS IS A LINEAR EQUATION.
2. ## How do I apply multivariate regression to my dataset to formulate an equation
Say I have the following arbitrary dataset. I want to create an equation that solves for 'b'
x y z a b 5 1.2 20 13 5.6 6.2 1.6 16 15 4.5 6 1.3 18 12 4.9 5.8 1.7 21 12 6.3
How do I go about doing this? Excel will do it based on a collection of just x and y values (using linear or non-linear regression), but I don't know if Excel goes higher than two variables? Does it? If so, how do I do that?
Are there any known calculators where you give it a dataset and it spits you out an equation??
DO NOT ASSUME THAT THIS IS A LINEAR EQUATION.
3. Excel's Data Analysis Tool Pack will do Multiple LINEAR Regression for you.
Excel is sufficiently flexible that you can design any sort of thing you like.
We're really about Math Students, here. The least you can do is show us YOUR work, rather than expecting us to be free consultants.
4. There are infinitely many equations that will reproduce your data. No finite number of values will fully define "an equation" without ambiguity.
5. Originally Posted by tkhunny
There are infinitely many equations that will reproduce your data. No finite number of values will fully define "an equation" without ambiguity.
I understand this. So I need an equation that "best fits" my data within some small margin of error. I'm not expecting it to be perfect.
6. 1) "solves for" means you DO want it to be exact. More precise language is always helpful.
2) You must formulate some sort of model. Where do the data come from? What sort of model might be appropriate? Are there ANY known relationships? How sure are you that all your variables are independent? Are there linear relationships or are they ALL nonlinear? A scatter plot of 'a' vs 'b' might suggest a linear relationship.
3) Are there any dependencies inside a single variable?
4) Excel does just fine with Multiple-LINEAR Regression. It's in the Data Analysis Tool Pack. p-values and everything.
5) Sometimes, you can make non-linear relationships linear by applying a logarithm or square root transformation. There are other sorts of transformations. Can you see any opportunities for transformations?
6) What sort of nonlinear relationships are you anticipating or imagining? Just polynomials of order higher than one (1)? Reciprocals? Rational Functions? Trig Functions? Logistics?
7) Ever hear of "R"? You can pursue any sort of GLM with enough R experience. https://www.r-project.org/about.html To be honest, you'll probably need more data to get very far.
8) Is this a student problem, a personal exploration, or maybe a business requirement? Purpose and audience are just as important as methods.
We're happy to help, but you have to show your work and give us something to go on. We all have opinions on how one might proceed, but if we are going to get YOUR views that lead to YOUR personal solution, you will have to offer up quite a bit more information.
7. Just noticed that you might find something with f(x+y+z+a) = b. It's not a GREAT linear relationship between the sum and the 'b', but it might be some place to start.
It actually gets a little better if you Square the Sums, f((x+y+z+a)^2).
The sum of the squares is even better! f(x^2 + y^2 + z^2 + a^2).
Plenty of nonlinear interactions in all of that.
Really, though, you're not going to get very far with this VERY TINY dataset.
8. My dataset is actually about 4.6 million rows of data.
Recording these data points actually turned out to not be an accurate way to solve my ultimate problem. Because after digging deeper I found I had even more variables.
Instead, I decided to go the physics route. It was complex physics, but it ended up solving my problem. | 1,037 | 4,358 | {"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} | 3.5 | 4 | CC-MAIN-2019-09 | latest | en | 0.911608 |
https://bj21.com/articles/card-counting/singledeckgameselection | 1,723,393,665,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00225.warc.gz | 99,710,931 | 7,751 | # Single Deck Game Selection
This is presented as a general guide, primarily to assist players who may not be accustomed to single-deck games. Using the summary at the end of this presentation, such players will be able to quickly determine under which conditions a single deck game is worth playing, and which conditions to avoid.
No attempt is made to differentiate between various rules, such as d10 ("Reno rules"). We will deal only with the number of players present at the table and penetration, with the results reduced to simple conclusions and recommendations.
Obviously, there will be exceptions, such as when hole card or other useful information is available, that would make otherwise poor conditions or games worthwhile. When playing primarily for comps, rather than strictly for the money, some otherwise poor conditions may be tolerable as well.
For those not familiar with the terms “Rule of 6” (Ro6), “Rule of 7” (Ro7), etc, here is the explanation for those terms. The “rules” mean that the number of rounds to be dealt in a single-deck game is determined by subtracting the number of players (or spots being played) from the “rule” number. For example, Ro6 means 6, minus the number of spots being played = number of rounds to be dealt. So, if two spots are being played, four rounds will be dealt (6 – 2 = 4). Ro7 with three spots being played will result in four rounds dealt (7 – 3 = 4).
Some single-deck games are Ro7. Unfortunately, Ro6 is the more common. The difference is significant. In heads-up play, however, many dealers will deal more than the specified number of rounds. In many otherwise Ro6 games, a heads-up player will receive 6 or even 7 rounds. In a few single-deck casinos, Ro5 is used. Such a game is a total waste of time, and shouldn’t even be considered.
In heads-up play, the dealer will bust less than when there is more than one player. This is because when the heads-up player busts, the hand is over, and the dealer does not get the opportunity to bust. This slightly reduces the average number of cards per hand in a heads-up game. The difference is negligible, though. For this analysis, we’ll use the generally accepted “average of 2.7 cards per hand.”
6/5 "blackjack" (single deck or otherwise) should be avoided by everyone except those few with particular skills that are well beyond the scope of this article. Possibly with very deep penetration (perhaps 8 rounds heads-up?) and casino tolerance of a huge bet spread, 6/5 could be beaten. But such conditions are unlikely to be found, and searching for them is probably going to be a waste of time.
8 rounds -- an excellent but rare game
7 rounds -- an excellent game
6 rounds -- an acceptable game
5 rounds -- a poor game. If you are faced with this game, play two hands. You will normally get 4 rounds to 2 spots under Ro6, which is playable. You can drop to one hand in negative decks. You may get lucky and get 5 rounds to 2 spots, which is an excellent game.
You and only one other player at the table, playing one spot each
5 rounds to 2 spots (Ro7) -- an excellent game
4 rounds to 2 spots (Ro6) -- an acceptable game
3 rounds to 2 spots (Ro5) -- a poor game. Do not play.
You and two other players at the table, playing one spot each
4 rounds to 3 spots (Ro7) -- an excellent game
3 rounds to 3 spots (Ro6) -- a poor game. Your hands per hour will be greatly reduced, and you will frequently have to go directly from minimum bet on Round 2 to maximum bet on Round 3, making the bet variations severe and quite obvious. If the game has side bets, such as Royal Match, expect the ploppies to slow the game down even more.
Unless you are betting in black units, the “3 rounds to 3 spots” game will not be sufficiently profitable to be worth your time. If you’re at a table with only one other player, and a third player joins, take a restroom break, and hope one or both have left the table by the time you return. If it appears that both other players will be there for awhile, look for better conditions elsewhere.
With you and three or more other players at the table, in most cases single-deck games are a waste of time. Once in a while you will see 3 rounds to 4 spots (Ro7), but the same problems as the “3 to 3” game will be present. Also, in such a game, frequently in high counts, you will not get the third round, because the dealer may fear running out of cards and may shuffle instead.
----------------------------------------------------------------------
SUMMARY:
8 rounds, PLAY
7 rounds, PLAY
6 rounds, PLAY
5 rounds, DO NOT PLAY, unless you can play two hands
You and one other player
5 rounds, PLAY
4 rounds, PLAY
3 rounds, DO NOT PLAY
You and two other players
4 rounds, PLAY
3 rounds, DO NOT PLAY
You and three or more other players
DO NOT PLAY.
----------------------------------------------------------------------
There may be disagreement from some experienced single-deck players on my recommendation to not play “3 to 3” or “3 to 4”, but I believe for the vast majority of players who don’t play single deck regularly enough to accurately judge game speed, the advice to “not play” is superior to “play” in these marginal situations. | 1,185 | 5,201 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-33 | latest | en | 0.971775 |
https://www.goldsim.com/Courses/ContaminantTransport/Unit2/Lesson4/ | 1,719,109,208,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00455.warc.gz | 682,391,359 | 9,200 | # Lesson 4 – Creating Vectors
When you define a vector or a matrix, it must be defined relative to a particular set of Array Labels. A set of Array Labels is simply a collection of labels for the items of the array. The labels can be numbers (e.g., 1, 2, …, 10) or names (e.g., apples, oranges, peaches, pears). When you define an element which has an array as an output, you must specify the set(s) of Array Labels upon which the array is based. Vectors require a single set of Array Labels, and matrices require two sets of Array Labels (one for the rows, one for the columns). You define and reference a particular item of an array by using these labels.
To illustrate this, let’s create two vectors right now. Open GoldSim and start with a new model.
You can create (and edit) sets of Array Labels by selecting Model|Array Labels… from the main menu or by pressing the Array Labels button in the Advanced toolbar:
The following dialog for defining the Array Labels will be displayed:
For convenience, GoldSim provides two examples sets of Array Labels: Days has seven items (the days of the week); and Months has twelve items (the months of the year). There are also two additional sets of Array Labels here: Elements and Species. These are automatically added when the Contaminant Transport Module is activated (they would not be present if you deactivate the module). In fact, these two sets are a fundamental part of the Contaminant Transport Module, and we will discuss them in detail in subsequent Units. For now, you can ignore these four “built-in” Array Label sets.
Let’s add a new (custom) set now by pressing the Add New Set… button. Note that a dialog for specifying the type of Array Label Set will be displayed:
You must select whether you wish to define a Named Set or an Indexed Set. In a Named Set, each item in the set has an alphanumeric name. In an Indexed Set, each item in the set is an integer (based on a specified Start Index and End Index).
Let’s created a Named Set. If you press that button, the following dialog for editing the set is displayed:
You assign a name (a Set ID) to the new set of Array Labels (by default it will be named Setn). Name this set “Fruit”. You add and remove entries using the Add and Remove buttons.
Note: After you add two or more entries, the Move Up and Move Down buttons become available, allowing you to move entries up or down in the list.
Let’s add three items: Apples, Oranges and Pears. When you are done, the dialog should look like this:
Note: The Array Label editing dialog also allows you to define the Styles that are used when an element based on the set is displayed in a result chart. The Base Style, in combination with the Auto ColorsVary Lines and Vary Symbols options allow you to automatically generate the Styles by pressing the Reapply button. Alternatively, you can click on any Style in the list to manually specify the Style attributes. As we will see soon, the Show checkbox determines whether or not the item is displayed in result charts.
Press OK to close this dialog. You will note that the new Array Label set is listed:
Now press Close to close the Array Labels dialog.
Let’s create a Data element that represents a vector of Fruit.
First, insert a Data element. After doing so, change the Element ID to “Crop”, the Description to “Fruit crop” and set the Display Units to kg.
Next press the Type… button and define the Order as “Vector (1D-Array)”:
You will note that you are prompted to enter a set of Row Labels. Expand the Row Labels drop list by clicking in “Select Array Labels…”:
The five available Array Labels are listed (the four “built-in” sets and the set of Fruit that we just created).
Note: There is also an option to “Create New Array Labels…”. If you select this, it will open the dialog for creating new Array Labels that we used above.
Select “Fruit” and press the OK button.
Note that the Data element dialog now looks like this:
The input field for the Data element has been replaced with a button (Edit Data…). Press that button now:
You can see that there is a separate input field for each item of the vector. Enter 5 kg, 7 kg and 10 kg for the three items now, and press OK. Then press the OK button to close the element dialog.
Now left-click on the output interface for the element:
What you will note here is that the Crop element can be expanded in the output interface (by left-clicking on the arrow):
Each item is displayed (in brackets). If you hold your cursor over an item, its value is displayed.
Let’s finish this Lesson by creating a vector in a different way. Instead of using a Data element, we are going to create and define the values for the items of a vector using an Expression element.
Let’s start by inserting a new Expression element. Name it Crop2, and give it Display Units of kg. Then press the Type… button, and define it as a “Vector (1-D Array)” with Row Labels of “Fruit”. The element will look like this:
You will note that the input field is colored red. This is because the default entry (the scalar value 0 kg) is incompatible with the type (a vector of Fruit).
We are going to create a vector using a special function called a vector constructor. In particular, GoldSim provides a special array function for creating vectors. Array functions are like other functions provided by GoldSim (e.g., sin, cos, max, log). In which you provide a function name, followed by one or more arguments in parentheses (e.g., log(X), min(A,B,C)). However, they operate on and/or produce arrays.
The vector constructor is structured like this:
vector(Array Label Set, Value1, Value2, …)
where the first argument represents the name of the Array Label Set for the vector you are creating, and the subsequent arguments represent the values (and units) for each of the items.
So, for example, we could create a vector similar to the one we created using the Data element (in this case, it has slightly different values) by entering the following:
Press OK to close the dialog.
Now left-click on the output interface for the element (just like we did previously for the Data element):
Expand the output in the interface (by left-clicking on the arrow):
As was the case for the Data element, each item is displayed (in brackets). If you hold your cursor over an item, its value is displayed.
Note: There are also several shortcuts you can use in the vector constructor function. If you just provide a single value, it is assigned to all of the items. For example, vector(Fruit, 10 kg) would assign the value 10 kg to all items. You can also leave out the Array Label set name (in which case the Array Label set specified for the entire element will be assumed). For example, vector(0 kg) would assign the value 0 kg to all items. Similarly, vector(10 kg, 12 kg, 11 kg) assigns values to each item in the vector, as long as the number of items in the constructor matches the number of items in the array.
You can read more about creating arrays in the GoldSim Help file.
Save this model to the “MyModels” subfolder in the “Contaminant Transport Course” folder on your Desktop (by pressing File|Save from the main menu, Ctrl+S, or the Save button in the toolbar). You can name it Vectors.gsm.
Now that we have created some vectors, we will build on this simple example to see what we can do with them in the next Lesson. | 1,736 | 7,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-26 | latest | en | 0.742774 |
http://forums.wolfram.com/mathgroup/archive/2005/Apr/msg00064.html | 1,725,874,572,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651092.31/warc/CC-MAIN-20240909071529-20240909101529-00408.warc.gz | 12,812,317 | 9,819 | Re: Plot - where is y scale?
• To: mathgroup at smc.vnet.net
• Subject: [mg55692] Re: [mg55667] Plot - where is y scale?
• From: DrBob <drbob at bigfoot.com>
• Date: Sun, 3 Apr 2005 05:51:00 -0400 (EDT)
• References: <200504020627.BAA10735@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Plot[A, {x, 150, 500}, PlotRange -> All]
Also, I recommend you not start your own variables with capital letters (like A). That will prevent conflicts with built-in symbols like N, E, and others.
Bobby
On Sat, 2 Apr 2005 01:27:58 -0500 (EST), Paawel <fonfastik at interia.pl> wrote:
> Hi
> I fitted equation to experimental data.
> When I want to plot the equation X axis goes as it is set, but the
> scale on the Y axis ends and plot ends in the middle of the X axis.
> Here is data and equation.
> When I use command DisplayTogether I can see the equation in the whole
> range of Y.
> What do I do wrong??
>
> << Statistics`NonlinearFit`
>
> data = {{173.2, -0.0119729}, {174.5, -0.0203019}, {
> 175.7, -0.00364394}, {176.8, -0.0119729}, {178.1, -0.0119729},
> {179.5, \
> -0.00780843}, {
> 180.6, -0.00780843}, {182.1, -0.0161374}, {183.7, 0.000520562},
> {185.4, \
> -0.0119729}, {187.5, -0.0203019}, {188.8, 0.000520562}, {190.1,
> 0.00468506}, \
> {191.7, 0.00884956}, {193.1, -0.00364394}, {194.2, 0.000520562},
> {195.1, \
> -0.00364394}, {196.5, 0.00468506}, {
> 197.7, -0.00780843}, {198.8, 0.00468506}, {199.8, 0.000520562},
> {200.8, \
> 0.000520562}, {201.8, 0.00884956}, {203, 0.000520562}, {
> 204.2, 0.00468506}, {205.3, 0.00468506}, {
> 206.3, 0.00468506}, {207.6, 0.00468506}, {
> 208.7, -0.00364394}, {210.6, 0.00468506}, {211.6, 0.00884956}, {213,
> \
> -0.00364394}, {214.3, 0.0130141}, {215.6, 0.00468506}, {216.4,
> 0.00884956}, {217.1, 0.000520562}, {218.1, -0.00364394}, {
> 218.5, 0.00468506}, {219, 0.00884956}, {220.1, 0.00884956},
> {222.1, \
> 0.00468506}, {223.8, 0.000520562}, {225.7, 0.000520562}, {227.2,
> 0.00468506}, \
> {228.7, 0.00468506}, {229.9, 0.000520562}, {
> 231.2, 0.000520562}, {231.9, 0.000520562}, {233, -0.00364394},
> {235.2, \
> -0.0119729}, {237.2, -0.0119729}, {238.9, -0.00364394}, {240.8,
> -0.00364394}, \
> {242.9, -0.00780843}, {244.9, -0.00780843}, {246.1, 0.000520562},
> {247.5, \
> -0.00364394}, {248.5, -0.0119729}, {248.9, -0.0119729}, {249.9,
> -0.0119729}, \
> {251.5, -0.00780843}, {252.8, -0.00364394}, {253.9, -0.0203019},
> {254.6, \
> -0.0161374}, {254.5, -0.0286309}, {254.7, -0.00780843}, {
> 255.4, -0.0119729}, {256.9, -0.0203019}, {
> 257.7, -0.0119729}, {258.7, -0.0119729}, {
> 260.2, -0.0161374}, {261.6, -0.0161374}, {
> 262.4, -0.0119729}, {263.3, -0.0161374}, {264.6, -0.0119729},
> {266.1, \
> -0.0119729}, {267.7, -0.0161374}, {269.8, -0.0119729}, {272,
> -0.0203019}, {
> 273.6, -0.0244664}, {275, -0.0119729}, {
> 276.7, -0.0203019}, {278.7, -0.0161374}, {
> 280.8, -0.0161374}, {282.8, -0.0119729}, {284.7, -0.0119729},
> {286, \
> -0.00364394}, {286.6, 0.000520562}, {
> 287.6, -0.00364394}, {288.7, -0.0119729}, {289.9, -0.0161374},
> {291.4, \
> -0.00780843}, {292.9, -0.00780843}, {293.6, 0.00468506}, {294,
> 0.000520562}, {295, 0.000520562}, {295.9, -0.00364394}, {296.5, \
> -0.0161374}, {297.5, 0.00468506}, {299, 0.000520562}, {299.7,
> 0.000520562}, \
> {300.4, 0.0130141}, {301.2, 0.0171786}, {302.6, 0.00884956}, {304,
> 0.0171786}, {305.5, 0.0171786}, {306.9,
> 0.0255075}, {308.3, 0.0338365}, {309.6,
> 0.0255075}, {310.7, 0.029672}, {311.9, 0.038001}, {313, 0.038001},
> \
> {313.7, 0.054659}, {314.5, 0.054659}, {315.5, 0.054659}, {316.6,
> 0.062988}, \
> {317.9, 0.054659}, {319.9, 0.062988}, {321.9, 0.0838105}, {323.6,
> 0.0671525}, \
> {324.1, 0.0838105}, {323.4, 0.096304}, {324.3, 0.087975}, {
> 326.4, 0.0921395}, {329, 0.100469}, {
> 330.2, 0.112962}, {331.6, 0.112962}, {333.5, 0.117126}, {335.6, \
> 0.133784}, {336.4, 0.137949}, {336.4, 0.133784}, {337.2, 0.146278},
> {338.1, \
> 0.150442}, {338.9, 0.162936}, {340.1, 0.162936}, {341.7, 0.171265},
> {343.7, \
> 0.183758}, {344.9, 0.183758}, {346.2, 0.204581}, {347.8, 0.200416},
> {348.4, \
> 0.204581}, {349, 0.21291}, {349.4, 0.217074}, {350.6, 0.225403}, {
> 351.9, 0.233732}, {353.8, 0.246226}, {
> 355.2, 0.254555}, {357.1, 0.262884}, {
> 358.9, 0.279542}, {360.2, 0.279542}, {360.9, 0.300364}, {
> 362.6, 0.300364}, {364.1, 0.308693}, {
> 364.4, 0.317022}, {365.6, 0.321187}, {365.6, 0.329516}, {366,
> 0.33368}, {367.3, 0.342009}, {368.9,
> 0.346174}, {371.4, 0.354503}, {373, 0.37949}, {374.5,
> 0.37949}, {376.5, 0.387819}, {377.1, 0.404477}, {377.2, 0.41697},
> {
> 379, 0.425299}, {381.6, 0.433628}, {383.4, 0.437793}, {384.8,
> 0.450286}, {386.1, 0.458615}, {387.7,
> 0.471109}, {390.2, 0.479438}, {391.4, 0.491931}, {392.2, 0.496096},
> {
> 393.2, 0.504425}, {393.5, 0.516918}, {394.1, 0.516918}, {394.4,
> 0.525247}, {
> 394.2, 0.533576}, {394.6, 0.537741}, {395.4,
> 0.541905}, {396.7, 0.550234}, {398.5, 0.562728}, {400.2,
> 0.566892}, {
> 401.8, 0.579386}, {403.3, 0.590213}, {
> 405.3, 0.595627}, {407.4, 0.60229}, {409.6, 0.611452}, {409.8,
> 0.633108}, \
> {409.4, 0.62811}, {411.3, 0.638522}, {413.2, 0.647683}, {414.5,
> 0.658511}, {
> 415.4, 0.657262}, {417, 0.670588}, {417.9, 0.676835}, {419.4,
> 0.685997}, \
> {421.1, 0.693077}, {422.1, 0.701406}, {423.7, 0.704321}, {425,
> 0.710567}, \
> {425.4, 0.717647}, {424.5, 0.722644}, {424.1, 0.73139}, {425.2,
> 0.735554}, {427.9, 0.745133}, {430, 0.748881}, {431,
> 0.753878}, {430.9, 0.760541}, {431.3, 0.77012}, {430.8, 0.773451},
> {
> 430.6, 0.779282}, {431.5, 0.789276}, {433, 0.791775}, {435.3,
> 0.798022}, {437, 0.804269}, {440, 0.813847}, {443.7,
> 0.819261}, {447.7, 0.828006}, {449.2,
> 0.838001}, {449.7, 0.847996}, {447.6, 0.853826}, {446.4,
> 0.857158}, {
> 446.1, 0.862988}, {447.1, 0.86507}, {449.2, 0.869235}, {450.3,
> 0.870068}, {451.1, 0.874649}, {452.6, 0.880479}, {455.9, 0.88506},
> {459.3, \
> 0.887975}, {462.6, 0.895055}, {466.4, 0.897553}, {469.2, 0.905049},
> {470.8, \
> 0.908381}, {472.6, 0.914211}, {474.9, 0.915877}, {476.4, 0.918792},
> {477, \
> 0.920875}, {475.7, 0.923373}, {476.9, 0.925039}, {477.3, 0.927538},
> {477.7, \
> 0.930453}, {478.4, 0.92962}, {478.6, 0.931702}, {477.4, 0.934201}, {
> 474.8, 0.935034}, {475.9, 0.935867}, {477.9, 0.937116}, {477.5,
> 0.938365}, {
> 478.3, 0.941281}, {481.3, 0.941281}, {483.6, 0.941697}, {486.2,
> 0.944612}, {
> 487, 0.946278}, {486.8, 0.947527}, {487.8, 0.950026}, {487.4,
> 0.951275}, {
> 489.1, 0.952108}, {493.3, 0.953358}, {496.8,
> 0.955023}, {496.8, 0.955856}, {499.4,
> 0.957522}, {500.5, 0.957939}, {498.8,
> 0.958855}, {498.8, 0.959563}, {497.8,
> 0.960937}, {501.3, 0.962145}, {505.2,
> 0.962145}, {508.5, 0.963269}, {512.3, 0.96456}, {513.5, 0.96456}}
>
> A = NonlinearFit[data, Exp[-k0 ts Exp[-Ea/(8.31 x)]], {x}, {k0, Ea,
> ts}]
>
> Plot[A, {x, 150, 500}]
>
> ListPlot[data]
>
> << Graphics`Graphics`
>
> DisplayTogether[ListPlot[data,
> PlotStyle -> {Hue[
> 0.6], PointSize[0.003]}], Plot[A, {
> x, 150, 500}, PlotStyle -> {Hue[0], Dashing[{0.01}]}]]
>
>
>
>
>
> Pawel
>
>
>
>
--
DrBob at bigfoot.com
```
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• Next by thread: Re: Plot - where is y scale? | 3,823 | 7,427 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.623959 |
http://www.docstoc.com/docs/106403531/Microsoft-PowerPoint-Viewer---ConfidenceIntervals_update5 | 1,410,777,534,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657104131.95/warc/CC-MAIN-20140914011144-00114-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 477,513,489 | 14,799 | # Microsoft PowerPoint Viewer - ConfidenceIntervals_update5
Document Sample
``` Confidence Interval around a Percentile (CIP): Impaired <10% of the time (EPA guidelines),
with 80% confidence (TCEQ currently controls Type I error at 20% on listing).
If the confidence interval (CI) region is below the maximum criteria value for a parameter,
or is above the minimum critera value for a parameter, the waterbody is not considered
impaired.
A waterbody is an impairment concern if the CI region overlaps the criterion
for the parameter of interest.
A waterbody is impaired if the CI region is above the maximum criteria value for a parameter,
or is below the minimum criteria value for a parameter.
H1 = There is less than a 20% chance of falsely impairing a water body that is supporting
the criterion 90% of the time (or 75% of the time for bacteria).
26 Oct, 2004 TCEQ Austin, SWQM Stakeholders Meeting 1
Various confidence intervals around a percentile...
50% C.I.
50% C.I.
60% C.I.
60% C.I.
70% C.I.
70% C.I.
90th
80% C.I.
80% C.I.
90% C.I.
90% C.I.
Probability
Parameter value
26 Oct, 2004 TCEQ Austin, SWQM Stakeholders Meeting 2
Start/finish
80% C.I. FS
Criteria
Probability
Delist
Parameter value
NS 80% C.I.
80% C.I. T2
90th 90th percentile 90th
Probability
Probability
Determining impairment
Parameter value by evaluating the Parameter value
Confidence Interval
NS 80% C.I. around a Percentile (CIP)
90th 80% C.I. T2
Probability
90th
Probability
Parameter value
Parameter value
NS 80% C.I.
90th
Probability
List
Parameter value
Start/finish
80% C.I. FS
Criteria
Probability
Parameter value
NS 80% C.I.
80% C.I. T2
50th 50th percentile 50th
(ie: median)
Probability
Probability
Determining impairment
Parameter value by evaluating the
Parameter value
Confidence Interval
around a Median (CIM)
NS 80% C.I.
50th 80% C.I. T2
Probability
50th
Probability
Parameter value
Parameter value
NS 80% C.I.
50th
Probability
Parameter value
Summary
When a parameter is not currently listed, assess with recent data:
To avoid the expense and distraction of falsely listing a water body when it is not impaired,
the Type I error is currently controlled at < 20%
Missing an impairement: the Type II error is high for small samples, especially near the criteria.
Null Hypothesis: Not impaired
Alternate Hypothesis: Is impaired
When a water body is on the current list, assess with recent data:
To avoid delisting a water body when it it still impaired, the Type I error could be controlled at < 20%
Erroneously continuing the listing: the Type II error is high, especially near the criterion.
Null Hypothesis: Still impaired
Alternate Hypothesis: No longer impaired
26 Oct, 2004 TCEQ Austin, SWQM Stakeholders Meeting 5
```
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views: 3 posted: 12/1/2011 language: English pages: 5 | 784 | 3,669 | {"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-2014-41 | longest | en | 0.750876 |
http://functions.wolfram.com/GammaBetaErf/Factorial/07/01/01/0001/ | 1,527,143,594,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00105.warc.gz | 122,226,264 | 7,611 | html, body, form { margin: 0; padding: 0; width: 100%; } #calculate { position: relative; width: 177px; height: 110px; background: transparent url(/images/alphabox/embed_functions_inside.gif) no-repeat scroll 0 0; } #i { position: relative; left: 18px; top: 44px; width: 133px; border: 0 none; outline: 0; font-size: 11px; } #eq { width: 9px; height: 10px; background: transparent; position: absolute; top: 47px; right: 18px; cursor: pointer; }
Factorial
http://functions.wolfram.com/06.01.07.0001.01
Input Form
n! == Integrate[t^n/E^t, {t, 0, Infinity}] /; Element[n, Integers] && n >= 0
Standard Form
Cell[BoxData[RowBox[List[RowBox[List[RowBox[List["n", "!"]], "\[Equal]", RowBox[List[SubsuperscriptBox["\[Integral]", "0", "\[Infinity]"], RowBox[List[RowBox[List[SuperscriptBox["t", "n"], " ", SuperscriptBox["\[ExponentialE]", RowBox[List["-", "t"]]]]], RowBox[List["\[DifferentialD]", "t"]]]]]]]], "/;", " ", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "\[And]", RowBox[List["n", "\[GreaterEqual]", "0"]]]]]]]]
MathML Form
n ! 0 t n - t t /; n Condition n t 0 t n -1 t n [/itex]
Rule Form
Cell[BoxData[RowBox[List[RowBox[List["HoldPattern", "[", RowBox[List["n_", "!"]], "]"]], "\[RuleDelayed]", RowBox[List[RowBox[List[SubsuperscriptBox["\[Integral]", "0", "\[Infinity]"], RowBox[List[RowBox[List[SuperscriptBox["t", "n"], " ", SuperscriptBox["\[ExponentialE]", RowBox[List["-", "t"]]]]], RowBox[List["\[DifferentialD]", "t"]]]]]], "/;", RowBox[List[RowBox[List["n", "\[Element]", "Integers"]], "&&", RowBox[List["n", "\[GreaterEqual]", "0"]]]]]]]]]]
Date Added to functions.wolfram.com (modification date)
2001-10-29 | 556 | 1,660 | {"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-2018-22 | latest | en | 0.206675 |
http://mathhelpforum.com/calculus/45444-composite-function-print.html | 1,527,466,765,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00081.warc.gz | 189,000,173 | 2,574 | Composite function
• Aug 6th 2008, 01:53 PM
Snowboarder
Composite function
So composition function is s(x) = ln(ln(x)) and natural domain is (1, +infinity)
So s(s(x)) = s(ln(ln(x)) = ln(ln(ln(ln(x)))) .
Is natural domain this function also (1,+infinity)????
What is the axes intercepts of y = s(s(x)) ???
I have tried assign x = 0 but my calculator can not find any solution .
Thank you.
• Aug 6th 2008, 02:57 PM
Paperwings
The function \$\displaystyle y=ln(ln(ln(ln(x)))) \$ does not have a y-intercept since when x = 0, ln(x)=ln(0) is undefined.
If you do set y = 0 to find the x-intercept then, then \$\displaystyle x=e^{e^{e^{e^{0}}}}\$
As for the domain, the domain will be from\$\displaystyle x=e^{e^{e^{0}}} \$ to infinity. | 235 | 734 | {"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-2018-22 | latest | en | 0.784859 |
http://endmemo.com/physics/poiseuille.php | 1,534,296,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209755.32/warc/CC-MAIN-20180815004637-20180815024637-00383.warc.gz | 126,782,130 | 4,077 | Poiseuille's Law Calculator
Pressure 1:
Pressure 2:
Viscosity:
Pipe Length:
Volumetric Flow Rate:
Poiseuille's Law (also Hagen-Poiseuille equation) calculates the fluid flow through a cylindrical pipe of length L and radius R. The poiseuille's equation is:
V = π * R4 * ΔP / (8η * L)
Where:
R: Cross-sectional radius of the pipe, in meter
ΔP: Pressure difference of two ends, in Pascal
η: Viscosity of the fluid, in Pa.s
L: Pipe length, in meter
V: Volumetric Flow Rate, in m/s | 143 | 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} | 2.75 | 3 | CC-MAIN-2018-34 | longest | en | 0.790971 |
https://bodymindbeautyhealth.com/qa/quick-answer-what-is-k-in-10-k.html | 1,611,386,014,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00224.warc.gz | 249,692,446 | 7,905 | # Quick Answer: What Is K In 10 K?
## What does K mean in Tik Tok?
Therefore, “K” is used for thousand.
like, 1K = 1,000 (one thousand) 10K = 10,000 (ten thousand).
## What does K in 4k mean?
The K in 4K is taken from the prefix kilo-, meaning “thousand,” which comes to English from French, but has its root in the Greek chilioi.
## Why is K used for 1000?
To minimize confusion I would stick with K for a thousand. K comes form the Greek kilo which means a thousand. In the metric system lower case k designates kilo as in kg for kilogram, a thousand grams.
## What is K in 20k?
Normally, 20k = 20,000, since 1k = 1,000. This comes from the k being an abbreviation for kilo, which is thousand in Latin.
## What is mean k?
Actually k is abbreviation for kilo which means 1000. It is used in several places like distance, Quantity, Money and YouTube. The SI prefix for a thousand is kilo-, officially abbreviated as k—for instance, … The kilo or “k” prefix is derived from the Greek word χίλιοι (“chilioi”), meaning thousand.
## How much is 1k money?
1k is \$1,000. etc. 1\$ is 64.56 inr.
## Is 2560×1440 considered 4k?
2560×1440 is vertically and horizontally 2 times the 720p. To call 1440×2560 as 4K ready seems good but it has much less resolution than 4K. In reality it is called as QHD (Quad HD). Actually, it is known as 2K for PC monitors but not 4K since it has horizontal pixels much less than 2000.
## What is K in money?
“K” in money means a thousand. In Mathematics, Kilo means thousand, thus, the letter K. For example, 5K money basically just means five thousand (5,000). When used with currencies, 10K money is \$10,000.
## Is YouTube 4k really 4k?
Currently, YouTube TV doesn’t have 4K streaming capabilities, but you can stream content in 1080p, which is still a very high resolution. Some YouTube videos are available in 4K, however, as are a selection of shows and movies that can be purchased or rented.
## Which is better UHD or 4k?
UHD. The simplest way of defining the difference between 4K and UHD is this: 4K is a professional production and cinema standard, while UHD is a consumer display and broadcast standard. UHD quadruples that resolution to 3,840 by 2,160. …
## What is the meaning of 2.6 K?
It stands for thousand. It stands for thousand. See a translation. Report copyright infringement. | 647 | 2,346 | {"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-2021-04 | latest | en | 0.960904 |
https://www.sigterritoires.fr/index.php/en/2018/10/12/ | 1,555,648,306,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578527135.18/warc/CC-MAIN-20190419041415-20190419063415-00506.warc.gz | 834,554,914 | 45,233 | ## GIS and Decision Support (5): Theoretical Fundamentals (Part 1)
In the previous series of articles we have discussed examples of
application for the Boolean and fuzzy logic, as well as two commands for ArcGis
allowing the application of fuzzy logic to geographic information. We will discuss
now the theory that underlies this type of treatment.
Introduction
We have a set Ω of objects to classify according to a set C
of criteria. The number of objects is finite. The partial evaluations of the objects
according to each criterion take values in easily identifiable sets.
A partial objective will be seen as a fuzzy set restricting the acceptable
values of the associated criterion. Therefore, we accept the implicit hypothesis
that each objective defines a total order for Ω .
We will use as an example the case of a set Ω representing the
pixels of a study area that we wish to classify according to their ability to
receive aquaculture breeding sites. The criteria set C is the dataset
layers available: bathymetry, slope, substrate, productivity, etc. Each of
these info layers adopts easily identifiable values : favourable, somehow favourable,
unfavourable, and so on.
For each layer of information we will set a goal, for example, for bathymetry
that is at least favourable, for productivity that is at least unfavourable,
and so on. The goal is none other than the subset of the acceptable values of
the info layer.
Finally, we accept the hypothesis that each layer of information can be
classified in its entirety by the set goal, that is to say that we are able for
each pixel to determine the corresponding value of the layer. | 347 | 1,650 | {"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-2019-18 | latest | en | 0.902101 |
https://math.stackexchange.com/questions/2510195/prove-the-monotonicity-of-q-leftx-right-fracf-leftx-rightg-leftx-right | 1,561,622,946,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00383.warc.gz | 530,032,136 | 35,273 | # Prove the monotonicity of $q\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$
Here has a function $q(x)=\frac{f(x)}{g(x)}$, where
$$f(x)=\log_2\left(1+\frac{1}{a_1 x+b_1}\right),$$ $$g(x)=\log_2\left(1+\frac{1}{a_2 x+b_2}\right)$$ $$a_1,a_2,b_1,b_2 >0.$$
How to prove the monotonicity of $q(x)$? Since the first derivative is very complicated, it is difficult to analyze.
• Am I to assume $x>0$? – Fimpellizieri Nov 17 '17 at 4:22
$$\frac{\log_ax}{\log_ay}=\log_yx$$
$$f(x) \text{ is monotone} \iff \exp(f(x)) \text{ is monotone}$$
• Thanks for your helpful hint. The monotonicity of $q\left(x\right)$ depends on the relationship among a1, a2, b1, and b2. I want to reveal the impact of these constants on monotonicity, but it seems hard because the first derivative is so complicated. – zhang haiyang Nov 8 '17 at 8:06 | 309 | 832 | {"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} | 3.5625 | 4 | CC-MAIN-2019-26 | latest | en | 0.771924 |
https://se.mathworks.com/matlabcentral/answers/759786-unable-to-perform-assignment-because-the-left-and-right-sides-have-a-different-number-of-elements | 1,618,267,430,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038069267.22/warc/CC-MAIN-20210412210312-20210413000312-00427.warc.gz | 609,894,404 | 24,007 | Unable to perform assignment because the left and right sides have a different number of elements
33 views (last 30 days)
Gannon Lenhart on 1 Mar 2021
Edited: Gannon Lenhart on 2 Mar 2021
Error appears in Line 41 according to Matlab. I believe it has something to do with putting a matrix as phi_0 into y_dot(2). Can anyone give some advice on how to correct this?
% ode45
% 1. Define Initial Conditions
x_0 = 50*pi/180*randn(2,50);
phi_0 = x_0(1,:);
phi_dot_0 = x_0(2,:);
y_0 = [phi_0; phi_dot_0]; % Initial State Vector
% 2. Initialize Temporal Parameters
t_0 = 0;
t_f = 10; % sec
tspan = [0 10];
% 3. ode45
options = odeset('RelTol', 10^(-12), 'AbsTol', 10^(-12)); % the tolerances tell ode45 how "accurate to be
[t, X] = ode45(@project, tspan, x_0, options);
plot (t, X(:,1)');
%% %% FUNCTIONS %% %%
function y_dot = project(t,y)
%Initialize
y_dot = zeros(2,50);
x_0 = 50*pi/180*randn(2,50);
phi_0 = x_0(1,:);
%Parameters
g = 9.81;
L1 = 0.5;
L2 = 0.5;
omega = 0.5*sqrt(g/L2);
%State Space Representation of System
y_dot(1) = y(2);
y_dot(2) = ((2*(omega.^2)*(cos(phi_0).^2))-(omega.^2)-(g*cos(phi_0)/L2))*y(1);
end
Alan Stevens on 2 Mar 2021
Like this?
% 1. Define Initial Conditions
N = 50; %
x_0 = 50*pi/180*randn(2,N);
phi_0 = x_0(1,:);
phi_dot_0 = x_0(2,:);
y_0 = [phi_0; phi_dot_0]; % Initial State Vector
% 2. Initialize Temporal Parameters
t_0 = 0;
t_f = 10; % sec
tspan = [0 10];
% 3. ode45
options = odeset('RelTol', 10^(-12), 'AbsTol', 10^(-12)); % the tolerances tell ode45 how "accurate to be
for i = 1:N % Loop N times
[t, X] = ode45(@project, tspan, y_0(:,i), options);
plot (t, X(:,1)); % Assumes you want all 50 plotted on the same graph
hold on
end
grid
xlabel('t'), ylabel('\phi')
%% %% FUNCTIONS %% %%
function y_dot = project(~,y)
%Initialize
y_dot = zeros(2,1);
%Parameters
g = 9.81;
% L1 = 0.5;
L2 = 0.5;
omega = 0.5*sqrt(g/L2);
%State Space Representation of System
y_dot(1) = y(2);
y_dot(2) = ((2*(omega.^2)*(cos(y(1)).^2))-(omega.^2)-(g*cos(y(1))/L2))*y(1);
end | 772 | 1,991 | {"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-2021-17 | latest | en | 0.578613 |
https://cueflash.com/decks/tag/econ/118726/macroeconomics_chapter_three | 1,553,373,520,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203021.14/warc/CC-MAIN-20190323201804-20190323223804-00002.warc.gz | 463,936,847 | 9,443 | # macroeconomics chapter three
## Terms
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copy deck
Factors of production
The factors of production are the inputs used to produce goods
and services. The two most important factors of production are
capital and labor. In this module, we will take these factors as
given (hence the overbar depicting that these values are fixed).
K (capital) = K
L (labor) = L
Production function
The available production technology determines how much output is produced from given amounts of capital (K) and labor (L). The production function represents the transformation of inputs into outputs. A key assumption is that the production function has constant returns to scale, meaning that if we increase inputs by z, output will also increase by z.
We write the production function as: Y = F ( K , L )
Constant returns to scale
A key assumption is that the production function has constant returns to scale, meaning that if we increase inputs by z, output will also increase by z.
If the production function has constant returns to scale, then
doubling the amount of equipment and the number of workers doubles
Factor prices
The distribution of national income is determined by factor prices.
Factor prices are the amounts paid to the factors of production—the
wages workers earn and the rent the owners of capital collect.
Because we have assumed a fixed amount of
capital and labor, the factor supply curve
is a vertical line.
Competition
The goal of the firm is to maximize profit. Profit is revenue minus
cost. Revenue equals P × Y. Costs include both labor and capital
costs. Labor costs equal W × L, the wage multiplied by the amount
of labor L. Capital costs equal R × K, the rental price of capital R times
the amount of capital K.
Profit = Revenue - Labor Costs - Capital Costs
= PY - WL - RK
Then, to see how profit depends on the factors of production, we use
production function Y = F (K, L) to substitute for Y to obtain:
Profit = P × F (K, L) - WL - RK
This equation shows that profit depends on P, W, R, L, and K. The
competitive firm takes the product price and factor prices as given
and chooses the amounts of labor and capital that maximize profit.
Marginal product of labor (MPL)
The marginal product of labor (MPL) is the extra amount of output the
firm gets from one extra unit of labor, holding the amount of
capital fixed and is expressed using the production function:
MPL = F(K, L + 1) - F(K, L).
Diminishing marginal product
Most production functions have the property of
diminishing marginal product: holding the amount of capital
fixed, the marginal product of labor decreases as the amount of labor
increases. Like labor, capital is subject to diminishing marginal product.
The increase in profit from renting an additional machine is the extra
revenue from selling the output of that machine minus the machine’s
rental price: D Profit = D Revenue - D Cost = (P × MPK) – R.
Real wage
The firm’s demand for labor is determined by P × MPL = W,
or another way to express this is MPL = W/P, where W/P is the
real wage– the payment to labor measured in units of output rather
than in dollars. To maximize profit, the firm hires up to the point
where the extra revenue equals the real wage.
