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www.homesnplots.com | 1,558,765,513,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257889.72/warc/CC-MAIN-20190525044705-20190525070705-00483.warc.gz | 288,210,292 | 6,154 | # User description
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https://finance.yahoo.com/news/why-hillenbrand-inc-nyse-hi-155050220.html | 1,563,525,352,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526153.35/warc/CC-MAIN-20190719074137-20190719100137-00047.warc.gz | 397,852,295 | 105,399 | U.S. Markets open in 4 hrs 55 mins
Why Hillenbrand, Inc. (NYSE:HI) Looks Like A Quality Company
Want to participate in a short research study? Help shape the future of investing tools and you could win a \$250 gift card!
While some investors are already well versed in financial metrics (hat tip), this article is for those who would like to learn about Return On Equity (ROE) and why it is important. To keep the lesson grounded in practicality, we'll use ROE to better understand Hillenbrand, Inc. (NYSE:HI).
Over the last twelve months Hillenbrand has recorded a ROE of 20%. That means that for every \$1 worth of shareholders' equity, it generated \$0.20 in profit.
How Do I Calculate Return On Equity?
The formula for ROE is:
Return on Equity = Net Profit ÷ Shareholders' Equity
Or for Hillenbrand:
20% = US\$147m ÷ US\$770m (Based on the trailing twelve months to March 2019.)
Most readers would understand what net profit is, but it’s worth explaining the concept of shareholders’ equity. It is all the money paid into the company from shareholders, plus any earnings retained. You can calculate shareholders' equity by subtracting the company's total liabilities from its total assets.
What Does Return On Equity Signify?
ROE measures a company's profitability against the profit it retains, and any outside investments. The 'return' is the amount earned after tax over the last twelve months. The higher the ROE, the more profit the company is making. So, all else being equal, a high ROE is better than a low one. That means ROE can be used to compare two businesses.
Does Hillenbrand Have A Good Return On Equity?
One simple way to determine if a company has a good return on equity is to compare it to the average for its industry. However, this method is only useful as a rough check, because companies do differ quite a bit within the same industry classification. As you can see in the graphic below, Hillenbrand has a higher ROE than the average (14%) in the Machinery industry.
That's clearly a positive. I usually take a closer look when a company has a better ROE than industry peers. One data point to check is if insiders have bought shares recently.
Why You Should Consider Debt When Looking At ROE
Virtually all companies need money to invest in the business, to grow profits. The cash for investment can come from prior year profits (retained earnings), issuing new shares, or borrowing. In the first two cases, the ROE will capture this use of capital to grow. In the latter case, the use of debt will improve the returns, but will not change the equity. In this manner the use of debt will boost ROE, even though the core economics of the business stay the same.
Hillenbrand's Debt And Its 20% ROE
Although Hillenbrand does use debt, its debt to equity ratio of 0.47 is still low. The fact that it achieved a fairly good ROE with only modest debt suggests the business might be worth putting on your watchlist. Careful use of debt to boost returns is often very good for shareholders. However, it could reduce the company's ability to take advantage of future opportunities.
The Key Takeaway
Return on equity is one way we can compare the business quality of different companies. A company that can achieve a high return on equity without debt could be considered a high quality business. If two companies have the same ROE, then I would generally prefer the one with less debt.
But when a business is high quality, the market often bids it up to a price that reflects this. The rate at which profits are likely to grow, relative to the expectations of profit growth reflected in the current price, must be considered, too. So you might want to take a peek at this data-rich interactive graph of forecasts for the company.
But note: Hillenbrand may not be the best stock to buy. So take a peek at this free list of interesting companies with high ROE and low debt.
We aim to bring you long-term focused research analysis driven by fundamental data. Note that our analysis may not factor in the latest price-sensitive company announcements or qualitative material.
If you spot an error that warrants correction, please contact the editor at editorial-team@simplywallst.com. This article by Simply Wall St is general in nature. It does not constitute a recommendation to buy or sell any stock, and does not take account of your objectives, or your financial situation. Simply Wall St has no position in the stocks mentioned. Thank you for reading. | 977 | 4,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-30 | latest | en | 0.945498 |
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# Florida: 4/1 - 4/30/2012
Topic closed. 991 replies. Last post 5 years ago by G26GRACE27.
Page 34 of 67
San Diego, CA
United States
Member #61467
May 24, 2008
28146 Posts
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Posted: April 18, 2012, 2:13 pm - IP Logged
04/18
Midday:
Pairs: 04 06 34 35 45 47 48 67 68 78
Due Digit: 5
Due Sum: 18
Evening:
Pairs: 05 13 24 34 39 47 48 58 68 78
Due Digit: 4
Due Sum: 3
Midday: 275
Pairs: 04 06 34 35 45 47 48 67 68 78
Due Digit: 5
Due Sum: 18
Congrats to all the winners!
