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# Finance problems - inventory, credit policies, cash budget
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See attachment.
I need help with problems 26.7,26.14,27.8 and 28.10.
26.7 Indicate whether the following company actions increase, decrease, or cause no change to
the cash cycle and the operating cycle.
a. The use of discounts offered by suppliers is decreased.
b. More finished goods are being produced for orders instead of for inventory.
c. A greater percentage of raw materials purchases is paid for with cash.
d. The terms of discounts offered to customers are made more favorable for the customers.
e. A larger than usual amount of raw materials is purchased as a result of a price decline.
f. An increased number of customers pay with cash instead of credit.
26.14 Below are some important figures from the budget of Pine Mulch Company for the second quarter of 20X2.
27.8 Garden Groves, Inc., a Florida-based company, has determined that a majority of its customers are located in the New York City area. Therefore, it is considering using a lockbox system offered by a bank located in New York. The bank has estimated that use of the system will reduce collection float by three days. Based on the following information, should the lockbox system be adopted?
Average number of payments per day: 150
Average value of payment: \$15,000
Fixed annual lockbox fee: \$80,000
Variable lockbox fee: \$0.50/transaction
Annual interest rate on money-market securities: 7.5 percent
28.10 Major Electronics sells 85,000 personal stereos each year at a price per unit of \$55. All sales are on credit; the terms are 3_15, net 40. The discount is taken by 40 percent of the customers. What is the investment in accounts receivable?
In reaction to a competitor, Major Electronics is considering changing its credit terms to
5_15, net 40, to preserve its sales level. Describe qualitatively how this policy change will affect the company's investment in accounts receivable.
#### Solution Preview
26.7 Indicate whether the following company actions increase, decrease, or cause no change to the cash cycle and the operating cycle.
Operating Cycle = AR days + Inventory Days
Cash Cycle = Operating Cycle - Payable Days
a. The use of discounts offered by suppliers is decreased.
This reduces the payables period since we would not be taking the discounts, so no effect on the operating cycle but the cash cycle will decrease since payable days increase
b. More finished goods are being produced for orders instead of for inventory.
This will decrease the inventory days and so will decrease the operating cycle and the cash cycle
c. A greater percentage of raw materials purchases is paid for with cash.
This will reduce the payable days. There is no effect on the operating cycle but cash cycle will increase
d. The terms of discounts offered to customers are made more favorable for the customers.
More customers would take the discount and so the AR days will decrease. The operating cycle and cash cycle will decrease.
e. A larger than usual amount of raw materials is purchased as a result of a price decline.
This will increase the inventory days. The operating cycle and cash cycle will increase
f. An ...
#### Solution Summary
The solution has various finance problems dealing with inventory, AR credit policies, cash budget, Lockbox and cash and operating cycle
\$2.19 | 741 | 3,482 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-34 | latest | en | 0.94939 |
https://www.hotukdeals.com/discussions/heaven-or-hell-explained-by-a-chemistry-student-242421 | 1,503,300,217,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886107720.63/warc/CC-MAIN-20170821060924-20170821080924-00630.warc.gz | 912,977,568 | 25,269 | Groups
EXPIRED
# Heaven or Hell ? .... Explained by a chemistry student.
.Heaven or Hell ? .... Explained by a chemistry student.
The following is an actual question given on a University of Washington chemistry mid term.
The answer by one student was so "profound" that the professor shared it with colleagues, which is why we now have the pleasure of enjoying it as well:
Bonus Question: Is Hell exothermic (gives off heat) or endothermic (absorbs heat)?
Most of the students wrote proofs of their beliefs using Boyle's Law (gas cools when it expands and heats when it is compressed) or some variant.
One student, however, wrote the following:
First, we need to know how the mass of Hell is changing in time. So we Need to know the rate at which souls are moving into Hell and the rate at
which they are leaving. I think that we can safely assume that once a soul gets to Hell, it will not leave. Therefore, no souls are leaving. As for
how many souls are entering Hell, let's look at the different religions that exist in the world today.
Most of these religions state that if you are not a member of their religion, you will go to Hell. Since there is more than one of these
religions and since people do not belong to more than one religion, we can project that all souls go to Hell. With birth and death rates as they are,
we can expect the number of souls in Hell to increase exponentially.
Now, we look at the rate of change of the volume in Hell because Boyle's Law states that in order for the temperature and pressure in Hell to stay
the same, the volume of Hell has to expand proportionately as souls are added.
This gives two possibilities:
1. If Hell is expanding at a slower rate than the rate at which souls enter Hell, then the temperature and pressure in Hell will increase until all Hell breaks loose.
2. If Hell is expanding at a rate faster than the increase of souls in Hell, then the temperature and pressure will drop until Hell freezes over.
So which is it?
If we accept the postulate given to me by Teresa during my Freshman year that, "It will be a cold day in Hell before I sleep with you," and take
into account the fact that I slept with her last night, then number two must be true, and thus I am sure that Hell is exothermic and has already frozen over.
The corollary of this theory is that since Hell has frozen over, it follows that it is not accepting any more souls and is therefore,
extinct.....leaving only Heaven, thereby proving the existence of a divine being which explains why, last night, Teresa kept shouting "Oh my God."
reminds me of the story I heard when a tutor posed this question for his students final exam
"what is courage?"
"this is"
he got an A+ lol
Original Poster
ChipSticks;3153742
reminds me of the story I heard when a tutor posed this question for his students final exam
"what is courage?"
"this is"
he got an A+ lol
I like his style ! :-D
ChipSticks;3153742
reminds me of the story I heard when a tutor posed this question for his students final exam
"what is courage?"
"this is"
he got an A+ lol
Apparently that's a true story of a student sitting a philosophy exam.
Q. Is this a question?
A. Yes, if this is an answer.
Original Poster
.... though not through an exam or anything, I saw this yesterday & thought it was quite sweet .. ..
BOY : May I hold your hand?
GIRL : No thanks, it isn't heavy
LOL...But I highly doubt the hell response was written during a mid-term.
Original Poster
& .....
WIFE : You tell a man something, it goes in one ear and comes out of the other.
HUSBAND : You tell a woman something: It goes in both ears and comes out of the mouth!
Where does Teresa live?
Original Poster
Jumpingphil;3154739
Where does Teresa live?
....With her 14 children
now that is funny!
Sorry, commenting is no longer available on this discussion.
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# Problem 43569. Avengers Assemble!
Solution 1980673
Submitted on 17 Oct 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
x = [0 0 0 0 0; 0 2 0 4 0; 0 0 0 0 0; 0 1 0 3 0] output = [2 4; 1 3]; assert(isequal(avengersAssemble(x),output))
x = 0 0 0 0 0 0 2 0 4 0 0 0 0 0 0 0 1 0 3 0 ans = 2
Assertion failed.
2 Fail
x = [0 0 0 0 0 0 0; 0 2 0 4 0 6 0; 0 0 0 0 0 0 0; 0 1 0 3 0 5 0; 0 0 0 0 0 0 0; 0 7 0 8 0 9 0; 0 0 0 0 0 0 0;] output = [2 4 6; 1 3 5; 7 8 9]; assert(isequal(avengersAssemble(x),output))
x = 0 0 0 0 0 0 0 0 2 0 4 0 6 0 0 0 0 0 0 0 0 0 1 0 3 0 5 0 0 0 0 0 0 0 0 0 7 0 8 0 9 0 0 0 0 0 0 0 0 ans = 3
Assertion failed. | 442 | 766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-16 | latest | en | 0.59947 |
https://askdev.io/questions/995398/given-series-xx-math-0-and-a-correlation-is-it-possible-to | 1,660,370,642,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571909.51/warc/CC-MAIN-20220813051311-20220813081311-00121.warc.gz | 134,548,581 | 8,276 | Given series $A$ and a correlation, is it possible to randomly calculate a fitting series $B$?
With reference to the , my question is as follows.
Usually, one would enter two value-series and a script or program calculates the correlation. For instance, with $x = 5,3,6,7,4,2,9,5$ and $y = 4,3,4,8,3,2,10,5$, the correlation is $0.93439982209434$.
For an educational website, I'm trying to find a way to let students:
• put in value series $x$, eg. $x = 5,3,6,7,4,2,9,5$
• put in the correlation, eg. $0.9344$
• put in upper and lower boundaries of $y$-series, eg. between $1$ and $10$
• give back a random series which fits the citeria, eg. $y = 4,3,4,8,3,2,10,5$
The PHP script I have written to calculate the correlation can be found in the referred-to post on stackexchange. However, it was suggested mine was much more a mathematical than a programmatical question, hence this post. Would it be possible to execute this "reverse correlation"?
2
2022-07-25 20:45:08
Source Share | 300 | 988 | {"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} | 3.109375 | 3 | CC-MAIN-2022-33 | latest | en | 0.931481 |
https://plainmath.net/7534/function-satisfies-hypotheses-rolles-theorem-interval-satisfy-conclusion | 1,656,376,399,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103344783.24/warc/CC-MAIN-20220627225823-20220628015823-00764.warc.gz | 509,458,812 | 13,732 | # The function f(x) = x(64-x^2)^1/2 satisfies the hypotheses of Rolle's Theorem on the interval [-8,8]. Find all values of that satisfy the conclusion o
The function $f\left(x\right)=x\frac{{\left(64-{x}^{2}\right)}^{1}}{2}$ satisfies the hypotheses of Rolles
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2abehn
Your derivative should have been $\sqrt{64-{x}^{2}}-\frac{{x}^{2}}{\sqrt{64-{x}^{2}}}$
The minus sign coming from the derivative of $\left(-{x}^{2}\right)$
Setting the derivative to zero,
$\sqrt{64-{x}^{2}}-\frac{{x}^{2}}{\sqrt{64-{x}^{2}}}=0$ multiply both terms by $\sqrt{64-{x}^{2}}$
$\left(64-{x}^{2}\right)-{x}^{2}=0$
$64=2{x}^{2}$
$32={x}^{2}$
$x=\frac{+}{-}4.\sqrt{2}$ | 297 | 853 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 24, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.726702 |
https://kids.classroomsecrets.co.uk/resource/year-3-arithmetic-quiz-4/ | 1,721,555,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517663.24/warc/CC-MAIN-20240721091006-20240721121006-00173.warc.gz | 309,130,903 | 55,134 | # Year 3 Arithmetic Quiz 4
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Use your knowledge and develop your arithmetic skills by playing this Year 3 Arithmetic Quiz. How quickly can you work through them?
### Teacher Specific Information
This Year 3 Arithmetic Quiz covers a range of national curriculum objectives over 10 questions and provides children with the opportunity to practise their arithmetic skills.
### National Curriculum Objectives
Number – number and place value
(1N1a)Â Count to and across 100, forwards and backwards, beginning with 0 or 1, or from any given number
(1N1b)Â Count in multiples of twos, fives and tens
(2N6)Â Use place value and number facts to solve problems
(2C2b) Add and subtract numbers using concrete objects and pictorial representations, including: a two-digit number and ones; a two-digit number and tens; two two-digit numbers; adding three one-digit numbers
(2C3)Â Recognise and use the inverse relationship between addition and subtraction and use this to check calculations and missing number problems
(3C1)Â Add and subtract numbers mentally, including: three-digit number and ones; three-digit number and tens; three-digit number and hundreds
(3C2)Â Add and subtract numbers with up to three digits, using formal written methods of columnar addition and subtraction
(3C4)Â Solve problems, including missing number problems, using number facts, place value, and more complex addition and subtraction
Number – multiplication and division
(2C6)Â Recall and use multiplication and division facts for the 2, 5 and 10 multiplication tables, including recognising odd and even numbers
(3C6)Â Recall and use multiplication and division facts for the 3, 4 and 8 multiplication tables
(3C8) Solve problems, including missing number problems, involving multiplication and division, including positive integer scaling problems and correspondence problems in which n objects are connected to m objects
Number – fractions
(1F1a) Recognise, find and name a half as one of two equal parts of an object, shape or quantity
(2F1a) Recognise, find, name and write fractions 1/3 , 1/4 , 2/4 and 3/4 of a length, shape, set of objects or quantity
(3F4)Â Add and subtract fractions with the same denominator within one whole [for example, 5/7 + 1/7 = 6/7 ] | 562 | 2,324 | {"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} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.862162 |
https://nrich.maths.org/public/leg.php?code=-99&cl=2&cldcmpid=964 | 1,527,259,344,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867094.5/warc/CC-MAIN-20180525141236-20180525161236-00074.warc.gz | 644,597,768 | 10,187 | # Search by Topic
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### Team Scream
##### Stage: 2 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides? | 2,364 | 10,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-22 | latest | en | 0.883503 |
https://fr.scribd.com/document/105680146/Assignment-SA-1 | 1,561,524,072,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000164.31/warc/CC-MAIN-20190626033520-20190626055520-00443.warc.gz | 447,514,910 | 65,393 | Vous êtes sur la page 1sur 1
# Q01. Write the set {2, 5,10,17,...} in set-builder form. Q02.
## Find the value of cos( 1710 )
.
Q03. Solve:
3
5 5 8
2
s s
x
. Q04. Find the range of ( ) 1 = f x x .
Q05. Find the multiplicative inverse of: i . Q06. Solve: 12 30 > x , when x is integer.
Q07. If A B c , then find the value of A B and A B . Q08. Evaluate:
1 2 3 + + +
+ + +
n n n n
i i i i .
Q09. If
1
1
1
+
| |
=
|
\ .
p
i
i
, find least positive integral value of p . Q10. Convert
11
16
in degree measure.
Q11. If
2
2
( )
2 1
+
+ =
+
x i
a ib
x
then, prove that
2 2
2 2
2 2
( 1)
(2 1)
+
+ =
+
x
a b
x
.
Q12. Let U {1, 2, 3, ..., 10} = , A {1, 2, 5, 6} = and B {2, 3, 4, 9} = . Verify that ( ) A B A B
'
' ' = .
Q13. By using principle of mathematical induction, prove the rule of exponents: ( ) N = e
n n n
xy x y n .
Q14. Find all the pair of consecutive even positive integers both of which are larger than 5 such that their sum
is less than 23.
Q15. Write the relation
( ) { }
3
R , : is a prime number less than 10 = x x x in roster form. Hence find its domain.
Q16. Prove that: (a) cos cos 2 cos
4 4
t t
| | | |
+ + =
| |
\ . \ .
x x x (b)
sec8 1 tan8
sec4 1 tan2
x x
x x
.
Q17. From a class of 25 students, 10 are to be chosen for excursion party. There are 3 students who decide that
either all of them will join or none of them will join. In how many ways can the party be chosen?
Q18. Let ( ) ( ) ( ) ( ) { }
1,1 , 2, 3 0, 1 , 1, 3 = f be a function from Z to Z defined by ( ) = + f x mx n for some
integers m and n. Find the function and hence its domain.
Q19. Show that:
2 2 2
2 2 3
cos A cos A cos A
3 3 2
t t
| | | |
+ + + =
| |
\ . \ .
.
Q20. Prove that (1 ) (1 ) + > +
n
x nx , for all natural numbers n, where 1 > x .
Q21. Find the real numbers a and b such that ( )(3 5 ) + a ib i is conjugate of 6 24 i .
Q22. How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the
resulting mixture will contain more than 25% but less than 30% acid content?
Q23. The function ( ) f x is defined by
1 , when 0
( ) 1, when 0
1 , when 0
<
= =
+ >
x x
f x x
x x
. Draw the graph of ( ) f x . Also find its
domain and the range.
Q24. Solve the equation: (1 tan)(1 sin2) 1 tan + = + .
Q25. If
1
sin , II
4
= e x quadrant, then find
sin , cos
2 2
and
tan
2
.
Q26. Using principle of mathematical induction, prove that:
( )( )
( )
( )( )
3 1 1 1
...
1.2.3 2.3.4 1 2 4 1 2
+
+ + + =
+ + + +
n n
n n n n n
.
Q27. Find the modulus and argument of
1
cos sin
3 3
t t
+
i
i
. Also find its polar form.
Q28. Solve graphically: 2 10, 1, 0, 0, 0 + s + > s > > x y x y x y x y .
Q29. There are 200 individuals with a skin disorder, 120 had been exposed to the chemical
1
C , 50 to chemical
2
C and 30 to both the chemical
1
C and
2
C . Find the number of individuals exposed to (a) chemical
1
C
but not chemical
2
C (b) chemical
2
C but not chemical
1
C (c) chemical
1
C or chemical
2
C .
Q30. Find the number of different 8 letter arrangements that can be made from the letters of the word
DAUGHTER so that the vowels never occur together.
KPS College Of Competitions
Mathematics for Class XI [20122013]
By OP Gupta (+91 9650 350 480)
Visit At:
www.scribd.com/theopgupta
www.theopgupta.blogspot.com
A
S
S
I
G
N
M
E
N
T
0
1
F
o
r
T
h
e
S
e
s
s
i
o
n
2
0
1
2
2
0
1
3 | 1,310 | 3,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2019-26 | latest | en | 0.796987 |
https://onebuilding.org/archive/energyplus-support/msg04377.html | 1,656,739,533,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103984681.57/warc/CC-MAIN-20220702040603-20220702070603-00444.warc.gz | 482,866,489 | 4,031 | # [EnergyPlus_Support] Re: EIR Chiller Performance Curve Fitting
```Hi Sam
I gott you. But I am not sure that the third curve you defined is
able to simulate the chiller performance. Since you have done some
work on the chiller module, does the results show the correctness?
Thank you very much for your help
charlie
--- In EnergyPlus_Support@xxxxxxxxxxxxxxx, wing sun lee
<samwslee@y...> wrote:
> Hi Charlie,
>
> As you know 1) the PRL is equal to the rated cooling load divided
by the available cooling load. e.g. from the manufacture, tha
standard rating for the specific leaving chilled water temperature
(6.7oC) and the entering air temperature (35oC).
> 2) the value ChillerEIRFPLR (Y) is EIRavailable / (EIRdesign x
PRL). You can refer to the EngineeringDoc P.338.
>
> SAMLEE
>
> applezealot <applezealot@y...> »¡¡G
> Dear sam
>
> calculation:
> 1. You used (Qr/Qa) as PLR and (Pa/Pr) as EIR output ratio for
the
me
> why you choose these parameters? P.S. Pa,Pr,Qa,Qr are power
> available, power rated, capacity available and capacity rated.
> 2. And I look into the curve fitting results in sheet1 of your
> file. Is it related to our discussion? I can not understand the
> purpose of this sheet. Please explain it for me.
>
> charlie
>
> --- In EnergyPlus_Support@xxxxxxxxxxxxxxx, wing sun lee
> <samwslee@y...> wrote:
> > Hi,
> >
>
http://groups.yahoo.com/group/EnergyPlus_Support/Files/Problem_Submit
> tals/Input Problem/Fit curve biquadratic d.xls
> >
> > It seems to be corrected and solved
> >
> > SAMLEE
> >
> >
> > donbshireyiii <shirey@f...> ¼¶¼g:
> > Charlie and SAMLEE,
> >
> > In the spreadsheet you are using, it looks like EIR as a function
> > of PLR is not being properly calculated.
> >
> > The spreadsheet seems to be using available capacity at the
current
> > operating conditions divided by the design (rated) capacity of
the
> > chiller. This is not correct.... is should be just as the
> > input/output reference says... PLR = (cooling load) divided by
> > (available capacity at the current operating conditions).
> >
> > Don
> >
> > --- In EnergyPlus_Support@xxxxxxxxxxxxxxx, wing sun lee
> > <samwslee@y...> wrote:
> > > Hi Charlie,
> > >
> > > I just use your advice that the PRL shoudl equal to the ratio
of
> > cooling load to the available capacity.
> > > Please see my attached file at
> > http://groups.yahoo.com/group/EnergyPlus_Support/F
> > iles/Problem_Submittals/Input
> > Problem/Fit curve biquadratic b.xls
> > >
> > > SAMLEE
> > >
> > > energypluser <energypluser@y...> ¼¶¼g:
> > > Dear Mike,
> > >
> > > I wanna simulate a electric chiller with the EIR module.
> And I
> > > found that you give a EXCEL example for the curve fitting. But
I
> > think
> > > the third curve in your example file may not be correct.
> > > According to the input/output reference, the third curve
is
> > the
> > > EIR output ratio in function of Part Load Ratio. The PLR
> > calculation in
> > > your file is the ratio of the available capacity to the
nominal
> > > capacity. In my opinion, PLR should equal to the ratio of the
> > cooling
> > > load to the available capacity. Am I right?
> > > If I am right, could you please tell me how to obtain the
> > third
> > > curve for my simulation based on the manufacturer catalog.
> > > Thank you in adavance
> > >
> > >
> charlie
> >
> >
> >
> >
> > The primary EnergyPlus web site is found at:
> > http://www.energyplus.gov
> >
> > The group web site is:
> > http://groups.yahoo.com/group/EnergyPlus_Support/
> >
> > Attachments are not allowed -- please post any files to the
> appropriate folder in the Files area of the Support Web Site.
> >
> >
> >
> >
> > ---------------------------------
> >
> > To visit your group on the web, go to:
> > http://groups.yahoo.com/group/EnergyPlus_Support/
> >
> > To unsubscribe from this group, send an email to:
> > EnergyPlus_Support-unsubscribe@xxxxxxxxxxxxxxx
> >
> > Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service.
> >
> >
> >
> > ---------------------------------
> > ·Q§Y®É¦¬¨ì·s email ³qª¾¡H
> > ¤U¸ü Yahoo! Messenger
> > http://messenger.yahoo.com.hk
> >
> > [Non-text portions of this message have been removed]
>
>
>
>
> The primary EnergyPlus web site is found at:
> http://www.energyplus.gov
>
> The group web site is:
> http://groups.yahoo.com/group/EnergyPlus_Support/
>
> Attachments are not allowed -- please post any files to the
appropriate folder in the Files area of the Support Web Site.
>
>
>
>
> ---------------------------------
>
> To visit your group on the web, go to:
> http://groups.yahoo.com/group/EnergyPlus_Support/
>
> To unsubscribe from this group, send an email to:
> EnergyPlus_Support-unsubscribe@xxxxxxxxxxxxxxx
>
> Your use of Yahoo! Groups is subject to the Yahoo! Terms of
Service.
>
>
>
>
> ---------------------------------
> ·Q§Y®É¦¬¨ì·s email ³qª¾¡H
> ¤U¸ü Yahoo! Messenger
> http://messenger.yahoo.com.hk
>
> [Non-text portions of this message have been removed]
The primary EnergyPlus web site is found at:
http://www.energyplus.gov
The group web site is:
http://groups.yahoo.com/group/EnergyPlus_Support/
Attachments are not allowed -- please post any files to the appropriate folder in the Files area of the Support Web Site.
<*> To visit your group on the web, go to:
http://groups.yahoo.com/group/EnergyPlus_Support/
<*> To unsubscribe from this group, send an email to:
EnergyPlus_Support-unsubscribe@xxxxxxxxxxxxxxx
<*> Your use of Yahoo! Groups is subject to:
http://docs.yahoo.com/info/terms/
``` | 1,563 | 5,607 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.894351 |
http://holani.net/error-in/error-in-cels-with-individual-nonmem.php | 1,539,667,099,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510019.12/warc/CC-MAIN-20181016051435-20181016072935-00173.warc.gz | 162,360,731 | 6,490 | # holani.net
Home > Error In > Error In Cels With Individual Nonmem
# Error In Cels With Individual Nonmem
If not, how do you compute the Log-Likelihood?? Best regards, Vladimir ***** From: [email protected] Date: Wed, 5 Apr 2000 09:43:25 +0200 Subject: RE: INITIAL OBJ FUNCTION ERROR Dear Vladimir, Raj & other NM users, I thought Here are a few possibilities: 1. Genom att använda våra tjänster godkänner du att vi använder cookies.Läs merOKMitt kontoSökMapsYouTubePlayNyheterGmailDriveKalenderGoogle+ÖversättFotonMerDokumentBloggerKontakterHangoutsÄnnu mer från GoogleLogga inDolda fältBöckerbooks.google.se - This is a second edition to the original published by Springer in
Of course I think in terms of MCPEM but I cannot understand how you can combine an exponential error with additive and compute a log-likelihood. best regards, Joost DeJongh Dr. Message issured from estimation step at initial obj function evaluation. He has worked at Genzyme, Hoechst Marion Roussel, Eli Lilly, and Quintiles.
An individual residual is the difference between a response and its individual-specific prediction, (when there are L2 records, the response may be multivariate, and the residual will be a vector difference). If anyone can help, it'll be much appreciated. He has served or currently serves on the editorial boards for the Journal of Clinical Pharmacology, Pharmaceutical Research, Journal of Pharmacokinetics and Pharmacodynamics, and The AAPS Journal. I had wanted to combine the proportional errors (eps 1,3 & eps 5,7) but wanted to leave the additive residual errors individually.
1. Since the "weight" for the sum of squares is the error variance, and this appears in the denominator of the calculation of the sum of weighted squared deviations, you're getting an
2. When you assume exponential error variance, the data are log-transformed and the error model is then assumed to follow a constant variance model.
3. In this case, you have proportional error model.
4. Best regards, Yuhong 0PROGRAM TERMINATED BY OBJ ERROR IN CELS WITH INDIVIDUAL 1 ID= 1.00E+00 SUM OF SQUARED WEIGHTED INDIVIDUAL RESIDUALS IS INFINITE MESSAGE ISSUED FROM ESTIMATION STEP AT INITIAL OBJ.
5. There are other possibilities, and there can be combinations of possibilities.
6. the prediction is a logarithm, whose argument is mistakenly coded so that it is very small.) 2.
7. In addition, many of the other chapters have been expanded and updated.
8. FUNCTION EVALUATION The control file is as follows \$PROBLEM PRODRUG AND DRUG IN PLASMA AND CSF \$INPUT ID TIME AMT DV RATE CMT MDV \$DATA MPHSMPPC1.CSV IGNORE=C \$SUBROUTINE ADVAN6 TRANS 1
9. When can it be infinite?
Joost DeJongh Leiden Advanced Pharmacokinetics & Pharmacodynamics (LAP&P) Consultants Archimedesweg 31 2333 CM Leiden The Netherlands Phone: + 71 568 6920 fax: + 71 568 6972 ***** Serge Guzy President POP-PHARM _______________________________________________________ From: Leonid Gibiansky [email protected] Subject: RE: [NMusers] Error when combining proportional residual error, Date: Wed, 02 Aug 2006 13:22:29 -0400 No, this is not the case If prediction is zero (or close to zero) at some point, you may get it. I would try to estimate > (rather than fix to zero) the additive part of the error model.
If prediction is zero (or close to zero) at some point, you may get it. Yuhong On Mon, Feb 13, 2012 at 5:00 PM, Leonid Gibiansky wrote: > Yuhong, > Usually this error points out to zero variance of the residual error. I guess something similar is used with NONMEM. FUNCTION EVALUATION= My control file is \$INPUT ID STUD STYP DOSE AMT EVID= MDV TIME DV HEIG WT BMI AGE SEX RACE \$DATA ../data/M_PK.NM.csv = IGNORE=# \$SUBROUTINE ADVAN6 TRANS1 TOL=3\$MODEL NCOMP=3=
From: Monica Grandison Subject: CELS error message Date: Fri, 11 May 2001 17:15:35 GMT While attempting to generate index plots to build a structural model I recieved the following Weighted sum of squared individuals residuals is infinite. One possible explanation is ALAG1 use: if the first observation is before ALAG1 time, you will get zero prediction (and this error) Thanks Leonid -- Leonid Gibiansky, Ph.D. FUNCTION EVALUATION >> >> My control file is >> >> \$INPUT ID STUD STYP DOSE AMT EVID MDV TIME DV HEIG WT BMI AGE SEX RACE >> >> \$DATA ../data/M_PK.NM.csv IGNORE=#
BonateUtgåva2, illustreradUtgivareSpringer Science & Business Media, 2011ISBN1441994858, 9781441994851Längd618 sidor Exportera citatBiBTeXEndNoteRefManOm Google Böcker - Sekretesspolicy - Användningsvillkor - Information för utgivare - Rapportera ett problem - Hjälp - Webbplatskarta - Googlesstartsida Leonid _______________________________________________________ RE: [NMusers] NONLINEAR PK MODEL ERROR This message: [ Message body ] [ More options (top, bottom) ] Related messages: [ Next message ] [ Previous message ] [ He received his PhD in 1996 from Indiana University in Medical Neurobiology with an emphasis on the pharmacokinetics of drugs of abuse. BonateIngen förhandsgranskning - 2011Visa alla »Vanliga ord och fraseralgorithm analysis approach approximation assumed assumption baseline Bayesian bootstrap clearance Clinical Pharmacology clinical trial coefficient compartment concentrations confidence interval correlation covariate data set
Förhandsvisa den här boken » Så tycker andra-Skriv en recensionVi kunde inte hitta några recensioner.Utvalda sidorSidan 23Sidan 12Sidan 30Sidan 32TitelsidaInnehållCHAPTER 1 The Art of Modeling1 CHAPTER 2 Linear Models and Regression61 Best regards, Yuhong 0PROGRAM TERMINATED BY OBJ ERROR IN CELS WITH INDIVIDUAL 1 ID= 1.00E+00 SUM OF SQUARED WEIGHTED INDIVIDUAL RESIDUALS IS INFINITE MESSAGE ISSUED FROM ESTIMATION STEP AT INITIAL OBJ. The comprehensive volume takes a textbook approach systematically developing the field by starting from linear models and then moving up to generalized linear and non-linear mixed effects models.Since the first edition The phrase "weighted sum" may be confusing.
There is an error in the coding for a prediction, so that the prediction is very large (in absolute value), and along with a very small intraindividual variance, the weighted squared I tried to run the nonlinear model but got error message as following, could someone point out what could be the error in my control file or my data? Can you try to write instead IF (STDY.EQ.1) THEN Y=IPRD* (1+EPS(1)) +EPS(5) ENDIF IF (STDY.EQ.2) THEN Y=IPRD* (1+EPS(2)) +EPS(6) ENDIF IF (STDY.EQ.3) THEN Y=IPRD* (1+EPS(3)) +EPS(7) ENDIF IF (STDY.EQ.4) THEN Y=IPRD*(1+EPS(4)) With a predicted value of 0, the proportional error variance will be 0, and you have the additive error variance fixed to 0.
when i did this, I received an error that states: 0PROGRAM TERMINATED BY OBJ ERROR IN CELS WITH INDIVIDUAL 934 ID=0.93400000E+03 WEIGHTED SUM OF "SQUARED" INDIVIDUAL RESIDUALS IS INFINITE MESSAGE ISSUED You might consider not fixing the additive error variance to 0.MarkMark Sale MDPresident, Next Level Solutions, LLCwww.NextLevelSolns.com 919-846-9185A carbon-neutral companySee our real time solar energy production at:http://enlighten.enphaseenergy.com/public/systems/aSDz2458 Original Message Subject: [NMusers] Thank you MK Grandison ***** Date: Mon, 14 May 2001 15:37:03 -0700 (PDT) From: [email protected] The actual error message is: WEIGHTED SUM OF "SQUARED" INDIVIDUAL RESIDUALS IS
## FUNCTION EVALUATION My control file is \$INPUT ID STUD STYP DOSE AMT EVID MDV TIME DV HEIG WT BMI AGE SEX RACE \$DATA ../data/M_PK.NM.csv IGNORE=# \$SUBROUTINE ADVAN6 TRANS1 TOL=3 \$MODEL NCOMP=3
As to the likelihood step evaluation, I think NONMEM uses Taylor expansion exp(eta)=1+eta effectively fitting the proportional rather than exponential error model. I tried to run the nonlinear model >> but got error message as following, could someone point out what could >> be the error in my control file or my data? BonateSpringer Science & Business Media, 1 juli 2011 - 618 sidor 0 Recensionerhttps://books.google.se/books/about/Pharmacokinetic_Pharmacodynamic_Modeling.html?hl=sv&id=kfS5oLko95kCThis is a second edition to the original published by Springer in 2006. To generate these index plots I used ADVAN3 TRANS4.
With a predicted value of 0, the proportional error vari= ance will be 0, and you have the additive error variance fixed to 0. = So, the total error variance will Nagaraja Post Doc Res Assoc Department of Pharmaceutics, University of Florida, Gainesville, FL-32610 Ph: 352-846-3257 Fax: 352-392-4447 ***** From: "Piotrovskij, Vladimir [JanBe]" Subject: RE: INITIAL OBJ From: "nelamangala nagaraja" Subject: INITIAL OBJ FUNCTION ERROR Date: Wed, 05 Apr 2000 01:00:15 EDT Dear NM Users, I am fitting a dataset with parent drug (actually a prodrug) and I tried to run the nonlinear model but got error message as following, could someone point out what could be the error in my control file or my data?
To avoid STDY problem, try Y=IPRD*EXP(EPS(1))+EPS(5) IF (STDY.EQ.2) THEN Y=IPRD*EXP(EPS(2))+EPS(6) ENDIF IF (STDY.EQ.3) THEN Y=IPRD*EXP(EPS(3))+EPS(7) ENDIF IF (STDY.EQ.4) THEN Y=IPRD*EXP(EPS(4))+EPS(8) ENDIF Leonid _______________________________________________________ From: "Serge Guzy" [email protected] Subject: RE: [NMusers] Error I tried to run the nonlinear model but got error message a= s following, could someone point out what could be the error in my control = file or my data? BonateBegränsad förhandsgranskning - 2006Pharmacokinetic-Pharmacodynamic Modeling and SimulationPeter BonateIngen förhandsgranskning - 2010Pharmacokinetic-Pharmacodynamic Modeling and SimulationPeter L. I've even tried increasing the initial estimates.
It is perfectly OK to combine exponential and additive errors. You might consider not fixing the addit= ive error variance to 0.Mark= Mark Sale MDPresident, Next Level So= lutions, LLCwww.NextLevelSoln= s.com 919-846-9185A carbon-neutral companySee= our real time solar energy production at:http://enlighten.enphaseen= | 2,628 | 9,945 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.803516 |
https://math.libretexts.org/Bookshelves/Algebra/Intermediate_Algebra_1e_(OpenStax)/03%3A_Graphs_and_Functions/3.02%3A_Graph_Linear_Equations_in_Two_Variables/3.2E%3A_Exercises | 1,696,069,842,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510671.0/warc/CC-MAIN-20230930082033-20230930112033-00035.warc.gz | 417,123,789 | 31,843 | # 3.2E: Exercises
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## Practice Makes Perfect
Plot Points in a Rectangular Coordinate System
In the following exercises, plot each point in a rectangular coordinate system and identify the quadrant in which the point is located.
1. a. $$(−4,2)$$ b. $$(−1,−2)$$ c. $$(3,−5)$$ d. $$(−3,0)$$
e. $$(53,2)$$
2. a. $$(−2,−3)$$ b. $$(3,−3)$$ c. $$(−4,1)$$ d. $$(4,−1)$$
e. $$(32,1)$$
3. a. $$(3,−1)$$ b. $$(−3,1)$$ c. $$(−2,0)$$ d. $$(−4,−3)$$
e. $$(1,145)$$
4. a. $$(−1,1)$$ b. $$(−2,−1)$$ c. $$(2,0)$$ d. $$(1,−4)$$
e. $$(3,72)$$
In the following exercises, for each ordered pair, decide
a. is the ordered pair a solution to the equation? b. is the point on the line?
5. $$y=x+2$$;
A: $$(0,2)$$; B: $$(1,2)$$; C: $$(−1,1)$$; D: $$(−3,−1)$$.
a. A: yes, B: no, C: yes, D: yes b. A: yes, B: no, C: yes, D: yes
6. $$y=x−4$$;
A: $$(0,−4)$$; B: $$(3,−1)$$; C: $$(2,2)$$; D: $$(1,−5)$$.
7. $$y=12x−3$$;
A: $$(0,−3)$$; B: $$(2,−2)$$; C: $$(−2,−4)$$; D: $$(4,1)$$.
a. A: yes, B: yes, C: yes, D: no b. A: yes, B: yes, C: yes, D: no
8. $$y=13x+2$$;
A: $$(0,2)$$; B: $$(3,3)$$; C: $$(−3,2)$$; D: $$(−6,0)$$.
Graph a Linear Equation by Plotting Points
In the following exercises, graph by plotting points.
9. $$y=x+2$$
10. $$y=x−3$$
11. $$y=3x−1$$
12. $$y=−2x+2$$
13. $$y=−x−3$$
14. $$y=−x−2$$
15. $$y=2x$$
16. $$y=−2x$$
17. $$y=12x+2$$
18. $$y=13x−1$$
19. $$y=43x−5$$
20. $$y=32x−3$$
21. $$y=−25x+1$$
22. $$y=−45x−1$$
23. $$y=−32x+2$$
24. $$y=−53x+4$$
Graph Vertical and Horizontal lines
In the following exercises, graph each equation.
25. a. $$x=4$$ b. $$y=3$$
a.
b.
26. a. $$x=3$$ b. $$y=1$$
27. a. $$x=−2$$ b. $$y=−5$$
a.
b.
28. a. $$x=−5$$ b. $$y=−2$$
In the following exercises, graph each pair of equations in the same rectangular coordinate system.
29. $$y=2x$$ and $$y=2$$
30. $$y=5x$$ and $$y=5$$
31. $$y=−12x$$ and $$y=−12$$
32. $$y=−13x$$ and $$y=−13$$
Find x- and y-Intercepts
In the following exercises, find the x- and y-intercepts on each graph.
33.
$$(3,0),(0,3)$$
34.
35.
$$(5,0),(0,−5)$$
36.
In the following exercises, find the intercepts for each equation.
37. $$x−y=5$$
$$x$$-int: $$(5,0)$$, $$y$$-int: $$(0,−5)$$
38. $$x−y=−4$$
39. $$3x+y=6$$
$$x$$-int: $$(2,0)$$, $$y$$-int: $$(0,6)$$
40. $$x−2y=8$$
41. $$4x−y=8$$
$$x$$-int: $$(2,0)$$, $$y$$-int: $$(0,−8)$$
42. $$5x−y=5$$
43. $$2x+5y=10$$
$$x$$-int: $$(5,0)$$, $$y$$-int: $$(0,2)$$
44. $$3x−2y=12$$
Graph a Line Using the Intercepts
In the following exercises, graph using the intercepts.
45. $$−x+4y=8$$
46. $$x+2y=4$$
47. $$x+y=−3$$
48. $$x−y=−4$$
49. $$4x+y=4$$
50. $$3x+y=3$$
51. $$3x−y=−6$$
52. $$2x−y=−8$$
53. $$2x+4y=12$$
54. $$3x−2y=6$$
55. $$2x−5y=−20$$
56. $$3x−4y=−12$$
57. $$y=−2x$$
58. $$y=5x$$
59. $$y=x$$
60. $$y=−x$$
Mixed Practice
In the following exercises, graph each equation.
61. $$y=32x$$
62. $$y=−23x$$
63. $$y=−12x+3$$
64. $$y=14x−2$$
65. $$4x+y=2$$
66. $$5x+2y=10$$
67. $$y=−1$$
68. $$x=3$$
## Writing Exercises
69. Explain how you would choose three x-values to make a table to graph the line $$y=15x−2$$.
70. What is the difference between the equations of a vertical and a horizontal line?
71. Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $$4x+y=−4$$? Why?
72. Do you prefer to use the method of plotting points or the method using the intercepts to graph the equation $$y=23x−2$$? Why?
## Self Check
a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
b. If most of your checks were:
Confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
With some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
No, I don’t get it. This is a warning sign and you must address it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
This page titled 3.2E: Exercises is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 2,296 | 5,826 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2023-40 | longest | en | 0.259323 |
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### P4 p5 p6 resource
1. 1. The Atom Nucleus Electron (negative) Proton (positive) Neutron (no charge) P4
2. 2. Charging by Friction Duster ends up with negative charge Rod ends up with positive charge Duster rubbed over surface of Glass - Friction causes outer electrons of atoms near the edge to be removed. Glass Rod - showing atoms (enlarged) with only outer electron showing. Electron Negative Charge Nucleus Positive Charge Some outer electrons rubbed onto the duster - leaving atoms behind with a net positive charge
3. 3. Charging by Friction Rubbing the Glass rod strips some of the __________ from some of the atoms at the edge of the rod. This leaves these atoms with a net ________ charge. This means that overall the Glass rod has a __________ charge as the other atoms have equal numbers of positive and __________ charges (called neutral). Perspex and Acetate also behave like this. In other materials such as Polythene the electrons are rubbed off the cloth onto the insulator, giving the material a net __________ charge. Electrostatic Law :- “ Like charges _______ ; ______ charges ________”.
4. 4. Charging by Friction Rubbing the Glass rod strips some of the __________ from some of the atoms at the edge of the rod. This leaves these atoms with a net ________ charge. This means that overall the Glass rod has a __________ charge as the other atoms have equal numbers of positive and __________ charges (called neutral). Perspex and Acetate also behave like this. In other materials such as Polythene the electrons are rubbed off the cloth onto the insulator, giving the material a net __________ charge. Electrostatic Law :- “ Like charges _______ ; ______ charges ________”. electrons positive positive negative negative repel Unlike attract
5. 5. Static Electricity : Charging Problem Planes become charged as dusty air rubs against them in flight They are earthed by connection to the ground when refuelling. This stops sparks jumping near the fuel nozzle.
6. 7. Static Electricity : Useful Charging The paint is charged as it comes out of the nozzle. The paint is attracted to the car. The car must be earthed or connected to a positive voltage.
7. 8. Another use of static: Electrostatic Precipitators.
8. 11. live earth neutral fuse cord grip
9. 12. Fuse <ul><li>Thin Tin wire in a glass or ceramic case (with metal ends) </li></ul><ul><li>Wire chosen so that it melts (‘blows’) if too much current flows </li></ul><ul><li>Disconnects live pin from the wire/appliance when it melts </li></ul><ul><li>Can be chosen from 3A, 5A, 10A , 13A </li></ul><ul><li>Choose fuse with next highest current value e.g. TV uses 4A – choose 5A fuse </li></ul>
10. 13. Circuit breakers and RCD’s <ul><li>Circuit breakers work like fuses </li></ul><ul><li>Disconnect the live (‘trip’) when current is too high </li></ul><ul><li>Can be reset (no need to change fuse!) </li></ul><ul><li>Work by using electromagnetic or thermal switches </li></ul><ul><li>R esidual C urrent D evices (RCD’s) are more sensitive and disconnect the live if any current is measured in the Earth wire- useful for lawn mowers etc. </li></ul>
11. 14. <ul><li>If you double the voltage, but the resistance stays the same, you double the current. </li></ul><ul><li>If you double the resistance, but the voltage stays the same, you halve the current. </li></ul><ul><li>V = I x R </li></ul><ul><li>Voltage = Current x Resistance </li></ul><ul><li>I = </li></ul><ul><li>R = </li></ul>
12. 15. What is it used for ? <ul><li>Pre-natal scanning </li></ul><ul><li>Cleaning – it can be used to dislodge dirt </li></ul><ul><li>Detecting flaws or cracks </li></ul><ul><li>Medical treatment e.g scans </li></ul>Ultrasound is sound of such a high frequency that we cannot hear it (i.e above 20 000 Hz)
13. 16. How does it work? The waves are reflected off the different layers
14. 17. Is this a stable (balanced) atom? Why? What is the MASS NUMBER? 5 (three Protons & two Neutrons)
15. 18. He 2 4 MASS NUMBER = number of protons + number of neutrons SYMBOL ATOMIC NUMBER = number of protons
16. 19. What do you notice? So, what is an alpha particle?
17. 20. What do you notice? The Atomic number decreases by a value of 1.
18. 21. What do you notice? The atomic structure doesn’t fundamentally change.
19. 22. Half Life
20. 23. The gamma radiation from the radioactive source is picked up above the ground, so the leak in the pipe can be detected. Gamma - tracers Radiotherapy Gamma source
21. 24. X-Rays and Gamma Rays <ul><li>Gamma rays come from the nucleus of an atom, x-rays come from a machine which fires electrons at metal targets. </li></ul><ul><li>Therefore x-rays are easier to control than gamma rays. </li></ul><ul><li>They both give out similar wavelengths </li></ul>(Both are electromagnetic waves)
22. 25. Nuclear Fission
23. 26. <ul><li>This energy is used to heat water into steam. </li></ul><ul><li>The steam drives round a turbine. </li></ul><ul><li>The turbine drives a generator (dynamo). </li></ul>
24. 27. The Earth and the Moon The Earth’s gravity attracts the Moon, in the same way the Moon’s gravity attracts the Earth. It is the gravitational attraction of the Moon which causes the tides in the oceans on Earth. gravitational force Provides centripetal force Faster Heavier Smaller circle/ ellipse BIGGER Centripetal force P5
25. 28. Resultant Force When two forces act at different angles on a body, the resultant force is calculated using the parallelogram of forces... Force from tug 1 Force from tug 2 Resultant force
26. 29. Constant Acceleration Formulas - Summary We have 4 formulas. These formulas will help you calculate any motion problem in which a body undergoes zero or constant acceleration. Whenever you have any 3 of the five ‘v-u-s-t-a’ unknowns, you can find out the remaining 2 unknown values by using one or more of the above formulas… acceleration time taken distance travelled initial velocity final velocity Meaning m/s v m/s u m/s 2 a s t m s Unit Symbol s = (u + v ) x t 2 s = ut + ½at 2 v 2 = u 2 + 2as v = u + at Formulas
27. 30. A tourist drops a croissant from the top of the Eiffel Tower. The croissant falls for 8 seconds. Ignoring air resistance, how high is the Eiffel Tower? Example Use... s = ut + ½ at 2 s = 0 x 8 + ½ x 10 x 8 2 The Eiffel Tower is 320m tall s = 320m v = u = s = t = a = 0 8 10
28. 31. Horizontal Vertical Ignoring air resistance , the only force acting on a projectile during the flight is gravity . Projectiles have a downward acceleration (due to gravity) and this only affects the vertical velocity . For a projectile there is no acceleration in the horizontal direction . Projectile Motion - Forces Acting
29. 32. Calculating Time Taken Example: Calculate the time taken, from firing, for the cannon ball to hit the target. t = d/s t = 48/24 Time taken is 2s t = 2 t = d/s is a formula that can be applied to solve problems, whenever velocity is constant... Velocity is constant in the horizontal vector...
30. 33. Calculate the final vertical velocity of the cannon ball as it hits the target. In this case a = g = 9.8ms -2 (9.8m/s 2 ) Calculating Final Velocity Example: v = u + at v = 0 + 9.8 x 2 v = 19.6m/s v = u + at is a formula that can be applied to solve problems, whenever acceleration is constant... Final vertical velocity is 19.6m/s v = 0 + 19.6
31. 34. Conservation of Momentum
32. 35. Impulse & Momentum - Summary An impulse is a sudden force applied for a short period of time. An impulse causes (and is equal to) the change in momentum. = Force x Time = Change in Momentum Impulse The units of impulse are Ns
33. 36. Geostationary - Equatorial Orbiting Satellites Geostationary satellites are used for telecommunication purposes, and they are also known as geosynchronous satellites. Geostationary satellites are used for telephone signals and satellite TV broadcasts. Three satellites can cover the whole surface of the Earth. Pacific Ocean Atlantic Ocean Indian Ocean
34. 37. Microwaves - Satellite Communications Microwaves pass through the Earth’s atmosphere with less interference than longer wavelengths (radio waves), so they are used to carry signals to satellites orbiting the Earth.
35. 38. Polarisation Polarising filters polarise light because they are composed of long-chain molecules, which are aligned within the filter in the same direction, (e.g. vertically). As unpolarised light strikes the polarising filter, the portion of the waves vibrating in any plane other than vertical, (e.g. horizontal waves) are absorbed by the filter. light beam beam of polarised light polarising filter
36. 39. Electromagnetic waves obey the wave formula: Wave Equation Write the 3 formulas... Wavelength ( ) (v) (f) ( ) Frequency (f) x = Wave Speed (v) (metre/second, m/s) (Hertz, Hz) (metre, m)
37. 40. Interference can be demonstrated using two speakers (1 metre apart), producing the same frequency sounds. Constructive and Destructive Interference The sound waves produced from these speakers cause interference. 1 metre Constructive interference Destructive interference Peak Trough
38. 41. Snell's law also states that the refractive index of a medium can be calculated from the angles of incidence and refraction: Refraction at a Boundary The refractive index of the glass block is 1.54 Glass Air n = sin i sin r n = sin 57 º sin 33 º n = 0.838 0.544
39. 42. Total Internal Reflection 42 º is known as the critical angle. Above this angle the light ray is reflected internally. This is known as total internal reflection and it is the phenomenon exploited by fibre optic cable technology. Use the angle slider to see what happens when the angle at which the light enters a glass block is increased...
40. 43. Convex Lenses - Light Ray Diagrams The principal focus of a biconvex lens...
41. 44. Convex Lenses - Light Ray Diagrams
42. 45. Resistance The resistance of a component can be calculated using Ohm’s Law: P6 Georg Simon Ohm 1789-1854 Resistance is anything that will RESIST a current. It is measured in Ohms, a unit named after me. Resistance = Voltage (in V) (in ) Current (in A) V R I
43. 46. Current-voltage graphs 1. Resistor 2. Bulb Explain the shape of each graph I V I V
44. 48. Motor Effect – left hand rule
45. 49. AC Generator <ul><li>Induced current can be increased in 4 ways: </li></ul><ul><li>Increasing the speed of movement </li></ul><ul><li>Increasing the magnetic field strength </li></ul><ul><li>Increasing the number of turns on the coil </li></ul><ul><li>Increasing the area of the coil </li></ul>
46. 50. 1) Alternating current into the primary coil 2) Current, passing through a coil, generates a magnetic field 3) Because the current is alternating, it generates a changing magnetic field 4) A changing magnetic field induces a current in the secondary coil
47. 51. Power <ul><li>Reminder P = IV </li></ul><ul><li>In a perfectly efficient transformer, the power in is equal to the power out. </li></ul><ul><li>Or, </li></ul><ul><li>So, if the Voltage doubles, the Current will halve. </li></ul>
48. 52. Full-wave rectified How does it work? +6V -6V
49. 55. AND GATE OR GATE 1 1 1 0 0 1 0 1 0 0 0 0 OUTPUT B A 1 1 1 1 0 1 1 1 0 0 0 0 OUTPUT B A
50. 56. A NOT gate swaps 0’s to 1’s and 1’s to 0’s. If it is placed after AND or OR gates, its turns them into NAND (NOT AND) and NOR (NOT OR) gates. It is sometimes called an inverter.
51. 57. NAND GATE NOR GATE
52. 58. The basis of a latch is this pattern of linked NOR gates. They can also be made with NAND gates in a similar pattern. NOR gates work with negative pulses, NAND gates work with positive pulses. A brief high (1) signal at one input results in a permanent high (1) signal at the latch output. A brief high (1) signal at the other input causes a low (0) signal at the latch output. A low (0) signal at both inputs leaves the latch output signal unchanged. | 3,259 | 12,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-22 | latest | en | 0.877296 |
https://disconnecteddog.com/qa/how-do-you-identify-a-variable-in-research.html | 1,604,061,823,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107910815.89/warc/CC-MAIN-20201030122851-20201030152851-00377.warc.gz | 278,307,566 | 9,083 | # How Do You Identify A Variable In Research?
## How do you identify an independent variable?
An easy way to think of independent and dependent variables is, when you’re conducting an experiment, the independent variable is what you change, and the dependent variable is what changes because of that.
You can also think of the independent variable as the cause and the dependent variable as the effect..
## What do you mean by a variable?
A variable is a quantity that may change within the context of a mathematical problem or experiment. Typically, we use a single letter to represent a variable. The letters x, y, and z are common generic symbols used for variables.
## What is variable and example?
more … A symbol for a value we don’t know yet. It is usually a letter like x or y. Example: in x + 2 = 6, x is the variable.
## What are the different kinds of variables?
A variable is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually has three kinds of variables: independent, dependent, and controlled.
## What are the characteristics of a variable?
Characteristics of a variable A variable is often denoted by a letter in an algebraic expression and represents a value which can be changed or varied. For example, in the expression x + 2, the letter x is a real variable and can take on the value of any real number.
## How do you explain variables to students?
A variable is something that can be changed. In computer programming we use variables to store information that might change and can be used later in our program. For example, in a game a variable could be the current score of the player; we would add 1 to the variable whenever the player gained a point.
## Why do we use variables in research?
The importance of dependent and independent variables is that they guide the researchers to per sue their studies with maximum curiosity. Dependent and independent variables are important because they drive the research process.
## What type of variable is age?
Mondal[1] suggests that age can be viewed as a discrete variable because it is commonly expressed as an integer in units of years with no decimal to indicate days and presumably, hours, minutes, and seconds.
## What are examples of variables in research?
In an experimental example, if a study is investigating the differences between males and females, gender would be a variable (some subjects in the study would be men, and others would be women). If a study has only female subjects, gender would not be a variable, since there would be only women.
## What are the 5 types of variables?
There are six common variable types:DEPENDENT VARIABLES.INDEPENDENT VARIABLES.INTERVENING VARIABLES.MODERATOR VARIABLES.CONTROL VARIABLES.EXTRANEOUS VARIABLES.
## What are the research variables?
A variable is defined as anything that has a quantity or quality that varies. The dependent variable is the variable a researcher is interested in. An independent variable is a variable believed to affect the dependent variable. Confounding variables are defined as interference caused by another variable.
## What are the two types of variables in research?
Discrete and continuous variables are two types of quantitative variables:Discrete variables represent counts (e.g. the number of objects in a collection).Continuous variables represent measurable amounts (e.g. water volume or weight).
## What is an independent variable easy definition?
Answer: An independent variable is exactly what it sounds like. It is a variable that stands alone and isn’t changed by the other variables you are trying to measure. For example, someone’s age might be an independent variable.
## How do you distinguish between dependent and independent variables?
You can think of independent and dependent variables in terms of cause and effect: an independent variable is the variable you think is the cause, while a dependent variable is the effect. In an experiment, you manipulate the independent variable and measure the outcome in the dependent variable.