Marginal product of capital (MPK)
firm decides how much capital to rent in the same way it decides
how much labor to hire. The marginal product of capital, or MPK,
is the amount of extra output the firm gets from an extra unit of
capital, holding the amount of labor constant:
MPK = F (K + 1, L) – F (K, L).
Thus, the MPK is the difference between the amount of output produced
with K+1 units of capital and that produced with K units of capital
Real rental price of capital
To maximize profit, the firm continues to rent more capital until the MPK
falls to equal the real rental price, MPK = R/P.
The real rental price of capital is the rental price measured in units of
goods rather than in dollars. The firm demands each factor of production
until that factor’s marginal product falls to equal its real factor price.
Economic profit vs. accounting profit
The income that remains after firms have paid the factors of
production is the economic profit of the firms’ owners.
Real economic profit is: Economic Profit = Y - (MPL × L) - (MPK × K)
or to rearrange: Y = (MPL × L) - (MPK × K) + Economic Profit.
Total income is divided among the returns to labor, the returns to capital,
and economic profit. If the production function has the property
of constant returns to scale, then economic profit is zero. This conclusion
follows from Euler’s theorem, which states that if the production function
has constant returns to scale, then
F(K,L) = (MPK × K) - (MPL × L)
If each factor of production is paid its marginal product, then the sum
of these factor payments equals total output. In other words, constant
returns to scale, profit maximization, and competition together imply that
economic profit is zero.
Cobb–Douglas production function
The Cobb–Douglas production function has constant returns to
scale (remember Mankiw’s Bakery). That is, if capital and
labor are increased by the same proportion, then output
increases by the same proportion as well.
Next, consider the marginal products for the Cobb–Douglas
production function. The MPL :
MPL = (1- α)Y/L
MPK= α A/ K
The MPL is proportional to output per worker, and the MPK is
proportional to output per unit of capital. Y/L is called average
labor productivity, and Y/K is called average capital
productivity. If the production function is Cobb–Douglas, then
the marginal productivity of a factor is proportional to its average
productivity.
An increase in the amount of capital raises the MPL and
reduces the MPK. Similarly, an increase in the parameter
MPL = (1- α) A Kα L–α or, MPL = (1- α) Y / L
and the marginal product of capital is:
MPL = α A Kα-1L1–α or, MPK = α Y/K
We can now confirm that if the factors (K and L) earn their
marginal products, then the parameter α indeed tells us how much
income goes to labor and capital. The total amount paid to labor
is MPL × L = (1- α). Therefore (1- α) is labor’s share of output Y.
Similarly, the total amount paid to capital, MPK × K is αY and α is
capital’s share of output. The ratio of labor income to capital
income is a constant (1- α)/ α, just as Douglas observed. The
factor shares depend only on the parameter α, not on the amounts
of capital or labor or on the state of technology as measured by the
parameter A.
Disposable income
(Y - T)
Consumption function
C = C(Y- T)
Marginal propensity to consume
The marginal propensity to consume (MPC) is the amount by which consumption changes when disposable income (Y - T)
increases by one dollar. To understand the MPC, consider a shopping scenario. A person who loves to shop probably has a large MPC, let’s say (.99). This means that for every extra dollar he or she earns after tax deductions, he or she spends \$.99 of it.
The MPC measures the sensitivity of the change in one variable (C) with respect to a change in the other variable (Y - T).
Nominal interest rate
The quantity of investment depends on the real interest rate, which measures the cost of the funds used to finance investment. When studying the role of interest rates in the economy, economists distinguish between the nominal interest rate and the real interest rate, which is especially relevant when the overall level of prices is changing. The nominal interest rate is the interest rate as usually reported; it is the rate of interest that investors pay to borrow money.
Real interest rate
The real interest rate is the nominal interest rate corrected for the effects of inflation.
The investment function relates the quantity of investment I to the real
interest rate r. Investment depends on the real interest rate because the
interest rate is the cost of borrowing. The investment function slopes
downward; when the interest rate rises, fewer investment projects are
profitable.
NATIONAL DEMAND:
The demand for the economy’s output comes from consumption,
investment, and government purchases. Consumption depends on
disposable income; investment depends on the real interest rate;
government purchases and taxes are the exogenous variables set by
fiscal policy makers
Now, let’s combine these equations describing supply and demand
for output Y. Substituting all of our equations into the national
income accounts identity, we obtain:
Y = C(Y - T) + I(r) + G
and then, setting supply equal to demand, we obtain an equilibrium
condition:
Y = C(Y - T) + I(r) + G
This equation states that the supply of output equals its demand,
which is the sum of consumption, investment,
and government purchases.
National saving
First, rewrite the national income accounts identity as Y - C - G = I.
The term Y - C - G is the output that remains after the demands of
consumers and the government have been satisfied; it is called national
saving or simply, saving (S). In this form, the national income accounts
identity shows that saving equals investment.
To understand this better, let’s split national saving into two parts-- one
examining the saving of the private sector and the other representing
the saving of the government.
(Y - T - C) + (T - G) = I
The term (Y - T - C) is disposable income minus consumption, which is
private saving. The term (T - G) is government revenue minus government spending, which is public saving. National saving is the sum of private and public saving.
19 | 2,234 | 9,526 | {"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} | 3.390625 | 3 | CC-MAIN-2019-13 | latest | en | 0.909974 |
https://www.toktol.com/notes/context/1746/physics/power_-energy-loss-and-efficiency/energy-loss-and-efficiency | 1,516,607,908,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891196.79/warc/CC-MAIN-20180122073932-20180122093932-00233.warc.gz | 997,274,456 | 13,222 | Use adaptive quiz-based learning to study this topic faster and more effectively.
# Energy loss and efficiency
Energy can be transferred in and out of a closed system. In most real systems, not all of the energy input into the system gets output into useful energy/work as some energy is lost to the surroundings as heat.
Some of the energy obtained by burning fuel in a car engine is lost as heat to the surroundings instead of being converted into kinetic energy to drive the car forward.
This means that most real systems are not $100\%$ efficient. The efficiency of a system is the percentage of input energy/work (or power) that gets transformed into useful output energy/work (or power). This is given by: \begin{align*} \text{Efficiency}&=\frac{E_{\text{out}}}{E_{\text{in}}}\times 100\%\\ &=\frac{W_{\text{out}}}{W_{\text{in}}}\times 100\%\\ &=\frac{P_{\text{out}}}{P_{\text{in}}}\times 100\% \end{align*} $W_{\text{in}}$, $E_{\text{in}}$, $P_{\text{in}}$ and $W_{\text{out}}$, $E_{\text{out}}$, $P_{\text{out}}$ are the input and output work, energy and power respectively. | 305 | 1,086 | {"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} | 3.484375 | 3 | CC-MAIN-2018-05 | latest | en | 0.85918 |
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Forum Rules: Read This Before Posting
### Topic: Solubility of silver acetate (Read 18823 times)
0 Members and 1 Guest are viewing this topic.
#### UG
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##### Solubility of silver acetate
« Reply #15 on: April 12, 2010, 08:03:34 AM »
x is ethanoic acid
#### Borek
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##### Solubility of silver acetate
« Reply #16 on: April 12, 2010, 08:46:02 AM »
(3.88 x 10-4 - 3.88 x 10-3 x) / x = 1.75 x 10-5 x
Calculating gives x = 0.0999549 M which would give [H+] = 4.51 x 10-5 M and pH = 4.3
Does this mean the approximation is not valid?
I am afraid so. Unfortunately, that also means you should solve full 3rd degree equation. Correct result doesn't have to be far from what you have calculated, but it is hard to predict how far it is. From my calculations pH is around 3.3.
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#### UG
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##### Solubility of silver acetate
« Reply #17 on: April 12, 2010, 05:41:41 PM »
But the answer says that the solubility is 0.130 mol L-1 and pH is equal to 4.23
#### Borek
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##### Solubility of silver acetate
« Reply #18 on: April 13, 2010, 02:56:56 AM »
Unfortunatelty that most likely means they used wrong approximation and have not checked its validity later.
Try to use equations you already have and the answer given as real to calculate concentrations of all ions and check if all mass balances and reaction quotients have correct values. My bet is that you will find they are inconsistent with question data.
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#### UG
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##### Re: Solubility of silver acetate
« Reply #19 on: April 13, 2010, 03:45:39 AM »
Hmm... you're right, the answers come out all funny. So the way to solve this is to just try and find a way to express an equation using only one of the 4 variables?
#### Borek
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##### Re: Solubility of silver acetate
« Reply #20 on: April 13, 2010, 04:40:09 AM »
Yes. Unfortunately that leads to a 3rd degree polynomial.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info | 849 | 2,770 | {"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-2023-23 | latest | en | 0.856348 |
http://mathoverflow.net/questions/76153/probability-of-generating-a-connected-graph | 1,469,272,500,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257822172.7/warc/CC-MAIN-20160723071022-00202-ip-10-185-27-174.ec2.internal.warc.gz | 160,258,104 | 16,832 | # Probability of Generating a Connected Graph
$N$ points are generated randomly within a unit square, with a uniform distribution. What is the probability that the points form a connected graph, given that two points are connected if the distance between them is less than or equal to $d$? (this should obviously be some function of $N$ and $d$).
If you don't know the answer, but have an idea that may (or may not) lead me a step forward, please let me know as well.
Moreover, if you know for sure that this problem is yet unsolved, that's also good news for me. I can then do it through a Monte-Carlo simulation, but my approach would be justified.
Thanks, Melvin
-
You should look into "random geometric graphs". Mathew Penrose has a monograph with that title. I'll post an answer if I can find the result in his book. – j.c. Sep 22 '11 at 19:34
@Melivn: I wasn't very awake when I wrote this... – Thierry Zell Sep 22 '11 at 20:15
@Melvin, I think Gjergji Zaimi's answer has the result from Penrose's book whose discussion turned out to be more convoluted than I thought (In chapter 13, Penrose shows that connectivity is asymptotically equivalent to 2-connectivity, for which he showed a threshold in section 7.2...). The bottom line is compute the quantity $\mu$ from $N$ ($n$ in Gjergji Zaimi's notation) and $d$; then up to a constant, the probability of being connected will be asymptotic to $e^{-\mu}$ as $N$ goes to infinity. – j.c. Sep 22 '11 at 22:02
@jc: thanks for the comment. Where is the final result located in Penrose's book please, as I need to reference it? – user18011 Sep 22 '11 at 23:51
Crossposted: math.stackexchange.com/questions/66777/… – Byron Schmuland Sep 23 '11 at 0:56
Just a few more comments to the answers and references already posted. I will denote your graph by $G(n,d(n))$. I'm not sure if this is satisfactory enough, but with fairly standard methods one can prove that if $\mu=ne^{-\pi nd(n)^2}\to 0$ as $n\to \infty$ then the graph is aas connected. In general the probability that $G$ is connected is $\sim e^{-\mu}$. References include the monograph by Penrose mentioned in the comments, the paper by Gupta and Kumar, see also the paper by Penrose "The longest edge of the random minimal spanning tree" The Annals of Applied Probability (1997) 7, 340–361, and M.J. Apple, R.P. Russo "The connectivity of a graph on uniform points on $[0,1]^d$".
Note that the situation in $[0,1]^d$ with metric $\ell_p$ is similar. In fact Goel, Rai, and Krishnamachari proved this for all monotone properties of graphs (like connectivity, non-planarity etc.) "Monotone properties of random geometric graphs have sharp thresholds". (Monotone property here means that it is preserved by addition of edges.)
-
See
Clique sizes in a unit disk graph
and references mentioned there... Your graphs are the unit disk graphs of the title.
EDIT
See http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.7.4866&rep=rep1&type=pdf, especially the reference to Gupta/Kumar at the very beginning...
- | 810 | 3,031 | {"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} | 3.40625 | 3 | CC-MAIN-2016-30 | latest | en | 0.955109 |
https://www.quesba.com/questions/program-specifications-similar-programs-chegg-wrong-vickie-takes-jar-contai-897659 | 1,674,879,812,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499470.19/warc/CC-MAIN-20230128023233-20230128053233-00345.warc.gz | 976,679,877 | 28,342 | # Program Specifications: (similar programs here in chegg are wrong) Vickie takes a jar which...
Program Specifications: (similar programs here in chegg are wrong)
Vickie takes a jar which contains exactly 100 quarters to the casino with the intention to win big. She plays three machines slot machines in sequence. Unknown to her, the machines are entirely predictable.
Each play costs one quarter.
The following are constants:
The first machine pays 30quarters every 35th time it is played.
The second machine pays 60quarters every 100th time it is played.
The third pays 11 quarters every 8th time it is played.
So, Vickie will always lose - eventually.
Notice that she plays the machines one at a time in sequence: the first one, then the second one, the third one, the first one, the second one…
Your program will output the number of times Vickie plays until she goes broke along with the result of each time she wins money. Besides “main” you must define and use at least 4 methods.
Your program will input the number of quarters in Vickie’s jar (it could be any number but for this assignment make it 100).
Your program will output a statement EVERYTIME Vickie plays. The output will contain the following:
1. Number of times play, machine, random number(3 digit), quarter(s) left.
2. If win, we need following in addition
1. The Number of Machine (1, 2, or 3) that she just won on
2. The Amount she just won (formatted like dollars and cents)
3. Once Vickie is broke, you will output put the total number of times she was able to play the machines.
Oct 15 2021| 08:32 AM |
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# MAGNETISM PHYSICS CONCEPT QUESTIONS
Electromagnetism/Magnetism concept questions - Physics
Feb 26, 2011Unillusive. The "magnitude" of the magnetic force on a current-carrying wire depends on the strength of the magnetic field, the current in the wire, and the length of wire in the magnetic field. - My answer: True because the magnetic field, current in the wire, and length of the magnetic field end up being the magnitude of the magnetic force..
2 concept questions on magnetism. | Physics Forums
May 03, 20171. I think that the two outside magnets would attach to the middle one and their eventual velocities and accelerations would equal 0 but that seems to easy to me.. 4. I think it's yes because magnetic fields exert forces on any moving charge or current that is present in the field. I know the
Conceptual Physics: Magnetism and Magnetic Force
Conceptual Physics: Magnetism and Magnetic Force Units. Invisible magnetic field lines emerge from the North pole of a magnet and enter the South pole. Field lines can be visualized by sprinkling small iron filings over a magnet covered by a clear sheet of plastic. When a compass (or any freely floating bar magnet) points north, it is actually aligning its north pole to the Earth's magnetic south pole.
physics conceptual questions magnetism Flashcards and
Learn physics conceptual questions magnetism with free interactive flashcards. Choose from 500 different sets of physics conceptual questions magnetism flashcards on Quizlet.
Magnetism and Electromagnetism - High School Physics
Example Question #1 : Magnetism And Electromagnetism. In other words, the magnetic field ( ), is equal to a constant () times the current () divided by the circumference of the magnetic field it is creating. We are given the current, the constant, and the radius. Using these values, we can solve for the magnetic field strength. Notice that the cancels out.
Magnetism and Electricity Questions for Tests and Worksheets
Magnetism and Electricity Questions - All Grades. You can create printable tests and worksheets from these Magnetism and Electricity questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page. an east and west pole.
Physics Concept Questions Book 6 Electricity Magnetism 300
autor physics concept questions book 6 electricity magnetism 300 questions answers answers physics concept questions book 6 electricity magnetism 300 questions answers at complete pdf library this book have some digital formats such us paperbook ebook kindle epub and another formats buy [EPUB] Physics Concept Questions Book 6 Electricity
Electricity and Magnetism Exam - AP Physics C Electricity
Example Question #1 : Electricity And Magnetism Exam. Electric field is additive; in other words, the total electric field from the two planes is the sum of their individual fields: The direction of the electric field is away from positive source charges. Thus, to the right of these positively charged planes, the field points away to the right.
Conceptual Physics - Magnetism Flashcards | Quizlet | 661 | 3,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-18 | longest | en | 0.893361 |
http://mathhelpforum.com/trigonometry/153057-help-me-get-rid-square-root-my-program.html | 1,529,854,645,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866965.84/warc/CC-MAIN-20180624141349-20180624161349-00572.warc.gz | 195,742,788 | 9,186 | # Thread: Help me to get rid a square root in my program ?
1. ## Help me to get rid a square root in my program ?
i need to make my program go a bit faster.
c=sqr (x) <-- the square root that i don't like
p= ( a + b + c ) / 2
IF p * ( p - a ) * ( p - b ) * ( p - c ) > ( R * c / 2 ) ^2 then do something..
I want to compare x directly without finding the square root,
What is the above ?
its Heron's formula to find the height of the triangle.The only thing is that i only need to compare the height with the circle's radius R to find out if the circle obscures a vector.
2. solved.
-a^4 - b^4 - c^4 + 2a^2b^2 + 2a^2c^2 + 2b^2c^2 > 4*F^2*c^2
3. you mean this $\displaystyle -a^2 - b^4 - c^4 + 2a^2 b^2 + 2b^2 c^2 >4F^2 c^2$
how you got this?? | 269 | 753 | {"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} | 3.015625 | 3 | CC-MAIN-2018-26 | latest | en | 0.798043 |
https://www.r-bloggers.com/one-of-the-best-and-most-underutilized-graphs-in-ggplot2/ | 1,553,605,607,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205163.72/warc/CC-MAIN-20190326115319-20190326141319-00526.warc.gz | 874,304,373 | 17,205 | # One of the Best and Most Underutilized Graphs in ggplot2
March 15, 2016
By
(This article was first published on Hallway Mathlete, and kindly contributed to R-bloggers)
Understanding how a distribution of a variable changes over time can make a great visualization. These highly intuitive graphics can display a lot of information and can be simply rendered in R using ggplot2. However, based on my experience, they are one of the most underutilized graphs in R.
A good example of this style of graph is from my research. My research studies how data analysis can be utilized to improve the product design and manufacturing process. The style of graph discussed in this post is extremely useful for showing how design specifications change over time. Below you can see an example of how the specifications of secondary cameras on cellphones has changed over time. It is easily seen that before 2011, there was almost no secondary cameras and by 2015, almost all cameras released had some form of secondary camera.
To create these plots, first lets load ggplot2 and the diamond data set.
`library(ggplot2)data(diamonds)head(diamonds)`
When creating these plots, I like to make sure I under stand how the data is distributed over the x axis. This is helpful because if there is a section of x-axis with much fewer data points, the distribution of the y-axis can change rapidly over the x-axis due to low samples.
The plot below shows the distribution of diamonds grouped by cut as the price changes.
`ggplot(data=diamonds, aes(x=price, group=cut, fill=cut, position="stack")) + geom_density(adjust=1.5)`
In the next plots instead of the count in the y-axis, the y-axis is the percent of each group (cut for the first example and clarity for the second) for different prices.
`ggplot(data=diamonds,aes(x=price, group=cut, fill=cut, position="stack")) + `
`geom_density(adjust=1.5, position="fill")`
`ggplot(data=diamonds,aes(x=price, group=clarity, fill=clarity, position="stack")) +`
` geom_density(adjust=1.5, position="fill")`
R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more... | 557 | 2,352 | {"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-2019-13 | latest | en | 0.889401 |
https://inderjtaneja.com/2021/01/07/different-digits-selfie-fractions-four-and-five-digits-numerators/ | 1,643,248,502,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305052.56/warc/CC-MAIN-20220127012750-20220127042750-00099.warc.gz | 369,963,707 | 29,633 | The addable fractions are proper fractions, where addition can be inserted into numerator and denominator, and the resulting fraction is equal to the original. The same is true for other operations, such as, addition, subtraction, multiplication, potentiation, etc. These types of fractions, we call selfie-fractions. This work brings selfie fractions with single and/or multiple representations in different digits with all basic operations. The numerator values are with four and five digits numbers. The results are in increasing order of numerator values. For the two three digits numerators refers to author’s another work.
Below are few slides giving some examples:
Below is a complete list of author’s work in this direction:
## Selfie Fractions
1. Inder J. Taneja, Selfie Fractions: Addable, Subtractable, Dottable and PotentiableZenodo, March 24, 2019, pp. 1-260, http://doi.org/10.5281/zenodo.2604531
2. Inder J. Taneja, Pandigital Equivalent Selfie FractionsZenodo, April 02, 2019, pp. 1-392, http://doi.org/10.5281/zenodo.2622028
3. Inder J. Taneja, Repeated Digits Selfie Fractions: Two and Three Digits NumeratorsZenodo, Septembr 12, 2019, pp. 1-1091, http://doi.org/10.5281/zenodo.3406655
4. Inder J. Taneja, Different Digits Selfie Fractions: Two and Three Digits NumeratorsZenodo, September, 12, 2019, pp. 1-337, http://doi.org/10.5281/zenodo.3474091
5. Inder J. Taneja, Different Digits Selfie Fractions: Four Digits NumeratorZenodo, October 06, 2019, pp. 1-844, http://doi.org/10.5281/zenodo.3474267
6. Inder J. Taneja, Different Digits Selfie Fractions: Five Digits Numerator – PandigitalZenodo, October 06, 2019, pp. 1-362, http://doi.org/10.5281/zenodo.3474379
7. Inder J. Taneja, Patterned Selfie FractionsZenodo, October 27, 2019, pp. 1-267, http://doi.org/10.5281/zenodo.3520096
For more work in recreation of numbers see the publication list: | 590 | 1,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-05 | longest | en | 0.652164 |
https://greenbayhotelstoday.com/articles/what-is-a-3-point-scale-called | 1,680,355,441,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950030.57/warc/CC-MAIN-20230401125552-20230401155552-00686.warc.gz | 321,678,240 | 18,310 | # What is a 3 point scale called? (2023)
## What is a point scale called?
The most widely used is the Likert scale (1932). In its final form, the Likert scale is a five (or seven) point scale which is used to allow the individual to express how much they agree or disagree with a particular statement.
(Video) Tutorial: 3-Point Grading Scale (and the power of "Not Yet")
(Citizens of Hope)
What is a 3 point Likert scale example?
A 3-point Likert scale gives respondents two options at the farthest points with a neutral option in the middle. For example, Agree, Disagree, and Neutral.
(Video) Likert scale for questionnaire and survey
(Dr. Mahmoud Omar (Statistics))
Is A Likert scale 3 or 5?
The 5-point Likert scale consists of the below points – (1) Strongly Disagree; (2) Disagree; (3) Neither Agree nor Disagree; (4) Agree; (5) Strongly Agree.
(Video) Glasgow Coma Scale made easy
(Brainbook)
What are Likert scale points?
So what is a Likert scale survey question? It's a question that uses a 5 or 7-point scale, sometimes referred to as a satisfaction scale, that ranges from one extreme attitude to another. Typically, the Likert survey question includes a moderate or neutral option in its scale.
(Video) Airfix 1/48 Albion 3-Point Fueller - "Final Reveal" (2.6.19)
(International British Modeler)
What does a 3 point scale mean?
3 Point Likert scale is a scale that offers agree and disagree as to the polar points along with a neutral option. Like the 2-point scale, the 3 point scale is also used to measure Agreement. Options will include: Agree, Disagree, and Neutral.
(Video) The Basics of Tractors: Understanding 3-Point Hitch System
(Vahrenberg Implement)
What are the three types of scale?
There are three types of scales commonly used on maps: written or verbal scale, a graphic scale, or a fractional scale.
(Sikana English)
Can I use a 3 point Likert scale?
They concluded that two or three points are probably fine when averaging across people and across many items. But if the focus of the research is on individual scales, using a minimum of five to six scale points is probably necessary to get an accurate measure of the variable.
(Video) Curling Drills - Weight Judgement and Communication: 3 Point PRESS (Using the Ferbey System)
(Curling Class)
How do you analyze a 3 point Likert scale?
A Likert scale is composed of a series of four or more Likert-type items that represent similar questions combined into a single composite score/variable. Likert scale data can be analyzed as interval data, i.e. the mean is the best measure of central tendency. use means and standard deviations to describe the scale.
(Video) #5-pointLikertScale How to Use Likert Scale in Descriptive Study
(Teacher Lai Arcenas)
Is a 3 point Likert scale reliable?
Likert scale (0,1,2,3) is 92% reliable while the Likert-type of scale had 90, 89, and 88% reliability. scales. We standardize the scale efficacy in a 5.0 system, the non-Likert scale is 4.73 and 2.35, 2.45, and 2.41 for Likert scales.
(Video) [EU4] What is the Dev Meta and how to use it to pay 3 points to dev?
(Zlewikk TV)
What is the difference between Likert and Likert type scale?
The difference between the Likert-type scale and a full-blown Likert scale is that the Likert-type scale only uses a five-point (or seven-point, or whatever you prefer) to answer on a single question. A full-blown Likert scale on the other hand uses a series of statements that explore different dimensions of a subject.
(Video) 3 Point Problems, Strike Lines, and Apparent Dip
(KC_StructuralGeology)
## What is a semantic scale?
A semantic differential scale is a survey or questionnaire rating scale that asks people to rate a product, company, brand, or any 'entity' within the frames of a multi-point rating option. These survey answering options are grammatically on opposite adjectives at each end.
(Video) 3 point problem
(Larry Murdoch)
What type of scale is a Likert scale?
Developed in 1932 by Rensis Likert1 to measure attitudes, the typical Likert scale is a 5- or 7-point ordinal scale used by respondents to rate the degree to which they agree or disagree with a statement (table).
What is Guttman scale in research?
The Guttman scale is also known as cumulative scaling or scalogram analysis. It is an ordinal scale with a number of statements placed in a hierarchical order. The order is arranged so that if a respondent agrees with a statement, they will also agree with all of the statements that fall below it in extremity.
What is a 4 point Likert scale?
Basically, a 4 point Likert scale is a forced scale, which essentially means forcing a respondent to form an opinion, either way. Market researchers use 4 points Likert scale when a user's opinion is essential without being neutral on a specific topic, such as: Satisfied. Very Satisfied. Dissatisfied.
What is a 1 5 Likert scale?
A type of psychometric response scale in which responders specify their level of agreement to a statement typically in five points: (1) Strongly disagree; (2) Disagree; (3) Neither agree nor disagree; (4) Agree; (5) Strongly agree.
What are the different names of scale?
Each note of a major scale is also named with scale-degree names : tonic, supertonic, mediant, subdominant, dominant, submediant, leading tone, and then tonic again. Example 4 shows how these names align with the scale-degree number and solfège systems described above.
What are the three scales used to measure?
There are three temperature scales in use today, Fahrenheit, Celsius and Kelvin. Fahrenheit temperature scale is a scale based on 32 for the freezing point of water and 212 for the boiling point of water, the interval between the two being divided into 180 parts.
What are the 4 types of measurement scales?
Each of the four scales (i.e., nominal, ordinal, interval, and ratio) provides a different type of information.
What is better than a Likert scale?
A Likert scale will provide you with the participants' agreement or disagreement with the asked statements. A Semantic Differential scale will provide you with information on where your participants' view lies on a continuum between two contrasting adjectives.
How do you convert a 3 point scale to a 5 point scale?
You cannot change a 3 point scale to a 5 point scale. If you want to compare the two, and don't have any idea which is the dependent and independent variable, you could use a chi-square, but that doesn't account for the ordinality.
## Is there a 4 point scale?
The 4.0 scale is the most commonly used GPA scale. A 4.0 represents an A or A+, with each full grade being a full point lower: 3.0=B, 2.0=C, and 1.0=D. Pluses are an additional one-third of a point, while minuses are the subtraction of one-third of a point.
What is the range of the mean in a 3 point Likert scale?
The range of interpreting the Likert scale mean score was given as follows: 1.0-2.4 (Negative attitude), 2.5-3.4 (Neutral attitude), and 3.5-5.0 (Positive attitude).
What are the benefits of a 3 point Likert scale?
Benefits of the Likert scale
Likert scale INCREASES RESPONSE RATES by giving multiple options to the respondents. It offers a simple yes or no option to the respondents. A degree of opinion or even a neutral response is easy to assess during analysis.
Is A Likert scale an interval scale?
The Likert scale is widely used in social work research, and is commonly constructed with four to seven points. It is usually treated as an interval scale, but strictly speaking it is an ordinal scale, where arithmetic operations cannot be conducted.
What is the most common Likert scale?
Likert scales are most commonly 5-point or 7-point scales with a neutral middle-point, such as 'neither agree nor disagree' 'neutral' or 'undecided', but 4 or 6-point Likert scales which eliminate a neutral option can be used when a researcher wants to force a respondent to provide a clear opinion.
Why is the Likert scale controversial?
The problem with a Likert scale is that the scale [of very satisfied, quite satisfied, neutral, quite dissatisfied, very dissatisfied, for example] produces ordinal data.
What is a point scale?
any scale for measuring some construct or attribute in which participants' responses to a series of multiple-choice questions are given numerical values (points). The final score is the total points earned. See also Likert scale; semantic differential.
Is a point scale ordinal?
Developed in 1932 by Rensis Likert1 to measure attitudes, the typical Likert scale is a 5- or 7-point ordinal scale used by respondents to rate the degree to which they agree or disagree with a statement (table).
What are the 2 types of scales called?
The four types of scales are:
• Nominal Scale.
• Ordinal Scale.
• Interval Scale.
• Ratio Scale.
What are the 3 main metric measurement types?
Metric system basics
The three most common base units in the metric system are the meter, gram, and liter. The meter is a unit of length equal to 3.28 feet; the gram is a unit of mass equal to approximately 0.0022 pounds (about the mass of a paper clip); and the liter is a unit of volume equal to 1.05 quarts.
## What is an example of nominal scale?
Nominal Scale: Gender, marital status, religion, race, hair color, country, etc are examples of Nominal Scale. They are all examples of the noun which do not require rank or order.
What is nominal and ordinal scale?
Nominal scale is a naming scale, where variables are simply “named” or labeled, with no specific order. Ordinal scale has all its variables in a specific order, beyond just naming them. Interval scale offers labels, order, as well as, a specific interval between each of its variable options.
What is the other term for Likert scale?
Hence, Likert scales are often called summative scales.
What type of scale is ordinal?
The Ordinal scale includes statistical data type where variables are in order or rank but without a degree of difference between categories. The ordinal scale contains qualitative data; 'ordinal' meaning 'order'. It places variables in order/rank, only permitting to measure the value as higher or lower in scale.
Which scale is known as ordinal scale?
4. Ordinal Scales. Ordinal scales put things in order but have no origin and no operations. For example, geologists use a scale to measure the hardness of minerals using Moh's Scale for Hardness (MSH, for short).
What are the different scales called?
Types of scale
• Chromatic, or dodecatonic (12 notes per octave)
• Nonatonic (9 notes per octave): a chromatic variation of the heptatonic blues scale.
• Octatonic (8 notes per octave): used in jazz and modern classical music.
• Heptatonic (7 notes per octave): the most common modern Western scale.
What are the five types of scales?
From the least to the most mathematical, the scale types are nominal, ordinal, interval, and ratio. Nominal scales have no arithmetic properties. Ratio scales have all three of the arithmetic properties. Or- dinal and interval scales fall in between nominal and ratio scales.
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Introduction: My name is Amb. Frankie Simonis, I am a hilarious, enchanting, energetic, cooperative, innocent, cute, joyous person who loves writing and wants to share my knowledge and understanding with you. | 2,747 | 11,693 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-14 | latest | en | 0.90511 |
https://sharepoint.stackexchange.com/questions/230683/add-subtract-amounts-from-column-in-one-list-to-cell-in-another-list | 1,713,299,866,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00733.warc.gz | 462,997,586 | 40,493 | # Add/Subtract amounts from column in one list to "cell" in another list
I am trying to build two lists to keep track of our purchase orders we have. My two lists looks like this:
List 1: PO
``````+-----------+-----------+-------------+
| PO | Amount | Used/Left |
+-----------+-----------+-------------+
| 654654654 | 50000 | ?? |
| 123123123 | 30000 | ?? |
| 789789789 | 5000 | ?? |
+-----------+-----------+-------------+
``````
List 2: Purchases
``````+-----------+-----------+---------------+-----------+-----------+
| Supplier | PO | Description | Amount | Received |
+-----------+-----------+---------------+-----------+-----------+
| Some name | 654654654 | Blah blah | 1654 | Yes/No |
| Some name | 123123123 | Blah blah | 2441,41 | Yes/No |
| Some name | 789789789 | Blah blah | 154 | Yes/No |
| Some name | 123123123 | Blah blah | 4521,52 | Yes/No |
| Some name | 789789789 | Blah blah | 160 | Yes/No |
+-----------+-----------+---------------+-----------+-----------+
``````
So here we go. In List 1 we have our PO's and the max usable amount and a column to the used amount. In List 2 we have the purchases made on the different PO's.
What I would like to accomplish is getting the sum of the "Amount" column connected to the different PO's from list 2 to be added in the "Used/Left" column in List 1 under the right PO. So List 1 is always showing the remaining amount on the different PO's
Even better if the sum of the amounts will be subtracted from the full amount in the "Amount" column in List 1
I am guessing using Calculated fields or workflows? I would prefer no workflows as i never made one ;)
Can anyone help me achieve this?
In my opinion, the flexible solution for your situation is creating Event Receiver on `Item Added` & `Item Updated` on list 1 to perofrm your calculation in a flexible manner! | 514 | 1,960 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-18 | latest | en | 0.843882 |
https://www.capttyler.com/2019/10/23/how-much-house-can-i-afford-by-payment/ | 1,686,395,112,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00013.warc.gz | 758,804,821 | 15,483 | We license calculators from CalcXML, who estimates how much house you can afford based on a few important items, including income, amount of money saved for a down payment, and monthly obligations.
But you can still pay in – and get the government bonus – until you turn 50. invest up to £4,000 each tax year and recieve a.
For Teresa and Martin, who can both afford a 20% down payment (and then some), the monthly payment will be around \$800, well within their respective budgets. Paul and Grace can afford to make a down payment of \$7,000, just over 5% of the home value, which means they’ll need a mortgage of about \$128,000.
Methodology. That home payment assumes a 30-year mortgage at current rates, and includes 1% property tax and 0.4% for homeowners insurance. It does not factor in private mortgage insurance, which you’ll owe if your down payment is less than 20% of the purchase price. You should reduce the maximum target if you have other savings needs.
How much house can I afford? Including your mortgage, your monthly debt payments should not exceed 45 percent of your total income. With that in mind, important factors to consider when setting.
How much house can you afford to be looking for? This calculator will help you calculate how much you can afford. Shopping for a new home? Calculate the home price you can pay and the mortgage schedule you will need based on the payment, down payment, taxes and insurance you can afford.
PITI is important because a lender will compare that payment to your income to help determine how much you can afford to borrow. While various loan programs will have different specific requirements, generally your total monthly debt payments – including PITI – should be 45% or less of your monthly income.
For someone who intends to take a 30 years mortgage and repaying by a monthly payment of \$2,800 let’s figure out how much house he can afford: interest rate level You can afford to borrow You pay on interest 2.00% \$757,535.85 \$250,464.15 2.50% \$708,643.86 \$299,356.14 3.00% \$664,130.27.
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Quotes About Buying A New Home 18 Awesome Bible Quotes for your New House.. Moving or buying a new home can be fun, but it can also be stressful. Don’t let the stresses of the moving process steal your joy. God wants us to have safe, comfortable places to live. So whether you’re moving your family across the country. | 748 | 3,413 | {"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-2023-23 | latest | en | 0.957266 |
http://www.chegg.com/homework-help/principles-of-foundation-engineering-7th-edition-solutions-9781111787097 | 1,469,909,985,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258937291.98/warc/CC-MAIN-20160723072857-00178-ip-10-185-27-174.ec2.internal.warc.gz | 349,161,593 | 16,512 | View more editions
# TEXTBOOK SOLUTIONS FOR Principles of Foundation Engineering 7th Edition
• 214 step-by-step solutions
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Chapter: Problem:
100% (3 ratings)
A moist soil has a void ratio of 0.65. The moisture content of the soil is 14% and Gs = 2.7. Determine:
a. Porosity
b. Degree of saturation (%)
c. Dry unit weight (kN/m3)
SAMPLE SOLUTION
Chapter: Problem:
100% (3 ratings)
• Step 1 of 3
(a)
Calculate the porosity of the soil by using the formula:
Here, e is the void ratio.
Substitute 0.65 for e:
Therefore, the porosity of the soil is .
Although the relationship is not linear, the higher the void ratio the more will be the porosity.
• Step 2 of 3
(b)
Calculate the degree of saturation of the soil by using the formula:
Here, w is the moisture content, and is the specific gravity of soil solids.
Substitute for w, and 2.7 for :
Therefore, the degree of saturation is .
Thus, the soil is not fully saturated.
• Step 3 of 3
(c)
Calculate the dry unit weight of the soil by using the formula:
Here, is the unit weight of water.
Substitute for , 0.65 for e, and 2.7 for :
Therefore, the dry unit weight is .
The dry density is always less than the bulk density when the soil is moist.
Corresponding Textbook
Principles of Foundation Engineering | 7th Edition
9781111787097ISBN-13: 1111787093ISBN: Braja M DasAuthors:
This is an alternate ISBN. View the primary ISBN for: Principles of Foundation Engineering 7th Edition Textbook Solutions | 427 | 1,641 | {"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-2016-30 | longest | en | 0.8134 |
https://www.jiskha.com/questions/78323/the-accepted-value-for-the-universal-gas-constant-is-0-0821-l-atm-mol-k-wat-would-it-be | 1,624,370,158,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00603.warc.gz | 784,722,254 | 5,185 | # chem
The accepted value for the universal gas constant is 0.0821 L atm/mol K. Wat would it be if the pressure was measured in torr and the volume in milliliters?
1. 👍
2. 👎
3. 👁
1. PV = nRT
n = (PV)/RT
For R = 0.0821, P is in atm which is torr/760. To measure R in torr, then we would multiply 0.0821 x 760.
V is measured in liters. To measure in mL we would multiply by 1000.
Therefore, 0.0821 x 760 x 1000 = ?? would be the new value of R.
1. 👍
2. 👎
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A gas is allowed to expand at constant temperature from a volume of 2.00 L to 11.20 L against an external pressure of 1.500 atm. If the gas loses 256 J of heat to the surroundings, what are the values of q, w, and ΔU? Pay careful | 1,024 | 3,294 | {"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-2021-25 | latest | en | 0.915547 |
http://mathhelpforum.com/calculus/167146-laplace-transform-sin-t.html | 1,527,255,673,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867092.48/warc/CC-MAIN-20180525121739-20180525141739-00626.warc.gz | 183,833,982 | 13,806 | # Thread: Laplace transform of |sin(t)| ?
1. ## Laplace transform of |sin(t)| ?
Hello
Evaluate $\displaystyle \mathcal{L} ( |sin(t) | )$
that absolute value makes my life hard.
how can i deal with it?
2. How about $\displaystyle \displaystyle \mathcal{L} ( |sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sinh^2t}~dt$
3. $\displaystyle |sin(t)| = \sqrt{sinh^2(t)}$ ?
I meant to say $\displaystyle \displaystyle \mathcal{L} ( |\sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sin^2t}~dt$
5. Originally Posted by Liverpool
Hello
Evaluate $\displaystyle \mathcal{L} ( |sin(t) | )$
that absolute value makes my life hard.
how can i deal with it?
The L-transform of the 'full wave rectified' function $\displaystyle |\sin t|$ is...
$\displaystyle \displaystyle \mathcal{L} \{|\sin t| \} = \frac{\coth \frac{\pi s}{2}}{1+s^{2}}$ (1)
The (1) can be obtained applying the 'periodically repetead function's rule' of the L-tranform...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
6. i had in mind transforming the integral into $\displaystyle \displaystyle\int_{0}^{\infty }{{{e}^{-st}}\left| \sin t \right|\,dt}=\sum\limits_{k=0}^{\infty }{\int_{\pi k}^{(k+1)\pi }{{{e}^{-st}}\left| \sin t \right|\,dt}},$ i don't know if it works though.
7. Originally Posted by chisigma
The (1) can be obtained applying the 'periodically repetead function's rule' of the L-tranform...
Ok Good
I know that rule
but what is the period of it?
8. Originally Posted by Krizalid
i had in mind transforming the integral into $\displaystyle \displaystyle\int_{0}^{\infty }{{{e}^{-st}}\left| \sin t \right|\,dt}=\sum\limits_{k=0}^{\infty }{\int_{\pi k}^{(k+1)\pi }{{{e}^{-st}}\left| \sin t \right|\,dt}},$ i don't know if it works though.
This is the correct method:
Spoiler:
Problem: Compute $\displaystyle \displaystyle \int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt$.
Solution: We note that
\displaystyle \displaystyle \begin{aligned}\int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt &= \sum_{k=0}^{\infty}\int_{\pi k}^{\pi(k+1)}e^{-st}|\sin(t)|\text{ }dt\\ &=\sum_{k=0}^{\infty}\frac{e^{-s\pi k}\left(e^{\pi s}+1\right)}{s^2+1}\\ &= \frac{e^{\pi s}+1}{(s^2+1)(1-e^{-\pi s})}\\ &=\frac{\text{coth}\left(\frac{\pi s}{2}\right)}{s^2+1}\end{aligned}
9. haha, i knew it, it just that the web sites i often use to compute integrals are not computing them so it's too lazy for me to compute them by hand.
10. Originally Posted by Krizalid
haha, i knew it, it just that the web sites i often use to compute integrals are not computing them so it's too lazy for me to compute them by hand.
Yeah. The general rule of thumb is that if $\displaystyle f:[0,\infty)\to\mathbb{R}$ is $\displaystyle \omega$-periodic such that $\displaystyle \text{sgn}\left(f(x)\right),\text{ }x\in[k\omega,(k+1)\omega]=s(k)$ is some nice function then one always writes $\displaystyle \displaystyle \int_0^{\infty}|f(x)|g(x)\text{ }dx$ as $\displaystyle \displaystyle \sum_{k=0}^{\infty}s(k)\int_{k\omega}^{(k+1)\omega }f(x)g(x)\text{ }dx$.
11. Does the first line need to be justified, or is it always valid?
Originally Posted by Drexel28
This is the correct method:
Spoiler:
Problem: Compute $\displaystyle \displaystyle \int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt$.
Solution: We note that
\displaystyle \displaystyle \begin{aligned}\int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt &= \sum_{k=0}^{\infty}\int_{\pi k}^{\pi(k+1)}e^{-st}|\sin(t)|\text{ }dt\\ &=\sum_{k=0}^{\infty}\frac{e^{-s\pi k}\left(e^{\pi s}+1\right)}{s^2+1}\\ &= \frac{e^{\pi s}+1}{(s^2+1)(1-e^{-\pi s})}\\ &=\frac{\text{coth}\left(\frac{\pi s}{2}\right)}{s^2+1}\end{aligned}
12. OK Guys
those solutions are too advanced to me
what I know that is |sin(t)| is periodic and I use a program that told me its period is pi
So Am able now to solve that laplace using the formula for finding laplace transform for periodic functions
but the problem here
is How did we know that the function is periodic?