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August 9, 2010
2913 Posts
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Posted: April 18, 2012, 3:19 pm - IP Logged
Yes very very easy !!!!
Example:
when a combo with a zero appears play doubles made up of the digits from the combo.
408 = 44x 99x 55x 33x 88x
then you sit and wait for the dummies to fall into ya trap. compare to other workouts to narrow down your plays...doubles usually will appear within 4 draws here in florida..
Doubles from the combo?
So how does 55,99,33 come from
408!
Baltimore
United States
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September 24, 2009
44043 Posts
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Posted: April 18, 2012, 4:26 pm - IP Logged
Doubles from the combo?
So how does 55,99,33 come from
408!
MIRRORS
408 has mirrors
44 00 88 first pairs
99 55 33 are the mirrors to the first set....
mirror chart in lottery is as follows (every number has a mirror in lottery world)
1=6
2=7
3=8
4=9
5=0
you now owe me 1 winning ticket valued at \$500 next time you win for the lessons !!!
Baltimore
United States
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September 24, 2009
44043 Posts
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Posted: April 18, 2012, 4:31 pm - IP Logged
966 fast cash
57 20 21 67 31 68
18 36 28 37 29 47
966 fast cash
57 20 21 67 31 68
18 36 28 37 29 47
gotcha 275
Baltimore
United States
Member #80332
September 24, 2009
44043 Posts
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Posted: April 18, 2012, 4:34 pm - IP Logged
275 fast cash
x42 x97 x43 x98 x08 x53
x90 x45 x19 x64 x92 x47
United States
Member #95409
August 9, 2010
2913 Posts
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Posted: April 18, 2012, 4:49 pm - IP Logged
MIRRORS
408 has mirrors
44 00 88 first pairs
99 55 33 are the mirrors to the first set....
mirror chart in lottery is as follows (every number has a mirror in lottery world)
1=6
2=7
3=8
4=9
5=0
you now owe me 1 winning ticket valued at \$500 next time you win for the lessons !!!
I know what mirrors are, just didnt realize what you meant.
I'll still pay up!! LOL
Baltimore
United States
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September 24, 2009
44043 Posts
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Posted: April 18, 2012, 4:53 pm - IP Logged
I know what mirrors are, just didnt realize what you meant.
I'll still pay up!! LOL
lmao...ok just jerkin ya chain but thats what i do.....in terms of making pairs from numbers containing zeros.
United States
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2913 Posts
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Posted: April 18, 2012, 5:15 pm - IP Logged
lmao...ok just jerkin ya chain but thats what i do.....in terms of making pairs from numbers containing zeros.
I know ya are!
DId you really think I would pay ya! haha
Ive never heard of that before, but I'll back test a little and
Sheck it out bra'
fort pierce
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1987 Posts
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Posted: April 18, 2012, 5:47 pm - IP Logged
lol!
Pittsburg, Ks
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81000 Posts
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Posted: April 18, 2012, 6:09 pm - IP Logged
4/18/12 M/E
-/+12 491
Total numbers : 52 Digits used (9) : 611323750 ----------------------------------- 611, 613, 612, 617, 615, 610, 632, 633, 637, 635, 630, 623, 627, 625, 620, 675, 670, 650, 113, 112, 117, 115, 110, 132, 133, 137, 135, 130, 123, 127, 125, 120, 175, 170, 150, 323, 327, 325, 320, 337, 335, 330, 375, 370, 350, 237, 235, 230, 275, 270, 250, 750
The -/+12 hit FL's mid 275 str8
God Bless America
South Florida
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389 Posts
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Posted: April 18, 2012, 7:10 pm - IP Logged
275 fast cash
x42 x97 x43 x98 x08 x53
x90 x45 x19 x64 x92 x47
Technics I will use your "fast cash" to narrow down some picks.
Filter 1
806 809 463 293 490 719 264 465 295 493 910 912 296 495 913 914 468 297 498 915 916 469 298 674 918 609 479 379 460 429 461 462 290 563
Filter 2
719 264 465 295 296 495 914 468 297 498 916 469 298 674 918 479 461 462
Filter 3
719 264 465 296 914 468 916 469 298 674 918 461 462
"Mr. Maths" - Today is a New Day, New Opportunity....Another Chance!