## What are two basic types of variables in statistics?
All variables can be classified as quantitative or categorical variables. Discrete variables are indeed a category of quantitative variables. Categorical variables, however, are not numeric.
## What are the characteristics of variables in research?
Variables represents the measurable traits that can change over the course of a scientific experiment. In all there are six basic variable types: dependent, independent, intervening, moderator, controlled and extraneous variables.
## What are 3 types of variables?
A variable is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually has three kinds of variables: independent, dependent, and controlled.
## Why do we need variables?
Variables can represent numeric values, characters, character strings, or memory addresses. Variables play an important role in computer programming because they enable programmers to write flexible programs. Rather than entering data directly into a program, a programmer can use variables to represent the data.
## Which is the dependent variable?
The dependent variable is the variable being tested and measured in an experiment, and is ‘dependent’ on the independent variable. An example of a dependent variable is depression symptoms, which depends on the independent variable (type of therapy). | 1,045 | 5,479 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2020-45 | latest | en | 0.925049 |
https://in.mathworks.com/matlabcentral/cody/problems/33-create-times-tables/solutions/529659 | 1,606,891,016,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00323.warc.gz | 318,664,639 | 17,105 | Cody
# Problem 33. Create times-tables
Solution 529659
Submitted on 16 Nov 2014
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = 2; y_correct = [1 2; 2 4]; assert(isequal(timestables(x),y_correct))
ans = []
2 Fail
%% x = 3; y_correct = [1 2 3; 3 6 9]; assert(isequal(timestables(x),y_correct))
ans = []
Assertion failed.
3 Pass
%% x = 5; y_correct = [1 2 3 4 5; 2 4 6 8 10; 3 6 9 12 15; 4 8 12 16 20; 5 10 15 20 25]; assert(isequal(timestables(x),y_correct))
ans = []
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Start Hunting! | 251 | 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} | 2.8125 | 3 | CC-MAIN-2020-50 | latest | en | 0.618322 |
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# Using Units of Measurement Worksheets - Year 4
• Updated: 20 Nov 2023
16 measurement worksheets linked to the Australian Curriculum.
• Non-Editable: PDF
• Pages: 32 Pages
• Year: 4
Tag #TeachStarter on Instagram for a chance to be featured!
teaching resource
# Using Units of Measurement Worksheets - Year 4
• Updated: 20 Nov 2023
16 measurement worksheets linked to the Australian Curriculum.
• Non-Editable: PDF
• Pages: 32 Pages
• Year: 4
16 measurement worksheets linked to the Australian Curriculum.
This teaching resource could be used in a variety of ways when teaching measurement. Some suggestions include:
• pre- and post-testing
• independent classwork
• revision
• homework.
This teaching resource pack includes worksheets addressing the following concepts:
• length – centimetres
• length – metres
• mass – grams and kilograms
• capacity – millilitres and litres
• temperature – degrees celsius
• area – square centimetres
• volume – cubic centimetres
• time.
• the year level displayed at the top of the sheet
• no year level displayed at the top of the sheet to use for differentiation.
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OK I do think that I should learn this!
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Hey Catherine, thank you for reaching out to us. Feel free to browse the resource to see if it suits your classroom needs.
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The capacity worksheet is a little confusing. It’s asking to calculate the capacity of each of the containers, but they are all the same size with differing amounts of water. Should it be asking ‘calculate the volume of water’ instead?
• ·
Hi Liz, Thanks for your question. In this case we feel the worksheet is correct. The attribute of volume is the amount of space occupied by an object or substance, usually measured in cubic units. Capacity is only used in relation to containers and generally refers to liquid measurement in millilitres or litres. In this case we are not asking the students to measure the potential capacity of the jugs, just the current capacity in millilitres. If there is anything else I can assist you with, please don't hesitate to contact me. | 516 | 2,285 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-18 | latest | en | 0.849419 |
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## Need to checkmate the Alarming Rate of Students’ Abductions in Nigeria
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## Teens! Learn from Thomas Edison’s Efficient Scheduling Method
We all have at most twenty four hours in a day. And most often, we complain about the time not being enough. Indeed, time is never enough; for several reasons. But if you learn how to effectively manage the little time you have on your hands, you might as well be able to […]
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## The Different Types of Friends you may have…
There are many characters in every close-knit group of friends. There are the funny ones, the annoying ones, the compassionate ones and the touché ones. There are also the religious ones, the loutish ones, the serious ones and the lucky go lucky ones. Indeed there are typically several characters, which make me wonder […]
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IF YOU A […]
## Have you Learnt the Algebraic Symbols?
Symbol
Symbol Name
Meaning / definition
Example
x
x variable
unknown value to find
when 2x = 4, then x = 2
equivalence
identical to
equal by definition
equal by definition
:=
equal by definition
equal by definition
~
approximately equal
weak approximation
11 ~ 10
approximately equal
approximation
sin(0.01) ≈ 0.01
proportional to
proportional to
y ∝ x when y = kx, k constant
lemniscate
infinity symbol
much less than
much less than
1 ≪ 1000000
much greater than
much greater than
1000000 ≫ 1
( )
parentheses
calculate expression inside first
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brackets
calculate expression inside […]
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## Important Geometry Symbols for Math Students
Symbol
Symbol Name
Meaning / definition
Example
angle
formed by two rays
∠ABC = 30°
measured angle
ABC = 30°
spherical angle
AOB = 30°
right angle
= 90°
α = 90°
°
degree
1 turn = 360°
α = 60°
deg
degree
1 turn = 360deg
α = 60deg
prime
arcminute, 1° = 60′
α = 60°59′
double prime
arcsecond, 1′ = 60″
α = 60°59′59″
line
infinite line
AB
line segment
line from point A to point B
ray
line that start from point A
arc
arc from point […]
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## Classwork Series and Exercises {Agricultural Science – SS1}: 6 Biotic Factors Affecting Agricultural Production
WEEK 4
Agricultural Science. S.S.S 1. Second Term
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A biotic factor is any living component that affects another organism, including animals that consume the organism in question, and the living food that the organism consumes. Biotic factors are the living components of […]
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## Classwork Series {Home Economics – JSS3}: Immunization
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Week 8
Topic: IMMUNIZATION
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TOPIC: Production and Factors of Production
Production
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## Are You There God? It’s Me, Margaret
Are You There God? It’s Me, Margaret is a 1970 book by Judy Blume, which according to Wikipedia is typically categorized as a young adult novel, about a girl in sixth grade who has grown up without a religious affiliation. Margaret’s mother is Christian and her father is Jewish, and the novel explores her quest for a single religion. Margaret also confronts […]
By |November 8th, 2016|Categories: features||1 Comment | 1,132 | 4,492 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.944982 |
http://www.zealot.com/threads/the-lights-on-my-layout-are-very-dim.105975/ | 1,531,980,666,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590559.95/warc/CC-MAIN-20180719051224-20180719071224-00037.warc.gz | 575,296,765 | 14,885 | The lights on my layout are very dim
Discussion in 'FAQs' started by Insey, Apr 1, 2005.
1. InseyMember
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They are much dimmer than what they should be..does this mean there's a short someone along the line? Or maybe a bad light bulb somewhere? My volt meter is reading 13 volts out from the power pack, yet it is rated for 20. Without the load, it's pushing 19.9. All my lights combined are only using 1/100 an amp of current. All my lights are wired in parallel to a main bus. What could be the problem?
Also, what does it mean if a light bulb lights up much brighter than others of the same type? Does that mean it shorted?
2. ezdaysOut AZ way
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Let's answer one question at a time. Firstly, if your rated for 20 volts no load, it's conceivable that your only going to get 13 volts with a load, but that is dependent on the load. I cannot believe you are only using one milliamp with all you lamps. One lamp along can draw 100 mA. I don't know how many lamps you have in parallel, and I don't know the current or wattage rating on your power pack, but I can conclude that you are probably overloading the power pack.
What voltage is your bulbs rated at? If you have one bulb much brigher than the others, I would say it is rated to run at maybe 12 volts while the other might be rated for 18 or even 24.
I don't think you have any shorts, you have more load than your power pack can supply. Try disconnecting all but one bulb and start there. Measure the voltage and current and then compare that with the power pack rating. Watts equal voltage times the current for a quick calculation.
Let us know how you're doing.
3. InseyMember
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I have my layout wiring divided in half right now and when I have only one of the halves running, my powerpack is pushing 19 volts. The problem, I believe, lies with the other half. Something in there is draining all my voltage. Is it possible for a lightbulb to drain voltage like that?
I'm not sure what my bulbs are rated at. I probably bought them about 5 years ago from a hobby shop. I have two different types on my layout - wheat bulbs and house bulbs. They are all exactly the same in regards to each other. Yet one of the wheat bulbs (I use them as street lights) is glowing brighter than the other. Even when disconnected, my layout is still dim.
Btw, I meant to say 1 amp instead of 1/100th. I miscalculated my volt meter's decimal place.
4. ezdaysOut AZ way
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What happens when you connect the second half with the first half disconnected? If it's still running aroun 19 volts, then you are overloading the supply. If it is significatly less, then I think you have a bulb that is rated lower than the voltage you are trying to run it at. It will indeed run brigher. Look at your power pack, it should give you an output rating somewhere, in terms of amps, or watts. Some just say "overall watts" and don't specify either the accessory output or the rail output.
Again, how many bulbs are you trying to run? Mini-bulb voltages are rated at as low as 1.5 volts, up to 24 volts, but the most common usage is 12 volts. If you are running them at 18 volts and they are still dim, I'm still convinced that you are overloading the amount of current you power pack can put out.
Keep isolating the different bulbs, and I'll bet you can run some of them, but not all of them at any time.
5. InseyMember
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Ok, when I just connected the second half, it was dim. My power pack is rated for 20 volts AC output and 7 amps AC.
I have 27 bulbs total on my layout. It never was dim like this. When I first built it, everything worked fine with no shorts and the lights were all very bright. I am thinking I will just redo the entire wiring on it, which would include replacing all the lights and rerunning the main bus. It would only cost me \$40 to do it.
6. ezdaysOut AZ way
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Well, seven amps AC at 20 volts is a rather large transformer and would be difficult to load down. Your power pack could be the problem, but one other thing, can you disconnect the lamp that is burning brighter that the others? Just a thought, but if it was pulling so much power that the rest of the lamps would go dim, you would know it, it would get extremely hot. You could have a high-resistance short, like something laying across the wires, if you do, you should be able to feel around for some heat. If it was working at one time and isn't now, then something changed and I guess you'll have to poke around until you find it, but right now, from what you're saying, I'm suspecting the power pack, and that one bulb that is running bright.
Keep at it. If you do buy new stuff, be sure that the voltage rating on them match the voltage output of your power pack, and also check the current draw. Some bulbs run brighter than others and use more current. You also might consider the use of LEDs.
7. InseyMember
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Ok I just disconnected the other half again and now neither half will light up. This is driving me nuts. I just went ahead and ordered all new bulbs and wiring. Came out to \$37. This is what I got:
Model Power Economy Lighting Set 16V (6)
Life-Like Grain Of Wheat Bulb 14V (2)
Model Power Peel/Stick Light 12-16V (20)
Is it bad to have those two 14V wheat bulbs running at 17V?
8. ezdaysOut AZ way
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Yes it is. They will burn brighter, run hotter and burn out a whole lot quicker. It is much better to run a 14 volt bulb at 12 volts, then their lifespan increases appreciably. Putting a resistor in series with them will drop the voltage enough that it doesn't become a problem, but you need to know their rated current draw to calculate the resistor values. You might consider running them at using a 12 volt transformer that usually puts out around 16 volts at no load.
I've done a lot of troubleshooting over the phone with customers, and I gotta say, without being there, it is difficult to actually tell what's going on, that's why I'm asking these different things. I'll keep working with you as much as you want, just don't get frustrated, that makes things a lot harder.
9. InseyMember
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Ok well I just re-ran the two wires for the main bus. Now I just have to wait for my shipment of lights to get here.
Do you think putting on resistors would be better than running them in a series? RadioShack sells quarter-watt resistors with an assortment of different ohm ratings. It's been forever since I've studied Ohm's law, but do you know off-hand what ohm rating I would need to go from 17 to 14?
10. InseyMember
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11. ezdaysOut AZ way
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I would need to know the bulbs current rating. Do you have a link to that lighbulb offhand? If not, I'll see if I can find it.
12. InseyMember
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13. ezdaysOut AZ way
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No, the 1/4 watt refers to the amount of power the resistor can handle before it burns up. I can find no information on the amount of current your grain of wheat lamps take, but I'm assuming that it is very low possibly in the 20 to 50 milliamps. Your series resistor will probably be around 100 ohms, and 1/4 watt should do fine. Try it and measure the voltage across the lamp, and it it's too low, then decrease the resistor value, if it's still too high, than increase it. It's going to depend on how much current the lamp takes, but maybe there's someone here that has used these and can tell you right off. There's nothing wrong with trial and error. Try a value and see it it works. The higer the resistor value, the lower the voltage across the lamp and the dimmer it will light, but the longer it will last.
14. InseyMember
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Ok thanks for the tips on the resistors. But it looks like I have even bigger problems now. I just redid all the wiring for my layout (took me 3 hours). I mocked up all the new wiring (I didn't solder) and I hooked it up to the powerpack. I got everything to light up, but it only stayed lit for about 15 seconds and then the "OVERLOAD" light came on and the entire thing shut off. What in the world is the problem?
15. ezdaysOut AZ way
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At one point I mentioned that I thought you may have power pack problems which was why I was trying to get you to reduce the number of lamps in the circuit. I really don't think you have a seven amp power pack. It could be .7, or 700 milliamps, in which case you don't have enough power to light all the lamps fully. "Overload" means just that, you have exceeded the capacity of the power pack output significantly. Let it cool down, some have a button reset, some thermal. Start off with just a few lamps and add them on one or two at a time. At one point you will reach the maximum you can put in the circuit. The solution: get a second power supply just to run the lights.
Good luck,
16. InseyMember
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Hmm...where can I get a power supply that just runs the lights? I haven't seen those around anywhere.
By the way, I just disconnected about 8 lights and now the power pack seems to be running without problems.
17. ezdaysOut AZ way
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Yeah, I think that comfirms it. I think you only have about 1/10 the power you thought you did. Radio shack has 12 volt transformers it you feel compfortable wiring them up. I think you can get them up to about 4 amps. You can also get something like a MRC power pack that gives you AC out for your accessories, 12 volt DC and variable rail power, all in one nice package. Or you can buy a few really cheap power packs and use the accessory outputs from them, but they only have a few watts of power so unless you have a few laying around the house, I wouldn't go out and buy any. You can even go a buy a fixed DC supply from Radio Shack, or even use an old power supply from a scrap computer. Lots of ways to get the power you need. Computer power supplies give you 5 VDC and 12VCD usually at around 5 amps.
18. InseyMember
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20. kf4jqdActive Member
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Fire Hazard!!
I want to mention one thing. You could have a potental fire hazard! I had the same problem. I disconnected the load (lights) and checked it with my meter. The voltage was at 14.5vdc. Which was correct output. Then I reconnected the load. The lights came on dim and the meter read around 12.5vdc. This wasn't good.
There had to be a short somewhere. I found it in a Brawa plug in street light. The wire inside the pole had no insulation and was shorting at the pole! I unpluged the lamp. The light went up to their normal brightness. The meter now read about 13.9vdc.
Hope this helps,
Andy
PS
The lamp pole was warm to the touch! | 2,915 | 11,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-30 | latest | en | 0.967699 |
https://study.com/academy/topic/prentice-hall-geometry-chapter-6-quadrilaterals.html | 1,561,376,450,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999482.38/warc/CC-MAIN-20190624104413-20190624130413-00204.warc.gz | 605,121,468 | 24,685 | # Ch 6: Prentice Hall Geometry Chapter 6: Quadrilaterals
The Quadrilaterals chapter of this Prentice Hall Geometry Textbook companion course helps students learn essential geometry lessons of quadrilaterals. Each of these simple and fun video lessons is about five minutes long and is sequenced to align with the Quadrilaterals textbook chapter.
## How it works:
• Identify the lessons in the Prentice Hall Geometry Quadrilaterals chapter with which you need help.
• Find the corresponding video lessons within this companion course chapter.
• Watch fun videos that cover the quadrilateral topics you need to learn or review.
• Complete the quizzes to test your understanding.
• If you need additional help, rewatch the videos until you've mastered the material or submit a question for one of our instructors.
## Students will learn:
• Properties of shapes
• Definition and properties of a rhombus
• Definition, properties and construction of rectangles
• Definition and properties of squares
• Definition, properties and proof theorems of parallelograms
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• Identifying special types of quadrilaterals
• Using properties of special parallelograms
• Definition and properties of kites in geometry
• Definition and properties of trapezoids
• Plotting points and graphing lines in the Cartesian coordinate system
• Naming coordinates
• Using coordinate geometry with parallelograms
Prentice Hall Geometry is a registered trademark of Prentice Hall, which is not affiliated with Study.com.
7 Lessons in Chapter 6: Prentice Hall Geometry Chapter 6: Quadrilaterals
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View Full Version : flopping a set
03-25-2003, 08:08 AM
It is my understanding that the probability of making a set IF u stay all the way to the river is just under 20%
My question is what is the probability of making a set on the FLOP
03-25-2003, 08:13 AM
I forgot to add while holding pocket pairs.....sorry
SunTzu68
03-25-2003, 09:09 AM
ie.
2/50 (4%) + 2/49 (4.08%) + 2/48 (4.17%) =12.25%
pudley4
03-25-2003, 12:14 PM
approx 11.75% (or 7.5-1)
pudley4
03-25-2003, 01:10 PM
Using your way (adding the probabilities of each card) we need to make a correction:
The chance of hitting the set on the first card = 2/50 = 4%.
The chance of hitting on the second = 2/49 has to be multiplied by the chance of missing on the first (48/50) = 3.918%
The chance of hitting on the third = 2/48 has to be multiplied by the chance of missing the first and second (48/50 * 47/49) = 3.836% Add them up to get 11.754%
Another way: There are 19600 total flops. 17296 (48*47*46/6) do not contain one of your pair. 2304 contain one or both of your cards. 2304/19600 = 11.755%
BruceZ
03-25-2003, 05:27 PM
This is correct for a set or quads. For just a set, you can do this just with fractions as (2/50)(48/49)(47/48)*3 = 11.510% or 7.7-1. The reason we multiply by 3 is because the set can be made on any of the 3 cards.
So this would say the probability of quads must be 11.755% - 11.510% = 0.245%. To check this, the probability of quads is (2/50)(1/49)(48/48)*3 = 0.245% so it checks. Here we multiply by 3 because the extra card can be in 3 places.
BruceZ
03-26-2003, 12:26 PM
This still isn't the probability of just a set, because we are still including flopping a full house. The probability of just a set is (2/50)(48/49)(44/48)*3 = 10.776% or 8.3-1. The probability of a full house is (2/50)(48/49)(3/48)*3 = 0.735%. The probability of quads as we said is (2/50)(1/49)(48/48)*3 = 0.245%. These add up to 11.756% or 7.5-1 for a set or better.
SunTzu68
04-01-2003, 02:25 PM
Pudley or Bruce,
Maybe you can help me with the remedial math. I understand how you got to the percentage, however the odds do not match up for me. Let me walk through what I did, and if you could tell me where I went astray I'd appreciate it.
I took the odds to make the set on each card and added them together (ie. 50 cards left in the deck minus 2 cards for outs gets me to 48-2)....(1/24)+(1/23.5)+(1/23) and I got 7.83-1 to make a set.
SExyBeast
04-01-2003, 08:28 PM
or put another way pp in the hole ur about 8-1 shot to make set /forums/images/icons/grin.gif gl hf
BruceZ
04-03-2003, 07:50 AM
I took the odds to make the set on each card and added them together (ie. 50 cards left in the deck minus 2 cards for outs gets me to 48-2)....(1/24)+(1/23.5)+(1/23) and I got 7.83-1 to make a set.
There are several problems here. First of all, 48-2 odds is not 1/24, it is 2/50 = 1/25. Second of all, this sum adds up to 1 IN 7.83, but that is 6.83-1. Third of all, the second term cannot be 2/47 = 1/23.5 because this is just the probability of making a set on the 2nd card when you miss it on the 1st card. You have to multiply that by the probability that we also didn't make it on the 1st card as pudley did above. This still would include full houses. You can also break it down as I did.
Here's another way to do this. Before the flop, the probability of making a set any given card is 2/50. If we take 3 times this or 2/50 + 2/50 + 2/50 = 6/50, that would be the probability of making it on any of the 3 cards, except that we would be double counting all the times we make quads because when we add the second 2/50, that counts all the times we make quads on cards 2 and 1 which we already counted, and when we add the third 2/50, that counts all the times we make quads on cards 3 and 1 and on cards 3 and 2 which we already counted. Since all quads are counted twice, we can subtract the probability of making quads, which we found above is 0.245%. So 6/50 - 0.245% = 11.76% or 7.5-1 for a set or better. | 1,338 | 3,999 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-17 | latest | en | 0.909159 |
http://mathoverflow.net/revisions/74027/list | 1,369,523,137,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706470784/warc/CC-MAIN-20130516121430-00069-ip-10-60-113-184.ec2.internal.warc.gz | 166,652,574 | 5,204 | Ramsey's Theorem
The theorem has two forms.
Finite form. For all nonzero $n, k, m \in \mathbb{N}$ there is a nonzero $r \in \mathbb{N}$ such that, for each set $R$ of size $r$ and each $k$-coloring of the $n$-element subsets of $R$, there is a homogeneous set of size $m$.
Ininite form. For all nonzero $n, k \in \mathbb{N}$, each infinite set $W$, and each $k$-coloring of the $n$-element subsets of $W$, there is an infinite homogeneous set.
Frank Ramsey's original paper was titled On a problem of formal logic, so it should be no surprise that the famous theorem and its generalizations have magical applications in logic. For example, Ramsey's Theorem, the Erdős-Rado Theorem, and their more esoteric extensions, are often used in model theory to establish the existence of indiscernible elements. | 219 | 806 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-20 | latest | en | 0.893415 |
http://ned.ipac.caltech.edu/level5/March02/Bertschinger/Bert4_6.html | 1,501,073,320,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426161.99/warc/CC-MAIN-20170726122153-20170726142153-00580.warc.gz | 220,270,356 | 5,135 | 4.6. Poisson gauge
Recall that our general perturbed Robertson-Walker metric (4.11) contains four extraneous degrees of freedom associated with coordinate invariance. In the synchronous gauge these degrees of freedom are eliminated from g00 (one scalar) and g0i (one scalar and one transverse vector) by requiring = wi = 0. There are other ways to eliminate the same number of fields. As we shall see, a good choice is to constrain g0i (eliminating one scalar) and gij (eliminating one scalar and one transverse vector) by imposing the following gauge conditions on eq. (4.11):
(4.46)
I call this choice the Poisson gauge by analogy with the Coulomb gauge of electromagnetism ( . A = 0). (2) More conditions are required here than in electromagnetism because gravity is a tensor rather than a vector gauge theory. Note that in the Poisson gauge there are two scalar potentials ( and ), one transverse vector potential (w), and one transverse-traceless tensor potential h.
A restricted version of the Poisson gauge, with wi = hij = 0, is known in the literature as the longitudinal or conformal Newtonian gauge (Mukhanov, Feldman & Brandenberger 1992). These conditions can be applied only if the stress-energy tensor contains no vector or tensor parts and there are no free gravitational waves, so that only the scalar metric perturbations are present. While this condition may apply, in principle, in the linear regime (| / | << 1), nonlinear density fluctuations generally induce vector and tensor modes even if none were present initially. Setting w = h = 0 is analogous to zeroing the electromagnetic vector potential, implying B = 0. In general, this is not a valid gauge condition - it is rather the elimination of physical phenomena. The longitudinal/conformal Newtonian gauge really should be called a "restricted gauge." The Poisson gauge, by contrast, allows all physical degrees of freedom present in the metric.
To prove the last statement, and to find out how much residual gauge freedom is allowed, we must find a coordinate transformation from an arbitrary gauge to the Poisson gauge. Using eq. (4.40) with hats indicating Poisson gauge variables, we see that a suitable transformation exists with
(4.47)
where w comes from the longitudinal part of w (w|| = - w), while h and hi come from the longitudinal and solenoidal parts of h in eq. (4.14). Because these conditions are algebraic in , , and (they are not differentiated, in contrast with the transformation to synchronous gauge of eq. 4.41), we have found an almost unique transformation from an arbitrary gauge to the Poisson gauge. One can still add arbitrary functions of time alone (with no dependence on xi) to and i. (Adding a function of time alone to has no effect at all because the transformation, eq. 4.39, involves only the gradient of .)
Spatially homogeneous changes in represent changes in the units of time and length, while spatially homogeneous changes in represent shifts in the origin of the spatial coordinate system. These trivial residual gauge freedoms - akin to electromagnetic gauge transformations generated by a function of time, the only gauge freedom remaining in Coulomb gauge - are physically transparent and should cause no conceptual or practical difficulty.
It is interesting to see the coordinate transformation from a synchronous gauge to the Poisson gauge. As an exercise the reader can show that this is given by
(4.48)
Comparing with eq. (4.43), we see that the two Poisson-gauge scalar potentials are = A and = - H. (Kodama & Sasaki 1984 call these variables = and = - .) The vector potential wi in Poisson gauge is related simply to the solenoidal potential hi of the synchronous gauge (eq. 4.31).
Thus, the metric perturbations in the Poisson gauge correspond exactly with several of the gauge-invariant variables introduced by Bardeen. By imposing the explicit gauge conditions (4.46), we have simplified the mathematical analysis of these variables.
Now that we have seen that the Poisson gauge solves the gauge-fixing problem, let us give the components of the perturbed Einstein equations. They are no more complicated than those of the synchronous gauge:
(4.49) (4.50) (4.51) (4.52) (4.53) (4.54) (4.55)
As in the synchronous gauge, the scalar and vector modes satisfy initial-value (ADM) constraints (eqs. 4.49-4.51) in addition to evolution equations. However, it is remarkable that in the Poisson gauge we can obtain the scalar and vector potentials directly from the instantaneous stress-energy distribution with no time integration required. This is clear for - and w, both of which obey elliptic equations with no time derivatives (eqs. 4.53 and 4.51, respectively). By combining the ADM energy and longitudinal momentum constraint equations we can also get an instantaneous equation for :
(4.56)
Bardeen (1980) defined the matter perturbation variable m ( + 3 f / and noted that it is the natural measure of the energy density fluctuation in the normal (inertial) frame at rest with the matter such that v + w = 0 (recall the discussion in section 4.3). However, for our analysis we will remain in the comoving frame of the Poisson gauge, in which case / and not m is the density fluctuation.
We can show that for nonrelativistic matter the field equations we have obtained reduce to the Newtonian forms. First, it is clear that in the non-cosmological limit ( = K = 0), eq. (4.56) reduces to the Poisson equation. For 0 the longitudinal momentum density f is also a source for , but it is unimportant for perturbations with | / |>> vHv / c2 where vH is the Hubble velocity across the perturbation. Next, consider the implications of the fact that the shear stress for any physical system is at most O( cs2) where cs is the characteristic thermal speed of the gas particles. (For a collisional gas the shear stress is much less than this.) Equation (4.53) then implies that the relative difference between and is no more than O(cs / c)2. Third, eq. (4.51) implies that the vector potential w ~ (vH / c)2v. Thus, the deviations from the Newtonian results are all O(v / c)2. Poisson gauge gives the relativistic cosmological generalization of Newtonian gravity.
There are still more remarkable features of the Poisson gauge. First, the Poisson gauge metric perturbation variables are almost always small in the nonrelativistic limit (|| << c2, v2 << c2), in contrast with the synchronous gauge variables hij, which become large when | / |> 1. (However, Bardeen 1980 shows that the relative numerical merits of these two gauges can reverse for isocurvature perturbations of size larger than the Hubble distance.) Second, if (, , w, h) are very small, they - but not necessarily their derivatives! - may be neglected to a good approximation, in which case the Poisson gauge coordinates reduce precisely to the Eulerian coordinates used in Newtonian cosmology. Finally, it is amazing that the scalar and vector potentials depend solely on the instantaneous distribution of stress-energy - in fact, only the energy and momentum densities and the shear stress are required. Only the tensor mode - gravitational radiation - follows unambiguously from a time evolution equation. In fact, it obeys precisely the same equation as in the synchronous gauge (with a factor of 2 difference owing to our different definitions) because tensor perturbations are gauge-invariant - coordinate transformations involving 3-scalars and a 3-vector cannot change a 3-tensor (leaving aside the special case of eq. 4.17 for a closed space).
2 The same gauge has been proposed recently by Bombelli, Couch & Torrence (1994), who call it "cosmological gauge." However, I prefer the name Poisson gauge because cosmology - i.e., nonzero - is irrelevant for the definition and physical interpretation of this gauge. Although I have seen no earlier discussion of Poisson gauge in the literature, its time slicing corresponds with the minimal shear hypersurface condition of Bardeen (1980). Back. | 1,788 | 8,012 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2017-30 | longest | en | 0.87051 |
https://kr.mathworks.com/matlabcentral/cody/problems/334-poker-series-03-isfullhouse/solutions/196048 | 1,582,502,065,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145859.65/warc/CC-MAIN-20200223215635-20200224005635-00465.warc.gz | 405,328,877 | 15,978 | Cody
# Problem 334. Poker Series 03: isFullHouse
Solution 196048
Submitted on 24 Jan 2013 by J.R.! Menzinger
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = flag: 1 usedCards: [4x13 logical]
2 Pass
%% hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = flag: 1 usedCards: [4x13 logical]
3 Pass
%% hm = [0 0 0 0 1 0 0 0 1 1 0 1 1 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = flag: 1 usedCards: [4x13 logical]
4 Pass
%% hm = [0 0 0 0 1 0 0 0 1 1 0 1 1 1 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = flag: 1 usedCards: [4x13 logical]
5 Pass
%% hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]; y_correct.flag = false; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = flag: 0 usedCards: [4x13 logical]
6 Pass
%% hm = [0 0 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0]; y_correct.flag = true; y_correct.usedCards = logical([0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0]) assert(isequal(isFullHouse(hm),y_correct))
y_correct = flag: 1 usedCards: [4x13 logical] | 1,621 | 2,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2020-10 | latest | en | 0.527333 |
https://brilliant.org/problems/rotational-analysis/ | 1,532,195,115,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592650.53/warc/CC-MAIN-20180721164755-20180721184755-00504.warc.gz | 602,907,832 | 11,534 | # Rotational analysis!
Classical Mechanics Level pending
A solid sphere of mass $$2 Kg$$ and Radius $$2$$metre is rotating clockwise as well as translating on a rough horizontal surface of friction coefficient ,$$\mu = 0.9$$.
A force at a height $$\large\ \frac{4}{3}$$ from centre of sphere is applied of magnitude $$84$$N. $$a$$ is the linear acceleration of centre of mass of sphere, its value is $$3$$m/$$s^2$$.
Friction is enough to prevent slipping.
Find sphere's angular acceleration in $$rad/(s^2)$$.
× | 136 | 516 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2018-30 | latest | en | 0.855384 |
http://csl.skku.edu/SSE2025/Overview?action=diff | 1,550,454,326,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00241.warc.gz | 65,064,238 | 4,947 | ## SSE2025.Overview History
March 03, 2011, at 06:49 AM by 115.145.212.168 -
!! SSE2025: Problem Solving (Spring 2010)
!!! [General information]
(:table border=0 width=90% align=left :)
(:cellnr width=15% valign=top :) '''When:'''
(:cell valign=top :) 13:30 - 14:45 (Monday)\\
16:30 - 17:45 (Wednesday)
(:cellnr valign=top :) '''Where:'''
(:cell valign=top :) Lecture room #400118, Semiconductor Bldg 1st Floor Rm 018.
(:cellnr valign=top :) '''Instructor:'''
(:cell valign=bottom :) [[http://csl.skku.edu/People/joon|Joonwon Lee]]
[[http://csl.skku.edu|Computer Systems Laboratory]]
(:cellnr valign=top :) '''Course'''\\
'''Description:'''
(:cell valign=top :) Let's solve problems
(:cellnr valign=top :) '''Textbook:'''
(:cell valign=top :)
(:cell valign=top :) (Subject to change)
* Lab Execercise : 20%
* Individual Programming Assignment: 20%
* Team Programming Assignment: 20%
* Midterm Exam: 20%
* Final Exam: 20%
(:cellnr valign=top :) '''Teaching Assistants:'''
(:cell valign=top :)
* 이장환 janghwan.lee at gmail.com
* 김상욱 ghost86 at nate dot com
* 이원재 leewonjae.sse at gmail dot com
* 김나현 nahyunkim.sse at gmail.com
(:cellnr valign=top :) '''실습실 분반'''
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* A 반(학번 00000000000 ~ 2009311700) PC 실습실 400202호
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(:tableend:)
March 03, 2011, at 06:47 AM by 115.145.212.168 -
Deleted lines 0-40:
!! SSE2025: Problem Solving (Spring 2011)
!!! [General information]
(:table border=0 width=90% align=left :)
(:cellnr width=15% valign=top :) '''When:'''
(:cell valign=top :) 13:30 - 14:45 (Monday)\\
16:30 - 17:45 (Wednesday)
(:cellnr valign=top :) '''Where:'''
(:cell valign=top :) Lecture room #400118, Semiconductor Bldg 1st Floor Rm 018.
(:cellnr valign=top :) '''Instructor:'''
(:cell valign=bottom :) [[http://csl.skku.edu/People/joon|Joonwon Lee]]
[[http://csl.skku.edu|Computer Systems Laboratory]]
(:cellnr valign=top :) '''Course'''\\
'''Description:'''
(:cell valign=top :) Let's solve problems
(:cellnr valign=top :) '''Textbook:'''
(:cell valign=top :)
(:cell valign=top :) (Subject to change)
* Lab Execercise : 20%
* Individual Programming Assignment: 20%
* Team Programming Assignment: 20%
* Midterm Exam: 20%
* Final Exam: 20%
(:cellnr valign=top :) '''Teaching Assistants:'''
(:cell valign=top :)
* 이장환 janghwan.lee at gmail.com
* 김상욱 ghost86 at nate dot com
* 이원재 leewonjae.sse at gmail dot com
* 김나현 nahyunkim.sse at gmail.com
(:cellnr valign=top :) '''실습실 분반'''
(:cell valign=top :)
* A 반(학번 00000000000 ~ 2009311700) PC 실습실 400202호
* B 반(학번 20093117000 ~ 9999999999) PC 실습실 400212호
(:tableend:)
March 03, 2011, at 06:47 AM by 115.145.212.168 -
Changed line 1 from:
!! SSE2025: Problem Solving (Spring 2010)
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!! SSE2025: Problem Solving (Spring 2011)
March 08, 2010, at 05:32 AM by 115.145.212.143 -
* 이장환 janghwan.lee at gmail.com
* 김나현 nahyunkim.sse at gmail.com
March 04, 2010, at 04:45 AM by 115.145.212.143 -
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* A 반(학번 0 ~ 2009311700) PC 실습실 400202호
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* A 반(학번 00000000000 ~ 2009311700) PC 실습실 400202호
March 04, 2010, at 04:45 AM by 115.145.212.143 -
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(:cellnr valign=top :) '''실습실 분반'''
March 04, 2010, at 04:44 AM by 115.145.212.143 -
(:cellnr valign=top :) ''실습실 분반'''
(:cell valign=top :)
* A 반(학번 0 ~ 2009311700) PC 실습실 400202호
* B 반(학번 20093117000 ~ 9999999999) PC 실습실 400212호
March 03, 2010, at 06:45 AM by 115.145.212.143 -
March 03, 2010, at 06:44 AM by 115.145.212.143 -
Changed line 20 from:
* Programming Challenges by Steven S. SKiena and Miguel A. Revilla, Springer [[(Attach:)ProgrammingChallenges.pdf| download]]
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* Class attendance: 10%
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March 03, 2010, at 05:50 AM by 115.145.212.143 -
March 03, 2010, at 05:50 AM by 115.145.212.143 -
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* Programming Challenges by Steven S. SKiena and Miguel A. Revilla, Springer
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* Programming Challenges by Steven S. SKiena and Miguel A. Revilla, Springer [[(Attach:)ProgrammingChallenges.pdf| download]]
March 03, 2010, at 05:49 AM by 115.145.212.143 -
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* A Book On C OR Programming Challenges
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* Programming Challenges by Steven S. SKiena and Miguel A. Revilla, Springer
February 25, 2010, at 04:50 AM by 115.145.212.143 -
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* A Book On C
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* A Book On C OR Programming Challenges
February 24, 2010, at 05:41 AM by 115.145.212.143 -
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* TBA
February 24, 2010, at 05:40 AM by 115.145.212.143 -
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* Midterm Exam: 20%
* Final Exam: 20
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February 24, 2010, at 05:38 AM by 115.145.212.143 -
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*이원재 leewonjae.sse at gmail dot com
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* 이원재 leewonjae.sse at gmail dot com
February 24, 2010, at 05:37 AM by 115.145.212.143 -
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* Embedded System Design: An Introduction to Processes, Tools, and Techniques, Arnold Berger.
(:cellnr valign=top :) '''References:'''
(:cell valign=top :)
* Embedded Systems Design, S. Heath.
* Embedded System Design: A Unified Hardware/Software Introduction, Frank Vahid and Tony Givargis.
* MDS lecture notes.
(:cellnr valign=top :) '''Prerequisites:'''
(:cell valign=top :)
* CSE2003: System programming
* ICE3003: Computer Architecture (Recommended)
* CSE3008 Operating Systems
to:
* A Book On C
* 김상욱 ghost86 at nate dot com
*이원재 leewonjae.sse at gmail dot com
February 17, 2010, at 09:04 AM by 115.145.212.143 -
February 17, 2010, at 09:04 AM by 115.145.212.143 -
!! SSE2025: Problem Solving (Spring 2010)
!!! [General information]
(:table border=0 width=90% align=left :)
(:cellnr width=15% valign=top :) '''When:'''
(:cell valign=top :) 13:30 - 14:45 (Monday)\\
16:30 - 17:45 (Wednesday)
(:cellnr valign=top :) '''Where:'''
(:cell valign=top :) Lecture room #400118, Semiconductor Bldg 1st Floor Rm 018.
(:cellnr valign=top :) '''Instructor:'''
(:cell valign=bottom :) [[http://csl.skku.edu/People/joon|Joonwon Lee]]
[[http://csl.skku.edu|Computer Systems Laboratory]]
(:cellnr valign=top :) '''Course'''\\
'''Description:'''
(:cell valign=top :) Let's solve problems
(:cellnr valign=top :) '''Textbook:'''
(:cell valign=top :)
* Embedded System Design: An Introduction to Processes, Tools, and Techniques, Arnold Berger.
(:cellnr valign=top :) '''References:'''
(:cell valign=top :)
* Embedded Systems Design, S. Heath.
* Embedded System Design: A Unified Hardware/Software Introduction, Frank Vahid and Tony Givargis.
* MDS lecture notes.
(:cellnr valign=top :) '''Prerequisites:'''
(:cell valign=top :)
* CSE2003: System programming
* ICE3003: Computer Architecture (Recommended)
* CSE3008 Operating Systems | 2,315 | 6,959 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-09 | longest | en | 0.556849 |
http://www.filehungry.com/product/windows_software/education/mathematics/function_grapher/ | 1,508,777,866,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826210.56/warc/CC-MAIN-20171023164236-20171023184236-00114.warc.gz | 447,118,792 | 9,481 | Download Function Grapher - Function Grapher is graph maker to create 2D, 2.5D and 3D function graphs, animations and...
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Function Grapher
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company name: graphnow license: Demo minimum requirements: functional limitations:
Function Grapher description
Function Grapher is graph maker to create 2D, 2.5D and 3D function graphs, animations and table graphs. 2D and 2.5D function graphs can be plotted in Cartesian and polar coordinate systems, and 3D function graphs can be plotted in Cartesian,cylindrical and spherical coordinate systems.
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Latest Reviews | 1,719 | 7,423 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-43 | longest | en | 0.819349 |
http://www.mathskey.com/question2answer/12997/solve-the-following-system-of-equations | 1,708,520,161,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00687.warc.gz | 54,567,294 | 10,761 | # Solve the following system of equations.
2x - y + z = -3
2x + 2y + 3z = 2
3x - 3y - z = -4?
Elimination method :
The system of equations are
2x - y + z = - 3 →(1)
2x + 2y + 3z = 2 →(2)
3x - 3y - z = - 4 →(3)
Write the equations (1) & (2) in column form, then subtact them, to eliminate the x - variable.
2x - y + z = - 3
2x + 2y + 3z = 2
(-)____________________
- 3y - 2z = - 5 ⇒ 3y + 2z = 5 →(4)
Multiply eq(2) by 3 and eq(3) by 2, and then write the equations in column form, then subtract them, to eliminate the x - variable.
6x + 6y + 9z = 6
6x - 6y - 2z = - 8
(-)____________________
12y + 11z = 14 →(5)
Multiply eq(4) by 4 ,then write the equations in column form, then subtract them, to eliminate the y - variable.
12y + 8z = 20
12y + 11z = 14
(-)_______________
- 3z = 6 ⇒ z = - 6/3 = - 2.
Substitute the z - value in eq (4), and solve for y.
3y + 2(- 2) = 5
3y - 4 = 5
3y = 5 + 4 = 9
⇒ y = 9/3 = 3.
Substitute the values y = 3, z = - 2 in eq(1), and solve for x.
2x - 3 + (- 2) = - 3
2x - 3 - 2 = - 3
2x - 5 = - 3
2x = - 3 + 5 = 2
⇒ x = 2/2 = 1.
The solution of the system is x = 1, y = 3, and z = - 2. | 517 | 1,158 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-10 | longest | en | 0.649329 |
http://upscfever.com/upsc-fever/en/test/rsagrawal/6.html | 1,556,049,313,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578613603.65/warc/CC-MAIN-20190423194825-20190423220825-00294.warc.gz | 185,844,335 | 6,179 | # CLOCKS
Ans .
$$47*1/2^{\circ}$$
1. Explanation :
angle traced by the hour hand in 12 hours = $$360^{\circ}$$
Angle traced by it in three hours 25 min (ie) 41/12 hrs=$$(360*41/12*12)^{\circ}$$
=$$102*1/2^{\circ}$$
angle traced by minute hand in 60 min. = $$360^{\circ}$$.
Angle traced by it in 25 min. = (360 X 25 )/60= $$150^{\circ}$$
Required angle = 1500 -102*$$1/2^{\circ}$$= 47*$$1/2^{\circ}$$
Ans .
120/11 min. past 2
1. Explanation :
At 2 o'clock, the hour hand is at 2 and the minute hand is at 12, i.e. they
are 10 min spaces apart.
To be together, the minute hand must gain 10 minutes over the hour hand.
Now, 55 minutes are gained by it in 60 min.
10 minutes will be gained in (60 x 10)/55 min. = 120/11 min.
The hands will coincide at 120/11 min. past 2.
Ans .
paste right option
1. Explanation :
At 4 o'clock, the minute hand will be 20 min. spaces behind the hour hand,
Now, when the two hands are at right angles, they are 15 min. spaces apart. So,
they are at right angles in following two cases.
Case I. When minute hand is 15 min. spaces behind the hour hand:
In this case min. hand will have to gain (20 - 15) = 5 minute spaces. 55 min. spaces are gained by it in 60 min.
5 min spaces will be gained by it in 60*5/55 min=60/11min.
They are at right angles at 60/11min. past 4.
Case II. When the minute hand is 15 min. spaces ahead of the hour hand:
To be in this position, the minute hand will have to gain (20 + 15) = 35 minute spa' 55 min. spaces are gained in 60 min.
35 min spaces are gained in (60 x 35)/55 min =40/11
They are at right angles at 40/11 min. past 4
Ans .
120/11 min
1. Explanation :
At 8 o'clock, the hour hand is at 8 and the minute hand is at 12, i.e. the two
hands_ are 20 min. spaces apart.
To be in the same straight line but not together they will be 30 minute spaces apart.
So, the minute hand will have to gain (30 - 20) = 10 minute spaces over the hour hand.
55 minute spaces are gained. in 60 min.
10 minute spaces will be gained in (60 x 10)/55 min. = 120/11min.
The hands will be in the same straight line but not together at 120/11 min.
Ans .
346/11 min. past 5
1. Explanation :
At 5 o'clock, the minute hand is 25 min. spaces behind the hour hand.
Case I. Minute hand is 3 min. spaces behind the hour hand.
In this case, the minute hand has to gain' (25 - 3) = 22 minute spaces. 55 min. are
gained in 60 min.
22 min. are gaineg in (60*22)/55min. = 24 min.
The hands will be 3 min. apart at 24 min. past 5.
Case II. Minute hand is 3 min. spaces ahead of the hour hand.
In this case, the minute hand has to gain (25 + 3) = 28 minute spaces. 55 min. are gained in 60 min.
28 min. are gained in (60 x 28_)/55=346/11
The hands will be 3 min. apart at 346/11 min. past 5.
Ans .
440/43 minutes
1. Explanation :
In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.
To be together again, the minute hand must gain 60 minutes over the hour hand. 55 min. are gained in 60 min.
60 min are gained in (60/55) x 60 min =720/11 min.
But, they are together after 65 min.
Gain in 65 min =720/11-65 =5/11min.
Gain in 24 hours =(5/11 * (60*24)/65)min =440/43
The clock gains 440/43 minutes in 24 hours.
Ans .
Wednesday
1. Explanation :
Time from 8 a.m. on Sunday to 8 p.m. on following Sunday = 7 days 12 hours = 180 hours
The watch gains (5 + 29/5) min. or 54/5 min. in 180 hrs.
Now 54/5 min. are gained in 180 hrs.
5 min. are gained in (180 x 5/54 x 5) hrs. = 83 hrs 20 min. = 3 days 11 hrs 20 min.
Watch is correct 3 days 11 hrs 20 min. after 8 a.m. of Sunday.
It will be correct at 20 min. past 7 p.m. on Wednesday
Ans .
11 p.m
1. Explanation :
Time from 5 a.m. on a day to 10 p.m. on 4th day = 89 hours.
Now 23 hrs 44 min. of this clock = 24 hours of correct clock.
356/15 hrs of this clock = 24 hours of correct clock.
89 hrs of this clock = (24 x 31556 x 89) hrs of correct clock.
= 90 hrs of correct clock.
So, the correct time is 11 p.m.
Ans .
48 min. past 12
1. Explanation :
Time from 8 a.m. on a day 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
145 /6 hrs of this clock = 24 hrs of the correct clock
29 hrs of this clock = (24 x 6/145 x 29) hrs of the correct clock
= 28 hrs 48 min. of correct clock
The correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12. | 1,432 | 4,408 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-18 | latest | en | 0.800914 |
https://www3.nd.edu/~steve/computing_with_data/8_Functions_list/functions_on_lists.html | 1,606,172,190,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141168074.3/warc/CC-MAIN-20201123211528-20201124001528-00284.warc.gz | 915,970,933 | 7,507 | # Applying a function to the components of a list
In doing practical problems, when there are many cases to consider or possible covariates, it's common to store all of the work into a list. Further anayleses of these results boils down to applying a function to each component of the list. R has a nice way of doing this that is similar to the `apply` function for matrices.
## Format of an `lapply`
The basic `lapply` function takes two arguments. The first argument is a vector, either an atomic vector or a list. The second argument is function. `lapply` returns a list as it's output. In the output list there is one component for each component of the input list and it's value is the result of applying the function to the input component.
``````output_list <- lapply(input_list, function)
``````
## Example problem with sensor readings
Following is simulated data from a network of 100,000 environmental sensors. They take a reading of CO2 concentrations each day. However, they are configured so that if the concentration goes above a certain threshold, it records a reading every 6 hours until it goes down. Find the mean CO2 concentrations for each sensor. Secondly, compute the variance of all of these readings
The challenge here is that there may be a varying number of readings for each sensor, so arranging these in a matrix won't work. instead, we are given a list with 100,000 components, each one containing the readings from one sensor.
Let's read in the list of readings over a 28 day period.
``````setwd("~/Documents/Computing with Data/8_Functions_lists/")
``````
The following generates the simulated data. This was only run once and the resulting objects saved to disk and then reloaded.
``````base_readings <- lapply(1:1e+05, function(z) {
sample(x = 292:335, size = 28, replace = T)
})
number_extras <- sapply(1:1e+05, function(z) {
sample(x = 0:15, size = 1, replace = T)
})
number_extras[1:5]
sample(x = 335:355, size = number_extras[z], replace = T)
})
})
names(co2_readings) <- paste("S", 1:1e+05, sep = "")
``````
``````load("./Data/co2_readings.RData")
``````
Finding the mean value of readings for one sensor at a time is easy:
``````mean(co2_readings[[1]])
``````
``````## [1] 322.8
``````
``````mean(co2_readings[[1000]])
``````
``````## [1] 326.9
``````
To compute the means for all of them we use `lapply` as follows:
``````all_means_L <- lapply(co2_readings, mean)
all_means_L[1:2]
``````
``````## \$S1
## [1] 322.8
##
## \$S2
## [1] 321.7
``````
``````class(all_means_L)
``````
``````## [1] "list"
``````
This is a list in which each component is a single number. It is more natural and useful to represent this as a numeric vector of length 100,000. The function `unlist` maps the list to an atomic vector by concatentating all of the component vectors.
``````all_means <- unlist(all_means_L)
all_means[1:2]
``````
``````## S1 S2
## 322.8 321.7
``````
``````class(all_means)
``````
``````## [1] "numeric"
``````
Now that we have a numeric vector of the means, we can answer the second question; i.e., what is the variance of these means?
``````var(all_means)
``````
``````## [1] 15.23
``````
## More complex functions
Seldom is it enough to use a basic pre-defined function with only default options. Here are examples of other situations.
Select the first reading for each sensor.
Do the first sensor as a sample.
``````sens1 <- co2_readings[[1]]
sens1[1]
``````
``````## [1] 312
``````
``````# Define a function that selects the first entry for a vector of length at
# least 1 and NA otherwise
sel_first <- function(x) {
if (length(x) > 0) {
y <- x[1]
} else {
y <- NA
}
y
}
``````
``````## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 292 302 314 314 325 335
``````
An example in which non-default options need to be set.
``````load("./Data/co2_readings2.RData")
``````
This set of readings is nearly identical but there are some missing values for some sensors. We just ignore these in computing the mean.
``````all_means2_L <- lapply(co2_readings2, mean)
all_means2 <- unlist(all_means2_L)
sum(is.na(all_means2))
``````
``````## [1] 1
``````
In this form we lose all readings for that sensor. We can recover the non-missing ones as follows.
``````all_means3_L <- lapply(co2_readings2, function(x) {
mean(x, na.rm = T)
})
all_means3 <- unlist(all_means3_L)
sum(is.na(all_means3))
``````
``````## [1] 0
``````
## `sapply` is a useful shortcut
We've seen several instances in which we want to collapse the returned list into a vector of outputs. Using the function `sapply` instead of `lapply` does this in one step.
``````means_vector <- sapply(co2_readings, mean)
means_vector[1:2]
``````
``````## S1 S2
## 322.8 321.7
``````
If for some reason, for one of the input components, the function doesnt return a single number or character, `sapply` will return a list.
## Execution time for `lapply`
It's tempting to simply iterate over the list and compute the means one component at a time. Let's see how the two alternatives compare in runtime.
``````means_v2 <- numeric(1e+05)
system.time(for (j in 1:1e+05) {
})
``````
``````## user system elapsed
## 0.820 0.004 0.823
``````
``````system.time(all_means_L <- lapply(co2_readings, mean))
``````
``````## user system elapsed
## 0.570 0.003 0.574
``````
``````system.time(means_vector <- sapply(co2_readings, mean))
``````
``````## user system elapsed
## 0.705 0.002 0.707
``````
The `lapply` version was considerably faster.
## Deeper analysis of flight arrival data
The `lapply` function facilitates deeper study of the arrival/delay data as it depends on airline, day of the week, type of delay, etc.
First load the data from the homework set.
``````arrival_df <- read.csv("../Homework/ontime_flight_data/ONTIME1.csv", stringsAsFactors = F)
sapply(arrival_df, class) # Quick way to survey variables and their classes. A data.frame is a list!
``````
``````## YEAR MONTH DAY_OF_MONTH
## "integer" "integer" "integer"
## AIRLINE_ID CARRIER ORIGIN_AIRPORT_ID
## "integer" "character" "integer"
## ORIGIN_AIRPORT_SEQ_ID ORIGIN_CITY_MARKET_ID DEST_AIRPORT_ID
## "integer" "integer" "integer"
## DEST_AIRPORT_SEQ_ID DEST_CITY_MARKET_ID ARR_TIME
## "integer" "integer" "integer"
## ARR_DELAY CARRIER_DELAY WEATHER_DELAY
## "numeric" "numeric" "numeric"
## NAS_DELAY SECURITY_DELAY LATE_AIRCRAFT_DELAY
## "numeric" "numeric" "numeric"
## X
## "logical"
``````
Problem: Compute the rates of delay for each airline
First add a variable indicating there is a delay of some kind.
``````arrival_df2 <- transform(arrival_df, DELAY_LOGICAL = !is.na(CARRIER_DELAY))
``````
What are the possible carriers?
``````unique(arrival_df2\$CARRIER)
``````
``````## [1] "9E" "AA" "AS" "B6" "DL" "EV" "F9" "FL" "HA" "MQ" "OO" "UA" "US" "VX"
## [15] "WN" "YV"
``````
``````length(unique(arrival_df2\$CARRIER))
``````
``````## [1] 16
``````
We saw in the homework how to subset to a single carrier.
``````united_flights_df <- subset(arrival_df2, CARRIER == "UA")
``````
The fraction of United flights with a delay is
``````sum(united_flights_df\$DELAY_LOGICAL)/nrow(united_flights_df)
``````
``````## [1] 0.1641
``````
Write a function that will go through these steps for an arbitrary carrier.