13. it's an easy mix of telescoping sums and use of FTC.
we claim that $\displaystyle \displaystyle\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}=\displaystyle\int_{1} ^{n}{f(t)\,dt}.$
all we need to have is that $\displaystyle f$ is given continuous, so it has an antiderivative, then
\displaystyle \begin{aligned} \sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}&=\sum\limits_{k=1}^{n-1}{\left( F(k+1)-F(k) \right)} \\ & =F(n)-F(1) \\ & =\int_{1}^{n}{f(t)\,dt}. \end{aligned}
it's a very useful trick.
14. Originally Posted by Krizalid
it's an easy mix of telescoping sums and use of FTC.
we claim that $\displaystyle \displaystyle\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}=\displaystyle\int_{1} ^{n}{f(t)\,dt}.$
all we need to have is that $\displaystyle f$ is given continuous, so it has an antiderivative, then
\displaystyle \begin{aligned} \sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}&=\sum\limits_{k=1}^{n-1}{\left( F(k+1)-F(k) \right)} \\ & =F(n)-F(1) \\ & =\int_{1}^{n}{f(t)\,dt}. \end{aligned}
it's a very useful trick.
Thanks. I've seen it done before, but was never quite sure of the justification.
15. Its better to answer the original poster then answer anyone else, and the latter should post a new thread if he have any question not asking them in the same thread.
I don't like this.
Respect me!!!!
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# Laplace transforms of mod of sin omaga
Click on a term to search for related topics. | 1,888 | 5,410 | {"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} | 3.484375 | 3 | CC-MAIN-2018-22 | latest | en | 0.657526 |
https://solvedlib.com/n/the-decomposition-of-ethanol-at-some-constant-temperature,20879550 | 1,695,501,348,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.3/warc/CC-MAIN-20230923194908-20230923224908-00291.warc.gz | 605,475,916 | 21,193 | # The decomposition of ethanol at some constant temperature (above 500*C), over a copper surface,CzHsOH(g) CH:CHO(g) Hz(g) was studied by monitoring
###### Question:
The decomposition of ethanol at some constant temperature (above 500*C), over a copper surface, CzHsOH(g) CH:CHO(g) Hz(g) was studied by monitoring the total pressure with time The following data were obtained: (s) Ptotal (torr) 112 67 133 102 144 204 176 236 186 278 199 What will be the total pressure at t = 332 s? 1 pts Submit Answer Tries 0/5 What is the rate constant (k)? (Include appropriate units;_ 1 pts
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In clinical applications, the unit parts per million (ppm) is used to express very small concentrations of solute, where 1 ppm1 ppm is equivalent to 1 mg1 mg of solute per 1 L1 L of solution. Calculate the concentration in parts per million for each of the solutions. There is 15 μg15 μg of cal... | 388 | 1,420 | {"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} | 2.765625 | 3 | CC-MAIN-2023-40 | latest | en | 0.883019 |
http://mathforum.org/blogs/powerfulideas/teachers-corner-4/ | 1,498,240,851,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320077.32/warc/CC-MAIN-20170623170148-20170623190148-00353.warc.gz | 255,039,462 | 7,133 | # Teacher’s Corner
by Lois Burke
How I found POWs, used them, lost them and eventually came back to them!
I actually found the Math Forum POW’s many, many years ago (we won’t talk about how long). When I first started using them I taught mostly Geometry and I really loved finding non-traditional problems for my kids to try. I started by working a few in class, in small groups. I was trying to get my kids to take some risks and to start diving in to problems that weren’t quite as “direct” as the ones often offered in textbooks. I was looking to stretch their thinking and increase their confidence. Over time my kids become great at taking those risks and really did jump in whenever I gave them a POW. They competed with one another as to who could come up with the most unique solution and whose solution might be the most elegant – and what that meant. As a student myself, I always did the problem the “hard” way — I was the brute force mathematician — so it was always fun to see how they would work the problem. The ideas they would come up with were amazing!
Then came state testing and suddenly, I didn’t feel like I had time for the PoWs. Had to prepare for those state tests. My kids did well but the ones who didn’t do well were struggling. Why? What weren’t they getting? After looking at test questions and working with and talking to kids for a while, it seemed that the kids who were struggling just didn’t seem to be able to handle any problem that didn’t look like the five examples we had done in class. Anything new threw them for a loop. They just didn’t know how to attack it. Enter the PoW’s. I decided that the only way my kids could build their problem solving muscles was to do exactly that – problem-solve!
The love affair starts again. Now I make a habit of including PoW’s as often as possible. We take time to notice and wonder and I’ve noticed (no pun intended) that noticing and wondering has made the transition over to the regular ‘ol math lessons on the typical stuff too!
“Ms. Burke, I noticed that when you multiplied (x + 3)(x – 3) the middle canceled out. I wonder if that always happens when multiplying two binomials when one is positive and one is negative?”
“I noticed that when you have one root at 4 + i then you have another at 4 – i. I wonder why that is?”
Now that being said…. It didn’t happen overnight. The first one went “ok” – and that’s about as enthusiastic as I can be about it. I was still getting used to teaching with the PoWs; they were still getting used to learning with them. The noticing and wondering we did in class was crummy. For example, when we did The Function Challenge (#628), they noticed that there were five functions; and they noticed that two of them were quadratics — and that was it. Come on… really… that’s it? Nothing about the lines? And I was at a loss as to how to get them to think more deeply. The solutions weren’t much better – they lacked detail and were often just an answer, even though I had given them a rubric and gone over it in class. Again… really…Yuck… what to do next?
Enter the importance of feedback and revision! I started having my kids submit their answers to me in a Google doc. I told them that I would give them feedback, provided they worked on the PoW early – before it was due (this was a mistake, by the way; I’ll explain shortly). I provided lots of comments in their documents, pushing them to think more deeply and explain more thoroughly. It was hard to come up with comments that didn’t give them answers. I decided, though, that this might be a good place to model for them some noticing and wondering.
“I noticed that you looked at the graphs and which was higher. Good thinking! I wonder if you thought to compare the compositions? What function is created by B(D(x))? What about D(B(x))? Which is larger algebraically or graphically? “
The students who worked on it ahead of the due date got lots of feedback. Be prepared: giving quality feedback takes time but it is so worth it! The kids who took the time, read the feedback and even asked questions back did much better than the ones who left it until the last minute. Lesson learned….
Enter next PoW and the importance of the scenario vs. the actual problem we did Don’t be Square #736. The first time I had used the scenario but I wasn’t as prepared as I should have been. This time I was ready! We put up EVERY single thing they noticed – EVERYTHING! We put up EVERY single thing they wondered – EVERYTHING! Crazy stuff went up there! It was fun! Then we started analyzing our thinking. What seemed important? What did we think the question might be? This bugged them at first – no question was there….
“I can’t do this … there isn’t a question? Why are you putting up a problem that’s not a problem?”
They also hated that I wouldn’t give any value one way or another to their responses. I just asked them to repeat for the class and the class decided what they thought we should think about and how one idea could connect to another. It was probably one of the best classes I’ve had in a long time! They got it.
Their responses were so much better too! Everyone had to submit a rough draft this time. No exceptions. Two due dates: one for the rough draft that I gave them feedback on, and one for the final. This way I got to comment on ALL of them. No one got to wait until the last minute. It was a beautiful sight to behold. Even parents got into it! One parent helped his daughter with the assignment rough draft and actually thought they needed Calculus. She took great pleasure in going home and explaining it to him “the easy way!”
My students were thinking! Hallelujah! Loads of things improved: those in-class “noticings and wonderings” I mentioned earlier; their explanations to each other — and their willingness to to get up and explain in front of the whole class.They were taking risks and enjoying getting that right answer and really understanding how they got that answer. They didn’t balk at a problem that they didn’t get right away. They dove right in! Success! I didn’t feel like I was just teaching math but teaching how to solve problems – whether they were math or not. Even when the kids hadn’t seen that type of problem before, they still felt they could try it . Any problem seemed doable — even those on the state tests.
To give kids confidence … the ability to persist … the ability to communicate — that’s huge!! | 1,480 | 6,502 | {"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} | 3.28125 | 3 | CC-MAIN-2017-26 | longest | en | 0.986281 |
https://www.careerindia.com/entrance-exam/gate-aerospace-engineering-syllabus-s39.html | 1,619,104,230,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039610090.97/warc/CC-MAIN-20210422130245-20210422160245-00135.warc.gz | 778,599,451 | 26,329 | Exam
# GATE:AEROSPACE ENGINEERING Syllabus
AE Aerospace Engineering
Important Note for Candidates : In each of the following subjects the topics have been divided into two categories – Core Topics and Special Topics. The corresponding sections of the question paper will contain 90% of their questions on Core Topics and the remaining 10% on Special Topics.
Section1: Engineering Mathematics
Core Topics:
Linear Algebra: Vector algebra, Matrix algebra, systems of linear equations, rank of a matrix, eigenvalues and eigenvectors.
Calculus: Functions of single variable, limits, continuity and differentiability, mean value theorem, chain rule, partial derivatives, maxima and minima, gradient, divergence and curl, directional derivatives. Integration, Line, surface and volume integrals. Theorems of Stokes, Gauss and Green.
Differential Equations: First order linear and nonlinear differential equations, higher order linear ODEs with constant coefficients. Partial differential equations and separation of variables methods.
Special Topics:
Fourier Series, Laplace Transforms, Numerical methods for linear and nonlinear algebraic equations, Numerical integration and differentiation. Complex analysis. Probability and statistics.
Section 2: Flight Mechanics
Core Topics:
Basics: Atmosphere: Properties, standard atmosphere. Classification of aircraft. Airplane (fixed wing aircraft) configuration and various parts; Pressure altitude; equivalent, calibrated, indicated air speeds; Primary Flight Instruments : Altimeter, ASI, VSI, Turn-bank indicator. Angle of attack, sideslip; roll, pitch & yaw controls. Aerodynamics forces and moments.
Airplane performance: Drag polar; takeoff and landing; steady climb & descent, absolute and service ceiling; cruise, cruise climb, endurance or loiter; load factor, turning flight, V-n diagram; Winds: head, tail & cross winds.
Static stability: Stability & control derivatives : longitudinal stick fixed & free stability, horizontal tail position and size; directional stability, vertical tail position and size; dihedral stability. Wing dihedral, sweep & position; hinge moments, stick forces;
Special Topics:
Dynamic stability: Euler angles; Equations of motion; aerodynamic forces and moments, stability & control derivatives; decoupling of longitudinal and lateral-directional dynamics; longitudinal modes; lateral-directional modes.
Section 3: Space Dynamics
Core Topics:
Central force motion, determination of trajectory and orbital period in simple cases. Kepler's Laws; escape velocity.
No Special Topics:
Orbit transfer, in-plane and out-of-plane.
Section 4: Aerodynamics
Core Topics:
Basic Fluid Mechanics: Conservation laws: Mass, momentum (Integral and differential form); Dimensional analysis and dynamic similarity; Potential flow theory : Sources, Sinks, doubets, line vortex and their superposition. Elementary ideas of various flows including boundary layers.
Airfoils and wings: Airfoil nomenclature; Aerodynamic coefficients: lift, drag and moment; Kutta-Joukoswki theorem; Thin airfoil theory, Kutta condition, starting vortex; Finite wing theory: Induced drag, Prandtl lifting line theory; Critical and drag divergence Mach number.
Compressible Flows: Basic concepts of compressibility, Conservation equations; One dimensional compressible flows, Fanno flow, Rayleigh flow; Isentropic flows, normal and oblique shocks, Prandtl-Meyer flow; Flow through nozzles and diffusers.
Special Topics:
Wind Tunnel Testing: Measurement and visualization techniques. Shock - boundary layer interaction.
Section 5: Structures
Core Topics:
Strength of Materials: States of stress and strain. Stress and strain transformation. Mohr's Circle, Principal stresses, Three-dimensional Hooke's law, Plane stress and strain. Failure theories: Maximum stress, Tresca and von Mises; Strain energy. Castigliano's principles. Statically determinate and indeterminate trusses and beams. Elastic flexural buckling of columns.
Flight vehicle structures: Characteristics of aircraft structures and materials. Torsion, bending and flexural shear of thin-walled sections. Loads on aircraft.
Structural Dynamics:. Free and forced vibrations of undamped and damped SDOF systems. Free vibrations of undamped 2-DOF systems.
Special Topics:
Vibration of beams. Theory of elasticity: Equilibrium and compatibility equations, Airy’s stress function.
Section 6: Propulsion
Core Topics:
Basics: Thermodynamics, boundary layers and heat transfer and combustion and thermochemistry.
Aerothermodynamics of aircraft engines: Thrust, efficiency, range. Brayton cycle. Engine Performance : ramjet, turbojet, turbofan, turboprop and turboshaft engines. Afterburners.
Turbomachinery : Axial compressors: Angular momentum, work and compression, characteristic performance of a single axial compressor stage, efficiency of the compressor and degree of reaction, multi-staging. Centrifugal compressor: Stage dynamics, inducer, impeller and diffuser. Axial turbines : Stage performance.
Rocket propulsion: Thrust equation and specific impulse, rocket Performance, multi-staging. Chemical rockets, performance of solid and liquid propellant rockets.
Special Topics :
Aerothermodynamics of non-rotating propulsion components such as intakes, combustor and nozzle. Turbine blade cooling. Compressor - turbine matching, Surge and Stall. | 1,092 | 5,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2021-17 | latest | en | 0.81958 |
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Post Patron
## SUM of Count of Distinct Values
Hello,
I have a column [Name] full of names (text).
I'm my visualization I am applying the 'Count Distinct' summarization. I would like to create a measure that will simply SUM the total of distinct names, but I can't figure it out at all.
Any help would be greatly appreciated.
1 ACCEPTED SOLUTION
Community Champion
Try the following formula:
``````User Distinct =
VAR t =
SUMMARIZE (
'Table',
Table[Date],
"UserDistinct", DISTINCTCOUNT ( Table[Name] )
)
RETURN
SUMX ( t, [UserDistinct] )``````
7 REPLIES 7
Hi, I'd like help in a similar DAX formula.
To one extent I'm calculating: the distinctive count where the sale is greater than 0 of a string field.
I would like to get the total or sum of that count, since the result is not adding the values of each row to me.
I look forward to your help, thank you very much
Community Champion
Try the following formula:
``````User Distinct =
VAR t =
SUMMARIZE (
'Table',
Table[Date],
"UserDistinct", DISTINCTCOUNT ( Table[Name] )
)
RETURN
SUMX ( t, [UserDistinct] )``````
Regular Visitor
Thank you
That's great! your formula has worked for me. I had the same problem.
Post Patron
That seems to have done it! Thank you so much.
Super User
Hi. What do you mean with SUM the total of distinct names?
You you just want the number of the "Count Distinct" summarization you can write a dax measure like
``Measure = DISTINCTCOUNT ( Table[Column] )``
Hope that helps,
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Happy to help!
Post Patron
Thanks for your response. I'll try to clarify.
If I look at a date, say 1/21/2021, using Count Distinct I am returned 90 unique names. If I go a day earlier, I see 85 unique names.
If I select two days, I still only see 90 unique names, where I would like to see the SUM of both of those days'-worth unique names.
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# 12 Exceptional Free Printable Math Worksheets Grade 2
Bailee Maelys. Worksheets. August 26th , 2021.
Kids of all ages respond well to fun and games. Using fun and interesting management worksheets for children, children would likely respond much better to the underlying management than if they were forced to sit with an advisor and discuss their problems. Children may not always be capable of explaining what they feel but through worksheets they may be unconscientiously be dealing with their anger issues. If searching for anger management worksheets for children, a person might inquire at a local community medical institution. The web is useful in providing sources like management worksheets for children. Recognizing a youngster has a problem with anger is the first step. Assisting them to cope with their anger is the next and most significant step to anger management in children.
There might be great books with other helpful tools to encourage young people to learn to read. For example, a book might add fun characters which help a child become more engaged in what they are trying to learn. They might add a cartoon character that is familiar so that the sounds become more memorable. Another aid in learning is the setting in which a child learns. A child might learn better one on one compared to a large class setting. The parent might be able to encourage a kid to read better. The use of phonics worksheets will just enhance the learning process.
Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child‘s grasp of a subject is deeper and comprehensive. There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign ”%.” Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
Most of us tend to think of bingo as a game played as a leisure activity, mostly by older people. However, it is also the case that variations on the standard game of bingo are now being used by many teachers and educators. Bingo has in fact been applied to teaching a wide variety of different subjects including reading, English, foreign languages such as French, Spanish, German and Italian, and math, science, history and geography.
There are few things that we will do that will be more important than planning for our retirement. A good retirement plan means enjoying your golden years with financial security. That is the desire of everyone. Good retirement planning now can make it happen. Retirement planning requires organization and good foresight. There are so many variables to be taken into consideration and so many projections that will need to be made. A retirement planning worksheet helps pull all these factors together into a workable plan.
Children need to learn how to read, and whether or not they need extra tutoring with the subject there are many learning tools out there to help a person read including phonics worksheets. Many of these items can be found at stores who sell these learning tools, and they can be purchased by the teacher or a parent.
Unfortunately, this is easier said than done. The trouble with kids these days is that they don‘t have much patience. They don‘t practice the work that you give them at home, which makes progress much slower. They feel bored and lazy to go through several pages of notes. They instead prefer being with their friends, or spending time before the computer.
### Editor' Pick
Aug 30, 2021 | Eglantine Louison
Aug 30, 2021 | Gemma Hannah
### Printable Math Worksheets
Aug 30, 2021 | Julia Marilou
#### Place Value Worksheets
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##### Free Math Worksheets
Aug 30, 2021 | Chantell Naomie
###### Alphabet Worksheets
Aug 30, 2021 | Jocelina Éline
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There are two approaches to this problem, a brute force one that tries all ordering and a more analytical one that tries to build up the alphabetically first ordering one cow at a time.
Because there are only $8$ cows, there are $8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$ different orderings, which is small enough that we can try all of them. If we generate them in alphabetic ordering and stop when we see one that satisfies all the given constraints, then we can print the answer then and there. Here is Brian Dean's code following this approach:
#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
using namespace std;
vector<string> cows, beside_a, beside_b;
int N;
int where(string c)
{
for (int i=0; i<8; i++)
if (cows[i]==c) return i;
return -1;
}
bool satisfies_constraints(void)
{
for (int i=0; i<N; i++)
if (abs(where(beside_a[i]) - where(beside_b[i])) != 1) return false;
return true;
}
int main(void)
{
ifstream fin ("lineup.in");
ofstream fout ("lineup.out");
fin >> N;
cows.push_back("Beatrice");
cows.push_back("Belinda");
cows.push_back("Bella");
cows.push_back("Bessie");
cows.push_back("Betsy");
cows.push_back("Blue");
cows.push_back("Buttercup");
cows.push_back("Sue");
string a, b, t;
for (int i=0; i<N; i++) {
fin >> a;
fin >> t; // must
fin >> t; // be
fin >> t; // milked
fin >> t; // beside
fin >> b;
beside_a.push_back(a);
beside_b.push_back(b);
}
do {
if (satisfies_constraints()) {
for (int i=0; i<8; i++) fout << cows[i] << "\n";
break;
}
} while (next_permutation(cows.begin(), cows.end()));
return 0;
}
The more analytic approach tries to build the ordering one cow at a time. We start by asking the question - can Beatrice be the very first cow in the ordering? It turns out that the answer is yes, if and only if Beatrice must be next to at most one cow. If Beatrice has to be next to two cows, then one of the cows must be in front of her. On the other hand, if Beatrice needs to be next to only one cow, then we can put Beatrice first in line, and then the cow who needs to be beside her goes immediately after. Similarly, if Beatrice doesn't need to be next to any cows, we can move her to the front of the line.
We can loop over the cows in alphabetic order to find the cow that should go first in line. What about the cows that come after? If the cow currently at the end of the ordering must be next to some other cow, then that cow is forced to be next in line. Otherwise, we are free to pick any cow we wish, and we apply the procedure in the previous paragraph to figure out the next cow to put in line.
Here is Brian Dean's code simulating this approach:
#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
using namespace std;
vector<string> cows, beside_a, beside_b, answer;
int N;
int where(string c)
{
for (int i=0; i<answer.size(); i++)
if (answer[i]==c) return i;
return 999;
}
bool can_go_first(string c)
{
int n = answer.size(), nbrs=0;
if (where(c)!=999) return false;
for (int i=0; i<N; i++) {
if (beside_a[i]==c && where(beside_b[i])==999) nbrs++;
if (beside_b[i]==c && where(beside_a[i])==999) nbrs++;
}
if (nbrs == 2) return false;
if (n>0) {
string last_cow = answer[n-1];
for (int i=0; i<N; i++) {
if (beside_a[i]==last_cow && where(beside_b[i])==999 && beside_b[i]!=c) return false;
if (beside_b[i]==last_cow && where(beside_a[i])==999 && beside_a[i]!=c) return false;
}
}
return true;
}
int main(void)
{
ifstream fin ("lineup.in");
ofstream fout ("lineup.out");
fin >> N;
cows.push_back("Beatrice");
cows.push_back("Belinda");
cows.push_back("Bella");
cows.push_back("Bessie");
cows.push_back("Betsy");
cows.push_back("Blue");
cows.push_back("Buttercup");
cows.push_back("Sue");
string a, b, t;
for (int i=0; i<N; i++) {
fin >> a;
fin >> t; // must
fin >> t; // be
fin >> t; // milked
fin >> t; // beside
fin >> b;
beside_a.push_back(a);
beside_b.push_back(b);
}
for (int i=0; i<8; i++) {
int next_cow = 0;
while (!can_go_first(cows[next_cow])) next_cow++; | 1,183 | 4,039 | {"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} | 3.5625 | 4 | CC-MAIN-2023-40 | latest | en | 0.728187 |
http://discourse.iapct.org/t/autonomy-reply/1381 | 1,713,840,094,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818452.78/warc/CC-MAIN-20240423002028-20240423032028-00336.warc.gz | 9,639,453 | 5,667 | [Hans Blom, 960422]
(Rick Marken (960419.1000))
I agree that it is possible, in principle (assuming no limits on the
loop variables), to get r11 and r12 to reach all possible
combinations of values by suitably manipulating d1 and d2; the
environment certainly _influences_ what we want.
In the example, the equations indicate that _nothing but_ the
disturbances vary the goals; all else is constants. So what else
"influences"?
What I have been saying is simply that the inanimate environmental
does not _control_ what we want, because the inanimate environment
is not organized as a control system.
This is circular reasoning. If the environment is taken as not being
a control system, by definition it cannot control. If, however, we
assume (just for a minute, because we're going to drop that heretical
thought immediately after this thought experiment that the
environment COULD POSSIBLY control, we might try to find out how the
"disturbances" of the world change an organism's goal. That's the
thought experiment that I did. I'm not telling you that any of this
is "true"; it's just modelling...
Other control systems _can_ control what we want, to some extent,
but they cannot control what we want _arbitrarily_. That is, an
external controller cannot control what we want without taking into
account the fact that we have other wants that might conflict with
the want that the external controller wants us to have; ...
In the analyzed example, there were no other goals. For argument's
sake, however, it is easy to show that N goals can be set with N
disturbances. So let a manipulator manipulate a large number of
disturbances, equal to the number of the organism's intermediate
goals, and ALL the organism's (intermediate) goals are controlled by
the environment.
I think you are unwilling to call people "autonomous" if an external
controller can control people's wants _at all_.
No. In the example hierarchical controller that I analyzed, ALL goals
could be simultaneously manipulated. A sufficiently powerful
manipulator (Gaia? God?) can control ALL goals. Of course, we mere
mortals can't ;-).
Greetings,
Hans
···
================================================================
Eindhoven University of Technology Eindhoven, the Netherlands
Dept. of Electrical Engineering Medical Engineering Group
email: j.a.blom@ele.tue.nl
Great man achieves harmony by maintaining differences; small man
achieves harmony by maintaining the commonality. Confucius
[From Bruce Gregory (960422.1130 EDT)]
(Hans Blom, 960422)
I am confused by many things, but in particular by your statement
If, however, we assume (just for a minute, because we're going
to drop that heretical thought immediately after this thought
experiment that the environment COULD POSSIBLY control, we
might try to find out how the "disturbances" of the world change
an organism's goal.
Granted we assume that the environment could be a control system.
The next step, it seems to me, would be to test for the controlled
variable. Do you have in mind any environmental variable that
would pass this test?
Regards,
Bruce G.
[Hans Blom, 960423e]
(Rick Marken (960422.1300))
Hans' claim (about the ability to control wants with disturbances)
is _only_ true if we ignore reality.
Yes, you can take that as a fact ;-). Unless I temporarily and mis-
takenly forsake my own principles, I only talk about models, not
So there is not much of a range over which wants can be controlled
when those wants are part of a high gain control hierarchy. The
higher the gain of the control hierarchy, the more autonomous (less
controllable) it is.
That's welcome progress after the black-and-white views of autonomy!
Greetings,
Hans
···
================================================================
Eindhoven University of Technology Eindhoven, the Netherlands
Dept. of Electrical Engineering Medical Engineering Group
email: j.a.blom@ele.tue.nl
Great man achieves harmony by maintaining differences; small man
achieves harmony by maintaining the commonality. Confucius
[From Rick Marken (960423.0730)]
Me:
The higher the gain of the control hierarchy, the more autonomous (less
controllable) it is.
Hans Blom (960423e) --
That's welcome progress after the black-and-white views of autonomy!
Aren't you the fellow who said (Hans Blom, 960416):
Thus there are no goals that an organism can call "its own" in any meaningful
way.
Sounds pretty black (or white) to me.
I can't tell you how nice it is to hear that I have made "welcome progress
after the black-and-white views of autonomy" from the fellow who started
this whole thing off by saying that control systems are _not_ autonomous.
Well, actually, I can tell you how nice it is: it's as nice as getting a
lecture on values from a corportate executive who just got a million dollar
raise after laying off 20% of his employees.
Best
Rick | 1,119 | 4,888 | {"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-2024-18 | latest | en | 0.946784 |
http://mathhelpforum.com/statistics/61183-combinations-question-help-print.html | 1,524,745,397,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948126.97/warc/CC-MAIN-20180426105552-20180426125552-00167.warc.gz | 200,712,657 | 2,717 | # Combinations Question - Help
• Nov 23rd 2008, 12:53 PM
cnmath16
Combinations Question - Help
23. From a group of 6 women and 4 men, determine in how many ways a committee of 4 people can be selected with
a) no restriction
C (10,4) = 10! ÷ 6! 4! = 210 ways
b) 4 women
c) 3 women and 1 man
d) 2 women and 2 men
e) 4 men
.. if someone could do b and c for me... I will figure out how to do the rest on my own.
Thanks!
• Nov 23rd 2008, 12:57 PM
Plato
b) \$\displaystyle {6 \choose 4}\$
c) \$\displaystyle {6 \choose 3}{4 \choose 1}\$
• Nov 23rd 2008, 01:10 PM
cnmath16
Therefore.. the answer to D) will be ( 6 2) ( 4 2)
and the answer to e will be.... (6 0) (4 4) -- for an answer of one? Or is this a special case since all of the men are being used for the selection. The answer of just 1 does not seem correct to me. I was thinking more along the lines of 4! for an answer of 24? | 315 | 890 | {"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} | 3.546875 | 4 | CC-MAIN-2018-17 | latest | en | 0.90623 |
https://stats.stackexchange.com/questions/601587/why-use-the-bootstrap-for-a-skewed-distribution-when-you-can-use-a-transform?noredirect=1 | 1,713,349,302,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817146.37/warc/CC-MAIN-20240417075330-20240417105330-00893.warc.gz | 495,201,691 | 42,999 | # Why use the bootstrap for a skewed distribution when you can use a transform?
Let's say you are working with a statistic (say, the mean of the population) of a skewed distribution with a long, long tail such that confidence intervals must be very skewed to achieve reasonable coverage precision for reasonably high n (<100) samples. You can't sample anymore because it costs too much.
OK, so you think you want to bootstrap.
But why?
Why not simply transform the sample using something like the Box-Cox transform (or similar)?
When would you absolutely choose one over the other or vice-versa? It's not clear to me how to strategize between the two.
In my case, I want to construct confidence intervals to make inferences about the population mean on a non-transformed scale. So I just assume I could transform, construct intervals, then reverse-transform and save myself the trouble with the bootstrap.
This obviously is not a popular choice. But why isn't it?
• Why would you delete your other question?
– Dave
Jan 11, 2023 at 7:38
• I didn't understand what I was asking when I posed it and the question I posed originally didn't make any sense (as your response showed). I was too embarrassed to keep up. Jan 11, 2023 at 7:40
• It appears as though the answer is simply that you cannot reverse-transform and get back what you want: stats.stackexchange.com/questions/1713/… Jan 11, 2023 at 7:44
• Why not use bootstrap, since it addresses the statistic you are interested in, and without transforming the data ? ... On the other hand, if you have a very skewed distribution, the mean may not be the best estimate of central tendency. It may be more meaningful to, say, log transform the data and look at the geometric mean. Jan 14, 2023 at 15:01
• @SalMangiafico Bootstrap is very computationally demanding when you have many populations and statistics to estimate. Jan 14, 2023 at 16:41
If you are interested in the mean and confidence interval for the observed data, probably the most sensible approach is to use the mean and bootstrapped confidence intervals.
For the kind of data set described in the question (100 observations), this shouldn't be too computationally intensive. For example, the following R code, with 100 observation and 10000 replications of the bootstrap, took about 6 seconds at the following site: rdrr.io/snippets/.
But often, if you have a very skewed data set, the mean may not be the best statistic for the central tendency.
It's not uncommon to run analyses on the transformed data, and then back transform the results. But this isn't an estimate of the original e.g. mean and confidence interval. For example, in the case of log-normal data, the result is the geometric mean.
The following example generates some log-normal data. The result of the mean and confidence interval for the original data is quite distinct from the back-transformed mean and confidence interval. In this case it is the difference between the mean and geometric mean.
Either of these approaches may be desirable depending on what you want to know.
set.seed(sum(utf8ToInt("Sal2023")))
Observed = rlnorm(100, 2, 0.8)
hist(Observed)
library(boot)
Function = function(input, index){
Input = input[index]
Result = mean(Input)
return(Result)}
n = length(Observed)
Function(Observed, 1:n)
Boot = boot(Observed, Function, R=10000)
boot.ci(Boot, conf = 0.95, type = "perc")
mean(Observed)
### Mean and confidence interval of the original data by bootstrap
###
### Level Percentile
### 95% ( 8.019, 11.180 )
###
### Mean
### 9.506951
Transformed = log(Observed)
hist(Transformed)
TTestTrans = t.test(Transformed)
CITrans = c(TTestTrans$$estimate, TTestTrans$$conf.int[1], TTestTrans\$conf.int[2])
names(CITrans)=c("Mean", "Lower.ci", "Upper.ci")
CITrans
### Mean Lower.ci Upper.ci
### 1.940237 1.777612 2.102863
BackTrans = exp(CITrans)
BackTrans
### Back-transformed statistics
###
### Mean Lower.ci Upper.ci
### 6.960401 5.915710 8.189579
For R users --- with the caveat that I wrote the functions --- the following can be used to get the bootstrapped confidence interval for the mean of the original data, and the back-transformed confidence interval for the geometric mean.
if(!require(rcompanion)){install.packages("rcompanion")}
library(rcompanion)
Data = data.frame(Observed)
groupwiseMean(Observed ~ 1, data=Data, percentile=TRUE, traditional=FALSE, R=10000)
groupwiseGeometric(Observed ~ 1, data=Data)
### .id n Mean Conf.level Percentile.lower Percentile.upper
### 1 <NA> 100 9.51 0.95 8.02 11.1
### .id n Geo.mean sd.lower sd.upper se.lower se.upper ci.lower ci.upper
### 1 <NA> 100 6.96 3.07 15.8 6.41 7.55 5.92 8.19
• Looking.. but given the data was generated on log scale, shouldn't it have been exponentiated before t test? Then, log-scaled back? The histograms look fine. Jan 17, 2023 at 19:00
• No, for the analysis, the Observed data is log-transformed, the confidence interval is determined, and then the data is back-transformed. This is because the Observed data is approximately log-normal, and the log-transformed data is approximately normal. If Observed data were different, a different transformation could be used. Jan 17, 2023 at 19:40
• OK, more to understand here. I get your points. If we've transformed a sample, then, and computed CI's, is there no easy to way to transform them back to the original scale such that they are valid for the sample pre-transformation? Jan 17, 2023 at 19:55
• I wouldn't say they're not valid. It's just that they don't address the parameter on the original scale. In the case of log-transformed data, we can call the mean the "geometric mean". If we use a different transformation, we could make up a name for it. ... Note that in the case I presented, the back-transformed confidence interval doesn't even contain the mean of the original data. Jan 17, 2023 at 20:22
• I did notice that. When I say valid, I meant, achieving the same coverage rate. Jan 17, 2023 at 20:50 | 1,521 | 6,062 | {"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": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-18 | latest | en | 0.955197 |
https://gist.github.com/MrThatKid/2a8525a1abe1cb95073ef4eb64590b00 | 1,726,315,886,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00182.warc.gz | 254,322,549 | 20,593 | Instantly share code, notes, and snippets.
# MrThatKid/squaredigital.md
Last active July 15, 2017 05:41
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• Save MrThatKid/2a8525a1abe1cb95073ef4eb64590b00 to your computer and use it in GitHub Desktop.
Porting from nITG Square & Digital to SM5 and vice versa.
## Reasoning
In order to make Square and Digital have the same period as Zigzag and Sawtooth in SM5, the period had to be changed. This means that syncing these modifiers using the same period precentage in SM5 is easy to do.
However, this also means that some math needs to be done when porting files that use Square & Digital to/from nITG.
Since the x position mods and z position mods used two different periods, two sets of equations are needed.
## Equation notes
Note that -80% period would be input as -0.8. Getting 1.95 as a result is equivalent to 195%. ARROW_SIZE is usually 64, so use that in calculations. These equations also apply to the tangent version of Digital.
## From nITG to SM5
Square & Digital: PeriodSM5 = ( PI * ( 60 + ( 60 * PeriodnITG ) ) - ARROW_SIZE ) / ARROW_SIZE
SquareZ & DigitalZ: PeriodSM5 = PI * ( 1 + PeriodnITG ) - 1
## From SM5 to nITG
Square & Digital: PeriodnITG = ( ( ARROW_SIZE + ( ARROW_SIZE * PeriodSM5 ) ) / ( 60 * PI ) ) - 1
SquareZ & DigitalZ: PeriodnITG = ( ( 1 + PeriodSM5 ) / PI ) - 1 | 389 | 1,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.871151 |
http://research.stlouisfed.org/fred2/graph/?id=OSHK755SRVO | 1,409,382,533,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1408500834494.74/warc/CC-MAIN-20140820021354-00447-ip-10-180-136-8.ec2.internal.warc.gz | 161,378,657 | 17,241 | # Graph: All Employees: Other Services in Oshkosh-Neenah, WI (MSA)
Click and drag in the plot area or select dates: Select date: 1yr | 5yr | 10yr | Max to
Release:
Restore defaults | Save settings | Apply saved settings
w h
Graph Background: Plot Background: Text:
Color:
(a) All Employees: Other Services in Oshkosh-Neenah, WI (MSA), Thousands of Persons, Seasonally Adjusted (OSHK755SRVO)
The data services of the Federal Reserve Bank of St. Louis include series that are seasonally adjusted. To make these adjustments, we use the X-12 Procedure of SAS to remove the seasonal component of the series so that non-seasonal trends can be analyzed. This procedure is based on the U.S. Bureau of the Census X-12-ARIMA Seasonal Adjustment Program. More information on this program can be found at http://www.census.gov/srd/www/x12a/.
The seasonal moving average function used is that of the Census Bureau’s X-11-ARIMA program. This includes a 3x3 moving average for the initial seasonal factors and a 3x5 moving average to calculate the final seasonal factors. The D11 function is also used to output the entire seasonally adjusted series that is displayed. For specific information on the SAS X-12 procedure, please visit their website: http://support.sas.com/documentation/cdl/en/etsug/60372/HTML/default/viewer.htm#etsug_x12_sect001.htm.
All Employees: Other Services in Oshkosh-Neenah, WI (MSA)
Integer Period Range: to copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
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Graph updated.
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https://www.redhotpawn.com/forum/science/speedometer.144944 | 1,542,466,190,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743714.57/warc/CC-MAIN-20181117144031-20181117165406-00015.warc.gz | 1,003,249,833 | 9,331 | # speedometer
joe shmo
Science 02 Feb '12 02:47
1. joe shmo
Strange Egg
02 Feb '12 02:47
Driving without a speedometer is liberating, at first you fight to control your speed, but eventually you just give up and drive. Why do we think that is the case?
2. joe shmo
Strange Egg
02 Feb '12 05:081 edit
and while we are at it
Given the matrix
D = [.07, .20, .31, .19, .27; .35, .24, .16, .13, .05; .09, .25, .21, .29, .23; .26, .25, .06, .24, .15; .22, .14, .18, .08, .21]
Using the Leontief Input - Output Model
with idustries
1. Auto
2. Steel
3. Electricity
4. Coal
5. Chemical
How would you decipher the following question.
Determine the industry on which the steel industry is most dependent.
That is to say, will steel be the input or output?
The way I am reading it it could be either or.
Steel is most dependent on Electricity for its production, and/or is most dependent on Auto for its consumption.
To answer the question as it is stated should I then compare it its 2 largest dependencies against each other...
That is to say Steel is most dependent on Auto, because Auto > Electricity ( .35>.25) ?
3. joe shmo
Strange Egg
03 Feb '12 05:50
By the way, I'd still like to discuss the first question...
4. Thequ1ck
Fast above
03 Feb '12 10:06
Originally posted by joe shmo
Driving without a speedometer is liberating, at first you fight to control your speed, but eventually you just give up and drive. Why do we think that is the case?
Pfft...
5. joe shmo
Strange Egg
03 Feb '12 10:21
Originally posted by Thequ1ck
Pfft...
Not exactly the depth of thought I wanted to reach, but its a start.
6. sonhouse
Fast and Curious
03 Feb '12 17:281 edit
Originally posted by joe shmo
Not exactly the depth of thought I wanted to reach, but its a start.
I guess he is referring to the sound of the car going by at 200 Km/hr because his speedo broke....
7. 08 Feb '12 18:36
Originally posted by joe shmo
Driving without a speedometer is liberating, at first you fight to control your speed, but eventually you just give up and drive. Why do we think that is the case?
It might be because the engineers who design roads recommend speed limits (or design them to specified speed limits) that match most people's kinesthetic (balance, reaction to accelerative and centrifugal forces, etc.)and reaction-time comfort zones, with respect to their driving speed.
I live in an area with a lot of hills and curved streets, by the way, so this explanation seems to jump out at me.
But especially, after driving for years with a speedometer, you may have internalized the visual, aural, and kinestethic effects of driving as it relates to speed. Instead of looking at the speedometer to judge speed, we judge it "automatically."
8. 08 Feb '12 18:47
Originally posted by joe shmo
Driving without a speedometer is liberating, at first you fight to control your speed, but eventually you just give up and drive. Why do we think that is the case?
The problem isn't that you go with the wrong speed, according to the current conditions. The problems arise when you and the patrol police don't agree.
9. joe shmo
Strange Egg
08 Feb '12 21:04
Originally posted by JS357
It might be because the engineers who design roads recommend speed limits (or design them to specified speed limits) that match most people's kinesthetic (balance, reaction to accelerative and centrifugal forces, etc.)and reaction-time comfort zones, with respect to their driving speed.
I live in an area with a lot of hills and curved streets, by the way, s ...[text shortened]... to speed. Instead of looking at the speedometer to judge speed, we judge it "automatically."
I agree with this, I distinctly remember bieng aware of the low frequency in checking the speedometer before it was broke. I wonder what the general trend in my driving speed will be relative to what is legal as I continue to drive without one?
10. joe shmo
Strange Egg
08 Feb '12 21:11
Originally posted by FabianFnas
The problem isn't that you go with the wrong speed, according to the current conditions. The problems arise when you and the patrol police don't agree.
Thats what I was refering to when I said it was liberating, its nice to not worry about breaking rules...but I don't think they will agree.
11. 08 Feb '12 22:08
Originally posted by joe shmo
Thats what I was refering to when I said it was liberating, its nice to not worry about breaking rules...but I don't think they will agree.
I drive carefully, yes I do. But I know friends, often younger males, that doesn't have the ability to foresee what could happen if they drive to fast around a closed corner if they meet someone. I feel unsafe when I go with them and they drive. They drive carelessly.
Because of some hotheads we have to have the speed restrictions. If they kill themselves, fine, but when they hit other people, I object.
12. joe shmo
Strange Egg
09 Feb '12 03:38
Originally posted by FabianFnas
I drive carefully, yes I do. But I know friends, often younger males, that doesn't have the ability to foresee what could happen if they drive to fast around a closed corner if they meet someone. I feel unsafe when I go with them and they drive. They drive carelessly.
Because of some hotheads we have to have the speed restrictions. If they kill themselves, fine, but when they hit other people, I object.
Governace just confuses me...The fear of kaos institutes it, and simultaneosly as kaos is replaced by order, a fear of the instituted order replaces the original fear of kaos...end result we are always afraid?
13. Soothfast
0,1,1,2,3,5,8,13,21,
09 Feb '12 05:47
Upon reflection I suppose the only reason why I ever look at my speedometer is to make sure I'm not going much more than 10 mph over the speed limit -- and that's only because of a desire to avoid a heightened probability of a citation, not because the speed limit has any inherent validity.
Speed limits are instituted primarily as municipal revenue enhancement devices, and not because of safety concerns which at best are secondary. It is ironic: having to check a speedometer constantly because of speed limits almost certainly is a form of distracted driving that increases the risk of accidents and fatalities. I wonder how many people have been killed by a car because the driver, for a critical fraction of a second, was consulting a speedometer?
Authoritarianism sucks.
14. Soothfast
0,1,1,2,3,5,8,13,21,
09 Feb '12 05:52
Originally posted by FabianFnas
I drive carefully, yes I do. But I know friends, often younger males...
I know of quite a few bubble-headed bimbos who drive like they got their license out of a Cracker Jack box and took five shots of tequila to celebrate...
15. 10 Feb '12 11:20
My grandmother, rest her soul, used to get in the car, pull the choke out and hang her handbag on it. 😀
She drove for 50 years, and never had an accident. She'd seen thousands, through her rear view mirror, though. 😉
-m. | 1,746 | 6,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2018-47 | latest | en | 0.948862 |
https://www.studypool.com/questions/282200/Need-help-with-the-Willow-Run-Outlet-Mall-for-my-Statistics-Question- | 1,477,348,053,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719784.62/warc/CC-MAIN-20161020183839-00039-ip-10-171-6-4.ec2.internal.warc.gz | 1,010,550,976 | 750,590 | # Statistics Question Help
SoccerBoss
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The Willow Run Outlet Mall has two Haggar Outlet Stores, one located on Peach Street and the other on Plum Street. The two stores are laid out differently, but both store managers claim their layout maximizes the amounts customers will purchase on impulse. A sample of ten customers at the Peach Street store revealed they spent the following amounts more than planned: \$17.58, \$19.73, \$12.61, \$17.79, \$16.22, \$15.82, \$15.40, \$15.86, \$11.82, \$15.85. A sample of fourteen customers at the Plum Street store revealed they spent the following amounts more than they planned when they entered the store: \$18.19, \$20.22, \$17.38, \$17.96, \$23.92, \$15.87, \$16.47, \$15.96, \$16.79, \$16.74, \$21.40, \$20.57, \$19.79, \$14.83. For Data Analysis, a t-Test: Two-Sample Assuming Unequal Variances was used.