United States
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Posted: April 18, 2012, 8:00 pm - IP Logged
111, 333
America
United States
Member #121867
January 18, 2012
2025 Posts
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Posted: April 18, 2012, 8:01 pm - IP Logged
The winner is 949 and 9336
Baltimore
United States
Member #80332
September 24, 2009
44043 Posts
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Posted: April 18, 2012, 8:15 pm - IP Logged
Yes very very easy !!!!
Example:
when a combo with a zero appears play doubles made up of the digits from the combo.
408 = 44x 99x 55x 33x 88x
then you sit and wait for the dummies to fall into ya trap. compare to other workouts to narrow down your plays...doubles usually will appear within 4 draws here in florida..
408 = 44x 99x 55x 33x 88x
i posted the winning pair in my dang example ..wow
United States
Member #95409
August 9, 2010
2913 Posts
Offline
Posted: April 18, 2012, 8:21 pm - IP Logged
408 = 44x 99x 55x 33x 88x
i posted the winning pair in my dang example ..wow
I was going to say!
"Where the heck is 408!" hahaha
Funny!
Page 34 of 67 | 2,034 | 5,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.728032 |
https://www.crazyengineers.com/threads/solve-the-ant-problem.803 | 1,726,826,338,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652246.93/warc/CC-MAIN-20240920090502-20240920120502-00444.warc.gz | 644,825,010 | 17,873 | • Kaustubh
MemberApr 11, 2007
## Solve The Ant Problem
Got this problem from a friend through IM. Solve it !
An ant is at a fixed end 'P' of a 1 meter-long rubber band.
Just as the ant starts walking along the rubber band at a constant 1 cm/sec,
the other end 'Q' is pulled away from 'P' at a steady speed of 100 cm/sec.
Assume a uniformly stretching, infinitely elastic rubber band is used in this puzzle.
Will the ant ever reach 'Q', and if yes, when?
-The Big K-
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Replies
• MemberApr 16, 2007
To the best of my knowledge, the problem was discussed in Mindsport section of TOI! heck I don't know the answer.
I think the ant should take infinite time to reach the other end.
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• MemberApr 16, 2007
Since the ant is located on the rubber, the total distance that the ant has to cover is just 100 cm. The total time taken will be 100 cm / 1 (cm/sec) = 100 sec.
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• MemberJun 26, 2007
@the_big_K
Is the answer given by rocker is correct ???
This is what i think.
The aunt is at a fixed point P.Now the other end,Q is moving away from aunt(or point P) at the speed of 100 cm/sec.So in one second,the point Q will be 100cm that means Q is now 200cm apart from P.And after one more second,it will be at 300cm apart.Thus the distacne PQ will go on increasing.And the speed of aunt is not sufficient to compensate for that.
Am i right or am i misinterpreting the puzzle.
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• MemberJun 27, 2007
This problem can be correlated to a person walking@1m/s on a 100m long train.the person will cover the length of the train in 100s irrespective of the speed of the train.
Now people will say that here the entire rubber band is not moving, it is getting stretched.Hence the ant which is moving at an absolute speed of 1cm/s will be assisted by the stretching of the rubber band (increasing its absolute speed) and will cover the distance in 100s.(the increasing of speed can be justified if you look at stretching on a microscopic level)
😁
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• MemberJul 27, 2008
The ant never reaches the end Q.....
bcoz the band is infinitely elastic rubber band......
So it keep on stretches......😁
Do u think, It is correct??????
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• MemberJul 31, 2008
The problem has not been interpreted correctly. Assuming uniform elasticity, the length of distance the ant has to cover is increasing by 100 cm/s ... but please note, that in the first second, the ant has covered 1cm, the band is 200cm long ... and the ant has covered 2cms (since that 1cm is now 2cm ... due to stretching!!).
Then the ant goes ahead by 1cm, the band becomes 300cm long .. the distance behind the ant is 5cms. Then it goes ahead by 1cm, the band becomes 400cm long ... the distance behind the ant is 8 cms ahead.
Time Distance Covered %age
----------------------------
1 100 1 1%
2 200 3 1.5%
3 300 5 1.66%
4 400 8 2%
Looks like the ant is making some progress after all 😀. I say it will reach the end. And you know what ...
Moral of the story - Determination gets you there.
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• MemberAug 11, 2008
no it won't reach Q
bcoz it will fall down as the band is stretched.
banu
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• MemberAug 11, 2008
Yeah i agree with the solution provided by Kidakaka...
Since the rubber is uniform ....so the distance travelled by the ant also get's streched.....
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• MemberNov 26, 2009
distance -time relation for ant:
(1)t*(100/100)*t
as per second increment of unit section rubber length is (100/100)*x where x is its orignal length
therefore increment in given section in time t is:
1*x*t
hence length reletion with time is =100t
left distance=100t-t*t=let's say "r"
ant rech at Q when
100t-t*t=0
therefore
t=100sec
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• MemberNov 26, 2009
Very nice answer Kida!!