``````delay_rate <- function(arriv_df, carr) {
carr_df <- subset(arriv_df, CARRIER == carr)
rt <- sum(carr_df\$DELAY_LOGICAL)/nrow(carr_df)
rt
}
``````
We want to apply this function to the arrival_df and each carrier. We need the vector of carriers for input to `lapply`.
``````carrier_vector <- unique(arrival_df2\$CARRIER)
``````
Now apply the delay_rate function to each entry in the vector. The function outputs a single number, so we can use `sapply`.
``````del_rates <- sapply(carrier_vector, function(c) delay_rate(arrival_df2, c))
del_rates
``````
``````## 9E AA AS B6 DL EV F9 FL HA
## 0.17994 0.18487 0.12388 0.21046 0.11825 0.25179 0.28360 0.09965 0.07355
## MQ OO UA US VX WN YV
## 0.21726 0.19408 0.16413 0.16075 0.06785 0.14385 0.16653
`````` | 2,475 | 8,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.871948 |
https://jp.maplesoft.com/support/help/view.aspx?path=DifferentialGeometry/Tensor/EpsilonSpinor&L=J | 1,670,405,597,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711151.22/warc/CC-MAIN-20221207085208-20221207115208-00171.warc.gz | 359,655,684 | 35,716 | DifferentialGeometry/Tensor/EpsilonSpinor - Maple Help
Tensor[EpsilonSpinor] - create an epsilon spinor
Calling Sequences
EpsilonSpinor(indexType, spinorType, fr)
Parameters
indexType - a string, either "cov" or "con"
spinorType - a string, either "spinor" or "barspinor"
fr - (optional) the name of a defined frame
Description
• The epsilon spinor is a rank 2 spinor which is fully skew-symmetric and whose component values are 1 or -1.
• The command EpsilonSpinor(indexType, spinorType) returns the epsilon symbol of the type specified by indexType and spinorType in the current frame unless the frame is explicitly specified.
• This command is part of the DifferentialGeometry:-Tensor package, and so can be used in the form EpsilonSpinor(...) only after executing the commands with(DifferentialGeometry); with(Tensor) in that order. It can always be used in the long form DifferentialGeometry:-Tensor:-EpsilonSpinor.
Examples
> $\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):$$\mathrm{with}\left(\mathrm{Tensor}\right):$
Example 1.
First create a vector bundle $M$ with base coordinates $\left(x,y,z,t\right)$ and fiber coordinates $\left(\mathrm{z1},\mathrm{z2},\mathrm{w1},\mathrm{w2}\right)$.
> $\mathrm{DGsetup}\left(\left[x,y,z,t\right],\left[\mathrm{z1},\mathrm{z2},\mathrm{w1},\mathrm{w2}\right],M\right)$
${\mathrm{frame name: M}}$ (2.1)
Here are the 4 epsilon spinors one can define:
M > $\mathrm{P1}≔\mathrm{EpsilonSpinor}\left("cov","spinor"\right)$
${\mathrm{P1}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right)$ (2.2)
M > $\mathrm{P2}≔\mathrm{EpsilonSpinor}\left("con","spinor"\right)$
${\mathrm{P2}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"con_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"con_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right)$ (2.3)
M > $\mathrm{P3}≔\mathrm{EpsilonSpinor}\left("cov","barspinor"\right)$
${\mathrm{P3}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{7}{,}{8}\right]{,}{1}\right]{,}\left[\left[{8}{,}{7}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{7}{,}{8}\right]{,}{1}\right]{,}\left[\left[{8}{,}{7}\right]{,}{-1}\right]\right]\right]\right)$ (2.4)
M > $\mathrm{P4}≔\mathrm{EpsilonSpinor}\left("con","spinor"\right)$
${\mathrm{P4}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"con_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"con_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right)$ (2.5)
Define some other manifold $N$.
M > $\mathrm{DGsetup}\left(\left[x,y,z,t\right],N\right)$
${\mathrm{frame name: N}}$ (2.6)
The current frame is $N$. Because there are no fiber variables, one cannot calculate an epsilon spinor in this frame. To now re-calculate the epsilon spinor $\mathrm{P1}$, either use the ChangeFrame command or pass EpsilonSpinor the frame name $M$ as a third argument.
N > $\mathrm{EpsilonSpinor}\left("cov","spinor",M\right)$
${\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right)$ (2.7)
Example 2.
The covariant and contravariant forms of the epsilon spinors are inverses of each other.
M > $\mathrm{DGsetup}\left(\left[x,y,z,t\right],\left[\mathrm{z1},\mathrm{z2},\mathrm{w1},\mathrm{w2}\right],M\right)$
${\mathrm{frame name: M}}$ (2.8)
M > $\mathrm{P1}≔\mathrm{EpsilonSpinor}\left("cov","spinor"\right)$
${\mathrm{P1}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"cov_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right)$ (2.9)
M > $\mathrm{P2}≔\mathrm{EpsilonSpinor}\left("con","spinor"\right)$
${\mathrm{P2}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"con_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"con_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{6}\right]{,}{1}\right]{,}\left[\left[{6}{,}{5}\right]{,}{-1}\right]\right]\right]\right)$ (2.10)
Contract the first index of $\mathrm{P1}$ with the first index of $\mathrm{P2}$. The result is the Kronecker delta spinor.
M > $\mathrm{P5}≔\mathrm{ContractIndices}\left(\mathrm{P2},\mathrm{P1},\left[\left[1,1\right]\right]\right)$
${\mathrm{P5}}{:=}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{5}\right]{,}{1}\right]{,}\left[\left[{6}{,}{6}\right]{,}{1}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{5}\right]{,}{1}\right]{,}\left[\left[{6}{,}{6}\right]{,}{1}\right]\right]\right]\right)$ (2.11)
M > $\mathrm{P5}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}&minus\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{KroneckerDeltaSpinor}\left("spinor"\right)$
${\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{5}\right]{,}{0}\right]\right]\right]\right){,}{\mathrm{_DG}}{}\left(\left[\left[{"tensor"}{,}{M}{,}\left[\left[{"con_vrt"}{,}{"cov_vrt"}\right]{,}\left[\right]\right]\right]{,}\left[\left[\left[{5}{,}{5}\right]{,}{0}\right]\right]\right]\right)$ (2.12) | 2,874 | 7,476 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 35, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2022-49 | latest | en | 0.42932 |
https://oeis.org/A014617 | 1,571,356,368,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986677230.18/warc/CC-MAIN-20191017222820-20191018010320-00401.warc.gz | 622,765,024 | 4,187 | This site is supported by donations to The OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A014617 Antidiagonals of the prime-composite array B(m,n) (see A067681) that are zeros from the first Borve conjecture. 1
4, 8, 12, 23, 30, 35, 46, 49, 70, 73, 88, 97, 102, 106, 118, 123, 146, 162, 167, 171, 195, 205, 236, 240, 242, 245, 254, 270, 272, 290, 292, 297, 320, 325, 332, 342, 355, 365, 374, 444, 453, 502, 508, 523, 532, 578, 585, 596, 599, 609, 634, 645, 663, 677, 687 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Let c(m) be the m-th composite and p(n) be the n-th prime. The prime-composite array, B, is defined such that each element B(m,n) is the highest power of p(n) that is contained within c(m). The m-th antidiagonal of the array consists of the m elements B(m,1), B(m-1,2), B(m-2,3),...,B(1,m). The First Borve Conjecture states that there is an infinite number of zero-only antidiagonals. LINKS EXAMPLE Each composite has its own row, consisting of the indices of its prime factors. For example, the 10th composite is 18 and 18 = 2^1 * 3^2 * 5^0 * 7^0 * 11^0 * ..., so the 10th row reads: 1, 2, 0, 0, 0, ... Similarly, B(6,2) = 1 because c(6) = 12, p(2) = 3 and the highest power of 3 contained within 12 is 3^1 = 3. And B(34,3) = 2 because c(34) = 50, p(3) = 5 and the highest power of 5 contained within 50 is 5^2 = 25. MATHEMATICA Composite[n_Integer] := FixedPoint[n + PrimePi[ # ] + 1 &, n + PrimePi[n] + 1]; m = 750; a = Table[0, {m}, {m}]; Do[b = Transpose[ FactorInteger[ Composite[n]]]; a[[n, PrimePi[First[b]]]] = Last[b], {n, 1, m}]; Do[ If[ Union[ Table[ a[[n - i + 1, i]], {i, 1, n} ]] == {0}, Print[n]], {n, 1, m}] CROSSREFS Sequence in context: A092108 A015781 A130643 * A239053 A272708 A157416 Adjacent sequences: A014614 A014615 A014616 * A014618 A014619 A014620 KEYWORD nonn AUTHOR Robert G. Wilson v, Feb 04 2002 STATUS approved
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Last modified October 17 19:44 EDT 2019. Contains 328128 sequences. (Running on oeis4.) | 829 | 2,275 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2019-43 | latest | en | 0.701492 |
https://efinancemanagement.com/investment-decisions/cost-of-equity-capital-asset-pricing-model-capm | 1,709,586,063,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476532.70/warc/CC-MAIN-20240304200958-20240304230958-00668.warc.gz | 234,059,462 | 72,302 | # Cost of Equity – Capital Asset Pricing Model (CAPM)
The cost of equity is estimated using Sharpe’s Model of Capital Asset Pricing Model. The model finds the cost of capital by establishing a relationship between risk and return. As per this model, at least a risk-free return is expected out of every investment and the expectation greater than that is dependent on the amount of risk associated with the respective investment. As per this model, the required rate of return is equal to the sum of the risk-free rate and a premium based on the systematic risk associated with the security.
## Systematic Risk and Unsystematic Risk
Systematic Risk is that risk that is unavoidable by diversifying the investments. This risk includes the factors which affect the overall market and not a particular stock in the market viz. changes in economic conditions, inflationary situations all over the world, etc. Beta (β) is the measurer of this risk.
Unsystematic Risk is the stock-specific risk that affects only a particular stock price and not the whole market. Such a risk can be diversified and reduced to a great extent by diversifying a portfolio.
## Capital Asset Pricing Model (CAPM)
The result of the model is a simple formula based on the explanation just given above.
ke = Rf + (Rm – Rf
ke = Required rate of return or cost of equity
Rf = Risk-free rate of return, normally the treasury interest rate offered by the government.
Rm = It is the expected return from the Market Portfolio.
β = Beta is a measure of risk in the equation.
Market Portfolio conceptually means the portfolio of shares containing all the shares trading in the market with their weights as their market capitalization. Practically, we can take a portfolio that has all the industries and has sufficient stock of each industry to diversify the stock-specific risks. The objective of taking the market portfolio is that we want a portfolio that has diversified all the avoidable risk and such a portfolio cannot be more than a market portfolio.
Beta is the measure of the responsiveness of an individual stock’s return due to a change in market return. The higher the beta of security, the higher would be the required rate of return by the investor. It impacts the required return because it has a direct multiplication impact on the premium. If the beta is 1, the stock return would be equal to the market return. If less than 1, a stock return would be less than the market return and if it is greater than 1, the required return of the stock is greater than the market return.
## Example of Cost of Equity with CAPM
Suppose Rm is 20%, Rf is 8%, β is 1.2, the Rj would be
Rj = Rf + (Rm – Rf
= 8% + (20% – 8%)1.2
= 8% + 14.4%
Rj = 22.4%
In the above example, the risk-free rate of return is 8% and the market risk premium of the stock is higher than 20% i.e. more than the market return. It is because the stock beta is greater than 1 i.e. 1.2.
Refer to Models for Calculating Cost of Equity to learn about different models.
Finding the cost of equity with the CAPM model is widely used. the primary advantage of this model is that it relates to return to the risk which is a general behavior of all rational investors. The disadvantage is that it is difficult to estimate the market return and the beta.
Who developed the Capital Asset Pricing Model?
CAPM was developed by Willian Forsyth Sharpe
Explain the impact of Beta
Beta impacts the required return because it has a direct multiplication impact on the premium. If the beta is 1, the stock return would be equal to the market return. If less than 1, a stock return would be less than the market return and if it is greater than 1, the required return of the stock is greater than the market return.
How do you calculate CAPM?
The formula for calculating CAPM is:
Rf + (Rm – Rf
Quiz on Cost of Equity – CAPM | 864 | 3,868 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | longest | en | 0.940556 |
https://europedriveguide.com/examples-567 | 1,669,650,136,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710533.96/warc/CC-MAIN-20221128135348-20221128165348-00762.warc.gz | 279,124,850 | 5,026 | # Solve geometric sequence
We can do your math homework for you, and we'll make sure that you understand how to Solve geometric sequence. We can help me with math work.
## Solving geometric sequence
These sites allow users to input a Math problem and receive step-by-step instructions on how to Solve geometric sequence. If the batteries seem fine, try resetting the camera. This can be done by taking out the batteries and then putting them back in. If neither of these fixes work, you may need to take your camera to a professional to have it repaired or replaced.
There are a number of ways to solve ln equations, depending on the complexity of the equation. For simple equations, taking the natural log of both sides of the equation will usually suffice. However, for more complex equations, more sophisticated methods may be necessary. In any case, it is always important to carefully consider the equation before attempting to solve it, in order to ensure that the chosen method is appropriate.
To solve for the hypotenuse of a right angled triangle, you can use the Pythagorean Theorem. This theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. So, in order to solve for the hypotenuse, you would need to square the other two sides and then add them together. Afterwards, you would need to take the square root of the result in order to find the value of the hypoten
There are many different types of math problem solving questions that can be asked. Some common examples include questions about finding a particular numerical answer, identifying a specific mathematical pattern, or determining the best way to solve a given problem. No matter what the specific question is, there are a few key steps that can be followed in order to solve it. First, it is important to read the question carefully and identify any key information that is necessary for solving the problem. Next, it is helpful to devise
## Help with math
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Williamina Richardson | 543 | 2,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-49 | latest | en | 0.946649 |
https://mathopenref.com/pythagoreantriples.html | 1,652,942,232,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662525507.54/warc/CC-MAIN-20220519042059-20220519072059-00517.warc.gz | 460,206,541 | 4,443 | # Pythagorean Triples
A right triangle where the sides are in the ratio of integers. (Integers are whole numbers like 3, 12 etc)
For example, the following are pythagorean triples:
There are infinitely many pythagorean triples. There are 50 with a hypotenuse less than 100 alone. Here are the first few: 3:4:5 , 6:8:10 , 5:12:13 , 9:12:15 , 8:15:17 etc...
If you multiply each side by an integer, the result will be another triple, demonstrating that there is an infinite number of them. Remember: it is the ratio of the lengths of the sides that counts, not the actual length. The units of measurement are thus irrelevant.
The smallest and perhaps best known triple, the 3:4:5 is explored in greater depth 3-4-5 Triangles. | 195 | 728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.910502 |
http://www.spss-tutorials.com/spss-syntax/ | 1,606,612,001,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195967.34/warc/CC-MAIN-20201129004335-20201129034335-00379.warc.gz | 151,051,018 | 15,044 | SPSS TUTORIALS BASICS DATA ANALYSIS CHI-SQUARE TESTS ANOVA T-TESTS
# SPSS Syntax Introduction
SPSS syntax is a language containing instructions for
analyzing and editing data and other SPSS commands.
SPSS users working directly from the menu may not actually see they syntax they're running. However, this is a terrible practice and we'll explain why in a minute. So let's download and open bank.sav -partly shown below- and jump right in.
## How to paste SPSS syntax?
Now let's suppose I'd like to gain some insight into the percentages of male and female respondents. I could first navigate to Analyze Descriptive statistics Frequencies as shown below.
I'll now move gender into the variable box and perhaps request a bar chart as well.
Now clicking may seem the obvious thing to do. A much better idea, however, is to click the button. Upon doing so, a new SPSS window opens which is known as the Syntax Editor. It's recognized by the orange icon in its left top corner.
The Syntax Editor contains a FREQUENCIES command which holds the instructions we just gave SPSS in the Frequencies dialog. However, we don't see the frequency distribution and bar chart we asked for. This is because we still need to run the command we just created.
## How to run SPSS syntax?
The simplest way to run syntax is to select the command(s) you'd like to run and click the “run selection” icon in your toolbar.
A faster way to run syntax is to use several shortkeys, especially
• F2 for selecting the command in which your mouse pointer is located;
• Ctrl + a for selecting all syntax;
• Ctrl + r for running all selected commands.
So let's now run our pasted syntax. On doing so, a new window will open, containing our frequency table and barchart. This is an output window which we'll discuss in our next tutorial.
## How to get SPSS syntax?
The right way to do basically anything in SPSS -editing and analyzing data, creating tables and charts and more- is by running syntax. So how to get syntax? First off, using the button from the menus adds syntax to your syntax window. If you don't have a syntax window open yet, it'll open one for you. Options for opening a syntax window are
• using the button from the menus;
• drag and drop a syntax file into the Data Editor window (shown below);
• clicking the New Syntax toolbar icon;
• Navigate to File New Syntax.
If you've a syntax window open, you still need the actual syntax. Options to get the syntax you need are
• use the button from SPSS’ menu;
• copy-paste syntax from our tutorials, online forums and elsewhere;
• type the commands you need into the syntax window.
Now, typing syntax may seem like a crazy thing to do at this point. However, typing syntax is much easier than it seems because most of it can be dramatically simplified.
## Writing simpler syntax
The syntax we just pasted from the menu was:
FREQUENCIES VARIABLES=gender
/BARCHART FREQ
/ORDER=ANALYSIS.
Now typing all that manually is a lot of work. However, we'll get the exact same results if we run:
frequencies gender
/barchart.
Just typing and running this is much faster and easier than clicking through all menu options. So if you want to get real good -and real fast- with SPSS, start learning short syntax. This will take some practice but it will save you tons of time and effort in the longer run.
## SPSS Syntax Files
We can now save all contents of our Syntax Editor as a syntax file by going to File Save as... The resulting syntax file has the .sps (for “SPSS syntax”) file extension and is a plain text file. You can open, edit and save it with SPSS or any text editor such as Notepad++.
When saving syntax in newer SPSS versions, something like * Encoding: UTF-8. may be added. Just leave and ignore this, it's not meant for you but, rather, some kind of “note to self” from SPSS.
## Why even use SPSS syntax?
The single best SPSS practice is doing everything from syntax. Some reasons for this are
• you'll always know exactly which steps you took in which order so you can prove that your results are correct;
• if you made some mistake -don't we all sometimes?- you can correct it and rerun everything you did in just seconds;
• you'll work way faster than from the menu and you never have to do things twice;
• some of the best SPSS tricks and time savers are available as syntax only.
So say you run 10 tables and charts from the menu. And then you realize you should have filtered out all respondents working in IT. Now you have to start all over again: remove the unwanted respondents and click your way through all the same menus and dialogs again... Sounds like a terrible idea. Doesn't it? Unfortunately, I see students having to do days of SPSS work all over again on a daily basis. Not working from syntax really is the very worst SPSS practice.
So say I run those 10 tables and charts and I saved all syntax. Then I realize I should have filtered out all respondents working in IT. Ok. No problem. I'll just add SELECT IF (jtype <> 3). to the top of my syntax and rerun all tables and charts in one go.
# Tell us what you think!
*Required field. Your comment will show up after approval from a moderator.
# THIS TUTORIAL HAS 49 COMMENTS:
• ### By Cynthia M. Terrell on May 30th, 2020
I like Syntax because you can allways to back and see what you missed and it gives you missing information
• ### By Betty D Primus on October 16th, 2020
Trying to visualise these all the steps. A video would be good. | 1,254 | 5,480 | {"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} | 2.5625 | 3 | CC-MAIN-2020-50 | latest | en | 0.934602 |
https://in.mathworks.com/matlabcentral/answers/142322-find-n-random-points-with-a-minimum-distance-r-inside-a-2d-rectangular-box | 1,675,802,098,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500641.25/warc/CC-MAIN-20230207201702-20230207231702-00039.warc.gz | 327,750,448 | 29,235 | # Find n random points with a minimum distance r inside a 2D rectangular box
108 views (last 30 days)
jaydeep on 18 Jul 2014
Edited: Jan on 21 Feb 2017
I would like to generate the (x, y) coordinates of N randomly distributed points within a 2D Square box having 2000 m X 2000m. the points must have a minimum distance of 200m from each other, means a second point should not lies between the area of circle covered by the first point and so on. Any idea for a uniform random distribution?
John BG on 12 Feb 2017
Edited: John Kelly on 17 Feb 2017
Hi Jaydeep
just completed a couple functions that may help with your question
1.
from
2.
run this
[X,Y,Nmax,Dmatrix]=scatter_points7
3.
key in the menu the rectangle size, the amount of points you want and the minimum distance.
4.
scatter_points7 returns:
X Y coordinates of the random points
Nmax the maximum amount of points that would fit in the rectangle if placing them orderly.
Dmatrix the distances between all combinations of points.
5.
Check the minimum distance requirement is met with
% test 1
Ap=20;L=combinator(Ap,2,'c');
relD2=((X(L(:,2))-X(L(:,1))).^2+(Y(L(:,2))-Y(L(:,1))).^2).^.5
R0=200;find(relD2<R0)
=
Empty matrix: 1-by-0
or
% test 2
Ap=20;L2=combinator(Ap,2);
D2=Dmatrix+NaN*eye(Ap);
R0=200;D2(D2<R0)
=
Empty matrix: 0-by-1
.
'
6.
Now try this
[X,Y,Nmax,Dmatrix]=scatter_points_saturate(2000,2000,200)
.
.
to know the amount of points (below Nmax) actually generated
numel(X)
John BG
Doug Hull on 18 Jul 2014
Edited: Doug Hull on 18 Jul 2014
Here is something stupid that just might work.
While NOT_DONE
generate a point
If NOT too close to existing points
Place point
end
check to see if enough points are placed.
end
Jan on 12 Feb 2017
Edited: Jan on 21 Feb 2017
function [X, Y, D] = GetPointsRandom(nWant, XWidth, YWidth, MinDist)
X = zeros(nWant, 1);
Y = zeros(nWant, 1);
dist_2 = MinDist ^ 2; % Squared once instead of SQRT each time
iLoop = 1; % Security break to yoid infinite loop
nValid = 0;
while nValid < nWant && iLoop < 1e6
newX = XWidth * rand;
newY = YWidth * rand;
if all(((X(1:nValid) - newX).^2 + (Y(1:nValid) - newY).^2) >= dist_2)
% Success: The new point does not touch existing points:
nValid = nValid + 1; % Append this point
X(nValid) = newX;
Y(nValid) = newY;
end
iLoop = iLoop + 1;
end
% Throw an error, if the area is filled too densely:
if nValid < nWant
error('Cannot find wanted number of points in %d iterations.', iLoop)
end
if nargout > 2
% D = pdist([X, Y]); % Faster with statistics toolbox
D = sqrt(bsxfun(@minus, X, X.') .^ 2 + bsxfun(@minus, Y, Y.') .^ 2);
end
end | 820 | 2,574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.796712 |
https://www.brainbalancemathematics.com/2021/09/a-woman-starts-shopping-with-rs-x-and-y-paise-spends-Rs.-3.50-and-is-left-with-Rs.-2Y-and-2X-paise.html?showComment=1632505031148 | 1,653,272,532,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662552994.41/warc/CC-MAIN-20220523011006-20220523041006-00510.warc.gz | 753,325,470 | 52,061 | # A woman starts shopping with Rs. X and Y paise, spends Rs. 3.50 and is left with Rs. 2Y and 2X paise. The amount she started with is
## Question:
A woman starts shopping with Rs. X and Y paise, spends Rs. 3.50 and is left with Rs. 2Y and 2X paise. The amount she started with is
1. Rs. 48.24
2. Rs. 28.64
3. Rs. 32.14
4. Rs. 23.42
If $Y>50$ then according to the information given,
$Y-50=2X$ and $X-3=2Y$
$\therefore 2X-Y=-50$ ............ (1)
and $X-2Y=3$ .................. (2)
equation (1) - 2$\times$equation (2) gives,
$3Y=-56$ which is absurd as Y is positive.
Now, if $Y<50$ then according to the information given,
$100+Y-50=2X$ and $X-1-3=2Y$
$\therefore 2X-Y=50$ ............. (3)
and $X-2Y=4$ ..................... (4)
equation (3) - 2$\times$equation (4) gives,
$3Y=42$
$\implies Y=14$
$\therefore$ from (4),
$X=32$
$\therefore$ The amount she started with is Rs. 32.14. | 347 | 887 | {"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-2022-21 | latest | en | 0.867473 |
https://max.pm/posts/hessian_ls/ | 1,726,727,118,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651995.50/warc/CC-MAIN-20240919061514-20240919091514-00194.warc.gz | 346,233,618 | 2,439 | Max R. P. Grossmann
The interpretation of the Hessian matrix in a least squares context
Posted: 2020-07-19 · Last updated: 2022-03-02
As you may know, in maximum likelihood estimation, the inverse of the Hessian matrix at the point of the maximum likelihood estimates is equal to the estimated variance of the estimates.
But what is the interpretation of the Hessian in a (ordinary) least squares context?
Remember that the solution to
$$\min_{\beta} \underbrace{(y-X\beta)'(y-X\beta)}_{L\left(\beta\right)}$$
is
$$\hat{\beta} = (X^T X)^{-1} X^T y.$$
We arrived at this conclusion by taking the first derivative of the objective function with respect to $\beta$, $L'(\beta) = -2 X^T y + 2 X^{T} X \beta$. If we take the second derivative, we should get the Hessian; and indeed we do. It is trivial to show that the Hessian turns out to be simply
$$H(\hat{\beta}) = \left.\frac{\partial^2 L\left(\beta\right)}{\partial \beta^2}\right|_{\hat{\beta}} = 2 X^T X.$$
Recognizing that in ordinary least squares, the variance of the estimates is estimated as $\widehat{V\left[\hat{\beta}\right]} = \hat{\sigma}^2 (X^T X)^{-1}$, with $\hat{\sigma}^2$ being the estimated error variance, we conclude that the Hessian matrix can be used to estimate the variance of the estimates in the following way:
$$\widehat{V\left[\hat{\beta}\right]} = \hat{\sigma}^2\, \text{diag}\left\{\left(\frac{1}{2}H(\hat{\beta})\right)^{-1}\right\}.$$
Pretty awesome, isn't it? | 436 | 1,458 | {"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} | 3.9375 | 4 | CC-MAIN-2024-38 | latest | en | 0.812672 |
https://electronics.stackexchange.com/questions/713846/why-do-electric-arcs-rise | 1,718,728,085,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861762.73/warc/CC-MAIN-20240618140737-20240618170737-00414.warc.gz | 203,045,100 | 42,439 | # Why do electric arcs rise?
I am a junior electrical engineering student, recently we were learning about the circuit breakers and professor showed us a video of an electrical arc which can be seen in the picture (Sorry for the quality). I wondered "if electrical arc is actually the electrons that are moving" why electrical arc didn't move in straight line but chose to rise and connect. I asked the lecturer about this but what I heard was just some gibberish. Then I thought about it and did some research but couldn't really find something directly about this. I believe it is caused by the Lorentz forces and also the rising of the warm air. But if warm air rises it should reduce the pressure of the air in the top side. Which will increase the distance between the particles in air and make it harder for air to ionize, right? Can you explain it to me or give me your thoughts about it?
• You don't need thoughts. It's not as though the topic hasn't been studied to death. Just glancing over at my library, I see three of the five books I have on just this topic alone: "The Art and Science of Lightning Protection" by Martin Uman; "Lightning Physics and Lightning Protection" by Bazelyan & Raizer; and this one on quantitative physical theory for plamas, "Principles of Plasma Discharges and Materials Processing" by Lieberman & Lichtenberg. The first two are filled with experimental data and experience. The last is pure plasma physics theory. But just get a good book. Commented May 20 at 18:55
## 3 Answers
When the potential between the two conductors reaches sufficient voltage, it causes a breakdown in the air - ionising the air into a hot plasma between the two conductors. It is this plasma through which a sustained current can flow because it is significantly lower resistance than the surrounding air.
Because the plasma channel is hotter than the surrounding air, it will start to rise - think convection currents. Remember though it is this plasma that is conducting the current and emitting light, and so as the plasma rises, so will the arc.
Up to a point, the plasma is still such a low resistance compared to the surrounding air and still ionised, the extra distance of this path doesn't initially disrupt the current flow. At some point there is not enough thermal energy or electric potential to keep the arc ionised and the plasma begins to both disperse (mix with the surrounding air) and recombine (ions return to ground state).
If there is still enough potential between the conductors a new arc will form and the process repeats. A great example of this is a Jacobs ladder (travelling arc). In this example the Lorentz force is not actually the driving factor of the arc rising - for example this answer links to a video of an upside down Jacobs Ladder where the arc no longer moves along the ladder as would be expected if the magnetic forces were dominant.
• The air that heats up and rises is actually ionized air. So another heated air does not merge with the ionized air above. Instead the ionized part heats up and rises. Do I understand you correctly?
– kaan
Commented May 20 at 17:52
• There is something else, that is, the discharge, having a certain energy, it is transformed partly into heat and sound energy, another fraction is transformed into electromagnetic energy radiated into space at a frequency around 100MHz, another part is transformed into light energy and finally a small fraction also into X-rays. Commented May 20 at 18:09
• @kaan pretty much. It takes quite a while to transfer heat from the arc into the surrounding air, so initially it is the plasma that is rising. Convection currents will be formed causing motion in the surrounding air (likely chaotic hence the weird shape of the arc). Everything will ultimately mix back together which is why the arcs tend to get quite "hazy" for lack of a better term as they move upwards. Commented May 20 at 18:12
The electrical arc is heated up quite a bit compared to the atmospheric air. Like a hot air balloon, the arc is at lower density than ambient air so the arc rises.
A large, high energy arc like you show in the photo will have a lot of heat, and thus a lot of swirling convection which stirs up the plasma into the random-looking shape you see. A smaller arc (like a Jacob’s ladder) won’t have as much heat or space so will tend to rise more regularly, without the swirling.
• Do you think Lorentz forces also effect arc to rise or not ?
– kaan
Commented May 20 at 17:53
But if warm air rises it should reduce the pressure of the air in the top side. Which will increase the distance between the particals in air and make it harder for air to ionize, right ?
Plasma also has resistance which is small, but you can get heating in plasma from a large current. The plasma is heated and the arc sustains the plasma, this process continues until the plasma gets pinched by air and the arc is extinguished. So it's really the self heating of the plasma that keeps the arc going, the heated air rises and also the lorentz forces push it upwards. | 1,101 | 5,072 | {"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} | 2.703125 | 3 | CC-MAIN-2024-26 | latest | en | 0.964593 |
https://www.studypool.com/questions/258800/Calculus-2 | 1,480,949,632,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541697.15/warc/CC-MAIN-20161202170901-00155-ip-10-31-129-80.ec2.internal.warc.gz | 1,006,059,950 | 750,652 | # Calculus 2
SoccerBoss
Category:
Calculus
Price: \$5 USD
Question description
```A team of biologists is interested in the ability of certain birds to migrate great distance with little rest. the biologists are monitoring a flock of birds known to migrate after spending the winter in the warm climates of the Okefenokee swamp. The location of the flock in x(t) hundreds of miles east and y(t) hundreds of miles north of the base camp of biologists, t days after their departure from Okefenokee swamp is given by: x(t)=3t+(1/2), y(t)=t^(3/2) +(1/5).
```
```a) What is Okefenokee swamp in relation to the base camp of biologists?
```
```b) Is there a time when the flock of birds is traveling due North East? If not, explain why not.
```
```c) Is the flock of birds moving at a constant rate throughout the first three days of their journey? Why or Why not?
```
`d) How far (to the nearest tenth of a mile) does the flock of birds travel in the first three days of their journey?`
(Top Tutor) Daniel C.
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https://us.metamath.org/mpeuni/2ndrn.html | 1,726,336,451,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00686.warc.gz | 550,757,174 | 4,535 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > 2ndrn Structured version Visualization version GIF version
Theorem 2ndrn 7744
Description: The second ordered pair component of a member of a relation belongs to the range of the relation. (Contributed by NM, 17-Sep-2006.)
Assertion
Ref Expression
2ndrn ((Rel 𝑅𝐴𝑅) → (2nd𝐴) ∈ ran 𝑅)
Proof of Theorem 2ndrn
StepHypRef Expression
1 1st2nd 7742 . . 3 ((Rel 𝑅𝐴𝑅) → 𝐴 = ⟨(1st𝐴), (2nd𝐴)⟩)
2 simpr 488 . . 3 ((Rel 𝑅𝐴𝑅) → 𝐴𝑅)
31, 2eqeltrrd 2853 . 2 ((Rel 𝑅𝐴𝑅) → ⟨(1st𝐴), (2nd𝐴)⟩ ∈ 𝑅)
4 fvex 6671 . . 3 (1st𝐴) ∈ V
5 fvex 6671 . . 3 (2nd𝐴) ∈ V
64, 5opelrn 5784 . 2 (⟨(1st𝐴), (2nd𝐴)⟩ ∈ 𝑅 → (2nd𝐴) ∈ ran 𝑅)
73, 6syl 17 1 ((Rel 𝑅𝐴𝑅) → (2nd𝐴) ∈ ran 𝑅)
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 ∈ wcel 2111 ⟨cop 4528 ran crn 5525 Rel wrel 5529 ‘cfv 6335 1st c1st 7691 2nd c2nd 7692 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2113 ax-9 2121 ax-10 2142 ax-11 2158 ax-12 2175 ax-ext 2729 ax-sep 5169 ax-nul 5176 ax-pr 5298 ax-un 7459 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3an 1086 df-tru 1541 df-fal 1551 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2557 df-eu 2588 df-clab 2736 df-cleq 2750 df-clel 2830 df-nfc 2901 df-ne 2952 df-ral 3075 df-rex 3076 df-rab 3079 df-v 3411 df-sbc 3697 df-dif 3861 df-un 3863 df-in 3865 df-ss 3875 df-nul 4226 df-if 4421 df-sn 4523 df-pr 4525 df-op 4529 df-uni 4799 df-br 5033 df-opab 5095 df-mpt 5113 df-id 5430 df-xp 5530 df-rel 5531 df-cnv 5532 df-co 5533 df-dm 5534 df-rn 5535 df-iota 6294 df-fun 6337 df-fv 6343 df-1st 7693 df-2nd 7694 This theorem is referenced by: gsumhashmul 30842 heicant 35394 mblfinlem1 35396
Copyright terms: Public domain W3C validator | 1,036 | 1,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2024-38 | latest | en | 0.220881 |
https://brainmass.com/statistics/variation-complaints-levels-wages-560481 | 1,481,284,700,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542695.22/warc/CC-MAIN-20161202170902-00349-ip-10-31-129-80.ec2.internal.warc.gz | 836,375,558 | 19,876 | Share
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# Need stepwise solution
2. The Postal Service is attempting to reduce the number of complaints made by the public against its workers. In order to facilitate this task, a staff analyst for the Service regresses the number of complaints lodged against an employee last year (Y) on the hourly wage of the employee for the year (X). She obtains the following results:
Y =10.2 - 1.9X
r2 = .73 sb=.87 n=348 t= slope/ sb
Based on these results, answer the following questions:
a) If wages increased by \$1.00, what is the expected effect on the number of complaints received per employee?
b) What percent of the variation in complaints can be explained by wages?
c) At what level of wages can the employee be expected to receive zero (0) complaints? (hint: recalculate the equation with Y=0)
d) Calculate the t-value using the formula above.
e) Can the null hypothesis be rejected at the .05 level of statistical significance that wages have no effect on the number of complaints (Recall the critical values for t-test for a two-tail test is 1.96 and for a one-tail test 1.65)?
f) Discuss the policy implications of the regression analysis.
#### Solution Preview
2. The Postal Service is attempting to reduce the number of complaints made by the public against its workers. In order to facilitate this task, a staff analyst for the Service regresses the number of complaints lodged against an employee last year (Y) on the hourly wage of the employee for the year (X). She obtains the following results:
Y =10.2 - 1.9X
r2 = .73 sb=.87 n=348 t= slope/ sb
Based on these results, answer the following questions:
a) If ...
#### Solution Summary
Variation complaints for levels of wages are given. The policy implications of the regression analysis is examined.
\$2.19 | 415 | 1,796 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2016-50 | longest | en | 0.915161 |
ieltsquangbinh.com | 1,725,903,453,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651133.92/warc/CC-MAIN-20240909170505-20240909200505-00413.warc.gz | 291,133,248 | 21,590 | # (Update 2024) Numeracy: Can animals tell numbers? | IELTS Reading Practice Test Free
## Numeracy: Can animals tell numbers?
### Passage
A. Prime among basic numerical faculties is the ability to distinguish between a larger and a smaller number, says psychologist Elizabeth Brannon. Humans can do this with ease – providing the ratio is big enough – but do other animals share this ability? In one experiment, rhesus monkeys and university students examined two sets of geometrical objects that appeared briefly on a computer monitor. They had to decide which set contained more objects. Both groups performed successfully but, importantly, Brannon’s team found that monkeys, like humans, make more errors when two sets of objects are close in number. The students’ performance ends up looking just like a monkey’s. It’s practically identical, ‘she says.
B. Humans and monkeys are mammals, in the animal family known as primates. These are not the only animals whose numerical capacities rely on ratio, however. The same seems to apply to some amphibians. Psychologist Claudia Uller’s team tempted salamanders with two sets of fruit flies held in clear tubes. In a series of trials, the researchers noted which tube the salamanders scampered towards, reasoning that if they had a capacity to recognise number, they would head for the larger number.
The salamanders successfully discriminated between tubes containing 8 and 16 flies respectively, but not between 3 and 4, 4 and 6, or 8 and 12. So it seems that for the salamanders to discriminate between two numbers, the larger must be at least twice as big as the smaller. However, they could differentiate between 2 and 3 flies just as well as between 1 and 2 flies, suggesting they recognise small numbers in a different way from larger numbers.
C. Further support for this theory comes from studies of mosquitofish, which instinctively join the biggest shoal they can. A team at the University of Padova found that while mosquitofish can tell the difference between a group containing 3 shoal-mates and a group containing 4, they did not show a preference between groups of 4 and 5. The team also found that mosquitofish can discriminate between numbers up to 16, but only if the ratio between the fish in each shoal was greater than 2:1. This indicates that the fish, like salamanders, possess both the approximate and precise number systems found in more intelligent animals such as infant humans and other primates.
D. While these findings are highly suggestive, some critics argue that the animals might be relying on other factors to complete the tasks, without considering the number itself. ‘Any study that’s claiming an animal is capable of representing number should also be controlling for other factors, ‘ says Brannon. Experiments have confirmed that primates can indeed perform numerical feats without extra clues, but what about the more primitive animals?
E. To consider this possibility, the mosquitofish tests were repeated, this time using varying geometrical shapes in place of fish. The team arranged these shapes so that they had the same overall surface area and luminance even though they contained a different number of objects. Across hundreds of trials on 14 different fish, the team found they consistently discriminated 2 objects from 3. The team is now testing whether mosquitofish can also distinguish 3 geometric objects from 4.
F. Even more primitive organisms may share this ability. Entomologist Jurgen Tautz sent a group of bees down a corridor, at the end of which lay two chambers – one which contained sugar water, which they like, while the other was empty. To test the bees’ numeracy, the team marked each chamber with a different number of geometrical shapes – between 2 and 6. The bees quickly learned to match the number of shapes with the correct chamber. Like the salamanders and fish, there was a limit to the bees’ mathematical prowess – they could differentiate up to 4 shapes, but failed with 5 or 6 shapes.
G. These studies still do not show whether animals learn to count through training, or whether they are born with the skills already intact. If the latter is true, it would suggest there was a strong evolutionary advantage to a mathematical mind. Proof that this may be the case has emerged from an experiment testing the mathematical ability of three-and four-day-old chicks. Like mosquitofish, chicks prefer to be around as many of their siblings as possible, so they will always head towards a large number of their kin. If chicks spend their first few days surrounded by certain objects, they become attached to these objects as if they were family.
Researchers placed each chick in the middle of a platform and showed it two groups of balls of paper. Next, they hid the two piles behind screens, changed the quantities and revealed them to the chick. This forced the chick to perform simple computations to decide which side now contained the biggest number of its “brothers’7. Without any prior coaching, the chicks scuttled to the larger quantity at a rate well above chance. They were doing some very simple arithmetic, claim the researchers.
H. Why these skills evolved is not hard to imagine, since it would help almost any animal forage for food. Animals on the prowl for sustenance must constantly decide which tree has the most fruit, or which patch of flowers will contain the most nectar. There are also other, less obvious, advantages of numeracy. In one compelling example, researchers in America found that female coots appear to calculate how many eggs they have laid – and add any in the nest laid by an intruder – before making any decisions about adding to them.
Exactly how ancient these skills are is difficult to determine, however. Only by studying the numerical abilities of more and more creatures using standardised procedures can we hope to understand the basic preconditions for the evolution of number.
### Questions
Questions 15-21
Choose NO MORE THAN THREE WORDS AND/OR A NUMBER from the passage for each answer.
Questions 22-27
Do the following statements agree with the information given in Reading Passage 2?
TRUE if the statement is true
FALSE if the statement is false
NOT GIVEN if the information is not given in the passage 22
22. Primates are better at identifying the larger of two numbers if one is much bigger than the other.
23 Jurgen Tautz trained the insects in his experiment to recognise the shapes of individual numbers.
24 The research involving young chicks took place over two separate days.
25 The experiment with chicks suggests that some numerical ability exists in newborn animals.
26 Researchers have experimented by altering quantities of nectar or fruit available to certain wild animals.
27 When assessing the number of eggs in their nest, coots take into account those of other birds. | 1,415 | 6,881 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2024-38 | latest | en | 0.967193 |
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# Segments from Chords
## Products of the segments of each of two intersecting chords are equal.
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Circles: Segments and Lengths Study Guide
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Theorems for chords, secants, and tangents (of a circle) are covered in this study guide. | 152 | 560 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2017-13 | latest | en | 0.717571 |
https://www.mathway.com/popular-problems/Trigonometry?page=1045 | 1,611,437,724,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00137.warc.gz | 863,488,530 | 29,972 | Popular Problems
Rank Topic Problem Formatted Problem
96049 Solve for s 7-4s=4(7-st)
96050 Solve for r t(r+s)=rs
96051 Solve for r S(r+t)=rt
96052 Solve for r r^2=36
96053 Solve for r 900pi=4pir^2
96054 Solve for r r=5
96055 Solve for r r=6cos(theta)
96056 Solve for r r=6sin(-( square root of 3)/3)
96057 Solve for r r^2-6r-83=-10
96058 Solve for R R=8/17
96059 Solve for r r = square root of (0.30)^2+(0.16)^2
96060 Solve for r r = square root of (-24)^2+(10)^2
96061 Solve for t log base 3 of 7t+6- log base 3 of t = log base 3 of 8
96062 Solve for t theta=arctan((2t)/(t^2))
96063 Solve for s tan(s)^2=1
96064 Solve for t 150=-16t^2+117t+2
96065 Solve for s cos(s)^2=3/4
96066 Solve for s cot(s)^2=3
96067 Solve for s s^3+27=0
96068 Solve for T ( log of 2)/( log of 1.65)=T
96069 Solve for t 1/2=5e^(-2.3t)cos(10pit)
96070 Solve for B (27^2-12^2-20^2)/(-2(12)(20))=cos(B)
96071 Solve for b (1.892071)^3=(b^(1/3))^3
96072 Solve for b 10/(sin(10))=9/(sin(b))
96073 Solve for B (sin(33))/23=(sin(B))/17
96074 Solve for b (2pi)/b=(8pi)/3
96075 Solve for b (2pi)/b=0.1
96076 Solve for b 11^2+b^2=61^2
96077 Solve for b sin(b)=11/18
96078 Solve for B sin(B)=(5sin(70))/7
96079 Solve for B (9.9sin(86.1 degrees ))/11.2=sin(B)
96080 Solve for b b/(sin(114.20))=400/(sin(42.12))
96081 Solve for b b/(sin(88.1))=648/(sin(0.8))
96082 Solve for b b = natural log of e^3
96083 Solve for b b=arccos(9/24)
96084 Solve for B cos(B)=11/18
96085 Solve for B tan(B)=15/9
96086 Solve for c (13.5sin(126.5))/17.2=sin(c)
96087 Solve for b 5^2+b^2=10^2
96088 Solve for c 5^2+13^2=c^2
96089 Solve for c 6^2+12^2=c^2
96090 Solve for c 6^2+4^2=c^2
96091 Solve for c 4^2+7^2=c^2
96092 Solve for c 3^2+8^2=c^2
96093 Solve for C 15^2+36^2=C^2
96094 Solve for c 2^2+4^2=c^2
96095 Solve for c 17.2/(sin(126.5))=13.5/(sin(c))
96096 Solve for c cos(c)=(22^2+19^2-19^2)/(2(22)(19))
96097 Solve for c 9^2+5^2=c^2
96098 Solve for c 9^2+7^2=c^2
96099 Solve for c 9^2+9^2=c^2
96100 Solve for c 6^2+9^2=c^2
96101 Solve for c 86.49=123.85-122.64cos(c)
96102 Solve for C 9^2+15^2=C^2
96103 Solve for d d=3sin(8pi*0)-2
96104 Solve for d tan(42 degrees )=79/d
96105 Solve for f 5(2f-31)=10f-155
96106 Solve for d tan(42 degrees )=79/d
96107 Solve for h csc(60 degrees )=26/h
96108 Solve for h h=22x^2+6.5
96109 Solve for h 114=8h
96110 Solve for h 373.75=h^2
96111 Solve for k 576=k*6^2
96112 Solve for k 6k-3z=12
96113 Solve for k k=(1+sin(x))/n
96114 Solve for L 4.7/9.9=sin(L)
96115 Solve for m tan(34 degrees )=m/400
96116 Solve for m 10pi=25pi*m/360
96117 Solve for n (1/36)^n=216^(n+5)
96118 Solve for m (8m^2+5m-3)/(5m+5)=1
96119 Solve for m log base 4 of m-3+ log base 4 of m+3=2
96120 Solve for n log base 3 of n=2
96121 Solve for n 3-|4-n|>1
96122 Solve for n a_25=-7n+17
96123 Solve for p 1/(p^2)+1/p=5/(2p^2)
96124 Solve for r 4/3pir^3=36pi
96125 Solve for r 16pi=pir^2
96126 Solve for r 169pi=pir^2
96127 Solve for r 18=2pir
96128 Solve for r -30r-45<=-5(5r+11)
96129 Solve for r 4^2+3^2=r^2
96130 Solve for r 4^(-r+3)=1
96131 Solve for p (4p-1)(p-4)=0
96132 Solve for p square root of p^2+1=5
96133 Solve for p 8^(1-p)=16^(2p-1)
96134 Solve for P P=(7,pi/3)
96135 Solve for p m^2n+3ps^3=2q
96136 Solve for A (6k^2-36k)/A*(k^2+12k+36)/(k^2-36)=1
96137 Solve for a a/(3x)*(3x+3)/(x^2-2x-3)=1
96138 Solve for A cos(A)=0.29
96139 Solve for a (sin(60 degrees ))/a=(sin(65 degrees ))/120
96140 Find the Coterminal Angle 420 degrees
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This website uses cookies to ensure you get the best experience on our website. | 1,592 | 3,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2021-04 | latest | en | 0.512175 |
https://www.unitsconverters.com/en/Gal-To-Picometer/Squaremicrosecond/Unittounit-3429-7447 | 1,643,298,102,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305266.34/warc/CC-MAIN-20220127133107-20220127163107-00668.warc.gz | 1,079,283,695 | 39,132 | Formula Used
1 Meter per Square Second = 100 Gal
1 Meter per Square Second = 1 Picometer per Square Microsecond
1 Gal = 0.01 Picometer per Square Microsecond
## Gals to Picometer per Square Microseconds Conversion
Gal stands for gals and pm/μs² stands for picometer per square microseconds. The formula used in gals to picometer per square microseconds conversion is 1 Gal = 0.01 Picometer per Square Microsecond. In other words, 1 gal is 100 times smaller than a picometer per square microsecond. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
## Convert Gal to Picometer per Square Microsecond
How to convert gal to picometer per square microsecond? In the acceleration measurement, first choose gal from the left dropdown and picometer per square microsecond from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from picometer per square microsecond to gal? You can check our picometer per square microsecond to gal converter.
How to convert Gal to Picometer per Square Microsecond?
The formula to convert Gal to Picometer per Square Microsecond is 1 Gal = 0.01 Picometer per Square Microsecond. Gal is 100 times Smaller than Picometer per Square Microsecond. Enter the value of Gal and hit Convert to get value in Picometer per Square Microsecond. Check our Gal to Picometer per Square Microsecond converter. Need a reverse calculation from Picometer per Square Microsecond to Gal? You can check our Picometer per Square Microsecond to Gal Converter.
How many Meter per Square Second is 1 Gal?
1 Gal is equal to 0.01 Meter per Square Second. 1 Gal is 100 times Smaller than 1 Meter per Square Second.
How many Kilometer per Square Second is 1 Gal?
1 Gal is equal to 0.01 Kilometer per Square Second. 1 Gal is 100 times Smaller than 1 Kilometer per Square Second.
How many Micrometer per Square Second is 1 Gal?
1 Gal is equal to 0.01 Micrometer per Square Second. 1 Gal is 100 times Smaller than 1 Micrometer per Square Second.
How many Mile per Square Second is 1 Gal?
1 Gal is equal to 0.01 Mile per Square Second. 1 Gal is 100 times Smaller than 1 Mile per Square Second.
## Gals to Picometer per Square Microseconds Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like acceleration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like Gal to pm/μs² through multiplicative conversion factors. When you are converting acceleration, you need a Gals to Picometer per Square Microseconds converter that is elaborate and still easy to use. Converting Gal to Picometer per Square Microsecond is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Gal to Picometer per Square Microsecond conversion along with a table representing the entire conversion.
Let Others Know | 740 | 3,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.792484 |
http://mathhelpforum.com/pre-calculus/138948-y-intercepts-print.html | 1,529,847,680,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866937.79/warc/CC-MAIN-20180624121927-20180624141927-00340.warc.gz | 212,242,599 | 2,574 | Y intercepts
• Apr 13th 2010, 09:54 AM
cristinaivelisse
Y intercepts
how do I find the Y intercept for the polynomial function?
P(x)= x^5-4x^4+3x^3
• Apr 13th 2010, 09:58 AM
masters
Quote:
Originally Posted by cristinaivelisse
How do I find the Y intercept for the polynomial function?
\$\displaystyle P(x)= x^5-4x^4+3x^3\$
Hi cristinaivelisse,
Just find P(0) because wherever the function crosses the y-axis, x = 0.
In your case, it's easy. P(0) = 0. | 162 | 459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2018-26 | latest | en | 0.773767 |
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#### Resources tagged with Factors and multiples similar to Fractions Made Faster:
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### There are 144 results
Broad Topics > Numbers and the Number System > Factors and multiples
### Flashing Lights
##### Stage: 2 Challenge Level:
Norrie sees two lights flash at the same time, then one of them flashes every 4th second, and the other flashes every 5th second. How many times do they flash together during a whole minute?
### Red Balloons, Blue Balloons
##### Stage: 2 Challenge Level:
Katie and Will have some balloons. Will's balloon burst at exactly the same size as Katie's at the beginning of a puff. How many puffs had Will done before his balloon burst?
### Cuisenaire Environment
##### Stage: 1 and 2 Challenge Level:
An environment which simulates working with Cuisenaire rods.
### Tom's Number
##### Stage: 2 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
### What's in the Box?
##### Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
### In the Money
##### Stage: 2 Challenge Level:
There are a number of coins on a table. One quarter of the coins show heads. If I turn over 2 coins, then one third show heads. How many coins are there altogether?
### What Two ...?
##### Stage: 2 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
### What Is Ziffle?
##### Stage: 2 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Path to the Stars
##### Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
### Factor-multiple Chains
##### Stage: 2 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
### Divide it Out
##### Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
### Mathematical Swimmer
##### Stage: 3 Challenge Level:
Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .
### Becky's Number Plumber
##### Stage: 2 Challenge Level:
Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
### Three Neighbours
##### Stage: 2 Challenge Level:
Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Being Determined - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that may require determination.
### Multiplication Series: Number Arrays
##### Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
### What Do You Need?
##### Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
### Factor Lines
##### Stage: 2 and 3 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### Which Is Quicker?
##### Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
### Multiplication Square Jigsaw
##### Stage: 2 Challenge Level:
Can you complete this jigsaw of the multiplication square?
##### Stage: 3 Challenge Level:
A mathematician goes into a supermarket and buys four items. Using a calculator she multiplies the cost instead of adding them. How can her answer be the same as the total at the till?
### Repeaters
##### Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Special Sums and Products
##### Stage: 3 Challenge Level:
Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
### A Dotty Problem
##### Stage: 2 Challenge Level:
Starting with the number 180, take away 9 again and again, joining up the dots as you go. Watch out - don't join all the dots!
### Three Spinners
##### Stage: 2 Challenge Level:
These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
### Multiples Grid
##### Stage: 2 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
### Tiling
##### Stage: 2 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
##### Stage: 2 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
### Venn Diagrams
##### Stage: 1 and 2 Challenge Level:
Use the interactivities to complete these Venn diagrams.
### Remainders
##### Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Biscuit Decorations
##### Stage: 1 and 2 Challenge Level:
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### Fractions in a Box
##### Stage: 2 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each disc. Use the information to find out how many discs of each colour there are in the box.
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Graphing Number Patterns
##### Stage: 2 Challenge Level:
Does a graph of the triangular numbers cross a graph of the six times table? If so, where? Will a graph of the square numbers cross the times table too?
### Scoring with Dice
##### Stage: 2 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Eminit
##### Stage: 3 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Down to Nothing
##### Stage: 2 Challenge Level:
A game for 2 or more people. Starting with 100, subratct a number from 1 to 9 from the total. You score for making an odd number, a number ending in 0 or a multiple of 6.