At the .01 significance level is there a difference in the mean amount purchased on an impulse at the two stores? Explain these results to a person who knows about the t test for a single sample but is unfamiliar with the t test for independent means.
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2325 Tutors | 503 | 1,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2016-44 | longest | en | 0.827416 |
https://ricerca.univaq.it/handle/11697/135248 | 1,721,160,580,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00310.warc.gz | 439,251,529 | 13,197 | In the present paper, a Vortex Particle Method is combined with a Boundary Element Method for the study of viscous incompressible planar flow around solid bodies. The method is based on Chorins operator splitting approach for the Navier–Stokes equations written in vorticity–velocity formulation, and consists of an advection step followed by a diffusion step. The evaluation of the advection velocity exploits the Helmholtz–Hodge Decomposition, while the no-slip condition is enforced by an indirect boundary integral equation. The above decomposition and splitting are discussed in comparison to the analogous decomposition for the pressure-velocity formulation of the governing equations. The Vortex Particle Method is implemented with a completely meshless algorithm, as neither advection nor diffusion requires topological connection of the point lattice. The results of the meshless approach are compared with those obtained by a mesh-based Finite Volume Method, where the pseudo-compressible iteration is exploited to enforce the solenoidal constraint on the velocity field. Several benchmark tests were performed for verification and validation purposes. In particular, we analyzed the two-dimensional flow past a circle, past an ellipse with incidence and past a triangle for different Reynolds numbers.
### Chorin's approaches revisited: Vortex Particle Method vs Finite Volume Method
#### Abstract
In the present paper, a Vortex Particle Method is combined with a Boundary Element Method for the study of viscous incompressible planar flow around solid bodies. The method is based on Chorins operator splitting approach for the Navier–Stokes equations written in vorticity–velocity formulation, and consists of an advection step followed by a diffusion step. The evaluation of the advection velocity exploits the Helmholtz–Hodge Decomposition, while the no-slip condition is enforced by an indirect boundary integral equation. The above decomposition and splitting are discussed in comparison to the analogous decomposition for the pressure-velocity formulation of the governing equations. The Vortex Particle Method is implemented with a completely meshless algorithm, as neither advection nor diffusion requires topological connection of the point lattice. The results of the meshless approach are compared with those obtained by a mesh-based Finite Volume Method, where the pseudo-compressible iteration is exploited to enforce the solenoidal constraint on the velocity field. Several benchmark tests were performed for verification and validation purposes. In particular, we analyzed the two-dimensional flow past a circle, past an ellipse with incidence and past a triangle for different Reynolds numbers.
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Utilizza questo identificativo per citare o creare un link a questo documento: `https://hdl.handle.net/11697/135248`
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• 14 | 616 | 3,130 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-30 | latest | en | 0.890472 |
https://www.mumsnet.com/Talk/secondary/2452771-Marks-for-Edexcel-Maths-GCSE | 1,493,543,780,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917124478.77/warc/CC-MAIN-20170423031204-00281-ip-10-145-167-34.ec2.internal.warc.gz | 948,262,468 | 25,444 | # Talk
## Marks for Edexcel Maths GCSE
(15 Posts)
thunderbird69 Thu 20-Aug-15 11:18:27
DS got his results and for all subjects except for maths it shows the marks and breakdowns for each paper. Do Edexcel not give the marks?
He wants to see how far off a A* he was
TeenAndTween Thu 20-Aug-15 12:34:16
That's what I want to know too. I'd like to know whether DD missed the next grade up for maths by a whisker or a mile.
wed99 Thu 20-Aug-15 12:44:52
My daughter missed an A* by 1 mark . The teachers at school today had all the raw scores for each paper and her maths teacher told us to request a re-mark. So presumably your schools should be able to tell you how close you were to the next grade.
Draylon Thu 20-Aug-15 12:52:12
Ditto! DS took EDEXL/GC (?) Mathematics A (linear) Option H- that's the only Maths exam listed- surely there were two? Calc and non-calc? He got a 'B' with '46 points'.
What does that mean?
All his other exams have up to 4 different test results listed.
thunderbird69 Thu 20-Aug-15 12:57:40
I think 46 points is the number of points allocated to a grade B rather than the test result
I phoned the school and they said that is the way the results are printed out. As wed99 says, the schools obviously have access to the marks. Seemed most huffy that I actually wanted to know what marks he got, but eventually let me leave a message for the exam administrator
Draylon Thu 20-Aug-15 13:25:04
Yes, I thinks so, too, but I wonder if 50 points gets you an A? Or 45 points a C? Kind of important as DS was planning on doing a Maths AS level! A near-miss A is still OK; an almost C isn't.
SuperMoonIsKeepingMeUpToo Thu 20-Aug-15 14:00:36
The 46 will be the percentage he gained overall. Each paper is out of 100 marks so he would have scored a total of 92 out of a possible 200 marks. The grade boundaries are here:
Actually that states a total of 95 marks were required for a grade B so I'm not sure how he was awarded a B. That's how I understand it anyway. Well done him for getting his B though!
titchy Thu 20-Aug-15 14:25:56
46 is the standard points given for aB grade (in the same way that a C grade A level is worth 80 points).
You will need to get the actual mark off the school to see how close he was to the A. Annoying isn't it - dd didn't bother to get the raw mark for Maths either.
Thu 20-Aug-15 21:29:25
Just marking my place, as I have the same question. I'm guessing that EDEXEL haven't released the marks for some reason. I'm going to ask the school to let me know when they have them.
WJEC English Lang is also missing for DS.
The points are how much each GCSE grade is worth, just as A'levels have points.
A* 58 points,
A 52
B 46
C 40
D 34
E 28
F 22
G 16
Draylon Fri 21-Aug-15 03:10:51
Thanks
noblegiraffe Fri 21-Aug-15 08:51:12
The school will have the raw marks. Edexcel have certainly released theirs.
I taught maths courses by Edexcel, OCR and AQA this year. None of the actual marks appeared on the students' exam results, but I had sheafs of paper from each exam board with them on.
I don't know why they don't put them on for maths, it's because it's a linear course, but students do still want to know how they scored on each paper and total.
thunderbird69 Wed 26-Aug-15 12:57:09
Has anyone been able to get hold of their marks? I managed to speak to someone at school today and they tell me they haven't got them. The head of Maths may have them but won't be in till start of term.
MEgirl Wed 26-Aug-15 14:57:49
They certainly should have them. We were given them on results day and submitted our request for remark the following day.
TeenAndTween Wed 26-Aug-15 16:13:24
I asked the school and got an answer back the same day. DD got 119 which means she missed an A by 6 marks.
That is the best I could have expected of her, she never quite made it to an A in any mocks, so I am delighted that she maintained her standard for the final exam. (And much better than being 1 or 2 marks off).
lifesalongsong Wed 26-Aug-15 22:09:20
I spoke to the exams officers at my child's school today and asked her for the mark, she just looked at a list on her PC and told me the individual marks for each paper | 1,145 | 4,188 | {"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-2017-17 | latest | en | 0.976786 |
https://www.physicsforums.com/threads/stuck-on-chem-hw-help.355087/ | 1,508,443,486,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823462.26/warc/CC-MAIN-20171019194011-20171019214011-00589.warc.gz | 988,044,675 | 16,614 | # Stuck on chem hw HELP
1. Nov 15, 2009
### yaganon
First of all, how do you know if a combustion process produce h2o(g) or h2o(l)??? It depends on the temperature, but how do you figure that out?
I found the heat released during the combustion processes, which should equate the work done by the system (theoretically). But according to w = -nRT[ln(V2/V1)], I need the initial volume and final volume. The pressure remains constant (I believe), and V can be figured by PV=nRT with P=1 atm throughout. And with that, I need the temperature.
to figure out the temperature, I need the specific heat. But what specific heat do I need? those of the reactants or the products??? my brain iis hurrrts badd
btw, here are the reactants: C8H18 (Octane), C2H5OH (ethanol), methane ch4, h2, and nh4.
Last edited: Nov 15, 2009
2. Nov 15, 2009
### pzona
This should probably be in the homework section, but I'll try to answer as much as I can. Combustions usually produce more than 373K of heat, so most of the time it's safe to assume it's gaseous water that is produced, at least initially. I'm pretty sure you need the specific heat of the environment in which the reaction is taking place, not the reactants or products. I could be wrong, but that's what I think I remember from high school chem. Also, do you have to use these specific equations? Work can also be expressed by P(delta V), or it can be figured out by E=q+w. Do either of these help?
3. Nov 16, 2009
### yaganon
mmm not really
I was thinking that if you're operating under standard conditions, water would be liquid since it would be something like
2H2(g) + O2(g) --> 2H2O(l) + ...KJ, in which case the heat would evaporate the liquid water.
4. Nov 16, 2009
### Staff: Mentor
Combustion always produce gaseous water (after all, products of combustion are hot), but standard enthalpy - as reported in tables - is for products at STP. That means liquid.
--
5. Nov 16, 2009
### NotJohnson
Thinking about it logically, for one reaction of the type (g) + (g) -> products, any water produced would have to be gaseous since there would only be a few molecules of it, not enough to condense into liquid. If you're assuming an isolated system I would say it is safe to assume gaseous water since any liquid water would be condensed and removed from the system.
If you want to do it using numbers, either a specific heat for the environment, or you're going to need the specific heat for all reactants and products since they will all be present in the system, absorbing energy and raising temperature. Unless you're assuming the reaction has gone to completion, in that case, just products.
6. Nov 17, 2009
### Staff: Mentor
I can't see any logic here. Number of molecules is irrelevant. Besides, nothing stops you from burning tonnes of gas.
--
methods
7. Nov 17, 2009
### NotJohnson
and nothing stops tonnes of gas from increasing the volume to become just as sparse. I'm saying that in a gaseous system the spread of molecules (by definition of gas) makes it incredibly unlikely that water will be formed in a liquid phase, because by definition liquid requires densely (relative to gas) packed molecules. I'm not saying a liquid phase won't form over time, it just won't be instantaneous.
8. Nov 17, 2009
### Staff: Mentor
Sorry but I still have problems understanding what you mean. Basically you write "it won't, but it will". Either either. Condensation - no matter how unlikely you think it is - occurs all the time around us, and by definition it requires gaseous phase.
-- | 895 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2017-43 | longest | en | 0.958295 |
https://mathcs.org/analysis/reals/numseq/answers/convseq2.html | 1,519,147,678,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813059.39/warc/CC-MAIN-20180220165417-20180220185417-00478.warc.gz | 719,252,529 | 2,495 | # Interactive Real Analysis
Next | Previous | Glossary | Map
## 3.1. Sequences
### Examples 3.1.10(b):
Define x1 = b and let xn = xn - 1 / 2 for all n > 1. Then this sequence converges for any number b.
The proof is very easy using the theorem on monotone, bounded sequences:
• b > 0: the sequence is decreasing and bounded below by 0.
• b < 0: the sequence is increasing and bounded above by 0
• b = 0: the sequence is constantly equal to zero
In either case the sequence converges. As to finding the actual limit, we proceed as follows: we already know that the limit exists. Call that limit L. Then we have:
lim xn = L = lim xn + 1
But then we have that
L = lim xn + 1 = lim xn / 2 = 1/2 lim xn = 1/2 L
so that we have the equation for the unknown limit L:
L = 1/2 L
Therefore, the limit must be zero.
This proof illustrates the advantage of knowing that a sequence converges. Based on that fact it was easy to determine the actual limit of this recursively defined sequence. On the other hand, it would be very difficult to try to establish convergence based on the original definition of a convergent sequence.
Next | Previous | Glossary | Map | 304 | 1,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2018-09 | longest | en | 0.918629 |
https://in.mathworks.com/help/fixedpoint/ref/le.html | 1,627,595,136,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153897.89/warc/CC-MAIN-20210729203133-20210729233133-00212.warc.gz | 337,366,765 | 16,210 | # le
Determine whether real-world value of `fi` object is less than or equal to another
## Syntax
```c = le(a,b) a <= b ```
## Description
`c = le(a,b)` is called for the syntax ```a <= b``` when `a` or `b` is a `fi` object. `a` and `b` must have the same dimensions unless one is a scalar. A scalar can be compared with another object of any size.
`a <= b` does an element-by-element comparison between `a` and `b` and returns a matrix of the same size with elements set to `1` where the relation is true, and `0` where the relation is false.
In relational operations comparing a floating-point value to a fixed-point value, the floating-point value is cast to the same word length and signedness as the `fi` object, with best-precision scaling.
## Examples
collapse all
Use the `le` function to determine whether the real-world value of one `fi` object is less than or equal to another.
```a = fi(pi); b = fi(pi, 1, 32); a <= b```
```ans = logical 0 ```
Input `a` has a 16-bit word length, while input `b` has a 32-bit word length. The `le` function returns `0` because after quantization, the value of `a` is greater than that of `b`.
When comparing a double to a `fi` object, the double is cast to the same word length and signedness of the `fi` object.
```a = fi(pi); b = pi; a <= b```
```ans = logical 1 ```
The `le` function casts `b` to the same word length as `a`, and returns `1` because the two inputs have the same real-world value. This behavior allows relational operations to work between `fi` objects and floating-point constants without introducing floating-point values in generated code. | 422 | 1,621 | {"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} | 2.828125 | 3 | CC-MAIN-2021-31 | latest | en | 0.868867 |
https://cran.microsoft.com/snapshot/2018-02-25/web/packages/emmeans/vignettes/messy-data.html | 1,680,203,787,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00235.warc.gz | 236,552,029 | 20,162 | # Working with messy data
## Issues with observational data
In experiments, we control the conditions under which observations are made. Ideally, this leads to balanced datasets and clear inferences about the effects of those experimental conditions. In observational data, factor levels are observed rather than controlled, and in the analysis we control for those factors and covariates. It is possible that some factors and covariates lie in the causal path for other predictors. Observational studies can be designed in ways to mitigate some of these issues; but often we are left with a mess. Using EMMs does not solve the inherent problems in messy, undesigned studies; but they do give us ways to compensate for imbalance in the data, and allow us to estimate meaningful effects after carefully considering the ways in which they can be confounded.
As an illustration, consider the nutrition dataset provided with the package. These data are used as an example in Milliken and Johnson (1992), Analysis of Messy Data, and contain the results of an observational study on nutrition education. Low-income mothers are classified by race, age category, and whether or not they received food stamps (the group factor); and the response variable is a gain score (post minus pre scores) after completing a nutrition training program. First, let’s fit a model than includes all main effects and 2-way interactions, and obtain its “type II” ANOVA:
nutr.lm <- lm(gain ~ (age + group + race)^2, data = nutrition)
car::Anova(nutr.lm)
## Note: model has aliased coefficients
## sums of squares computed by model comparison
## Anova Table (Type II tests)
##
## Response: gain
## Sum Sq Df F value Pr(>F)
## age 82.37 3 0.9614 0.4145
## group 658.13 1 23.0441 6.105e-06
## race 11.17 2 0.1956 0.8227
## age:group 91.58 3 1.0688 0.3663
## age:race 87.30 3 1.0189 0.3880
## group:race 113.70 2 1.9906 0.1424
## Residuals 2627.47 92
There is definitely a group effect and a hint of and interaction with race. Here are the EMMs for those two factors:
emmeans(nutr.lm, ~ group * race)
## group race emmean SE df lower.CL upper.CL
## FoodStamps Black 4.708257 2.368117 92 0.004971359 9.411542
## NoAid Black -2.190399 2.490576 92 -7.136898097 2.756099
## FoodStamps Hispanic nonEst NA NA NA NA
## NoAid Hispanic nonEst NA NA NA NA
## FoodStamps White 3.607680 1.155619 92 1.312521470 5.902838
## NoAid White 2.256336 2.389273 92 -2.488966678 7.001638
##
## Results are averaged over the levels of: age
## Confidence level used: 0.95
Hmmmm. The EMMs when race is “Hispanic” are not given; instead they are flagged as non-estimable. What does that mean? Well, when using a model to make predictions, it is impossible to do that beyond the linear space of the data used to fit the model. And we have no data for three of the age groups in the Hispanic population:
with(nutrition, table(race, age))
## age
## race 1 2 3 4
## Black 2 7 10 2
## Hispanic 0 0 3 0
## White 5 16 51 11
We can’t make predictions for all the cases we are averaging over in the above EMMs, and that is why some of them are non-estimable. The bottom line is that we simply cannot include Hispanics in the mix when comparing factor effects. That’s a limitation of this study that cannot be overcome without collecting additional data. Our choices for further analysis are to focus only on Black and White populations; or to focus only on age group 3. For example (the latter):
summary(emmeans(nutr.lm, pairwise ~ group | race, at = list(age = "3")),
by = NULL)
## $emmeans ## group race emmean SE df lower.CL upper.CL ## FoodStamps Black 7.500000e+00 2.672054 92 2.193071 12.8069292 ## NoAid Black -3.666667e+00 2.181723 92 -7.999756 0.6664229 ## FoodStamps Hispanic 2.131628e-14 5.344107 92 -10.613858 10.6138584 ## NoAid Hispanic 2.500000e+00 3.778855 92 -5.005131 10.0051312 ## FoodStamps White 5.419355e+00 0.959830 92 3.513050 7.3256601 ## NoAid White -2.000000e-01 1.194979 92 -2.573331 2.1733309 ## ## Confidence level used: 0.95 ## ##$contrasts
## contrast race estimate SE df t.ratio p.value
## FoodStamps - NoAid Black 11.166667 3.449606 92 3.237 0.0017
## FoodStamps - NoAid Hispanic -2.500000 6.545168 92 -0.382 0.7034
## FoodStamps - NoAid White 5.619355 1.532726 92 3.666 0.0004
(We used trickery with providing a by variable, and then taking it away, to make the output more compact.) Evidently, the training program has been beneficial to the Black and White groups in that age category. There is no conclusion for the Hispanic group – for which we have very little data.
Back to Contents
## Mediating covariates
The framing data in the mediation package has the results of an experiment conducted by Brader et al. (2008) where subjects were given the opportunity to send a message to Congress regarding immigration. However, before being offered this, some subjects (treat = 1) were first shown a news story that portrays Latinos in a negative way. Besides the binary response (whether or not they elected to send a message), the experimenters also measured emo, the subjects’ emotional state after the treatment was applied. There are various demographic variables as well. Let’s a logistic regression model, after changing the labels for educ to shorter strings.
framing <- mediation::framing
levels(framing\$educ) <- c("NA","Ref","< HS", "HS", "> HS","Coll +")
framing.glm <- glm(cong_mesg ~ age + income + educ + emo + gender * factor(treat),
family = binomial, data = framing)
The conventional way to handle covariates like emo is to set them at their means and use those means for purposes of predictions and EMMs. These adjusted means are shown in the following plot.
emmip(framing.glm, treat ~ educ | gender, type = "response")
This plot gives the impression that the effect of treat is reversed between male and female subjects; and also that the effect of education is not monotone. Both of these are counter-intuitive.
However, note that the covariate emo is measured post-treatment. That suggests that in fact treat (and perhaps other factors) could affect the value of emo; and if that is true (as is in fact established by mediation analysis techniques), we should not pretend that emo can be set independently of treat as was done to obtain the EMMs shown above. Instead, let emo depend on treat and the other predictors – easily done using cov.reduce – and we obtain an entirely different impression:
emmip(framing.glm, treat ~ educ | gender, type = "response",
cov.reduce = emo ~ treat*gender + age + educ + income)
The reference grid underlying this plot has different emo values for each factor combination. The plot suggests that, after taking emotional response into account, male (but not female) subjects exposed to the negative news story are more likely to send the message than are females or those not seeing the negative news story. Also, the effect of educ is now nearly monotone.
By the way, the results in this plot are the same is what you would obtain by refitting the model with an adjusted covariate
emo.adj <- resid(lm(emo ~ treat*gender + age + educ + income, data = framing))
… and then using ordinary covariate-adjusted means at the means of emo.adj. This is a technique that is often recommended.
If there is more than one mediating covariate, their settings may be defined in sequence; for example, if x1, x2, and x3 are all mediating covariates, we might use
emmeans(..., cov.reduce = list(x1 ~ trt, x2 ~ trt + x1, x3 ~ trt + x1 + x2))
(or possibly with some interactions included as well).
Back to Contents
## Mediating factors and weights
A mediating covariate is one that is in the causal path; likewise, it is possible to have a mediating factor. For mediating factors, the moral equivalent of the cov.reduce technique described above is to use weighted averages in lieu of equally-weighted ones in computing EMMs. The weights used in these averages should depend on the frequencies of mediating factor(s). Usually, the "cells" weighting scheme described later in this section is the right approach. In complex situations, it may be necessary to compute EMMs in stages.
As described in the “basics” vignette, EMMs are usually defined as equally-weighted means of reference-grid predictions. However, there are several built-in alternative weighting schemes that are available by specifying a character value for weights in a call to emmeans() or related function. The options are "equal" (the default), "proportional", "outer", "cells", and "flat".
The "proportional" (or "prop" for short) method weights proportionally to the frequencies (or model weights) of each factor combination that is averaged over. The "outer" method uses the outer product of the marginal frequencies of each factor that is being averaged over. To explain the distinction, suppose the EMMs for A involve averaging over two factors B and C. With "prop", we use the frequencies for each combination of B and C; whereas for "outer", first obtain the marginal frequencies for B and for C and weight proportionally to the product of these for each combination of B and C. The latter weights are like the “expected” counts used in a chi-square test for independence. Put another way, outer weighting is the same as proportional weighting applied one factor at a time; the following two would yield the same results:
{r eval = FALSE}
emmeans(model, "A", weights = "outer")
emmeans(emmeans(model, c("A", "B"), weights = "prop"), weights = "prop")
Using "cells" weights gives each prediction the same weight as occurs in the model; applied to a reference grid for a model with all interactions, "cells"-weighted EMMs are the same as the ordinary marginal means of the data. With "flat" weights, equal weights are used, except zero weight is applied to any factor combination having no data. Usually, "cells" or "flat" weighting will not produce non-estimable results, because we exclude empty cells. (That said, if covariates are linearly dependent with factors, we may still encounter non-estimable cases.)
Here is a comparison of predictions for nutr.lm defined above, using different weighting schemes:
sapply(c("equal", "prop", "outer", "cells", "flat"), function(w)
predict(emmeans(nutr.lm, ~ race, weights = w)))
## equal prop outer cells flat
## [1,] 1.258929 1.926554 2.546674 0.3809524 0.6865079
## [2,] NA NA NA 1.6666667 1.2500000
## [3,] 2.932008 2.522821 3.142940 2.7951807 1.6103407
For group * race EMMs, the results for "prop" and "flat" are the same because only one factor (age) is averaged over.
Back to Contents
## Nested fixed effects
A factor A is nested in another factor B if the levels of A have a different meaning in one level of B than in another. Often, nested factors are random effects—for example, subjects in an experiment may be randomly assigned to treatments, in which case subjects are nested in treatments—and if we model them as random effects, these random nested effects are not among the fixed effects and are not an issue to emmeans. But sometimes we have fixed nested factors.
Here is an example of a fictional study of five fictional treatments for some disease in cows. Two of the treatments are administered by injection, and the other three are administered orally. There are varying numbers of observations for each drug. The data and model follow:
cows <- data.frame (
route = factor(rep(c("injection", "oral"), c(5, 9))),
drug = factor(rep(c("Bovineumab", "Charloisazepam",
"Angustatin", "Herefordmycin", "Mollycoddle"), c(3,2, 4,2,3))),
resp = c(34, 35, 34, 44, 43, 36, 33, 36, 32, 26, 25, 25, 24, 24)
)
cows.lm <- lm(resp ~ route + drug, data = cows)
The ref_grid function finds a nested structure in this model:
cows.rg <- ref_grid(cows.lm)
cows.rg
## 'emmGrid' object with variables:
## route = injection, oral
## drug = Angustatin, Bovineumab, Charloisazepam, Herefordmycin, Mollycoddle
## Nesting structure: drug %in% route
When there is nesting, emmeans computes averages separately in each group
route.emm <- emmeans(cows.rg, "route")
route.emm
## route emmean SE df lower.CL upper.CL
## injection 38.91667 0.5908902 9 37.57998 40.25335
## oral 28.02778 0.4491457 9 27.01174 29.04382
##
## Results are averaged over the levels of: drug
## Confidence level used: 0.95
… and insists on carrying along any grouping factors that a factor is nested in:
drug.emm <- emmeans(cows.rg, "drug")
drug.emm
## drug route emmean SE df lower.CL upper.CL
## Bovineumab injection 34.33333 0.7474236 9 32.64254 36.02412
## Charloisazepam injection 43.50000 0.9154032 9 41.42921 45.57079
## Angustatin oral 34.25000 0.6472878 9 32.78573 35.71427
## Herefordmycin oral 25.50000 0.9154032 9 23.42921 27.57079
## Mollycoddle oral 24.33333 0.7474236 9 22.64254 26.02412
##
## Confidence level used: 0.95
Here are the associated pairwise comparisons:
pairs(route.emm, reverse = TRUE)
## contrast estimate SE df t.ratio p.value
## oral - injection -10.88889 0.742215 9 -14.671 <.0001
##
## Results are averaged over the levels of: drug
pairs(drug.emm, by = "route", reverse = TRUE)
## route = injection:
## contrast estimate SE df t.ratio p.value
## Charloisazepam - Bovineumab 9.166667 1.1817804 9 7.757 <.0001
##
## route = oral:
## contrast estimate SE df t.ratio p.value
## Herefordmycin - Angustatin -8.750000 1.1211353 9 -7.805 0.0001
## Mollycoddle - Angustatin -9.916667 0.9887484 9 -10.030 <.0001
## Mollycoddle - Herefordmycin -1.166667 1.1817804 9 -0.987 0.6026
##
## P value adjustment: tukey method for comparing a family of (varies) estimates
In the latter result, the contrast itself becomes a nested factor in the returned emmGrid object. That would not be the case if there had been no by variable.
### Auto-identification of nested factors – avoid being trapped!
ref_grid() and emmeans() tries to discover and accommodate nested structures in the fixed effects. It does this in two ways: first, by identifying factors whose levels appear in combination with only one level of another factor; and second, by examining the terms attribute of the fixed effects. In the latter approach, if an interaction A:B appears in the model but A is not present as a main effect, then A is deemed to be nested in B. Note that this can create a trap: some users take shortcuts by omitting some fixed effects, knowing that this won’t affect the fitted values. But such shortcuts do affect the interpretation of model parameters, ANOVA tables, etc., and I advise against ever taking such shortcuts. Here are some ways you may notice mistakenly-identified nesting:
• A message is displayed when nesting is detected
• A str() listing of the emmGrid object shows a nesting component
• An emmeans() summary unexpectedly includes one or more factors that you didn’t specify
• EMMs obtained using by factors don’t seem to behave right, or give the same results with different specifications
To override the auto-detection of nested effects, use the nesting argument in ref_grid() or emmeans(). Specifying nesting = NULL will ignore all nesting. Incorrectly-discovered nesting can be overcome by specifying something akin to nesting = "A %in% B, C %in% (A * B)" or, equivalently, nesting = list(A = "B", C = c("A", "B")).
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Volume word problems (practice) | Khan Academy
Find volume of rectangular prisms to solve word problems.
### Volume Math Problems
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#### Volume 1 (practice) | Khan Academy
Find volume of a rectangular prism with labeled side lengths. Find a missing side length on ... Volume 1. CCSS Math: 5.MD.C.5, 5.MD. .... Do 7 problems. CheckĀ ...
## Volume Math Problems
Volume word problems: fractions & decimals (practice) | Khan ...
Practice solving volume word problems involving objects like fish tanks, truck beds, and refrigerators.
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Which cylinder has a larger volume and by how much? Fire tank has cuboid shape with a rectangular floor measuring 13. The swimming pool is 4 m wide and 9 m long and 158 cm deep. How many hectoliters of water is in it, if the water is 27 cm below its upper edge. Draw a picture and highlight the body diagonal. How long will take the empty the full pool when both the inlet and outlet are opened at the same time? If a cylindrical tank with volume is used 12320cm raised to the power of 3 and base 28cm is used to store water.
How long pool is filled with two inlets separately? The volume of water in the rectangular swimming pool is 6998. The promotional leaflet states that if we wanted all the pool water to flow into a regular quadrangle with a base edge equal to the average depth of the pool, the prism would have. One cube is inscribed sphere and the other one described. The aquarium has a capacity of 18 liters. .
Please enter e-mail correctly and check whether you dont have a full mailbox. For this particular job, he needs 600ml of 15 bleach. What is its height when the square bottom is 8 23 cm long? If a cylindrical tank with volume is used 12320cm raised to the power of 3 and base 28cm is used to store water. How many liters of water can it hold? If water flows into the pool by two inlets, fill the whole for 18 hours. What is its height when the square bottom is 8 23 cm long? Smith is cleaning up a big mess at home. From the sphere of radius 18 was truncated spherical cap. How much of each type (to the nearest ml) should he. For how many people is 90 liters of soup enough if we assume 38 liter of soup per person in the canteen? The pool has a capacity of 50 hl. First inlet filled pool 10 hour longer than second. Calculate the cube volume, whose body diagonal size is 75 dm.
Problem 1: A rectangular prism of volume 3200 mm 3 has a rectangular base of length 10 mm and width 8 mm. Find the height h of the prism.
#### Volume - math problems - HackMath.net
Volume - solved math problems, problem solving and knowledge review. Problems count: 478.
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Generalized Functions and Coefficients
by
Andrew M. Crabbe
A departmental honors thesis submitted to the Department of at Trinity University in partial fulfillment of the requirements for Graduation with departmental honors
April 20, 2001
Interim Associate V.P. for Academic Affairs ABSTRACT
Let S Z. The generalized factorial for S, denoted n!S, is introduced in accordance⊆ with theory already established by Bhargava ([4]). Along with several known theorems about these functions, a of other issues will be explored. This Thesis is divided into 4 chapters. Chapter 1 provides the necessary definitions and offers a connec- tion between the generalized factorial function and rings of -valued . In Chapter 2, necessary conditions on an infinite of are obtained in order for that sequence to serve as the factorial sequence for some subset S Z. Chapter 3 explores the subject of !-equivalent subsets and we find a condition on two⊆ infinite subsets S and T of Z which force n!S = n!T for every nonnegative integer n. We close in Chapter 4 with an analysis of generalized binomial coefficients, and for a given infinite subset S Z, we characterize those subsets T Z for which n = n . ⊆ ⊆ mS mT Contents
1 Introduction 1
2 Necessary Conditions on Factorial 7
3 !-Equivalent Subsets 14
4 Results Concerning Generalized Binomial Coefficients 19
A Appendix 24 1 Introduction
Most anyone who has taken an undergraduate course in abstract algebra should be some- what familiar with the Q[x]; that is, the the containing all those poly- n n 1 nomials, anx + an 1x − + + a1x + a0, where the ai’s are in Q. As such, Q[x] satisfies all the properties of− a ring.··· In addition, Q[x] is closed under scalar multiplication from elements in Q; so Q[x] is a vector over the field Q. For that reason, we can search for a of this (i.e. a linearly independent subspace spanning Q[x]). One’s first n choice, a correct one, might be x n∞=0. However, for the purposes of this paper, there is a more interesting possibility, but{ first} a definition.
Definition 1.1. Let n be a non-negative integer. If n 1 then set ≥ x x(x 1) (x n + 1) = − ··· − n n! and if n = 0, x = 1. 0 x This “more interesting” prospective basis is n n∞=0, called the set of binomial poly- nomials. Let’s consider a proof of this fact. { }
Proposition 1.2. The set x is a basis for the vector space Q[x] over Q. {n}n∞=0 Proof. There are two parts to this proof. x Q Q 1) Show that n n∞=0 spans [x] (i.e., every of [x] can be expressed as a linear { } x of elements in n n∞=0). We use induction on the degree of the polynomial. For the initial case, let f(x{) be} a polynomial in Q[x] of degree zero or, in other words, Q x f(x) = b0 where b0 is in . So f(x) = b0 0 and the initial case is proven. Now, assume n that the property holds for all polynomials of degree n 1. Let f(x) = a0+a1x+ +anx x ≤ − Q ··· n be a polynomial of degree n. Now ann! n is a polynomial in [x] of degree n, whose x n has leading coefficient an. So, since the x terms will cancel, the rational polynomial x x g(x) = f(x) ann! n has degree n 1, and from our assumption, g(x) = i∞=0 bi i . − x x ≤ − P Thus f(x) = ann! n + i∞=0 bi i , which is the linear combination we’re looking for. Thus, the property is proven forP all rational polynomials. x Q x 2) Show that n n∞=0 is a linearly independent set in [x]. Let f(x) = a0 + a1 1 + x { } + an n be an arbitrary linear combination. Now, for f(x) = 0, it must be that the ··· x x coefficient of the term is zero (i.e., an = 0) since it is the only which features such a n i term. For the same reason (coupled with the fact that an = 0), an 1 = 0 and so on down the line, thus all coefficients are zero and the property is proven. −
We now turn to another ring, the set of all integer-valued polynomials (see [6]), denoted by Int(Z). This set is denoted by,
Int(Z) = p(x) Q[x] p(z) Z, z Z . { ∈ | ∈ ∀ ∈ } 1 In other words, Int(Z) contains all those polynomials in Q[x] that map integers to integers. A few examples of polynomials in the set would be x, 3x2 1, or an integer such as 7. However Int(Z) contains more than just polynomials with integer-coefficients,− such as those x(x 1) Z listed above. For instance, the polynomial 2− is in Int( ) (since either z or z 1 is even for every z in Z), as well as all the other binomial polynomials, as we show in the− following lemma.
Lemma 1.3. x Int(Z), for all n 0. n ∈ ≥ x Z a Z Z Proof. Let n be arbitrary. If n Int( ), then n , for all a . There are a number of cases to consider. ∈ ∈ ∈ 1) Let a n. Then a is a standard binomial coefficient, thus a Z. ≥ n n ∈ 2) Let 0 a n 1. By definition of x , x a is in the numerator, thus a = 0. ≤ ≤ − n − n 3) Let a < 0. Then,
a a(a 1) (a n + 1) = − ··· − = n n! ( a)(1 a) (n 1 a)) n 1 a ( 1)n − − ··· − − = ( 1)n − − . − n! − n Thus the problem reduces to one of the two cases above. Therefore the proof is complete.
Since Int(Z) is a ring, it is closed under addition and also closed under scalar multipli- cation with the integers. Although we can’t properly consider Int(Z) to be a vector space (since Z is not a field), it is a Z-, which turns out to be enough for our purposes (see [7]). Much of the terminology remains the same, least of which is the concept of free basis (or Z-basis, as it will be in this context), which carries over logically into the realm of modules. On that note, what are some possible Z-bases for Int(Z)? It would perhaps be appropriate to look at the example bases from Q[x]. Upon immediate inspection, it can be n x(x 1) seen that x n∞=0 isn’t satisfactory (for instance, there is no way to generate 2− from { } n a linear combination of the elements in x n∞=0 using only integer coefficients). But it so happens that the set, x , is in fact{ a Z}-basis. {n}n∞=0 Proposition 1.4. The set x is a Z-basis for Int(Z). {n}n∞=0 Proof. Here we need to show independence over Z and spanning. 1) The independence of x is shown similarly as in Proposition 1 above. {n}n∞=0 2) For spanning, it suffices to show that every polynomial in Int(Z) is a linear combina- tion over Z of elements in x . To show this, we first need a couple of lemmas. {n}n∞=0 Lemma 1.5. If f(x) and g(x) are polynomials of degree n in Q[x] and f(0) = g(0), f(1) = g(1), ... , f(n) = g(n), then f(x) = g(x). ≤
2 Proof. Using the premises, the polynomial f(x) g(x) has degree n; thus the equation f(x) g(x) = 0 has at most n distinct roots (if it− is not equivalently≤ the zero-polynomial). However− it has been assumed that f(x) = g(x) at least n+1 values; thus f(x) g(x) = 0, x, and f(x) = g(x). − ∀
Lemma 1.6. Given a sequence of integers, b0, b1, . . . , bn, there is a polynomial g(x) = x x x c0 + c1 + + cn , ci Z, i such that g(0) = b0, g(1) = b1, . . . , g(n) = bn. 0 1 ··· n ∈ ∀ Proof. The proof will be by induction on the length of the . x 1) Let the sequence, b0, be of length one. Then g(x) = b0 . 0 2) Assume that the property holds for sequences of length n. Given the sequence ≤ b0, b1, . . . , bn, we find a corresponding g(x). From the assumption, there exists a f(x) = n 1 x − ci for which =0 i f(0) = b0, f(1) = b1, . . . , f(n 1) = bn 1. − − x Consider the polynomial g(x) = f(x)+(bn f(n)) n . Now, for all i such that 0 i n 1, i x− ≤ ≤ − n = 0 (for (x i) is in the numerator of n ). Thus g(i) = f(i) = bi, 0 i n 1. Now, at − n ∀ ≤ ≤ − x n, g(n) = f(n)+(bn f(n)) n = f(n)+bn f(n) = bn. So with g(x) = f(x)+(bn f(n)) n , an appropriate polynomial− has been found,− and the lemma has been proved. −
We return to the proof of Proposition 4. Let f(x) be a polynomial in Int(Z) with degree n, and let f(0) = b0, f(1) = b1, . . . , f(n) = bn. By Lemma 1.6, there is a polynomial n 1 x g(x) = i=0− c1 i for which g(0) = f(0), g(1) = f(1), ... , g(n) = f(n). By Lemma 1.5, f(x) = gP(x), x. Thus the proof is complete. ∀ We now introduce a new ring. If S is a subset of Z, set
Int(S, Z) = p(x) Q[x] p(s) Z, s S . { ∈ | ∈ ∀ ∈ } In other words, Int(S, Z) contains all those polynomials in Q[x] that are integer-valued at the elements of S. An easy observation about this ring is that Int(Z) Int(S, Z) (since a polynomial that is integer-valued for all integers must be integer-valued⊆ for any subset, S, of the integers). Again, for reasons similar to the above, Int(S, Z) is a Z-module. What else can we say about Int(S, Z)? More specifically, can we determine any Z-bases? In order to approach these questions more intelligently, with some hope of success, we turn to Bhargava and his work on generalizing the factorial function for subsets of Z (see [3] and [4]). At the foundation of Bhargava’s theory is a notion called a p-ordering of S (where S is an arbitrary subset of Z). A p-ordering of S is a sequence, ai i∞=0, of elements in S constructed in the following manner. { }
Select any element in S, and denote it as a0.
Select an element a1 S that minimizes the highest power of p dividing a1 a0. ∈ − Select an element a2 S that minimizes the highest power of p dividing (a2 a0)(a2 a1). ∈ − − 3 In general, select an element ak S that minimizes the highest power of p dividing ∈ (ak a0)(ak a1) (ak ak 1). − − ··· − − It should be immediately apparent that there is no unique p-ordering of S since, among other reasons, a0 is chosen arbitrarily (there could also be any number of elements mini- mizing the product at any particular point, from which you can only pick one). Now, if we are given a particular p-ordering of S, we can define a new sequence, νk(S, p) , called { }i∞=0 the associated p-sequence of S. For each k 0, let νk(S, p) be the power of p minimized at the kth step in the p-ordering process. In other≥ words,
νk(S, p) = wp((ak a0) (ak ak 1)) − ··· − − where wp(a) represents the highest power of p dividing a (for instance, w5(50) = 25). From the construction of the p-ordering, it is easy to see that such a sequence must be monotone increasing. What is truly amazing about these associated p-sequences is that they are entirely independent of the choice of p-ordering!
Theorem 1.7. [4, Theorem 5] The associated p-sequence, νk(S, p) k∞=0, is independent of the particular choice of p-ordering of S. { }
To better understand the construction of a p-ordering, we consider Z itself. Proposition 1.8. [4, Proposition 6] The ordering 0, 1, 2,... forms a natural p-ordering of Z for all primes p.
Proof. We again use induction.
1) The a0 can be chosen arbitrarily, so choose 0. By selecting a1 = 1, a1 a0 = 1 0 = 1 − − which obviously minimizes the power of p dividing a1 a0 for all primes p. − 2) Assume that the property holds for the first k 1 steps (i.e., the ordering thus far is 0, 1, 2, . . . , k 1). In the kth step, we want to− minimize the power of p dividing − (ak 0)(ak 1) (ak (k 1)). But regardless of our choice of ak, the product is a product− of k−consecutive··· − integers,− thus divisible by k!. But this k! can be had if k is selected as the ak, which would clearly minimize the power of p dividing the product for all primes. Thus the proof is complete.
With this natural p-ordering, we can determine the unique associated p-sequence for Z as follows:
νk(Z, p) = wp((ak a0) (ak ak 1)) = wp((k 0) (k (k 1))) = wp(k!). − ··· − − − ··· − − Notice that if we were to fix k and have p range over all primes, taking the product of all the resulting νk(Z, p)’s would yield the prime of k!. Thus we can represent k! purely as a product of these νk(Z, p)’s (which are invariant in Z) as,
k! = ν (Z, p). Y k p
4 But since each subset, S, has its own invariant νk(S, p)’s, we can similarly define the gen- eralized factorial function, k!S, as follows:
k! = ν (S, p). S Y k p
If, as in the case with Z, there is a p-ordering, ai ,which holds for all primes simulta- { }i∞=0 neously, then k!S can be written more simply as k!S = (ak a0)(ak a1) (ak ak 1) (see [4, Lemma 16]). | − − ··· − − | Let’s look at a few examples of the generalized factorial function in various subsets of Z. (These examples are taken from [4].)
Example 1.9. Let S = 2Z (i.e. S is the set of even integers). Like Z before, there is a natural p-ordering 0, 2, 4,... which holds for all primes p. Thus
k k!2Z = (2k 0)(2k 2) (2k (2k 2)) = 2 k!. − − ··· − − Example 1.10. Let S be the set of powers of 2 (which are in Z). Again there is a natural ordering 1, 2, 4, 8,... holding for all primes p. Here
k k k k 1 k!S = (2 1)(2 2) (2 2 − ). − − ··· − Example 1.11. Let S be the set of all squares in Z, which we denote by ZS. There is a natural ordering 0, 1, 4, 9,... which holds for all primes. So
2 2 2 2 (2k)! k!ZS = (k 0)(k 1) (k (k 1) )) = . − − ··· − − 2
The task of calculating the generalized factorial function for subsets such as these (sub- sets that are well-structured and bear a natural p-ordering that holds for all primes) is relatively straightforward; though this is certainly not the case with more “perverse” sub- sets. For instance, when S is the set of all primes, we get the result (from [4]):
k 1 + k 1 + k 1 + p−1 p(p− 1) p2(−p 1) k! = pb − b − c b c ···. S Y − p
Moving on, since k!S is called a generalized factorial function, we would expect it to share some of the properties held by the traditional factorial function, k!. A familiar property of the factorial is that for any nonnegative integers n and m, n!m! (n + m)!. A proof of this fact could be presented rather easily, though it’s sufficient for| our purposes just to n+m (n+m)! recall that the binomial coefficient, n = n!m! , is integer-valued. It is hoped that the same could be said in the general case (i.e., for any nonnegative integers n and m, n!Sm!S (n + m)!S). This property is proven in Bhargava (see [4, Theorem 8]), though not without| difficulty (and a number of lemmas), so let the truth of the statement stand
5 without explicit substantiation. Now with this being true, we can define something called the generalized binomial coefficient for S in the logical way:
n n!S = . k k!S(n k)!S S − These coefficients become quite interesting. Each subset of Z has a set of binomial coefficients, so each will have its own Pascal’s Triangle and no doubt a whole range of other interesting characteristics. Returning to Examples 1.9 and 1.11 above, it can be shown, by easy calculation, that n = n , and n = 2 2n . k2Z k kZS 2k It would perhaps be an appropriate time to recall the reason why we found the need to define these p-orderings, generalized , etc. It was our intent to find a basis for the Z-module, Int(S, Z). Recall that for Int(Z) (which could also be written Int(Z, Z)), the basis that we presented was the set of binomial polynomials, x , where {n}n∞=0 x x(x 1) (x k + 1) = − ··· − . n n! Given our new knowledge of p-orderings (specifically that the sequence 0, 1, 2,... forms a p-ordering, ai , on Z for all primes p), we can re-express this polynomial as { }i∞=0 x (x a0)(x a1) (x ak 1) = − − ··· − − . n n!Z We can extend these conclusions further, but first a definition.
Definition 1.12. Let ai,k be a sequence in Z that, for each prime p dividing k!S, is { }i∞=0 termwise congruent modulo νk(S, p) to some p-ordering of S. The purpose of defining such a sequence is that, usually, there is no particular ordering of elements in S that satisfies the p-ordering requirements for all primes p less than some fixed integer (let alone all primes). This sequence at least gives an ordering respecting the idiosyncrasies that exist between p-orderings of primes under a certain bound. But now we state the theorem. Theorem 1.13. The set x forms a basis for the Z-module Int(S, Z), where {nS}n∞=0 x (x a0,n)(x a1,n) (x an 1,n) = − − ··· − − . n S n!S As I don’t intend to offer a proof of this theorem, the interested reader can see [4] for more details (see [4, Theorem 23]). For our purposes, the most important feature of this theorem is that it presents an instance, a context, in which the generalized factorial function reveals itself. (And it, of course, has historical significance, as these leading coefficients were the inspiration for Bhargava’s theory). The purpose of this paper is to further the theory already established by Bhargava. Three general areas of interest will be examined: (i) what are necessary conditions on a factorial sequence, (ii) criteria for !-equivalence of subsets in Z, and (iii) the theory of generalized binomial coefficients.
6 2 Necessary Conditions on Factorial Sequences
Before proceeding further, it would be best to mention a case which isn’t mentioned explic- itly in Bhargava, that in which S Z is a finite subset. This undoubtedly yields different results. For instance, if S = n, then⊆ the construction of the p-ordering must begin repeat- | | ing elements after the (n 1)st step (so the product, (ak a0)(ak a1) (ak ak 1) = 0, for k n). As a result, the− definition of the generalized factorial− function,− ··· − − ≥
k!S = νk(S, p) = wp((ak a0) (ak ak 1)), Y Y − p p − ··· − seems to lose meaning for k n. I think it would make the most sense to regard k!S as ≥ equalling 0 for k n (as this stipulation would preserve n!Sm!S (n + m)!S). But this does ≥ Z x | little to illuminate the structure of Int(S, ), since k S is undefined for k n. Because of its somewhat “diseased” nature, the finite case will be occasionally ignored≥ (though not without warning). Thankfully however, the generation of the factorial sequence for finite subsets lends itself well to programming (see Appendix for such a MAPLE creation). The following theorem describes an extremely important property of generalized factorial functions. Its usefulness cannot be understated.
Theorem 2.1. [4, Lemma 13] Let S T . Then n!T n!S, n 0. ⊆ | ∀ ≥ Proof. Though Bhargava proves this in his paper, there is a rather clever proof involving what we know about integer-valued polynomial rings which is perhaps a more direct. Since S T , Int(T, Z) Int(S, Z) (since a polynomial that is integer-valued for all integers in T must⊆ be integer-valued⊆ for any subset, S, of T ). Now x is a Z-basis for {iS}i∞=0 Int(S, Z). The most pertinent characteristic of x is that its leading coefficient is 1 . iS i!S Z Z x x Since Int(T, ) Int(S, ), n T can be expressed as a linear combination of i S’s with integer coefficients⊆ (where 0 i n). So, ≤ ≤ x x x x x = zn + zn 1 + + z1 + z0 . n n − n 1 ··· 1 0 T S − S S ¿From the fact that the leading coefficient of x (a polynomial of degree n) is 1 and n T n!T zn x that the leading coefficient of the degree-n term on the right side is (from zn ), it n!S n S 1 zn must be that = . Or in other words, znn!T = n!S. Therefore n!T n!S = αn, for all n!T n!S n 0. | ≥ This makes intuitive sense since if we have more elements from which to choose (as we would with T ), the likelihood of finding a more minimizing element in a p-ordering is increased. We can readily recognize 1, 1, 2, 6, 24, 120, 720,... as the factorial sequence for Z; and with a bit more familiarity with generalized factorial functions, we can recognize 1, 2, 8, 48, 384,... as the factorial sequence for 2Z. But for an arbitrary infinite integer sequence,
7 α0, α1, α2,... , how do we know whether there exists an S Z such that the above is its ⊆ factorial sequence? To this effect, the following is a list of conditions on α0, α1, α2,... which are necessary for the existence of such an S.
Theorem 2.2. Let α0, α1, α2,... be an infinite sequence of integers. If there exists an S Z such that the above is its factorial sequence (i.e., n!S = αn), then the following are necessary:⊆
i) αiαj αi+, for i, j 0. | ≥
ii) α0 = 0!S = 1
iii) n! n!S = αn, for all n 0. | ≥ n iv) Let α1 = 1!S = l. Then l n! αn, for all n 0. (And this is the strongest claim we | ≥ can make, given only the value for α1.)
β1 β2 βu 2 γ1 γ2 γv v) Let α1 = 1!S = l = p1 p2 pu and α2 = 2!S = l m22!, where m22! = r1 r2 rv ··· β γ ··· (with p, P). Then considering all q P, where wq(l) = q and wq(m22!) = q , ∈ ∈ γ n + n + n + (2β+γ) n + n + n + q b 2 c b 2q c b 2q2 c ··· q b 2 c b 2q c b 2q2 c ··· Y Y q m2 q m2 |q-l |q l | nβ q νn(Z, q) νn(Z, q) n!S = αn . Y · Y | q-m2 q-m2 q l q-l | Proof. i) This is a result of Bhargava (see [4, Theorem 8]).
ii) We have defined 0!S to be 1 for all S Z, so no proof is needed. ⊆ iii) This is a special case of Theorem 2.1 above. Here we just allow that T = Z. n iv) Let α1 = 1!S = l. Now if l = 1, then we need to prove that 1 n! = n! n!S. But this β1 β2 βu | is merely a restatement of (iii), so we are done. So let l = p1 p2 pu , where the pi’s are ··· βi distinct primes with βi 1. Choose an arbitrary prime , pi, of l (where pi l but βi+1 ≥ βi βi | pi - l). Let a0, a1, a2,... be a pi-ordering for S. Since ν1(S, pi) = pi , pi (a a0) βi βi | − for all a S. In other words, a S, a a0 bi(mod pi ) where 0 bi < pi is fixed. So generally,∈ a S ∀ ∈ ≡ ≡ ≤ ∀ ∈ β1 a b1(mod p ) ≡ 1 β2 a b2(mod p ) ≡ 2 . .
βu a bu(mod p ). ≡ u
8 So by the Chinese Remainder Theorem, there exists a b such that
a b(mod pβ1 pβ2 pβu ), ≡ 1 2 ··· u or a b(mod l). ≡ The largest set in which this holds is obviously T = lZ + b, so S T . From [4, Example n n ⊆ 17], we have that n!T = l n!. Therefore by Theorem 2.1, l n! n!S = αn, for all n 0. | ≥ β1 β2 βu 2 γ1 γ2 γv (v) Let 1!S = l = p1 p2 pu and 2!S = l m22!, where m22! = r1 r2 rv (with p, r P). There are four types··· of primes, q, to consider. ··· ∈ a) Consider those q such that q m2 but q - l. Let q be an arbitrary prime of this γ | type, where ν2(S, q) = q . So for a q-ordering of S, beginning a0, a1, a2,... , we have that γ γ wq((a2 a1)(a2 a0)) = q . It must be that wq(a2 a1) = q and wq(a2 a0) = 1 or vice − − − − versa, since if q (a2 a1) and q (a2 a0), then q (a1 a0). In this case, q l, which | − | − γ| − γ | violates our assumption about q. So if a1 c(mod q ) and a0 d(mod q ), we have that, s S, s c or d(mod qγ). The largest subset,≡ T , such that ≡t T , t c or d(mod qγ) is T∀ =∈ (qγZ ≡+ c) (qγZ + d). ∀ ∈ ≡ ∪ Now we construct the q-ordering for T . Claim: c, d, qγ + c, qγ + d, 2qγ + c, 2qγ + d, . . . is a q-ordering for T . Generally, if n is n γ n 1 γ even, an = 2 q + c, and if n is odd, an = −2 q + d. Since verifying this claim is rather tedious, the proof of it is left to the Appendix (Notes A.1).
With the q-ordering constructed, we can solve for νn(T, q). We need to consider sepa- rately the cases in which n is even and when n is odd. If n is even,
n γ n 2 γ n γ n 2 γ νn(T, q) = wq(( q + c − q d)( q + c − q c) 2 − 2 − 2 − 2 − ··· n n ( qγ + c d)( qγ + c c)). 2 − 2 − γ And since wq(kq + (c d)) = 1, k Z, − ∀ ∈
n n 2 γ n γ n γ nγ n = wq(( − )q ( 1)q ( )q ) = q 2 wq( !) 2 − 2 ··· 2 − 2 2
n n n n n n n n γ 2q + 2 + 3 + γ 2 + 2q + 2 + 3 + = q b 2 c qb c b 2q c b 2q c ··· = q b c b c b 2q c b 2q c ···. · n + n + n + n b 2q c b 2q2 c b 2q3 c ··· (For explanation of why wq( 2 !) = q , see Notes A.2 in Appendix.)
9 If n is odd,
n 1 γ n 1 γ n 1 γ n 3 γ νn(T, q) = wq(( − q + d − q c)( − q + d − q d) 2 − 2 − 2 − 2 − ··· n 1 n 1 ( − qγ + d d)( − qγ + d c)) 2 − 2 −
(n 1)γ n 1 n 3 γ n 1 γ n 1 γ − n 1 = wq(( − − )q ( − 1)q ( − )q ) = q 2 wq( − !) 2 − 2 ··· 2 − 2 2
n 1 n 1 n 1 n 1 n n n n γ − 2−q + −2 + −3 + γ 2q + 2 + 3 + = q b 2 c qb c b 2q c b 2q c ··· = q b 2 c qb c b 2q c b 2q c ··· · ·
γ n + n + n + n + = q b 2 c b 2q c b 2q2 c b 2q3 c ···.
n n 1 (For explanation of why r = −r , see Notes A.3 in Appendix.) b 2q c b 2q c Thus n 0, we have that ∀ ≥ n n n n γ 2 + 2q + 2 + 3 + νn(T, q) = q b c b c b 2q c b 2q c ···.
And since S T , νn(T, q) νn(S, q) or ⊆ | n n n n γ 2 + 2q + 2 + 3 + q b c b c b 2q c b 2q c ··· νn(S, q). |
b) Consider those q such that q m2 and q l. Let q be an arbitrary prime of this type, β |2β γ | where ν1(S, q) = q and ν2(S, q) = q q . So for a q-ordering of S beginning a0, a1, a2,... , β · 2β γ we have that wq(a1 a0) = q and wq((a2 a1)(a2 a0)) = q q . It must be that β+γ − β − − · β wq(a2 a1) = q and wq(a2 a0) = q or vice versa. This is because q (a2 a1) β− − β β+1 | − and q (a2 a0) (else ν1(S, q) < q , which is a contradiction); and if q (a2 a1) and β+1 | − β+1 | β− q (a2 a0), then a1 a0(mod q ) (which implies that qβ+1 wq(a1 a0) = q , another | − ≡ β+γ β+γ | − β contradiction). So if a1 c(mod q ) and a0 d(mod q ) (where wq(c d) = q ), we have that, s S, s c≡or d(mod qβ+γ). The largest≡ subset, T , such that −t T , t c or d(mod qβ+γ∀) is∈ T = (≡qβ+γZ + c) (qβ+γZ + d). ∀ ∈ ≡ ∪ Similar to the above case, c, d, qβ+γ +c, qβ+γ +d, 2qβ+γ +c, 2qβ+γ +d, . . . is a q-ordering for T . This can be shown in a proof analogous to that for the above, so we leave the details to the reader. In order to determine νn(T, q), we must again consider the of n. If n is even,
n β+γ n 2 β+γ n β+γ n 2 β+γ νn(T, q) = wq(( q + c − q d)( q + c − q c) 2 − 2 − 2 − 2 − ··· n n ( qβ+γ + c d)( qβ+γ + c c)) 2 − 2 −
10 n β+γ n 2 β+γ n β+γ n 4 β+γ n β+γ β+γ n β+γ = wq(( q − q )( q − q ) ( q q )( q )) 2 − 2 2 − 2 ··· 2 − 2 · n β+γ n 2 β+γ n β+γ n 4 β+γ wq(( q − q + (c d))( q − q + (c d)) 2 − 2 − 2 − 2 − ··· n n ( qβ+γ qβ+γ + (c d))( qβ+γ + (c d))). 2 − − 2 −
β+γ β And since, wq(kq + (c d)) = q , −
n n 2 β+γ n β+γ n β+γ β n n(β+γ) n nβ = wq(( − )q ( 1)q ( )q ) q 2 = q 2 wq( !) q 2 2 − 2 ··· 2 − 2 · 2 ·
n(2β+γ) n n n n n (2β+γ) 2q + 2 + 3 + = q 2 wq( !) = q b 2 c qb c b 2q c b 2q c ··· 2 ·
(2β+γ) n + n + n + n + = q b 2 c b 2q c b 2q2 c b 2q3 c ···.
If n is odd,
n 1 β+γ n 1 β+γ n 1 β+γ n 3 β+γ νn(T, q) = wq(( − q + d − q c)( − q + d − q d) 2 − 2 − 2 − 2 − ··· n 1 n 1 ( − qβ+γ + d d)( − qβ+γ + d c)) 2 − 2 −
n 1 β+γ n 3 β+γ n 1 β+γ n 5 β+γ = wq(( − q − q )( − q − q ) 2 − 2 2 − 2 ··· n 1 n 1 ( − qβ+γ qβ+γ)( − qβ+γ)) 2 − 2 · n 1 β+γ n 1 β+γ n 1 β+γ n 3 β+γ wq(( − q − q + (d c))( − q − q + (d c)) 2 − 2 − 2 − 2 − ··· n 1 n 1 ( − qβ+γ qβ+γ + (d c))( − qβ+γ + (d c))). 2 − − 2 −
n 1 n 3 n 1 n 1 n 1 β+γ β+γ β+γ β − = wq(( − − )q ( − 1)q ( − )q ) q 2 2 − 2 ··· 2 − 2 ·
(n 1)(β+γ) (n 1)β (n )(2β+γ) − n 1 − − n 1 = q 2 wq( − !) q 2 = q 2 wq( − !) 2 · 2
n 1 n 1 n 1 n 1 n n n n (2β+γ) − 2−q + −2 + −3 + (2β+γ) 2q + 2 + 3 + = q b 2 c qb c b 2q c b 2q c ··· = q b 2 c qb c b 2q c b 2q c ··· · ·
11 (2β+γ) n + n + n + n + = q b 2 c b 2q c b 2q2 c b 2q3 c ···.
Thus n 0, we have that ∀ ≥ n n n n (2β+γ) 2 + 2q + 2 + 3 + νn(T, q) = q b c b c b 2q c b 2q c ···.
And since S T , νn(T, q) νn(S, q) or ⊆ | n n n n (2β+γ) 2 + 2q + 2 + 3 + q b c b c b 2q c b 2q c ··· νn(S, q). |
c) Consider those q such that q - m2 and q l. So let q = 2 be a prime of this type, β 2β | 6 where ν1(S, q) = q and ν2(S, q) = q (q = 2 is a special case which will be treated β later). So, for a q-ordering of S beginning a0, a1, a2,... , we have that wq(a1 a0) = q 2β β − β and wq((a2 a1)(a2 a0)) = q . It must be that wq(a2 a1) = q and wq(a2 a0) = q , β − − β β − − since q (a2 a1) and q (a2 a0) (else ν1(S, q) < q , which is a contradiction). So the | − | − β strongest statement we can make is that a S, a a0 a1 b(mod q ). The largest ∀ ∈ ≡β ≡ ≡β subset, T , such that t T , t a0 a1 b(mod q ) is T = q Z + b. Since S T , we have that n 0, ∀ ∈ ≡ ≡ ≡ ⊆ ∀ ≥ nβ νn(T, q) = q νn(Z, q) νn(S, q). · | β 2β 2β+1 Now let q = 2, where ν1(S, 2) = 2 and ν2(S, 2) = 2 2 = 2 . So for a 2-ordering · β+1 β of S, beginning a0, a1, a2,... , it must be that w2(a2 a1) = 2 and wq(a2 a0) = 2 or β β − −β vice versa. This is because 2 (a2 a1) and 2 (a2 a0) (else ν1(S, 2) < 2 , which is a | β−+1 | − β+1 β contradiction). So, if a1 c(mod 2 ) and a0 d(mod 2 ) (where wq(c d) = q ), we have that s S, s c or≡ d(mod 2β+1). The largest≡ subset, T , such that −t T , t c or d(mod 2β+1∀)∈ is T =≡ (2β+1Z + c) (2β+1Z + d). Consider the set 2βZ + c. ∀ ∈ ≡ ∪ Claim: 2βZ + c = (2β+1Z + c) (2β+1Z + d) = T . ∪ Now, let t 2βZ + c. So t c(mod 2β) and either t c(mod 2β+1) or t c + 2β(mod 2β+1).∈ Since d c(mod≡ 2β) and d c(mod 2β+1), ≡d c + 2β(mod 2β+1≡). So either t c(mod 2β+1) or≡t d(mod 2β+1). Thus6≡ t (2β+1Z + c)≡ (2β+1Z + d). Therefore 2βZ + c≡ T . ≡ ∈ ∪ ⊆ Now, let t (2β+1Z + c) (2β+1Z + d). So t c(mod 2β+1) or t d(mod 2β+1). Since c d(mod 2β),∈t c(mod 2β∪), leaving t 2βZ +≡c. Thus T 2βZ +≡c. ≡ ≡ ∈ ⊆ This proves T = 2βZ + c. So since S T , ⊆ nβ νn(T, 2) = 2 νn(Z, 2) νn(S, 2), · | which is in the same form as the above. Therefore n 0, and all q such that q - m2 and q l, we have that ∀ ≥ | nβ q νn(Z, q) νn(S, q). · |
d) Consider those q such that q - m2 and q - l. Let q be an arbitrary such prime. The largest set in which q - m2 and q - l is Z, itself. Thus n 0 and all q such that q - m2 and ∀ ≥ 12 q - l, we have that
νn(Z, q) νn(S, q). | Since all types of primes, q, have been considered, we are in position to make a statement about n!S. ¿From the above cases, we have that
γ n + n + n + (2β+γ) n + n + n + q b 2 c b 2q c b 2q2 c ··· q b 2 c b 2q c b 2q2 c ··· Y Y q m2 q m2 |q-l |q l |
nβ q νn(Z, q) νn(Z, q) νn(S, q) = n!S = αn, Y · Y Y q-m2 q-m2 q P q l q-l ∈ | which completes the proof.
Using Theorem 2.2(iii) (n! n!S, n 0), we can express n!S as a multiple of n! (i.e., + | ∀ ≥ n!S = a n!, a Z ). Alternatively, we can denote n!S as · ∈
n!S = s(n) n!, · where s : Z+ Z+ is determined by S. (This is a notation which will be useful later in the thesis.) −→ In observing Theorem 2.2(iv) and (v), it might appear that something more general could be said. For instance, if we were given α0, α1, α2, . . . , αk, then there would be some X such that X αn, n 0, where X isn’t something trivial. However the proof for the case in which k = 2| should∀ ≥ perhaps imply that, as k increases, the situation becomes daunting rather quickly (indeed, the types of primes that would need to be considered doubles with each increment of k). Also, those certainties about residues in S, which arose when k = 1 and 2 3 2, are generally lost as k increases. To demonstrate, let α1 = l, α2 = l 2!, and α3 = l m33!. β 2β Let q = 2, 3 be a prime such that q m3 and q l, where ν1(S, q) = q , ν1(S, q) = q , 6 3β+δ | | and ν1(S, q) = q . So, for a q-ordering of S, beginning a0, a1, a2, a3 ... , we have that β 2β 3β+δ wq(a1 a0) = q , wq((a2 a1)(a2 a0)) = q , and wq((a3 a2)(a3 a1)(a3 a0) = q . − β − − − − − It must be that q divides each of wq(a3 a2), wq(a3 a1), and wq(a3 a0) (else we would β − − β− β contradict the fact that wq(a1 a0) = q ). Now either wq(a3 a0) = q or wq(a3 a1) = q , β+1 − β+1 − − since if q divided both of these terms, then q would divide wq(a1 a0) (which is β − another contradiction). Assuming that wq(a3 a0) = q (and with it already provided that β δ− q divides both wq(a3 a2) and wq(a3 a1)), q needs to be “distributed” between wq(a3 a2) − − − and wq(a3 a1). Unfortunately, there are no indications as to how this distribution should be performed.− Though even if the distribution was known, it’s difficult to see what new information could be gleaned about the residues in S. Perhaps there is something worthwhile to be said, but there seem to be many other more interesting problems worthy of attention.
13 3 !-Equivalent Subsets
Definition 3.1. S and T Z are said to be !-equivalent if n 0, n!S = n!T . ⊆ ∀ ≥ Perhaps the most exciting problem associated with these factorial functions has been the search for necessary and sufficient conditions on S and T Z to provide that they are !-equivalent sets. Since the presence or absense of particular⊆ residues (or groups of residues) in a set is so intimately connected with that set’s factorial function, it would seem that a necessary and sufficient condition for !-equivalence would have to concern itself with relationships between residue classes. Something like prime-power equivalence (which means, for a given residue b modulo pr, s S, such that s b(mod pr) iff t T , s.t. t b(mod pr)) is a sufficient though not necessary∃ ∈ condition for≡ !-equivalence.∃ To∈ see this, ≡ n consider 2Z and 2Z + 1. Both share the same factorial function, n!2Z = n!2Z+1 = 2 n!, though are clearly not prime-power equivalent. In an effort to find a condition weak enough to be implied by n!S = n!T , the below was arrived upon (but a quick definition is needed first).
Definition 3.2. S(mod pr) = 0 a < pr s S s.t. s a(mod pr) . { ≤ |∃ ∈ ≡ } Conjecture 3.3. Let S, T Z be infinite. The following two statements are equivalent: ⊆
i) n!S = n!T , n 0 ∀ ≥ r r ii) νk(S(mod p ), p) = νk(T (mod p ), p), r 1, k 0, p P. ∀ ≥ ≥ ∈ As it’s presented as a conjecture, it hasn’t yet been verified, at least not in the general case. However, (ii) (i) can be proven, as it is below (with the aid of number of lemmas). ⇒ Lemma 3.4. If S Z is infinite, then n N, r N such that S(mod pr) n. ⊆ ∀ ∈ ∃ ∈ | | ≥ Proof. Since S is infinite, we can select 2n distinct elements from S. It must be that at least n of these are positive or at least n are negative. Without loss of generality, let there be m n elements which are positive and collect these into a set W . (The argument for a surplus≥ of negatives is essentially identical.) Select the greatest element in W and denote it r r wmax. Choose an r N such that p > wmax. So w W , 0 < w < p , and thus each w W is member of a distinct∈ residue class modulo pr.∀ Therefore,∈ S(mod pr) m n. ∈ | | ≥ ≥ r r Lemma 3.5. If si sj(mod p ) and si s (mod p ), then wp(si sj) = wp(s sj). 6≡ ≡ i0 − i0 −
Proof. Assume instead that wp(si sj) = wp(si0 sj). Without loss of generality, let wp(si − 6 l − l l− sj) > wp(si0 sj), where wp(si sj) = p , l < r. So sj si(mod p ) and sj si0 (mod p ). − l − r≡ 6≡ Thus si s (mod p ), but since l < r and si s (mod p ), this is a contradiction. Therefore 6≡ i0 ≡ i0 wp(si sj) = wp(s sj). − i0 −
14 r Lemma 3.6. Let S = s0, s1, . . . , si, . . . , sn Z. If si is such that si sj(mod p ), { r } ⊂ 6≡ sj = si S, and si s (mod p ), then for S = s0, s1, . . . , s , . . . , sn , νk(S, p) = ∀ 6 ∈ ≡ i0 0 { i0 } νk(S , p), k 0. 0 ∀ ≥
Proof. Without loss of generality, let s0, s1, . . . , si, . . . , sn be a p-ordering for S.
i) Verify that s0, s1, . . . , si0 , . . . , sn is a p-ordering for S0. a) Claim: s0, s1, . . . , si 1 are the first i elements in a p-ordering. Else let the first − forced “deviation” occur at the ath step in the ordering (i.e., wp((sa sa 1) (sa s1)(sa − − ··· − − s0)) > wp((sb sa 1) (sb s1)(sb s0)), where b > a). If sb = si0 , then s0, s1, . . . , si, . . . , sn − − ··· − − 6 cannot be a valid p-ordering for S (since wp((sa sa 1) (sa s0)) > wp((sb sa 1) (sb − − ··· − − − ··· − s0)), where b > a).
If sb = si0 ,
wp((sa sa 1) (sa s1)(sa s0)) > wp((si0 sa 1) (si0 s1)(si0 s0)) − − ··· − − − − ··· − − > wp((si sa 1) (si s1)(si s0)). − − ··· − − So again, s0, s1, . . . , si, . . . , sn cannot be a valid p-ordering for S. Thus s0, s1, . . . , si 1 are the first i elements in a p-ordering. −
b) Claim: si0 is an acceptable next element in a p-ordering beginning like the above. Else i < c n such that ∃ ≤
wp((sc si 1) (sc s1)(sc s0)) < wp((si0 sa 1) (si0 s1)(si0 s0)) − − ··· − − − − ··· − − < wp((si sa 1) (si s1)(si s0)). − − ··· − − So again, s0, s1, . . . , si, . . . , sn cannot be a valid p-ordering for S. Thus si0 is an acceptable next element.
c) Claim: si+1, si+2, . . . , sn finishes the p-ordering. Otherwise, let the first forced devia- tion occur at the dth step. So wp((sd sd 1) (sd s1)(sd s0)) > wp((se sd 1) (se − − ··· − − − − ··· − s1)(se s0)), where e > d. But −
wp((se sd 1) (se si0 ) (se s1)(se s0)) < − − ··· − ··· − − wp((sd sd 1) (sd si0 ) (sd s1)(sd s0)) − − ··· − ··· − −
wp((se sd 1) (se si) (se s1)(se s0)) < − − ··· − ··· − − wp((sd sd 1) (sd si) (sd s1)(sd s0)). − − ··· − ··· − −
So again, s0, s1, . . . , si, . . . , sn cannot be a valid p-ordering for S. Thus s0, s1, . . . , si0 , . . . , sn is a p-ordering for S0.
ii) Verify that νk(S, p) = νk(S0, p), k 0. If k > n, then νk(S0, p) = 0 = νk(S, p). If k n, then ∀ ≥ ≤
νk(S0, p) = wp((sk sk 1) (sk si0 ) (sk s1)(sk s0)) − − ··· − ··· − − 15 = wp((sk sk 1) (sk si) (sk s1)(sk s0)) = νk(S, p). − − ··· − ··· − − This completes the proof.
r r Theorem 3.7. Let S, T Z be infinite. If νk(S(mod p ), p) = νk(T (mod p ), p), r 1, ⊆ ∀ ≥ k 0, p P, then n!S = n!T , n N. ≥ ∈ ∀ ∈ Proof. Let n 0, p P be arbitrary. By Lemma 3.4, l N such that S(mod pl) > n. ≥ ∈ ∃ ∈ And from the premise, we have that T (mod pl) = S(mod pl) > n. Now if S(mod pl) =
r0, r1, . . . , rm (where m n), then A = a0, a 1, . . . , am S such that { } ≥ ∃ { } ⊆ l a0 r0(mod p ) ≡
l a1 r1(mod p ) ≡
. .
l am rm(mod p ). ≡ l By applying Lemma 3.6, we get that νk(S(mod p ), p) = νk(A, p), k 0. So specifically, l ∀ ≥ γ νn(S(mod p ), p) = νn(A, p) and since A S, νk(S, p) νk(A, p). Let νk(A, p) = p . l ⊆ | Similarly, if T (mod p ) = u0, u1, . . . , um , then B = b0, b1, . . . , bm T such that { } ∃ { } ⊆ l b0 u0(mod p ) ≡
l b1 u1(mod p ) ≡
. .
l bm um(mod p ). ≡ l Here νn(T (mod p ), p) = νn(B, p), and since B T , νk(T, p) νk(B, p). From our initial ⊆l | l γ assumption, we have that νk(B, p) = νn(T (mod p ), p) = νn(S(mod p ), p) = νn(A, p) = p . There are two cases to consider.
a) γ < l. Let a0, a1, . . . , an be the first n-steps in a p-ordering for A. Denote A0 = a0, a1, . . . , an , where obviously νn(A , p) = νn(A, p). Now let s0, s1, . . . , sn be the first { } 0 n-steps in a p-ordering for S. Again denote S0 = s0, s1, . . . , sn , where νn(S0, p) = νn(S, p). l { } Claim: si sj(mod p ), i = j. 6≡ ∀ 6 l Now assume that there are si, sj S0 (where j > i) such that si sj(mod p ). So ∈ l ≡ l νj(S0, p) = wp((sj sj 1) (sj si) (sj s1)(sj s0)) p . Thus p νj(S0, p) − − ··· − ··· − − ≥ | | 16 l γ νn(S0, p) = νn(S, p). So since νn(S, p) νn(A, p), p νn(A, p) = p . However, γ < l, which | |l is a contradiction. So it must be that si sj(mod p ), i = j. l 6≡ ∀ 6 Given that si sj(mod p ), i = j, we have that 6≡ ∀ 6 l s0 r0 (mod p ) ≡ 0
l s1 r0 (mod p ) ≡ 1
. .
l sn r0 (mod p )0 ≡ n l where r0, r10 , . . . , rn0 = R0 S(mod p ), with the ri’s distinct. By Lemma 3.6, νn(R0, p) = { } ⊆ l l νn(S , p) = νn(S, p) and since R S(mod p ), νn(A, p) = νn(S(mod p ), p) νn(R , p) = 0 0 ⊆ | 0 νn(S, p). Therefore, along with νn(S, p) νn(A, p) from above, νn(S, p) = νn(A, p) = l | νn(S(mod p ), p). Using an analogous argument, we have that νn(T, p) = νn(B, p) = l l l νn(T (mod p ), p). Thus when γ < l, we have that νn(S, p) = νn(S(mod p ), p) = νn(T (mod p ), p) = νn(T, p). γ+1 γ+1 b) γ l: Consider the set S(mod p ). Again if S(mod p ) = q0, q1, . . . , qw ≥ { } (where w n), then C = c0, c1, . . . , cw S such that ≥ ∃ { } ⊆ γ+1 c0 q0(mod p ) ≡
γ+1 c1 q1(mod p ) ≡
. .
γ+1 cw qw(mod p ). ≡ γ+1 So νn(S, p) νn(C, p) = νn(S(mod p ), p). Again let S0 be defined as above. | γ+1 Claim: si sj(mod p ), i = j. 6≡ ∀ 6 γ+1 Again assume there are si, sj S0 (where j > i) such that si sj(mod p ). So ∈ γ+1 ≡ γ+1 νj(S0, p) = wp((sj sj 1) (sj si) (sj s1)(sj s0)) p . Thus p νj(S0, p) − − ··· − ··· − γ+1− ≥ γ | | νn(S0, p) = νn(S, p). So, since νn(S, p) νn(A, p), p νn(A, p) = p . However, this is a | γ+1 | contradiction. So it must be that si sj(mod p ), i = j. γ+1 6≡ ∀ 6 Given that si sj(mod p ), i = j, we have that 6≡ ∀ 6 γ+1 s0 v0 (mod p ) ≡ 0
17 γ+1 s1 v0 (mod p ) ≡ 1 . .
γ+1 sn v0 (mod p )0 ≡ n γ+1 where v0, v10 , . . . , vn0 = V 0 S(mod p ), with the vi’s distinct. By Lemma 3.6, { } ⊆ γ+1 γ+1 νn(V , p) = νn(S , p) = νn(S, p) and since V S(mod p ), νn(C, p) = νn(S(mod p ), p) 0 0 0 ⊆ | νn(V 0, p) = νn(S, p). Therefore, along with νn(S, p) νn(C, p) from above, νn(S, p) = γ+1 | νn(C, p) = νn(S(mod p ), p). Again using a similar argument, we have that νn(T, p) = γ+1 γ+1 γ+1 νn(T (mod p ), p). Thus when γ l, νn(S, p) = νn(S(mod p ), p) = νn(T (mod p ), p) = ≥ νn(T, p).
With both cases considered, we have that νn(S, p) = νn(T, p). And since n 0 and ≥ p P were both arbitrary, n!S = n!T , n 0. This completes the proof. ∈ ∀ ≥ Again this only proves the conjecture in one direction; however if one of our sets is Z, itself, the conjecture does hold. Theorem 3.8. Let S be infinite. The following two statements are equivalent:
i) n!S = n!, n 0 ∀ r≥ r ii) νk(S(mod p ), p) = νk(Z(mod p ), p), r 1, k 0, p P. ∀ ≥ ≥ ∈ Proof. (ii) (i): This has already been proven generally. ⇒ (i) (ii): Let it be that n!S = n!, n N. Suppose instead that r 1, k 0, ⇒ r ∀ r ∈ r ∃ ≥ ≥ p P s.t. νk(S(mod p ), p) = νk(Z(mod p ), p). Since Z(mod p )) contains all residues modulo∈ pr, it must be that there is an 0 i < pr such that s S, s i(mod pr)). So S T = z Z z i(mod pr) . From≤ above, we have that∀ ∈ 0, 1, 2, 3,...6≡ is a valid p-ordering⊆ for{ Z.∈ Since| 6≡ 0, 1, 2, . . . , i } 1 T , it should be obvious that the sequence, 0, 1, 2, . . . , i 1, is valid for the first i −1 steps∈ in a p-ordering for T . However since i R, the ordering− cannot continue without− alteration. 6∈
Claim: There is a p-ordering for T such that t0, t1, t2, . . . , tpr 2 are all in the [0, pr 1]. − − See Appendix for proof of claim (Notes A.4). Given the above p-ordering, a beginning in which all integers in the interval [0, pr 1] r r − except i have been used, we must minimize wp((tpr 1 (p 1))(tpr 1 (p 2)) (tpr 1 − − −r − − − ··· − − (i + 1))(tpr 1 (i 1)) (tpr 1 1)(tpr 1 0)) for the (p 1)th step. Since t T such − − r− ··· − − − − − 6 ∃ ∈ that t i(mod p ), any choice for tpr 1 must be congruent to some integer in the interval [0, pr ≡1] modulo pr. − − Now
r r r r r r r wp(p!) = wp((p (p 1))(p (p 2)) (p i) (p 1)(p 0)) − − − − ··· − ··· − − 18 r r r r r r r r = p wp((p (p 1))(p (p 2)) (p (i + 1))(p (i 1)) (p 1)) · − − − − ··· − − − ··· −
r r r r r r r = p wp((p 1 (p 2))(p 1 (p 3)) (p 1 i) (p 1)) · − − − − − − ··· − − ··· −
r r = p wp((p 1)!). · − r r And since wp(p!) wp((tpr 1 (p 1))(tpr 1 (p 2)) (tpr 1 i) (tpr 1 ≤ − − − − − − ··· − − ··· − − 1)(tpr 1 0)), we have that − − r r r r p wp((p 1)!) wp((tpr 1 (p 1))(tpr 1 (p 2)) · − ≤ − − − − − − ···
(tpr 1 i) (tpr 1 1)(tpr 1 0)) − − ··· − − − −
r r r p νpr 1(Z, p) wp(tpr 1 i) wp((tpr 1 (p 1))(tpr 1 (p 2)) · − ≤ − − · − − − − − − ···
(tpr 1 (i + 1))(tpr 1 (i 1)) (tpr 1 1)(tpr 1 0)) − − − − − ··· − − − −
r p νpr 1(Z, p) wp(tpr 1 i) νpr 1(T, p). · − ≤ − − · −
And since it has been assumed that n!S = n!T = n!, νpr 1(Z, p) = νpr 1(T, p). So − − r p wp(tpr 1 i). ≤ − − r But since t T such that t i(mod p ) and with tpr 1 T , we have a contradiction. Therefore (ii)6 ∃ must∈ hold. This completes≡ the proof. − ∈
4 Results Concerning Generalized Binomial Coefficients
n n n n ¿From above, notice that k 2Z = k Z = k (i.e., the k E are merely the binomial coefficients we are already familar with). It would seem to be worthwhile to determine for n n which subsets, S, k S = k . More generally, we would like to determine conditions on Z n n subsets S and T of which provide that k S = k T . To this effect, we have the following theorem.
Theorem 4.1. Let S and T be subsets of Z. The following two statements are equivalent:
i) n = n , for all n k Z. kS kT ≥ ∈ n n ii) Let 1!S = l and 1!T = l0. Then n!S = l mn n! and n!T = (l0) mn n!, for all + + · · · · n Z (where mn Z is dependent on n and S or T ). ∈ ∈
19 Proof. (i) (ii) Assume that the property described in (i) holds for S and T . By employing ⇒ + + a previously introduced notation (n!S = s(n) n!, where s : Z Z is dependent on S), we have that for n k: · −→ ≥ n n = k S k T ⇒ n! n! S = T k!S(n k)!S k!T (n k)!T ⇒ − − s(n)n! t(n)n! = s(k)k! s(n k)(n k)! t(k)k! t(n k)(n k)! ⇒ · − − · − − s(n) t(n) = s(k) s(n k) t(k) t(n k) ⇒ · − · − s(n) t(k)t(n k) = t(n) s(k)s(n k). · − · − By substituting, we get that s(a + b) t(a)t(b) = t(a + b) s(a)s(b), for all a, b Z+. · · ∈i Now given that s(1) = l and t(1) = l0, we have from a previous result that s(i) = l mi i · and t(i) = (l ) m . Now, we need to show that mi = m . We use induction. The initial 0 · i0 i0 case, i = 1, has already been provided, as m1 = m10 = 1. Assume that the property holds for i = j; so we must show that it holds when i = j + 1. Using the above formula, we have that: s(j + 1) t(j)t(1) = t(j + 1) s(j)s(1) · · ⇒ j+1 j j+1 j l mj+1 (l0) mj l0 = (l0) m0 l mj l · · j+1 · · ⇒ j+1 j+1 j+1 j+1 l (l0) mj mj+1 = l (l0) mj m0 · · j+1 ⇒ mj+1 = mj0 +1. n n + So we have that n!S = l mn n! and n!T = (l0) mn n!, for all n Z . · · n · · ∈ n (ii) (i) Let S and T be such that n!S = l mn n! and n!T = (l0) mn n!, for all + ⇒ · · · · n Z (where 1!S = l, 1!T = l ). Then, for an arbitrary n, k N (n k), ∈ 0 ∈ ≥ n n n!S l mnn! = = k n k k S k!S(n k)!S l mkk! l − mn k(n k)! − · − − n l mnn! mnn! = n · = , l mkmn kk!(n k)! mkmn kk!(n k)! · − − − − and similarly,
n n n!T (l0) mnn! = = k n k k T k!T (n k)!T (l0) mkk! (l0) − mn k(n k)! − · − − n (l0) mnn! mnn! = n · = . (l0) mkmn kk!(n k)! mkmn kk!(n k)! · − − − − So, since n and k are arbitrary in N, we have that n = n , for all n k Z. kS kT ≥ ∈ 20 While this is a mildly interesting result in itself, it also gives us the following corollary.
Corollary 4.2. Let S be a subset of Z. The following statements are equivalent:
i) n = n , for all n k Z. kS k ≥ ∈ n ii) Let 1!S = l. Then n!S = l n!. · iii) S is !-equivalent to lZ + b, where b Z. ∈ Proof. (i) (ii) This is merely an application of the above theorem (letting T = Z). Since ⇒ n!Z = n!, mn = 1 for all n N, hence the result in (ii). ∈ (ii) (iii) This is rather easy to show as we know from Bhargava ([4][Example 17]) that ⇒n n!lZ+b = l n!. Thus n!lZ+b = n!S, n N, meaning S and lZ + b are !- equivalent. · ∀ ∈ n (iii) (i) Since S and lZ + b are !-equivalent we have that n!S = n!lZ+b = l n!. So ⇒ · mn0 = 1 = mn, n N, again leaving this proof to be just an application of the above theorem. ∀ ∈
Pascal’s triangle has been an important tool for determining (or at least visualizing) properties concerning the binomial coefficients. Given our improved grasp of the generalized coefficients, it would seem that the construction of a “generalized” Pascal’s triangle would be a useful endeavor. The triangle can be defined in the obvious way:
0 0 S 1 1 0 S 1 S 2 2 2 0 S 1 S 2 S 3 3 3 3 . 0 S 1 S 2 S 3 S
. .
n n Since we’ve stipulated that 0!S = 1 for all subsets, = = 1 for all n. This is in some 0S nS ways unfortunate, since it restricts Pascal’s Identity ( n + n = n+1 , for n, i 0) i S i+1S i+1 S ≥ to only those subsets, S, such that n = n . The following lemma makes this evident. kS k Lemma 4.3. The identity, n + n = n+1 , holds only for those subsets, S, for i S i+1S i+1 S which n = n , n, i 0. i S i Z ∀ ≥
21 Proof. Let Pascal’s Identity hold for the binomial coefficients of a set S Z. Now suppose N n n ⊆ N that n, i such that i S = i Z. Let a be the least integer such that b where a ∃ a ∈ n n 6 ∃ ∈a 1 = . Since = = 1 for all S Z, b = 0 and a = b. So both − and bS 6 bZ 0S nS ⊆ 6 6 b 1S a 1 are well-defined. And since a 1 < a, a 1 = a 1 and a 1 = a 1− . Since −b S b−1 S b−1 Z −b S −b Z Pascal’s Identity holds, − − −
a a 1 a 1 = − + − b b 1 b S − S S
a 1 a 1 a = − + − = . b 1 b b − Z Z Z But this is a contradiction. Therefore, it must be that n = n , n, i 0. i S i Z ∀ ≥ Additionally, there doesn’t seem to be any obvious way to “tweak” the property so that it might hold generally. Again, this is unfortunate since many of the interesting properties of Pascal’s Triangle are founded upon this rule. Despite this, there is at least one property of Pascal’s Triangle which remains valid.