### Odds and Threes
##### Stage: 2 Challenge Level:
A game for 2 people using a pack of cards Turn over 2 cards and try to make an odd number or a multiple of 3.
### Money Measure
##### Stage: 2 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### A First Product Sudoku
##### Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
### Spelling Circle
##### Stage: 2 Challenge Level:
Find the words hidden inside each of the circles by counting around a certain number of spaces to find each letter in turn.
### GOT IT Now
##### Stage: 2 and 3 Challenge Level:
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
### Crossings
##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest? | 2,257 | 9,588 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2016-50 | longest | en | 0.919849 |
https://onlineguitarbooks.com/basic-bar-chords/ | 1,718,838,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861853.72/warc/CC-MAIN-20240619220908-20240620010908-00780.warc.gz | 395,980,664 | 47,687 | # Basic Bar Chords – Master Root 6 and Root 5 Bar Chords
In today’s lesson, we’re going to jump right into bar chords (also written as barre chords). If guitar chords could be placed neatly into two boxes, one would be called open chords, the other would be called bar chords.
You’ve learnt 15 open chords and you know that with them, you have the ability to play thousands of songs. There are certain chords though, that can’t be played as open chords. For this, we need bar chords.
## What Are Bar Chords?
Before we explain exactly what bar chords are, and their uses, it’s worth talking a bit more about open chords. Open chords, by definition, are chords that have open strings in them. There are technically hundreds of open chords (because any possible shape that you form that includes an open string is technically an open chord), but there are a bunch that we use most of the time.
In the lesson on intervals and the fretboard, we learnt that we can move a note up a semitone by moving it up a fret. We can do that with chords too. Technically, if we move every note of a D Major chord up one tone (2 frets), it becomes an E Major chord. Easy, right?
## Why Can’t You Move Open Chords?
The problem with applying this theory to open chords, is that by moving the shape of an open chord up (or down) the fretboard, only the notes that are being fingered move, while the open notes stay the same. Remember, when we play an open string, we are technically playing a ‘0’. If we simply shift an open chord shape up one fret, the zeros stay as zeros, while the other numbers move. This effectively changes the structure of the chord and usually results in something that sounds weird.
## Bar Chords Are Moveable
The point of this little analysis of open chords is that it highlights one of the benefits of bar chords. Bar chords, by definition include a finger that is ‘barring’ most of the strings. There are no open notes with standard bar chords. What this means is that when we learn a bar chord shape, we can play it anywhere along the fretboard. They are said to be ‘moveable’, in the same way that scale shapes are moveable.
## How To Play A Bar Chord
We’ll talk more about using bar chords shortly. What we are going to do now is look at a bar chord shape and discuss how to play it. The following is a ‘Major chord’, played as a bar chord:
As you can see from the diagram, the 1st finger is playing multiple strings. This is achieved by literally ‘barring’ your finger, so that it straightens out and can cover multiple strings.
Have a go at playing the chord. Remember, it is a moveable chord, so you can try playing it anywhere. I recommend starting around the 5th fret, where the frets aren’t too wide and aren’t too narrow either. Playing the bar chord on the 5th fret would result in an A major chord being played (we’ll discuss how to figure this out soon). From a notes/tabs perspective, it would look like this:
## Don’t Be Too Fussy
Unless you are just naturally gifted with strength and technique, you will probably find that it feels weird and sounds pretty bad. This is normal. Most people struggle to get a clear sound with bar chords when they first attempt to play them. Being able to play bar chords cleanly and effectively requires a steady build up of strength and technique, which of course, only comes about with consistent and steady practice over time.
## A Closer Look At The Bar
It’s worth having a closer look at the finger that performs the ‘barre’. Have a look at the following chord diagram, which shows just the first finger performing the bar, on the 5th fret:
This image shows what the 1st finger is doing when playing the bar chord that we looked at. This is worth looking at for a few reasons.
Firstly, you can see what the 1st finger should look like. It is straight, with even pressure being applied along the length of the finger. This is what the 1st finger should be doing. The other fingers come in and effectively play over the top of this bar, but the bar shouldn’t change.
Secondly, playing just the bar by itself is a good strength exercise. It’s good to practise because it allows you to focus on the most difficult part of the bar chord (the bar). It is a good idea to form the bar by itself and then play each string individually, to see if you can get each note to ring out cleanly. It’s hard, but it’s a great exercise in building strength. You will most likely feel your hands and fingers fatiguing pretty quickly when practising in this way.
## Introducing The Minor Chord
Before we start putting the bar chord that we have just learnt to good use, let’s look at another shape. The following shape is for the minor chord, played as a bar chord.
As you can see, the minor chord is very similar to the Major chord. In fact, to play the minor chord all you have to do is play the major chord shape and then take your 2nd finger off. An ‘A minor’ chord (minor bar chord shape played on the 5th fret) would sound like this:
Just like with the Major chord shape that we did, practise playing the minor chord, and don’t get disheartened if it doesn’t sound perfect in the beginning.
## Why Bar Chords Are So Useful
The main benefit of bar chords is that they are moveable. What this means is that by learning one shape, you are effectively learning twelve chords. By playing the the Major chord shape starting on the 5th fret, you are playing an A Major chord. By playing the major chord starting on the 10th fret, you are playing a D major chord. The same applies to the minor chord. So how do you know which chord is which? By being familiar with the 6th string. With each of the two bar chords that we looked at, the root note is played on the 6th string. In fact, it is also played on the 4th string (up an octave) and on the 1st string (up another octave), but the 6th string is a better reference point. Have a look at the chords again. The root notes are darker.
Because the root note is played on the 6th string, we just need to know the notes along the 6th string to determine which chord we are playing.
Of course, you already know how to figure that out, using what you learnt in the lesson on intervals, but for the sake of this lesson, here is a chart displaying the notes along the 6th string:
The sharps and flats aren’t written, but they are obviously the ones in between. If you’re not too sure how sharps and flats work, read this lesson.
So for example, if you wanted to play a G Major chord, you would play the major shape, starting on the 3rd fret (because the 3rd fret of the 6th string is G).
If you wanted to play a C minor chord, you would play the minor shape, starting on the 8th fret of the 6th string (because the 8th fret of the 6th string is C).
It means that to use bar chords effectively, you need to be familiar with the notes along the 6th string. We already discussed the importance of knowing the fretboard, and it’s obviously extra important for bar chords. The good news is, playing songs that use bar chords is a great way of learning the notes along the 6th string. There are only so many times that you can play a D Major bar chord before you remember where D is. Just get playing!
## How To Use Bar Chords
Introducing bar chords into your playing can be tough at first. Like I said, bar chords don’t sound that great to start with. This presents a bit of a catch-22 situation; people avoid practising them because they don’t sound good initially. Because they’re not practising them, they never improve. You get the picture. You have to avoid this trap. The sooner you start practising bar chords, the sooner you will get over that initial, painful stage and get to the point where you can use bar chords as easily and enjoyably as open chords.
Here is a good process to follow for integrating bar chords into your playing. The time spent in each phase will vary for each individual, so be patient and enjoy the process:
## Start Out Just Playing The Shapes
This might seem obvious, but the first step is to memorise the shapes, and then practise playing them. The important thing with this stage is that you don’t need to use them in a musical context. Don’t get out the song book just yet. All you need to do is play the shapes, observe how they sound, and where the weaknesses may be. You will naturally improve your strength and technique by doing this. If you set yourself the goal of using them seamlessly in a song before you have had a chance to simply ‘hang out’ with the chords, you will get discouraged.
## Move From One Chord To Another
The next step is to practise moving between two bar chords. Start off with just two bar chords and practise going from one to another. Here are a few examples. In each of the following exercises, whole notes are being used. This gives you more time in between chord changes and allows you to not worry about strumming (just yet).
##### Bar Chord Exercise 2
That is only two exercises, but of course you can make your own combinations.
## Use Bar Chords And Open Chords Together
Once you feel like you have made progress with moving from one bar chord to another, you should start using bar chords in combination with open chords. Using bar chords with open chords allows you to effectively ‘ease’ bar chords into your playing, while using open chords that you are hopefully already comfortable with.
It’s important to remember that there is no functional difference between a chord that is played as an open chord and one that is played as a bar chord. That is to say that there is not much difference between a G major chord (for example) played as an open chord and a G major chord played as a bar chord. Both versions have a slightly different sound, but for all intents and purposes, they are the exact same chord.
What this means is that you can take a chord progression that you can already play using open chords and substitute one of the chords with a bar chord. Because only one of the chords will be played as a bar chord, it should make the task of playing the bar chord a bit easier. You can effectively relax through the open chords and apply extra focus when you get to the bar chord.
Here is an example of a chord progression being played using open chords.
Now try playing the same chord progression still using open chords, but use a bar chord for the G major chord. Both versions, played correctly should sound very similar. Listen to the version with the G major chord played as a bar chord:
## Play Songs And Exercises Using Only Bar Chords
The ultimate test of bar chord prowess is to be able to play entire chord progressions using only bar chords. Being able to do so means that you are comfortable with the bar chord shapes, can move comfortably from one to another and have a sound knowledge of the 6th string (so that you know which chord you are playing). Let’s have a look at another chord progression. Try to play this exercise using only bar chords.
## Introducing Root-5 Bar Chords
The bar chords that we have looked at up until this point come under the category of ‘root-6 bar chords’. This basically means that the root note for the two shapes that we looked at, falls on the 6th string. As you may have guessed, there are other bar chords as well. We are now going to look at ‘root-5 bar chords’.
We are going to look at two more shapes. The chord types are exactly the same as the two root-6 shapes that we learnt – Major and Minor. The difference is that for the following two shapes, the root note is found on the 5th string. Let’s have a look at the two shapes:
##### Root 5 Minor Chord
As you may have guessed, if we want to put our root-5 bar chords to good use, we need to be familiar with the notes along the 5th string. Here is a chart that tells us where the notes are along the 5th string:
## Root 6 and Root 5 Bar Chords Together
With a root-6 bar chord, playing G major requires you to play the major shape on the 3rd fret, because that’s where G is on the 6th string. As you can see from the above chart, if we want to play a G major using the root-5 bar chord shape, we need to go to the 10th fret, because that’s where G is on the 5th string.
This means that we now have even more options. Some chords can be played as an open chord, a root-6 bar chord and a root-5 bar chord.
So how do you know which one to use? There is no real definitive answer. Often it just depends on the context. Often it doesn’t really matter. While it can seem daunting to have options (especially if you’re still trying to deal with one set of chords), having options can often make things easier. Let’s look at the following chord progression as an example:
Let’s assume that we are going to play the four chords using bar chords.
If we only used root-6 chords, we would get the following:
• A Major (5th fret) – D Major (10th Fret) – B Major (7th Fret) – E Major (12th Fret)
As you can see, using only root-6 chords requires us to jump around a fair bit. This might be the effect that you are after, but it seems like too much work.
Let’s now look at where we would need to play the chords if we only used root-5 bar chords:
• A Major (12th fret) – D Major (5th Fret) – B Major (2nd Fret) – E Major (7th Fret)
As you can see, there is even more jumping around now. Not great.
When we use a combination of root-6 and root-5 bar chords, we are able to arrange things so that we don’t have to move around as much. This makes things easier, but it can also make things sound more organised and flowing. Let’s look at the following combination example:
• A Major (5 fret, root 6) – D Major (5th Fret, root 5) – B Major (7th Fret, root 6) – E Major (7th Fret, root 5)
As you can see, we barely have to move when using the right combinations.
## How To Introduce Root 5 Chords
It’s a good idea to not worry about playing root-5 bar chords until you start becoming comfortable with root-6 chords. You will find that root-5 chords aren’t that hard if you can already play root-6 chords, because after all, you will effectively just be learning a few more bar chords. You will not be learning a new technique like you are when you play bar chords for the first time.
When it’s time to focus on root-5 bar chords, it’s a good idea to play a few songs and exercises using only root-5 bar chords, just so you become familiar with the shapes and positions. Once you are comfortable with them, use them in combination with root-6 chords and open chords.
## Give It Time
The purpose of this lesson is to introduce you to bar chords by showing you how to play them, how they are used, and giving you some approaches to practising. Of course, mastering them can take many weeks (or months), so you should not rush things. This lesson is similar in approach to the lesson on reading notation – there is more in this one lesson than you can probably master in one sitting, but the concepts are relatively straight forward.
What this means is that you should read the whole lesson, take in and understand all the information and principles, but give yourself time to develop your bar chord playing.
Perhaps you are still trying to master open chords. That’s fine. You now have an understanding of bar chords and will be more prepared when it comes time to getting stuck into bar chords. Just like any of the lessons in this series, feel free to come back to this lesson as often as you need. | 3,522 | 15,422 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-26 | longest | en | 0.958888 |
https://it.mathworks.com/matlabcentral/cody/problems/55355-chain-multiplication-05 | 1,713,590,699,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817474.31/warc/CC-MAIN-20240420025340-20240420055340-00677.warc.gz | 291,095,423 | 17,849 | # Problem 55355. Chain multiplication - 05
Following up on the problem in 55305, you found the optimal way of multiplying a chain of matrices.
In problem 55315, you had to calculate the number of multiplications required based on the positions of parenthesis.
This problem is a combination of the previous two. You have to place the parenthesis in the proper places so that minimum number of multiplications are required.
For instance, array= [1,2,3,2].
So there are three matrices A, B, and C.
you can multiply in two ways - A(BC) or (AB)C.
A(BC) - requires total 16 multiplications, while (AB)C requires total 12 multiplications. So the later one is the answer.
### Solution Stats
100.0% Correct | 0.0% Incorrect
Last Solution submitted on Aug 25, 2022
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Start Hunting! | 215 | 885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-18 | latest | en | 0.878539 |
https://en.wikipedia.org/wiki/Eversion_of_the_sphere | 1,490,801,738,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190753.92/warc/CC-MAIN-20170322212950-00234-ip-10-233-31-227.ec2.internal.warc.gz | 789,331,984 | 14,324 | # Sphere eversion
(Redirected from Eversion of the sphere)
A Morin surface seen from "above"
In differential topology, sphere eversion is the process of turning a sphere inside out in a three-dimensional space. (The word eversion means "turning inside out".) Remarkably, it is possible smoothly and continuously to turn a sphere inside out in this way (with possible self-intersections) without cutting or tearing it or creating any crease. This is surprising, both to non-mathematicians and to those who understand regular homotopy, and can be regarded as a veridical paradox; that is something that, while being true, on first glance seems false.
More precisely, let
${\displaystyle f\colon S^{2}\to \mathbb {R} ^{3}}$
be the standard embedding; then there is a regular homotopy of immersions
${\displaystyle f_{t}\colon S^{2}\to \mathbb {R} ^{3}}$
such that ƒ0 = ƒ and ƒ1 = −ƒ.
## History
An existence proof for crease-free sphere eversion was first created by Stephen Smale (1957). It is difficult to visualize a particular example of such a turning, although some digital animations have been produced that make it somewhat easier. The first example was exhibited through the efforts of several mathematicians, including Arnold S. Shapiro and Bernard Morin who was blind. On the other hand, it is much easier to prove that such a "turning" exists and that is what Smale did.
Smale's graduate adviser Raoul Bott at first told Smale that the result was obviously wrong (Levy 1995). His reasoning was that the degree of the Gauss map must be preserved in such "turning"—in particular it follows that there is no such turning of S1 in R2. But the degree of the Gauss map for the embeddings f and −f in R3 are both equal to 1, and do not have opposite sign as one might incorrectly guess. The degree of the Gauss map of all immersions of S2 in R3 is 1, so there is no obstacle. The term "veridical paradox" applies perhaps more appropriately at this level: until Smale's work, there was no documented attempt to argue for or against the eversion of S2, and later efforts are in hindsight, so there never was a historical paradox associated with sphere eversion, only an appreciation of the subtleties in visualizing it by those confronting the idea for the first time.
See h-principle for further generalizations.
## Proof
Smale's original proof was indirect: he identified (regular homotopy) classes of immersions of spheres with a homotopy group of the Stiefel manifold. Since the homotopy group that corresponds to immersions of ${\displaystyle S^{2}\,}$ in ${\displaystyle \mathbb {R} ^{3}}$ vanishes, the standard embedding and the inside-out one must be regular homotopic. In principle the proof can be unwound to produce an explicit regular homotopy, but this is not easy to do.
There are several ways of producing explicit examples and beautiful mathematical visualization:
• Half-way models: these consist of very special homotopies. This is the original method, first done by Shapiro and Phillips via Boy's surface, later refined by many others. The original half-way model homotopies were constructed by hand, and worked topologically but weren't minimal. The movie created by Nelson Max, over a seven-year period, and based on Charles Pugh's chicken-wire models (subsequently stolen from the Mathematics Department at Berkeley), was a computer-graphics 'tour de force' for its time, and set the bench-mark for computer animation for many years. A more recent and definitive graphics refinement (1980s) is minimax eversions, which is a variational method, and consist of special homotopies (they are shortest paths with respect to Willmore energy). In turn, understanding behavior of Willmore energy requires understanding solutions of fourth-order partial differential equations, and so the visually beautiful and evocative images belie some very deep mathematics beyond Smale's original abstract proof.
• Thurston's corrugations: this is a topological method and generic; it takes a homotopy and perturbs it so that it becomes a regular homotopy. This is illustrated in the computer-graphics animation 'Outside In' (see link below).
• Aitchison's 'holiverse' (2010): this uses a combination of topological and geometric methods, and is specific to the actual regular homotopy between a standardly embedded 2-sphere, and the embedding with reversed orientation. This provides conceptual understanding for the process, revealed as arising from the concrete structure of the 3-dimensional projective plane and the underlying geometry of the Hopf fibration. Understanding details of these mathematical concepts is not required to conceptually appreciate the concrete eversion that arises, which in essence only requires understanding a specific embedded circle drawn on a torus in 3-space. George Francis suggested the name "holiverse", derived from the word "holistic", since (after some thought) the complete eversion can be conceptually grasped from beginning to end, without the visual aids provided by animation. In spirit, this is closer to the ideas originally suggested by Shapiro, and in practice provides a concrete proof of eversion not requiring the abstraction underlying Smale's proof. This is partially illustrated in a Povray computer-graphics animation, again easily found by searching YouTube. | 1,167 | 5,335 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-13 | latest | en | 0.962604 |
https://arkadiusz-jadczyk.eu/blog/2017/04/eshers-limit-circle-iv-rendered-complex-upper-half-plane/ | 1,726,001,804,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651318.34/warc/CC-MAIN-20240910192923-20240910222923-00664.warc.gz | 85,511,491 | 15,562 | # Esher’s Limit Circle IV rendered on the complex upper half-plane
In the recent note “SL(2,R) generators and vector fields on the half-plane” we got acquainted with two particular simple symmetries of the non-Euclidean hyperbolic geometry of the complex upper half-plane These two symmetries are horizontal translation and uniform dilation. The horizontal translational symmetry means that all geometric properties of objects are invariant when all parts of the object keep the same height (constant ), and only the coordinate is translated by a fixed amount. The symmetry with respect dilations means that when we multiply all coordinates by a certain positive number, then we should also multiply the coordinates by the same number. We obtain this way not only an object that is similar to the original, but, moreover, it also has the same size!
In Hyperbolic angels and demonswe were playing with “equal sizes” of Esher’s angels and demons on the Poincare disk:
Today we will use the Cayley transform and move the Esher’s art piece to the upper half-plane.
The exact name of the image we will be playing with is Circle Limit IV. We want to render this image on the upper half-plane using Cayley transform. The problem that we need to address is this: to render the whole circle, we would need the whole upper half-plane. But we can only show a finite part of the plane. Therefore we need to decide how to cut?
I created a little Mathematica program to help me with this decision:
I decided to use and The part of the disk that is not being used is then small enough not to worry about (on the right in the image, near on the circle boundary ). That gives the ratio 3:2 of the image. I decided to create 1440×960 pixels image. For this I had to find on the net Esher’s Limit Circle with high enough resolution. I have found it here. I cropped it to remove border and ended up with 2272×2276 image. In principle it should be a square, but few pixels should not make a difference. So, here is my Mathematica code: importing the image, converting to table, converting pixels to coordinates, making Cayley transform to the disk, converting disk coordinates to pixels, reading the color, using it for coloring the pixel on the half-plane;
And here is the end result:
All these angels have the same size, when measured with the hyperbolic stick, that was possibly created by the hyperbolic devils, that are all of the same size as well.
## 5 thoughts on “Esher’s Limit Circle IV rendered on the complex upper half-plane”
1. Bjab says:
Ark,
we talk about Cayley transform but what exactly is the Cayley transform that we talk about?
Is Cayley transform a member of a set of transforms or rather is it exactly one transform with one formula function, with exact domain?
And, if in the second case, what is that formula and its domain?
1. I would say that “Cayley transform” is one formula (or even: one idea) and its variations.
Wolfram for instance gives only this formula:
f(z)=(i-z)/(i+z)
Wikipedia, on the opther hand, is more talkative. It tells us that
“In mathematics, the Cayley transform, named after Arthur Cayley, is any of a cluster of related things.”
And then discusses several variations on the theme.
1. Bjab says:
So this is a mess:
Wolfram gives f(z)=(i-z)/(i+z)
on contrary
Wikipedia gives f(z)=(z-i)/(z+i)
and on contrary you used f(z)=i(z-i)/(z+i) or something like that for getting your “Esher’s Limit Circle IV developed onto upper half-plane.” (what is seen from the (lack of the) eyes of the “main” demon).
2. Bjab says:
I think I like your choice
[f(z)=i(z-i)/(z+i)] because it transposes 1 into 1 and -1 into -1 and I don’t have to bend my head. | 857 | 3,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-38 | latest | en | 0.911409 |
https://certifiedcalculator.com/body-weight-set-point-calculator/ | 1,716,081,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057631.79/warc/CC-MAIN-20240519005014-20240519035014-00608.warc.gz | 143,029,158 | 14,134 | # Body Weight Set Point Calculator
Maintaining a healthy body weight is crucial for overall well-being. Your body has a natural weight range it tends to maintain, known as the set point weight. Understanding your set point weight can help you make informed decisions about your health and fitness goals. To make this process easier, we have developed the Body Weight Set Point Calculator.
Formula:
The Body Weight Set Point Calculator uses a straightforward formula. It calculates your set point weight based on the assumption that your body will tend to return to its set point weight if you deviate from it. The formula is as follows:
Set Point Weight = Current Weight + (Set Point Weight – Current Weight)
How to Use:
Using the Body Weight Set Point Calculator is simple:
1. Enter your current weight in pounds in the “Current Weight” field.
2. Input your desired set point weight in pounds in the “Set Point Weight” field.
3. Click the “Calculate” button.
The calculator will process your inputs and display your estimated set point weight.
Example:
Let’s say your current weight is 160 lbs, and your desired set point weight is 140 lbs. You would enter 160 in the “Current Weight” field and 140 in the “Set Point Weight” field, then click “Calculate.” The calculator will then provide you with an estimated set point weight.
FAQs:
1. What is a body weight set point?
• The body weight set point is the weight range your body naturally maintains without constant dieting or intense exercise.
2. Why is knowing my set point weight important?
• Understanding your set point weight helps you make realistic and sustainable weight management goals.
3. Is the set point weight accurate for everyone?
• No, the set point weight varies from person to person. This calculator provides an estimate based on a general assumption.
4. Can my set point weight change over time?
• Yes, it can change due to factors like aging, lifestyle, and health conditions.
5. How often should I calculate my set point weight?
• It’s not necessary to calculate it frequently; it’s more about understanding your body’s tendencies.
6. What if my current weight is below my set point weight?
• The calculator will still provide an estimate based on the formula.
7. Is this calculator a substitute for medical advice?
• No, always consult a healthcare professional for personalized guidance.
8. How can I use my set point weight for weight management?
• It can help you set realistic goals and understand what your body considers its natural weight.
9. Is there a way to influence my set point weight?
• It may change over time with healthy habits, but it’s primarily genetically determined.
10. Is the set point weight the same as the ideal weight?
• They are related, but the ideal weight might be a personal preference, while the set point weight is more about your body’s natural balance.
Conclusion:
Understanding your body weight set point is a valuable step in your journey to better health. The Body Weight Set Point Calculator simplifies this process, providing you with an estimate of your set point weight. Remember that this estimate is based on a formula and may not be entirely accurate for everyone. Use it as a tool to set realistic and sustainable weight management goals, and always consult a healthcare professional for personalized advice. | 671 | 3,346 | {"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} | 3.09375 | 3 | CC-MAIN-2024-22 | latest | en | 0.911238 |
https://www.learnmathsonline.org/cbse-class-8-maths/cbse-class-8-mathematics-linear-equations-in-one-variable-mcq/ | 1,718,271,293,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00193.warc.gz | 765,153,272 | 17,453 | # CBSE Class 8 Mathematics/Linear Equations in One Variable/ MCQ
CBSE Class 8 Mathematics /Linear Equations in One Variable/MCQ is about the multiple choice questions that you can expect for Yearly examination. Here you can find out practice problems for Class 8 Mathematics.
CBSE Class 8 Mathematics/ Linear Equations in One Variable – Chapter 2/MCQ
Extra Questions for Practice/Multiple Choice Questions.
Choose the correct answer from the options given below:
1. Which of the following is a linear equation in one variable?
A. 3x
B. 3x + 2y
C. 3x + 2y + z
2. Solve 3(x + 2) = 6
A. 0
B. 1
C. -1
3. Solve 5p = 40
A. 5
B. 8
C. 35
4. Sum of two numbers is 120. One of the numbers is 20 more than the other. What are the numbers?
A. 50, 70
B. 60, 60
C. 80, 40
5. The sum of three consecutive multiples of 9 is 297. Find these multiples?
A. 81, 90, 99
B. 90, 99, 108
C. 99, 108, 117
6. Solve 3x – 4 = x + 2
A. 2
B. 3
C. 4
7. “5 is added to 7 times a number gives 54”Write this as an equation?
A. 5x + 7 = 54
B. 7x + 5 = 54
C. 5 – 7x = 54
8. 5 times a number gives 50. The number is ———–
A. 5
B. 10
C. 25
9. The standard form of linear equations in one variable is written as ————
A. ax + b = 0
B. ax + by = 0
C. ax + by + c = 0
10. A linear equation in one variable has ——– solution.
A. One and unique
B. Infinite solutions
C. No solution
11. The perimeter of a rectangle is 30 cm and its width is 5 cm. Find its length?
A. 5
B. 10
C. 15
12. Two numbers are in the ratio 7:4.If they differ by 18, find the numbers?
A. 12, 24
B. 42, 24
C. 36, 18
13. If 2(x + 3) = x + 18, find the value of x?
A. 10
B. 12
C. 14
14. When a number is divided by 5, the result is -2. Find the number?
A. -5/2
B. -10
C. 10
15. If the sum of two consecutive numbers is 31, form an equation to find the numbers?
A. 2x + 1 = 31
B. 2x = 31
C. 2x – 1 = 31
1. 3x
2. 0
3. 8
4. 50, 70
5. 90, 99, 108
6. 3
7. 7x + 5 = 54
8. 10
9. ax + b = 0
10. One and unique
11. 10
12. 42, 24
13. 12
14. -10
15. 2x + 1 = 31 | 820 | 1,978 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.86757 |
https://dlmf.nist.gov/search/search?q=as%20Bernoulli%20or%20Euler%20polynomials | 1,726,791,413,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652073.91/warc/CC-MAIN-20240919230146-20240920020146-00351.warc.gz | 186,510,495 | 9,256 | # as Bernoulli or Euler polynomials
(0.005 seconds)
## 1—10 of 148 matching pages
##### 1: 24.1 Special Notation
###### Bernoulli Numbers and Polynomials
The origin of the notation $B_{n}$, $B_{n}\left(x\right)$, is not clear. …
###### Euler Numbers and Polynomials
The notations $E_{n}$, $E_{n}\left(x\right)$, as defined in §24.2(ii), were used in Lucas (1891) and Nörlund (1924). …
##### 2: 24.18 Physical Applications
###### §24.18 Physical Applications
Bernoulli polynomials appear in statistical physics (Ordóñez and Driebe (1996)), in discussions of Casimir forces (Li et al. (1991)), and in a study of quark-gluon plasma (Meisinger et al. (2002)). Euler polynomials also appear in statistical physics as well as in semi-classical approximations to quantum probability distributions (Ballentine and McRae (1998)).
##### 5: 24.16 Generalizations
###### §24.16 Generalizations
For $\ell=0,1,2,\dotsc$, Bernoulli and Euler polynomials of order $\ell$ are defined respectively by …When $x=0$ they reduce to the Bernoulli and Euler numbers of order $\ell$ : …
###### §24.16(iii) Other Generalizations
In no particular order, other generalizations include: Bernoulli numbers and polynomials with arbitrary complex index (Butzer et al. (1992)); Euler numbers and polynomials with arbitrary complex index (Butzer et al. (1994)); q-analogs (Carlitz (1954a), Andrews and Foata (1980)); conjugate Bernoulli and Euler polynomials (Hauss (1997, 1998)); Bernoulli–Hurwitz numbers (Katz (1975)); poly-Bernoulli numbers (Kaneko (1997)); Universal Bernoulli numbers (Clarke (1989)); $p$-adic integer order Bernoulli numbers (Adelberg (1996)); $p$-adic $q$-Bernoulli numbers (Kim and Kim (1999)); periodic Bernoulli numbers (Berndt (1975b)); cotangent numbers (Girstmair (1990b)); Bernoulli–Carlitz numbers (Goss (1978)); Bernoulli–Padé numbers (Dilcher (2002)); Bernoulli numbers belonging to periodic functions (Urbanowicz (1988)); cyclotomic Bernoulli numbers (Girstmair (1990a)); modified Bernoulli numbers (Zagier (1998)); higher-order Bernoulli and Euler polynomials with multiple parameters (Erdélyi et al. (1953a, §§1.13.1, 1.14.1)).
##### 6: 24.13 Integrals
###### §24.13(i) BernoulliPolynomials
24.13.3 $\int_{x}^{x+(1/2)}B_{n}\left(t\right)\,\mathrm{d}t=\frac{E_{n}\left(2x\right)}% {2^{n+1}},$
###### §24.13(ii) EulerPolynomials
24.13.8 $\int_{0}^{1}E_{n}\left(t\right)\,\mathrm{d}t=-2\frac{E_{n+1}\left(0\right)}{n+% 1}=\frac{4(2^{n+2}-1)}{(n+1)(n+2)}B_{n+2},$
##### 9: 24.14 Sums
###### §24.14 Sums
24.14.4 $\sum_{k=0}^{n}{n\choose k}E_{k}E_{n-k}=-2^{n+1}E_{n+1}\left(0\right)=-2^{n+2}(% 1-2^{n+2})\frac{B_{n+2}}{n+2}.$
24.14.5 $\sum_{k=0}^{n}{n\choose k}E_{k}\left(h\right)B_{n-k}\left(x\right)=2^{n}B_{n}% \left(\tfrac{1}{2}(x+h)\right),$
Bernoulli and Euler numbers and polynomials occur in: number theory via (24.4.7), (24.4.8), and other identities involving sums of powers; the Riemann zeta function and $L$-series (§25.15, Apostol (1976), and Ireland and Rosen (1990)); arithmetic of cyclotomic fields and the classical theory of Fermat’s last theorem (Ribenboim (1979) and Washington (1997)); $p$-adic analysis (Koblitz (1984, Chapter 2)). | 1,089 | 3,153 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 55, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-38 | latest | en | 0.72185 |
http://mathhelpforum.com/advanced-algebra/54233-abstract-algebra-print.html | 1,524,770,602,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948464.28/warc/CC-MAIN-20180426183626-20180426203626-00325.warc.gz | 199,254,939 | 2,596 | # Abstract Algebra
• Oct 17th 2008, 09:36 AM
GreenandGold
Abstract Algebra
What does the order of an element in a group tell you about its centralizer? So if the order is lets say 5 what does that say about the centralizer?
• Oct 17th 2008, 10:13 AM
ThePerfectHacker
Quote:
Originally Posted by GreenandGold
What does the order of an element in a group tell you about its centralizer? So if the order is lets say 5 what does that say about the centralizer?
Remember that $\displaystyle \left< a \right> \subseteq C(a)$.
Therefore if $\displaystyle \text{ord}(a)=5$ then it means $\displaystyle |C(a)|$ is a multiple of 5. | 176 | 624 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2018-17 | latest | en | 0.87522 |
http://www.dreamincode.net/forums/topic/77516-increment-order/ | 1,444,162,903,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443736679145.29/warc/CC-MAIN-20151001215759-00014-ip-10-137-6-227.ec2.internal.warc.gz | 535,612,691 | 21,237 | increment order
Page 1 of 1
2 Replies - 991 Views - Last Post: 19 December 2008 - 10:32 AMRate Topic: //<![CDATA[ rating = new ipb.rating( 'topic_rate_', { url: 'http://www.dreamincode.net/forums/index.php?app=forums&module=ajax§ion=topics&do=rateTopic&t=77516&s=2cfc7eb93a8e70e728e92259e7b136a9&md5check=' + ipb.vars['secure_hash'], cur_rating: 0, rated: 0, allow_rate: 0, multi_rate: 1, show_rate_text: true } ); //]]>
#1 amitdhar
Reputation: 3
• Posts: 49
• Joined: 01-February 08
increment order
Posted 19 December 2008 - 09:45 AM
int a=10,c;
c=++a + a + ++a/a++ + a;
cout<<c;
the above segment theoritcaly gives 36, but practicaly it is giving 37.
Could anyone please make me understand the sequence of execution performed in the above calculation?
Is This A Good Question/Topic? 0
Replies To: increment order
#2 GWatt
Reputation: 292
• Posts: 3,092
• Joined: 01-December 05
Re: increment order
Posted 19 December 2008 - 10:13 AM
The first thing that happens is the pre-increment is evaluated.
++a happens twice which makes a 12.
second, a is divided by a which yields 1.
third the additive operators are evaluated. you have three 12's and one 1, so the answer is 37.
finally, the post-increment (a++) is evaluated bringing the value of a to 13.
This post has been edited by GWatt: 19 December 2008 - 10:14 AM
#3 Bench
• D.I.C Lover
Reputation: 859
• Posts: 2,343
• Joined: 20-August 07
Re: increment order
Posted 19 December 2008 - 10:32 AM
Note that this is all undefined behaviour. Theoretically, there is no correct value, because there is no defined evaluation order for these kinds of expressions (The outcome depends entirely on your compiler)
This post has been edited by Bench: 19 December 2008 - 10:33 AM | 544 | 1,750 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-40 | longest | en | 0.862829 |
https://voiceofwave.com/the-is-the-top-of-a-wave | 1,638,913,365,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363418.83/warc/CC-MAIN-20211207201422-20211207231422-00074.warc.gz | 644,986,347 | 9,508 | # The is the top of a wave?
Asked By: Emily Cruickshank
Date created: Mon, Jun 7, 2021 11:42 PM
Content
FAQ
Those who are looking for an answer to the question «The is the top of a wave?» often ask the following questions:
### ☑️ A micro wave wave?
Microwaves are electromagnetic waves with wavelengths longer than those of terahertz (THz) wavelengths, but relatively short for radio waves… A microwave oven works by passing microwave radiation, usually at a frequency of 2450 MHz (a wavelength of 12.24 cm), through the food.
Question from categories: infared waves radio wave microwaves waves examples microwaves waves diagram radio waves
### ☑️ Does wave integrate with wave?
Can I connect Novo to Wave? Although Novo does not have a formal integration with Wave, you can connect Novo to Wave by logging into your Wave account, navigating to Banking > Connected Accounts, and searching for Novo. From there you should be able to log into your Novo account securely and activate the connection!
### ☑️ What does wave wave mean?
literary. large quantities or groups of something, one after another: Allied planes launched wave after wave of air attacks on the city.
Crest
We've handpicked 20 related questions for you, similar to «The is the top of a wave?» so you can surely find the answer!
### Which wave does most damage p wave or s wave?
A p wave because it can move so fast with more force. P waves cause little damage, they are compressional waves. They are at a higher frequency and a lower amplitude than surface waves which are shear waves. S-waves cause the most damage because they are slow moving and have an amplitude much greater than P-waves.
### A gravitational wave is a ___________ wave?
A gravitational wave is an invisible (yet incredibly fast) ripple in space. Gravitational waves travel at the speed of light (186,000 miles per second). These waves squeeze and stretch anything in their path as they pass by. A gravitational wave is an invisible (yet incredibly fast) ripple in space.
### A plane wave is a wave?
• A plane wave is a wave that propagates through space as a plane, rather than as a sphere…
### Are electromagnet wave is radio wave?
yes, electromagnetic wave is a radio wave
ewan k0e !
No....
### Does wave type affect wave velocity?
• The data convincingly show that wave frequency does not affect wave speed. An increase in wave frequency caused a decrease in wavelength while the wave speed remained constant. The last three trials involved the same procedure with a different rope tension.
### Does wave type effect wave velocity?
Both Wave A and Wave B travel at the same speed. The speed of a wave is only altered by alterations in the properties of the medium through which it travels… The wavelength of a wave does not affect the speed at which the wave travels. Both Wave C and Wave D travel at the same speed.
### How does the ocean wave wave?
gravity of course!
### Incoming wave is the incident wave?
The incoming wave is called the incident wave.
### Is a transverse wave light wave?
it sure is!!yes, light waves is a transverse wave.
### Is electromagnetic wave a longitudinal wave?
Electromagnetic waves are transverse waves… Only transverse waves can be polarized, because in a longitudinal wave, the disturbance is always parallel to the direction of wave propagation. So, you can classify electromagnetic waves based on the direction of disturbance in them (polarization).
### Is light wave a periodic wave?
Light, and other electromagnetic waves, do not require a medium; we'll deal with those later in the semester. Waves can be broadly separated into pulses and periodic waves. A pulse is a single disturbance while a periodic wave is a continually oscillating motion.
### Is light wave an electromagnetic wave?
The terms light, electromagnetic waves, and radiation all refer to the same physical phenomenon: electromagnetic energy… Radio and microwaves are usually described in terms of frequency (Hertz), infrared and visible light in terms of wavelength (meters), and x-rays and gamma rays in terms of energy (electron volts).
### Is ocean wave a mechanical wave?
Ocean waves and water waves are mechanical waves (because they move through the medium of water).
### Is primary wave the fastest wave?
i think so.. :/ Travel fastest through rock material causing rock particles in the rock to move back or forth
### Is radio wave a light wave?
No, it is actually a microwave wave.
### Is radio wave an electromagnetic wave?
Radio waves have the longest wavelengths in the electromagnetic spectrum. They range from the length of a football to larger than our planet… The radio "receives" these electromagnetic radio waves and converts them to mechanical vibrations in the speaker to create the sound waves you can hear. | 1,013 | 4,825 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.934989 |
https://numberworld.info/root-of-685 | 1,628,150,165,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00236.warc.gz | 438,234,939 | 2,965 | # Root of 685
#### [Root of six hundred eighty-five]
square root
26.1725
cube root
8.8152
fourth root
5.1159
fifth root
3.6909
In mathematics extracting a root is known as the determination of the unknown "x" in the following equation $y=x^n$ The result of the extraction of the root is known as a so-called root. In the case of "n is equivalent to 2", one talks about a square root or sometimes a second root also, another possibility could be that n = 3 at that time one would consider it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 685 is represented as this: $$\sqrt[]{685}=26.172504656605$$
Also it is possible to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 685 is 26.172504656605. The cube root of 685 is 8.815159819493. The fourth root of 685 is 5.1159070218882 and the fifth root is 3.6909495945887.
Look Up | 283 | 976 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-31 | latest | en | 0.932147 |
https://schoolencasa.com/2016/03/08/this-weeks-plans/?lang=en | 1,580,229,881,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251779833.86/warc/CC-MAIN-20200128153713-20200128183713-00263.warc.gz | 649,501,208 | 15,668 | # This Week´s Plans!
Share It!
I’ve found a moment to share what we’re going to be studying this week…although you should know that our plans almost always change, and these are just the “responsibilities” that I’ve decided we need to work on this week. The children will also be choosing their own work to go along with this.
Here are the sheets I’ll give to them tomorrow..although of course Miss Braveheart doesn’t care at all about hers, but it helps me to have one for her, too! The plan are in Spanish; I’ll give you the translation below!
Mr. Scientist is seven and a half, in second grade:
Miss Adventuress is five, in preK:
And Miss Braveheart is 21 months.
And here is the English translation of the plans shown in the images! If you want to see the activities as we do them, you can on my Tumblr!
Mr. Scientist (7.5 years old, second grade):
• presentation: winds, with experiments
• polygons game
• paint habitat map
• presentation: lines
• presentation: angles
• presentation: antonyms
• presentation: using the large bead frame paper
• presentation: fraction equivalence with circle insets
• practice handwriting
• presentation: the leaf
• copy and memorize Bible verse(s) of the week
• presentation: circles
• First Great Lesson: God with No Hands
• presentation: time zones
• make hierarchical material together
• longitude and latitude mysteries
• choose a continent to study
• presentation: Roman numerals
• continue sewing his quilt
• make a box for Bible verses/passages memorized
• chemistry set
• work on arithmetic tables (math facts)
• Lent footprints
• make a meal plan for the week
• make geometric art with the metal insets
• guess the letter
• division with racks and tubes
Miss Adventuress (5 years old, pre-K)
• subtraction with golden beads, under 1000
• presentation: walking the line
• presentation: drawing contours
• sew a pillow for her doll
• presentation: paper for the large bead frame
• presentation: brown stair (pieces on table across room)
• presentation: constructive triangles box 2
• build binomial and trinomial cubes
• copy and practice Bible verse
• sandpaper tablets with blindfold
• writing practice with the salt tray
• presentation: geometric solids
• cut out continent shapes in construction paper to cover continents on globe
• South America puzzle
• South American flags
• art lesson
• Lent booklet
• snake reading game with two-syllable words
• presentation: fractions
• South American animals
• build words with moveable alphabet
• make South America folder
• make geometric design art using metal insets
Miss Braveheart (21 months)
• make playdough
• brown stair work
• put discs in slot on coffee can lid
• cylinder blocks
• three-solids shape sorter
• pour water into glasses
• basket of red and blue objects
• open and close containers
• rough and smooth boards
• build the tower
• build with Duplo legos
• puzzles
• cleaning
• wash hands in a bowl
• name animals
Linking up with a Peek at my Week party at these blogs:
Esta entrada también está disponible en: Spanish
Share It!
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https://www.pokerstrategy.com/forum/thread.php?threadid=142864 | 1,529,660,483,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00537.warc.gz | 891,599,708 | 20,871 | • Bronze
Joined: 15.07.2011
How exactly can you commonly calculate the expected value (EV) during a hand? I have read different answers on numerous websites but I just can't seem to get my head round it.
Would anybody be kind enough to explain it in dummy terms please?
Thanks.
• 11 replies
Joined: 19.04.2006
he ytsoP,
unfortunately there won't be an easy solution to calculate the EV.
Why? Let's assume you hold AA, your opponent holds KK and the board is AA8. You will win 100% of a time, so your EV (expected value) is 100% (of the Pot ). This should be somehow obvious since your opponent can not beat you.
But in reality we usually face two problems:
• we do not know what hand our opponent has
• we thus usually do not know if we are ahead or behind, do not know exactly what cards are good for us on turn and river
What does a good poker player do now ?
• He learns a lot about how certain hands perform. For example do you know how many outs an open ended straight draw has, or a flushdraw? You should therefore read and understand Mathematics of Poker: Odds and Outs . This will enable you to calculate how often your hand will improve if on a draw.
• He learns to assume what hands his oponent could possibly have. The combination of all hands is called a 'range' and determines your course of actions.
Explaining with an Example:
Let us go trough the following hand to demonstrate how you can define an opponents range. I played this hand in my yesterday's german coaching, I got rid of all unimportend information:
[quote]
NL \$0.02/\$0.05
Hero (\$7.07)
BB (\$2.07)
UTG+1 (\$4)
Dealt to Hero K A
UTG calls \$0.05, UTG+1 raises to \$0.20, fold, fold, fold, Hero raises to \$0.60, fold, fold, UTG+1 calls \$0.40
[COLOR=green][b]I head no specific read on this oponent, so we have to go with general assumptions. According to our Starting hand chart AK is a reraise here because we think better hands could fold (pocket pairs who indeed are in front =>they have a pair,we don't) but also worse hands like AJs, KQs could call.
We now have to consider what type of hand UTG+1 raises preflop = define his range. I assumed that he openraises every pocket pair bigger then 77, means 88+. Also he will openraise any Ace with a kicker better then 9 = means AT+. In addition he will raise some other hands like KQ,KJ,76s+.
[/color][COLOR=blue]His raisingrange is: 88+,KQ,KJ,76s,AT+[/color][COLOR=green]
But the situation changes: I reraise. I assume that his range changes now because he will go allin with AA,KK. However he will fold a lot of hands: let's say he folds 88,76s+,KJ,AT. His range is now narrowed down to 99+,AJ+ and KQ.
[/color][COLOR=blue]His callingrange to our reraise is: 99-QQ,KQ,AJ+[/color]
We are ahead of his range with 60% to 40%. How did I calculate that? Have a look at what our great software, the PokerStrategy.com Equilab does for me:[/b]
FLOP (\$1.30) K 9 A
[COLOR=green][b]Now I hit a great flop. Just bet out big and hope you get a call? NO! We now have to take the very same range that we made up before: QQ-99,AJs+,KQs,AJo+,KQo and see how our opponent performs now: we are ahead 85%. But it also means that our opponent did not hit that big. He will have no flushdraws in his range (AJs+ and KQs can not have a Flushdraw since they the A and K for the Flushdraw are already on the board), a lot of Ax with top pair but also hands like TT,JJ and QQ that would probably fold. I could now bet pretty small to get calls out of all hands of his range, which I actually decided to do - so his range does not change.
[/color][COLOR=blue]His range stays: 99+,KQ,AJ+[/color] [/b]
Hero bets \$0.75, UTG+1 calls \$0.75
TURN (\$2.80) K 9 A 2
[COLOR=green][b]The turn card let's the Flushdraw arrive, but hey, we are sure he did not had one. But a fourth club could certainly hit on the river and we now have to commit us, bring in most of the money. Unfortunately this means that hands like TT,JJ,QQ will fold to our bet, even KQ will be likely to fold. Therefore if we bet the hands that will continue are now only: 99,AJ+.
[/color][COLOR=blue]His calling- and maybe Allin-range is: 99, AJ+[/color]
Good news, we still have 76% against this range. If we knew our oponent had only 99 we would only have 9%, but again: he plays a lot of hands like seen to this point so we should always assume a reasonable range. We bet and our oponent goes allin, we just have to bring another 79cent and think that he will do it with any ace,too - so 'easy call'
[/b]
Hero bets \$1.86, UTG+1 raises to \$2.65 (AI), Hero calls \$0.79
RIVER (\$8.10) K 9 A 2 8
UTG+1 shows Q A
(Pre 25%, Flop 2.2%, Turn 0.0%)
Hero shows K A
(Pre 75%, Flop 97.8%, Turn 100.0%)
Hero wins \$7.71
[COLOR=green]We see here that we in fact had the oponent crushed troughout the hand. However I am pretty certain that he would have played 99 or AK similar to the actual hand. The hand he showed was indeed included in our calculations![/COLOR]
[/quote]Now what is your course of action?
• try the PokerStrategy.com Equilab, try the hand analysis with ranges
• confirm your ranges => you can do it by two courses of action:
• a) have a look if the hands that show up at showdown are included in your calculations
• b) use our hand evaluation forum to confirm with our hand judges if your assumptions are right
• read our articles and make odds&outs your 2nd nature. You should and will know as a good poker player that an Open ended Straight and Flush draw has 9(for the flush)+6(for the straight,2 outs already counted for the flush)=15 outs
• ask in the forum if you are unsure about anything poker related. There are no stupid questions, just stupid answers.
• use the equity trainer, it is part of the Equilab
• Bronze
Joined: 11.07.2010
Awesome post SvenBe thanks
• Bronze
Joined: 31.12.2010
Really great explanation !! thankyou for posting this
Roo
• Bronze
Joined: 06.10.2008
Originally posted by matel17
Awesome post SvenBe thanks
+1
Added to favorites so we can direct others to it in future.
• Bronze
Joined: 30.01.2010
• Bronze
Joined: 06.12.2009
great post, thanks
• Bronze
Joined: 02.03.2011
wow! amazing post Sven! Thank you for helping out!
• Bronze
Joined: 06.12.2009
I re- read it again this morning and learnt so much from it, it should be turned into an article
• Bronze
Joined: 01.06.2011
Great post SvenBe.
This should be made sticky or as Fishlock says turned into an article.
Its too good to get lost amongst all the other posts.
SvenBe can we request that you post more of these little treasures soon???????
If so look forward to reading them.
Great work and hope I dont see you at the tables LOL
Joined: 19.04.2006
he guys - guess what - I am just playing NL25 myself. It is not a rocket science to review your game - just use the free tools you have at PokerStrategy.com.
If you don't have a clue about equity 'n stuff - participate in the Beginner Course and you will learn it all.
Also Articles, Hand evaluations, coachings and videos will surely help you to understand all those topics.
Maybe I have a task for YOU guys!
Try to analyse a hand the way I did it - and post it in the hand evaluation forum (+crosslink it here) - veriz and all other hand judges will surely help you in finding out the proper ranges.
I'll promise I'll care about a news to publish the best evaluations once we gathered a decent amount of evaluations here...Deal?
• Bronze
Joined: 25.03.2011
yeah,i guess you can say it's a good post,but that's explaining only a part of the whole equation,namely your equity.
it's a wee bit more complicated.
I guess range assumptions are ok-ish for NL 10,but you assume villain is never calling a 3bet with a speculative hand OOP (remember that fishes love SC type hands).Since you assume he is a fish,because no SS reg or even wSS will call 3bet OOP with that wide of a range and will never open to min in UTG with the same range.
SO, WHAT YOU WANT TO DO IS:
determine a range for the opponent.remember,you must take into account all information available and also,if you make a wrong assumption about his range,you must fix this as fast as you can or it will lead to disastrous decisions from your part.
NOW you can calculate the equity of your hand against a specific part of your opponents range (bluffs,semibluffs,mediums,monsters).
then make a decision based on the sum of all possibilities.
Formula:
EV = Pfold * Pot + ( 1 – Pfold ) * ( Equity * ( Pot + Bet ) - ( 1- Equity ) * Bet )
where
- Pfold is the likelihood of your opponent folding,
- Pot,the potsize WITHOUT your input,
- equity,your share of the pot
to estimate profitability,EV=0
Pfold,in a rough,general sense is Bet/Pot+Bet or in percentages Bet%(Bet% +1) with Bet%=Bet/Pot
there are situations where will your equity will be zero or Pfold will be zero.
using zeroes as general values are good for exercising
any further discussions generated will be appreciated.any corrections and feedback also. | 2,385 | 8,936 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-26 | latest | en | 0.951876 |
https://www.e-iceblue.com/Tutorials/Python/Spire.XLS-for-Python/Program-Guide/Formula/Python-Add-or-Read-Formulas-in-Excel.html | 1,708,483,510,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473360.9/warc/CC-MAIN-20240221002544-20240221032544-00268.warc.gz | 795,818,454 | 19,542 | Formulas in Excel are equations or expressions that perform calculations on data within a spreadsheet. They allow you to perform basic arithmetic operations like addition, subtraction, multiplication, and division, as well as more advanced functions like statistical analysis, date and time calculations, and logical evaluations. By incorporating formulas into your Excel spreadsheets, you can save time, eliminate errors, and gain valuable insights from your data. In this article, we will demonstrate how to add or read formulas in Excel in Python using Spire.XLS for Python.
Install Spire.XLS for Python
This scenario requires Spire.XLS for Python and plum-dispatch v1.7.4. They can be easily installed in your VS Code through the following pip command.
`pip install Spire.XLS`
If you are unsure how to install, please refer to this tutorial: How to Install Spire.XLS for Python in VS Code
Add Formulas to Excel in Python
Spire.XLS for Python offers the Worksheet.Range[rowIndex, columnIndex].Formula property to add formulas to specific cells in an Excel worksheet. The detailed steps are as follows.
• Create an object of the Workbook class.
• Get a specific worksheet by its index using the Workbook.Worksheets[sheetIndex] property.
• Add some text and numeric data to specific cells of the worksheet using the Worksheet.Range[rowIndex, columnIndex].Text and Worksheet.Range[rowIndex, columnIndex].NumberValue properties.
• Add text and formulas to specific cells of the worksheet using the Worksheet.Range[rowIndex, columnIndex].Text and Worksheet.Range[rowIndex, columnIndex].Formula properties.
• Save the result file using the Workbook.SaveToFile() method.