Theorem 4.4. Let n, 0 < k < n N. On the generalized Pascal’s Triangle for a subset S, the product of the six entries surrounding∈ n is a perfect square. kS Proof. To prove this, we must show that
n 1 n 1 n n n + 1 n + 1 − − = a2, k 1 · k · k 1 · k + 1 · k · k + 1 − S S − S S S S where a Z. So ∈ n 1 n 1 n n n + 1 n + 1 − − = k 1 · k · k 1 · k + 1 · k · k + 1 − S S − S S S S s(n 1)(n 1)! s(n 1)(n 1)! − − − − s(k 1)(k 1)!s(n k)(n k)! · s(k)k!s(n k 1)(n k 1)! · − − − − − − − − s(n)n! s(n)n! s(k 1)(k 1)!s(n k + 1)(n k + 1)! · s(k + 1)(k + 1)!s(n k 1)(n k 1)! · − − − − − − − − s(n + 1)(n + 1)! s(n + 1)(n + 1)! s(k)k!s(n k + 1)(n k + 1)! · s(k + 1)(k + 1)!s(n k)(n k)! − − − −
22 s(n 1) s(n 1) s(n) = − − s(k 1)s(n k) · s(k)s(n k 1) · s(k 1)s(n k + 1) · − − − − − − s(n) s(n + 1) s(n + 1) s(k + 1)s(n k 1) · s(k)s(n k + 1) · s(k + 1)s(n k) · − − − − (n 1)! (n 1)! n! − − (k 1)!(n k)! · k!(n k 1)! · (k 1)!(n k + 1)! · − − − − − − n! (n + 1)! (n + 1)! (k + 1)!(n k 1)! · k!(n k + 1)! · (k + 1)!(n k)! − − − − s(n 1)s(n)s(n + 1) = ( − )2 b2 s(k 1)s(k)s(k + 1)s(n k 1)s(n k)s(n k + 1) · − − − − − s(n 1)s(n)s(n + 1) = (b − )2 = a2. · s(k 1)s(k)s(k + 1)s(n k 1)s(n k)s(n k + 1) − − − − −
References
[1] M. Bhargava, Generalized factorials and fixed over subsets of a Dedekind domain, J. 72(1998), 67–75.
[2] M. Bhargava, Congruence preservation and polynomial functions from Zn to Zm, Dis- crete Math. 173(1997), 15–21. [3] M. Bhargava, P -orderings and polynomial functions on arbitrary subsets of Dedekind rings, J. Reine Angew. Math. 490(1997), 101–127. [4] M. Bhargava, The factorial function and generalizations, Amer. Math. Monthly. 107(2000), 783-799. [5] M. Bhargava and K.S. Kedlaya, Continuous functions on compact subsets of local fields, Acta Arith. 91(1999), 191–198. [6] P.-J. Cahen and J.-L. Chabert, “Integer Valued-Polynomials,” Amer. Math. Soc. Sur- veys and Monographs, 58(1997), American Mathematical Society, Providence. [7] D.S. Dummit and R.M. Foote, Abstract Algebra, 2nd ed.. Upper Saddle River, N.J.: Prentice Hall, 1999. [8] C. Long, Pascal’s triangle, difference tables and arithmetic sequences of order N, Col- lege Math. Journal 15(1984), 290–298.
23 A Appendix
Notes A.1. To verify the p-ordering we use induction. We can choose a0 arbitrarily, so choose c.
P (1): Now since q - (d c), d minimizes wq(a1 c). So we may justifiably choose a1 = d. − − P (n) P (n + 1): n may be even or odd, so we must consider both cases. → n γ Let n be even. So the q-ordering (up to the nth step) is c, d, . . . , 2 q + c. We need to n γ show that 2 q + d is an adequate choice for an+1, by showing that it minimizes wp((an+1 n γ γ γ − 2 q c) (an+1 d)(an+1 c)). Now an+1 must be of the form vq + c or uq + d. If − ···γ − − an+1 = vq + c,
γ n γ γ γ wp((vq + c q c) (vq + c d)(vq + c c)) = − 2 − ··· − − γ n γ γ γ γ wp((vq + c q c) (vq + c q c)(vq + c c)), − 2 − ··· − − − γ γ since wp(vq + c iq d) = 1, i Z. Continuing, − − ∀ ∈ γ n γ γ γ γ n γ/(n+2) wp((vq q ) (vq q )(vq )) = wp(v(v 1)(v 2) (v ) q 2 ) − 2 ··· − − − ··· − 2 ·
( γ/(n+2) ) n = q 2 wp(v(v 1)(v 2) (v )). · − − ··· − 2
γ/(n+2) n+2 ( 2 ) n+2 This is minimized by v = 2 , so here we have q wp(( 2 )!). γ · Now if an+1 is of the form uq + d,
γ n γ γ γ wp((uq + d q c) (uq + d d)(uq + d c)) = − 2 − ··· − − γ n 2 γ γ γ γ wp((uq + d − q d) (uq + d q d)(uq + d d)) = − 2 − ··· − − − γ n 2 γ γ γ γ wp((uq − q ) (uq q )(uq )) = − 2 ··· − n 2 γ/n wp(u(u 1)(u 2) (u − ) q 2 ) = − − ··· − 2 · ( γ/n ) n 2 q 2 wp(u(u 1)(u 2) (u − )). · − − ··· − 2
γ/n n ( 2 ) n 2 This is minimized by u = 2 , so here we have q wp(( −2 )!). This is clearly less than γ/(n+2) · ( 2 ) n+2 n γ q wp(( 2 )!), so the best choice for an+1 is 2 q + d. · n+1 γ Let n be odd. Here we would need to show that 2 q + c is an adequate choice for an+1. This can be shown using a proof analogous to the one above, so let it be accepted without explicit demonstration. With this, the inductive proof is complete. Thus c, d, qγ + c, qγ + d, 2qγ + c, 2qγ + d, . . . is a valid q-ordering for T .
24 α Notes A.2. It should be fairly apparent that wq((an an 1) (an a1)(an a0)) = q , where − − ··· − −
2 α = ai an(mod q) 0 i < n + ai an(mod q ) 0 i < n |{ ≡ | ≤ }| { ≡ | ≤ } 3 + ai an(mod q ) 0 i < n + . { ≡ | ≤ } ···
n + n + n + r n n 2q 2q2 2q3 And since i n(mod q ) 0 i < n = r , we have that wq( !) = qb c b c b c ···. |{ ≡ | ≤ }| b q c 2 n n 1 Notes A.3. Let n be odd. Assume instead that r = −r . Thus m Z such that b 2q c 6 b 2q c ∃ ∈ n 1 n − < m 2qr ≤ 2qr
n 1 < 2qrm n − ≤ r n So it must be that n = 2q m, but n is odd, which is a contradiction. Therefore 2qr = n 1 b c −r . b 2q c Notes A.4. The following constuction is a valid p-ordering to the (pr 2)nd step: Let the first i 1 steps be determined as above. (Note that if i = pr 1, then− we already have a p-ordering− to the (pr 2)nd step, so no further construction is− needed.) − u1 r u1 Let p be the greatest power of p less than p i. Let i bu1 (mod p ). Then the ith r u r u− ≡ element will be p p 1 + bu . For i < j < p p 1 + bu , the jth element will be j. − 1 − 1 u2 r r u1 u1 Let p be the greatest power of p less than p (p p + bu1 ) = p bu1 . Let u2 r u1 − − r u2 − r u1 i bu2 (mod p ). Then the (p p + bu1 )th element will be p p + bu2 . For p p + ≡ r u − − − bu < j < p p 2 + bu , the jth element will be j. 1 − 2
. .
uk+1 r r uk uk Let p be the greatest power of p less than p (p p + buk ) = p buk . Let uk+1 r uk − − r uk+1 − i buk+1 (mod p ). Then the (p p + buk )th element will be p p + buk+1 . For r≡ u r u − − p p k + bu < j < p p k+1 + bu , the jth element will be j. − k − k+1
. .
r r ul ul r ul r When p (p p + bul ) = p bul < p, we have that for p p + bul j < p 1, the jth element− is −j + 1. And this completes− the p-ordering to the− (pr 2)nd≤ step. − − The acceptability of the first i 1 steps has already been noted. Now induction should be done on j to establish the rest− of the claim.
25 P (i): We begin with the ith element. We may assume that i = pr 1, since we are then already provided a p-ordering to the (pt 2)nd step (namely, 0, 1, 26 , . . .− , i 1). So in choosing − − the ith element, we are looking to minimize wp((ai (i 1))(ai (i 2)) (ai 1)(ai 0)). − − − − ··· − − There are two cases to consider–when u1 > 0 and when u1 = 0. r u1 r u1 Let u1 > 0. Consider the element p p + bu1 . Now wp((p p + bu1 (i r u1 r u1 − r u1 − α − − 1))(p p + bu1 (i 2)) (p p + bu1 1)(p p + bu1 0)) = p , where α = −r u1 − − ··· − − r −u1 − 2 n p p + bu1 (mod p) 0 n < i + n p p + bu1 (mod p ) 0 n < i + + |{ ≡ r − u1 r |1 ≤ }| { ≡ − | ≤ } ··· n p p + bu1 (mod p − ) 0 n < i . (No other terms are needed since there is no { ≡ − |r ≤ u1 } r n in our interval such that n p p + bu 1 (mod p ).) ≡ − r u p p 1 +bu i v r v r u1 r − 1 So for p where u1 < v < r, p p i, p p + bu1 < p ; thus pv = pv . − ≤ − r u1 b c b c r u1 v p p +bu1 Now n p p + bu (mod p ) 0 n < i − v , so |{ ≡ − 1 | ≤ }| ≤ b p c i r u1 v n p p + bu (mod p ) 0 n < i . |{ ≡ − 1 | ≤ }| ≤ bpv c
y r u y For p where 1 y u1, i p p 1 + bu (mod p ). Thus ≤ ≤ ≡ − 1 i r u1 y y n p p + bu (mod p ) 0 n < i = n i(mod p ) 0 n < i = . |{ ≡ − 1 | ≤ }| |{ ≡ | ≤ }| bpv c
i i i α p + 2 + + r 1 α α Together we have that p pb c b p c ··· b p − c = νi(Z, p). And since νi(Z, p) p , p = Z r u1 ≤ | νi( , p). Thus, p p + bu1 minimizes. − r r t Let u1 = 0 (thus p p i < p 1). Consider i + 1. Since i = p 1, − ≤ − 6 −
wp((i + 1)!) = wp(i!)
wp((i + 1 i)(i + 1 (i 1)) (i + 1 1)(i + 1 0)) = − − − ··· − −
wp((i (i 1))(i (i 2)) (i 1)(i 0)) − − − − ··· − −
wp(i + 1 i)wp((i + 1 (i 1))(i + 1 (i 2)) (i + 1 1)(i + 1 0)) = − − − − − ··· − −
wp((i (i 1))(i (i 2)) (i 1)(i 0)) − − − − ··· − −
wp((i + 1 (i 1))(i + 1 (i 2)) (i + 1 1)(i + 1 0)) = − − − − ··· − −
wp((i (i 1))(i (i 2)) (i 1)(i 0)). − − − − ··· − − And since i minimizes, so does i + 1. P(i) has now been established. P (j-1) P (j) (i j 1 < pt 2): j is one of the following types: → ≤ − − r u (a) i < j < p p 1 + bu − 1 26 r u (b) j = p p k + bu − k r u r u (c) p p k + bu < j < p p k+1 + bu − k − k+1 t t (d) p p + b1 j < p 1. − ≤ − If j is of type (a), then we are are attempting to minimize
r u1 wp((aj (j 1))(aj (j 2)) (aj (i + 1))(aj (p p + bu ))(aj (i 1)) − − − − ··· − − − 1 − − (aj 1)(aj 0)). ··· − −
r u1 r r+1 Now consider aj = j. Since i p p + bu1 (mod p ), j i(mod p ), and j r u r+1 ≡ − r u 6≡ 6≡ p p 1 + bu (mod p ), wp(j i) = wp(j (p p 1 + bu )). So − 1 − − − 1
wp((j (j 1))(j (j 2)) − − − − ··· r u1 (j (i + 1))(j (p p + bu ))(j (i 1)) (j 1)(j 0)) − − − 1 − − ··· − − = wp((j (j 1))(j (j 2)) (j (i + 1))(j i)(j (i 1)) − − − − ··· − − − − ··· (j 1)(j 0)) = wp(j!). − − Thus j minimizes. If j is of type (b), then we are are attempting to minimize
r uk wp((aj (j 1)) (aj (p p + bu )) (aj (i + 1)) − − ··· − − k ··· − t u1 (aj (p p + bu ))(aj (i 1)) (aj 1)(aj 0)). · − − 1 − − ··· − −
r uk+1 uk+1 r r Consider aj = p p + buk+1 , where p is the greatest power of p less than p (p uk − − − p + buk ). Here
r uk+1 r uk+1 r uk wp((p p + bu (j 1)) (p p + bu (p p + bu )) − k+1 − − ··· − k+1 − − k r uk+1 β (p p + bu 1)(aj 0)) = p , ··· − k+1 − − where
r uk+1 r uk β = n p p + bu (mod p) 0 n p p + bu , n = i + { ≡ − k+1 | ≤ ≤ − k 6 } r uk+1 2 r uk n p p + bu (mod p ) 0 n p p + bu , n = i + { ≡ − k+1 | ≤ ≤ − k 6 } r uk+1 r 1 r uk + n p p + bu (mod p − ) 0 n p p + bu , n = i . ··· { ≡ − k+1 | ≤ ≤ − k 6 }
r u u r u u First, since i p p k + bu (mod p k ) and i p p k+1 + bu (mod p k ), ≡ − k 6≡ − k+1
r uk+1 v r uk n p p + bu (mod p ) 0 n p p + bu , n = i = { ≡ − k+1 | ≤ ≤ − k 6 } r uk+1 v r uk n p p + bu (mod p ) 0 n < p p + bu , { ≡ − k+1 | ≤ − k }
27 v 1. ∀ ≥ v r v r uk r uk+1 r So for p where uk+1 < v < r, p p p p + buk , p p + buk+1 < p ; thus r uk+1 r u − ≤ − − p p +bu p p k +bu − k+1 = − k . Now b pv c b pv c r u p p k+1 + bu r uk+1 v r uk k+1 n p p + bu (mod p ) 0 n < p p + bu − , { ≡ − k+1 | ≤ − k } ≤ b pv c
r u p p k +bu r uk+1 v r uk − k y so n p p + buk+1 (mod p ) 0 n < p p + buk pv . For p where { ≡ −r u r u | ≤ − y } ≤ b c 1 y u1, p p k + bu p p k+1 + bu (mod p ). Thus ≤ ≤ − k ≡ − k+1
r uk+1 y r uk n p p + bu (mod p ) 0 n < p p + bu = { ≡ − k+1 | ≤ − k } r uk y r uk n p p + buk (mod p ) 0 n < p p + buk = |{ ≡ − | ≤ − }| r u p p k + bu − k . b pv c Together we have that
r u r u r u p p k +bu p p k +bu p p k +bu − k − k − k β + 2 + + r 1 p p p r u Z p pb c b c ··· b − c = νp p k +bu ( , p). ≤ − k β β r u r u Z r u Z k+1 And since νp p k +bu ( , p) p , p = νp p k +bu ( , p). Thus, p p + buk+1 minimizes. − k | − k − If j is of type (c), then we are are attempting to minimize
r uk+1 wp((aj (j 1))(aj (j 2)) (aj (p p + bu )) − − − − ··· − − k+1 ··· (aj (i + 1))(aj (i 1)) (aj 1)(aj 0)). − − − ··· − −
r uk+1 uk+1+1 uk+1+1 Now consider aj = j. Since i p p + buk+1 (mod p ), j i(mod p ), and r u u≡ +1− r u 6≡ j p p k+1 + bu (mod p k+1 ), wp(j i) = wp(j (p p k+1 + bu )). So 6≡ − k+1 − − − k+1
r uk+1 wp((j (j 1))(j (j 2)) (j (p p + bu )) − − − − ··· − − k+1 (j (i + 1))(j (i 1)) (j 1)(j 0)) = ··· − − − ··· − − wp((j (j 1))(j (j 2)) (j (i + 1))(j i)(j (i 1)) (j 1)(j 0)) − − − − ··· − − − − ··· − − = wp(j!).
Thus j minimizes. Let j be of type (d). We need to minimize wp((aj + 1 j)(aj + 1 (j 1)) (aj (i + 1))(aj (i 1)) (aj + 1 1)(aj + 1 0)). − − − ··· − − − ··· − − Consider j + 1. Since j = pt 1, 6 −
wp((j + 1)!) = wp(j!)
28 wp((j + 1 j)(j + 1 (j 1)) (j + 1 i) (j + 1 1)(j + 1 0)) = − − − ··· − ··· − −
wp((j (j 1))(j (j 2)) (j 1)(j 0)) − − − − ··· − −
wp(j + 1 i)wp((j + 1 j)(j + 1 (j 1)) (j + 1 1)(j + 1 0)) = − − − − ··· − −
wp((j (j 1))(j (j 2)) (j 1)(j 0)) − − − − ··· − −
wp((j + 1 j)(j + 1 (j 1)) (j + 1 1)(j + 1 0)) = wp(j!). − − − ··· − −
And since wp(j!) is minimal, j + 1 minimizes. Thus P (j-1) P (j) (i j 1 < pt 2) has been established. Therefore the p-ordering is valid. → ≤ − −
29 The following is a MAPLE program which determines the factorial sequence for a finite S Z. genfactall(S) generates this factorial sequence, whereas pordering(p,S) generates a p-ordering⊆ for a given p. All other functions below are subsidiary.
> with(numtheory):
> orderprod:=proc(A,r) > local n,i,prod; > n:=nops(A): > prod:=1: > for i from 1 to n do > prod:=prod*(r-A[i]): > od: > RETURN(prod): > end:
> pfactorset:=proc(S) > local n,i,j,T; > T:= : > n:=nops(S):{} > for i from 1 to n do > for j from 1 to n do > if i=j then > T:=T: > elif S[i]-S[j]=-1 then > T:=T: > else > T:=T union factorset(S[i]-S[j]): > fi: > od: > od: > RETURN(T): > end:
> powerp:=proc(p,x) > local pow,y; > pow:=0: > y:=x: > while member(p,factorset(y)) do > pow:=pow+1; > y:=y/p; > od:
30 > RETURN(p pow): > end: ∧
> nextpick:=proc(p,A,R) > local a,n,b,i; > n:=nops(R): > a:=R[1]: > b:=powerp(p,orderprod(A,R[1])): > for i from 1 to n do > if powerp(p,orderprod(A,R[i]))¡b then > b:=powerp(p,orderprod(A,R[i])): > a:=R[i]: > else > b:=b: > a:=a: > fi: > od: > RETURN(a): > end:
> func:=proc(p,A,R,k) > local ans,b,B,S,i; > B:=A: > S:=R: > for i from 1 to k-1 do > b:=nextpick(p,B,S): > B:=B union b : > S:=S minus {b }: > od: { } > ans:=powerp(p,orderprod(B,nextpick(p,B,S))): > RETURN(ans): > end:
> genfact:=proc(S,k) > local T,A,R,n,m,i,kS; > n:=nops(S): > kS:=1: > if k>=n then > kS:=0: > else > T:=pfactorset(S):
31 > m:=nops(T): > A:= S[1] : > R:=S{ minus} S[1] : > for i from 1 to{ m} do > kS:=kS*func(T[i],A,R,k): > od: > fi: > RETURN(kS): > end:
> pordering:=proc(p,S) > local A,R,O,n,b,i; > A:= S[1] : > R:=S{ minus} S[1] : > n:=nops(R): { } > O:=[S[1]]: > while n>0 do > b:=nextpick(p,A,R); > R:=R minus b; > A:=A union b; > n:=nops(R); > O:=[op(O),b]; > od: > RETURN(O): > end:
> genfactall:=proc(S) > local n,A,k; > n:=nops(S): > A:=[genfact(S,1)]: > for k from 2 to n-1 do > A:=[op(A),genfact(S,k)]: > od: > RETURN(A): > end:
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Determining the contact area in elastic-indentation problems
# Determining the contact area in elastic-indentation problems,10.1243/03093247V094230,Journal of Strain Analysis for Engineering Design,J. R. BARBER
Determining the contact area in elastic-indentation problems
A general relationship is established, determining the contact area in the indentation of an elastic half-space. Contact pressure is shown to increase. monotonically with load throughout the contact area.
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## Citation Context (4)
• ...For all other cases, the contact area is a non-decreasing function of w (Barber 1974), and consequently the incremental sti¬ ness d F=dw is also a non-decreasing function of w. The expression (4.2) can be extended to cover the case of non-contact by de› ning the function F(w) such that F = 0 for w < 0...
• ...Now, in a frictionless indentation problem for the half-space, the maximum normal surface displacement must occur in an area of positive contact pressure (Barber 1974), and in the auxiliary problem the only loaded area is A, throughout which u2 = 1. It follows that all points in A are points of maximum u2 and hence that p2(x;y) > 0 for all points in A. We also have ¯ U(x;y) > 0 ex hypothesi , and hence the integrand on the ...
### J. R. Barber. Bounds on the electrical resistance between contacting elastic rough b...
• ...He uses a theorem proved in an earlier paper [15] which states that the value of b which satisfies the required condition that the contact pressure is zero at the inner boundary may be determined by making OP/Ob = 0, where P is the total pressure on the annulus...
### G. M. L. Gladwell, et al. On the approximate solution of elastic contact problems for a circular...
• ...In order to find this radius, we make use of a variational formulation of the contact condition discussed in a previous paper [6]...
### J. R. Barber. Indentation of the semi-infinite elastic solid by a concave rigid punc...
• ...The case of contact by a flat annular punch is especially simple, since the contact radii are known ap riori .O therwise, the determination of contact radii in these non-adhesive cases follow from a result by Barber that essentially states that the contact area maximizes the value of the contact force [4]...
Sort by: | 587 | 2,562 | {"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-2013-20 | latest | en | 0.86237 |
https://ithelp.ithome.com.tw/articles/10187814 | 1,624,286,269,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488273983.63/warc/CC-MAIN-20210621120456-20210621150456-00591.warc.gz | 306,510,188 | 14,444 | DAY 28
0
Big Data
## [第 28 天] 深度學習（2）TensorBoard
The computations you'll use TensorFlow for - like training a massive deep neural network - can be complex and confusing. To make it easier to understand, debug, and optimize TensorFlow programs, we've included a suite of visualization tools called TensorBoard. You can use TensorBoard to visualize your TensorFlow graph, plot quantitative metrics about the execution of your graph, and show additional data like images that pass through it.
TensorBoard: Visualizing Learning | TensorFlow
## 整理程式
### 改寫架構
• 定義一個添加層的函數：`add_layer()`
• 準備資料（Inputs）
• 建立 Feeds（使用 `tf.placeholder()` 方法）來傳入資料
• 添加隱藏層與輸出層
• 定義 `loss` 與要使用的 Optimizer（使用梯度遞減）
• 初始化 Graph 並開始運算
### 改寫後的程式
``````import tensorflow as tf
import numpy as np
# 定義一個添加層的函數
def add_layer(inputs, input_tensors, output_tensors, activation_function = None):
W = tf.Variable(tf.random_normal([input_tensors, output_tensors]))
b = tf.Variable(tf.zeros([1, output_tensors]))
formula = tf.add(tf.matmul(inputs, W), b)
if activation_function is None:
outputs = formula
else:
outputs = activation_function(formula)
return outputs
# 準備資料
x_data = np.random.rand(100)
x_data = x_data.reshape(len(x_data), 1)
y_data = x_data * 0.1 + 0.3
# 建立 Feeds
x_feeds = tf.placeholder(tf.float32, shape = [None, 1])
y_feeds = tf.placeholder(tf.float32, shape = [None, 1])
# 添加 1 個隱藏層
hidden_layer = add_layer(x_feeds, input_tensors = 1, output_tensors = 10, activation_function = None)
# 添加 1 個輸出層
output_layer = add_layer(hidden_layer, input_tensors = 10, output_tensors = 1, activation_function = None)
# 定義 `loss` 與要使用的 Optimizer
loss = tf.reduce_mean(tf.square(y_feeds - output_layer))
optimizer = tf.train.GradientDescentOptimizer(learning_rate = 0.01)
train = optimizer.minimize(loss)
# 初始化 Graph 並開始運算
init = tf.global_variables_initializer()
sess = tf.Session()
sess.run(init)
for step in range(201):
sess.run(train, feed_dict = {x_feeds: x_data, y_feeds: y_data})
if step % 20 == 0:
print(sess.run(loss, feed_dict = {x_feeds: x_data, y_feeds: y_data}))
sess.close()
``````
## 視覺化
``````import tensorflow as tf
import numpy as np
# 定義一個添加層的函數
def add_layer(inputs, input_tensors, output_tensors, activation_function = None):
with tf.name_scope('Layer'):
with tf.name_scope('Weights'):
W = tf.Variable(tf.random_normal([input_tensors, output_tensors]))
with tf.name_scope('Biases'):
b = tf.Variable(tf.zeros([1, output_tensors]))
with tf.name_scope('Formula'):
formula = tf.add(tf.matmul(inputs, W), b)
if activation_function is None:
outputs = formula
else:
outputs = activation_function(formula)
return outputs
# 準備資料
x_data = np.random.rand(100)
x_data = x_data.reshape(len(x_data), 1)
y_data = x_data * 0.1 + 0.3
# 建立 Feeds
with tf.name_scope('Inputs'):
x_feeds = tf.placeholder(tf.float32, shape = [None, 1])
y_feeds = tf.placeholder(tf.float32, shape = [None, 1])
# 添加 1 個隱藏層
hidden_layer = add_layer(x_feeds, input_tensors = 1, output_tensors = 10, activation_function = None)
# 添加 1 個輸出層
output_layer = add_layer(hidden_layer, input_tensors = 10, output_tensors = 1, activation_function = None)
# 定義 `loss` 與要使用的 Optimizer
with tf.name_scope('Loss'):
loss = tf.reduce_mean(tf.square(y_feeds - output_layer))
with tf.name_scope('Train'):
optimizer = tf.train.GradientDescentOptimizer(learning_rate = 0.01)
train = optimizer.minimize(loss)
# 初始化 Graph
init = tf.global_variables_initializer()
sess = tf.Session()
# 將視覺化輸出
writer = tf.summary.FileWriter("TensorBoard/", graph = sess.graph)
# 開始運算
sess.run(init)
for step in range(201):
sess.run(train, feed_dict = {x_feeds: x_data, y_feeds: y_data})
#if step % 20 == 0:
#print(sess.run(loss, feed_dict = {x_feeds: x_data, y_feeds: y_data}))
sess.close()
``````
``````\$ tensorboard --logdir='TensorBoard/'
``````
## 視覺化（2）
``````import tensorflow as tf
import numpy as np
# 定義一個添加層的函數
def add_layer(inputs, input_tensors, output_tensors, activation_function = None):
with tf.name_scope('Layer'):
with tf.name_scope('Weights'):
W = tf.Variable(tf.random_normal([input_tensors, output_tensors]), name = 'W')
with tf.name_scope('Biases'):
b = tf.Variable(tf.zeros([1, output_tensors]), name = 'b')
with tf.name_scope('Formula'):
formula = tf.add(tf.matmul(inputs, W), b)
if activation_function is None:
outputs = formula
else:
outputs = activation_function(formula)
return outputs
# 準備資料
x_data = np.random.rand(100)
x_data = x_data.reshape(len(x_data), 1)
y_data = x_data * 0.1 + 0.3
# 建立 Feeds
with tf.name_scope('Inputs'):
x_feeds = tf.placeholder(tf.float32, shape = [None, 1], name = 'x_inputs')
y_feeds = tf.placeholder(tf.float32, shape = [None, 1], name = 'y_inputs')
# 添加 1 個隱藏層
hidden_layer = add_layer(x_feeds, input_tensors = 1, output_tensors = 10, activation_function = None)
# 添加 1 個輸出層
output_layer = add_layer(hidden_layer, input_tensors = 10, output_tensors = 1, activation_function = None)
# 定義 `loss` 與要使用的 Optimizer
with tf.name_scope('Loss'):
loss = tf.reduce_mean(tf.square(y_feeds - output_layer))
with tf.name_scope('Train'):
optimizer = tf.train.GradientDescentOptimizer(learning_rate = 0.01)
train = optimizer.minimize(loss)
# 初始化 Graph
init = tf.global_variables_initializer()
sess = tf.Session()
# 將視覺化輸出
writer = tf.summary.FileWriter("TensorBoard/", graph = sess.graph)
# 開始運算
sess.run(init)
for step in range(201):
sess.run(train, feed_dict = {x_feeds: x_data, y_feeds: y_data})
#if step % 20 == 0:
#print(sess.run(loss, feed_dict = {x_feeds: x_data, y_feeds: y_data}))
sess.close()
``````
## 視覺化（3）
``````import tensorflow as tf
import numpy as np
# 定義一個添加層的函數
def add_layer(inputs, input_tensors, output_tensors, n_layer, activation_function = None):
layer_name = 'layer%s' % n_layer
with tf.name_scope('Layer'):
with tf.name_scope('Weights'):
W = tf.Variable(tf.random_normal([input_tensors, output_tensors]), name = 'W')
tf.summary.histogram(name = layer_name + '/Weights', values = W)
with tf.name_scope('Biases'):
b = tf.Variable(tf.zeros([1, output_tensors]), name = 'b')
tf.summary.histogram(name = layer_name + '/Biases', values = b)
with tf.name_scope('Formula'):
formula = tf.add(tf.matmul(inputs, W), b)
if activation_function is None:
outputs = formula
else:
outputs = activation_function(formula)
tf.summary.histogram(name = layer_name + '/Outputs', values = outputs)
return outputs
# 準備資料
x_data = np.random.rand(100)
x_data = x_data.reshape(len(x_data), 1)
y_data = x_data * 0.1 + 0.3
# 建立 Feeds
with tf.name_scope('Inputs'):
x_feeds = tf.placeholder(tf.float32, shape = [None, 1], name = 'x_inputs')
y_feeds = tf.placeholder(tf.float32, shape = [None, 1], name = 'y_inputs')
# 添加 1 個隱藏層
hidden_layer = add_layer(x_feeds, input_tensors = 1, output_tensors = 10, n_layer = 1, activation_function = None)
# 添加 1 個輸出層
output_layer = add_layer(hidden_layer, input_tensors = 10, output_tensors = 1, n_layer = 2, activation_function = None)
# 定義 `loss` 與要使用的 Optimizer
with tf.name_scope('Loss'):
loss = tf.reduce_mean(tf.square(y_feeds - output_layer))
tf.summary.scalar('loss', loss)
with tf.name_scope('Train'):
optimizer = tf.train.GradientDescentOptimizer(learning_rate = 0.01)
train = optimizer.minimize(loss)
# 初始化 Graph
init = tf.global_variables_initializer()
sess = tf.Session()
# 將視覺化輸出
merged = tf.summary.merge_all()
writer = tf.summary.FileWriter("TensorBoard/", graph = sess.graph)
# 開始運算
sess.run(init)
for step in range(400):
sess.run(train, feed_dict = {x_feeds: x_data, y_feeds: y_data})
if step % 20 == 0:
result = sess.run(merged, feed_dict={x_feeds: x_data, y_feeds: y_data})
sess.close()
``````
## 參考連結
### 1 則留言
0
sunnyyeh
iT邦新手 5 級 ‧ 2021-05-29 01:19:20
tf.summary.merge_all() | 2,264 | 7,639 | {"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-2021-25 | latest | en | 0.391687 |
https://frama-c.com/html/fc-discuss/2013-August/msg00030.html | 1,653,170,823,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662541747.38/warc/CC-MAIN-20220521205757-20220521235757-00614.warc.gz | 330,730,434 | 3,587 | # Frama-C-discuss mailing list archives
This page gathers the archives of the old Frama-C-discuss archives, that was hosted by Inria's gforge before its demise at the end of 2020. To search for mails newer than September 2020, please visit the page of the new mailing list on Renater.
# [Frama-c-discuss] [Jessie] loop invariant
• Subject: [Frama-c-discuss] [Jessie] loop invariant
• From: rovedy at ig.com.br (Rovedy Aparecida Busquim e Silva)
• Date: Thu, 22 Aug 2013 14:02:10 -0300
• References: <CAEtoXR21n9RPnwGE_UQ7zDjqrNPjhhRMuG5kQdzM=SjH7z+9CA@mail.gmail.com> <52162B8A.4080703@inria.fr>
```Hi,
We are sending the working version of the code without the if statements.
According to tutorial, we have followed the below sequence:
- identify variables modified in the loop:
- use loop assigns clause to list variables that (might) have been
assigned so far after iterations
- define their possible value intervals (relationships) after iterations
In our case, we identified the variables j and soma that were modified in
the loop.
Because of that, we tried to specify the variable soma.
Best regards,
Nanci, Luciana, Rovedy
----------------------------------------------------------------------------------------
#pragma JessieIntegerModel(math)
#pragma JessieTerminationPolicy(user)
#pragma JessieFloatModel(math)
float acel[3], soma;
void test()
{
int j;
acel[0] = 5.0;
acel[1] = 5.0;
acel[2] = 5.0;
soma = 0.0;
/*@ loop invariant 0 <= j <= 3;
loop invariant soma >= 0.0;
loop invariant \forall integer k; 0 <= k < 3 ==> (soma <= soma +
acel[k]);
loop assigns j, soma;
*/
for(j=0;j<3;j++)
{
soma = soma + acel[j];
}
}
2013/8/22 Claude Marche <Claude.Marche at inria.fr>
>
>
> On 08/22/2013 04:55 PM, Rovedy Aparecida Busquim e Silva wrote:
> > Hi,
> >
> > We have a function to be proved (attached).
> >
> > To include the code annotations we followed the tutorial
> > (www.acm.org/conferences/sac/sac2013/T4-Handout.pdf
> > <http://www.acm.org/conferences/sac/sac2013/T4-Handout.pdf>).
> >
> > First, we have included the "loop assigns".
> >
> > After that, we have tried to include one "loop invariant" for each
> > variable of the "loop assigns".
> >
> > The analysis of the loop has run without errors when there were not if
> > clauses in the code. After including the if clauses the analysis has
> > given four errors.
>
> I don't understand what you mean by "without the if clauses". could you
> send exactly your other version that has no error ?
>
> Apart from that, there are dubious annotations in your code, see remark
> below
>
> > We are using the plugin Jessie, Carbon version.
>
> more than 2 years old, 3 versions appeared after, you should consider
>
> > /*@ loop invariant 0 <= j <= 3;
> > loop invariant soma >= 0.0;
> > loop invariant \forall integer k; 0 <= k < 3 ==> (soma <= soma +
> > acel[k]);
>
> this formula is equivalent to
>
> \forall integer k; 0 <= k < 3 ==> (0 <= acel[k]);
>
> is it really what you want to specify ? It doesn't even depend on the
> loop variables.
>
> - Claude
>
> _______________________________________________
> Frama-c-discuss mailing list
> Frama-c-discuss at lists.gforge.inria.fr
> http://lists.gforge.inria.fr/cgi-bin/mailman/listinfo/frama-c-discuss
>
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``` | 982 | 3,437 | {"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 | 0.67493 |
https://www.media4math.com/MAFS.912.A-APR.2.2 | 1,716,622,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058789.0/warc/CC-MAIN-20240525065824-20240525095824-00139.warc.gz | 765,309,905 | 12,525 | ## MAFS.912.A-APR.2.2: Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x).
There are 170 resources.
Title Description Thumbnail Image Curriculum Topics
## Math Examples Collection: Synthetic Division
Polynomial Expressions and Polynomial Functions and Equations
## Math Worksheet Collection: Dividing Polynomials
Polynomial Expressions
## Math Examples Collection: Polynomial Expansion
Polynomial Expressions
## Math Video Definitions Collection: Polynomials (Spanish)
Polynomial Expressions, Polynomial Functions and Equations, The FOIL Method and Factoring Polynomials
## Math Examples Collection: Polynomial Long Division
Polynomial Expressions and Polynomial Functions and Equations
## Math Definitions Collection: Polynomials
Polynomial Functions and Equations, Factoring Polynomials, Polynomial Expressions and The FOIL Method
## Math Video Collection: Polynomials
Polynomial Expressions, The FOIL Method, Polynomial Functions and Equations and Factoring Polynomials
## Math Video Definitions Collection: Polynomials
Polynomial Expressions, Polynomial Functions and Equations, The FOIL Method and Factoring Polynomials
## Definition--Polynomial Concepts--Area Models with Polynomials
### Definition--Polynomial Concepts--Area Models with Polynomials This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Functions and Equations
## Definition--Polynomial Concepts--Binomial
### Definition--Polynomial Concepts--Binomial This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Expressions
## Definition--Polynomial Concepts--Binomial Factor
### Definition--Polynomial Concepts--Binomial Factor This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
## Definition--Polynomial Concepts--Binomial Theorem
### Definition--Polynomial Concepts--Binomial Theorem This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Expressions
## Definition--Polynomial Concepts--Coefficients
### Definition--Polynomial Concepts--Coefficients This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Expressions
## Definition--Polynomial Concepts--Degree of a Polynomial
### Definition--Polynomial Concepts--Degree of a Polynomial This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Expressions
## Definition--Polynomial Concepts--Difference of Cubes
### Definition--Polynomial Concepts--Difference of Cubes This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
## Definition--Polynomial Concepts--Difference of Squares
### Definition--Polynomial Concepts--Difference of Squares This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
## Definition--Polynomial Concepts--Factor Theorem
### Definition--Polynomial Concepts--Factor Theorem This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
## Definition--Polynomial Concepts--Factored Cubic
### Definition--Polynomial Concepts--Factored Cubic This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
## Definition--Polynomial Concepts--Factored Polynomial
### Definition--Polynomial Concepts--Factored Polynomial This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
### Definition--Polynomial Concepts--Factored Quadratic This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Factoring Polynomials
## Definition--Polynomial Concepts--Factorial
### Definition--Polynomial Concepts--Factorial This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Expressions
The FOIL Method
## Definition--Polynomial Concepts--Graphs of Polynomials
### Definition--Polynomial Concepts--Graphs of Polynomials This is a collection of definitions related to polynomials and similar topics. This includes general definitions for polynomials and polynomial functions, as well as terms related to factoring, roots, different polynomial types, and polynomial operations.
Polynomial Functions and Equations | 1,261 | 6,713 | {"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} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.801761 |
http://www.slidesearchengine.com/slide/3-3-estimating-decimals-audio | 1,502,964,059,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886103167.97/warc/CC-MAIN-20170817092444-20170817112444-00408.warc.gz | 665,830,314 | 6,611 | # 3.3 Estimating Decimals Audio
82 %
18 %
Information about 3.3 Estimating Decimals Audio
Education
Published on August 20, 2008
Author: mrwilliams
Source: authorstream.com
Slide 1: Objective: Estimate decimal sums, differences, products, and quotients. 3.3 Estimating with Decimals Slide 2: Number of the Day Estimate the product of your age in months and the number of days in a month. What does your answer represent? a x 12 x 30 a = age Slide 3: Problem of the Day Randall, Kira, and Sean have 100 baseball cards. Randall has twice as many as Kira and 10 more than Sean. How many cards does each have? Slide 4: DeAnn made a one-minute call to India, Jordan, and Pakistan. About how much did the three calls cost? Slide 5: \$1.79 \$1.87 +\$2.17 Slide 6: \$1.79 \$1.87 +\$2.17 \$2.00 \$2.00 +\$2.00 Slide 7: Practice Problems Estimate then find the answer. 6.7 + 9.4 + 15.82 31.5 x 2.8 82.5 ÷ 9.3 185.32 – 101.99 Slide 8: Image Attributions Slide 1: Image: 'Loosing money' www.flickr.com/photos/46425925@N00/213430779 Slide 3: Image: 'Baseball Season' www.flickr.com/photos/12692384@N00/136472519
User name: Comment:
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August 17, 2017
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August 17, 2017
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Start studying 7th Grade Math - Decimals, Fractions, Estimating, Integers. Learn vocabulary, ... 7th Grade Math - Decimals, Fractions, Estimating, Integers
### Chapter 2 unit 1 B. Estimate products of fractions. - YouTube
Chapter 2 unit 1 B. Estimate products of fractions. ... Lesson 3-3 Estimating ... Estimating Products of Whole Numbers and Decimals ... | 683 | 2,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2017-34 | longest | en | 0.851784 |
https://www.mrexcel.com/board/threads/the-find-function.172253/ | 1,685,787,962,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649193.79/warc/CC-MAIN-20230603101032-20230603131032-00013.warc.gz | 978,840,255 | 19,513 | # The Find Function..
#### Ste_Moore01
##### Active Member
Hi,
I'm trying to use the Find function in my spreadsheet and I've come across a problem.
When trying to find a space you use
=FIND(" ",A1,1)
If you wanted to find the first thing that isn't a space what would you use?
Thanks!
### Excel Facts
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#### Oaktree
##### MrExcel MVP
=FIND(LEFT(SUBSTITUTE(A1," ","")),A1)
#### Ste_Moore01
##### Active Member
Thanks for the reply Oaktree but that doesn't seem to work for me.
I have a cell which I'm trying to break up. It has about 8 spaces before a word and then more spaces and a price.
It is basically the way our accounts system creates invoices etc.
I'm trying to Import them into a spreadsheet, without using macros so I was going to use formulas to work out the info rather than using VBA as I find formulas easy to work with when I learn them.
#### Oaktree
##### MrExcel MVP
Please provide a sample record and how you'd like the resultant to be parsed.
Are the records all spaced the same way? (so that the first word always starts at the 9th position, etc.)?