• Python
```from spire.xls import *
from spire.xls.common import *
# Create an object of the Workbook class
workbook = Workbook()
# Get the first worksheet
sheet = workbook.Worksheets[0]
# Declare two variables: currentRow, currentFormula
currentRow = 1
currentFormula = ""
# Add text to the worksheet and set cell style
sheet.Range[currentRow, 1].Text = "Test Data:"
sheet.Range[currentRow, 1].Style.Font.IsBold = True
sheet.Range[currentRow, 1].Style.FillPattern = ExcelPatternType.Solid
sheet.Range[currentRow, 1].Style.KnownColor = ExcelColors.LightGreen1
sheet.Range[currentRow, 1].Style.Borders[BordersLineType.EdgeBottom].LineStyle = LineStyleType.Medium
currentRow += 1
# Add some numeric data to the worksheet
sheet.Range[currentRow, 1].NumberValue = 7.3
sheet.Range[currentRow, 2].NumberValue = 5
sheet.Range[currentRow, 3].NumberValue = 8.2
sheet.Range[currentRow, 4].NumberValue = 4
sheet.Range[currentRow, 5].NumberValue = 3
sheet.Range[currentRow, 6].NumberValue = 11.3
currentRow += 2
# Add text to the worksheet and set cell style
sheet.Range[currentRow, 1].Text = "Formulas"
sheet.Range[currentRow, 2].Text = "Results"
sheet.Range[currentRow, 1, currentRow, 2].Style.Font.IsBold = True
sheet.Range[currentRow, 1, currentRow, 2].Style.KnownColor = ExcelColors.LightGreen1
sheet.Range[currentRow, 1, currentRow, 2].Style.FillPattern = ExcelPatternType.Solid
sheet.Range[currentRow, 1, currentRow, 2].Style.Borders[BordersLineType.EdgeBottom].LineStyle = LineStyleType.Medium
currentRow += 1
# Add text and formulas to the worksheet
# Str
currentFormula = "=\"Hello\""
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Int
currentFormula = "=300"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Float
currentFormula = "=3389.639421"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Bool
currentFormula = "=false"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Expressions
currentFormula = "=1+2+3+4+5-6-7+8-9"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
currentFormula = "=33*3/4-2+10"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Cell reference
currentFormula = "=Sheet1!\$B\$2"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Functions
# AVERAGE
currentFormula = "=AVERAGE(Sheet1!\$D\$2:F\$2)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# COUNT
currentFormula = "=COUNT(3,5,8,10,2,34)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# NOW
currentFormula = "=NOW()"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
sheet.Range[currentRow, 2].Style.NumberFormat = "yyyy-MM-DD"
currentRow += 1
# SECOND
currentFormula = "=SECOND(0.503)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# MINUTE
currentFormula = "=MINUTE(0.78125)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# MONTH
currentFormula = "=MONTH(9)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# DAY
currentFormula = "=DAY(10)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# TIME
currentFormula = "=TIME(4,5,7)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# DATE
currentFormula = "=DATE(6,4,2)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# RAND
currentFormula = "=RAND()"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# HOUR
currentFormula = "=HOUR(0.5)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# MOD
currentFormula = "=MOD(5,3)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# WEEKDAY
currentFormula = "=WEEKDAY(3)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# YEAR
currentFormula = "=YEAR(23)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# NOT
currentFormula = "=NOT(true)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# OR
currentFormula = "=OR(true)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# AND
currentFormula = "=AND(TRUE)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# VALUE
currentFormula = "=VALUE(30)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# LEN
currentFormula = "=LEN(\"world\")"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# MID
currentFormula = "=MID(\"world\",4,2)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# ROUND
currentFormula = "=ROUND(7,3)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# SIGN
currentFormula = "=SIGN(4)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# INT
currentFormula = "=INT(200)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# ABS
currentFormula = "=ABS(-1.21)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# LN
currentFormula = "=LN(15)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# EXP
currentFormula = "=EXP(20)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# SQRT
currentFormula = "=SQRT(40)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# PI
currentFormula = "=PI()"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# COS
currentFormula = "=COS(9)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# SIN
currentFormula = "=SIN(45)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# MAX
currentFormula = "=MAX(10,30)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# MIN
currentFormula = "=MIN(5,7)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# AVERAGE
currentFormula = "=AVERAGE(12,45)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# SUM
currentFormula = "=SUM(18,29)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# IF
currentFormula = "=IF(4,2,2)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# SUBTOTAL
currentFormula = "=SUBTOTAL(3,Sheet1!A2:F2)"
sheet.Range[currentRow, 1].Text = "'" + currentFormula
sheet.Range[currentRow, 2].Formula = currentFormula
currentRow += 1
# Set width of the 1st, 2nd and 3rd columns
sheet.SetColumnWidth(1, 32)
sheet.SetColumnWidth(2, 16)
sheet.SetColumnWidth(3, 16)
# Create a cell style
# Set the horizontal alignment as left
style.HorizontalAlignment = HorizontalAlignType.Left
# Apply the style to the worksheet
sheet.ApplyStyle(style)
# Save the result file
workbook.Dispose()```
Read Formulas in Excel in Python
To read formulas in an Excel worksheet, you need to loop through all the cells in the worksheet, after that, find the cells containing formulas using the Cell.HasFormula property, and then get the formulas of the cells using the CellRange.Formula property. The detailed steps are as follows.
• Create an object of the Workbook class.
• Get a specific worksheet by its index using the Workbook.Worksheets[sheetIndex] property.
• Get the used range of the worksheet using the Worksheet.AllocatedRange property.
• Create an empty list.
• Loop through all the cells in the used range.
• Find the cells containing formulas using the Cell.HasFormula property.
• Get the names and the formulas of the cells using the CellRange.RangeAddressLocal and CellRange.Formula properties.
• Append the cell names and formulas to the list.
• Write the items in the list into a text file.
• Python
```from spire.xls import *
from spire.xls.common import *
# Create an object of the Workbook class
workbook = Workbook()
# Get the first worksheet
sheet = workbook.Worksheets[0]
# Get the used range of the worksheet
usedRange = sheet.AllocatedRange
# Create an empty list
list = []
# Loop through the cells in the used range
for cell in usedRange:
# Check if the cell has a formula
if(cell.HasFormula):
# Get the cell name
# Get the formula
formula = cell.Formula
# Append the cell name and formula to the list
list.append(cellName + " has a formula: " + formula)
# Write the items in the list into a text file
with open("Formulas.txt", "w", encoding = "utf-8") as text_file:
for item in list:
text_file.write(item + "\n")
workbook.Dispose()``` | 3,294 | 12,300 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-10 | longest | en | 0.805631 |
https://www.coursehero.com/file/5636875/Homework-17/ | 1,490,478,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189083.86/warc/CC-MAIN-20170322212949-00130-ip-10-233-31-227.ec2.internal.warc.gz | 897,745,610 | 28,598 | Homework 17
# Homework 17 - homework 17 BAUTISTA, ALDO Due: Mar 3 2006,...
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1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. A particle moving along the x -axis has a potential energy U ( x ), as shown in the ac- companying graph. Scale factors: a = 4 . 6 m, b = 28 J. x(meter) b 2b 3b 4b a 2a 3a 4a 5a 6a U(x) (joules) A B C What is the force exerted on the particle when x = 3 a (point A , the lowest point on the curve)? Correct answer: 0 N (tolerance ± 1 %). Explanation: From the de±nition of Potential Energy, we know F = - d U dx . In other words, the force on a particle at any particular point is given by the negative of the slope of the potential energy function at that point. At x = 3 a , the potential energy has a min- imum; the slope of the function is zero at that point. Therefore there are no forces on a par- ticle at that point. ( x = 3 a is an equilibrium point.) Question 2 Part 2 of 2. 10 points. If the particle has a mass of 8 . 8 kg and is released from rest at x = 5 a , what is the particle’s velocity when it reaches x = a ? Correct answer: 2 . 52262 m / s (tolerance ± 1 %). Explanation: Apply Conservation of Energy: U init + K init = U final + K final U B + 0 = U C + 1 2 m v 2 Solving for v gives v = r 2 m ( U B - U C ) = r 2 m (3 b - 2 b ) = r 2 m b = s 2 (8 . 8 kg) (28 J) = 2 . 52262 m / s . Question 3 Part 1 of 1. 10 points. A 4 kg block moving along the x axis is acted upon by a single horizontal force that varies with the block’s position according to the equation F x = a x 2 + b , where a = 7 N / m 2 , and b = - 2 . 9 N. At 1 . 3 m, the block is moving to the right with a speed of 3 . 4 m / s. Determine the speed of the block at 2 . 4 m. Correct answer: 4 . 85076 m / s (tolerance ± 1 %). Explanation: The change in the potential energy is Δ U = U f - U i = - Z f i F x dx = - Z f i ( a x 2 + b ) d x = ± 1 3 a x 3 i + b x i ² - ± 1 3 a x 3 f + b x f ² = 23 . 9397 J . From the conservation of energy,
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## This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
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Homework 17 - homework 17 BAUTISTA, ALDO Due: Mar 3 2006,...
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Ask a homework question - tutors are online | 857 | 2,627 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2017-13 | longest | en | 0.842114 |
https://www.mail-archive.com/gimp-developer@lists.xcf.berkeley.edu/msg01047.html | 1,477,526,911,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721008.78/warc/CC-MAIN-20161020183841-00274-ip-10-171-6-4.ec2.internal.warc.gz | 963,727,542 | 5,093 | # Re: [Gimp-developer] Thin lines using pencil
On Wed, 29 Aug 2001, Simon Budig wrote:
> Sven Neumann ([EMAIL PROTECTED]) wrote:
> > Instead of drawing brush pixmaps onto the canvas at
> > equidistant spots along the line as the paintbrush does, the pencil
> > tool could use a real line-drawing algorithm (Bresenham). This would
> > imply that our brushes couldn't be used with the pencil tool any
> > longer since this algorithm would only work for rounded or square
> > pencil tips (or am I wrong here?).
>
> I think you are wrong there :-)
>
> You could calculate the intermediate points with Bresenham and then
> place the brushes on these points only (or a subset, according to the
> spacing). So the Pencil would place the brushes always at integer
> positions which results in perfect lines for 1-pixel brushes and
> usable lines with ornamental brushes.
>
> So if we have a 10-pixel wide brush with a spacing of 50 you could
> place it on every fifth pixel of the bresenham-calculated coordinates.
>
> Does this sound good?
Thank You guys. It seems drawing a good thin non-antialiased line is so
foar not implemented, I hope it will be. I could do it, but it would take
a long time for me to understand code and the whole conception of GIMP. By
now, I have the following idea to do it:
[c-like pseudocode]
typedef int (*tool_t)(double x, double y, double intensity, ...);
/* uses tool on x,y
for pencil: puts a single pixel, the nearest to (x,y)
i.e. at ( (int)(x+0.5), (int)(y+0.5) )
e.g. for (26.25,123.75) it's (26,124)
for paintbrush: alters four pixel nearest to x,y, with intensity
negative-correlated to distance from (x,y)
i.e. altered are pixels:
((int)x,(int)y) with intensity (1-fx)*(1-fy)*intensity
((int)x+1,(int)y) with intensity fx*(1-fy)*intensity
((int)x,(int)y+1) with intensity (1-fx)*fy*intensity
((int)x+1,(int)y+1) with intensity fx*fy*intensity
where fx=x-(int)x, fy=y-(int)y
(these are fractional parts of x,y)
(for paintbrush, if intensity>1.0, more adjacent pixels may be
altered)
*/
void draw_slash_line_using_a_tool(int x1, int y1, int x2, int y2, tool_t
tool, double intensity, ...)
{
assume: x1!=x2, y1!=y2 (in these cases line is not slash)
assume: x1<x2, y1<y2, x2-x1>=y2-y1 (all other seven possible cases are
convertible to this by reasonable replacing +/-, x/y, 1/2)
int dx=x2-x1;
int dy=y2-y1;
int x;
for (x=x1; x<=x2; x++) {
tool(x,y1+x*((double)dy/dx),intensity,...);
}
/* x2-x1+1 tool hits, all of them with integer x values and
not-necessarily-integer y values, lying along line,
more precisely on crosings of line and "meridians" of grid */
}
As you see, it's universal, and would draw line with any tool, if this
tool is correctly implemented to handle non-integer coordinates.
There could be problem if brush size is big and line has to be
half-transparent. E.g. brush radius is 10 and intensity is 0.5. There
would be 1-pixel-close overlapping brush hits, whose intensities would
accumulate. But this problem can easily be compensated if we use more
sophiscated tool_t function, which would reasonably decrease its intensity
(simply dividing nominal intensity by radius size and then by sine/cosine
value of line angle) for all pixels except two ends, and these two ends
would be painted with full nominal intensity, but only half of brush would
be painted (this half which does not overlap with any mid-line brush hit).
How do you all feel about it? Is it a good idea, or not? I think you will
be more able to implement it and insert in proper place in GIMP's source.
If would use a lot of time to analyze sources (in fact, I would have to
learn how to use GIMP sources and thouroughly study it) and find this
proper place, and although this I would however have doubts if this is
correct place and if it does not conflict with GIMP's internal
architecture, assumptions and whatever.
If you have a bit of time to do it, do it. If no, tell me, I will however
try to do it for myself and I will send you results of my work, and I hope
this will be useful for you and will not make incompatibility forks.
_____________
Grzes \_______________________|_______________________/
[EMAIL PROTECTED] (_)
http://gnu.univ.gda.pl/~grzes/
'Nam chodzi o odglos paszczowo.' 'Patataj patataj patata...'
<annoy echelon> fbi militia bomb action president mossad delta force
militia action anthrax operation fbi codes top secret </annoy echelon>
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### Author Topic: Digital Target score (Read 2337 times)
0 Members and 1 Guest are viewing this topic.
#### cheerio
• Sr. Member
• Posts: 306
##### Digital Target score
« on: April 08, 2012, 19:52:42 PM »
Hi,
i shoot Airguns and stuff. I was wondering how i could count the target score with a circuit. While searching the web i found this:
Does anyone have an idea how they do this?
The projectile travels 0.15mm/us on impact.
#### kam
• Hero Member
• Posts: 1849
##### Re: Digital Target score
« Reply #1 on: April 09, 2012, 08:13:49 AM »
this target detects and measures the position where you shoot with the rifle and then tells you the score? Or detects the shooting projectile when moving?
#### cheerio
• Sr. Member
• Posts: 306
##### Re: Digital Target score
« Reply #2 on: April 09, 2012, 11:26:55 AM »
It must detect the projectile while moving. Because otherwise it won't detect an impact on the same location twice.
edit:
they use 3 sensors. so it must be a triangular positioning algorithm. But if they use laser distance measurement it would be pretty complex. isn't there a better/cheaper/simpler way?
« Last Edit: April 09, 2012, 19:59:35 PM by cheerio »
#### kam
• Hero Member
• Posts: 1849
##### Re: Digital Target score
« Reply #3 on: April 10, 2012, 21:53:08 PM »
i cannot even think of other way, nor could i believe that there could be a way to do it with lasers. I mean, they must be bloody fast to scan the whole area for the projectile!
#### cheerio
• Sr. Member
• Posts: 306
##### Re: Digital Target score
« Reply #4 on: April 11, 2012, 02:40:43 AM »
#### George
• Jr. Member
• Posts: 73
##### Re: Digital Target score
« Reply #5 on: April 12, 2012, 13:20:01 PM »
Each laser unit will have a detector so that each particular laser will be able to detect a penetration of it's beam and the distance to the point of penetration.
http://en.wikipedia.org/wiki/Laser_rangefinder
Each laser will scan an arc that covers the target area.
The by trigonometry, the location of where the projectile penetrated the three beams can be calculated. Basically, knowing the location of the three lasers equals three circles, where they intersect, is the spot where the projectile passed.
http://en.wikipedia.org/wiki/Trilateration
http://en.wikipedia.org/wiki/Multilateration
got this from a photonics dude, hope it helps
Cheers | 671 | 2,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-43 | longest | en | 0.754266 |
https://stats.libretexts.org/Courses/Saint_Mary's_College%2C_Notre_Dame/MATH_345__-_Probability_(Kuter)/3%3A_Discrete_Random_Variables/3.1%3A_Introduction_to_Random_Variables | 1,579,963,835,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251672537.90/warc/CC-MAIN-20200125131641-20200125160641-00389.warc.gz | 663,775,233 | 20,571 | 3.1: Introduction to Random Variables
Now that we have formally defined probability and the underlying structure, we add another layer: random variables. Random variables allow characterization of outcomes, so that we do not need to focus on each outcome specifically. We begin with the formal definition.
Definition $$\PageIndex{1}$$
A random variable is a function from a sample space $$S$$ to the real numbers $$\mathbb{R}$$. We denote random variables with capital letters, e.g., $$X: S \rightarrow \mathbb{R}.\notag$$
Informally, a random variable assigns numbers to outcomes in the sample space. So, instead of focusing on the outcomes themselves, we highlight a specific characteristic of the outcomes.
Example $$\PageIndex{1}$$
Consider again the context of Example 1.1.1, where we recorded the sequence of heads and tails in two tosses of a fair coin. The sample space for this random experiment is given by
$$S = \{hh, ht, th, tt\}.\notag$$
Suppose we are only interested in tosses that result in heads. We can define a random variable $$X$$ that tracks the number of heads obtained in an outcome. So, if outcome $$hh$$ is obtained, then $$X$$ will equal 2. Formally, we denote this as follows:
\begin{align*}
X: S & \rightarrow \mathbb{R} \\
s & \mapsto\ \text{number of}\ h\text{'s in}\ s
\end{align*}
Since there are only four outcomes in $$S$$, we can list the value of $$X$$ for each outcome individually:
\begin{align*}
\text{inputs:}\ S\ &\stackrel{\text{function:}\ X}{\longrightarrow}\ \text{outputs:}\ \mathbb{R} \\
\end{align*}
We can also write the above as follows:
$$X(hh) = 2,\quad X(ht) = X(th) = 1,\quad X(tt) = 0.\notag$$
The advantage to defining the random variable $$X$$ in this context is that the two outcomes $$ht$$ and $$th$$ are both assigned a value of $$1$$, meaning we are not focused on the actual sequence of heads and tails that resulted in obtaining one heads.
In Example 3.1.1, note that the random variable we defined only equals one of three possible values: $${0, 1, 2}$$. This is an example of what we call a discrete random variable. We will also encounter another type of random variable: continuous. The next definitions make precise what we mean by these two types.
Definition $$\PageIndex{2}$$
A discrete random variable is a random variable that has only a finite or countably infinite (think integers or whole numbers) number of possible values.
Definition $$\PageIndex{3}$$
A continuous random variable is a random variable with infinitely many possible values (think an interval of real numbers, e.g., $$[0,1]$$).
In this chapter, we take a closer look at discrete random variables, then in Chapter 4 we consider continuous random variables. | 687 | 2,716 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2020-05 | latest | en | 0.807869 |
https://www.coursehero.com/file/6787211/UCR-Econ-103-Hwk-3Answers/ | 1,529,622,051,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864300.98/warc/CC-MAIN-20180621211603-20180621231603-00076.warc.gz | 781,441,785 | 99,806 | # UCR--Econ 103--Hwk 3+Answers - Econ 103 Hwk 3(100 points...
Econ 103 Hwk 3 (100 points) Name: __________________________ TA Name: Due Date: Wednesday June 1 st . Place: TA office / drop box (By 7:00 pm) Multiple Choice (3 pts each) Use the following to answer questions 1-2: Exhibit: IS-LM Fiscal Policy 1. (Exhibit: IS-LM Fiscal Policy) Based on the graph, starting from equilibrium at interest rate r 1 and income Y 1 , a decrease in government spending would generate the new equilibrium combination of interest rate and income: A) r 2 , Y 2 B) r 3 , Y 2 C) r 2 , Y 3 D) r 3 , Y 3 2. (Exhibit: IS-LM Fiscal Policy) Based on the graph, starting from equilibrium at interest rate r 1 and income Y 1 , an increase in government spending would generate the new equilibrium combination of interest rate and income: A) r 2 , Y 2 B) r 3 , Y 2 C) r 2 , Y 3 D) r 3 , Y 3 Page 1
3. If MPC = 0.75 (and there are no income taxes) when G increases by 100, then the IS curve for any given interest rate shifts to the right by:
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https://mathoverflow.net/questions/421191/is-every-virtually-free-group-residually-finite | 1,701,210,252,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100016.39/warc/CC-MAIN-20231128214805-20231129004805-00188.warc.gz | 441,270,275 | 31,725 | # Is every virtually free group residually finite?
Question: Is every (finitely generated) virtually free group residually finite?
A well-known question asks whether every hyperbolic group is residually finite (Mladen Bestvina. Questions in geometric group theory. http://www.math.utah.edu/ ~bestvina/eprints/questions- updated.pdf.). My question is a very speciall case, and I wonder if it has been done.
It is known that residual finiteness is not a quasi isometry invariant (Is residual finiteness a quasi isometry invariant for f.g. groups?).
• Yes, every virtually (residually finite) group is residually finite (clear), and every free group is residually finite (O. Schreier). They are also linear, by the way.
– YCor
Apr 27, 2022 at 7:55
• A virtually residually finite group is always residually finite. Apr 27, 2022 at 11:39
• Eventually this seems to become a big-list question, namely "provide proofs that free groups are residually finite".
– YCor
Apr 29, 2022 at 13:20
• There is already a big list question on that Apr 29, 2022 at 21:51
• mathoverflow.net/questions/20471/… Apr 29, 2022 at 21:52
As already mentioned, it is an easy exercise to show that a group containing a residually finite subgroup of finite index is itself residually finite. But proving that free groups are residually finite is (standard but) not so easy. There exist several possible arguments. Here are two of my favourites.
First proof. As motivated in Stallings' excellent article Topology of finite graphs, one can prove many properties of free groups of finite ranks by doing some elementary algebraic topology on finite graphs.
Let $$g$$ be a non-trivial element of a free group $$F$$ of finite rank. Let $$B$$ be a bouquet of $$\mathrm{rk}(F)$$ oriented circles, each one labelled by a generator of $$F$$. Write $$g$$ as a reduced word and let $$X$$ be a (subdivided) path whose oriented edges are labelled by generators in such that the reading $$X$$ yields $$g$$. There is a map $$X \to B$$ that sends each edge of $$X$$ to the edge of $$B$$ having the same label. Because our word is reduced, the map $$X \to B$$ is locally injective. Now, add edges to $$X$$ to get a covering $$X' \to B$$. This is possible because, in order to get a covering of $$B$$, it suffices to have, for each vertex $$v$$ and for each generator $$s$$, one (half-)edge labelled $$s$$ leaving $$v$$ and one (half-)edge labelled $$s$$ arriving at $$v$$. Then the image of $$\pi_1(X')$$ in $$\pi_1(B)$$ yields a finite-index subgroup $$H$$ in $$F$$ that does not contain $$g$$ (because $$g$$, thought of as a loop in $$B$$, lifts to a path in $$X'$$ that is not a loop).
Second proof. Instead of proving that free groups are residually finite, we can prove that the Coxeter group $$W_n:= \underset{n \text{ free factors}}{\underbrace{\mathbb{Z}/2\mathbb{Z} \ast \cdots \ast \mathbb{Z}/2 \mathbb{Z}}}.$$ is residually finite (since it is virtually free). The key observation is that the Cayley graph of $$W_n$$ with respect to its canonical generating set is a tree $$T_n$$, that vertex-stabilisers are trivial, and that $$W_n$$ acts on $$T_n$$ by reflections: each edge is flipped by an element of order two. Now, let $$g \in W_n$$ be a non-trivial element. Fix a path $$P$$ between the vertices $$1$$ and $$g$$ in $$T_n$$, and let $$E$$ denote the collection of all the edges of $$T_n \backslash P$$ with one endpoints in $$P$$. By an easy ping-pong argument, we can prove that $$H:= \langle \text{reflection along } e, e \in E \rangle$$ acts on $$T_n$$ with $$P$$ as a fundamental domain. It follows that $$H$$ has finite-index (since $$P$$ is finite) and that $$g$$ does not belong to $$H$$ (since $$h P \cap P= \emptyset$$ for every non-trivial $$h \in H$$).
Remark. Observe that, in the two arguments, the construction of the finite-index subgroup $$H$$ is explicit once $$g$$ is given. In particular, its index is controlled by the length of $$g$$.
• +1, I for one am all for hijacking this question for a big-list of favorite proofs of Schreier's result, if one is missing... Apr 29, 2022 at 6:09
Here is a different, purely group-theoretical proof that free groups are residually finite, following [J. Rotman, Advanced Modern Algebra (Second Edition), Proposition 4.80 p. 279].
Proposition. Given a free group $$F$$ and a non-trivial element $$g \in F$$, there exists a finite permutation group $$S$$ and a homomorphism $$\varphi \colon F \to S$$ such that $$\varphi(g) \neq 1$$.
Proof. Let $$X$$ a basis of $$F$$, and write $$g$$ as a reduced word $$g=x^{e_n}_{i_n}\ldots x^{e_1}_{i_1},$$ where $$x_{i_k} \in X$$ and $$e_k \in \{\pm 1\}$$. There are $$m \leq n$$ distinct base elements occurring in this word, say $$x_1, \ldots, x_m$$. For each of these basis elements $$x_j$$ we are now going to construct a permutation $$\alpha_j \in S_{n+1}$$.
Consider the set of all positions $$k$$ where $$x_j$$ or $$x_j^{-1}$$ occurs. If $$x_j$$ occurs, define $$\alpha_j(k)=k+1$$; if $$x_j^{-1}$$ occurs define $$\alpha_j(k+1)=k$$, namely, $$\alpha_j^{-1}(k)=k+1$$. This specifies $$\alpha_j$$ on a subset of $$\{1, \ldots, n+1\}$$, and we can complete it to a permutation of $$S_{n+1}$$, that we denote again by $$\alpha_j$$.
Finally, let us define a group homomorphism $$\varphi \colon F \to S_{n+1}$$. Since $$X$$ is a basis for $$F$$, by the universal property of free groups it suffices to define $$\varphi$$ on the elements of $$X$$. If $$x \in X$$ is an $$x_j$$ occurring in the spelling of $$g$$, define $$\varphi(x)=\alpha_j$$; if $$x$$ is not involved in the expression of $$g$$, define $$\varphi(x)=1$$. With such a definition, we have $$\varphi(g)= \alpha^{e_n}_{i_n}\ldots \alpha^{e_1}_{i_1}.$$ This is not the trivial permutation, since it sends $$1$$ to $$n+1$$. $$\square$$
• That's nice! It needs virtually nothing beyond the universal property of the free group, which is often used as the definition. Apr 29, 2022 at 13:34
• @DerekHolt: On the contrary, it uses that non-trivial elements of free groups are represented by reduced words, which is equivalent to the topological description as the fundamental group of a rose.
– HJRW
Apr 29, 2022 at 14:06
• I think this is essentially O. Schreier's original proof (in the 1920s).
– YCor
Apr 29, 2022 at 16:05
• (By the way it doesn't really use any group theory result, I'd rather view it as combinatorial or "permutation-theoretic" if it makes any sense. Of course this point of view is subjective. This proof is, by the way, used in some quantitative versions of residual finiteness since for a given nontrivial group word it gives a finite set of reasonable size for which this word acts nontrivially.)
– YCor
Apr 29, 2022 at 16:16
• @YCor It is Schreier's proof indeed, with the caveat that "residually finite" would not be coined for another 20 years after Schreier's paper (this paper is the same as the one containing the famous subgroup theorem). All that Schreier was out to prove is that a non-empty reduced word is not equal to $1$ in a free group, and his method of doing so is the one in the answer. Apr 29, 2022 at 23:50
Continuing the suggested hijacking, here is another proof that free groups are residually finite. It suffices to prove $$F_2$$ is residually finite. As in Francesco's answer, given a nontrivial word $$g \in F_2$$ we need to find a finite group $$G$$ and a homomorphism $$f : F_2 \to G$$ such that $$f(g) \neq 1$$.
Take $$G = \mathrm{SL}_2(p)$$ for prime $$p$$ larger than the length of $$g$$. If the generators are $$x, y$$, define $$f$$ by $$f(x) = \begin{pmatrix} 1 & X \\ 0 & 1 \end{pmatrix},\qquad f(y) = \begin{pmatrix} 1 & 0 \\ Y & 1\end{pmatrix}$$ for $$X, Y \in \mathbf{F}_p$$ to be determined. Suppose without loss of generality that $$g = x^{a_1} y^{b_1} \cdots x^{a_k} y^{b_k}$$ where $$a_1 b_1 \cdots a_k b_k \neq 0$$. Then $$f(g) = \begin{pmatrix} a_1 b_1 \cdots a_k b_k X^k Y^k + \cdots & a_1 b_1 \cdots a_k X^k Y^{k-1} + \cdots \\ a_2 \cdots a_k b_k X^{k-1} Y^k + \cdots & b_1 a_2 b_2 \cdots a_k X^{k-1} Y^{k-1} + \cdots \end{pmatrix},$$ where lower-order terms are hidden. Now put $$Y = 1$$ and note that we have polynomials in one variable with nonzero leading coefficient and degree less than $$p$$, so we can choose $$X$$ so that any one of them is nonzero, which is what we wanted.
• On reflection, I guess this is the same as my answer, except that you check the non-triviality in the quotient without first checking it in what I have hear Alex Lubotzky call “the mother group”. So you simultaneously prove the more non-trivial of my two “standard facts”.
– HJRW
Apr 29, 2022 at 15:04
To add to the list, here's a proof that is probably actually longer than the others, but is immediate "modulo standard facts".
Standard fact 1: All finitely generated free groups embed into $$SL_2(\mathbb{Z})$$. Indeed, famously the congruence subgroup $$\Gamma(2)$$ of $$PSL_2(\mathbb{Z})$$ is free of rank 2, all finitely generated free groups embed in $$F_2$$, and by the universal property we can embed into $$SL_2(\mathbb{Z})$$.
Standard fact 2: $$SL_2(\mathbb{Z})$$ is residually finite. Indeed, given a non-identity matrix, map to $$SL_2(\mathbb{Z}/n)$$ for large enough $$n$$. | 2,722 | 9,183 | {"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": 124, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-50 | latest | en | 0.927043 |
https://phys.org/news/2012-01-mathematicians-minimum-sudoku-solution-problem.html | 1,563,319,815,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524972.66/warc/CC-MAIN-20190716221441-20190717003441-00391.warc.gz | 503,774,756 | 18,777 | # Mathematicians use computer to solve minimum Sudoku solution problem
(PhysOrg.com) -- Over the past several years, Sudoku, as most people know, has become wildly popular. Where once mainstream newspapers carried only crossword puzzles, they now also carry a Sudoku puzzle as well. But along with that popularity, has come increased scrutiny and competition between people to see if certain properties of the puzzle can be found. For example, in any given Sudoku puzzle, how many clues must be given in order to have just one unique solution to the problem? Most Sudoku enthusiasts will answer 17, because nobody has ever been able to find one with 16 or less; which is fine, except that people as a general rule like some sort of proof of such things. Thus, it should not come as much of a surprise to anyone that a team of mathematicians have not only set out to prove what everyone thinks they know, but have succeeded in their endeavor.
Sudoku, for the uninitiated, is a whereby a square is created with 9x9 rows and columns of boxes to be filled in with numbers (1 through 9). The puzzle is further divided into nine 3x3 sections. The trick to solving the puzzle is that each row and column cannot have repeating numbers and neither can any of the 3x3 sections. Also, when a puzzle is created, a certain number of the boxes are prefilled, creating in essence, a unique puzzle each time. Thus, to solve the puzzle, all a person has to do is figure out how to fill in the rest of the boxes.
The thing about Sudoku puzzles though, is that those that make them can pre-fill as many boxes as they choose, the more clues they give, generally, the easier the puzzle will be to solve. Thus, creators who want to make their puzzles as hard as possible want to fill in the fewest possible clues that will still allow for a unique single solution.
To prove that 17 is the magic number, Gary McGuire and colleagues at University College, Dublin, took the brute force approach. After all with every there is a solution, to find it, all a computer would have to do is try every possible scenario for every possible configuration. Unfortunately, that approach would take too long, so the team had to figure out a way to trim down the number of possibilities they’d have to check for.
Throwing out grids that are equivalent helps, that reduces the number of tests dramatically. But that’s still not enough, so the team wrote a software routine that tests to see if certain subsets of the puzzle are equivalent to others, which would mean not having to test for the others if the first are found. This reduced the amount of testing as well. But even so, it took almost all of a full year of computing to test all of the possible scenarios.
But when the program stopped, the researchers had their answer. To create a Sudoku puzzle that is unique, you have to give at least 17 clues.
Explore further
Discovery could lead to more difficult Sudoku puzzles
More information: There is no 16-Clue Sudoku: Solving the Sudoku Minimum Number of Clues Problem, arXiv:1201.0749v1 [cs.DS] arxiv.org/abs/1201.0749
Abstract
We apply our new hitting set enumeration algorithm to solve the sudoku minimum number of clues problem, which is the following question: What is the smallest number of clues (givens) that a sudoku puzzle may have? It was conjectured that the answer is 17. We have performed an exhaustive search for a 16-clue sudoku puzzle, and we did not find one, thereby proving that the answer is indeed 17. This article describes our method and the actual search.
The hitting set problem is computationally hard; it is one of Karp's twenty-one classic NP-complete problems. We have designed a new algorithm that allows us to efficiently enumerate hitting sets of a suitable size. Hitting set problems have applications in many areas of science, such as bioinformatics and software testing.
via ArxivBlog
Feedback to editors
Jan 08, 2012
Despite proofs of this type, related ultimately to the ancient Greek proof that the square root of 2 is not a rational number, someone of your acquaintance within the next day or so is sure to mouth that demonstrably false cliché to the effect that «you can't prove a negative»....
Henri
Jan 08, 2012
Despite this proof and others like it - ultimately related to the ancient Greek proof that the square root of 2 is not a rational number, within the next couple of days someone of your acquaintance is likely to mouth that demonstrably false cliché to the effect that «you can't prove a negative»....
Henri
Jan 08, 2012
Despite this proof and others like it - ultimately related to the ancient Greek proof that the square root of 2 is not a rational number, within the next couple of days someone of your acquaintance is likely to mouth that demonstrably false cliché to the effect that «you can't prove a negative»....
Henri
Jan 08, 2012
ultimately related to the ancient Greek proof that the square root of 2
this has nothing to do with it.
The adage that you can't prove a negative is only relevant if the entirety of your searchspace cannot be tested. With the brute force approach used here they did test it.
That said I would have guessed that information theory might have sufficed to show that the amount contained in 16 clues is not enough to contain all the information inherent in a completed sudoku puzzle. | 1,162 | 5,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2019-30 | longest | en | 0.973075 |
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# HW#2 - AE 322 Homework#2 Due Monday February 8 2010 NOTE If...
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AE 322 Homework #2 Due Monday, February 8, 2010 NOTE: If you use MATLAB ® , show command sequence and output. 1. Find the components of the traction n T on planes defined by n 1 = 1 2 , n 2 = 1 2 , n 3 = 0 and n 1 = 1 2 , n 2 = − 1 2 , n 3 = 0 for the following states of stress: (a) σ 11 = σ σ 12 = σ 21 = 0 σ 13 = σ 31 = 0 σ 22 = σ σ 23 = σ 32 = 0 σ 33 = σ (b) σ 11 = σ σ 12 = σ 21 = σ σ 13 = σ 31 = 0 σ 22 = σ σ 23 = σ 32 = 0 σ 33 = 0 2. The state of stress at a point P in a material is given by: σ ij [ ] = 20 2 1 2 15 2 1 2 3 KPa (a) Compute the components of traction n T on the plane passing through P whose outward normal vector n makes equal angles with the coordinate axes. Note: this is the octahedral plane, 1 2 3 1 3, 1 3, 1/ 3 n n n = = . (b) Compute the normal nn σ and tangential ns σ components of traction on this plane. (c) Repeat the above exercise for the stress state: 10 2 1 2 15 5 1 5 3 ij KPa σ =
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3. Determine the body forces for which the following stress field describes a state of equilibrium: σ ij [ ] = yz + 4
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Ask a homework question - tutors are online | 552 | 1,651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-13 | latest | en | 0.816491 |
https://quiztutors.com/home/physical-sciences/statistics/population-standard-deviation-sample-variance-formula/ | 1,723,391,635,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00322.warc.gz | 364,026,315 | 24,171 | ### Population Standard Deviation, Standard Error, Sample Variance Formula – Statistic
The key terms in these statistic chapters include Estimated Standard Error, Population Standard Deviation, Rejecting the Null Hypothesis, Sample Variance Formula & Sample Variance.
A sample of n=4 scores has SS=48. What is the estimated standard error for the sample mean?
2 (find the sample variance SS/n-1
estimated standard error= under square root (sample variance/ n) = 2)
A sample of n=4 scores has SS=60 what is the variance for this sample?
20 sample variance = SS/n-1
Why are t statistics more variable than z scores?
The extra variability is caused by variations in the sample variance
What is a fundamental difference between the t statistic and z-score
The t statistic uses the sample variance in place of the population variance.
What is the sample variance and the estimated standard error for a sample of n=9 scores with SS=72
s^2 = 9 and Sm =1 sample variance= SS/n-1 and estimated standard error= under square root (sample variance/ n)
A repeated-measures experiment and a matched-subjects experiment each produce a t statistic with df = 10. How many individuals participated in each study?
11 for repeated, and 22 for matched
A researcher plans to conduct a research study comparing two treatment conditions with a total of 20 scores in each treatment. Which of the following designs would require only 20 participants?
##### A repeated-measures design
A sample of n=25 scores has a mean of M=40 and variance of s^2 =100. What is the estimated standard error for the sample mean?
2 (estimated standard error= under square root (sample variance/ n) = 2.041)
Which set of characteristics will produce the smallest value for the estimated standard error?
A large sample size and small sample variance
Which of the following statements are true of the t statistic?
The t statistic could be considered as an estimated z statistic
The t statistic provides a relatively poor estimate of z with small sample size
when the population standard deviation is unknown you can use the t statistic, assuming all relevant assumptions are satisfied.
What is formula for for t statistic
#### t = (M-u) /Sm
A sample of n=4 scores has SS=48. What is the estimated standard error for the sample mean?
2 (find the sample variance SS/n-1
estimated standard error= under square root (sample variance/ n) = 2)
A sample of n=4 scores has SS=60 what is the variance for this sample?
20 sample variance = SS/n-1
Why are t statistics more variable than z scores?
The extra variability is caused by variations in the sample variance
What is a fundamental difference between the t statistic and z-score
The t statistic uses the sample variance in place of the population variance.
What is the sample variance and the estimated standard error for a sample of n=9 scores with SS=72
s^2 = 9 and Sm =1 sample variance= SS/n-1 and estimated standard error= under square root (sample variance/ n)
With the exception of whether the population standard deviation is known the necessary assumptions for hypothesis tests with the t statistic and with the z statistic are
Essentially the same
What is degree of freedom n=35
#### 34n-1= 34
What is the variance for the sample for n=31 that has an SS of 120
Sample for variance= SS/n-1 = 4
A repeated measures study using a sample of n = 20 participants would produce a t statistic with df = ____.
19
For which of the following situations would a repeated-measures research design be appropriate?
Comparing pain tolerance with and without acupuncture needles
What is estimated standard error for the sample for n=16 that has a sample variance of 400
Under the square root (sample variance/sample size)= 5
Sample Variance formula
#### s^2= SS/n-1
Estimated standard error for sample formula
estimated standard error= square root (sample variance/sample size)
Sm (big S and small M)
estimated standard error
formula= Sm= s/square root n
t statistic formula
#### t = (M-u) / Sm
Estimated standard error (Sm)
used as an estimate of the real standard error, σM, when the value of σ is unknown. It is computed from the sample variance or sample standard deviation and provides an estimate of the standard distance between a sample mean, M, and the population mean, μ.
Two reasons for making this shift from standard deviation to variance:
1. The sample variance is an unbiased statistic; on average, the sample variance (s2) provides an accurate and unbiased estimate of the population variance (σ2). Therefore, the most accurate way to estimate the standard error is to use the sample variance to estimate the population variance. 2. The t statistic that require variance (instead of standard deviation) in the formulas for estimated standard error. To maximize the similarity from one version to another, we use variance in the formula for all of the different t statistics. Thus, whenever we present a t statistic, the estimated standard error is computed as
t statistic:
used to test hypotheses about an unknown population mean, μ, when the value of σ is unknown. The formula for the t statistic has the same structure as the z-score formula, except that the t statistic uses the estimated standard error in the denominator.
A t statistic
is used instead of a z-score when the population standard deviation and variance are not known.
Degrees of freedom:
##### Describe the number of scores in a sample that are independent and free to vary. Because the sample mean places a restriction on the value of one score in the sample, there are n − 1 degrees of freedom for a sample with n scores.
The t Distribution
the complete set of t values computed for every possible random sample for a specific sample size (n) or a specific degrees of freedom (df). The t distribution approximates the shape of a normal distribution.
Like the normal distribution, t distributions bell-shaped and symmetrical and have a mean of zero. However, t distributions have more variability, indicated by the flatter and more spread-out shape. The larger the value of df is, the more closely the t distribution approximates a normal distribution.
The Unknown Population
The mean value is known before the the treatment. The question is whether the treatment influences the scores and causes the mean to change. In this cause, the unknown population is the one that exists after the treatment is administered, and null hypothesis simply states that the value of the mean is not changed by the treatment.
Influence of Sample Size and Sample Variance
Because the estimated standard error, sM, appears in the denominator of the formula, a larger value for sM produces a smaller value (closer to zero) for t. Thus, any factor that influences the standard error also affects the likelihood of rejecting the null hypothesis and finding a significant treatment effect
Large variance
##### Large variance is bad for inferential statistics. It means that the scores widely scattered, which makes it difficult to see any consistent patterns or trends in the data. In general, high variance reduces the likelihood of rejecting the null hypothesis.
Estimated d
When the population values not known and you must substitute the population values are not know and you must substitute the corresponding sample values in their place. the population mean with treatment and the standard deviation are both unknown. Therefore, we use the mean for the treated sample and the standard deviation for the sample after treatment as estimates of the unknown parameters.
Standard Error of M:
The standard deviation of the distribution of sample means, σM, is called the standard error of M. This standard error provides a measure of how much distance is expected on average between a sample mean (M) and the population mean (μ). The standard error serves the same two purposes for the distribution of the sample means.
The Alpha Level:
The alpha (α) value is a small probability that is used to identify the low-probability samples. By convention, commonly used alpha levels are α = .05 (5%), α = .01 (1%), and α = .001 (0.1%). For example, with α = .05, we separate the most unlikely 5% of the sample means (the extreme values) from the most likely 95% of the sample means (the central values)…….The alpha level, or the level of significance, is a probability value that is used to define the concept of “very unlikely” in a hypothesis test.
S
#### Sample variance
Population variance:
Is the square root of the variance and provides a measure of the standard, or average, distance from the mean.
The t statistic
Allows researchers to use sample data to test hypotheses about an unknown population mean and not require any knowledge of the population.
t distribution small vs. large values
For large values of df, the t distribution will be nearly normal, but with small values for df, the t distribution will be flatter and more spread out than a normal distribution.
Degrees of freedom and shape
The exact shape of the t distribution changes with degrees of freedom
df gets very large, t distribution gets closer to normal z-score distribution
Under what circumstances is a t statistic used instead of a z-score for a hypothesis test?
A t statistic is used instead of a z-score when the population standard deviation and variance are not known.
In general, a distribution of t statistics is flatter and more spread out than the standard normal distribution.
##### TRUE
For df = 15, find the value(s) of t associated with each of the following:
a . The top 5% of the distribution.
b. The middle 95% of the distribution.
c . The middle 99% of the distribution.
A. +1.753 B. ±2.131 C. ±2.947
Estimated standard error
An estimate of the standard error that uses the sample variance (or standard deviation) in place of the corresponding population value.
t statistic
A statistic used to summarize sample data in situations where the population standard deviation not known. The t statistic is similar to a z-score for a sample mean, but the t statistic uses an estimate of the standard error.
A sample of n=4 scores has SS=48. What is the estimated standard error for the sample mean?
2 (find the sample variance SS/n-1
estimated standard error= under square root (sample variance/ n) = 2)
A sample of n=4 scores has SS=60 what is the variance for this sample?
##### 20sample variance = SS/n-1
Why are t statistics more variable than z scores?
The extra variability caused by variations in the sample variance
What is a fundamental difference between the t statistic and z-score
The t statistic uses the sample variance in place of the population variance.
What is the sample variance and the estimated standard error for a sample of n=9 scores with SS=72
s^2 = 9 and Sm =1 sample variance= SS/n-1 and estimated standard error= under square root (sample variance/ n)
t distribution
The distribution of t statistics is symmetrical and centered at zero like a normal distribution. A t distribution is flatter and more spread out than the normal distribution, but approaches a normal shape as df increases.
Percentage of variance accounted for by the treatment (r2)
A measure of effect size that determines what portion of the variability in the scores can be accounted for by the treatment effect.
Confidence interval
An interval estimate that described in terms of the level (percentage) of confidence in the accuracy of the estimation.
Homepage | 2,418 | 11,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.902846 |
https://kr.mathworks.com/matlabcentral/answers/1706685-how-can-i-find-out-the-indices-of-the-largest-10-values-within-the-matrix?s_tid=prof_contriblnk | 1,675,470,923,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00659.warc.gz | 364,822,579 | 26,530 | # How can I find out the indices of the largest 10 values within the matrix?
조회 수: 48(최근 30일)
Ashfaq Ahmed 2022년 4월 27일
답변: Matt J 2022년 4월 27일
Hey guys, I am facing a problem and I am sensing the solution should be relatively easy.
Suppose I have a 100x100 matrix with random values -
A = rand(100,100);
and I have stored the LARGEST 10 values from the matrix in this variable -
stored = maxk(A(:),10);
My question is - how can I find out the indices of these top 10 values from matrix A?
Any feedback will be highly appreciated!
댓글을 달려면 로그인하십시오.
### 답변(1개)
Matt J 2022년 4월 27일
[stored,indices] = maxk(A(:),10);
댓글을 달려면 로그인하십시오.
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#### Resources tagged with Speed and acceleration similar to On the Road:
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### There are 36 results
Broad Topics > Measuring and calculating with units > Speed and acceleration
##### Age 14 to 16 Challenge Level:
Four vehicles travelled on a road. What can you deduce from the times that they met?
##### Age 14 to 16 Challenge Level:
Can you adjust the curve so the bead drops with near constant vertical velocity?
##### Age 14 to 16 Challenge Level:
Four vehicles travel along a road one afternoon. Can you make sense of the graphs showing their motion?
### Graphical Interpretation
##### Age 14 to 16 Challenge Level:
This set of resources for teachers offers interactive environments to support work on graphical interpretation at Key Stage 4.
### Motion Sensor
##### Age 14 to 16 Challenge Level:
Looking at the graph - when was the person moving fastest? Slowest?
### There and Back
##### Age 14 to 16 Challenge Level:
Brian swims at twice the speed that a river is flowing, downstream from one moored boat to another and back again, taking 12 minutes altogether. How long would it have taken him in still water?
### Hike and Hitch
##### Age 14 to 16 Challenge Level:
Fifteen students had to travel 60 miles. They could use a car, which could only carry 5 students. As the car left with the first 5 (at 40 miles per hour), the remaining 10 commenced hiking along the. . . .
### Lap Times
##### Age 14 to 16 Challenge Level:
Two cyclists, practising on a track, pass each other at the starting line and go at constant speeds... Can you find lap times that are such that the cyclists will meet exactly half way round the. . . .
### Walk and Ride
##### Age 7 to 14 Challenge Level:
How far have these students walked by the time the teacher's car reaches them after their bus broke down?
### Up and Across
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its vertical and horizontal movement at each stage.
### An Unhappy End
##### Age 11 to 14 Challenge Level:
Two engines, at opposite ends of a single track railway line, set off towards one another just as a fly, sitting on the front of one of the engines, sets off flying along the railway line...
### N Is a Number
##### Age 11 to 14 Challenge Level:
N people visit their friends staying N kilometres along the coast. Some walk along the cliff path at N km an hour, the rest go by car. How long is the road?
### One and Three
##### Age 14 to 16 Challenge Level:
Two motorboats travelling up and down a lake at constant speeds leave opposite ends A and B at the same instant, passing each other, for the first time 600 metres from A, and on their return, 400. . . .
##### Age 7 to 14
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Speedy Sidney
##### Age 11 to 14 Challenge Level:
Two trains set off at the same time from each end of a single straight railway line. A very fast bee starts off in front of the first train and flies continuously back and forth between the. . . .
### How Far Does it Move?
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects the distance it travels at each stage.
### Alternative Record Book
##### Age 14 to 18 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Swimmers
##### Age 14 to 16 Challenge Level:
Swimmers in opposite directions cross at 20m and at 30m from each end of a swimming pool. How long is the pool ?
##### Age 11 to 14 Challenge Level:
Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
### Speeding Boats
##### Age 14 to 16 Challenge Level:
Two boats travel up and down a lake. Can you picture where they will cross if you know how fast each boat is travelling?
### Speed-time Problems at the Olympics
##### Age 14 to 16 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### How Do You React?
##### Age 14 to 16 Challenge Level:
To investigate the relationship between the distance the ruler drops and the time taken, we need to do some mathematical modelling...
### Triathlon and Fitness
##### Age 11 to 14 Challenge Level:
The triathlon is a physically gruelling challenge. Can you work out which athlete burnt the most calories?
### John's Train Is on Time
##### Age 11 to 14 Challenge Level:
A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station?
### An Average Average Speed
##### Age 14 to 16 Challenge Level:
My average speed for a journey was 50 mph, my return average speed of 70 mph. Why wasn't my average speed for the round trip 60mph ?
### Jumping Gerbils
##### Age 14 to 16 Challenge Level:
A conveyor belt, with tins placed at regular intervals, is moving at a steady rate towards a labelling machine. A gerbil starts from the beginning of the belt and jumps from tin to tin.
### Bus Stop
##### Age 14 to 16 Challenge Level:
Two buses leave at the same time from two towns Shipton and Veston on the same long road, travelling towards each other. At each mile along the road are milestones. The buses' speeds are constant. . . .
### Around and Back
##### Age 14 to 16 Challenge Level:
A cyclist and a runner start off simultaneously around a race track each going at a constant speed. The cyclist goes all the way around and then catches up with the runner. He then instantly turns. . . .
### Speeding Up, Slowing Down
##### Age 11 to 14 Challenge Level:
Experiment with the interactivity of "rolling" regular polygons, and explore how the different positions of the red dot affects its speed at each stage.
### Nutrition and Cycling
##### Age 14 to 16 Challenge Level:
Andy wants to cycle from Land's End to John o'Groats. Will he be able to eat enough to keep him going?
### Circuit Training
##### Age 14 to 16 Challenge Level:
Mike and Monisha meet at the race track, which is 400m round. Just to make a point, Mike runs anticlockwise whilst Monisha runs clockwise. Where will they meet on their way around and will they ever. . . .
### Take a Message Soldier
##### Age 14 to 18 Challenge Level:
A messenger runs from the rear to the head of a marching column and back. When he gets back, the rear is where the head was when he set off. What is the ratio of his speed to that of the column?
### Illusion
##### Age 11 to 16 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Motion Capture
##### Age 11 to 16 Challenge Level:
Explore displacement/time and velocity/time graphs with this mouse motion sensor.
### In Constantly Passing
##### Age 14 to 16 Challenge Level:
A car is travelling along a dual carriageway at constant speed. Every 3 minutes a bus passes going in the opposite direction, while every 6 minutes a bus passes the car travelling in the same. . . .
### Escalator
##### Age 14 to 16 Challenge Level:
At Holborn underground station there is a very long escalator. Two people are in a hurry and so climb the escalator as it is moving upwards, thus adding their speed to that of the moving steps. . . . | 1,759 | 7,758 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-47 | latest | en | 0.921775 |
https://www.handlebar-online.com/writing-tips/what-is-clocking-in-vlsi/ | 1,716,072,349,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057516.1/warc/CC-MAIN-20240518214304-20240519004304-00064.warc.gz | 730,270,114 | 10,152 | # What is clocking in VLSI?
## What is clocking in VLSI?
Definition of clock signal: We can define a clock signal as the one which synchronizes the state transitions by keeping all the registers/state elements in synchronization. In common terminology, a clock signal is a signal that is used to trigger sequential devices (flip-flops in general).
### Why do we use VLSI clocking?
The whole reason that we need clocks is that we want the output to depend on more than just the inputs, we want it to depend on previous outputs too. These previous outputs are the state bits of the FSM, and are the signals that cause lots of problems.
What are the advantage of two phase clock?
Two-phase clock Because the two phases are guaranteed non-overlapping, gated latches rather than edge-triggered flip-flops can be used to store state information so long as the inputs to latches on one phase only depend on outputs from latches on the other phase.
What is ideal clock in VLSI?
An “ideal” clock has no physical distribution tree, it just shows up magically on time at all the clock pins. It refers to the delay that is specified to exist between the source of the clock signal and the flip-flop clock pin. This is a delay specified by the user – not a real, measured thing.
## What kind of clock is used for VLSI 6?
•The clock is a simple pulsating signal alternating between 0 and 1. •Digital systems use a number of clocking schemes: 1. Single-phase clocking with latches 2. Single-phase clocking with flip-flops 3. Two-phase clocking Clock period t CLK CAD for VLSI 6 Single-phase Clocking with Latches •The latch opens when the clock goes high.
### Why is clock synchronization important in a VLSI circuit?
•Clock synchronization is one of the most critical considerations in designing high-performance VLSI circuits. –Data transfer between functional elements is synchronized by the clock. –It is desirable to design a circuit with the fastest possible clock. •The clock signal is typically generated external to the chip.
What is internal logic of CLK tcycle CLK?
Tuncertainty Clk Tcycle Clk’ 9/27/18 3 DFF DFF Din Delay COMB Dout Delay CLK Internal logic & wire delay from input pin to register Input(s) CLK’ Internal logic & wire delay from register to output pin Output(s) Internal logic bound by F2F timing Clock Skew= CLK -CLK’ | 528 | 2,344 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.927866 |
https://advansta.net/richmond/exact-permutation-big-sample.php | 1,653,657,127,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662647086.91/warc/CC-MAIN-20220527112418-20220527142418-00417.warc.gz | 129,592,019 | 11,848 | # Exact Permutation Big Sample
Exact testing with random permutations arxiv.org. Opsis: Literary Arts Journal at Montana State University (MSU) Permutation testing for the 2 sample mean situation. Home > Table of Contents > Current. Second, we can calculate the exact number of permuted results that were larger than what we observed. To calculate the proportion of the 1,000 values that were larger than what we observed, Apr 03, 2019 · Permutation tests, which I'll be discussing in this post, aren't that widely used by econometricians. However, they shouldn't be overlooked.Let's begin with some background discussion to set the scene. This might seem a bit redundant, but it will help us to see how permutation tests differ from the sort of tests that we usually use in econometrics.Background ….