#### fairwinds
##### MrExcel MVP
Hi,
Try:
=FIND(LEFT(TRIM(A1)),A1)
#### Ste_Moore01
##### Active Member
It appears like
Print Proforma Invoice.txt
ABCD
26QUANTITY DESCRIPTION REFERENCE PRICE PER * TOTAL DISCOUNT NET VALUE
27
281 40 x 25 Mini Trunking
2910L X 3M = 30M FTS4W 165.60 30M 3 165.60 93.50 % 10.76
304 25 x 16 S/A Trunking
3125L X 3M = 75M FSA2W 378.00 75M 3 1512.00 93.50 % 98.28
321 20mm Heavy Gauge Conduit Black
3330L X 3M = 90M FHG20/3BL 223.20 90M 3 223.20 93.70 % 14.06
Print Proforma Invoice
and I'd like it to appear
Book2
ABCDEFGHI
1QUANTITYDESCRIPTIONREFERENCEPRICEPER*TOTALDISCOUNTNET VALUE
2
3140 x 25 Mini Trunking
410L X 3M = 30MFTS4W165.630M3165.693.50%10.76
5425 x 16 S/A Trunking
625L X 3M = 75MFSA2W37875M3151293.50%98.28
7120mm Heavy Gauge Conduit Black
830L X 3M = 90MFHG20/3BL223.290M3223.293.70%14.06
Sheet1
#### fairwinds
##### MrExcel MVP
This seems very hard as I cannot find any consistansy in how to separate.
Even a macro would be hard to write.
I would try to get better output from account system to work with.
#### Ste_Moore01
##### Active Member
The files are saved in .txt format and I can set fixed width columns but there is info above the stock details that gets split up.
It's ok though I can sort it out with ease, it just takes a couple of mintues rather than done straight away.
Thanks for the help though!
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# Should one focus only on short RC passages?
Author Message
Retired Moderator
Joined: 27 Aug 2012
Posts: 1184
Kudos [?]: 1985 [0], given: 153
Should one focus only on short RC passages? [#permalink]
### Show Tags
24 Jul 2013, 23:10
Hi Kaplan Verbal team,
I'd seek your feedback on the following:
There has been some news moving around in the forums that GMAT is having trend shift towards short RC passages and probably doing away with the long passages gradually...
Could you please share your thoughts on this and let me know that whether do we need to practice ONLY short RC passages now on-wards?
_________________
Kudos [?]: 1985 [0], given: 153
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Location: United States (WA)
GMAT 1: 750 Q49 V42
Re: Should one focus only on short RC passages? [#permalink]
### Show Tags
26 Jul 2013, 18:51
Interesting!!!!
Can you provide some links to the discussions mentioning? I would like to have a few for reference.
One thought as it relates to your question - if the gmac provides shorter passages, they will provide harder questions in order to keep the integrity of the test.
So I think you will have a bit of a catch 22 since most of the shorter passages are easier ones.
My suggestion would be to practice the harder passages regardless of the length.
Posted from my mobile device.
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Kudos [?]: 29429 [0], given: 5351
Retired Moderator
Joined: 27 Aug 2012
Posts: 1184
Kudos [?]: 1985 [0], given: 153
Re: Should one focus only on short RC passages? [#permalink]
### Show Tags
26 Jul 2013, 20:33
Kudos [?]: 1985 [0], given: 153
Re: Should one focus only on short RC passages? [#permalink] 26 Jul 2013, 20:33
Display posts from previous: Sort by | 663 | 2,590 | {"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-2017-51 | latest | en | 0.908786 |
https://www.physicsforums.com/threads/does-x-4-1-have-a-sollution.469406/ | 1,591,339,633,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348493151.92/warc/CC-MAIN-20200605045722-20200605075722-00528.warc.gz | 837,652,177 | 15,428 | # Does x^4=-1 have a sollution
## Main Question or Discussion Point
I am trying to find a sollution to this in the complex plane. One that seems to work is sqrt(i), but is this valid or not?
Related Linear and Abstract Algebra News on Phys.org
phyzguy
Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
$$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i$$
yup,it is valid
x= +/- sqrt(i)
sqrt(i) = +/- (1+i)/sqrt(2)
so you get four solutions
Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
$$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i$$
We can also write them in exponential form.
$-1 = e^{j180} = e^{j540} = e^{j900} = e^{j1260}$
Solutions are:
$e^{j\frac{180}{4}} = e^{j45} = +\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}$
$e^{j\frac{540}{4}} = e^{j135} = -\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}$
$e^{j\frac{900}{4}} = e^{j225} = e^{-j135} = -\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}$
$e^{j\frac{1260}{4}} = e^{j315} = e^{-j45} = +\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}$ | 509 | 1,303 | {"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} | 4.15625 | 4 | CC-MAIN-2020-24 | longest | en | 0.75648 |
https://mycollegebag.in/problems-on-trains-questions-and-answers-quiz-1 | 1,620,437,204,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988831.77/warc/CC-MAIN-20210508001259-20210508031259-00459.warc.gz | 461,871,002 | 19,356 | # Problems on Trains – Questions and Answers – Quiz 1
1. A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
120 metres
180 metres
324 metres
150 metres
2. A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
45 km/hr
50 km/hr
54 km/hr
55 km/hr
3. The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
200 m
225 m
245 m
250 m
4. Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
1 : 3
3 : 2
3 : 4
None of these
5. A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
120 m
240 m
300 m
None of these
6. A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
65 sec
89 sec
100 sec
150 sec
7. Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
50 m
72 m
80 m
82 m
8. A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
40 sec
42 sec
45 sec
48 sec
9. Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
36
45
48
49
10. A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
3.6 sec
18 sec
36 sec
72 sec
11. A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
230 m
240 m
260 m
320 m
Score = | 605 | 2,124 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-21 | latest | en | 0.915039 |
http://oeis.org/A228926 | 1,619,199,074,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039596883.98/warc/CC-MAIN-20210423161713-20210423191713-00414.warc.gz | 67,868,219 | 3,785 | 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!)
A228926 Sum(m^(n+1), m=1...n-1) modulo n. 0
0, 1, 2, 0, 0, 3, 0, 0, 6, 5, 0, 0, 0, 7, 7, 0, 0, 9, 0, 0, 14, 11, 0, 0, 0, 13, 18, 0, 0, 15, 0, 0, 22, 17, 23, 0, 0, 19, 26, 0, 0, 21, 0, 0, 30, 23, 0, 0, 0, 25, 34, 0, 0, 27, 44, 0, 38, 29, 0, 0, 0, 31, 42, 0, 0, 33, 0, 0, 46, 35, 0, 0, 0, 37, 35, 0, 66 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 LINKS FORMULA a(n) = A121706(n) mod (n-1). - T. D. Noe, Sep 16 2013 MATHEMATICA f[n_] := Mod[Sum[PowerMod[i, n+1, n], {i, 1, n}], n]; Table[f[n], {n, 100}] PROG (PARI) a(n)=lift(sum(m=1, n-1, Mod(m, n)^(n+1))) \\ Charles R Greathouse IV, Dec 27 2013 CROSSREFS Cf. A121706, A228919, A204187. Sequence in context: A182797 A212163 A212195 * A321414 A268865 A024159 Adjacent sequences: A228923 A228924 A228925 * A228927 A228928 A228929 KEYWORD nonn AUTHOR José María Grau Ribas, Sep 08 2013 STATUS approved
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Last modified April 23 13:25 EDT 2021. Contains 343204 sequences. (Running on oeis4.) | 606 | 1,393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-17 | latest | en | 0.646702 |
https://documen.tv/question/the-function-f-negative-startroot-endroot-is-shown-on-the-graph-on-a-coordinate-plane-an-absolut-28278375-2/ | 1,675,268,516,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499946.80/warc/CC-MAIN-20230201144459-20230201174459-00767.warc.gz | 225,520,365 | 20,126 | # The function f(x) = Negative Startroot x EndRoot is shown on the graph. On a coordinate plane, an absolute value graph
Question
The function f(x) = Negative Startroot x EndRoot is shown on the graph.
On a coordinate plane, an absolute value graph starts at (0, 0) and goes down and to the left through (4, negative 2).
Which statement is correct?
The domain of the function is all real numbers less than or equal to −1.
The range of the function is all real numbers greater than or equal to 0.
The range of the function is all real numbers less than or equal to 0.
The domain of the function is all real numbers less than or equal to 0.
in progress 0
3 weeks 2023-01-14T17:51:21+00:00 1 Answer 0 views 0
1. From that, we conclude that the correct statement is:
“The range of the function is all real numbers less than or equal to 0.”
### Which statement is correct about the given function?
Here we have the function:
f(x) = -√x
Now, remember that the argument of a square root can’t be a negative number, so the domain of our function is such that:
x ≥ 0.
Now, the maximum of the function (which is a decreasing function) is what we get when we evaluate on the minimum of the domain, so the maximum is:
f(0) = -√0 = 0
Then the range is the set of all real values equal or smaller than zero:
R: y ≤ 0.
From that, we conclude that the correct statement is:
“The range of the function is all real numbers less than or equal to 0.” | 370 | 1,439 | {"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-2023-06 | latest | en | 0.850636 |
https://www.theaverageteacher.com/tag/spring/ | 1,656,796,794,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104204514.62/warc/CC-MAIN-20220702192528-20220702222528-00352.warc.gz | 1,067,014,290 | 33,827 | • Weather Activities for Kids Spring is upon us, and it is the perfect time to teach your weather unit in science! Depending on where you live, spring likely showcases a lot of diverse weather patterns which makes this the prime time for weather observations and experiments. If you’re teaching weather right now (or about to), check out these fun weather activities for kids to really engage your students. 5 Weather Activities for Kids 1. Cloud Demonstration In this simple science experiment, students will observe what causes clouds to rain. For this activity, you will need: A jar Water Food coloring Pipette or dropper Shaving cream First, fill the jar with…
• Fun Statistics Activities for Kids I know you probably read the title of this post and asked yourself why elementary students would ever need to know statistics. Well, they don’t exactly, but that doesn’t mean that they wouldn’t still benefit from some subtle exposure to the concept! April is National Statistics & Mathematics Awareness Month, and is the perfect opportunity to add in some fun statistics activities to your instruction. If this is something you would like to do, here are a few (age appropriate) ideas! 1. Play SNAKE. SNAKE is a fun math dice game that loosely involves probability, as well as a little bit of risk! This game…
• Easter Activities for Kids Easter is one of my favorite times in the classroom! I am 100% here for all the Easter eggs and candy and Peeps. I know Peeps are controversial, but I LOVE them. Speaking of Peeps…did you know you can use them in several Easter activities for kids? Yep! So if you’re gifted a bunch in the next couples weeks but don’t want to eat them, you can turn them into a fun activity instead! If you do like Peeps, don’t worry – there are activities you can do that don’t involve Peeps, as well! So you can have your Peeps and eat them, too. 🙂 Anyway,…
• Spring is just around the corner, and it’s a wonderful time of year filled with blooming flowers, butterflies, and (depending on where you live) lots and lots of rain! Spring brings so many GREAT opportunities to do some really fun activities in your science class. So, today, I wanted to share with a few of my favorite spring science activities! Butterfly Craft This is a simple craft that helps students learn about the anatomy of a butterfly! Students will have fun making beautiful butterflies they can take home, while also being able to identify the parts of a butterfly’s body. Here’s what you need for each student: 1 clothespin 3…
• Math Easter Egg Hunt Okay, y’all know how much I love seasonal activities! Hands down one of my favorite holiday activities is a math Easter egg hunt. I’m a big fan of scavenger hunt-type games to begin with, so of course I had to come up with a math version to celebrate Easter. Math Easter egg hunts are literally SO easy to do! I’ll walk you through step by step how to set it up and run the hunt. It’s such a fun activity that your students are sure to love! Since most teachers will not physically be in the classroom before Easter due to Coronavirus, I will also run… | 670 | 3,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-27 | longest | en | 0.938122 |
http://www.brainia.com/essays/Fin-515-Week-4-Homework-Assignment/429831.html | 1,495,886,582,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608953.88/warc/CC-MAIN-20170527113807-20170527133807-00192.warc.gz | 547,993,688 | 5,328 | # FIN 515 WEEK 4 HOMEWORK ASSIGNMENT
## FIN 515 WEEK 4 HOMEWORK ASSIGNMENT
FIN 515 WEEK 4 HOMEWORK ASSIGNMENT
http://www.activitymode.com/product/fin-515-week-4-homework-assignment/
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FIN 515 WEEK 4 HOMEWORK ASSIGNMENT
FIN 515 Week 4 Homework Assignment
Week 4 Homework Assignment
Complete the following graded homework assignment in a Word document named “FIN515_Homework4_yourname“. Show the details of your calculation/work in your answer to the Problems.
• Problems (p. 297)
o 7-2 Constant Growth Valuation
o 7-4 Preferred Stock Valuation
o 7-5 Non-constant Growth Valuation
• Problems (p. 371)
o 9-2 After-Tax Cost of Debt
o 9-4 Cost of Preferred Stock with Flotation Costs
o 9-5 Cost of Equity – DCF
o 9-6 Cost of Equity – CAPM
o 9-7 WACC
Activity mode aims to provide quality study notes and tutorials to the students of FIN 515 Week 4 Homework Assignment in order to ace their studies.
FIN 515 WEEK 4 HOMEWORK ASSIGNMENT
http://www.activitymode.com/product/fin-515-week-4-homework-assignment/
SUPPORT@ACTIVITYMODE.COM
FIN 515 WEEK 4 HOMEWORK ASSIGNMENT
FIN 515 Week 4 Homework Assignment
Week 4 Homework Assignment
Complete the following graded homework assignment in a Word document named “FIN515_Homework4_yourname“. Show the details of your calculation/work in your answer to the Problems.
• Problems (p. 297)
o 7-2 Constant Growth Valuation
o 7-4 Preferred Stock Valuation
o 7-5 Non-constant Growth Valuation
• Problems (p. 371)
o 9-2 After-Tax Cost of Debt
o 9-4 Cost of Preferred Stock with Flotation Costs
o 9-5 Cost of Equity – DCF
o 9-6 Cost of Equity – CAPM
o 9-7 WACC
Activity mode aims to provide quality study notes and tutorials to the students of FIN 515 Week 4 Homework Assignment in order to ace their studies.
FIN 515 WEEK 4 HOMEWORK ASSIGNMENT | 507 | 1,798 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2017-22 | longest | en | 0.821693 |
http://www.manpagez.com/man/1/rpntutorial/ | 1,620,523,097,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00471.warc.gz | 83,751,088 | 4,786 | manpagez: man pages & more
man rpntutorial(1)
Home | html | info | man
```rpntutorial(1) rrdtool rpntutorial(1)
```
## NAME
``` rpntutorial - Reading RRDtool RPN Expressions by Steve Rader
```
## DESCRIPTION
``` This tutorial should help you get to grips with RRDtool RPN expressions
as seen in CDEF arguments of RRDtool graph.
```
``` The LT, LE, GT, GE and EQ RPN logic operators are not as tricky as they
appear. These operators act on the two values on the stack preceding
them (to the left). Read these two values on the stack from left to
right inserting the operator in the middle. If the resulting statement
is true, then replace the three values from the stack with "1". If the
statement if false, replace the three values with "0".
as "is two greater than one?" The answer to that question is "true".
So the three values should be replaced with "1". Thus the RPN
expression 2,1,GT evaluates to 1.
Now consider "2,1,LE". This RPN expression could be read as "is two
less than or equal to one?". The natural response is "no" and thus
the RPN expression 2,1,LE evaluates to 0.
```
``` The IF RPN logic operator can be straightforward also. The key to
reading IF operators is to understand that the condition part of the
traditional "if X than Y else Z" notation has *already* been evaluated.
So the IF operator acts on only one value on the stack: the third value
to the left of the IF value. The second value to the left of the IF
corresponds to the true ("Y") branch. And the first value to the left
of the IF corresponds to the false ("Z") branch. Read the RPN
expression "X,Y,Z,IF" from left to right like so: "if X then Y else Z".
For example, consider "1,10,100,IF". It looks bizarre to me. But when
I read "if 1 then 10 else 100" it's crystal clear: 1 is true so the
answer is 10. Note that only zero is false; all other values are true.
"2,20,200,IF" ("if 2 then 20 else 200") evaluates to 20. And
"0,1,2,IF" ("if 0 then 1 else 2) evaluates to 2.
Notice that none of the above examples really simulate the whole "if X
then Y else Z" statement. This is because computer programmers read
this statement as "if Some Condition then Y else Z". So it's important
to be able to read IF operators along with the LT, LE, GT, GE and EQ
operators.
```
## Some Examples
``` While compound expressions can look overly complex, they can be
considered elegantly simple. To quickly comprehend RPN expressions,
you must know the algorithm for evaluating RPN expressions: iterate
searches from the left to the right looking for an operator. When it's
found, apply that operator by popping the operator and some number of
values (and by definition, not operators) off the stack.
For example, the stack "1,2,3,+,+" gets "2,3,+" evaluated (as "2+3")
during the first iteration and is replaced by 5. This results in the
stack "1,5,+". Finally, "1,5,+" is evaluated resulting in the answer
6. For convenience, it's useful to write this set of operations as:
1) 1,2,3,+,+ eval is 2,3,+ = 5 result is 1,5,+
2) 1,5,+ eval is 1,5,+ = 6 result is 6
3) 6
Let's use that notation to conveniently solve some complex RPN
expressions with multiple logic operators:
1) 20,10,GT,10,20,IF eval is 20,10,GT = 1 result is 1,10,20,IF
read the eval as pop "20 is greater than 10" so push 1
2) 1,10,20,IF eval is 1,10,20,IF = 10 result is 10
read pop "if 1 then 10 else 20" so push 10. Only 10 is left so 10 is
multiplication operator:
1) 128,8,*,7000,GT,7000,128,8,*,IF eval 128,8,* result is 1024
2) 1024 ,7000,GT,7000,128,8,*,IF eval 1024,7000,GT result is 0
3) 0, 7000,128,8,*,IF eval 128,8,* result is 1024
4) 0, 7000,1024, IF result is 1024
Now let's go back to the first example of multiple logic operators, but
replace the value 20 with the variable "input":
1) input,10,GT,10,input,IF eval is input,10,GT ( lets call this A )
Read eval as "if input > 10 then true" and replace "input,10,GT" with
"A":
2) A,10,input,IF eval is A,10,input,IF
read "if A then 10 else input". Now replace A with it's verbose
description again and--voila!--you have an easily readable description
of the expression:
if input > 10 then 10 else input
Finally, let's go back to the first most complex example and replace
the value 128 with "input":
1) input,8,*,7000,GT,7000,input,8,*,IF eval input,8,* result is A
where A is "input * 8"
2) A,7000,GT,7000,input,8,*,IF eval is A,7000,GT result is B
where B is "if ((input * 8) > 7000) then true"
3) B,7000,input,8,*,IF eval is input,8,* result is C
where C is "input * 8"
4) B,7000,C,IF
At last we have a readable decoding of the complex RPN expression with
a variable:
if ((input * 8) > 7000) then 7000 else (input * 8)
```
## Exercises
``` Exercise 1:
Compute "3,2,*,1,+ and "3,2,1,+,*" by hand. Rewrite them in
3*2+1 = 7 and 3*(2+1) = 9. These expressions have
different answers because the altering of the plus and
times operators alter the order of their evaluation.
Exercise 2:
One may be tempted to shorten the expression
input,8,*,56000,GT,56000,input,*,8,IF
by removing the redundant use of "input,8,*" like so:
input,56000,GT,56000,input,IF,8,*
Use traditional notation to show these expressions are not the same.
Write an expression that's equivalent to the first expression, but uses
the LE and DIV operators.
if (input <= 56000/8 ) { input*8 } else { 56000 }
input,56000,8,DIV,LE,input,8,*,56000,IF
Exercise 3:
Briefly explain why traditional mathematic notation requires the use of
parentheses. Explain why RPN notation does not require the use of
parentheses.
Traditional mathematic expressions are evaluated by
doing multiplication and division first, then addition and
subtraction. Parentheses are used to force the evaluation of
addition before multiplication (etc). RPN does not require
parentheses because the ordering of objects on the stack
can force the evaluation of addition before multiplication.
Exercise 4:
Explain why it was desirable for the RRDtool developers to implement
The algorithm that implements traditional mathematical
notation is more complex then algorithm used for RPN.
So implementing RPN allowed Tobias Oetiker to write less
code! (The code is also less complex and therefore less
likely to have bugs.)
```
## AUTHOR
``` Steve Rader <rader@wiscnet.net>
1.4.8 2013-05-23 rpntutorial(1)
```
rrdtool 1.4.8 - Generated Sat Jun 1 18:07:32 CDT 2013
```© manpagez.com 2000-2021 | 1,957 | 6,692 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-21 | latest | en | 0.850195 |
https://optimization.cbe.cornell.edu/index.php?title=Job_shop_scheduling&diff=4317&oldid=3738 | 1,670,430,719,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711200.6/warc/CC-MAIN-20221207153419-20221207183419-00371.warc.gz | 468,736,047 | 8,918 | # Difference between revisions of "Job shop scheduling"
Authors: Carly Cozzolino, Brianna Christiansen, Elaine Vallejos, Michael Sorce, Jeff Visk (SYSEN 5800 Fall 2021)
## Introduction
The Job-Shop Scheduling Problem (JSSP) is an optimization problem, which mainly concerns the operations industry. The aim of the problem is to optimize the schedule for allocation of shared resources over time to competing activities[1]. Resources within a JSSP are usually signified as machines (M), while jobs (J) are the activities that will need to be completed by those machines. Jobs may require the use of a specific machine, or multiple machines in a certain order. For these problems it is assumed that machines can only process one job at a time.
## Theory/Methodology/Algorithms
Figure 1: A Gantt-Chart representation of the job shop scheduling problem[1]
The job shop scheduling problem is NP-hard meaning it’s complexity class for non-deterministic polynomial-time is least as hard as the hardest of problems in NP. As an input, there is a finite set J of jobs and a finite set of M of machines. Some jobs may need to be processed through multiple machines giving us an optimal potential processing order for each job. The processing time of the job j on machine m can be defined as the nonnegative integer . In order to minimize total completion time, the job shop can find the optimal schedule of J on M[2].
The assumptions made for solving the problem are as follows. The number of operations for each job is finite. The processing time for each operation in a particular machine is defined. There is a pre-defined sequence of operations that has to be maintained to complete each job. Delivery times of the products are undefined. No setup cost and tardiness cost. A machine can process only one job at a time. Each job visits each machine only once. No machine can deal with more than one type of task. The system cannot be interrupted until each operation of each job is finished. No machine can halt a job and start another job, before finishing the previous one[3]. Given these assumptions, the problem can be solved at complexity class NP-hard.
One way of visualizing the job shop scheduling problem is using a Gantt-Chart. Given a problem of size 3 x 3 (J x M), a solution can be represented as shown in Figure 1. Depending on the job, the operating time on each machine could vary.
## At least one numerical example
encouraged to have more than one
## Applications
There are several ways in which the job-shop scheduling problem can be modified, often to simplify the problem, for a variety of applications.
For example, in the manufacturing industry, it is common for some jobs to require certain machines to perform tasks, due to the proper capabilities or equipment of a given machine. This adds an additional layer of complexity to the problem, because not any job can be processed on any machine. This is known as flexible manufacturing.
This problem can also be applied to many projects in the technology industry. In computer programming, it is typical that instructions can only be executed one at a time on a single processor, sequentially. In this example of multiprocessor task scheduling, the instructions are the jobs to be performed and the processors required for each task can be compared to the machines. Here we would want to schedule the order of instructions such that the number of operations performed is maximized.
With the progression of automation in recent years, robotic tasks such as moving objects from one location to another are similarly optimized. In this application, the extent of movement of the robot is minimized while conducting the most amount of transport jobs.
## References
[1] T. Yamada and R. Nakano. "Job Shop Scheduling," IEEE Control Engineering Series, 1997.
[2] D. Applegate and W. Cook. "A computational study of the job-shop scheduling problem," ORSA J. on Comput, 1991.
[3] K. Hasan. "Evolutionary Algorithms for Solving Job-Shop Scheduling Problems in the Presence of Process Interruptions," Rajshahi University of Engineering and Technology, Bangladesh. 2009.
1. T. Yamada and R. Nakano. "Job Shop Scheduling," IEEE Control Engineering Series, 1997.
2. D. Applegate and W. Cook. "A computational study of the job-shop scheduling problem," ORSA J. on Comput, 1991.
3. K. Hasan. "Evolutionary Algorithms for Solving Job-Shop Scheduling Problems in the Presence of Process Interruptions," Rajshahi University of Engineering and Technology, Bangladesh. 2009. | 979 | 4,538 | {"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} | 3.28125 | 3 | CC-MAIN-2022-49 | latest | en | 0.932123 |
http://www.fixya.com/support/t4768117-aethra_vega_x7_problem | 1,508,754,117,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187825889.47/warc/CC-MAIN-20171023092524-20171023112524-00805.warc.gz | 442,410,661 | 32,408 | Question about Aethra 790209631 [JN37352] VEGA X5/N IP ONLY SET TOP-DEMO [790209631]
# Aethra Vega X7 problem
Hi.
I´me havinh this problem on a Aethra Vega X7: when i have a videocall between my two offices, one equipment is making a noise of deep that makes we to not listen what people are saying on other side. We have changes the micro, the cable and the noise persists. It´s not a continuous noise but irritating. Can you give me a help?
Posted by on
• Level 1:
An expert who has achieved level 1.
• Contributor
Dear sir
can i reset the ip in aethra vega x7 unit.i don,t have a remote
but now i unable to take access from the pc ??
Posted on Aug 06, 2013
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!
Posted on Jan 02, 2017
×
my-video-file.mp4
×
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### 180/900 in the lowest terms
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## Open Questions:
#### Related Topics:
239 people viewed this question
Level 3 Expert
Level 3 Expert | 834 | 3,041 | {"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-2017-43 | latest | en | 0.908381 |
http://mathhelpforum.com/advanced-statistics/115650-sum-independent-random-variables-moments.html | 1,527,068,973,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865468.19/warc/CC-MAIN-20180523082914-20180523102914-00150.warc.gz | 189,803,805 | 10,846 | # Thread: Sum of Independent Random Variables - Moments
1. ## Sum of Independent Random Variables - Moments
Let $\displaystyle X_1,...,X_n$ be independent, each with mean 0, and each with finite third moments. Show that:
$\displaystyle E((\sum_{i=1}^{n}X_i)^3)=\sum_{i=1}^n E(X_i^3)$
It gives a hint to use characteristic functions, so here is what I tried doing. I used $\displaystyle S_n$ to represent the sum of the Xi's from 1 to n.
$\displaystyle E(X_i^3)$ = $\displaystyle i \phi_{X_i}^{(3)} (0)$ and $\displaystyle E((\sum_{i=1}^{n}X_i)^3)=i \phi_{S_n}^{(3)}(0)$
Then, we want to show that $\displaystyle \phi_{S_n}^{(3)}(0)=\sum_{j=1}^n\phi_{X_i}^{(3)} (0)$.
My first question is if what I have done so far is okay. My second question is, where would I go from here? I can't seem to see it. Thank you!
2. Originally Posted by azdang
Let $\displaystyle X_1,...,X_n$ be independent, each with mean 0, and each with finite third moments. Show that:
$\displaystyle E((\sum_{i=1}^{n}X_i)^3)=\sum_{i=1}^n E(X_i^3)$
It gives a hint to use characteristic functions, so here is what I tried doing. I used $\displaystyle S_n$ to represent the sum of the Xi's from 1 to n.
$\displaystyle E(X_i^3)$ = $\displaystyle i \phi_{X_i}^{(3)} (0)$ and $\displaystyle E((\sum_{i=1}^{n}X_i)^3)=i \phi_{S_n}^{(3)}(0)$
Then, we want to show that $\displaystyle \phi_{S_n}^{(3)}(0)=\sum_{j=1}^n\phi_{X_i}^{(3)} (0)$.
My first question is if what I have done so far is okay. My second question is, where would I go from here? I can't seem to see it. Thank you!
a) You have $\displaystyle \phi_{S_n}(t)=\prod_{i=1}^n\phi_{X_i}(t)$, so you can derivate three time and let $\displaystyle t=0$, using $\displaystyle \phi_{X_i}'(0)=0$ (zero mean) and $\displaystyle \phi_{X_i}(0)=1$. But this is unnecessarily complicated...
b) The "good" proof goes by expanding the cube in $\displaystyle E[(X_1+\cdots+X_n)^3]$ and noting that $\displaystyle E[X_i^2X_j]=0$ for any $\displaystyle i\neq j$, and $\displaystyle E[X_iX_jX_k]=0$ for any distinct $\displaystyle i,j,k$. Then the formula is straightforward. I let you try that.
3. Thank you, Laurent. That second proof is definitely straightforward. However, the problem does specifically say to use characteristic functions, so I'm not sure I should be using this. I will continue trying to work on part a, so if you or anyone else has any suggestion, I would appreciate it
4. So, I tried to take the derivatives, using the fact that the $\displaystyle \phi_{S_n}(t)=(\phi_{X_i}(t))^n$ since the Xis are i.i.d. (I hope this is right?).
What I got down to was:
$\displaystyle \phi_{S_n}^{(3)}(0)=n((n-1)\phi_{X_i}''(0) + \phi_{X_i}^{(3)}(0))$
5. Originally Posted by azdang
So, I tried to take the derivatives, using the fact that the $\displaystyle \phi_{S_n}(t)=(\phi_{X_i}(t))^n$ since the Xis are i.i.d. (I hope this is right?).
What I got down to was:
$\displaystyle \phi_{S_n}^{(3)}(0)=n((n-1)\phi_{X_i}''(0) + \phi_{X_i}^{(3)}(0))$
If indeed the Xi's are i.i.d. (which was not specified in your first post), then you indeed have $\displaystyle \phi_{S_n}(t)=(\phi_{X_i}(t))^n$.
You did mistakes in your computation.
You should find (I simplify notations):
$\displaystyle \phi_S'=n\phi'\phi^{n-1}$,
$\displaystyle \phi_S''=n\phi''\phi^{n-1}+n(n-1)(\phi')^2\phi^{n-2}$,
$\displaystyle \phi_S'''=n\phi'''\phi^{n-1}+n\phi'\phi''\phi^{n-2}+2n(n-1)\phi'\phi''\phi^{n-2}+n(n-1)(n-2)(\phi')^3\phi^{n-3}$.
and conclude from there using what I said.
6. Yes, I ended up actually redoing the problem and getting the answer just fine. I should have updated here. THank you though! | 1,231 | 3,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-22 | latest | en | 0.799653 |
https://www.physicsforums.com/threads/finding-an-absolute-magnitude.815532/ | 1,527,451,333,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870082.90/warc/CC-MAIN-20180527190420-20180527210420-00527.warc.gz | 786,624,048 | 17,103 | # Homework Help: Finding an absolute magnitude
1. May 24, 2015
### shanepitts
1. The problem statement, all variables and given/known data
2. Relevant equations
m1-m2[/SB]=2.5log(ι21)
m-M=2.5log (d/10)2
3. The attempt at a solution
Not sure if my approach and answers are correct
2. May 24, 2015
### nrqed
First question: In your very first line with an equation, you changed the factor of 2.5 to a factor of 5 in front of the log. Why did you do this? This seems to be a mistake.
3. May 24, 2015
### shanepitts
I forgot the exponential: m-M=2.5log(d/10)2
4. May 24, 2015
### nrqed
AH yes, Ok.
EDIT: you seem to have made a sign mistake. In the exponential for the calculation of the luminosity, you should have
M_1 - M_2 = M_1 - ( m +1.99) = 5 -m - 1.99
Then your work looks good. You just need to plug in the value of m=2. The absolute magnitude of the star is smaller than the Sun's absolute magnitude (3.99 versus 5) so the star has a larger luminosity than the Sun's and your final expression agrees with this. All the steps look good.
5. May 24, 2015
### shanepitts
Thanks a bunch and sorry for the typo
6. May 24, 2015
### nrqed
You are welcome. And no problem about the typo, I make typos all the time :-)
Patrick
Last edited: May 24, 2015 | 402 | 1,274 | {"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-2018-22 | latest | en | 0.909052 |
http://www.docstoc.com/docs/74615708/DIMENSIONING | 1,432,541,604,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928423.12/warc/CC-MAIN-20150521113208-00327-ip-10-180-206-219.ec2.internal.warc.gz | 420,034,907 | 40,940 | DIMENSIONING by xiangpeng
VIEWS: 223 PAGES: 20
• pg 1
``` DIMENSIONING
• The dimension line should
be approximately 3/8” from
the object and the
extension lines should
extend 1/8” beyond the
dimension line.
• The actual measurement
is placed on the dimension
line with a break in the line
to allow for placement of
the text.
• The text is also 1/8” high.
• The arrows on each end of
the dimension line are very
slender and approximately
1/8” long.
Notation Errors
• For fractional inches
before the decimal point.
As an example, a quarter
of an inch becomes .25 Com m on errors using inc hes.
and not 0.25. Avoid
inserting a closed
quotation mark for an
inch measurement such
as .25”, and using the
notation of “mm” for
millimeter measurements
such as 68mm. Com m on errors using m illim eters.
Continuous Dimensioning
• Continuous dimensioning
starts on one edge of an
object and continues across
the object. Unless the object
has a finished surface, the
dimensions can start on
either end or the top or
bottom.
• Omit one dimension in a
series since the overall
dimension will provide the
missing measurement.
Visibility of Features
• Always place dimensions
where the shape is most
clearly visible. The cuts are
more easily seen in the front
view. The corners of the cuts
in the front view also appear
to be formed with “L’s” where
lines intersect.
• The same cuts, when
projected to the top view,
appear to form corner
intersections with “T’s”.
• In order to determine the most
visible view, always dimension
to the “L’s” and not the “T’s”.
Hidden Lines
• Never dimension to a
hidden line unless a
peculiarity of the object
forces the location of an
extension line on a hidden
line.
Angles
• For an angled surface,
use either two coordinates
or an angle and one
coordinate.
Size Dimensions
• In order to logically place
dimensions, approach the
object with the intent of
dimensioning in the order
of size, location and
overall dimensions.
• Size dimensions - Start
with the smaller parts of
the object and work
toward the larger and
finally the overall
dimensions.
Linear Dimension Placement
• For very small
dimensions, the
arrows and text are
placed outside the
extension lines. As
room permits, text is
placed between the
extension lines and
finally the text and
arrows.
• The last illustration
shows a series of
small measurements
where the “interior”
arrows are omitted.
Arc Dimension Placement
• For small arcs, omit
the center mark and
place the arrow and
text on the outside of
the curve. As the
arrow moves inside
the curve and the
enough, the
measurement also
moves inside the
curve.
Holes
• Holes are represented by
a circle in one view and
hidden lines in the
• The hole is dimensioned
where it appears round or
circular. The arrow
touches the outside of the
circle and points to the
center of the hole.
• Note that the
measurement is a
diameter by using the
symbol of the circle with a
slash.
Cylinders
• The cylinder is
dimensioned where
it appears
rectangular. Since
the cylinder is a
complete circle,
always provide the
diameter and never
• Note that the
measurement is a
diameter by using
the symbol of the
circle with a slash.
Machined Holes
DRILLED HOLE DRILLED HOLE COUN TERSIN K
(BLIN D)
Locating Rectangular Shapes
• Locate the
rectangular prism in
the top view where
two dimensions
(height and width)
can be used.
• Notice that the
position of the
rectangular prism is
more clearly visible
in the top view.
Locating Circular Shapes
• Location dimensions
for circular features
such as holes and
cylinders are always
placed on the object
where the hole or
cylinder appears
round (circular) and
not in the view
where the hole
appears as hidden
lines.
Finished Surfaces
• The finished surface
is important while
dimensioning as the
linear dimensions
typically begin from
the finished surface.
The finished surface
is noted by the
finished surface
mark (“V”) appearing
where the finished
surface appears as
an edge view.
Slotted Holes
• Slotted holes can be dimensioned several different ways.
• a. Center points on the longitudinal axis are located and the radii
are indicated with a note.
• b. Linear measurements indicate the overall size of the slot and
• c. A note indicates the two linear dimensions of the slot and
• Choose the most appropriate technique for sizing the slotted hole.
The technique used for sizing the slot determines how the slot is
located.
Bolt Circle
• Holes or cylinders can be located by referencing the diameter of the circle
or the radius of the arc. This technique is known as a circle of centers
or bolt circle because the holes would typically allow for a bolt to pass
through and fasten this part to another.
• Slotted holes can be dimensioned several different ways.
• a. Center points on the longitudinal axis are located and the radii are
indicated with a note.
• b. Linear measurements indicate the overall size of the slot and the radii
are specified.
• c. A note indicates the two linear dimensions of the slot and another note
Overall Dimensions
• After the size and location
dimensions have been
positioned on the object
and necessary notes have
dimensions are placed.
• The overall dimensions
provide the overall height,
width, and depth for the
object. They are typically
placed between the views
of the object. Avoid
indicating each of the
overall dimensions more
than once.
Overall Dimensions
• Avoid giving an overall
measurement for the
rounded-end object
since a combination of
the location dimension
(rounded end) or
diameter (cylindrical
feature) provides the
overall dimension.
Rounded End Objects
• The first technique is to provide
the location dimension between
the center points. The
given in a note. Omit the overall
dimension as location | 1,402 | 5,782 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2015-22 | longest | en | 0.882577 |
https://www.mersenneforum.org/showthread.php?s=51d77935f9d7ed131dcf86ce855e6f38&t=22061&page=3 | 1,606,821,066,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00605.warc.gz | 769,823,380 | 10,290 | mersenneforum.org Fulsorials
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2017-02-24, 05:07 #23 a1call "Rashid Naimi" Oct 2015 Remote to Here/There 2×7×139 Posts Note: The allocatemem()s are commented out. Code: print("\nBMT-300-A-Alternative-Factorials=Falsorials-PRPs.gp\n") /* allocatemem() allocatemem() allocatemem() allocatemem() allocatemem() allocatemem() allocatemem() allocatemem() allocatemem() allocatemem() allocatemem()*/ for(n= 3,3,{ forstep (i=3,3,1, falsorial=i; while(falsorial<10^1000000, falsorial=falsorial*(falsorial-1);\\print(falsorial); print(#digits(falsorial)," decimal digits."); ); ) }) print("**** End of Run ****") Code: BMT-300-A-Alternative-Factorials=Falsorials-PRPs.gp 1 decimal digits. 2 decimal digits. 3 decimal digits. 6 decimal digits. 12 decimal digits. 24 decimal digits. 48 decimal digits. 95 decimal digits. 189 decimal digits. 377 decimal digits. 753 decimal digits. 1505 decimal digits. 3010 decimal digits. 6020 decimal digits. 12040 decimal digits. 24079 decimal digits. 48157 decimal digits. 96314 decimal digits. 192628 decimal digits. *** at top-level: ...ial*(falsorial-1);print(#digits(falsorial)," *** ^-------------------- *** digits: the PARI stack overflows ! current stack size: 8000000 (7.629 Mbytes) [hint] you can increase GP stack with allocatemem() **** End of Run ****
2017-02-24, 05:10 #24 a1call "Rashid Naimi" Oct 2015 Remote to Here/There 36328 Posts My 80k dd estimate was way off, but the point is Pari can handle arithmetic on far larger integers than its #digits() function can be used to count digits for.
2017-02-24, 12:11 #25
science_man_88
"Forget I exist"
Jul 2009
Dumbassville
202618 Posts
Quote:
Originally Posted by a1call My 80k dd estimate was way off, but the point is Pari can handle arithmetic on far larger integers than its #digits() function can be used to count digits for.
you could use length(Str(number))
2017-02-24, 12:24 #26
paulunderwood
Sep 2002
Database er0rr
66608 Posts
Quote:
Originally Posted by science_man_88 you could use length(Str(number))
Code:
? allocatemem(100000000)
*** Warning: new stack size = 100000000 (95.367 Mbytes).
? length(Str(10^26000000))
26000001
2017-02-24, 12:30 #27 science_man_88 "Forget I exist" Jul 2009 Dumbassville 8,369 Posts another possibility is using digits with a base other than 10 in theory you could square a number and square the base the digits are counted in and end up with the same amount of digits roughly.
2017-02-25, 01:35 #28
a1call
"Rashid Naimi"
Oct 2015
Remote to Here/There
36328 Posts
Quote:
Originally Posted by science_man_88 you could use length(Str(number))
Thank you SM. Will give that a try.
2017-03-02, 15:03 #29 a1call "Rashid Naimi" Oct 2015 Remote to Here/There 36328 Posts The title of this thread is too huge to let it sink. Wouldn't n-1 and n+1 test be ideal for getting large primes using falserials? You start with known prime factors for 30, see if 30-1 is prime and get 30*29., Know it's prime factors, see if 869 is prime,... All using n-1 lucas test.
2019-03-25, 03:44 #30
a1call
"Rashid Naimi"
Oct 2015
Remote to Here/There
36328 Posts
The following 36716 dd Falserial has been proven Prime using PFGW:
There is currently no established way of showing the integer in a reduced form, but it would be quite easy to invent one.
Attached Files
36716 dd Falserial By Rashid Naimi - Proven Prime using PFGW.txt (35.9 KB, 53 views)
2019-03-25, 04:07 #31 Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 23CA16 Posts Are you sure that this is a Falserial and not a Ulshmartragorian? We don't know what either of these words mean. Your attached document shows nothing pertaining to PFGW. Could you please attach the full output of PFGW, where it says "is prime"?
2019-03-25, 04:23 #32 a1call "Rashid Naimi" Oct 2015 Remote to Here/There 111100110102 Posts The number is added to pfgw-prime.log using the -t flag which I believe it means it is a deterministic Prime. I could PM you the the helper file where every Prime is proven Prime by smaller found primes. The reason I did not post it in the open is to make a point that it can be proven prime if you have the structure in seconds but without it would take months or years. If the proper notation is invented/used it can also be proven prime in seconds since the structure would be known. If you would like to verify the primality please add the primes in the PM to a helper file as they are proven prime and move up to the last term.
2019-03-25, 04:32 #33
a1call
"Rashid Naimi"
Oct 2015
Remote to Here/There
2·7·139 Posts
Turns out I don't know how to add an attachment to the PM so I will attach it to this post.
You will have to prove each lower prime 1st and then add it to the helper file for proving the subsequent primes using the N-1 method.
The last 2 primes are of the same order.
Attached Files
pfgw-prime.log (107.9 KB, 60 views) | 1,468 | 4,935 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-50 | latest | en | 0.450673 |
https://www.coursehero.com/file/6289472/5/ | 1,498,650,133,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323604.1/warc/CC-MAIN-20170628101910-20170628121910-00202.warc.gz | 862,255,772 | 244,798 | # 5 - Example Acetone vapor is considered toxic to the...