### Permutation testing for the 2 sample mean situation
Ernst Permutation Methods A Basis for Exact Inference. Apr 03, 2019 · Permutation tests, which I'll be discussing in this post, aren't that widely used by econometricians. However, they shouldn't be overlooked.Let's begin with some background discussion to set the scene. This might seem a bit redundant, but it will help us to see how permutation tests differ from the sort of tests that we usually use in econometrics.Background …, Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall.
Apr 03, 2019 · Permutation tests, which I'll be discussing in this post, aren't that widely used by econometricians. However, they shouldn't be overlooked.Let's begin with some background discussion to set the scene. This might seem a bit redundant, but it will help us to see how permutation tests differ from the sort of tests that we usually use in econometrics.Background … The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001).
Calculator Use. Like the Combinations Calculator the Permutations Calculator finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders. The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001).
What is the Permutation Formula, Examples of Permutation Word Problems involving n things taken r at a time, How to solve Permutation Problems with Repeated Symbols, How to solve Permutation Problems with restrictions or special conditions, items together or not together or are restricted to the ends, how to differentiate between permutations and combinations, examples … Permutation tests with ANOVA have an advantage over traditional non-parametric techniques which are often not very powerful (with the exception of Kruskal-Wallis). You can adapt permutation tests to many different ANOVA designs. Here is an example of a two way ANOVA done with a permutation test (there is some
Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that
Sep 09, 2014 · What would be the big O notation of the length of the list of permutations of the characters of a list of words of lenght n? I just do not know how to express that because it would be like n! for each word where n is the characters of that word but O(n!) is just the complexity of a single word, not a list of n words. One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0)
Two-sample permutation tests for earthquake magnitudes before and after a damaging quake. Ask Question vertical lines show the sample medians and dots show their sample means. The outlier at 2.76 among those before the main quake might be called a foreshock. There are \${24, 12} = 2,704,156\$ possible permutations. To find the exact Apr 03, 2019В В· What we have here is a testing method that is "non-parametric", or "distribution free". However, it's more than that. We can show that the the test will be "exact", in the sense that it will have exactly the significance level that we decide we want, no matter how small the sample is! Now, let's be clear. All of this comes at a cost.
Minimum sample size for t-test?? that you perform Fisher's Exact test or more generally a permutation test, which will permute through all [n! / (n-k)!] … One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0)
Nov 30, 2017В В· However, most theoretical literature assumes that the whole permutation group is used, and methods based on random permutations tend to be seen as approximate. There exists a very limited amount of literature on exact testing with random permutations, and only recently a thorough proof of exactness was given. Calculator Use. Like the Combinations Calculator the Permutations Calculator finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders.
Permutation test is known as the only exact test pro-cedure in statistics. However, it is not exact in practice and only an approximate method due to the computa-tional bottleneck of enumerating every possible permu-tation. When dealing with big imaging data, it may take hundred years to do permutation in a desktop computer for converging results. Details. This function performs a two sample permutation test. If the mean is permuted, then the test assumes exchangability between the two samples. if the t-statistic is used, the test assumes either exchangability or a sufficiently large sample size.
Permutation test is known as the only exact test pro-cedure in statistics. However, it is not exact in practice and only an approximate method due to the computa-tional bottleneck of enumerating every possible permu-tation. When dealing with big imaging data, it may take hundred years to do permutation in a desktop computer for converging results. Exact testing with random permutations 813 p values based on random transformation are given in Sects. 3.3 and 3.4. In Sect. 4, some additional applications of these results are mentioned. 2 Fixed transformations Here we discuss tests that use the full group of transformations.
Opsis: Literary Arts Journal at Montana State University (MSU) Permutation testing for the 2 sample mean situation. Home > Table of Contents > Current. Second, we can calculate the exact number of permuted results that were larger than what we observed. To calculate the proportion of the 1,000 values that were larger than what we observed PERMUTATION TESTS FOR LINEAR MODELS MARTI J. ANDERSON1 AND JOHN ROBINSON2∗ UniversityofSydney Summary Several approximate permutation tests have been proposed for tests of partial regression coefficients in a linear model based on sample partial correlations. This paper begins with an explanation and notation for an exact test.
Ernst Permutation Methods A Basis for Exact Inference. University of Minnesota, Twin Cities School of Statistics Stat 5601 Rweb Stat 5601 (Geyer) Examples (Permutation Tests and Related Procedures) The Name of the Game Permutation tests are also called randomization tests.Different name, same concept., What is the Permutation Formula, Examples of Permutation Word Problems involving n things taken r at a time, How to solve Permutation Problems with Repeated Symbols, How to solve Permutation Problems with restrictions or special conditions, items together or not together or are restricted to the ends, how to differentiate between permutations and combinations, examples ….
### python Sampling Permutations of [123...N] for large
algorithm Big O Notation for the permutations of a list. The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001)., Details. This function performs a two sample permutation test. If the mean is permuted, then the test assumes exchangability between the two samples. if the t-statistic is used, the test assumes either exchangability or a sufficiently large sample size..
### statistics Two-sample permutation tests for earthquake
Alternatives of Fisher's exact test for more than 2 groups?. Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall https://en.wikipedia.org/wiki/Identity_permutation Details. This function performs a two sample permutation test. If the mean is permuted, then the test assumes exchangability between the two samples. if the t-statistic is used, the test assumes either exchangability or a sufficiently large sample size..
• Statistical Tests SAS
• Permutation testing for the 2 sample mean situation
• Permutation Analysis using unequal samples. Ask Question Asked 5 years, 3 months ago. With your sample sizes, a full permutation test would usually be impractical (unless the sample difference is fairly extreme, in which case a complete enumeration of the tail may be feasible). The corresponding exact 1- and 2- tailed p-values here are One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0)
One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0) Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall
Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that Given independent samples from P and Q, two-sample permu-tation tests allow one to construct exact level tests when the null hypothesis is P =Q. On the other hand, when comparing or testing particular parameters θ of P and Q, such as their means or medi-ans, permutation tests need not be level α, or even approximately level α in large samples.
The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001). Permutation tests The two-sample t-test The big picture Viewing our study as balls in an urn The same concept that we encountered in the two-sample Fisher’s exact test can be used for continuous data also If the smelter had no e ect on children’s neurological development, then it wouldn’t matter if the child lived near it or not
\$\begingroup\$ That's how I usually work out how big a sample to take for a randomization test - to work out how accurate I want my p-values and then get the original number of permutations from that. (If it turns out to be close to as many as doing all possible permutations, I do that instead - once or twice I caught myself randomly sampling a million permutations when the … The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001).
PERMUTATION TESTS FOR LINEAR MODELS MARTI J. ANDERSON1 AND JOHN ROBINSON2∗ UniversityofSydney Summary Several approximate permutation tests have been proposed for tests of partial regression coefficients in a linear model based on sample partial correlations. This paper begins with an explanation and notation for an exact test. Permutation Analysis using unequal samples. Ask Question Asked 5 years, 3 months ago. With your sample sizes, a full permutation test would usually be impractical (unless the sample difference is fairly extreme, in which case a complete enumeration of the tail may be feasible). The corresponding exact 1- and 2- tailed p-values here are
Permutation tests with ANOVA have an advantage over traditional non-parametric techniques which are often not very powerful (with the exception of Kruskal-Wallis). You can adapt permutation tests to many different ANOVA designs. Here is an example of a two way ANOVA done with a permutation test (there is some The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001).
What is the Permutation Formula, Examples of Permutation Word Problems involving n things taken r at a time, How to solve Permutation Problems with Repeated Symbols, How to solve Permutation Problems with restrictions or special conditions, items together or not together or are restricted to the ends, how to differentiate between permutations and combinations, examples … 1.2 Statistical testing by permutation The role of a statistical test is to decide whether some parameter of the reference population may take a value assumed by hypothesis, given the fact that the corresponding statistic, whose value i s estimated from a sample of objects, may have a somewhat different value.
Calculator Use. Like the Combinations Calculator the Permutations Calculator finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders. University of Minnesota, Twin Cities School of Statistics Stat 5601 Rweb Stat 5601 (Geyer) Examples (Permutation Tests and Related Procedures) The Name of the Game Permutation tests are also called randomization tests.Different name, same concept.
Statistical Tests The following section discusses the statistical tests performed in the MULTTEST procedure. For continuous data, a t-test for the mean is available.For discrete variables, available tests are the Cochran-Armitage (CA) linear trend test, the Freeman-Tukey (FT) double arcsine test, the Peto mortality-prevalence test, and the Fisher exact test. One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0)
Permutation Test is a good choice for hypothesis test of unknown distribution. It works regardless of the shape and size of the population gives exact p value Monte Carlo Sampling is introduced to permutation test when it is impossible to complete enumeration the data. Monte Carlo Method can well approximate the In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions. The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are …
## Combinations and permutations Rosetta Code
Exact testing with random permutations. Sep 09, 2014 · What would be the big O notation of the length of the list of permutations of the characters of a list of words of lenght n? I just do not know how to express that because it would be like n! for each word where n is the characters of that word but O(n!) is just the complexity of a single word, not a list of n words., Apr 03, 2019 · Permutation tests, which I'll be discussing in this post, aren't that widely used by econometricians. However, they shouldn't be overlooked.Let's begin with some background discussion to set the scene. This might seem a bit redundant, but it will help us to see how permutation tests differ from the sort of tests that we usually use in econometrics.Background ….
### Exact testing with random permutations
Permutations and Combinations Problems GMAT GRE Maths. Details. This function performs a two sample permutation test. If the mean is permuted, then the test assumes exchangability between the two samples. if the t-statistic is used, the test assumes either exchangability or a sufficiently large sample size., In our R packages the permutation tests are estimated by a sampling procedure, and not computed exactly (or deterministically). It turns out this is likely a necessary concession; a complete exact permutation test procedure at scale would be big news. Please read on for my comments on this issue..
non-parametric, and is usually approximated by Monte Carlo methods for large sample sizes where the exact permutation generation is computationally impractical. 2 The Jackknife: Introduction and Basic Properties The Jackknife was proposed by M.H. Quenouille in 1949 and later re ned and given its current Nov 30, 2017В В· However, most theoretical literature assumes that the whole permutation group is used, and methods based on random permutations tend to be seen as approximate. There exists a very limited amount of literature on exact testing with random permutations, and only recently a thorough proof of exactness was given.
In our R packages the permutation tests are estimated by a sampling procedure, and not computed exactly (or deterministically). It turns out this is likely a necessary concession; a complete exact permutation test procedure at scale would be big news. Please read on for my comments on this issue. The Permutation Approach - Big Picture • History and introduction to the permutation approach. • Possible situations where permutation methods can be used. 1 History • 1930’s: R. A. Fisher discovered a general exact method of testing hypotheses based on permuting the data in ways that do not change its distribution under H0. • This permutation method does not require …
The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001). In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions. The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are …
Statistical Tests The following section discusses the statistical tests performed in the MULTTEST procedure. For continuous data, a t-test for the mean is available.For discrete variables, available tests are the Cochran-Armitage (CA) linear trend test, the Freeman-Tukey (FT) double arcsine test, the Peto mortality-prevalence test, and the Fisher exact test. Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that
So Fisher's exact test is an exact test in the same sense that a permutation test is exact. Moreover it's the permutation test based on all permutations leaving the margins fixed. In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions. The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are …
One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0) Statistical Tests The following section discusses the statistical tests performed in the MULTTEST procedure. For continuous data, a t-test for the mean is available.For discrete variables, available tests are the Cochran-Armitage (CA) linear trend test, the Freeman-Tukey (FT) double arcsine test, the Peto mortality-prevalence test, and the Fisher exact test.
Given independent samples from P and Q, two-sample permu-tation tests allow one to construct exact level tests when the null hypothesis is P =Q. On the other hand, when comparing or testing particular parameters Оё of P and Q, such as their means or medi-ans, permutation tests need not be level О±, or even approximately level О± in large samples. Nov 30, 2014В В· The permutation test is based on the idea that under the null hypothesis, 4 Relevance to Adaptive Methods. the trial [1,16]. We exploit the connection between results about permutation and t-tests to show how to construct an exact, level О± test with adaptive sample size modification.
University of Minnesota, Twin Cities School of Statistics Stat 5601 Rweb Stat 5601 (Geyer) Examples (Permutation Tests and Related Procedures) The Name of the Game Permutation tests are also called randomization tests.Different name, same concept. 1.2 Statistical testing by permutation The role of a statistical test is to decide whether some parameter of the reference population may take a value assumed by hypothesis, given the fact that the corresponding statistic, whose value i s estimated from a sample of objects, may have a somewhat different value.
Permutation Analysis using unequal samples. Ask Question Asked 5 years, 3 months ago. With your sample sizes, a full permutation test would usually be impractical (unless the sample difference is fairly extreme, in which case a complete enumeration of the tail may be feasible). The corresponding exact 1- and 2- tailed p-values here are Minimum sample size for t-test?? that you perform Fisher's Exact test or more generally a permutation test, which will permute through all [n! / (n-k)!] …
Permutation Test is a good choice for hypothesis test of unknown distribution. It works regardless of the shape and size of the population gives exact p value Monte Carlo Sampling is introduced to permutation test when it is impossible to complete enumeration the data. Monte Carlo Method can well approximate the Given independent samples from P and Q, two-sample permu-tation tests allow one to construct exact level tests when the null hypothesis is P =Q. On the other hand, when comparing or testing particular parameters Оё of P and Q, such as their means or medi-ans, permutation tests need not be level О±, or even approximately level О± in large samples.
Minimum sample size for t-test?? that you perform Fisher's Exact test or more generally a permutation test, which will permute through all [n! / (n-k)!] … Exact testing with random permutations 813 p values based on random transformation are given in Sects. 3.3 and 3.4. In Sect. 4, some additional applications of these results are mentioned. 2 Fixed transformations Here we discuss tests that use the full group of transformations.
If the labels are exchangeable under the null hypothesis, then the resulting tests yield exact significance levels; Permutation tests. Significance tests tell us whether an observed effect, such as a difference between two means or a correlation between two variables, could reasonably occur just by chance in selecting a random sample Permutation tests The two-sample t-test The big picture Viewing our study as balls in an urn The same concept that we encountered in the two-sample Fisher’s exact test can be used for continuous data also If the smelter had no e ect on children’s neurological development, then it wouldn’t matter if the child lived near it or not
The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001). In our R packages the permutation tests are estimated by a sampling procedure, and not computed exactly (or deterministically). It turns out this is likely a necessary concession; a complete exact permutation test procedure at scale would be big news. Please read on for my comments on this issue.
Permutation tests The two-sample t-test The big picture Viewing our study as balls in an urn The same concept that we encountered in the two-sample Fisher’s exact test can be used for continuous data also If the smelter had no e ect on children’s neurological development, then it wouldn’t matter if the child lived near it or not 1.2 Statistical testing by permutation The role of a statistical test is to decide whether some parameter of the reference population may take a value assumed by hypothesis, given the fact that the corresponding statistic, whose value i s estimated from a sample of objects, may have a somewhat different value.
Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall Opsis: Literary Arts Journal at Montana State University (MSU) Permutation testing for the 2 sample mean situation. Home > Table of Contents > Current. Second, we can calculate the exact number of permuted results that were larger than what we observed. To calculate the proportion of the 1,000 values that were larger than what we observed
The Permutation Approach - Big Picture • History and introduction to the permutation approach. • Possible situations where permutation methods can be used. 1 History • 1930’s: R. A. Fisher discovered a general exact method of testing hypotheses based on permuting the data in ways that do not change its distribution under H0. • This permutation method does not require … Permutation tests with ANOVA have an advantage over traditional non-parametric techniques which are often not very powerful (with the exception of Kruskal-Wallis). You can adapt permutation tests to many different ANOVA designs. Here is an example of a two way ANOVA done with a permutation test (there is some
Nov 30, 2017В В· However, most theoretical literature assumes that the whole permutation group is used, and methods based on random permutations tend to be seen as approximate. There exists a very limited amount of literature on exact testing with random permutations, and only recently a thorough proof of exactness was given. Nov 30, 2017В В· However, most theoretical literature assumes that the whole permutation group is used, and methods based on random permutations tend to be seen as approximate. There exists a very limited amount of literature on exact testing with random permutations, and only recently a thorough proof of exactness was given.
Apr 03, 2019 · Permutation tests, which I'll be discussing in this post, aren't that widely used by econometricians. However, they shouldn't be overlooked.Let's begin with some background discussion to set the scene. This might seem a bit redundant, but it will help us to see how permutation tests differ from the sort of tests that we usually use in econometrics.Background … Given independent samples from P and Q, two-sample permu-tation tests allow one to construct exact level tests when the null hypothesis is P =Q. On the other hand, when comparing or testing particular parameters θ of P and Q, such as their means or medi-ans, permutation tests need not be level α, or even approximately level α in large samples.
mne.stats.permutation_t_test¶ mne.stats.permutation_t_test (X, n_permutations=10000, tail=0, n_jobs=1, seed=None, verbose=None) [source] ¶ One sample/paired sample permutation test based on a t-statistic. This function can perform the test on one variable or simultaneously on multiple variables. The Permutation Approach - Big Picture • History and introduction to the permutation approach. • Possible situations where permutation methods can be used. 1 History • 1930’s: R. A. Fisher discovered a general exact method of testing hypotheses based on permuting the data in ways that do not change its distribution under H0. • This permutation method does not require …
mne.stats.permutation_t_test¶ mne.stats.permutation_t_test (X, n_permutations=10000, tail=0, n_jobs=1, seed=None, verbose=None) [source] ¶ One sample/paired sample permutation test based on a t-statistic. This function can perform the test on one variable or simultaneously on multiple variables. Nov 30, 2014 · The permutation test is based on the idea that under the null hypothesis, 4 Relevance to Adaptive Methods. the trial [1,16]. We exploit the connection between results about permutation and t-tests to show how to construct an exact, level α test with adaptive sample size modification.
algorithm Big O Notation for the permutations of a list. Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that, Calculator Use. Like the Combinations Calculator the Permutations Calculator finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders..
### python Sampling Permutations of [123...N] for large
The Jackknife Estimation Method Avery I. McIntosh. The use of permutation methods for exact inference dates back to Fisher in 1935. Since then, the practicality of such methods has increased steadily with computing power. They can now easily be employed in many situations without concern for computing difficulties., Combinations and permutations A sample of permutations from 1 to 12 and Combinations from 10 to 60 using exact Integer arithmetic. A sample of permutations from 5 to 15000 and Combinations from 100 to 1000 using approximate Floating point arithmetic. p 15000. big_permutation (73) #=> 6.004137561717704e+304.
### mne.stats.permutation_t_test — MNE 0.19.0 documentation
mne.stats.permutation_t_test — MNE 0.19.0 documentation. Opsis: Literary Arts Journal at Montana State University (MSU) Permutation testing for the 2 sample mean situation. Home > Table of Contents > Current. Second, we can calculate the exact number of permuted results that were larger than what we observed. To calculate the proportion of the 1,000 values that were larger than what we observed https://en.wikipedia.org/wiki/Circular_permutation_in_proteins University of Minnesota, Twin Cities School of Statistics Stat 5601 Rweb Stat 5601 (Geyer) Examples (Permutation Tests and Related Procedures) The Name of the Game Permutation tests are also called randomization tests.Different name, same concept..
Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that
Apr 03, 2019В В· What we have here is a testing method that is "non-parametric", or "distribution free". However, it's more than that. We can show that the the test will be "exact", in the sense that it will have exactly the significance level that we decide we want, no matter how small the sample is! Now, let's be clear. All of this comes at a cost. Permutation test is known as the only exact test pro-cedure in statistics. However, it is not exact in practice and only an approximate method due to the computa-tional bottleneck of enumerating every possible permu-tation. When dealing with big imaging data, it may take hundred years to do permutation in a desktop computer for converging results.
Permutation tests with ANOVA have an advantage over traditional non-parametric techniques which are often not very powerful (with the exception of Kruskal-Wallis). You can adapt permutation tests to many different ANOVA designs. Here is an example of a two way ANOVA done with a permutation test (there is some Permutation test is known as the only exact test pro-cedure in statistics. However, it is not exact in practice and only an approximate method due to the computa-tional bottleneck of enumerating every possible permu-tation. When dealing with big imaging data, it may take hundred years to do permutation in a desktop computer for converging results.
The article summarizes permutation testing in models with one and two main effects, and notes that in a model with two main effects and an interaction term there is no exact permutation method for testing the interaction term. For tests of interactions, even with categorical G and Z no exact permutation method is available (Anderson 2001). Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that
So Fisher's exact test is an exact test in the same sense that a permutation test is exact. Moreover it's the permutation test based on all permutations leaving the margins fixed. \$\begingroup\$ That's how I usually work out how big a sample to take for a randomization test - to work out how accurate I want my p-values and then get the original number of permutations from that. (If it turns out to be close to as many as doing all possible permutations, I do that instead - once or twice I caught myself randomly sampling a million permutations when the …
Two-sample permutation tests for earthquake magnitudes before and after a damaging quake. Ask Question vertical lines show the sample medians and dots show their sample means. The outlier at 2.76 among those before the main quake might be called a foreshock. There are \${24, 12} = 2,704,156\$ possible permutations. To find the exact Nov 30, 2014В В· The permutation test is based on the idea that under the null hypothesis, 4 Relevance to Adaptive Methods. the trial [1,16]. We exploit the connection between results about permutation and t-tests to show how to construct an exact, level О± test with adaptive sample size modification.
non-parametric, and is usually approximated by Monte Carlo methods for large sample sizes where the exact permutation generation is computationally impractical. 2 The Jackknife: Introduction and Basic Properties The Jackknife was proposed by M.H. Quenouille in 1949 and later re ned and given its current Details. This function performs a two sample permutation test. If the mean is permuted, then the test assumes exchangability between the two samples. if the t-statistic is used, the test assumes either exchangability or a sufficiently large sample size.
Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that Apr 08, 2019 · A demonstration that the permutation test is "exact". That it, its significance level is exactly what we assign it to be. A comparison between a permutation test and the usual t-test for this problem. A demonstration that the permutation test remains "exact", even when the regression model is mi-specified by fitting it through the origin.
What is the Permutation Formula, Examples of Permutation Word Problems involving n things taken r at a time, How to solve Permutation Problems with Repeated Symbols, How to solve Permutation Problems with restrictions or special conditions, items together or not together or are restricted to the ends, how to differentiate between permutations and combinations, examples … PERMUTATION TESTS FOR LINEAR MODELS MARTI J. ANDERSON1 AND JOHN ROBINSON2∗ UniversityofSydney Summary Several approximate permutation tests have been proposed for tests of partial regression coefficients in a linear model based on sample partial correlations. This paper begins with an explanation and notation for an exact test.
Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall Sampling Permutations of [1,2,3,…,N] for large N. Ask Question Asked 7 years, 6 months ago. Active 1 year, 11 months ago. @senderle random.sample works without replacement do it does indeed produce a permutation. When the sample size is the same as the population size, it is equivalent to random.shuffle() on the population (except that
In our R packages the permutation tests are estimated by a sampling procedure, and not computed exactly (or deterministically). It turns out this is likely a necessary concession; a complete exact permutation test procedure at scale would be big news. Please read on for my comments on this issue. Given independent samples from P and Q, two-sample permu-tation tests allow one to construct exact level tests when the null hypothesis is P =Q. On the other hand, when comparing or testing particular parameters Оё of P and Q, such as their means or medi-ans, permutation tests need not be level О±, or even approximately level О± in large samples.
Apr 08, 2019 · A demonstration that the permutation test is "exact". That it, its significance level is exactly what we assign it to be. A comparison between a permutation test and the usual t-test for this problem. A demonstration that the permutation test remains "exact", even when the regression model is mi-specified by fitting it through the origin. PERMUTATION TESTS FOR LINEAR MODELS MARTI J. ANDERSON1 AND JOHN ROBINSON2∗ UniversityofSydney Summary Several approximate permutation tests have been proposed for tests of partial regression coefficients in a linear model based on sample partial correlations. This paper begins with an explanation and notation for an exact test.
The Permutation Approach - Big Picture • History and introduction to the permutation approach. • Possible situations where permutation methods can be used. 1 History • 1930’s: R. A. Fisher discovered a general exact method of testing hypotheses based on permuting the data in ways that do not change its distribution under H0. • This permutation method does not require … Minimum sample size for t-test?? that you perform Fisher's Exact test or more generally a permutation test, which will permute through all [n! / (n-k)!] …
Apr 03, 2019 · Permutation tests, which I'll be discussing in this post, aren't that widely used by econometricians. However, they shouldn't be overlooked.Let's begin with some background discussion to set the scene. This might seem a bit redundant, but it will help us to see how permutation tests differ from the sort of tests that we usually use in econometrics.Background … One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0)
Minimum sample size for t-test?? that you perform Fisher's Exact test or more generally a permutation test, which will permute through all [n! / (n-k)!] … Permutation tests work by resampling the observed data many times in order to determine a p-value for the test. Recall that the p -value is defined as the probability of getting data as extreme as the observed data when the null hypothesis is true.
The Permutation Approach - Big Picture • History and introduction to the permutation approach. • Possible situations where permutation methods can be used. 1 History • 1930’s: R. A. Fisher discovered a general exact method of testing hypotheses based on permuting the data in ways that do not change its distribution under H0. • This permutation method does not require … Apr 08, 2019 · A demonstration that the permutation test is "exact". That it, its significance level is exactly what we assign it to be. A comparison between a permutation test and the usual t-test for this problem. A demonstration that the permutation test remains "exact", even when the regression model is mi-specified by fitting it through the origin.
In this article you’ll learn about Permutation and Combination problems: Definition, formulas, solved examples and a quiz with practice questions. The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are … What is the Permutation Formula, Examples of Permutation Word Problems involving n things taken r at a time, How to solve Permutation Problems with Repeated Symbols, How to solve Permutation Problems with restrictions or special conditions, items together or not together or are restricted to the ends, how to differentiate between permutations and combinations, examples …
Nov 30, 2014В В· The permutation test is based on the idea that under the null hypothesis, 4 Relevance to Adaptive Methods. the trial [1,16]. We exploit the connection between results about permutation and t-tests to show how to construct an exact, level О± test with adaptive sample size modification. Nov 30, 2014В В· The permutation test is based on the idea that under the null hypothesis, 4 Relevance to Adaptive Methods. the trial [1,16]. We exploit the connection between results about permutation and t-tests to show how to construct an exact, level О± test with adaptive sample size modification.
Permutation Analysis using unequal samples. Ask Question Asked 5 years, 3 months ago. With your sample sizes, a full permutation test would usually be impractical (unless the sample difference is fairly extreme, in which case a complete enumeration of the tail may be feasible). The corresponding exact 1- and 2- tailed p-values here are So Fisher's exact test is an exact test in the same sense that a permutation test is exact. Moreover it's the permutation test based on all permutations leaving the margins fixed.
Minimum sample size for t-test?? that you perform Fisher's Exact test or more generally a permutation test, which will permute through all [n! / (n-k)!] … One sample/paired sample permutation test based on a t-statistic. It’s the exact test, that can be untractable when the number of samples is big (e.g. > 20). If n_permutations >= 2**n_samples then the exact test is performed. tail-1 or 0 or 1 (default = 0)
Statistical Tests The following section discusses the statistical tests performed in the MULTTEST procedure. For continuous data, a t-test for the mean is available.For discrete variables, available tests are the Cochran-Armitage (CA) linear trend test, the Freeman-Tukey (FT) double arcsine test, the Peto mortality-prevalence test, and the Fisher exact test. Exact testing with random permutations Jesse Hemerik and Jelle Goeman Radboudumc, The Netherlands November 5, 2017 Abstract The way in which random permutations have been used in various permutation-based methods leads to anti-conservativeness, especially in multiple testing contexts. Problems arise in particular for Westfall | 10,212 | 47,161 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.931711 |
www.fluidmechanics.co.uk | 1,726,211,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00052.warc.gz | 702,316,816 | 13,137 | # Water Hammer Analysis
Water hammer is part of the larger subject of transient flow or surge analysis. It is the special case when there is a sudden change in flow velocity. Usually this occurs when a valve closes quickly. Water hammer can generate very high pressure transients which could burst a pipe and can generate pipeline vibrations. The magnitude of the water hammer pressure rise can be calculated using the Joukowsky equation which is
P=ρCU (Pa)
Where
P is the change in pressure
ρ is the fluid density
U is the change in fluid velocity
C is the sonic velocity in the pipe
The sonic velocity is the speed of sound in the pipe and is determined by a modified hooks law formula which takes into account the stiffness of the fluid and the pipe wall.
Where
K Bulk modulus of fluid
E Young’s modulus of pipe material
e Wall thickness of pipe
The sonic velocity is also the speed at which the pressure waves generated by water hammer travel in the pipe.
For water in very stiff pipes the sonic speed could be as high as 1480 m/s. But in some plastic pipe the wave speed can be lower than 200 m/s.
The bulk modulus (k) of water is 2.19x109 Pa however this assumes that the water has no air bubbles in it. Often microscopic size bubbles can be seen suspended in the fluid. This can make a significant difference to the effective bulk modulus and so to the sonic speed. Often with water hammer sub atmospheric pressure and cavitation can also occur (as explained below). This can liberate dissolved air from the water which forms air bubbles reducing the effective bulk modulus and so reducing the wave speed.
Valve Closure Example of Water Hammer.
Figure 1 shows the initial conditions in the pipe. The pipe inlet at position A is connected to a header tank which provides the pressure P1 to drive the flow in the pipe. The other end of the pipe at position B is open to atmosphere and its pressure is P0
P1=ρgh+P0
The length of the pipe is L.
Figure 2 shows the pipe and flow conditions just after the pipe end at B has been instantaneously closed at time t0. A pressure wave at position X is traveling up the pipe with velocity C (the sonic velocity). The pressure rise across the wave is ρCU (Joukowsky equation). Upstream of position X the velocity is the initial velocity Ui. Downstream of X the velocity is 0.
Between X and B the fluid will be compressed and the pipe will be expanded. The rate of pipe volume change and fluid compression is the same as the flow rate upstream of X.
Figure 3 shows the conditions when the pressure wave reaches position A at time t1. The pipeline pressure has been raised by ρCUi and the fluid velocity is 0 throughout. This condition is unstable as the pipe inlet pressure is set by the head of fluid in the inlet tank h. So now the fluid needs to move in the reverse direction from the high pressure pipe into the lower pressure tank. This induces the first wave reflection and it occurs at time t1.
Where t1=t0+L/C
Figure 4 shows the conditions after the first reflection. The pressure wave is at position X and is traveling down the pipe with velocity C. The fluid between positions A and X is traveling in the reverse direction with velocity –Ui. The drop in pressure across the wave front is ρC(-Ui).
Figure 5 shows the conditions when the pressure wave reaches position B at time t2. The whole of the pipeline pressure has been reduced and the fluid velocity is -Ui throughout.
It should be noted there will be a small negative pressure gradient between A & B. This is required to overcome friction between the fluid and pipe as the flow is in the opposite direction. The magnitude of this pressure gradient will usually be significantly smaller than that generated by the change in velocity (Joukowsky equation). This friction gradient is exaggerated in the figure in comparison to water hammer effect for demonstration purposes.
As the end of the pipe at B is closed, this condition is unstable as there is fluid available to sustain the flow. This induces the second wave reflection at time t2.
Where t2=t1+L/C or t2=t0+2L/C.
Figure 6 shows the conditions after the second reflection. The pressure wave is at position X and it is traveling up the pipe with velocity C’. The fluid between positions A and X is still traveling in the reverse direction with velocity –Ui. The fluid between X and B has been stopped.
The drop in pressure across the wave front is ρC’(-Ui). It should be noted that in this case the wave speed or sonic velocity has been changed from C to C’. C’ may be the same or less than C, it depends on the minimum allowable pressure P3. Negative absolute pressures are not possible. The minimum pressure in the pipe line cannot be less than the vapour pressure of fluid and often the minimum pressure is higher than the vapour pressure because there is dissolved gas in the fluid which comes out of solution as the pressure is reduced. When cavitation occurs or gas comes out of solution the bulk modulus of fluid is reduced form K to K’. It is this reduction bulk modulus that allows the sonic speed to reduce from C to C’ .So all depending on minimum possible pressure the magnitude of C’ will adjust its self so that P3 is not lower than minimum possible pressure.
The formulas for wave speed and the Joukowsky equation are still valid when cavitation occurs but the bulk modulus will have reduced so ensuring consistency in the equations and no impossible pressures.
Figure 7 shows the conditions when the pressure wave reaches position A at time t3. The whole of the pipeline pressure has been reduced to P3 and the fluid velocity is 0 throughout. This condition is unstable as the pipe inlet pressure set by the head of fluid in the inlet tank h is higher than the pipe pressure so now the fluid needs to move into the pipe from the header tank. This induces the third wave reflection and it occurs at time t3.
Where t3=t2+L/C’ or t3=t0+2L/C+L/C’
Figure 8 shows the conditions after the third reflection. The pressure wave is at position X, its traveling down the pipe with velocity C’. The fluid between positions A and X is still traveling in the normal direction with velocity Ui’. The fluid between X and B is stopped. The velocity in section A to X is shown as Ui’ where Ui’ is slightly less than Ui. By this stage the process has undergone 3 reflections and at every stage some energy is lost so over time the magnitude of the pressure waves and velocities are reducing.
The sonic wave velocity is still the reduced velocity C’ as shown in figure 8. But if previously air or vapour had been liberated from the fluid then the vapour will be re-condense and gas bubbles will reduce in size and may start to go back into solution.
Figure 9 shows the conditions when the pressure wave reaches position B at time t4. As can be seen this is almost identical to the conditions shown in figure 1. The main difference is the velocity Ui’ is a little lower than the original Ui. As the end of the pipe is closed there is nowhere for the fluid at B to go. So this will induce the final reflection at time t4 and then the whole process is repeated.
Where t4=t3+L/C’ or t4=t0+2L/C+2L/C’
Figure 10 shows the condition after the forth reflection. The cycle has now began to repeat however the water hammer pressure is now reduced a little from P2 (figure 2) to P2’. This reduction in pressure has two causes. The fluid velocity has been reduced due to energy losses.
The sonic velocity C’’ may be less than the original sonic velocity C. If during the previous stages shown in figures 8 and 9 any gas was liberated from the fluid then this gas volume will have been reduced, however it takes time for the gas to be completely reabsorbed so there is likely to be small residual gas bubbles in the fluid. These gas bubbles will reduce the fluid bulk modulus so reducing the sonic speed.
Over a number of cycles the water hammer will eventually peter out. The pipe pressure will end up at P1 and the flow will stop oscillating. | 1,755 | 8,124 | {"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} | 3.515625 | 4 | CC-MAIN-2024-38 | latest | en | 0.903511 |
https://www.thegamerator.com/the-science-of-slot-machines-what-you-need-to-know/ | 1,716,171,172,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00161.warc.gz | 917,322,225 | 24,077 | Casino
# The Science of Slot Machines: what You need to Know
Online slots are among the most popular casino games since they are easy to learn and easy to win. Place your wagers and press the Spin button to begin the game. Time seemed to have slowed down on the reels at one point. If you can come up with a winning combination, you will be victorious. Isn’t it what we’re all saying? In both cases, the answer is yes and no.
Every aspect of a slot machine from its design and feel to its random number generator (RNG), which ensures a completely random outcome for every spin—is simple to a player; however, the science involved is extensive, including the psychological aspect that makes slots so popular with so many people.
This article examines the science of online slot machines and how they function in practise. Prior to delving into the topic of aesthetics, we’ll speak about how random numbers are generated.
Software that generates apparently random numbers is known as a random number generator (RNG) (RNG)
Online slot gacor hari ini games must function effectively, which is particularly critical given how far away they are from the gamers. To meet this condition, an online casino with slot machines uses a random number generator (RNG). Online casinos rely on the Random Number Generator (RNG) to ensure the outcomes of spins are as random as possible (RNG). Random numbers are generated by a computer algorithm. Each reel sign has its own unique sequence of integers created in order to do this. The only way to predict this outcome is to rely on chance.
RNGs may be classified in the following ways:
It’s called a hardware random number generator (HRNG): This RNG is responsible for generating really random numbers. They aren’t even calculated numbers; rather, they are the outcomes of an algorithm that was repeatedly run to collect data. Their unpredictability is enhanced by the fact that there are no repeating numbers or a predetermined method for cracking them. A true random number generator (TRNG) may produce really random numbers, as the name implies (TRNG).
A pseudo-random number generator (PRNG) creates seemingly random numbers, but they aren’t really. If you know the current state of the PRNG, you can replicate the produced random number sequence.
Inputs and outputs of the Random Number Generator (RNG)
As a first step in understanding how the Random Number Generator (RNG) works, when the algorithm is executed for the first time, a seed value is utilised. Because the seed value serves as a beginning point for every repeatable process, this is the only plausible explanation for this. The seed value may be retrieved from any of the computer’s many recursive processes and activities. A computer’s internal clock or a continuing piece of software might be to blame in slot gacor malam ini.
Conclusion
In terms of slots, this implies that each symbol on a reel is given a certain value, as previously indicated. Reels are spun hundreds of times per second at random, and the RNG algorithm generates a random series of numbers. After each spin, a unique set of numbers is generated that decides the symbols that display on your computer monitor. | 649 | 3,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.936678 |
https://physics.com.hk/2021/03/23/lightlike-compatification/ | 1,675,527,588,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00319.warc.gz | 476,020,683 | 15,180 | # Lightlike compatification
A First Course in String Theory
.
2.9 Lightlike compatification
(a) Rewrite this identification using light-cone coordinates.
\begin{aligned} \begin{bmatrix} x \\ ct \end{bmatrix} &\sim \begin{bmatrix} x \\ ct \end{bmatrix} + 2 \pi \begin{bmatrix} R \\ -R \end{bmatrix} \end{aligned}
~~~
\begin{aligned} \begin{bmatrix} x^+ \\ x^- \end{bmatrix} &= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \end{bmatrix} \\ \end{aligned}
\begin{aligned} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \\ \end{bmatrix} &\sim \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \\ \end{bmatrix} + \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} - 2 \pi R \\ 2 \pi R \\ \end{bmatrix} \\ \end{aligned}
\begin{aligned} \begin{bmatrix} x^+ \\ x^- \\ \end{bmatrix} &\sim \begin{bmatrix} x^+ \\ x^- \\ \end{bmatrix} + \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ - 4 \pi R \\ \end{bmatrix} \\ \end{aligned}
— Me@2021-03-22 06:06:10 PM
.
. | 469 | 1,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2023-06 | latest | en | 0.151623 |
http://de.metamath.org/mpegif/df-cid.html | 1,618,634,221,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00361.warc.gz | 22,575,326 | 6,612 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > df-cid Structured version Visualization version Unicode version
Definition df-cid 15653
Description: Define the category identity arrow. Since it is uniquely defined when it exists, we do not need to add it to the data of the category, and instead extract it by uniqueness. (Contributed by Mario Carneiro, 3-Jan-2017.)
Assertion
Ref Expression
df-cid comp
Distinct variable group: ,,,,,,,
Detailed syntax breakdown of Definition df-cid
StepHypRef Expression
1 ccid 15649 . 2
2 vc . . 3
3 ccat 15648 . . 3
4 vb . . . 4
52cv 1451 . . . . 5
6 cbs 15199 . . . . 5
75, 6cfv 5589 . . . 4
8 vh . . . . 5
9 chom 15279 . . . . . 6
105, 9cfv 5589 . . . . 5
11 vo . . . . . 6
12 cco 15280 . . . . . . 7 comp
135, 12cfv 5589 . . . . . 6 comp
14 vx . . . . . . 7
154cv 1451 . . . . . . 7
16 vg . . . . . . . . . . . . . 14
1716cv 1451 . . . . . . . . . . . . 13
18 vf . . . . . . . . . . . . . 14
1918cv 1451 . . . . . . . . . . . . 13
20 vy . . . . . . . . . . . . . . . 16
2120cv 1451 . . . . . . . . . . . . . . 15
2214cv 1451 . . . . . . . . . . . . . . 15
2321, 22cop 3965 . . . . . . . . . . . . . 14
2411cv 1451 . . . . . . . . . . . . . 14
2523, 22, 24co 6308 . . . . . . . . . . . . 13
2617, 19, 25co 6308 . . . . . . . . . . . 12
2726, 19wceq 1452 . . . . . . . . . . 11
288cv 1451 . . . . . . . . . . . 12
2921, 22, 28co 6308 . . . . . . . . . . 11
3027, 18, 29wral 2756 . . . . . . . . . 10
3122, 22cop 3965 . . . . . . . . . . . . . 14
3231, 21, 24co 6308 . . . . . . . . . . . . 13
3319, 17, 32co 6308 . . . . . . . . . . . 12
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3634, 18, 35wral 2756 . . . . . . . . . 10
3730, 36wa 376 . . . . . . . . 9
3837, 20, 15wral 2756 . . . . . . . 8
3922, 22, 28co 6308 . . . . . . . 8
4038, 16, 39crio 6269 . . . . . . 7
4114, 15, 40cmpt 4454 . . . . . 6
4211, 13, 41csb 3349 . . . . 5 comp
438, 10, 42csb 3349 . . . 4 comp
444, 7, 43csb 3349 . . 3 comp
452, 3, 44cmpt 4454 . 2 comp
461, 45wceq 1452 1 comp
Colors of variables: wff setvar class This definition is referenced by: cidfval 15660 cidffn 15662
Copyright terms: Public domain W3C validator | 1,071 | 2,249 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.137394 |
https://classroom.synonym.com/importance-handson-manipulatives-math-13601.html | 1,721,858,782,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518454.54/warc/CC-MAIN-20240724202030-20240724232030-00695.warc.gz | 143,598,122 | 27,459 | # Importance of Hands-on Manipulatives in Math
... Ryan McVay/Digital Vision/Getty Images
Math manipulatives range from simple counting blocks to geoboards and tangram puzzles. Manipulatives work well to solve problems, as a way to introduce new math skills and during free play to explore math concepts. The use of manipulatives varies based on the teacher's philosophy of math instruction, but these math materials offer several benefits to students.
## 1Concrete Representations
Manipulatives give the math student a concrete object to represent the concept he is learning. Instead of reading about a math concept or working out a problem on paper, he works with a physical object to better understand what he is learning. Diagrams in math textbooks often fall short because the student can't physically interact with them. The concrete representation is useful at all levels of math, from a preschooler using blocks to strengthen counting skills to an older student using fraction models to understand equivalent fractions.
## 2Engaged Senses
A worksheet or textbook assignment is limited in the senses it engages. The child only moves slightly to use his pencil. Manipulatives give him more freedom to move and get physically involved in solving the math problems. The manipulatives reach a wider range of learners, such as those who don't perform well on paper-and-pencil tasks. The manipulatives engage the sense of sight and touch. Discussions about the manipulatives -- either with the class or with a partner -- builds communication skills. You can also use these math tools to write about the concepts. Students can draw pictures and describe what they did with the manipulatives in a math journal.
## 3Problem Solving
Physical objects in front of the learner give him tools to solve problems that are complex of difficult to understand. Manipulating the objects can lead the child to the answer. For example, if he struggles to reduce a fraction to lowest terms, fraction strips can help him solve the problem. He sees that one-half matches up with three-sixths on the strips. A student learning division uses counters to solve the problem. For 42 divided by 7, he gathers 42 counters and divides them into 7 groups. Instead of staring at the paper trying to figure out the answer, he solves it with the counters. Learners also get confirmation on answers that they don't get on paper. With a worksheet, he won't know until the teacher checks the work if he was correct. With manipulatives, he can see right away that he is correct.
## 4Enjoyment
Manipulatives make math more enjoyable for most students. Completing paper-and-pencil assignments is often boring and tedious. Students lose interest quickly or struggle to get through the assignment. Manipulatives feel more like playing than learning, particularly when the students are allowed to experiment and explore with the tools outside of assignments. Even when a worksheet or written assignment is required, the manipulatives can make the problems easier and more interesting to solve.
Based in the Midwest, Shelley Frost has been writing parenting and education articles since 2007. Her experience comes from teaching, tutoring and managing educational after school programs. Frost worked in insurance and software testing before becoming a writer. She holds a Bachelor of Arts in elementary education with a reading endorsement. | 669 | 3,410 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2024-30 | latest | en | 0.942839 |
https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/chapter-6-section-6-1-the-greatest-common-factor-and-factoring-by-grouping-exercise-set-page-427/41 | 1,537,332,494,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155924.5/warc/CC-MAIN-20180919043926-20180919063926-00174.warc.gz | 754,801,698 | 14,069 | ## Introductory Algebra for College Students (7th Edition)
The terms of the given expression, $11x^2-23 ,$ are relatively prime. That is, the $GCF=1.$ Hence, it cannot be factored. | 50 | 181 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-39 | longest | en | 0.840877 |
https://biz.libretexts.org/Bookshelves/Business/Introductory_Business/Book%3A_Introduction_to_Business_(Lumen)/11%3A_Teamwork_and_Communication/11.05%3A_Communication_Channels_Flows_and_Networks | 1,656,828,275,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00501.warc.gz | 184,036,805 | 33,187 | 11.5: Communication Channels, Flows, and Networks
• Boundless
• Boundless
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What you’ll learn to do: describe typical communication channels, flows, and networks within an organization, and explain when different channels are appropriate
In this section, we’ll look more closely at the patterns of communication in business—who sends the messages, who receives them, and the different types of messages businesses typically use.
Learning Objectives
• Differentiate between appropriate and inappropriate uses of different communication channels
• Differentiate between formal and informal communication networks
Communication Channels
In communications, a channel is the means of passing information from a sender to a recipient. Determining the most appropriate channel, or medium, is critical to the effectiveness of communication. Channels include oral means such as telephone calls and presentations, and written modes such as reports, memos, and email.
Communication channels differ along a scale from rich to lean. Think about how you would select a steak—some have more fat than others; they are rich and full of flavor and body. If, however, you are on a diet and just want the meat, you will select a lean steak. Communication channels are the similar: rich channels are more interactive, provide opportunities for two-way communication, and allow both the sender and receiver to read the nonverbal messages. The leanest channels, on the other hand, trim the “fat” and present information without allowing for immediate interaction, and they often convey “just the facts.” The main channels of communication are grouped below from richest to leanest:
• Richest channels: face-to-face meeting; in-person oral presentation
• Rich channels: online meeting; video conference
• Lean channels: teleconference; phone call; voice message; video (e.g., Facetime)
• Leanest channels: blog; report; brochure; newsletter; flier; email; phone text; social media posts (e.g., Twitter, Facebook)
Bill Gates speaking at a school. A speaker giving a large presentation is an example of oral communication.
Oral communications tend to be richer channels because information can be conveyed through speech as well as nonverbally through tone of voice and body language. Oral forms of communication can range from a casual conversation with a colleague to a formal presentation in front of many employees. Richer channels are well suited to complex (or potentially unsettling) information, since they can provide opportunities to clarify meaning, reiterate information, and display emotions.
While written communication does not have the advantage of immediacy and interaction, it can be the most effective means of conveying large amounts of information. Written communication is an effective channel when context, supporting data, and detailed explanations are necessary to inform or persuade others. One drawback to written communications is that they can be misunderstood or misinterpreted by an audience that doesn’t have subsequent opportunities to ask clarifying questions or otherwise respond.
The following are some examples of different types of communication channels and their advantages:
• Web-based communication, such as video conferencing, allows people in different locations to hold interactive meetings. Other Web-based communication, such as information presented on a company Web site, is suited for sharing transaction details (such as order confirmation) or soliciting contact information (such as customer phone number and address)
• Emails provide instantaneous written communication; effective for formal notices and updates, as well as informal exchanges.
• Letters are a more formal method of written communication usually reserved for important messages such as proposals, inquiries, agreements, and recommendations.
• Presentations are usually oral and usually include an audiovisual component, like copies of reports, or material prepared in Microsoft PowerPoint or Adobe Flash.
• Telephone meetings/conference calls allow for long-distance interaction.
• Message boards and Forums allow people to instantly post information to a centralized location.
• Face-to-face meetings are personal, interactive exchanges that provide the richest communication and are still the preferred method of communication in business.
So, we have written and oral channels, channels that range from rich to lean, and then, within those, multiple channels from which the sender can choose. How do you decide the best channel for your message? When deciding which communication channel to use, the following are some of the important factors to consider:
• the audience and their reaction to the message;
• the length of time it will take to convey the information;
• the complexity of the message;
• the need for a permanent record of the communication;
• the degree to which the information is confidential; and
• the cost of the communication.
If you choose the wrong channel—that is, if the channel is not effective for the type of message and meaning you want to create—you are likely to generate misunderstanding and possibly end up making matters worse. Using the wrong channels can impede communication and can even create mistrust. For example, a manager wants to compliment an employee for his work on a recent project. She can use different approaches and channels to do this. She could send the an employee a text: “Hey, nice work on the project!” Or she could send him an email containing the same message. She could also stop by his desk and personally compliment him. She could also praise him in front of the whole department during a meeting. In each case the message is the same, but the different channels alter the way the message is perceived. If the employee spent months working on the project, getting a “Hey, nice work on the project!” text message or email might seem like thin praise—insulting even. If the employee is shy, being singled out for praise during a departmental meeting might be embarrassing. A face-to-face compliment during a private meeting might be received better. As you can see, getting the channel right is just as important as sending the right message.
Communication Flows
Communication within a business can involve different types of employees and different functional parts of an organization. These patterns of communication are called flows, and they are commonly classified according to the direction of interaction: downward, upward, horizontal, diagonal, external. As you learn about each of these, we will discuss how these flows function at Little Joe’s Auto.
Downward Communication
When leaders and managers share information with lower-level employees, it’s called downward, or top-down communication. In other words, communication from superiors to subordinates in a chain of command is a downward communication. This communication flow is used by the managers to transmit work-related information to the employees at lower levels. Ensuring effective downward communication isn’t always easy. Differences in experience, knowledge, levels of authority, and status make it possible that the sender and recipient do not share the same assumptions or understanding of context, which can result in messages being misunderstood or misinterpreted. Creating clearly worded, unambiguous communications and maintaining a respectful tone can facilitate effective downward communication.
Little Joes's auto: downward communication
Little Joe holds a meeting every morning with his entire sales staff. In this meeting he gives them information on new cars on the lot, current interest rates available to customers, and how close they are to meeting the company’s monthly sales goals. The most important information shared is a “hot sheet” that lists the cars that need to be sold ASAP because they have been on the lot for more than forty-five days. Every car sold from the hot sheet earns the salesperson a $500 bonus, adding more than a little motivation to the mix. As Little Joe goes through his morning briefing, the sales staff listen, take notes, and sometimes ask a few clarifying questions, but clearly the purpose of this daily pow-wow is for Little Joe to convey the information his staff need to perform their jobs and meet the expectations of management. Upward Communication Upward communication is the transmission of information from lower levels of an organization to higher ones; the most common situation is employees communicating with managers. Managers who encourage upward communication foster cooperation, gain support, and reduce frustration among their employees. The content of such communication can include requests, estimations, proposals, complaints, appeals, reports, and any other information directed from subordinates to superiors. Upward communication is often made in response to downward communication; for instance, when employees answer a question from their manager. In this respect, upward communication is a good measure of whether a company’s downward communication is effective. The availability of communication channels affects employees’ overall satisfaction with upward communication. For example, an open-door policy sends the signal to employees that the manager welcomes impromptu conversations and other communication. This is likely to make employees feel satisfied with their level of access to channels of upward communication and less apprehensive about communicating with their superiors. For management, upward communication is an important source of information that can inform business decisions. It helps to alert management of new developments, levels of performance, and other issues that may require their attention. little joe's auto: upward communication One afternoon, Frances knocks at Little Joe’s office door, which is always open. Frances wants Little Joe to know that he has a couple interested in one of the new cars on the hot sheet, a 2015 Sonata, but the car is out of their price range by just a hair. Frances knows the couple from his church and really wants to help them get reliable transportation, but he also knows he needs to get the deal past the finance manager. Frances wants to know if it’s possible for him to cut the price to his customers and give up his$500 bonus for selling the car. Little Joe agrees, since it really makes no difference who gets the $500—Frances or the customer. Horizontal Communication Horizontal communication, also called lateral communication, involves the flow of messages between individuals and groups on the same level of an organization, as opposed to up or down. Sharing information, solving problems, and collaborating horizontally is often more timely, direct, and efficient than up or down communication, since it occurs directly between people working in the same environment. Communication within a team is an example of horizontal communication; members coordinate tasks, work together, and resolve conflicts. Horizontal communication occurs formally in meetings, presentations, and formal electronic communication, and informally in other, more casual exchanges within the office. When there are differences in style, personality, or roles among coworkers, horizontal communication may not run smoothly. According to Professor Michael Papa, horizontal communication problems can occur because of territoriality, rivalry, specialization, and simple lack of motivation. Territoriality occurs when members of an organization regard other people’s involvement in their area as inappropriate or unwelcome. Rivalry between individuals or teams can make people reluctant to cooperate and share information. Specialization is a problem that occurs when there is a lack of uniform knowledge or vocabulary within or between departments. Finally, horizontal communication often fails simply because organization members are unwilling to expend the additional effort needed to reach out beyond their immediate team. little joe's auto: horizontal communication Little Joe picks up his phone and calls Brian, the finance manager. He explains that Frances is going to send a deal through on a hot-sheet car that is$500 less than the bottom line, but if the rest of the deal is solid, Brian should approve it. Brian immediately begins to object, when Little Joe cuts him off and says that Frances is waiving his hot-sheet bonus. When Little Joe hangs up with Brian, he tells Frances he’s set—now go sell that car!
Diagonal Communication
Diagonal communication is the sharing of information among different structural levels within a business. This kind of communication flow is increasingly the norm in organizations (in the same way that cross-functional teams are becoming more common), since it can maximize the efficiency of information exchange. The shortest distance between two points is a straight line. Diagonal communication routes are the straight lines that speed communications directly to their recipients, at the moment communication is necessary. Communications that zigzag along horizontal and vertical routes, on the other hand, are vulnerable to the schedules and availability of the individuals who reside at each level.
little joe's auto: diagonal communication
Frances returns to his customers and tells him he thinks he’s got a way to make the deal work. Brian, the finance manager, approves the deal per his conversation with Little Joe, so all that’s left is the final inspection in the service department. The customers have told Frances they need to be home by 3 pm, but when Frances sees the time and looks over at the line of cars waiting for final inspection, his stomach drops. There’s no way he is going to get them out of the dealership by three, and he’s afraid he’ll lose the sale. He heads over to the service department to find Marcie, the service manager. He finds her in one of the service bays and explains his situation, asking if there’s any way his customer can be moved ahead in the line. Marcie checks her clipboard, does some quick calculations, and calls over one of the service techs. She tells him to locate the 2015 Sonata and get it up on the lift next. Smiling, she turns to Frances and says, “Mission accomplished.”
External Communication
Another type of communication flow is external, when an organization communicates with people or organizations outside the business. Recipients of external communication include customers, lawmakers, suppliers, and other community stakeholders. External communication is often handled by marketing and sales. Annual reports, press releases, product promotions, financial reports are all examples of external communication.
little joe's auto: external communication
The last thing Frances does before he hands the keys to his customers is to affix a Little Joe’s Auto license plate frame to the front and back of the Sonata. Now everyone who sees his customers driving their new car will know where they bought it. He hopes this sale will generate more business for himself and the dealership, so along with the keys to the car, he gives them several business cards and a coupon for a free oil change. At 2:30, Frances waves good-bye to his customers as they drive their new Sonata off the lot.
In order to close this deal, the communication at Little Joe’s Auto has flowed in every direction—upward, downward, horizontally, diagonally, and externally.
Communication Networks
By now you know that business communication can take different forms and flow between different kinds of senders and receivers. Another way to classify communication is by network.
An organization’s formal communication network is comprised of all the communication that runs along its official lines of authority. In other words, the formal network follows reporting relationships. As you might expect, when a manager sends an email to her sales team describing the new commission structure for the next set of sales targets, that email (an example of downward communication) is being sent along the company’s formal network that connects managers to their subordinates.
An informal communication network, on the other hand, doesn’t follow authority lines and is established around the social affiliation of members of an organization. Such networks are also described as “grapevine communication.” They may come into being through the rumor mill, social networking, graffiti, spoof newsletters, and spontaneous water-cooler conversations.
Informal versus Formal Networks
• Formal communication follows practices shaped by hierarchy, technology systems, and official policy.
• Formal communication usually involves documentation, while informal communication usually leaves no recorded trace for others to find or share.
• Formal communications in traditional organizations are frequently “one-way”: They are initiated by management and received by employees.
• Formal Communication content is perceived as authoritative because it originates from the highest levels of the company.
• Informal communication occurs in any direction and takes place between individuals of different status and roles.
• Informal communication frequently crosses boundaries within an organization and is commonly separate from work flows. That is, it often occurs between people who do not work together directly but share an affiliation or a common interest in the organization’s activities and/or a motivation to perform their jobs well.
• Informal communication occurs outside an organization’s established channels for conveying messages and transmitting information.
In the past, many organizations considered informal communication (generally associated with interpersonal, horizontal communication) a hindrance to effective organizational performance and tried to stamp it out. This is no longer the case. The maintenance of personal networks and social relationships through information communication is understood to be a key factor in how people get work done. It might surprise you to know that 75 percent of all organizations’ practices, policies, and procedures are shared through grapevine communication.[1]
While informal communication is important to an organization, it also may have disadvantages. When it takes the form of a “rumor mill” spreading misinformation, informal communication is harmful and difficult to shut down because its sources cannot be identified by management. Casual conversations are often spontaneous, and participants may make incorrect statements or promulgate inaccurate information. Less accountability is expected from informal communications, which can cause people to be indiscreet, careless in their choice of words, or disclose sensitive information.
1. Keith Davis, "Grapevine Communication Among Lower and Middle Managers," Personnal Journal, April, 1969, p. 272.
11.5: Communication Channels, Flows, and Networks is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Boundless via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. | 3,869 | 20,142 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2022-27 | latest | en | 0.673417 |
https://whatisconvert.com/207-liters-in-imperial-quarts | 1,656,664,627,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00417.warc.gz | 645,079,811 | 7,215 | # What is 207 Liters in Imperial Quarts?
## Convert 207 Liters to Imperial Quarts
To calculate 207 Liters to the corresponding value in Imperial Quarts, multiply the quantity in Liters by 0.87987699319635 (conversion factor). In this case we should multiply 207 Liters by 0.87987699319635 to get the equivalent result in Imperial Quarts:
207 Liters x 0.87987699319635 = 182.13453759164 Imperial Quarts
207 Liters is equivalent to 182.13453759164 Imperial Quarts.
## How to convert from Liters to Imperial Quarts
The conversion factor from Liters to Imperial Quarts is 0.87987699319635. To find out how many Liters in Imperial Quarts, multiply by the conversion factor or use the Volume converter above. Two hundred seven Liters is equivalent to one hundred eighty-two point one three five Imperial Quarts.
## Definition of Liter
The liter (also written "litre"; SI symbol L or l) is a non-SI metric system unit of volume. It is equal to 1 cubic decimeter (dm3), 1,000 cubic centimeters (cm3) or 1/1,000 cubic meter. The mass of one liter liquid water is almost exactly one kilogram. A liter is defined as a special name for a cubic decimeter or 10 centimeters × 10 centimeters × 10 centimeters, thus, 1 L ≡ 1 dm3 ≡ 1000 cm3.
## Definition of Imperial Quart
The quart (abbreviation qt.) is an English unit of volume equal to a quarter gallon. It is divided into two pints or four cups. The imperial quart, used for both liquid or dry capacity, is equal to one quarter of an imperial gallon, or exactly 1.1365225 liters.
## Using the Liters to Imperial Quarts converter you can get answers to questions like the following:
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• How to convert 207 Liters to Imperial Quarts?
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• How many uk qt are in 207 L?
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• How much is 207 L in uk qt? | 562 | 2,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.836267 |
http://www.livmathssoc.org.uk/cgi-bin/sews.py?Bearing | 1,632,341,946,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057388.12/warc/CC-MAIN-20210922193630-20210922223630-00090.warc.gz | 107,498,317 | 2,167 | This needs a diagram. We take bearings by measuring the angle between two points.
There are several types of bearings, but there are two main types:
• Bearing from North
• Relative bearing.
Bearings from North are always given as an angle (usually in degrees) going clockwise. This due East is at a bearing of 90 degrees, due South is 180 degrees, and due West is at 270 degrees. North is very occasionally given as 360.0 degrees, but this is unusual.
Relative bearings are given from one to another, and again, are given as an angle measured clockwise.
Bearings always have a direction, they go from one place, to another, always clockwise (rightwards). One can speak of the angle between two points, but that is not a bearing. | 162 | 732 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.982617 |
https://rdrr.io/cran/quest/man/change.html | 1,638,811,165,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363309.86/warc/CC-MAIN-20211206163944-20211206193944-00177.warc.gz | 543,028,279 | 9,171 | # change: Change Score from a Numeric Vector In quest: Prepare Questionnaire Data for Analysis
## Description
`change` creates a change score (aka difference score) from a numeric vector. It is assumed that the vector is already sorted by time such that the first element is earliest in time and the last element is the latest in time.
## Usage
`1` ```change(x, n, undefined = NA) ```
## Arguments
`x` numeric vector. `n` integer vector with length 1. Specifies how the change score is calculated. If `n` is positive, then the change score is calculated from lead - original; if `n` is negative, then the change score is calculated from original - lag. The magnitude of `n` determines how many elements are shifted for the lead/lag within the calculation. If `n` is zero, then `change` simply returns a vector or zeros. See details of `shift`. `undefined` atomic vector with length 1 (probably makes sense to be the same typeof as `x`). Specifies what to insert for undefined values after the shifting takes place. See details of `shift`.
## Details
It is recommended to use `L` when specifying `n` to prevent problems with floating point numbers. `shift` tries to circumvent this issue by a call to `round` within `shift` if `n` is not an integer; however that is not a complete fail safe. The problem is that `as.integer(n)` implicit in `shift` truncates rather than rounds. See details of `shift`.
## Value
an atomic vector of the same length as `x` that is the change score. If `x` and `undefined` are different typeofs, then the return will be coerced to the most complex typeof (i.e., complex to simple: character, double, integer, logical).
`changes` `change_by` `changes_by` `shift`
``` 1 2 3 4 5 6 7 8 9 10``` ```change(x = attitude[[1]], n = -1L) # use L to prevent problems with floating point numbers change(x = attitude[[1]], n = -2L) # can specify any integer up to the length of `x` change(x = attitude[[1]], n = +1L) # can specify negative or positive integers change(x = attitude[[1]], n = +2L, undefined = -999) # user-specified indefined value change(x = attitude[[1]], n = -2L, undefined = -999) # user-specified indefined value change(x = attitude[[1]], n = 0L) # returns a vector of zeros ## Not run: change(x = setNames(object = letters, nm = LETTERS), n = 3L) # character vector returns an error ## End(Not run) ``` | 598 | 2,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-49 | latest | en | 0.819804 |
https://byjus.com/us/math/trapezoids/ | 1,669,693,281,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710685.0/warc/CC-MAIN-20221129031912-20221129061912-00259.warc.gz | 182,302,289 | 28,852 | What is a Trapezoid in Math? (Definition, Shape, Examples) - BYJUS
# Trapezoids
We know that a quadrilateral is a polygon with four sides and that there are various types of quadrilaterals such as a square, a rectangle, a trapezoid, a rhombus, and so on. Here we will learn about trapezoids, their properties and some related formulas....Read MoreRead Less
## What is a Trapezoid?
A trapezoid is a quadrilateral with exactly one pair of opposite sides that are parallel. The pair of opposite sides that are parallel are usually the top and the bottom sides of a trapezoid.
## Properties of a Trapezoid
• The parallel sides are called bases
• The non-parallel sides are called legs
• The perpendicular distance between the bases is called height or altitude
• The two consecutive angles with a common side as the base are called base angles
• The angles formed on the same leg, between the parallel sides, are supplementary
• The sum of all interior angles is 360°
## Area of a Trapezoid
The area A of trapezoid is one half of the product of its height h and sum of its bases b$$_1$$ and b$$_2$$.
A = $$\frac{1}{2}$$ x h x (b$$_1$$ + b$$_2$$ )
It must be noted that the area of geometric shapes is measured in square units.
## Types of Trapezoids
There are three types of trapezoids:
• Right trapezoid
• Isosceles trapezoid
• Scalene trapezoid
1. Right trapezoid : A trapezoid with one pair of adjacent right angles is called a right trapezoid.
2. Isosceles trapezoid : A trapezoid whose non parallel sides, called legs, are congruent or equal in length is called an isosceles trapezoid. The base angles of an isosceles trapezoid are equal.
3. Scalene trapezoid : A trapezoid that has no sides that are equal in length is a scalene trapezoid.
## Solved Trapezoid Examples
Example 1: Find the area of the given trapezoid.
Solution:
A = $$\frac{1}{2}$$ x h x (b$$_1$$ + b$$_2$$) [Write the formula]
A = $$\frac{1}{2}$$ x 4 x (5 + 6) [Substitute]
A = 22
So, the area of the given trapezoid is 22 square inches.
Example 2: A garden is in the shape of a trapezoid as shown in the image. The area of the garden is 10000 square meters. Find the shortest distance between the parallel sides of the garden?
Solution: The shortest distance between the parallel sides is the height of the trapezoid.
We can use the formula for the area of a trapezoid to find the height.
A = $$\frac{1}{2}$$ x h x (b$$_1$$ + b$$_2$$) [Write the formula]
10000 = $$\frac{1}{2}$$ x h x (150 + 100) [Substitute 10000 for A, 150 for b$$_1$$ and 100 for b$$_2$$]
10000 x 2 = h x 250 [Multiply by 2 on each side.]
$$\frac{20000}{250}$$ = h [Divide each side by 250]
80 = h [Solve]
So, the shortest distance between parallel sides of the trapezoid shaped garden is 80 meters.
Example 3: Find the value of x.
Solution :
The given trapezoid shows that AD = BC, so this is an isosceles trapezoid.
∠ADC = ∠BCD [Base angles of an isosceles trapezoid are equal]
65 = x + 15 [Substitute 65 for ∠ADC and x + 15 for ∠BCD.]
65 – 15 = x + 15 – 15 [Subtract 15 on both sides]
50 = x [Solve]
So, the value of x is 50. | 986 | 3,368 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2022-49 | latest | en | 0.893888 |
https://handwiki.org/wiki/Uniformly_most_powerful_test | 1,701,789,840,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100551.2/warc/CC-MAIN-20231205140836-20231205170836-00392.warc.gz | 343,988,974 | 18,386 | # Uniformly most powerful test
Short description: Hypothesis test
In statistical hypothesis testing, a uniformly most powerful (UMP) test is a hypothesis test which has the greatest power $\displaystyle{ 1 - \beta }$ among all possible tests of a given size α. For example, according to the Neyman–Pearson lemma, the likelihood-ratio test is UMP for testing simple (point) hypotheses.
## Setting
Let $\displaystyle{ X }$ denote a random vector (corresponding to the measurements), taken from a parametrized family of probability density functions or probability mass functions $\displaystyle{ f_{\theta}(x) }$, which depends on the unknown deterministic parameter $\displaystyle{ \theta \in \Theta }$. The parameter space $\displaystyle{ \Theta }$ is partitioned into two disjoint sets $\displaystyle{ \Theta_0 }$ and $\displaystyle{ \Theta_1 }$. Let $\displaystyle{ H_0 }$ denote the hypothesis that $\displaystyle{ \theta \in \Theta_0 }$, and let $\displaystyle{ H_1 }$ denote the hypothesis that $\displaystyle{ \theta \in \Theta_1 }$. The binary test of hypotheses is performed using a test function $\displaystyle{ \varphi(x) }$ with a reject region $\displaystyle{ R }$ (a subset of measurement space).
$\displaystyle{ \varphi(x) = \begin{cases} 1 & \text{if } x \in R \\ 0 & \text{if } x \in R^c \end{cases} }$
meaning that $\displaystyle{ H_1 }$ is in force if the measurement $\displaystyle{ X \in R }$ and that $\displaystyle{ H_0 }$ is in force if the measurement $\displaystyle{ X\in R^c }$. Note that $\displaystyle{ R \cup R^c }$ is a disjoint covering of the measurement space.
## Formal definition
A test function $\displaystyle{ \varphi(x) }$ is UMP of size $\displaystyle{ \alpha }$ if for any other test function $\displaystyle{ \varphi'(x) }$ satisfying
$\displaystyle{ \sup_{\theta\in\Theta_0}\; \operatorname{E}[\varphi'(X)|\theta]=\alpha'\leq\alpha=\sup_{\theta\in\Theta_0}\; \operatorname{E}[\varphi(X)|\theta]\, }$
we have
$\displaystyle{ \forall \theta \in \Theta_1, \quad \operatorname{E}[\varphi'(X)|\theta]= 1 - \beta'(\theta) \leq 1 - \beta(\theta) =\operatorname{E}[\varphi(X)|\theta]. }$
## The Karlin–Rubin theorem
The Karlin–Rubin theorem can be regarded as an extension of the Neyman–Pearson lemma for composite hypotheses.[1] Consider a scalar measurement having a probability density function parameterized by a scalar parameter θ, and define the likelihood ratio $\displaystyle{ l(x) = f_{\theta_1}(x) / f_{\theta_0}(x) }$. If $\displaystyle{ l(x) }$ is monotone non-decreasing, in $\displaystyle{ x }$, for any pair $\displaystyle{ \theta_1 \geq \theta_0 }$ (meaning that the greater $\displaystyle{ x }$ is, the more likely $\displaystyle{ H_1 }$ is), then the threshold test:
$\displaystyle{ \varphi(x) = \begin{cases} 1 & \text{if } x \gt x_0 \\ 0 & \text{if } x \lt x_0 \end{cases} }$
where $\displaystyle{ x_0 }$ is chosen such that $\displaystyle{ \operatorname{E}_{\theta_0}\varphi(X)=\alpha }$
is the UMP test of size α for testing $\displaystyle{ H_0: \theta \leq \theta_0 \text{ vs. } H_1: \theta \gt \theta_0 . }$
Note that exactly the same test is also UMP for testing $\displaystyle{ H_0: \theta = \theta_0 \text{ vs. } H_1: \theta \gt \theta_0 . }$
## Important case: exponential family
Although the Karlin-Rubin theorem may seem weak because of its restriction to scalar parameter and scalar measurement, it turns out that there exist a host of problems for which the theorem holds. In particular, the one-dimensional exponential family of probability density functions or probability mass functions with
$\displaystyle{ f_\theta(x) = g(\theta) h(x) \exp(\eta(\theta) T(x)) }$
has a monotone non-decreasing likelihood ratio in the sufficient statistic $\displaystyle{ T(x) }$, provided that $\displaystyle{ \eta(\theta) }$ is non-decreasing.
## Example
Let $\displaystyle{ X=(X_0 ,\ldots , X_{M-1}) }$ denote i.i.d. normally distributed $\displaystyle{ N }$-dimensional random vectors with mean $\displaystyle{ \theta m }$ and covariance matrix $\displaystyle{ R }$. We then have
\displaystyle{ \begin{align} f_\theta (X) = {} & (2 \pi)^{-MN/2} |R|^{-M/2} \exp \left\{-\frac 1 2 \sum_{n=0}^{M-1} (X_n - \theta m)^T R^{-1}(X_n - \theta m) \right\} \\[4pt] = {} & (2 \pi)^{-MN/2} |R|^{-M/2} \exp \left\{-\frac 1 2 \sum_{n=0}^{M-1} \left (\theta^2 m^T R^{-1} m \right ) \right\} \\[4pt] & \exp \left\{-\frac 1 2 \sum_{n=0}^{M-1} X_n^T R^{-1} X_n \right\} \exp \left\{\theta m^T R^{-1} \sum_{n=0}^{M-1}X_n \right\} \end{align} }
which is exactly in the form of the exponential family shown in the previous section, with the sufficient statistic being
$\displaystyle{ T(X) = m^T R^{-1} \sum_{n=0}^{M-1}X_n. }$
Thus, we conclude that the test
$\displaystyle{ \varphi(T) = \begin{cases} 1 & T \gt t_0 \\ 0 & T \lt t_0 \end{cases} \qquad \operatorname{E}_{\theta_0} \varphi (T) = \alpha }$
is the UMP test of size $\displaystyle{ \alpha }$ for testing $\displaystyle{ H_0: \theta \leqslant \theta_0 }$ vs. $\displaystyle{ H_1: \theta \gt \theta_0 }$
## Further discussion
Finally, we note that in general, UMP tests do not exist for vector parameters or for two-sided tests (a test in which one hypothesis lies on both sides of the alternative). The reason is that in these situations, the most powerful test of a given size for one possible value of the parameter (e.g. for $\displaystyle{ \theta_1 }$ where $\displaystyle{ \theta_1 \gt \theta_0 }$) is different from the most powerful test of the same size for a different value of the parameter (e.g. for $\displaystyle{ \theta_2 }$ where $\displaystyle{ \theta_2 \lt \theta_0 }$). As a result, no test is uniformly most powerful in these situations.
## References
1. Casella, G.; Berger, R.L. (2008), Statistical Inference, Brooks/Cole. ISBN:0-495-39187-5 (Theorem 8.3.17) | 1,795 | 5,804 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2023-50 | latest | en | 0.583044 |
https://testbook.com/question-answer/in-this-question-three-statements-are-given-foll--6375dfa540350aabb584b2e8 | 1,716,419,197,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00286.warc.gz | 472,827,096 | 46,882 | In this question, three statements are given, followed by two conclusions numbered I and II. Assuming the statements to be true, even if they seem to be at variance with commonly known facts, decide which of the conclusions logically follows/follow from the statements.Statements:All pens are books.All books are novels.All novels are good.Conclusions:I. Some novels are pens.II. All books are good.
This question was previously asked in
SSC CPO 2022 Tier-I Official Paper (Held On : 9 Nov 2022 Shift 2) [Answer Key]
View all SSC CPO Papers >
1. Both conclusions I and II follow.
2. Neither conclusion I nor II follows.
3. Only conclusion I follows.
4. Only conclusion II follows.
Answer (Detailed Solution Below)
Option 1 : Both conclusions I and II follow.
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Detailed Solution
Explanation:
We draw a Venn diagram using the given statements:
All pens are books.
All books are novels.
All novels are good.
Now we consider the given conclusions:
I. Some novels are pens → Follows → Since 'all pens are books' and 'all books are pens' we can say that all pens are novels which can also be referred as some novels are pens and therefore the conclusion follows.
II. All books are good → Follows → We have 'all books are novels' and 'all novels are good' so there is an indirect relation between books and good and it is safe to say that all books are good thus the conclusion follows.
So both conclusions I and II follows.
Hence the correct answer is option 1.
Latest SSC CPO Updates
Last updated on May 15, 2024
-> The SSC CPO Marks have been released.
-> Candidates can check their marks obtained in the official portal by registration number and password.
-> The facility for the same will be available on the website from 14th to 28th May 2024.
-> The SSC CPO Notification 2024 has been released by the Staff Selection Commission.
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-> Prepare well for the exam by solving SSC CPO Previous Year Papers. Also, attempt the SSC CPO Mock Tests | 552 | 2,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.93551 |
https://blog.wuruihong.com/2019/01/30s-dart-count-by/ | 1,652,894,165,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00728.warc.gz | 172,989,684 | 9,958 | # 30秒学会 Dart 片段 – countBy
Groups the elements of a list based on the given function and returns the count of elements in each group.
Use Iterable.map() to map each element to the value returned by fn, Iterable.toSet() to get the unique values of the list.
Use Map.fromIterable(), Iterable.where() and Iterable.length to generate a map with the unique values as keys and their frequencies as values.
#### 代码实现
Map<Y, int> countBy<T, Y>(Iterable<T> itr, Y Function(T) fn) {
return Map.fromIterable(itr.map(fn).toSet(),
value: (i) => itr.where((v) => fn(v) == i).length);
}
#### 使用样例
countBy(['one', 'two', 'three'], (v) => v.length); // [{3: 2, 5: 1}] | 187 | 655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.515352 |
https://www.cpalms.org/Public/PreviewStandard/Preview/5669 | 1,624,349,508,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488512243.88/warc/CC-MAIN-20210622063335-20210622093335-00423.warc.gz | 619,350,300 | 16,966 | # MAFS.912.S-MD.2.5
Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
1. Find the expected payoff for a game of chance. For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant.
2. Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident.
General Information
Subject Area: Mathematics
Grade: 912
Domain-Subdomain: Statistics & Probability: Using Probability to Make Decisions
Cluster: Level 2: Basic Application of Skills & Concepts
Date Adopted or Revised: 02/14
Date of Last Rating: 02/14
Status: State Board Approved
## Related Courses
This benchmark is part of these courses.
1210300: Probability and Statistics Honors (Specifically in versions: 2014 - 2015, 2015 - 2019, 2019 - 2022 (current), 2022 and beyond)
1298310: Advanced Topics in Mathematics (formerly 129830A) (Specifically in versions: 2014 - 2015, 2015 - 2022 (course terminated))
1200500: Advanced Algebra with Financial Applications (Specifically in versions: 2014 - 2015 (course terminated))
1200387: Mathematics for Data and Financial Literacy (Specifically in versions: 2016 - 2022 (current), 2022 and beyond)
## Related Access Points
Alternate version of this benchmark for students with significant cognitive disabilities.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this benchmark.
## Lesson Plans
Modeling Conditional Probabilities 2:
This lesson unit is intended to help you assess how well students understand conditional probability, and, in particular, to help you identify and assist students who have the following difficulties representing events as a subset of a sample space using tables and tree diagrams and understanding when conditional probabilities are equal for particular and general situations.
Type: Lesson Plan
Probability:
This lesson is designed to develop students' understanding of probability in real life situations. Students will also be introduced to running experiments, experimental probability, and theoretical probability. This lesson provides links to discussions and activities related to probability as well as suggested ways to integrate them into the lesson. Finally, the lesson provides links to follow-up lessons designed for use in succession with the current one.
Type: Lesson Plan
Phalangelpodscribitis? - Analysis with Probability:
Have you ever had a cold or some other ailment that was just a nuisance to you? You tried this medication and that medication in order to treat your self-diagnosis. However, when you have exhausted all your avenues, you find yourself at the Physician's office: paying the co-pay, getting a prescription, paying more to fill the prescription with hopes of not experiencing any of the side effects associated with the medicine, and if that particular medicine doesn't work, you are back at the doctor's office and switched to another.
Well, Phalangelpodscribitis is a recently diagnosed ailment that will put a person's feet in motion. It isn't contagious but the treatment can be intense. In this lesson students will be presented with seven (7) medications that will help cure an individual of Phalangelpodscribitis. Students will be given the effectiveness of each medication, the cost to patients with and without insurance, and the possible side effects of each. Each team will be tasked with ranking these medications for a client in order to help him decide the pros and cons of the medications that should be used in treating Phalangelpodscribitis (PPS).
Each team will be responsible for recording the procedure they used to rank the medications and to calculate the expected cost for the client when two medications must be administered since the first will prove ineffective for treatment alone. The team's suggestion brings results and the patient is cured!!
Time has passed and Phalangelpodscribitis, currently known as PPS, has returned. Oh no! What will your team suggest when the doctor begins to discuss the patient's mortality rate as it is associated with the medication?
Type: Lesson Plan
## Perspectives Video: Expert
How Math Models Help Insurance Companies After a Hurricane Hits:
Hurricanes can hit at any time! How do insurance companies use math and weather data to help to restore the community?
Download the CPALMS Perspectives video student note taking guide.
Type: Perspectives Video: Expert
## STEM Lessons - Model Eliciting Activity
Phalangelpodscribitis? - Analysis with Probability:
Have you ever had a cold or some other ailment that was just a nuisance to you? You tried this medication and that medication in order to treat your self-diagnosis. However, when you have exhausted all your avenues, you find yourself at the Physician's office: paying the co-pay, getting a prescription, paying more to fill the prescription with hopes of not experiencing any of the side effects associated with the medicine, and if that particular medicine doesn't work, you are back at the doctor's office and switched to another.
Well, Phalangelpodscribitis is a recently diagnosed ailment that will put a person's feet in motion. It isn't contagious but the treatment can be intense. In this lesson students will be presented with seven (7) medications that will help cure an individual of Phalangelpodscribitis. Students will be given the effectiveness of each medication, the cost to patients with and without insurance, and the possible side effects of each. Each team will be tasked with ranking these medications for a client in order to help him decide the pros and cons of the medications that should be used in treating Phalangelpodscribitis (PPS).
Each team will be responsible for recording the procedure they used to rank the medications and to calculate the expected cost for the client when two medications must be administered since the first will prove ineffective for treatment alone. The team's suggestion brings results and the patient is cured!!
Time has passed and Phalangelpodscribitis, currently known as PPS, has returned. Oh no! What will your team suggest when the doctor begins to discuss the patient's mortality rate as it is associated with the medication?
## Student Resources
Vetted resources students can use to learn the concepts and skills in this benchmark.
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this benchmark. | 1,371 | 6,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-25 | latest | en | 0.857301 |
https://math.stackexchange.com/questions/891169/conjecture-tract-version-of-gauss-lucas-theorem-for-higher-derivatives | 1,713,675,219,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817729.0/warc/CC-MAIN-20240421040323-20240421070323-00557.warc.gz | 363,765,872 | 36,037 | # Conjecture: Tract version of Gauss--Lucas Theorem for higher derivatives.
The Gauss--Lucas Theorem states that all zeros of a degree $n$ complex polynomial $p(z)$ are contained in the convex hull of the zeros of $p$. By iteration, this implies that the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$ are contained in the convex hull of the zeros of $p$.
The Riemann--Hurwitz Theorem (among others) implies that if a tract $D$ of $p$ (namely a component of the set $\{z:|p(z)|<\epsilon\}$ for some $\epsilon>0$) contains all the zeros of $p$ in its bounded face, then all the critical points of $p$ are contained in $D$.
My conjecture is that in fact, if $D$ is a tract of $p$ and contains all the zeros of $p$, then $D$ also contains all the zeros of $p',p^{(2)},\ldots,p^{(n-1)}$.
This certainly does not follow by straight-forward iteration, since in general there need not be a tract of $p'$ containing all the zeros of $p'$ which is contained in $D$. It seems that the tracts and level curves of $p'$ do not interact very nicely with the tracts and level curves of $p$ (even worse for $p'',p''',\ldots$).
I have taken a look at attempting to apply the Cauchy Integral Formula (some sort of integration by parts application perhaps?), but don't seem to be able to make progress there. Any ideas for proof or counter-example?
• The assumption "$D$ is a tract of $p$ and contains all the zeros of $p$" amounts to saying $\{z: |p(z)|<\epsilon\}$ is connected. There is some research on polynomials with connected lemniscates, but I haven't seen anything about higher derivatives.
– user147263
Aug 12, 2014 at 18:32
• Also, a lemniscate is connected iff it contains all of the critical points of the polynomial. Thus, your question is equivalent to the following: let $M=\max \{|p(z)| : p'(z)=0\}$; is it true that $|p|\le M$ at all zeros of all derivatives of $p$?
– user147263
Aug 12, 2014 at 18:50
• @Thursday Yes, I think your second comment is very helpful, and may be the way to go. It may allow more integral methods to be brought to bear. Sep 5, 2014 at 15:59
• Now posted to MO, mathoverflow.net/questions/189245/… Dec 9, 2014 at 4:00
• A counter-example has been posted on the M.O. page (mathoverflow.net/questions/189245/…). May 4, 2015 at 15:33
This is a community wiki answer to remove this question from the unanswered list: Bobby Ocean posted a counter-example at Mathoverflow: Set $p(z) = (z^4+2z^2+2) (z-1)$. Then $D:= \{ z : |p(z)|<1.45\}$ is connected, but some of the zeroes of $p'''$ and $p''''$ are outside $D$. | 744 | 2,533 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2024-18 | latest | en | 0.925039 |
https://engineering.stackexchange.com/questions/21374/how-do-i-measure-max-oscillation-of-building-using-accelerometer/21383 | 1,726,403,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00856.warc.gz | 210,129,981 | 42,099 | # How do I measure max oscillation of building using accelerometer?
For my school project, I am trying to measure the effect of a mass damper by changing the length of the string (that carries the mass), my setup is similar to the following video: https://youtu.be/f1U4SAgy60c?t=4m4s
I am trying to find a relationship between the length of the string and the oscillation of the "building". But I don't know exactly what the guy in the video is trying to measure and if it is related to what I'm trying to do.
Is it possible to use an accelerometer to map the oscillation of the "building" and if so, is there a relationship between the accelerometer data and the displacement of the oscillation?
My aim is to find the maximum amplitude for different lengths of string, so does a larger amplitude from the accelerometer data correspond to a larger amplitude in the displacement time graph?
I am desperate for help as I would much rather use accelerometer data rather than use a ruler in the background to measure the displacement of the oscillation.
• You should look at the relationship between distance, time and acceleration, any decent physics book or google can remind you of those. Commented Apr 20, 2018 at 4:51 | 264 | 1,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-38 | latest | en | 0.932366 |
https://en.m.wikibooks.org/wiki/Basic_Physics_of_Nuclear_Medicine/Dual-Energy_Absorptiometry | 1,713,521,344,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817382.50/warc/CC-MAIN-20240419074959-20240419104959-00897.warc.gz | 210,215,608 | 17,474 | # Basic Physics of Nuclear Medicine/Dual-Energy Absorptiometry
## Introduction
This is a developing chapter of a wikibook entitled Basic Physics of Nuclear Medicine.
Dual-energy radiography is an imaging technique which can be used to eliminate bone information in an radiograph, so that an image displaying tissues only can be displayed. Alternatively, the technique can be used to generate the reverse effect where tissue information is eliminated and an image displaying bone only is generated. This latter option ideally allows indicators of bone density to be assessed. A theoretical background to this technique will first be developed below with the discussion leading towards Dual-Energy X-Ray Absorptiometry (DEXA).
Dual-energy imaging is based on exploiting the difference in the attenuation of tissue and bone at different X- ray energies – see the following figure:
It generally involves acquiring images at two X-ray energies and processing these images to suppress either the bone or the tissue information. A simple mathematical model assumes that monoenergetic radiation is used and no scattered radiation is detected, so that the transmitted radiation intensity through a region of bone and tissue, acquired at a low X-ray energy and following logarithmic transformation, is given by:
Ilo = μtlo xt + μblo xb ,
where:
• μtlo is the linear attenuation coefficient of tissue at the low X-ray energy;
• xt is the tissue thickness;
• μblo is the linear attenuation coefficient of bone at the low X-ray energy; and
• xb is the bone thickness.
Similarly, the transmitted radiation intensity for the same region of an image acquired at a higher X-ray energy is given by:
Ihi = μthi xt + μbhi xb ,
where:
• μthi is the linear attenuation coefficient of tissue at the higher X-ray energy; and
• μbhi is the linear attenuation coefficient of bone at the higher X-ray energy.
When these images are multiplied by separate weighting factors, klo and khi, and the result combined to form a composite image, the output image is given by:
I = klo Ilo + khi Ihi .
Therefore:
I = (klo μtlo + khi μthi) xt + (klo μblo + khi μbhi) xb , (1)
which indicates that tissue cancellation can be achieved by setting the coefficient of xt equal to zero, i.e.
klo μtlo + khi μthi = 0 .
Thus,
klo μtlo = - khi μthi ,
and
${\displaystyle {\frac {k_{hi}}{k_{lo}}}=-{\frac {\mu _{tlo}}{\mu _{thi}}}}$
which indicates that tissue can be eliminated from the composite image when the ratio of weighting factors in equation (1) above is chosen to equal the negative of the ratio of the attenuation coefficients of tissue at the two X-ray energies. A similar approach can be used to cause bone cancellation by setting the coefficient of xb in equation (1) to zero.
This form of image data processing is illustrated in the following figure:
A chest radiograph acquired at 56 kVp is shown in the top left panel of the figure. This is referred to as a low energy image. In the top right panel is a radiograph of the same patient's chest acquired at a high energy – 120 kVp, with 1 mm copper filtration. Results of the dual-energy processing are shown on the bottom row. The bone-subtracted image is shown in the bottom left panel and the tissue-subtracted image in the bottom right panel. Notice that the tissue-subtracted image demonstrates that the lesion in the patient's left lung is a calcified nodule, since it doesn't appear in the bone-subtracted image.
There are at least three new technologies used in radiography which offer numerous advantages over the traditional screen/film image receptors. An overview of these technologies is provided below to support our treatment of dual-energy imaging – and also our consideration of the registration and fusion of planar images in another chapter of this wikibook.
This method of recording a projection radiograph is based on exploiting the photostimulable luminescent properties of barium fluorohalide compounds. The phosphor is layered on an imaging plate which when mounted in a cassette is used in place of the film/screen cassette used traditionally in radiography. The imaging plate is sometimes referred to as a storage phosphor because it stores a latent image following the radiation exposure which can be scanned afterwards using a readout device – as illustrated in the following figure:
The figure shows the radiation exposure in the top left where the cassette is placed anterior to the patient to record the projected image. The cassette is then placed in a readout device (follow the blue downward arrow!), where it is scanned by a laser beam to digitize the latent image. The cassette can then be prepared for re-use by exposing it to intense incandescent light to erase any residual latent image information (top right in the figure above).
The readout stage is shown in more detail in our next figure:
The figure illustrates one mechanism used to scan the imaging plate where a narrow laser beam strikes a rotating mirror causing it to scan a single line across the plate. The plate is then moved so that the scanning laser beam can read the next line of the latent image information.
The latent image is formed through radiation absorption processes in the photostimulable phosphor where electrons are knocked into higher energy states where they remain until they're later stimulated to return to their ground states using the red laser beam – a process similar to thermoluminescence where light is used instead of heat. The electrons emit blue light – called a photostimulable luminescence (PSL) – as they return to their ground states with the amount of light being proprotional to the radiation exposure.
The emitted light is guided by a light guide, as shown in the figure above, so that its intensity can be measured using a photomultiplier tube (PMT). The output of the PMT is digitized using an analogue-to-digital converter (ADC) before computer processing is applied.
One major advantage of computed radiography (CR) over traditional methods is its excellent linearity and dynamic range, as illustrated by the graph. The vertical axis on the left of the graph refers to the response of a CR plate, while the one on the right refers to a film/screen system. Its seen that the film/screen response (red curve) contains regions which can generate under- and over-exposures of the radiographic film. The linear region between these two is ideally used for recording a patient's image. Notice that the range of exposures equivalent to this linear region is substantially smaller than that provided by CR technology, i.e. its got a smaller dynamic range (also called exposure latitude in photography).
The important point to appreciate here is that CR images present data which is a superior record of the radiation exposure pattern incident on the cassette. They do not contain regions of over- and/or under-exposure, as is common is film/screen radiography. An improved ability to discriminate within regions of high (or indeed low) transmission results, so that, for example, a renal calculus can be readily depicted within a contrast filled renal pelvis in an IVP. Note that the dynamic range of CR is rather broad and that contrast enhancement techniques are typically applied in a post-processing mode to window through the image data so that particular features of interest can be discerned. Example CR images are shown below:
This linear, wide dynamic range characteristic is also shared by digital radiographic image receptors.
Considerable research has been conducted in recent years into the development of flat panel image receptors for digital radiography. This research has extended from the development of active-matrix liquid-crystal flat panel displays (AMLCDs) for application in, for example, portable computers. The underlying technology of AMLCDs is a large area integrated circuit called an active matrix array which consists of many millions of identical semiconductor elements deposited on a substrate material. An intensifying screen or a photoconductor coupled to such an active matrix array forms the basis of flat panel X-ray image receptors.
Such a receptor is illustrated in the following figure, where the active matrix array and associated electronic circuitry is mounted in a device which replaces the X-ray cassette in screen-film radiography:
Array sizes of up to 43 cm x 43 cm have been constructed with more than 9 million pixels (pixel size ~150 μm). Operation of the array is controlled by a digital image processor which also stores and displays the resultant images.
The operation is illustrated in more detail in the right panel of the above figure. Each pixel of the array has a switch (typically made from a thin film transistor) which is connected to switching control circuitry in a manner which allows all switches in a row of the array to be operated simultaneously. The output from each pixel is connected in columns with individual pre-amplifiers.
All switches are kept in the off position during the X-ray exposure. Following the exposure, the switches in the first row are turned on and the signal from each pixel is amplified by the pre-amplifiers, digitised in an analogue-to-digital converter (ADC) and stored in the image memory of the digital image processor. These switches are then turned off and the switches in the second row are turned on to acquire signals from the second row of pixels. This process is repeated for the whole array so that an image is acquired in a sequential, line-by-line manner.
Research developments have resulted in two distinct types of digital receptor:
• Indirect Image Receptors
Indirect receptors are based on coupling an fluorescent screen to the active matrix array. Phosphors such as Gd2O2S:Tb and CsI:Tl have been used, and the light produced following X-ray interaction is detected by an array of pixels consisting of photodetectors – see the following figure:
Each photodetector generates an electric charge that is proportional to the amount of light striking it, and this charge is stored until it is read-out by the switching control circuitry. The detection process is referred to as indirect since the detected X-rays are first converted to light, which is subsequently converted to electric charge.
• Direct Image Receptors
Direct receptors are based on coupling a photoconductor to the active matrix array. Photoconductors such as amorphous selenium (a-Se) have been used, and the electric charge produced following X-ray interaction is detected by an array of pixels each consisting of an electrode and a capacitor – see the following figure:
This charge is stored in each capacitor until it is read-out by the electronic switching circuitry. The photoconductor requires a voltage of the order of 5,000 V to be applied, using a surface electrode, so that the charge produced can be attracted to the pixel electrodes. Other photoconductors under investigation for this application include PbI2, PbO, TlBr and CdTe.
The photoconductive properties of amorphous selenium (a-Se) have been known for many years and have been widely applied in photocopier, facsimile and laser printer devices. This direct imaging technique has evolved from xeroradiography, which was a popular medical imaging technique many years ago mainly for soft tissue imaging. Here, X-ray absorption in the photoconductor results in changes in the distribution of the electric charge reflecting local X-ray absorption, i.e. a latent image is recorded in terms of the distribution of electric charge on the surface of an a-Se plate. Methods of rendering the latent image visible include:
• spraying charged toner particles onto the plate and transferring this image to paper;
• scanning the plate with electric charge transducers (e.g. electrometers) and digitising the output for storage in computer memory.
The toner approach was that adopted in the original xeroradiography systems and is no longer in widespread clinical use, mainly due to developments in film-screen technology. The scanning electrometer approach has been applied to the clinical imaging of chests with the Philips Thoravision system. This involves use of an a-Se layer deposited on a cylinder. The latent image is recorded on the surface of the layer and the drum is subsequently rotated past a linear array of tiny electrometers. In comparison, the direct image receptors use an a-Se layer in intimate contact with a two-dimensional array of pixel-sized electrodes, as we have seen above.
Important design features of flat panel image receptors include the pixel size and the fill factor. Pixel size undoubtedly affects spatial resolution and sizes of the order of 100 – 200 μm are typical. The fill factor is the percentage of a pixel area which is sensitive to the image signal – be it electric charge or light photons. It is never 100%, given the need to accommodate conductors (~10 μm wide) which input switching signals and which output image signals, as well as the thin film transistor in each pixel.
Digital image processing techniques are typically applied to acquired images before they are displayed. These include logarithmic transformation, to correct for exponential X-ray attenuation, and contrast enhancement, to optimise the displayed gray values. Other digital processing techniques are also necessary to overcome limitations associated with the manufacturing process for individual image receptors. These include flat field (i.e. uniformity) corrections, to overcome spatial sensitivity variations, and median filtering, to remove the effects of defective pixels.
Advantages of the indirect and direct image receptors purport to include a large, linear dynamic range, similar to photostimulable phosphors, and superior modulation transfer function (MTF) and detective quantum efficiency (DQE) compared with both film-screen and photostimulable phosphor systems. Such receptors are currently (i.e. in 2006) being introduced by major medical imaging companies and it is likely that their clinical application will become a significant feature in coming years.
A final point to note is that both CR and DR receptors can be used for dual-energy radiography in either of the following configurations:
• Two exposures: where two separate exposures are used in applications where patient movement isn't an issue; and
• Single exposure: where two imaging plates separated by a filter are mounted in a dual-energy cassette to record the low energy image on the anterior plate and the high energy image on the other.
## Dual-Energy X-Ray Absorptiometry (DEXA)
This technique has its origin in a nuclear medicine procedure where transmission for two gamma ray energies was used to determine bone mineral density. The procedure is referred to as Dual Photon Absorptiometry and typically used the isotope 153Gd, which emits gamma rays at 44 and 100 keV. As a result of limitations in photon flux and practical considerations, the radioactive source has been replaced by an X-ray tube (XRT) in the DEXA technique – just like scanning radioisotope sources have been replaced by XRTs for attenuation correction in SPECT imaging. This method has found widespread clinical application for the assessment and monitoring of osteoporosis and has surpassed the main alternative technique, Quantitative Computed Tomography (QCT) in terms of accuracy, precision and radiation dose.
Two general approaches to generating the appropriate X-ray energies have been developed. In one approach, the kilovoltage and filtration are switched rapidly during image acquisition, e.g. from 70 kVp and 4 mm Al filtration to 140 kVp with an additional 3 mm Cu filter. In the second approach, a single X-ray energy with two different filters is used, e.g. 80 kVp with alternatively no added filtration and a cerium or samarium filter. Cerium has a K absorption edge at 40.4 keV, and samarium at 46.8 keV, and both materials generate a hardened beam compared with the unfiltered spectrum.
The DEXA technique typically involves an X-ray tube and scintillation detector mounted on a C-arm (see the figure below) so that the patient is exposed to a pencil X-ray beam scanning in a rectilinear fashion.
Diagram of a DEXA scanning apparatus. DEXA scanning apparatus with an under-couch XRT and pencil-beam detector.
The pencil beam is used to reduce the detection of scattered radiation and the scintillation detector typically consists of a CdWO4 or NaI(Tl) scintillator coupled to a photomultiplier tube. The filter assembly is used to switch filters and calibration standards into and out of the pencil beam at appropriate intervals. Scan times with this approach are of the order of 2 to 5 minutes, depending on the examination, which are reduced in second generation instruments where a fan X-ray beam and a detector array are used. By rotating the C-arm around the patient during the scan in the second generation devices, a CT image can be formed. The output of the scintillation detector is fed to a computer for dual-energy data processing and image display. A host of body composition parameters can be derived from the image data, e.g. bone mineral density and soft tissue composition. | 3,493 | 17,217 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-18 | latest | en | 0.895139 |
https://community.adobe.com/t5/after-effects-discussions/ordinal-numerals-in-after-effects/td-p/13024490 | 1,660,175,229,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571222.74/warc/CC-MAIN-20220810222056-20220811012056-00641.warc.gz | 199,735,941 | 98,137 | • ordinal numerals in after effects
# ordinal numerals in after effects
New Here ,
Jun 23, 2022 Jun 23, 2022
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Hi,
is it possible to do ordinal numerals in after effects,
I mean like if a text source layer is linked to a slider control and goes 21, 22, 23 ... etc
another text layer shows "st" "nd" "rd" else "th"
Thank you,
R
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Adobe Community Professional , Jun 27, 2022 Jun 27, 2022
There's a pattern of 3s (ST,ND,RD) which a modulus (%3) can capture easily and you can run through your original numbers via their respective array index numbers.
You can use the javascript Map Method to run a function involving modulus across the index of the array and then just read of the newly created array. It's complex but also not too difficult. Unfortunately, it's going to take quite a while to go through this, for me. Hopefully, someone is quicker to the task or has a different approach
...
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Jun 23, 2022 Jun 23, 2022
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Lots of things are possible in AE and so is this. Mostly, it depends on how much time and knowledge you have to find the most efficient way to do it. At the basic level, you can link an Expression Slider to an array within a Text Layer such that each point (use Math.round) on the slider is accompanied by an array made up of your different text bits.
HTH
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New Here ,
Jun 27, 2022 Jun 27, 2022
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Thanks for replying, do you know if there is a better way than this
if(parseInt(thisComp.layer("Number").text.sourceText) == 1) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 2) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 3) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 21) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 22) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 23) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 31) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 32) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 33) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 41) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 42) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 43) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 51) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 52) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 53) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 61) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 62) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 63) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 71) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 72) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 73) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 81) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 82) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 83) "rd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 91) "st"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 92) "nd"; else if(parseInt(thisComp.layer("Number").text.sourceText) == 93) "rd";
else "the";
so this one works until 100, if there's no better solution, do you know how I can do it like
if(parseInt(thisComp.layer("Number").text.sourceText) == 1 OR 21 OR 31) "st"; else ..... etc
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Jun 27, 2022 Jun 27, 2022
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There's a pattern of 3s (ST,ND,RD) which a modulus (%3) can capture easily and you can run through your original numbers via their respective array index numbers.
You can use the javascript Map Method to run a function involving modulus across the index of the array and then just read of the newly created array. It's complex but also not too difficult. Unfortunately, it's going to take quite a while to go through this, for me. Hopefully, someone is quicker to the task or has a different approach.
Info on the Javascript MAP method can be found here - https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map
HTH
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Getting started with After Effects | 1,325 | 5,244 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-33 | latest | en | 0.883432 |
https://physics.stackexchange.com/questions/444240/how-to-derive-the-formula-for-the-radius-of-a-fermi-sphere | 1,716,257,382,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00185.warc.gz | 412,731,732 | 39,260 | # How to derive the formula for the radius of a Fermi sphere?
I'm trying to figure out how the radius of a Fermi sphere $$p_F = \hbar (3 \pi^2 \frac{N}{V})^{1/3}$$ is derived from the formula $$dN_{spatial}=\frac{V \ d^3p}{\hbar^3}.$$ The solution states that I should arrive at the following $$\frac{2V}{(2 \pi \hbar)^3} \cdot \frac{4\pi}{3}p_F^3.$$ However I am not quite sure how I should go about after the following $$2\int dN_{spatial} = \frac{2V}{\hbar^3}\int d^3p= \ ?$$ I would love some help to solve this, and if you have any web resources I'd gladly accept that too (I've tried googling my way for a while now).
I think it should be $$h$$ in the denominator of your second equation,
$${\rm d}N = \frac{V}{h^3}{\rm d}^3{\bf p}$$
Integrating at both sides, and taking into account the degeneracy
$$N = \frac{2 V}{h^3} \int{\rm d}^3 {\bf p} = \frac{2 V}{h^3} \int {\rm d}\Omega\int_0^{p_F}{\rm d}p ~p^2 = \frac{2V}{h^3} \frac{4\pi}{3} p_F^3$$
Using $$2\pi\hbar = h$$ you get
$$N = \frac{2V}{(2\pi \hbar)^3} \frac{4\pi}{3}p_F^3$$
and from here is just a matter of getting $$p_F$$
$$p_F = \hbar \left(3\pi^2 \frac{N}{V} \right)^{1/3} = \hbar k_F$$
• You are correct, seems like a typo in the exercise. However could you explain the part of $\int d\Omega$ with degeneracy? Nov 30, 2018 at 15:10
• That's the integral of the solid angle $d \Omega = \sin(\theta) d\theta d\phi$ Nov 30, 2018 at 15:13
• @PhyCoMath It is just a short version of the integration over the angle . $\int {\rm d}\Omega = \int_0^{2\pi}{\rm d}\phi \int_0^{\pi}{\rm d}\theta\sin \theta$ Nov 30, 2018 at 15:13
• @caverac Oh, never seen that notation before. Thanks for enlightening me :) Nov 30, 2018 at 15:19
• @PhyCoMath :) Happy to help Nov 30, 2018 at 15:19 | 665 | 1,748 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-22 | latest | en | 0.855853 |
https://www.physicsforums.com/threads/twos-compliment-of-signed-binary.662257/ | 1,723,554,664,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00396.warc.gz | 719,255,964 | 19,045 | Two's compliment of SIGNED binary
• fractal01
In summary: You are right. He just looked at a different question when he was writing the problem I think.In summary, the twos complement for regular binary numbers involves replacing all 1's with 0's and all 0's with 1's, then adding 1. This applies to "signed" numbers as well. There are other representations such as 1's complement and a representation with a separate sign bit, but these are not considered regular binary numbers.
fractal01
The twos compliment for regular binary numbers- you just replace all 1's by 0's and 0's by 1's and then +1. So how would you go about finding out for signed...I have got as far as changing the first digit then that's it! Does anyone know what to do?
Hi fractal01!
"Signed" is the same thing (at least in 2's complement, which is pretty standard).
Suppose you take the number 1.
The binary representation is: 0001
The 2's complement is 1110+1=1111.
This is the same as the representation for -1.
0001 + 1111 = (1)0000.
Yes! It is zero! (disregarding the carry over)
I like Serena said:
Hi fractal01!
"Signed" is the same thing (at least in 2's complement, which is pretty standard).Yes! It is zero! (disregarding the carry over)
Hey! Thank you so much! My lecturer gave us some questions and in one of the answers he came up with something completely different and it was the only example so I have been struggling over it for at least half an hour! Everyone is human I guess and makes mistakes...but are you sure that this is the only way of calculating this?
fractal01 said:
...but are you sure that this is the only way of calculating this?
This is the way the 2's complement representation and its associated "signed" works.
There are other representations, such as 1's complement (unusual), and a representation with a separate sign bit (used for floating point), but those are not "regular" binary numbers.
So what did your lecturer come up with?
It could be sign-magnitude representation.
aralbrec said:
It could be sign-magnitude representation.
That's what I said or what I at least intended:
"a representation with a separate sign bit (used for floating point)"
I like Serena said:
That's what I said or what I at least intended:
"a representation with a separate sign bit (used for floating point)"
Thanks for all of your help! You are right. He just looked at a different question when he was writing the problem I think.
aralbrec said:
It could be sign-magnitude representation.
Thanks
1. What is the purpose of using Two's Compliment in Signed Binary?
In signed binary, the most significant bit (MSB) represents the sign of the number, with 0 being positive and 1 being negative. Two's compliment allows for a more efficient way to represent negative numbers in binary, by inverting all the bits in a positive number and adding 1. This simplifies the mathematical operations involved in working with negative numbers in binary.
2. How is the Two's Compliment of a Signed Binary number calculated?
To calculate the two's compliment of a signed binary number, the bits are inverted (0 becomes 1 and 1 becomes 0) and then 1 is added to the result. This is equivalent to subtracting the original number from 2^n, where n is the number of bits in the binary number. For example, to find the two's compliment of -5 in an 8-bit binary:
-5 = 11111011 (inverted)
+1 = 11111100 (two's compliment)
3. Can a signed binary number be represented without using Two's Compliment?
Yes, a signed binary number can also be represented using a sign-magnitude system, where the MSB represents the sign and the remaining bits represent the magnitude of the number. However, this method requires additional operations for basic arithmetic and does not have a unique representation for 0, making it less efficient compared to the two's compliment system.
4. What is the range of numbers that can be represented using Two's Compliment in signed binary?
In an n-bit binary, the range of numbers that can be represented using Two's Compliment is -2^(n-1) to 2^(n-1)-1. For example, in an 8-bit binary, the range is -128 to 127. The maximum positive number that can be represented is one less than the maximum negative number, due to the inclusion of 0 as a positive number.
5. How is the Two's Compliment of a negative number converted back to its original value?
To convert the Two's Compliment of a negative number back to its original value, the bits are inverted and 1 is added to the result. This is equivalent to subtracting 1 from the two's compliment and then inverting the bits. For example, to convert the two's compliment of -5 back to its original value:
11111100 (inverted)
-1 = 11111011 (original value)
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1K | 1,287 | 5,307 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.967311 |
https://stats.stackexchange.com/questions/17890/what-is-the-difference-between-n-and-n-1-in-calculating-population-variance?noredirect=1 | 1,718,681,623,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861746.4/warc/CC-MAIN-20240618011430-20240618041430-00002.warc.gz | 489,377,862 | 46,764 | # What is the difference between N and N-1 in calculating population variance?