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Example Acetone vapor is considered toxic to the environment. As an engineer in the chemical plant, you are asked to design an acetone recovery system having the flow sheet illustrated in the next page. To make the calculation simpler, all concentrations of both gases and liquids are in presented in weight (mass) percent (normally, for gas phase, the concentration is presented in “volume” and “mole” basis) Calculate the values of A , F , W , B , and D in kg per hour, if given G = 1,400 kg/h
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Absorber column Distillation Column Condenser Water (100%) W kg/h Air A kg/h Air 99%) Water 0.5% Feed G kg/h Acetone 3.0 % Air 95.0% Water 2.9 % F kg/h Acetone 19.0 % Water 81.0% Bottom B kg/h Acetone 4.0 % Water 96.0% Distillate D kg/h Acetone 99.0 % Water 1.0%
Basis: 1 h of operation Hence, from the basis, G = 1,400 kg , and the “ Feed ” stream contains Acetone: 3.0/100 x (1,400) = 42 kg Air: 95.0/100 x (1,400) = 1,330 kg Water: 2.0/100 x (1,400) = 28 kg
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Consider the I unit – Absorber column Water (100%) W kg/h Feed G kg/h Acetone 3.0 % Air 95.0% Water 2.9 % Air A kg/h Air 99%) Water 0.5% F kg/h Acetone19.0 % Water 81.0% Absorber Column
G + W = A + F 1,400 + W = A + F xF x 100 19 1400 100 0 . 3 = Acetone balance Performing a species balance for acetone yields (y a0 ) (G) = ( y af ) (F) Over all balance F = 221.1 kg Air Balance (ya0) (G) = ( yaf ) (F) xA x 100 5 . 99 1400 100 0 . 95 = A = 1,336.7 kg
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Substituting the values of F and A in overall equation 1400 + W = 1336.7 + 221.1 W = 157.8 kg Balance around Absorber unit are W = 157.8 kg A = 1336.7 kg F = 221.1 kg Still D and B are unknown ?????
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## This note was uploaded on 06/06/2011 for the course CHEM 3040 taught by Professor Reddy during the Spring '10 term at Taylor's.
### Page1 / 22
5 - Example Acetone vapor is considered toxic to the...
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Ask a homework question - tutors are online | 775 | 2,466 | {"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-2017-26 | longest | en | 0.795098 |
https://denizen.io/en/energy-diagram-worksheet.html | 1,726,830,942,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00382.warc.gz | 178,596,361 | 6,716 | # Energy Diagram Worksheet
Energy Diagram Worksheet - Where the transition states for each step (ts1 and ts2). Determine the heat of reaction, δh, (enthalpy change) for this reaction. A simple diagram for the energy flow in a mobile phone is shown below. Potential energy diagrams and stability. A diagram of potential energy illustrates how a system's potential energy changes as reactants are converted into products. A bouncy ball is dropped from a height. A fossil fuel made from the remains of plants. Web an energy flow diagram is a visual way to show the uses of energy in a system. Does the graph represent an endothermic or exothermic reaction? A crisp is set on fire.
Ks3 physics ks3 energy changes and transfers extra resources. A potential energy diagram is a graph of the potential energy u ( x) vs. The forward reaction is (endothermic or exothermic). 4.8 (75 reviews) energy transfers can you guess? Using the first diagram to the right, record the letter that describes each statement: Web this handy worksheet asks students to complete an energy transfer diagram for the changes to four energy systems. Support and answer sheets are included.
Web students will use this resource to identify energy stores and transfers (pathways) from energy transfer diagrams and use the information provided to calculate the efficiency of each system. They can provide power to a circuit. Draw the pe diagram showing the pe changes that occur during a successful collision of the exothermic reaction: The pe of the reactants = 400 kj the activation energy of. A potential energy diagram, also known as a reaction progress curve, is a visual representation of the energy changes that take place during a chemical reaction.
It is also often useful to indicate the total energy of a system. Explain the connection between stability and potential energy. Web potential energy diagrams worksheet. Answer the following questions based on the potential energy diagram shown here: 4.8 (29 reviews) energy stores worksheet. Web an energy flow diagram is a visual way to show the uses of energy in a system.
Energy changes in chemical reactions for ks3 science Is the diagram above depicting an endothermic or exothermic reaction? Use the potential energy diagram to answer the questions below: Support and answer sheets are included. Web potential energy diagrams worksheet.
Web an energy diagram provides us a means to assess features of physical systems at a glance. The forward reaction is (endothermic or exothermic). The activation energy of the reaction is about kilojoules. How can i use this resource?
### This Shows How Energy Is Converted From One Form Into Another.
They can provide power to a circuit. The activation energy of the reaction is about kilojoules. The enthalpy of the reactants is (less/greater) than the enthalpy of the products. The activation energy for each step (εa1 and εa2) the gibbs free energy for each step (∆g1 and ∆g2) the overall free energy (∆goverall).
### Draw The Pe Diagram Showing The Pe Changes That Occur During A Successful Collision Of The Exothermic Reaction:
Web potential energy diagram worksheets 2024. The pe of the reactants = 400 kj the activation energy of. Is the diagram above depicting an endothermic or exothermic reaction? A diagram of potential energy illustrates how a system's potential energy changes as reactants are converted into products.
### Web Chapter 14, Worksheet 1 & Video Notes “Rates Chem Rxns”.
Determine the heat of reaction, δh, (enthalpy change) for this reaction. 4.8 (17 reviews) energy changes and transfers: Web an energy flow diagram is a visual way to show the uses of energy in a system. A potential energy diagram is a graph of the potential energy u ( x) vs.
### 4.8 (75 Reviews) Energy Transfers Can You Guess?
It is also often useful to indicate the total energy of a system. Web ks3 energy in the home homework worksheet. Web potential energy diagram worksheet. The energy unit for physics covers a lot of different forms of energy, and it can be hard to keep track of all the details. | 863 | 4,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2024-38 | latest | en | 0.905417 |
https://socratic.org/precalculus/real-zeros-of-polynomials/long-division-of-polynomials | 1,701,800,691,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100555.27/warc/CC-MAIN-20231205172745-20231205202745-00760.warc.gz | 601,884,752 | 23,853 | # Long Division of Polynomials
## Key Questions
• You can use the Factor Theorem or Synthetic Division on polynomials to find the quotient using long division.
Please see the tutorial for step-by-step instructions.
Here are a couple of examples...
#### Explanation:
Here's a sample animation of long dividing ${x}^{3} + {x}^{2} - x - 1$ by $x - 1$ (which divides exactly).
Write the dividend under the bar and the divisor to the left. Each is written in descending order of powers of $x$. If any power of $x$ is missing, then include it with a $0$ coefficient. For example, if you were dividing by ${x}^{2} - 1$, then you would express the divisor as ${x}^{2} + 0 x - 1$.
Choose the first term of the quotient to cause leading terms to match. In our example, we choose ${x}^{2}$, since $\left(x - 1\right) \cdot {x}^{2} = {x}^{3} - {x}^{2}$ matches the leading ${x}^{3}$ term of the dividend.
Write the product of this term and the divisor below the dividend and subtract to give a remainder ($2 {x}^{2}$).
Bring down the next term ($- x$) from the divisor alongside it.
Choose the next term ($2 x$) of the quotient to match the leading term of this remainder, etc.
Stop when there is nothing more to bring down from the dividend and the running remainder has lower degree than the divisor.
In our example, the division is exact. We are left with no remainder.
Instead of writing out all of the terms in full, you can just write out and divide the coefficients. For example:
Here we divide $3 {x}^{4} + 2 {x}^{3} - 11 {x}^{2} - 2 x + 5$ by ${x}^{2} - 2$ to get $3 {x}^{2} + 2 x - 5$ with remainder $2 x - 5$.
#### Explanation:
Given: What is long division of polynomials?
Long division of polynomials is very similar to regular long division. It can be used to simplify a rational function $\frac{N \left(x\right)}{D \left(x\right)}$ for integration in Calculus, to find a slant asymptote in PreCalculus, and many other applications. It is done when the denominator polynomial function has a lower degree than the numerator polynomial function. The denominator can be a quadratic.
Ex. $y = \frac{{x}^{2} + 12}{x - 2}$
" "ul(" "x + 2" ")
$x - 2 | {x}^{2} + 0 x + 12$
$\text{ } \underline{{x}^{2} - 2 x}$
$\text{ } 2 x + 12$
" "ul(2x -4" ")
$\text{ } 16$
This means $y = \frac{{x}^{2} + 12}{x - 2} = x + 2 + \frac{16}{x - 2}$
The slant asymptote in the above example is $y = x + 2$ | 714 | 2,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 27, "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} | 4.8125 | 5 | CC-MAIN-2023-50 | latest | en | 0.885 |
https://mrdardy.mtbos.org/2019/02/09/a-residue-of-professional-development/?replytocom=1049 | 1,653,418,880,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00103.warc.gz | 457,355,875 | 17,288 | # A Residue of Professional Development
So, the session I wrote about a few days ago (you can find that post here ) continues to pay dividends. Yesterday my Precalc Honors kiddos had a test. Today we were to begin discussing vectors. I had what felt like a pretty clever idea this morning. I started off by posting this image (stolen from the opening evening problem that Amy and Allyson shared with us )
I got a quick question back asking if the dots were equidistant. I confirmed and then my students began to quietly count. I encouraged them – as I always do – to chat with each other and I was hearing things about medium sized squares, big squares, etc. I suggested that some more formal classification might be helpful. A couple of kids quickly concluded that there are 30 squares to be formed. This is a correct answer under certain restrictions. unfortunately, these restrictions were not placed on the question. A student named Max said 40 out loud, then said 50. This shook up the crowd a bit and people began to dig in. However, they were hesitant to debate Max because he has a reputation (well deserved) for being pretty on point with questions like this one. SPOILER ALERT: I AM ABOUT TO UNVEIL OUR SOLUTION. IF YOU WOULD LIKE TO AVOID THAT AND THINK ABOUT IT FOR AWHILE FIRST, COME BACK LATER.
Still with me? Good, happy to have you. I went to the board and drew a square of side length sqrt(2) and got two great reactions right away. One person called this a diamond but acknowledged it is also a square. Another said we should redefine squares to avoid this. I then stepped out of the way to encourage discussion about sizes of diamonds that could be formed. We had a list on one side of the diagram listing number and size of ‘squares’ and developed a list on the other side of the number of, and size of, the different diamonds. We had some great debates about the parameters here. We decided that the only diamonds had size lengths of sqrt2, sqrt5, sqrt8, and sqrt10. We were unsatisfied with the seeming lack of a clear pattern here. You will see in the picture below how I tried to impose a little bit of order on the counting by making sure that I identified groups of diamonds or squares in numbered sets that were all perfect square integers in their count. What you will also see in the picture (coming soon, I promise!) is that I pivoted the conversation soon to vectors. My Precalc Honors kiddos took a test yesterday and we are prepared to start a new chapter on vectors. I did not particularly advertise that this was the next topic, but it felt like I could pivot in that direction. Many of the kids in this class took Geometry at our upper school with a text I wrote. In that text, I intentionally introduce some vector language early in the year. When I got to school today, I did not intend to pivot from this diagram straight into talking about vectors, but when we were discussing diamonds of length sqrt5 I realized that it was meaningful to distinguish between a horizontal change of 2 with a vertical change of 1 versus a horizontal change of 1 and a vertical change of 2. Time for the photo now and then a little more explanation.
So, in the photo above, a bit of glare there unfortunately, you see a green side of delta x = 1 and delta y = 2. I drew an arrowhead and one student muttered ‘vectors!’ It felt like such a natural trigger and frame to discuss vector notation. Almost instantly kids were discussing magnitude, direction, remembering notation, etc. Man, it was a great way to start the day!
I ended up sharing this problem with a couple of other classes during the day and each time I confessed that my partner and I only found 45 squares and were VERY confident of our answer. Each class figured out where we had gone wrong and they seemed pretty proud that we worked through this all together.
Another opportunity here to thanks Amy and Allyson for the great PD session and I know that I will be pulling some other tricks out of the bag of tools that they provided for us last week.
## 2 thoughts on “A Residue of Professional Development”
1. Luis Acosta says:
Hi. This was great fun working on it. Especially when you threw in the diamonds and the hypotenuse (root 2) side. I got to 48 squares. What is the total your kids came up with? Gonna try this with my 8th-graders.
Thanks.
1. Luis
Many apologies for such a late reply. My gang ended up with 50 as I remember it. I’ll dig around my notes again. How did your kiddos do with the problem? | 990 | 4,509 | {"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} | 3.28125 | 3 | CC-MAIN-2022-21 | longest | en | 0.980717 |
http://www.southernglowtans.com/b9a6hu6l/6ae1d3-mass-of-solvent-formula | 1,620,558,063,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00025.warc.gz | 92,865,843 | 13,189 | # mass of solvent formula
Given: Mass of solute (benzene) = 22 g, Mass of solvent (carbon tetrachloride) = 122 g. To Find: Mass percentage of benzene and carbon tetrachloride. Calculate the mass percent of sodium hypochlorite in commercial bleach, if 1.00 grams of NaOCl (the active ingredient in bleach) is dissolved in 19.05 grams of solution. A solid understanding of molality helps you to calculate changes in boiling and freezing points. Ask Question Asked 5 years, 10 months ago. Add the mass of the solute to the mass of the solvent to find your final volume. Calculate mass of solvent when mass percent and mass of solute is given, molar mass and molecular formula from freezing point, electron question on a metal ion, M3+ which has 5 electrons in the 3d subshell, What is the percent composition by mass of HC2H3O2 in the vinegar, Find the component mass using the mass percent, Molarity, molality calculation from mass and density, Molecular Mass from Mole Fraction and vapor pressure change, Find equilibrium constant given 2 initial concentrations and an equilibrium concentration. refers to the tendency of a solvent’s freezing point to decrease when an impurity is added. CH 3 COOH 33% w/w, and H 2 SO 4 98.0% w/w. About the Book Author . The formula for mass percentage is given as follows. {\displaystyle \rho =\sum _ {i}\rho _ {i}\,} Thus, for pure component the mass concentration equals the density of … For example, if you want to find the concentration of 10 g of cocoa powder mixed with 1.2 L of water, you would find the mass of the water using the density formula. The total mass (N + H) adds to 100 grams. So, i think now you got what is the exact procedure to calculate the Density of a Mixture, if you have any queries please feel free to contact us, Sum of mass concentrations - normalizing relation. 3 To determine the freezing point of a solution with a known mass of unknown solute, accurately weigh about 0.37 g of your unknown organic solid on the analytical balance (to 0.0001 g). 4 ), you can determine the molar mass of the unknown solute using the equation below. Answer: k instead of doing any of that stuff i just decided since 16.2% is supposed to be urea that I would assume 100grams. density = mass/volume mass = density x volume. Linear Formula: CH 3 C 6 H 3 Cl 2. Therefore, the mass of our solvent = 0.25 moles of hydrochloride / a molality of 1.5 moles, which equals 0.17 kilograms. The mass of the solvent is 0.17 kg. Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms. 0 0. Enter appropriate values in all cells except the one you wish to calculate. Each mass must be expressed in the same units to determine the proper concentration. The formula of mass concentration is as follows. Mass percentage of A = $\frac{\text{Mass of component A}}{\text{Total mass of solution}}\times 100$ e.g. Look up the Kb or Kf of the solvent (refer to the tables following this list). Remember that solubility refers to the maximum mass of solute that can be dissolved in a given mass of solvent at a specified temperature. A common organic solvent has an empirical formula of CH and a molecular mass of 78 g/mole. Subtract out the mass of the solute (5.00 g) from the total mass of the solution to get the mass of solvent: Plug moles value and the mass of the solvent into the molality formula. In the same way, a solid understanding of boiling point elevation and freezing point depression can help you determine the molecular mass of a mystery compound that’s being added to a known quantity of solvent. 83.8/3.24=25.9grams which is the answer in the back of the book. Do another determination of the freezing point of the pure solvent before adding the unknown solute (i.e., take time-temperature measurements for this sample of pure solvent). So you know all the numbers but m. Solve for m. m = 5g/molwt/0.1kg solve for molwt. mass percent = (mass of element in 1 mole of compound / mass of 1 mole of compound) x 100 The formula for a solution is: mass percent = (grams of solute / grams of solute plus solvent) x … Percentage by Mass - formula The fraction of a solute in a solution multiplied by 100. Peter J. Mikulecky, PhD, teaches biology and chemistry at Fusion Learning Center and Fusion Academy. Assume we want to dissolve 70.128 grams of salt in 1.5 kg of water. Calculate the molecular formula for this compound and name it. Lesson Summary. Mass of solution = 22 g + 122 g = 144 g. Percentage by mass = (Mass of solute/Mass of solution) x 100. Everyone who receives the link will be able to view this calculation. mass of solution = mass of solute + mass solvent. ok they are asking you basically how to make a 16.2 % by mass urea solution, % mass = mass of solute / (mass of solute + mass of solvent) x 100, urea is the solute, solve for mass of solvent, ok I get it now, and I found the moles of urea, but I do not understand how to solve for the mass of the solvent, where you see mass of solute put in 5.0 g urea. Weight the masses using a lab scale or convert the volume of the solvent to mass by using the density formula D = m/V. i will be one for the alcohol. Molecular Weight: 161.03. If you know the freezing point, subtract the freezing point of the pure solvent to it to get the. You can use a proportion, but I can't follow what you did above. Answer: 2. To further clarify this let us change N and H to numbers. The solvent is the chemical that is present in the larger amount, ... And the formula to find the mass of a solute from the molar concentration is: URL copied to clipboard. Now if there were 16.2grams urea there needs to be 83.8grams of H2O. A solvent (from the Latin solvō, "loosen, untie, solve") is a substance that dissolves a solute, resulting in a solution.A solvent is usually a liquid but can also be a solid, a gas, or a supercritical fluid.The quantity of solute that can dissolve in a specific volume of solvent varies with temperature.Major uses of solvents are in paints, paint removers, inks, dry cleaning. Result: 0.454 m; Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes. Each mass must be expressed in the same units to determine the proper concentration. Through the procedure called ebullioscopy, a known constant can be used to calculate an unknown molar mass. I don't get what it is asking or maybe not even how to do it. Here‘s an example: 97.30 g of a mystery compound is added to 500.0 g of water, raising its boiling point to 100.78 degrees C. What is the molecular mass of the mystery compound? Formula for mass percent: g solute % = (100) g solution Mass of solute Mass of solvent Mass of solution 2 Calculate the percent by mass of (NH 4 ) 2 CO 3 : Answer = 8.76% B. Mole fraction of (NH 4 ) … Find this as the difference between the mass of the solution and the mass of the solute.The mass of the solution is 1 L × (1000 mL / 1 L) × ( 1.02 g / mL) × (1 kg / 1000 g) = 1.02 kg.The mass of solute is 3.00 mol glucose × (180 g glucose / 1 mol glucose) × ( 1 kg / 1000 g ) = 0.54 kg.The mass of the solvent is 1.02 kg - 0.54 = 0.48 kg. Use the formula moles = mass of solute / molar mass. of solute * mass of solvent (in g) Molality is independent of temperature. Result: 1.65 °C; Determine the new boiling point from the boiling point of the pure solvent and the change. Percentage of benzene by mass = (22 g/144 g) x 100 = 15.28% . Step 2 - Determine mass of solvent. Note: The notation for mass concentration as well as density is ρ . Use the density of the water to find the mass. Subtract out the mass of the solute (5.00 g) from the total mass of the solution to get the mass of solvent: 30.9g - 5.0 g = 25.9 g ... of magnessium choride in preperation of 1.50% by mass solution. Calculate the number of moles of solute in the solution by multiplying the molality calculated in Step 3 by the given number of kilograms of solvent. Each calculator cell shown below corresponds to a term in the formula presented above. (vii) Normality (N) The number of gram equivalents of solute present in 1 L of solution. mass of solution = mass of solute + mass solvent If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. refers to the tendency of a solvent’s boiling point to increase when an impurity (a solute) is added to it. i.e., Density of Binary Mixture = ( ( Volume % of solvent 1 * Density of solvent 1) + (Volume % of Solvent 2 * Density of solvent 2 ) ) , Simply, D = ( ( ( V1 / V ) * D1) + ( ( V2 / V ) * D2 ) ). Calculate the grams of NaOCl (5.25% by mass) in 245 grams of a commercial bleach solution. You could set it up like this: Solve for the molality of the solution using the equation for. share my calculation. Molality Formula - Molality is defined as the number of moles of solute present in 1000 gm of the solvent. From this information, you then follow a set of simple steps to determine the molecular mass: Find the boiling point elevation or freezing point depression. so moles NaCl = 70.128 g / (58.44 g/mol) = 1.2 mol. ΔHvap is the molar enthalpy of vaporization. Step 1: First, calculate the empirical mass for CH g 12.01 1.01 13.04 mol Next, simplify the ratio of the molecular mass: empirical mass. Active 2 years ago. Convenient—calibration solutions, standards and solvent blends require little or no preparation High purity —all reagents provided in non-leachable containers Validated —all products have been manufactured in a ISO 9001 certified facility and fully tested using Thermo Scientific Mass … Calculate the amount of water (in grams) that must be added to 5.00g of urea [(NH2)2CO] in the preparation of 16.2% by mass solution. 30.9g - 5.0 g = 25.9 g. © 2020 Yeah Chemistry, All rights reserved. if mass of solute and mass of solvent is given then what is the formula of mass percent - 17483860 M is the molar mass of the solvent. Copy link. Solution: Mass of solution = mass of solute + mass of solvent. [Whcn solvent used is water, a molar (1 M) solution is more concentrated than a molal (1 M) solution.] Molality = mass of solute in gram * 1000 / mol. molecular mass 78. 100 grams solution/ 16.2 grams urea = x grams solution / 5.00 grams urea, Solving for x then gives you the mass of the solution: 30.9 g First subtract the boiling point of water from this new boiling point: Then plug this value and a Kb of 0.512 into the equation for boiling point elevation and solve for molality: Next, take this molality value and multiply it by the given mass of the solvent, water, in kilograms: Last, divide the number of grams of the mystery solute by the number of moles, giving you the molecular mass of the compound: The molecular mass of the mystery compound is 130 g/mol. Mass Percent Formula Questions: 1. Molecular mass of solute on the basis of depression in freezing point - definition M = Δ T f × w 1 1 0 0 0 × K f × w 2 w 2 = weight of solute w 1 = weight of solvent K f = molal depression constant Δ T f = depression in freezing point m solution = m solute + m solvent. so 16.2grams/5.00grams=3.24. Calculate Molecular Masses Using Boiling and Freezing Points of Solvents, How to Perform Mole-Mole Conversions from Balanced Equations, Calculate Limiting Reagents, Excess Reagents, and Products in Chemical Reactions, How to Calculate Percent Yield in a Chemical Reaction, How to Use Empirical Formulas to Find Molecular Formulas. mass = 0.975 g/ml x 350 ml mass solvent = 341.25 g Step 3 - Determine the total mass of the solution. CAS Number: 95-73-8 wt. Percentage by mass = (mass of solute/ mass of solution) x 100 Can you find the mass of solvent with mass of solute, volume of solution, and solution density? The sum of the mass concentrations of all components (including the solvent) gives the density ρ of the solution: ρ = ∑ i ρ i. Here, ρ i is the mass concentration of the i -th solute, m i is the mass of the i -th solute, and V is the volume of the solution. MM solute = m solute n solute (5) You will be working with cyclohexane as your solvent. i You look up the K for benzene and the freezing point of benzene. Note that V is the final or total volume of solution after the solute has been added to the solvent. Next, take this molality value and multiply it by the given mass of the solvent, water, in kilograms: Last, divide the number of grams of the mystery solute by the number of moles, giving you the molecular mass of the compound: The molecular mass of the mystery compound is 130 g/mol. The term ebullioscopy comes from the Latin language and means "boiling measurement". If you dissolve N grams of salt in water of mass H to make the combined mass 100 grams then the solution is N% salt solution. This is your molecular mass, or number of grams per mole, from which you can often guess the identity of the mystery compound. Log in or register to post comments; Similar Questions. Divide the given mass of solute by the number of moles calculated in Step 4. If you can measure the masses of the solute and the solution, determining the mass/mass percent is easy. Mass per volume (mass / volume) solution concentration calculator . When you’re asked to solve problems of this type, you’ll always be given the mass of the mystery solute, the mass of solvent, and either the change in the freezing or boiling point or the new freezing or boiling point itself. Freezing point depression. The solvent is the 80 °C water. In this case the solute is sodium chloride (NaCl (s)) and the solvent is 100 g of water.. Each point on the curve in the graph above tells how much solute we can add to 100 g of water at that temperature in order to form a saturated solution. If you’ve been given the boiling point, calculate the, by subtracting the boiling point of the pure solvent from the number you were given. Quick learn from Vedantu.com by using our free study materials like Sample Papers, Previous Year Question Papers and Textbook Solutions for CBSE & ICSE Boards. solvent (kg) K f (4) Because the mass of unknown solute is known (measured on the balance) and the number of moles has been calculated (by Eq. Volume Percentage (V/V) It is expressed in terms of volume percentage of solute to the solvent. Christopher Hren is a high school chemistry teacher and former track and football coach. Find your final volume there were 16.2grams urea there needs to be 83.8grams of H2O i do n't get it! * 1000 / mol at Fusion Learning Center and Fusion Academy, in kilograms Question. Terms of volume percentage of benzene the link will be working with cyclohexane as your solvent J.. Calculate an unknown molar mass moles NaCl = 70.128 g / ( 58.44 g/mol ) 1.2... Or register to post comments ; Similar Questions mass of solvent formula total mass of the to. Increase when an impurity is added grams of NaOCl ( 5.25 % by mass ) in 245 grams NaOCl! The solution from the boiling point to decrease when an impurity is added it... Of molality helps you to calculate an unknown molar mass a given mass of solute + solvent... ( a solute ) is added ) = 1.2 mol what it is expressed in the units. It to get the freezing points of doing any of that stuff i just decided since 16.2 % is to! Decrease when an impurity ( a solute ) is added of our solvent = moles... Of 1.5 moles, which equals 0.17 kilograms solution from the number of moles calculated Step. S boiling point from the number of moles calculated in Step 4 to get the / molar mass of that... D = m/V D = m/V in all cells except the one you wish calculate. And Fusion Academy one you wish to calculate an unknown molar mass comments ; Questions. Vii ) Normality ( N ) the number of moles of solute + mass 78! Expressed in the back of the solvent concentration as well as density is.! Result: 1.65 °C ; determine the total mass ( N + H ) adds to 100 grams the! Mikulecky, PhD, teaches biology and chemistry at Fusion Learning Center and Fusion Academy organic... A molecular mass of solvent ( refer to the tendency of a commercial bleach solution numbers! Expressed in the back of the solvent the answer in the same units to determine the molality 1.5! A commercial bleach solution of NaOCl ( 5.25 % by mass ) in grams... Of 1.5 mass of solvent formula, which equals 0.17 kilograms and solution density ) will... ) molality is defined as the number of gram equivalents of solute molar. You will be working with cyclohexane as your solvent H 2 so 4 98.0 %,... In kilograms moles NaCl = 70.128 g / ( 58.44 g/mol ) = mol. The solvent to find your final volume of hydrochloride / a molality of the solvent 1.65... Point of the solute and the mass of solvent molecular formula for mass concentration as well as density is.... For benzene and the change weight the masses of the solute to tables... Log in or register mass of solvent formula post comments ; Similar Questions hydrochloride / a molality of solvent! H 2 so 4 98.0 % w/w solution: mass of solvent, in kilograms impurity a... Get what it is expressed in the same units to determine the molality formula to dissolve 70.128 grams of solvent. At Fusion Learning Center and Fusion Academy ebullioscopy comes from the boiling point to when. Want to dissolve 70.128 grams of NaOCl ( 5.25 % by mass = ( 22 g. Ch 3 COOH 33 % w/w, and solution density boiling point of the solvent... Link will be able to view this calculation of solution = mass of solvent with mass of g/mole! Expressed in the same units to determine the proper concentration to further clarify this let change... Ch 3 COOH 33 % w/w, and H to numbers dissolve 70.128 grams of a ’. = ( 22 g/144 g ) molality is independent of temperature calculated in Step 4 the solute to tendency... Moles NaCl = 70.128 g / ( 58.44 g/mol ) = 1.2 mol to... A specified temperature mass / volume ) solution concentration calculator or maybe not how! What it is asking or maybe not even how to do it ), you can the... A molecular mass of solvent, in kilograms know the freezing point benzene! The procedure called ebullioscopy, a known constant can be used to calculate 4 98.0 % w/w gram equivalents solute. + H ) adds to 100 grams remember that solubility refers to the solvent 16.2grams urea there needs to 83.8grams! Density of the book Latin language and means boiling measurement '' ), you can use a proportion but! Each calculator cell shown below corresponds to a term in the mass of solvent formula units to determine proper! ( N ) the number of gram equivalents of solute in gram * 1000 / mol freezing points us! Years, 10 months ago is easy 245 grams of salt in 1.5 kg of water ch and a mass... Bleach solution the mass of solvent, in kilograms independent of temperature and Fusion.. You know the freezing point to decrease when an impurity is added PhD, teaches biology and chemistry Fusion. Normality ( N + H ) adds to 100 grams N and H to.... ) = 1.2 mol compound and name it specified temperature 100 = 15.28 % the solvent to find your volume... Of 78 g/mole 4 ), you can determine the proper concentration can determine the proper concentration 33... Track and football coach to a term in the formula moles = mass 78... Register to post comments ; Similar Questions the masses using a lab scale or convert the volume of =. But m. solve for m. m = 5g/molwt/0.1kg solve for the molality formula - molality is defined as the of... The total mass of the solution from the number of moles of solute mass. Solvent and the mass of our solvent = 0.25 moles of solute mass. Change N and H 2 so 4 98.0 % w/w, and solution density unknown... Is given as follows = 0.975 g/ml x 350 ml mass solvent or not! Is easy solute and the change m. solve for the molality of the solute to the tendency of solvent!, a known constant can be dissolved in a given mass of solvent tables following list! Mass/Mass percent is easy to decrease when an impurity ( a solute ) is added to it you... Lab scale or convert the volume of the solvent °C ; determine the proper concentration ebullioscopy, a known can... 5 ) you will be working with cyclohexane as your solvent divide the given mass solution. Find your final volume Mikulecky, PhD, teaches biology and chemistry at Learning! To determine the molar mass equals 0.17 kilograms high school chemistry teacher and former track and football.! Equals 0.17 kilograms the number of gram equivalents of solute that can be dissolved in a given of! ’ s freezing point of the solute and the mass of solute and the of! Stuff i just decided since 16.2 % is supposed to be 83.8grams of.. % w/w, and solution density when an impurity is added mass percentage is given as follows needs be! 341.25 g Step 3 - determine the new boiling point of the solvent the. The link will be able to view this calculation as mass of solvent formula solvent solution density 5 years, 10 months.! In gram * 1000 / mol want to dissolve 70.128 grams of a solvent ’ s freezing point benzene! The mass of the solvent to mass by using the density of the water to find your volume! Concentration calculator masses using a lab scale or convert the volume of the solvent to mass by the! Teaches biology and chemistry at Fusion Learning Center and Fusion Academy Fusion Academy 245 grams of salt in kg. A specified temperature g/mol ) = 1.2 mol + mass of solution, and H to.... Formula D = m/V refers to the tendency of a commercial bleach solution the numbers m.. Nacl = 70.128 g / ( 58.44 g/mol ) = 1.2 mol benzene the! Through the procedure called ebullioscopy, a known constant can be dissolved in a mass... W/W, and H to numbers the answer in the back of the water to find final., the mass a given mass of solvent with mass of solute in... 0.975 g/ml x 350 ml mass solvent a high school chemistry teacher and former track and coach! Presented above cell shown below corresponds to a term in the back of solution. Tendency of a commercial bleach solution per volume ( mass / volume ) solution calculator! Increase when an impurity ( a solute ) is added 83.8grams of H2O per volume ( mass / volume solution! S freezing point to decrease when an impurity ( a solute ) mass of solvent formula! To numbers in 1 L of solution = mass of solute that can be to..., determining the mass/mass percent is easy solution: mass of solvent at a specified temperature adds to grams! Do n't get what it is asking or maybe not even how to do it the term ebullioscopy from... H 2 so 4 98.0 % w/w solute, volume of the solvent doing any of that stuff just. The water to find the mass of solute + mass solvent = 0.25 moles of solute mass of solvent formula molar mass boiling! % by mass = ( 22 g/144 g ) molality is defined as the number of calculated. Us change N and H to numbers - molality is independent of temperature )! Except the one you wish to calculate term ebullioscopy comes from the Latin language and means boiling ''! Of benzene the maximum mass of solvent at a specified temperature is a high school chemistry teacher and former and... Name it is a high school chemistry teacher and former track and football coach ;... Solve for the molality formula, subtract the freezing point, subtract the freezing point of the and. | 5,773 | 23,148 | {"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} | 3.375 | 3 | CC-MAIN-2021-21 | latest | en | 0.870578 |
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# euler velocities to angular velocity vector
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### #1xDan Members - Reputation: 193
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Posted 11 July 2010 - 11:05 PM
How can I convert velocities specified about the x,y,z axes to the conventional way (a vector with magnitude as the speed, as used in physics engines like ODE)?
I want to do this to fit in with the 3d engine I use, which has the convention of using euler angles for most things. Also it's more user friendly to specify angular velocity as euler speeds.
Also, the opposite conversion of the vector-to-euler would be nice too.
Anyone have *code* for this? My maths and understanding of formulas is very bad.
### #2Alrecenk Members - Reputation: 400
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Posted 12 July 2010 - 05:12 PM
http://www.euclideanspace.com/maths/geometry/rotations/conversions/eulerToAngle/index.htm
The opposite conversion is there as well.
### #3xDan Members - Reputation: 193
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Posted 12 July 2010 - 09:21 PM
Correct me if I'm wrong, but since that is for rotations not velocities won't it wrap around at 360 degrees?
and I should have mentioned, my rotation is in order pitch,yaw,roll (XYZ).
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July 27, 2010, 04:50 Temperature dependent Stick-to-Wall #1 Senior Member Josef Runsten Join Date: Jul 2010 Location: Gothenburg, Sweden Posts: 188 Rep Power: 17 Hello, We are simulating a turbine guide vane with ash particles present in the flow. We would like to know if there is a way to put a condition on the Stick-to-Wall Model, so that it will be used only if the temperature of the particle exceeds a certain value? A particle with lower temperature is supposed to bounce of the surface. How do we write the code (if there is a way), specifying the temperature and wall behaviour? We are completely new to Fortran and CEL so any guidance would be greatly appreciated.
July 27, 2010, 07:26 #2 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 14,337 Rep Power: 110 Not sure but hopefully you can do this using CEL and therefore not require fortran. It is just a matter of putting the coeff. of restitution as a function of particle temperature.
July 30, 2010, 05:13 #3 Senior Member Josef Runsten Join Date: Jul 2010 Location: Gothenburg, Sweden Posts: 188 Rep Power: 17 Thank you, but I'm still a bit lost. Would this mean creating an expression close to if( particle.Temperature>1000 [K], restitution coefficient==0, restitution coefficient==1) ? I'm trying to find the answer in the CFX help, but haven't been able to. It says that the restititution coefficient is an available variable in CEL, but I can't see it anywhere.
July 30, 2010, 06:27 #4 Senior Member Josef Runsten Join Date: Jul 2010 Location: Gothenburg, Sweden Posts: 188 Rep Power: 17 or would it be as simple as to create an expression MyExpression = if(particle.Temperature>1000[K], 1, 0) and then writing MyExpression in the restitution dialog? I tried this but got the following error: "The parameter "Parallel Coefficient od Restitution' in object .... is defined to be "Single valued" but it depends on the following field valued variables: , particle.Temperature".
July 30, 2010, 06:43 #5 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 14,337 Rep Power: 110 Some variables can only be fixed values. I fear you have found one which can only be a fixed value. I would talk to your support office if there is away around this, otherwise you might need to do this in fortran.
August 2, 2010, 06:57 #6 New Member CCTech Join Date: Jul 2010 Posts: 20 Rep Power: 9 CFX allows CEL for coefficient of restitution!
August 2, 2010, 07:12 #7 Senior Member Josef Runsten Join Date: Jul 2010 Location: Gothenburg, Sweden Posts: 188 Rep Power: 17 CCTech_Pune, please explain more if you know. I tried if(particle.Temperature>1000[K], 1, 0), but that didn't work. Then I get the error message written in my previous post. However, it works if I instead write something like ave(particle.Temperature)@BLADE or something like that, but that doesn't give us what we want I think? We want the temperature of each single particle, not an average (or max, min etc..). Also, do you know if it's possible to do the same kind of thing in a transient run, using Stick-to-Wall? Stick-to-Wall forces all particles that hit the wall to become part of the wall film, but can this also be conditioned somehow? Not familiar with fortran so I don't really know how much we are allowed to modify.
August 2, 2010, 08:04 #8 New Member CCTech Join Date: Jul 2010 Posts: 20 Rep Power: 9 Hi I am checking for the same. I will update you. Thanks
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3.7. A GENERALIZED APPROACH FOR MODELING—THE PRINCIPLES OF CONSERVATION AND ANALOGY
Because of the large variety of control-system components that occur in practice, a generalized approach is useful for obtaining their mathematical model. Therefore, rather than pursue the presentation of further specific control-system components, this section will provide a generalized approach to deriving mathematical models.
There are several general principles that can be useful in serving as guides. The most important are the principle of conservation and the concept of analogous circuits.
A. Principle of Conservation
The principle of conservation is a very important guideline for the derivation of a mathematical model. A statement of this concept is that
In terms of rates, the principle of conservation is stated as follows:
Exactly what is being conserved depends on the application. However, this principle is usually used to establish a balance or inventory of mass, energy, momentum, or charge.
The principle of conservation has been used several times throughout this chapter. For example, let us reconsider the hydraulic motor and pump power transmission system illustrated in Figure 3.21 and the electric hot-water heating system of Figure 3.27. In the case of the hydraulic motor and pump power ...
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Statistics can be a complex and challenging subject for many people. Whether you’re a student struggling with statistics coursework or a professional needing to analyze data for work, getting help with statistics can make a significant difference.
In this blog post, we will explore different ways to get assistance with statistics, including learning resources, tutoring options, and online tools. We’ll also discuss common challenges in statistics and strategies to overcome them, so you can build your statistical skills with confidence.
## Importance of Statistics in Various Fields
Statistics plays a crucial role in various fields, as it involves the collection, analysis, interpretation, presentation, and organization of data. Here are some areas where statistics is of significant importance:
1. Business and Economics: Statistics is widely used in business and economics for market research, data analysis, financial analysis, forecasting, and decision-making.
2. Medicine and Healthcare: Statistics is essential in medical research, clinical trials, epidemiology, and health policy analysis to make evidence-based decisions and improve patient outcomes.
3. Social Sciences: Statistics is used in fields such as psychology, sociology, political science, and education to study human behavior, analyze social trends, and make policy decisions.
4. Environmental Science: Statistics is used in environmental research and monitoring to analyze data on climate, pollution, and natural resource management.
5. Sports and Recreation: Statistics is used in sports to analyze performance data, make strategic decisions, and evaluate player performance.
6. Education: Statistics is used in educational research to assess student performance, evaluate teaching methods, and analyze educational outcomes.
## Common Challenges in Statistics
Statistics can be challenging for many students and professionals due to various reasons. One common challenge is understanding complex concepts, such as intricate mathematical formulas and statistical methods.
Additionally, data analysis and interpretation can be daunting as it requires proficiency in statistical software and data manipulation. Applying the appropriate statistical methods to different data sets and situations can also be challenging, as it demands familiarity with various statistical techniques and their assumptions.
Furthermore, performing accurate calculations by hand or using statistical software can pose challenges that require attention to detail.
Lastly, interpreting statistical results and drawing meaningful conclusions from data may also be complex, as it necessitates critical thinking and statistical reasoning skills.
## Benefits of Seeking Help with Statistics
When seeking help with statistics, there are several benefits that can be obtained. One of the main advantages is an improved understanding of the subject matter. Statistics can be complex, and seeking assistance from experts can clarify concepts, methods, and interpretations, leading to a better grasp of the subject material.
Another benefit is the assurance of accuracy and quality in data analysis, interpretation, and presentation. Professional help can ensure that results are reliable and trustworthy, which is crucial in statistics. Additionally, seeking help can save time, as statistics assignments or projects can be time-consuming. This allows students or professionals to focus on other tasks or responsibilities, resulting in time savings.
Moreover, seeking help can potentially lead to higher grades or improved performance in statistics-related tasks or projects, which can positively impact academic or professional performance.
Lastly, overcoming challenges and obtaining assistance can boost confidence and self-esteem in dealing with statistics-related tasks or projects, contributing to overall personal growth and development.
## Types of Statistics Help Available
There are various types of statistics help available, depending on the specific needs of individuals. Some common types of statistics help include:
1. Tutoring Services: In-person or online tutoring services can provide one-on-one guidance, clarification of concepts, and assistance with data analysis, interpretation, and statistical methods.
2. Online Courses or Videos: Online courses, instructional videos, or tutorials can provide step-by-step explanations, demonstrations, and practice exercises for different statistical concepts or methods.
3. Statistical Software Assistance: Help with statistical software such as SPSS, R, or Excel can be available through online tutorials, documentation, or forums.
4. Consultation or Guidance from Experts: Consulting with statisticians or subject-matter experts can provide guidance, advice, and expertise in dealing with statistics-related tasks or projects.
5. Homework or Assignment Help: Online platforms or services that provide assistance with statistics homework or assignments can offer guidance, explanations, and solutions for specific problems.
## Tips for Finding Reliable Statistics Help Online
When seeking reliable statistics help online, it’s important to consider the following tips:
1. Research and Compare: Conduct thorough research and compare different statistics help services or platforms to evaluate their credibility, reputation, and expertise. Look for reviews, testimonials, and ratings from previous users to get an idea of their reliability.
2. Expertise and Experience: Look for statistics help services that have qualified and experienced statisticians or subject-matter experts who can provide accurate and reliable assistance. Check their credentials, qualifications, and experience in the field of statistics.
3. Range of Services: Consider the range of services offered by the statistics help provider. Ensure they offer the specific type of help you need, whether it’s tutoring, software assistance, consultation, or homework help.
4. Availability and Responsiveness: Check the availability and responsiveness of the statistics help service. Ensure they are responsive to your queries, available during your preferred time, and able to provide timely assistance.
5. Privacy and Security: Consider the privacy and security measures of the statistics help service. Ensure they maintain confidentiality of your data, protect your personal information, and follow ethical guidelines in handling data.
6. Pricing and Payment: Compare the pricing and payment options of different statistics help services. Look for transparent pricing, reasonable rates, and secure payment methods.
7. Customer Support: Consider the level of customer support provided by the statistics help service. Look for prompt and helpful customer support that can address any issues or concerns you may have.
8. Samples or Examples: Look for statistics help services that provide samples or examples of their work, such as sample solutions or case studies, to assess their quality and accuracy.
9. Flexibility and Customization: Consider the flexibility and customization options offered by the statistics help service. Ensure they can tailor their assistance to your specific needs and requirements.
10. Recommendations and Referrals: Seek recommendations or referrals from trusted sources, such as professors, peers, or professional networks, to find reliable statistics help services.
## Conclusion
Statistics is a crucial field that plays a significant role in many areas of life, from business and economics to healthcare and social sciences. If you find yourself struggling with statistics, don’t hesitate to seek help. With the right resources, support, and practice, you can develop a solid foundation in statistics and unlock its potential for informed decision-making and data-driven insights. | 1,323 | 7,821 | {"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} | 3.5 | 4 | CC-MAIN-2023-23 | longest | en | 0.920058 |
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