I did not get the why there are N and N-1 while calculating population variance. When we use N and when we use N-1?
It says that when population is very big there is no difference between N and N-1 but it does not tell why is there N-1 at the beginning.
Edit: Please don't confuse with n and n-1 which are used in estimating.
Edit2: I'm not talking about population estimation.
• You can find an answer there: stats.stackexchange.com/questions/16008/…. Basically, you should use N-1 when you estimate a variance, and N when you compute it exactly. Commented Nov 3, 2011 at 15:10
• @ocram, as far as I know when we estimate a variance we use either n or n-1. Commented Nov 3, 2011 at 16:06
• None of the answers below are written in terms of finite population inference. The word finite is absolutely crucial here; that's what Kish's book is about (and whoever was saying "The book is wrong" simply don't know enough about finite population surveys and samples). The quotient $N-1$ instead of $N$ just makes computations nicer and obviates the need to haul around factors like $1-1/N$. The full answer to this question would have to introduce the sampling inference where the sample indicators are random, and the values of observed characteristics $y$ are FIXED. Non-random. Set in stone. Commented May 16, 2013 at 11:25
• This doesn't really add to the other answers. That different divisors give different answers, or even that the difference diminishes with N, is not at issue. The question is when and why to use either divisor. Commented Aug 8, 2013 at 9:49
• I would just like to mention to the other readers that this issue is called "Bassel's correction". You can check it out on Wikipedia en.wikipedia.org/wiki/… Commented Aug 11, 2018 at 9:07
Instead of going into maths I'll try to put it in plain words. If you have the whole population at your disposal then its variance (population variance) is computed with the denominator N. Likewise, if you have only sample and want to compute this sample's variance, you use denominator N (n of the sample, in this case). In both cases, note, you don't estimate anything: the mean that you measured is the true mean and the variance you computed from that mean is the true variance.
Now, you have only sample and want to infer about the unknown mean and variance in the population. In other words, you want estimates. You take your sample mean for the estimate of population mean (because your sample is representative), OK. To obtain estimate of population variance, you have to pretend that that mean is really population mean and therefore it is not dependent on your sample anymore since when you computed it. To "show" that you now take it as fixed you reserve one (any) observation from your sample to "support" the mean's value: whatever your sample might have happened, one reserved observation could always bring the mean to the value that you've got and which believe is insensitive to sampling contingencies. One reserved observation is "-1" and so you have N-1 in computing the variance estimate. The unbiased estimate is called sample variance (not to be confused with the sample's variance) which is an argot; it is better call what it is: sample unbiased estimate of population variance estimated with the sample's mean.
[Pasting here from my below comments: Imagine you are taking repeatedly samples of N=3 size. Of the 3 values in a sample, only 2 values express random deviatedness of observations from the population mean, but the left one expresses (takes on itself) the shift of the sample's mean from the population mean. Thus the "degree of free" observational variability is 2 of the 3, in each separate sample. When we estimate variability on a sample but want it to be an unbiased (unshifted) estimate of populational variability, we "believe" only those 2 free observations. We "pay" for the decision to measure variability off the sample mean as if it were the population mean, for we need to infer about the population variability. This "fee" (N-1 denominator, the Bessel correction) makes the variability wider, incorporating the oscillation of sample means within the variance, but it makes such variance an unbiased estimator.]
But imagine now that you somehow know the true population mean, but want to estimate variance from the sample. Then you will substitute that true mean into the formula for variance and apply denominator N: no "-1" is needed here since you know the true mean, you didn't estimate it from this same sample.
• But my question has nothing to do with estimation. It is about computing population variance; with N and N-1. I'm not talking about n and n-1. Commented Nov 3, 2011 at 16:12
• @ilhan, in my reply, I used N for both N and n. N is a size of a totality at hand, either population or sample. To compute population variance, you must have population at your disposal. If you have only sample you can either compute this sample's variance or compute population estimate variance. No other way round. Commented Nov 3, 2011 at 16:21
• I have a complete information about my population; all the values are know. I'm not interested in estimation. Commented Nov 3, 2011 at 16:35
• If you do have your population then use N. N-1 would be illogical to use. Commented Nov 3, 2011 at 17:00
• @ilhan - Couldn't comment directly on your comment to ttnphns post, but here is an explanation of what you see in the book and how you should infer it. The symbol 'S' when used to imply variance always refers to sample variance. The Greek letter sigma is used to refer to the population variance. That is the reason why you see the book mention S = N * sigma / (N - 1) Commented Nov 13, 2011 at 9:23
$N$ is the population size and $n$ is the sample size. The question asks why the population variance is the mean squared deviation from the mean rather than $(N-1)/N = 1-(1/N)$ times it. For that matter, why stop there? Why not multiply the mean squared deviation by $1-2/N$, or $1-17/N$, or $\exp(-1/N)$, for instance?
There actually is a good reason not to. Any of these figures I just mentioned would serve just fine as a way to quantify a "typical spread" within the population. However, without prior knowledge of the population size, it would be impossible to use a random sample to find an unbiased estimator of such a figure. We know that the sample variance, which multiplies the mean squared deviation from the sample mean by $(n-1)/n$, is an unbiased estimator of the usual population variance when sampling with replacement. (There is no problem with making this correction, because we know $n$!) The sample variance would therefore be a biased estimator of any multiple of the population variance where that multiple, such as $1-1/N$, is not exactly known beforehand.
This problem of some unknown amount of bias would propagate to all statistical tests that use the sample variance, including t-tests and F-tests. In effect, dividing by anything other than $N$ in the population variance formula would require us to change all statistical tabulations of t-statistics and F-statistics (and many other tables as well), but the adjustment would depend on the population size. Nobody wants to have to make tables for every possible $N$! Especially when it's not necessary.
As a practical matter, when $N$ is small enough that using $N-1$ instead of $N$ in formulas makes a difference, you usually do know the population size (or can guess it accurately) and you would likely resort to much more substantial small-population corrections when working with random samples (without replacement) from the population. In all other cases, who cares? The difference doesn't matter. For these reasons, guided by pedagogical considerations (namely, of focusing on details that matter and glossing over details that don't), some excellent introductory statistics texts don't even bother to teach the difference: they simply provide a single variance formula (divide by $N$ or $n$ as the case may be).
There has, in the past been an argument that you should use N for a non-inferential variance but I wouldn't recommended that anymore. You should always use N-1. As sample size decreases N-1 is a pretty good correction for the fact that the sample variance gets lower (you're just more likely to sample near the peak of the distribution---see figure). If sample size is really big then it doesn't matter any meaningful amount.
An alternative explanation is that the population is a theoretical construct that's impossible to achieve. Therefore, always use N-1 because whatever you're doing you're, at best, estimating the population variance.
Also, you're going to be seeing N-1 for variance estimates from here on in. You'll likely not ever encounter this issue... except on a test when your teacher might ask you to make a distinction between an inferential and non-inferential variance measure. In that case don't use whuber's answer or mine, refer to ttnphns's answer.
Note, in this figure the variance should be close to 1. Look how much it varies with sample size when you use N to estimate the variance. (this is the "bias" referred to elswhere)
• Please, tell me why N "not recommended anymore" with true population at hand? Population is not always a theoretical construct. Sometimes your sample is a bona fide population for you. Commented Nov 3, 2011 at 17:10
• ilhan, N can be used for your sample, or it can be used for the population size, if one exists. In most cases the distinction between big N and small n is dependent upon the topic. For example, n might be the number of cases in each condition in an experiment while N might be the number for the experiment. They're both samples. There is no global rule.
– John
Commented Nov 3, 2011 at 17:39
• ttnphns, it depends on what you mean by population. I would argue that if your whole population is so small that N-1 matters then it's questionable whether calculating a mean squared deviation is remotely useful at all. Show all the values, their shape and range. Furthermore, the whole old argument that you actually have N degrees of freedom if you're not making an inference is questionable. You lost one when you calculated the mean, that you needed to calculate the variance.
– John
Commented Nov 3, 2011 at 17:42
• @John, if you calculate mean within population you just state the fact about the parameter, so you spend no degrees of freedom. If you calculate it in sample and want to infer about the population, then you do spend one. Also, I can have population with N=1. With denominator N-1, it appeares that such parameter as variance does not exist for it. It is nonsense. Commented Nov 3, 2011 at 17:55
• @ilhan Please, consider updating your question (as you did) and point to the updated version rather than leaving such non-constructive comments. Everything is debatable, especially when the question itself lacks some context. Here it seems that the problem stands from defining what a population really is.
– chl
Commented Nov 3, 2011 at 22:41
Generally, when one has only a fraction of the population, i.e. a sample, you should divide by n-1. There is a good reason to do so, we know that the sample variance, which multiplies the mean squared deviation from the sample mean by (n−1)/n, is an unbiased estimator of the population variance.
You can find a proof that the estimator of the sample variance is unbiased here: https://economictheoryblog.com/2012/06/28/latexlatexs2/
Further, if one were to apply the estimator of the population variance, that is the version of the variance estimator that divides by n, on a sample of instead of the population, the obtained estimate would biased.
• This seems to answer a different question concerning estimating the population variance. It looks circular: isn't this answer predicated on assuming a specific convention for defining the population variance in the first place?
– whuber
Commented Sep 1, 2016 at 16:16
The population variance is the sum of the squared deviations of all of the values in the population divided by the number of values in the population. When we are estimating the variance of a population from a sample, though, we encounter the problem that the deviations of the sample values from the mean of the sample are, on average, a little less than the deviations of those sample values from the (unknown) true population mean. That results in a variance calculated from the sample being a little less than the true population variance. Using an n-1 divisor instead of n corrects for that underestimation.
• @ Bunnenburg, If you got answer to your question. Please clear to me now, what you got? It is a big confusion to me as well. Commented Jul 20, 2017 at 6:23
• to compensate for that little less variance we get, why can't one use n-2, n-3, etc.? why n-1 in particular? why not a constant...??? Commented May 15, 2018 at 19:02
• @SaravanabalagiRamachandran The discrepancy varies with the sample size and so a constant will not serve. The correction using n-1 is closer works better than the others you mention. Commented May 15, 2018 at 20:18 | 3,014 | 13,203 | {"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} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.960974 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/980/2/k/a/ | 1,716,522,133,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00586.warc.gz | 757,460,329 | 59,113 | Properties
Label 980.2.k.a Level $980$ Weight $2$ Character orbit 980.k Analytic conductor $7.825$ Analytic rank $0$ Dimension $2$ CM discriminant -4 Inner twists $4$
Related objects
Show commands: Magma / PariGP / SageMath
Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [980,2,Mod(687,980)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(980, base_ring=CyclotomicField(4))
chi = DirichletCharacter(H, H._module([2, 1, 0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("980.687");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$980 = 2^{2} \cdot 5 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 980.k (of order $$4$$, degree $$2$$, minimal)
Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$7.82533939809$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{2} + 1$$ x^2 + 1 Coefficient ring: $$\Z[a_1, a_2]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 20) Sato-Tate group: $\mathrm{U}(1)[D_{4}]$
$q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - i - 1) q^{2} + 2 i q^{4} + ( - i + 2) q^{5} + ( - 2 i + 2) q^{8} - 3 i q^{9} +O(q^{10})$$ q + (-i - 1) * q^2 + 2*i * q^4 + (-i + 2) * q^5 + (-2*i + 2) * q^8 - 3*i * q^9 $$q + ( - i - 1) q^{2} + 2 i q^{4} + ( - i + 2) q^{5} + ( - 2 i + 2) q^{8} - 3 i q^{9} + ( - i - 3) q^{10} + ( - i + 1) q^{13} - 4 q^{16} + ( - 3 i - 3) q^{17} + (3 i - 3) q^{18} + (4 i + 2) q^{20} + ( - 4 i + 3) q^{25} - 2 q^{26} + 4 i q^{29} + (4 i + 4) q^{32} + 6 i q^{34} + 6 q^{36} + ( - 7 i - 7) q^{37} + ( - 6 i + 2) q^{40} + 8 q^{41} + ( - 6 i - 3) q^{45} + (i - 7) q^{50} + (2 i + 2) q^{52} + ( - 9 i + 9) q^{53} + ( - 4 i + 4) q^{58} - 12 q^{61} - 8 i q^{64} + ( - 3 i + 1) q^{65} + ( - 6 i + 6) q^{68} + ( - 6 i - 6) q^{72} + ( - 11 i + 11) q^{73} + 14 i q^{74} + (4 i - 8) q^{80} - 9 q^{81} + ( - 8 i - 8) q^{82} + ( - 3 i - 9) q^{85} + 16 i q^{89} + (9 i - 3) q^{90} + ( - 13 i - 13) q^{97} +O(q^{100})$$ q + (-i - 1) * q^2 + 2*i * q^4 + (-i + 2) * q^5 + (-2*i + 2) * q^8 - 3*i * q^9 + (-i - 3) * q^10 + (-i + 1) * q^13 - 4 * q^16 + (-3*i - 3) * q^17 + (3*i - 3) * q^18 + (4*i + 2) * q^20 + (-4*i + 3) * q^25 - 2 * q^26 + 4*i * q^29 + (4*i + 4) * q^32 + 6*i * q^34 + 6 * q^36 + (-7*i - 7) * q^37 + (-6*i + 2) * q^40 + 8 * q^41 + (-6*i - 3) * q^45 + (i - 7) * q^50 + (2*i + 2) * q^52 + (-9*i + 9) * q^53 + (-4*i + 4) * q^58 - 12 * q^61 - 8*i * q^64 + (-3*i + 1) * q^65 + (-6*i + 6) * q^68 + (-6*i - 6) * q^72 + (-11*i + 11) * q^73 + 14*i * q^74 + (4*i - 8) * q^80 - 9 * q^81 + (-8*i - 8) * q^82 + (-3*i - 9) * q^85 + 16*i * q^89 + (9*i - 3) * q^90 + (-13*i - 13) * q^97 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q - 2 q^{2} + 4 q^{5} + 4 q^{8}+O(q^{10})$$ 2 * q - 2 * q^2 + 4 * q^5 + 4 * q^8 $$2 q - 2 q^{2} + 4 q^{5} + 4 q^{8} - 6 q^{10} + 2 q^{13} - 8 q^{16} - 6 q^{17} - 6 q^{18} + 4 q^{20} + 6 q^{25} - 4 q^{26} + 8 q^{32} + 12 q^{36} - 14 q^{37} + 4 q^{40} + 16 q^{41} - 6 q^{45} - 14 q^{50} + 4 q^{52} + 18 q^{53} + 8 q^{58} - 24 q^{61} + 2 q^{65} + 12 q^{68} - 12 q^{72} + 22 q^{73} - 16 q^{80} - 18 q^{81} - 16 q^{82} - 18 q^{85} - 6 q^{90} - 26 q^{97}+O(q^{100})$$ 2 * q - 2 * q^2 + 4 * q^5 + 4 * q^8 - 6 * q^10 + 2 * q^13 - 8 * q^16 - 6 * q^17 - 6 * q^18 + 4 * q^20 + 6 * q^25 - 4 * q^26 + 8 * q^32 + 12 * q^36 - 14 * q^37 + 4 * q^40 + 16 * q^41 - 6 * q^45 - 14 * q^50 + 4 * q^52 + 18 * q^53 + 8 * q^58 - 24 * q^61 + 2 * q^65 + 12 * q^68 - 12 * q^72 + 22 * q^73 - 16 * q^80 - 18 * q^81 - 16 * q^82 - 18 * q^85 - 6 * q^90 - 26 * q^97
Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/980\mathbb{Z}\right)^\times$$.
$$n$$ $$101$$ $$197$$ $$491$$ $$\chi(n)$$ $$1$$ $$i$$ $$-1$$
Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
687.1
1.00000i − 1.00000i
−1.00000 1.00000i 0 2.00000i 2.00000 1.00000i 0 0 2.00000 2.00000i 3.00000i −3.00000 1.00000i
883.1 −1.00000 + 1.00000i 0 2.00000i 2.00000 + 1.00000i 0 0 2.00000 + 2.00000i 3.00000i −3.00000 + 1.00000i
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
4.b odd 2 1 CM by $$\Q(\sqrt{-1})$$
5.c odd 4 1 inner
20.e even 4 1 inner
Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 980.2.k.a 2
4.b odd 2 1 CM 980.2.k.a 2
5.c odd 4 1 inner 980.2.k.a 2
7.b odd 2 1 20.2.e.a 2
7.c even 3 2 980.2.x.c 4
7.d odd 6 2 980.2.x.d 4
20.e even 4 1 inner 980.2.k.a 2
21.c even 2 1 180.2.k.c 2
28.d even 2 1 20.2.e.a 2
28.f even 6 2 980.2.x.d 4
28.g odd 6 2 980.2.x.c 4
35.c odd 2 1 100.2.e.b 2
35.f even 4 1 20.2.e.a 2
35.f even 4 1 100.2.e.b 2
35.k even 12 2 980.2.x.d 4
35.l odd 12 2 980.2.x.c 4
56.e even 2 1 320.2.n.e 2
56.h odd 2 1 320.2.n.e 2
84.h odd 2 1 180.2.k.c 2
105.g even 2 1 900.2.k.c 2
105.k odd 4 1 180.2.k.c 2
105.k odd 4 1 900.2.k.c 2
112.j even 4 1 1280.2.o.g 2
112.j even 4 1 1280.2.o.j 2
112.l odd 4 1 1280.2.o.g 2
112.l odd 4 1 1280.2.o.j 2
140.c even 2 1 100.2.e.b 2
140.j odd 4 1 20.2.e.a 2
140.j odd 4 1 100.2.e.b 2
140.w even 12 2 980.2.x.c 4
140.x odd 12 2 980.2.x.d 4
280.c odd 2 1 1600.2.n.h 2
280.n even 2 1 1600.2.n.h 2
280.s even 4 1 320.2.n.e 2
280.s even 4 1 1600.2.n.h 2
280.y odd 4 1 320.2.n.e 2
280.y odd 4 1 1600.2.n.h 2
420.o odd 2 1 900.2.k.c 2
420.w even 4 1 180.2.k.c 2
420.w even 4 1 900.2.k.c 2
560.r even 4 1 1280.2.o.j 2
560.u odd 4 1 1280.2.o.g 2
560.bm odd 4 1 1280.2.o.j 2
560.bn even 4 1 1280.2.o.g 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
20.2.e.a 2 7.b odd 2 1
20.2.e.a 2 28.d even 2 1
20.2.e.a 2 35.f even 4 1
20.2.e.a 2 140.j odd 4 1
100.2.e.b 2 35.c odd 2 1
100.2.e.b 2 35.f even 4 1
100.2.e.b 2 140.c even 2 1
100.2.e.b 2 140.j odd 4 1
180.2.k.c 2 21.c even 2 1
180.2.k.c 2 84.h odd 2 1
180.2.k.c 2 105.k odd 4 1
180.2.k.c 2 420.w even 4 1
320.2.n.e 2 56.e even 2 1
320.2.n.e 2 56.h odd 2 1
320.2.n.e 2 280.s even 4 1
320.2.n.e 2 280.y odd 4 1
900.2.k.c 2 105.g even 2 1
900.2.k.c 2 105.k odd 4 1
900.2.k.c 2 420.o odd 2 1
900.2.k.c 2 420.w even 4 1
980.2.k.a 2 1.a even 1 1 trivial
980.2.k.a 2 4.b odd 2 1 CM
980.2.k.a 2 5.c odd 4 1 inner
980.2.k.a 2 20.e even 4 1 inner
980.2.x.c 4 7.c even 3 2
980.2.x.c 4 28.g odd 6 2
980.2.x.c 4 35.l odd 12 2
980.2.x.c 4 140.w even 12 2
980.2.x.d 4 7.d odd 6 2
980.2.x.d 4 28.f even 6 2
980.2.x.d 4 35.k even 12 2
980.2.x.d 4 140.x odd 12 2
1280.2.o.g 2 112.j even 4 1
1280.2.o.g 2 112.l odd 4 1
1280.2.o.g 2 560.u odd 4 1
1280.2.o.g 2 560.bn even 4 1
1280.2.o.j 2 112.j even 4 1
1280.2.o.j 2 112.l odd 4 1
1280.2.o.j 2 560.r even 4 1
1280.2.o.j 2 560.bm odd 4 1
1600.2.n.h 2 280.c odd 2 1
1600.2.n.h 2 280.n even 2 1
1600.2.n.h 2 280.s even 4 1
1600.2.n.h 2 280.y odd 4 1
Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(980, [\chi])$$:
$$T_{3}$$ T3 $$T_{13}^{2} - 2T_{13} + 2$$ T13^2 - 2*T13 + 2
Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2} + 2T + 2$$
$3$ $$T^{2}$$
$5$ $$T^{2} - 4T + 5$$
$7$ $$T^{2}$$
$11$ $$T^{2}$$
$13$ $$T^{2} - 2T + 2$$
$17$ $$T^{2} + 6T + 18$$
$19$ $$T^{2}$$
$23$ $$T^{2}$$
$29$ $$T^{2} + 16$$
$31$ $$T^{2}$$
$37$ $$T^{2} + 14T + 98$$
$41$ $$(T - 8)^{2}$$
$43$ $$T^{2}$$
$47$ $$T^{2}$$
$53$ $$T^{2} - 18T + 162$$
$59$ $$T^{2}$$
$61$ $$(T + 12)^{2}$$
$67$ $$T^{2}$$
$71$ $$T^{2}$$
$73$ $$T^{2} - 22T + 242$$
$79$ $$T^{2}$$
$83$ $$T^{2}$$
$89$ $$T^{2} + 256$$
$97$ $$T^{2} + 26T + 338$$ | 4,418 | 8,499 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-22 | latest | en | 0.525772 |
http://essay.helpstudents.xyz/exclusive/Math-worksheets-for-grade-3-long-division.html | 1,611,339,824,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703531335.42/warc/CC-MAIN-20210122175527-20210122205527-00556.warc.gz | 39,989,227 | 4,906 | # Math worksheets for grade 3 long division
Free 3rd grade division worksheets, including the meaning of division, division facts, dividing by 10 and 100, division by whole tens and whole hundreds, division with remainders and long division (within 100). No login required.
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Long Division Grade 3. Long Division Grade 3 - Displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are Grade 3 division work, Grade 3 division work, Division work, Division work, Division, Long division bingo, Long division, Name class division.
Grade 3 Division Worksheet - Long Division: Basic Division Facts Author: K5 Learning Subject: Grade 3 Division Worksheet Keywords: Grade 3 Divison Worksheet - Long Division: Basic Division Facts math practice printable elementary school Created Date: 20151213153204Z.
Welcome to the division worksheets page at Math-Drills.com! Please give us your undivided attention while we introduce this page. Our worksheets for division help you to teach students the very important concept of division.
Addition, Subtraction, Multiplication and Division problems are given. The other sections of Math are under construction. Our team is working on a new methodology for preparing engaging, colorful worksheets. Grade 3 worksheets are free for download. Print them and Practice.
## Long Division Grade 3 Worksheets - Kiddy Math.
Free Printable Math Worksheets for Grade 3 This is a comprehensive collection of math worksheets for grade 3, organized by topics such as addition, subtraction, mental math, regrouping, place value, multiplication, division, clock, money, measuring, and geometry. They are randomly generated, printable from your browser, and include the answer key.
Division worksheets for grades 3, 4, and 5 These are free, printable division worksheets, randomly generated, for grades 3-5. Topics include division facts, mental division, long division, division with remainders, order of operations, equations, and factoring.
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## Grade 3 math worksheet - Long division: basic division.
DadsWorksheets.com delivers thousands of printable math worksheets, charts and calculators for home school or classroom use on a variety of math topics including multiplication, division, subtraction, addition, fractions, number patterns, order of operations, standard form, expanded form, rounding, Roman numerals and other math subjects.Grade 3 Long Division Without Remainder. Grade 3 Long Division Without Remainder - Displaying top 8 worksheets found for this concept. Some of the worksheets for this concept are Grade 4 division work, Grade 4 division work, Division, Long division without remainder, Division witho ut remainder 2 digit by 1 digit s1, Long division, Division with remainders homework, Division work long division.An unlimited supply of worksheets for division with remainders (grades 3-5)! Some of the worksheets practice finding the remainder using mental math, some are for long division. The worksheets can be made in html or PDF format - both are easy to print. You can also customize them using the generator.
Long Division Worksheets Long Division Worksheets: Long Division Worksheet - Easy 1 Long Division Worksheet - Easy 2 Long Division Worksheet - Easy 3.Spelling Grade 3. Spelling Grade 4. Spelling Grade 5. More Spelling Worksheets. Chapter Books. Bunnicula.. THis crypto-code math worksheet has 3-digit dividends and 2-digit quotients. Each problem has a remainder.. Create your own long division worksheets! You choose the number of digits in the dividend and the divisor. | 1,187 | 5,582 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-04 | latest | en | 0.886198 |
https://www.physicsforums.com/threads/ap-physics-question.9346/#post-100187 | 1,685,912,423,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650264.9/warc/CC-MAIN-20230604193207-20230604223207-00744.warc.gz | 1,034,045,000 | 18,725 | # Ap Physics Question.
• pezzang
#### pezzang
Hi, thank you so much for making this kind of post for students like me. I have tried the quesotins but i want to make sure that i was right. if worng, please correct me.
THANK YOU SO MUCH. Also i had attached pictures of the question to give you enough information. PLEASE REFER TO THE PICTURE ATTACHED.
Q) The two uniform disks shown above have equal mass, and each can rotate on frictionless bearings about a fixed axis through its center. The smaller disk has a radius R and moment of inertia I about its axis.
(a) determine the moment of inertia of the larger disk about tis axis in terms of I.
-> I used the equation Inertia = mass*radius to solve this question,
Let's say that smaller disk has a subscript "A" and larger; "B".
Then, I(B) = m(2R) = mR(2).
since mR is equal to I, we have I(B) = I*2 = 2I
so the answer is 2*I. Am I right?
The two disks are then linked as shown below by a light chain that cannot slip. They are at rest when, at time t = 0, a student applies a torque to the smaller disk, and it rotates counterclockwise with constant angular acceleration alpha. Assume that the mass of the chain and the tension in teh lower part of the chain are negligible.
In terms of I, R, alpha and t, determine each of the following.
(b) The angular acceleration of the larger disk (alpha(B))
-> torque(A) = I*alpha
torque(B) = 2I*alpha(B)
torque(A) = torque(B) <- IS IT POSSIBLE THAT I ASSUME THIS?
I*alpha = 2I*alpha(B)
alpha(B) = alpha / 2. <- AM I RIGHT?
(c) The tension in the upper part of the chain
->
T(Tension) = (I*alpha) / R.
T(A) = (I*alpha) / R
T(B) = (2I)(alpha(B)) / (2R) = (2I*alpha/2) / R = (I*(alpha) / (2R)
TO FIND TENSION, DO I SUBTRACT THE SMALLER ONE FROM LARGER ONE??
IF I DO, I GET:
T(A) = T(B) = (I*alpha) / R - (I*alpha) / (2R) = (I*alpha) / (2R)
Therefore, the tension in teh upper part of the chain equals to (I*(alpha) / (2R).
(d) The torque that the student applied to the smaller disk
->
torque(A) = I*alpha <- I found this equation in my physics book. Is right? I kind of doubt it though. Please help me.
(e) The rotational kinetic energy of the smaller disk as a function of time
->
RKE = (1/2)*I*w^2 (omega)
to find an equation of w(omega) in terms of t and alpha, we take antiderivative of the equation alpha = (dw)/(dt), then we get:
alpha*t = w
IF we substitute alpha*t for w(omega) in the equation RKE = (1/2)*I*w^2 (omega), we get
RKE = (1/2)I*(alpha*t)^2.
So please correct me and thank you so much. Have a wonderful day, people!
#### Attachments
• untitled.jpg
13.5 KB · Views: 671
Last edited:
Originally posted by pezzang
PLEASE REFER TO THE PICTURE ATTACHED.
I don't see the picture.
can i send you email so that you can see the pictures?
Use the attachment feature.
i tried but it does not work..ok i will explain instead
in the first picture for question part (a) only, there are two regular circles. circle A has a radius of R and circle B has a radius 2R. They have the same mass and are indeendent of each other.
in the second picture which will be used from part (b) to (e), the two circles described in the first picture are in a chain. the circle B on the right and Circle A on the left. the chain moves from left to right and the tension on the bottom part at first is zero.
i hope this helps to solve the question.
Thank you!
Last edited:
Originally posted by pezzang
-> I used the equation Inertia = mass*radius to solve this question,
Let's say that smaller disk has a subscript "A" and larger; "B".
Then, I(B) = m(2R) = mR(2).
since mR is equal to I, we have I(B) = I*2 = 2I
so the answer is 2*I. Am I right?
First things first: What's the moment of inertia of a disk about a central axis? It's I = 1/2 M R2
Redo the first part, then we'll continue.
Let me give you some tips on the other parts.
Originally posted by pezzang
(b) The angular acceleration of the larger disk (alpha(B))
-> torque(A) = I*alpha
torque(B) = 2I*alpha(B)
torque(A) = torque(B) <- IS IT POSSIBLE THAT I ASSUME THIS?
I*alpha = 2I*alpha(B)
alpha(B) = alpha / 2. <- AM I RIGHT?
Forget about torques for this part. αA is a given. What's the relationship between αA and αB? Hint: the two disks are tied by a chain. What does that mean?
Also, no, you cannot assume that the torque is the same on both disks! Think about it. DiskA has two forces acting on it: the force applied by the student in giving his torque, and the tension in the chain pulling the other way. DiskB has only a single force on it.
(c) The tension in the upper part of the chain
->
T(Tension) = (I*alpha) / R.
T(A) = (I*alpha) / R
T(B) = (2I)(alpha(B)) / (2R) = (2I*alpha/2) / R = (I*(alpha) / (2R)
TO FIND TENSION, DO I SUBTRACT THE SMALLER ONE FROM LARGER ONE??
IF I DO, I GET:
T(A) = T(B) = (I*alpha) / R - (I*alpha) / (2R) = (I*alpha) / (2R)
Therefore, the tension in teh upper part of the chain equals to (I*(alpha) / (2R).
(d) The torque that the student applied to the smaller disk
->
torque(A) = I*alpha <- I found this equation in my physics book. Is right? I kind of doubt it though. Please help me.
The way to solve c & d is to write down the torque equations for both disks and solve the two equations together.
Torquenet=Iα, for each disk. (You'll need the results from parts b and a.)
(e) The rotational kinetic energy of the smaller disk as a function of time
->
RKE = (1/2)*I*w^2 (omega)
to find an equation of w(omega) in terms of t and alpha, we take antiderivative of the equation alpha = (dw)/(dt), then we get:
alpha*t = w
IF we substitute alpha*t for w(omega) in the equation RKE = (1/2)*I*w^2 (omega), we get
RKE = (1/2)I*(alpha*t)^2.
Yes!
Than you so much. Here is what I have so far after getting your advice.
(a) Moment of Inertia of a disk is equal to I = (MR^2) / 2.
So, I(B) = (1/2) (M)(2R)^2 = 2MR^2.
(b) Since the disks are connected by a chain, they have the same angular acceleration. Thus the angular acceleration of the larger disk is equal to just ¥áA, which is given in the problem.
(c) I do not know how to find tension. I guess the tension on the upper part of the chain would equal to torque(B) - torque(a). Since torque is equal to I*¥á, and ¥á is same for disk A and B,
Tension = (1/2)*M(2R^2)*¥á - (1/2)(MR^2)*¥á = (3/2)MR^2*¥á.
So, the answer is equal to (3/2)MR^2*¥á.
(d) torque of the smaller disk is equal to (1/2)(MR^2)*¥á, which we found in part(c).
How do they look now? Much closer?
Originally posted by pezzang
Than you so much. Here is what I have so far after getting your advice.
(a) Moment of Inertia of a disk is equal to I = (MR^2) / 2.
So, I(B) = (1/2) (M)(2R)^2 = 2MR^2.
Yes. Now put it in terms of IA= I;
IB= 4*(1/2MR2)= 4I
(b) Since the disks are connected by a chain, they have the same angular acceleration. Thus the angular acceleration of the larger disk is equal to just ¥áA, which is given in the problem.
No, they have the same tangential acceleration, not angular acceleration. Think about it: the same length of chain must pass along each disk. Imagine the disk B was 1000 times bigger than disk A: the same bit of chain that turns A half way around would barely budge disk B.
(c) I do not know how to find tension. I guess the tension on the upper part of the chain would equal to torque(B) - torque(a). Since torque is equal to I*¥á, and ¥á is same for disk A and B,
Tension = (1/2)*M(2R^2)*¥á - (1/2)(MR^2)*¥á = (3/2)MR^2*¥á.
So, the answer is equal to (3/2)MR^2*¥á.
(d) torque of the smaller disk is equal to (1/2)(MR^2)*¥á, which we found in part(c).
Write out the torque equations for each disk. Tension is force; TensionXradius is the torque exerted by the tension.
How do they look now? Much closer?
You're getting closer. Keep going!
I have found part (b) but i really can't solve part (c) and (d).
Meanwhile here is part (b)
(b) a(tan) = (dv)/(dt) = r*(dw)/(dt) = r*alpha.
a(tanA) = a(tanB)
R*alpha = (2R)*alpha(B)
alpha(b) = (R*alpha)/(2R) = alpha / R.
thus, the answer is alpha / R.
PLEASE TELL ME PART (C) AND (D). I AM REALLY REALLY REALLY CURIOUS~!
Originally posted by pezzang
Meanwhile here is part (b)
(b) a(tan) = (dv)/(dt) = r*(dw)/(dt) = r*alpha.
a(tanA) = a(tanB)
R*alpha = (2R)*alpha(B)
alpha(b) = (R*alpha)/(2R) = alpha / R.
thus, the answer is alpha / R.
PLEASE TELL ME PART (C) AND (D). I AM REALLY REALLY REALLY CURIOUS~!
You messed up the algebra above: αB= α/2
OK, OK, you've suffered enough...
The torque equation for Disk B:
FTRB = IB αB
Which, when we use the results of a & b, becomes:
FT = Iα/R(That's the answer to part C!)
The torque equation for Disk A:
Τ-FTR = I α (where FT is the tension; Τ is the the applied torque)
Now we can solve for Τ (using the answer from part C)
Τ = 2Iα (That's the answer to part D.)
Make sense? Go over this until you understand it. (And you'd better check my algebra.)
I TRULY APPRECIATE YOUR HELP ALONG THE WAY.
EVERYONE OF US WHO GETS HELP FROM YOU ARE GRATEFUL TO YOU!
THANK YOU SO SO SO SO SO MUCH! | 2,742 | 9,027 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-23 | latest | en | 0.908834 |
dialogue.usaee.org | 1,516,488,662,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889736.54/warc/CC-MAIN-20180120221621-20180121001621-00185.warc.gz | 96,503,171 | 14,520 | ## .
Fairly Pricing Net Intervals While Keeping the Utility
Financially Healthy
Mark B. Lively
Consulting Engineer
Utility Economic Engineers
Gaithersburg, MD
MbeLively@aol.com
Recent events have made electricity economics more complicated. Previously, input prices were predictable, as was the mix of generating fuels. The cost of the utility could thus be well determined ahead of time. Further, the load patterns of customers simplified rate design. Prices were often invariant over long time periods. The price invariance allowed utilities to read inexpensive watt-hour meters once a month, instead of installing more expensive interval meters.
The passage of the Public Utility Regulatory Policy Act (PURPA) of 1978 began a change in the industry, allowing Qualifying Facilities (QFs) to require its local utility to buy the output of their generation. PURPA also created the concept of net metering, where a single meter would measure the net flow between the customer and the utility, instead of different meters being used for customer consumption versus customer generation. Net metering requirements changed the customer load patterns experienced by the utility, invalidating traditional rate design.
NET METERING
Electrical engineers often wince when they hear associates use KW and KWH interchangeably. One is power, one is energy. The two are related in that energy is the integral of power, as is shown in Figure 1. Alternatively, power is the average amount of energy over a given period of time. Figure 1 is a "stylized" daily usage curve for a residential customer. The height of the curve is power, measured in KW.
The area under the curve (shaded in color) is energy, measured in KWH. The usage curve is "stylized" by making it into a straight line, just so some of the calculations, including calculus, is easier later on, for anyone who wishes to replicate the calculations. The maximum demand is 10 KW. The minimum demand is 4 KW. The total energy consumption is 168 KWH. For a residential customer, this is a large load.
Figure 2 replicates Figure 1 but overlays a daily distributed generation production curve, such as solar, wind, or a home cogeneration plant.
Again the production curve is stylized as a straight line to make the calculus easier later on. In contrast to the load curve, the production curve slopes in the other direction going from a low of 2 KW to a maximum of 8 KW, for a total production 120 KW. The opposing slopes were chosen just to make the differences starker.
The concept of net metering is using a single meter to measure the power going into or out of the customer's premises. Generally, the measurement is the integral of that power over an interval of time. This integral of power is energy. Historically, the high cost of metering meant that the interval of time was a month in most parts of the U.S., though some utilities had a standard reading interval of two months, or even a year for some utilities around the world. The electronic revolution has reduced the cost of interval metering and also reduced the standard interval from a month. In some places the standard interval is one hour, other places 15 minutes, and even as short as 6 seconds, as is used for Australia's Frequency Control Ancillary Services (FCAS).
Short billing intervals are important if loads change rapidly while prices change rapidly. If neither price nor load changes, then long billing intervals can be used. For instance, prior to restructuring of the U.S. electric industry, prices changed infrequently. Rate cases occurred less than once a year. Fuel clause adjustments (FCAs) were occasionally unchanged month to month. So metering intervals shorter than a month were not needed. The desire for increased economic efficiencies suggested time varying prices, though such efficiency gains have been disputed.
The alternative to net metering is gross metering, which involves two meters. One meter determines the customer load and the other meter determines the customer production. Besides the added cost to the utility of owning and processing two meters, there is the added cost to the customer to have the production separately wired back to the meter point. Net metering reduces these expenses but at the confusion of the utility not knowing the load it was meeting and the production that was being supplied by the customer. From simple algebra in Figure 2, the utility measures 64 KWH as going to the consumer and measures 16 KWH as coming to it, or a net of 48 KWH during this 24 hour period. Unmeasured is the 104 KWH that the consumer self generated and used itself.
EFFECT OF SMALLER PRICING INTERVALS
The issue associated with net metering begins to appear in Figure 3, which overlays a pricing curve on top of the power curves of Figure 2.
The pricing curves slopes in the same direction as the load curve, going from a maximum value of \$90/MWH to a low of \$10/MWH, for a simple average of \$50/MWH over the period. Given the slope of the pricing curve, a price responsive generator would have a slope in the opposite direction. On a simple average basis, the customer could pay the utility \$3.20, or \$50/MWH for the 64 KWH delivered by the utility, and receive \$0.80, or \$50/MWH for the 16 KWH delivered by the customer to the utility. This is summarized in the first column of Table 1 which uses a simpler calculation. This simpler calculation looks at the 48 KWH of net deliveries by the utility and has the customer simply pay a net of \$2.40. But this looks at the entire 24 hour period as a single metering interval, and does not look at the small time intervals.
Breaking the day into three metering intervals of 8 hours each results in the analysis shown to the right four columns of Table 1.
Again, the first column reproduces the analysis for treating the entire day as a single time interval. The second column presents data for the first 8 hours of the day. The simple average price during those 8 hours is \$76.66/MWH. During those 8 hours the customer takes 48 KWH and pays \$3.68. During the middle 8 hours, the simple average price is \$50/MWH, the customer consumption is 16 KWH, and the payment is \$0.80. During the last 8 hours, the customer delivers 16 KWH at a price of \$23.33/MWH and receives \$0.37. The net payment by the customer is \$4.11 or \$1.71 more that the \$2.40 calculated using the simple average over the 24 hour period. The \$1.71 represents an increase of 71% as a result of using finer billing intervals, at least in this example where the prices and loads are both varying, though admittedly slowly. Again, the example is for generation that varies contra-cyclically to the price of energy.
The "hourly" line near the bottom of Table 1 presents a summary of similar calculations for each hour within each of the 8 hour periods. The net payment increases again for this use of even finer billing intervals, though only by another \$0.21. The "continuous" line at the bottom of Table 1 presents a summary of the calculations using calculus, which essentially uses vanishingly small measurement intervals, similar to the 6 second intervals associated with FCAS. For this set of data, there is little additional benefit to sub-hourly pricing, essentially not more than a rounding difference.
SETTING REAL TIME PRICES
Utilities long ago realized that economic principles could be used to lower the fuel and other operating costs of generating plants by equalizing the marginal costs of all operating generating plants. Marginal costs roughly monotonically increase over the operating range of generators. The resulting optimization protocol is to increase the output of generators with low marginal costs and to decrease the output of generators with high marginal costs. This protocol moves the marginal costs of the various units closer together. The protocol also slowly adjusts the operating levels of all generators as load changes throughout the day, week, and year. This equalization process for marginal costs reflects geographic issues such as line losses (on a marginal basis) and transmission constraints.
The equalized marginal cost concept is a continuous process applicable to those units that are operating. Turning a unit on or off is a non-continuous unit commitment process. Each unit commitment action changes the marginal cost equilibrium point, requiring an adjustment to the operating levels of the various units. The unit commitment process greatly affects the profitability of individual generating units and contributes to the infra-marginal cost of the entire generating system. For generating systems that include generators with non-uniform ownership interests, the unit commitment process and the equalized marginal cost calculations are greatly scrutinized. This scrutiny has been addressed by the establishment of Independent System Operators (ISOs). Most of the continental U.S. is in the footprint of an ISO.
Just as the dispatch of generators is optimized by operating them at an equalized marginal cost level, economic theory suggests that the optimal price to charge purchasers for this electricity is a price equal to the equalized marginal cost. The ISO dispatch prices for generation provides appropriate marginal cost prices for this purpose. These dispatch prices are typically developed every 15 minutes, though often averaged over an hour. The example presented previously in Table 1 suggests that little economic efficiency is lost by using an hourly average price instead of minute by minute prices. But Table 1 does not have rapidly changing prices or rapidly changing loads. Rapidly changing prices, such as when generators are switched on or off or when transmission lines are opened or closed can cause great variations in the value of electricity.
Engineers have designed operating systems to respond to sudden changes in the configuration of the generation and transmission system, providing signals to increase or decrease generating levels as appropriate in the various parts of the network. In extreme cases, under-frequency relays will also interrupt load. The driver for these rapid response operating systems is frequency and Area Control Area (ACE). ACE is a combination of frequency error with inadvertent interchange, the unscheduled flow of electricity into or out of a utility. To some extent, ACE can be viewed as a free storage device, absorbing excess electricity or supplying electricity during a deficit. Though there is no price for ACE, NERC has mandatory reliability standards that can results in fines when a utility abuses ACE.
Fred Schweppe, the MIT professor who during the 1980s developed some of the early theories for ISO operation and pricing, pointed out that the marginal cost theories were only applicable when the system was in balance. This balance requirement suggests that the presence of ACE (or of frequency error for an isolated utility) reduces the robustness of ISO pricing mechanisms. One approach to addressing this imbalance issue is to treat the imbalance as an economic resource. This concept was presented in "Tie Riding Freeloaders--The True Impediment to Transmission Access,"[1] “Microgrids And Financial Affairs - Creating A Value-Based Real-Time Price For Electricity,”[2] and “The WOLF in Pricing: How the Concept of Plug, Play, and Pay Would Work for Microgrids.”[3]
For instance, a storage device makes money by absorbing electricity when prices are low and should be producing electricity when prices are high. Thus, when there is excess electricity, as is indicated by a positive frequency error and/or an export of inadvertent interchange, which are physical manifestations of using the grid to store the excess, the system price should be low. Conversely, when there is a shortage of electricity, as is indicated by a negative frequency error and/or an import of inadvertent interchange, which are physical manifestations of using the grid to take electricity out of storage, the system price should be high. It is unclear whether NERC mandatory reliability standards would need to be eliminated to accommodate such an automated, dynamic pricing mechanism. The above mentioned papers suggest using such a pricing mechanism for unscheduled flows of electricity between and among ISOs and/or utilities, for ISO internal pricing within the normal 15 minute dispatch intervals, and for independent utilities that are dealing with independent power generators within their footprint, such as the net metering loads described in Figure 2.
UPLIFT
A big issue with net metering is Uplift, a term common among the independent system operators (ISOs) for the cost not recovered by the hourly calculation of revenue from loads. For a vertically integrated utility, Uplift can be considered to be the difference between the total installed and operating cost of the generation function versus the revenue associated with the generation function (or perhaps the generation and transmission function). There can also be Uplift associated with the other functions. Most utilities in the U.S., are distribution utilities, either because they were originally constituted as distribution utilities (such as many rural electric cooperatives and some municipalities) or because of the wave of restructuring that occurred near the turn of the millennium. For distribution utilities, Uplift is the difference between purchased power costs and the associated "generation related" revenue.
In the example so far, the revenue from gross load would have been \$9.36 or \$55.72/MWH, probably much less than the utility's generation costs. The revenue from net load would only be \$4.32, as shown in Table 1. Hourly or sub-hourly pricing of net metered deliveries goes only a little way toward collecting Uplift costs. Since Uplift costs are net of collected revenue, net metering can increase Uplift costs, though in some cases metering can decrease uplift costs. But, sizeable uplift costs will remain.
Under general economic theory, a producer will build new productive capacity when the expected marginal price for sales will exceed the average cost of new productive capacity. For electric utilities operating in competitive market such as an ISO, the marginal sales price is generally set at the marginal cost of production. Studies of the Australian ISO have claimed that the marginal cost of production never reaches the cost of new production capacity, requiring non-market motivations for producers to add capacity. Some have speculated that part of the under-collection of operating costs partially relates to a practice of restraint on how high real time prices are allowed to go.
Non-market motivations may include requiring utilities to purchase sufficient capacity to achieve a defined reserve margin. Some reserve margins are defined as installed capacity versus project load. Other reserve margins have been defined as operating reserves, such as a real time requirement for spinning reserves. In the U.S., some of these reserve and operating requirements have been promulgated as Mandatory Reliability Standards under the Electricity Policy Act of 2005. The Australian studies have been confirmed by internal staff studies at the Federal Energy Regulatory Commission and at the U.S. Department of Energy (DOE) Energy Information Agency. The studies show that ISO hourly markets do not collect enough money to pay for new capacity. Further, a major portion of the loads within the ISOs are actually traded outside of the ISO hourly markets through bilateral contracts.
DEMAND CHARGES TO COVER UPLIFT COSTS
Electric utilities typically functionalize their costs among distribution, transmission, and distribution. Each function has a peculiar mix of revenue drivers, such as short run marginal costs for the generation function and to an extent for the transmission function.
Distribution Uplift
Utilities incur distribution costs for both metering the customer's use of electricity and for the wires that historically moved electricity from the transmission grid to the customer. The growth of distributed generation now has some of the electricity moving away from the customer. Generally this electricity can be considered as going to other customers. It is rare that electricity moves off a distribution feeder back into the transmission grid, though the growth of rooftop solar in Hawaii has been reputed to have resulted in some distribution feeders delivering electricity to the transmission grid.
The cost of customer metering and customer connection equipment is highly dependent on the size of the customer. Larger customers require more expensive equipment, though the cost of the equipment does not go up linearly with the size of the customer's demand. Indeed, these costs are very class dependent, sometimes being treated as being uniform within the residential class, within the commercial class, and within the industrial class. Determining the monthly customer charge necessary to recover these costs is outside the scope of this paper. The distribution uplift can be considered to be the total distribution revenue requirement after netting the revenue associated with monthly customer charge.
Utilities typically classify costs among customer, energy, and demand. The distribution grid incurs relatively minor costs that vary with the amount of energy that goes through the distribution wires. Indeed, about the only "energy-related" costs that a utility incurs on the distribution grid are associated with the electrical losses, and these are typically minor and are treated as generation related costs. The vast majority of the uplift costs after the netting of revenue from the monthly customer charge are determined the size of the wires that the utility has installed for the distribution grid.[4] The size of the wires are determined by the demands that the utility expects its customers to impose on the system, whether for bringing electricity from the transmission system or for delivering electricity from the distributed generation to other customers.
In the context of Figure 1, the customer imposes a demand on the utility of 10 KW, at the extreme left side of Figure 1. Under a gross metering context, the utility would use 10 KW as the billing demand for the customer. Most utilities use the maximum demand during a month as the primary billing demand. Some utilities impose a ratchet on the billing demand, which carries the billing demand forward to subsequent months. A demand of 10 KW imposed on the utility in January would then be used for the billing demand each month afterwards until superseded by a higher actual demand or the expiration of the ratchet. A demand rate of \$5.00/KW-Mo would result in the customer being billed a monthly demand charge of \$50. A one year ratchet would result in annual demand charges of \$600.
In the context of Figure 2, the customer imposes a net demand of only 8 KW on the utility, the difference between the gross demand of 10 KW and the gross simultaneous generation of 2 KW. The reduced demands associated with net metering will necessarily increase the unit price imposed on customers, perhaps \$5.60/KW-Mo instead of the aforementioned \$5.00/KW-Mo.
The implications of using a demand charge to collect the utility's distribution uplift costs are discussed in "Curing the Death Spiral"[5] and "Demand a Better Utility Charge During Era of Renewables: Getting Renewable Incentives Correct With Residential Demand Charges."[6] ATCO Gas, a distribution only utility in Alberta, Canada, uses a customer charge and a demand charge to bill it large customers with no commodity charge, the gas equivalent of an energy charge[7]. ATCO Gas eliminated its commodity charge for large use customers as of September 1, 2007. Thus, the use of a customer charge and a demand charge without a commodity charge is a relatively new phenomenon, at least for ATCO Gas.
Generation Uplift
Some ISOs use a coincident demand variation of the demand charge. Under a coincident demand charge customers are billed a demand charge based on their consumption during periods of stress on the system, generally when the system achieves it maximum demand. A period of stress could also be when generation is most out of balance relative to excess load, such as has been described in the California Duck Curve or the Hawaii Nessie Curve.
Figure 4 builds on Figure 2 by adding two coincident demand indicators, one at the 1 hour mark, the other at the 18 hour mark.
Table 2 develops coincident demand charges under the assumption that these two demand indicators are the only ones used by the utility for charging the uplift fee.
At the 1 hourmark, the customer has a load of 9.75 KW and generation of 2.25 KW. The customer's net metered amount is 7.5 KW. At the 18 hour mark, the customer has a load of 5.5 KW and generation of 6.5 KW. The customer's net metered amount is 1.0 KW, leaving the customer. Assuming generation uplift costs of \$20/KW-Mo, the customer would be billed \$150/mo for the coincident demand it imposed on the utility at the 1 hour mark and would receive a credit of \$20/mo for the relaxation of the coincident demand at the 18 hour mark. Some ISOs use more than two coincident demands for the recovery of uplift costs.
The bottom section of Table 2 provides an advanced coincident demand mechanism, with uneven rates for the two coincident demands. The difference between the two rates is a factor of two (2), with the coincident demand at the 1 hour mark being billed at half of the rate at which the coincident demand is billed at the 18 hour mark. The bottom section of Table 2 is just meant to show that the two coincident demands can be billed at different rates, not to establish the mechanism by which the different rates are determined.
Transmission Uplift
The real time price for the use of the transmission system is determined by geographically differentiating the real time price for generation. The geographic differentiation reflects marginal electrical losses and transmission constraints. The sudden and often unpredictable occurrence of transmission constraints makes the real time price for the use of the distribution system highly variable, even more so than the real time price for generation. Some ISOs may even collect more revenue for the use of their transmission systems than the associated revenue requirement. A coincident demand uplift fee such as that presented in Table 2 may be appropriate for collecting the transmission uplift fee. To the extent that the transmission costs are less than the real time price, the transmission uplift fee could be a credit to the generation uplift fee.
CONCLUSION
PURPA resulted in greater variation in the perceived value of electricity by forcing utilities to buy wholesale electricity from Qualifying Facilities. As the market grew, ISOs were created to manage the transactions and provided better quantification of the real time variability of the value of electricity.
PURPA also allowed customers to be both consumers and producers through a new process known as net metering. The power flow across the interface between the utility and its customers became more variable and less certain.
The combined variability of value and the variability of power flow made finer measurements of the power across the interface more important. This interval metering concept was facilitated by the advances in the electronics industry, which reduced the installed cost of interval meters as well as the cost of processing the data produced by interval meters.
Marginal cost pricing does not provide sufficient revenue to cover the costs of the utility, a point made most clearly by ISOs that impose Uplift charges to recover the difference between costs and revenue. The concept of Uplift charges are equally applicable to utilities other than ISOs and can be collected using demand charges. Coincident demand charges seem to be most applicable for the recovery of generation and transmission Uplift costs. Customer demand charges have traditionally been found to be most applicable for recovering distribution costs not recovered through monthly customer charges.
Notes:
[1] Public Utilities Fortnightly, 1989 December 21
[2] Cogeneration and On-Site Power Production, September, 2007. http://www.cospp.com/articles/article_display.cfm?ARTICLE_ID=307889&p=122
[3] IEEE Power & Energy Magazine, January/February 2009
[4] It should be noted that sometimes a portion of the cost of wires is treated as a customer charge.
[6] USAEE Dalogue, United States Association for Energy Economics, 2015 January. http://dialog.usaee.org/index.php/volume-23-number-1-2015/271-lively
[7] High Use Delivery Service Rates--ATCO Gas North
Fixed Delivery Charge (FDC): \$5.584/day
Variable Delivery Charge (VDC): \$0.000/GJ
Total Demand Charge: \$0.383/GJ/day of 24 Hour Billing Demand
http://www.atcogas.com/Rates/Current_Rates/ | 5,119 | 25,139 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-05 | longest | en | 0.931427 |
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