Its beyond imagination!!!
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• MemberJan 2, 2011
no it won't reach the Q
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• MemberJan 2, 2011
ya it's beyond the imagination.
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• MemberJan 2, 2011
wat could be the answer then
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• MemberJan 2, 2011
Very very interesting question!! I strongly believe that the ant will reach the end at one stage as Kida explains, but it will take a very very long amount of time.. The fact we have to consider here is the ant's position relative to the stretching of the rubber band. I dont know how to explain it, but i'm pretty sure it will make it! I would love to see someone bring out any solid proof..
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• MemberJan 3, 2011
The ant will surely reach the end at some point of time. As the band is being stretched, even the ant's position from the starting point increases at the rate of 100 cm/sec. Now in the same 1sec, the ant has covered a dist. of 1 cm. So actually the ant is covering a distance of 101 cm/sec. I don't know the exact time but it should reach.
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• MemberJan 4, 2011
The ant will reach the end, albeit after a very long time. Will post a detailed mathematical solution once I complete it.
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• MemberJan 6, 2011
Its simple.The ant would take same amount of time if the rubberband had not been streched.
Because,speed of ant would be (100+1)cm/sec.
For example:If we are walking inside a in a moving train our net velocity is algebraic addition of{velocity of train+our own velocity};-)
Are you sure? This action cannot be undone. | 1,552 | 5,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-38 | latest | en | 0.938157 |
http://mathhelpforum.com/algebra/125252-help-solving-z.html | 1,480,766,922,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540928.63/warc/CC-MAIN-20161202170900-00102-ip-10-31-129-80.ec2.internal.warc.gz | 189,371,621 | 10,994 | # Thread: Help solving for z
1. ## Help solving for z
(3/z - 3) + (5/z - 1) = (8/z - 2) Is z negative or positive in this case?
2. Originally Posted by jay1
(3/z - 3) + (5/z - 1) = (8/z - 2) Is z negative or positive in this case?
solve it and see.
first step here. multiply both sides by z. see what you can do from there.
3. Originally Posted by jay1
(3/z - 3) + (5/z - 1) = (8/z - 2) Is z negative or positive in this case?
is the equation ...
$\left(\frac{3}{z} - 3\right) + \left(\frac{5}{z} - 1\right) = \left(\frac{8}{z} - 2\right)$
or
$\frac{3}{z-3} + \frac{5}{z-1} = \frac{8}{z-2}$
???
4. Originally Posted by skeeter
is the equation ...
$\left(\frac{3}{z} - 3\right) + \left(\frac{5}{z} - 1\right) = \left(\frac{8}{z} - 2\right)$
or
$\frac{3}{z-3} + \frac{5}{z-1} = \frac{8}{z-2}$
???
Skeeter, the equation is the latter. Thanks!
5. $\frac{3}{z-3} + \frac{5}{z-1} = \frac{8}{z-2}$
you need a common denominator
6. Originally Posted by jay1
Skeeter, the equation is the latter. Thanks!
well, that makes a world of difference. next time, use parentheses! or better yet, learn LaTeX (see my signature).
to start you off: multiply both sides of the equation by the LCD. in this case, that's (z - 1)(z - 2)(z - 3)
7. Originally Posted by Jhevon
well, that makes a world of difference. next time, use parentheses! or better yet, learn LaTeX (see my signature).
to start you off: multiply both sides of the equation by the LCD. in this case, that's (z - 1)(z - 2)(z - 3)
Does z = -4 ???
8. Originally Posted by jay1
Does z = -4 ???
no.
(you can check this by plugging that value for z into the original equation. you will see that the equation does not make sense with that z-value)
what was your next step after doing what i said? | 607 | 1,757 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2016-50 | longest | en | 0.876562 |
http://cn.metamath.org/mpeuni/bj-rexcom4.html | 1,657,083,234,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00767.warc.gz | 10,640,614 | 3,745 | Mathbox for BJ < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > bj-rexcom4 Structured version Visualization version GIF version
Theorem bj-rexcom4 33191
Description: Remove from rexcom4 3365 dependency on ax-ext 2740 and ax-13 2391 (and on df-or 384, df-tru 1635, df-sb 2047, df-clab 2747, df-cleq 2753, df-clel 2756, df-nfc 2891, df-v 3342). This proof uses only df-rex 3056 on top of first-order logic. (Contributed by BJ, 13-Jun-2019.) (Proof modification is discouraged.)
Assertion
Ref Expression
bj-rexcom4 (∃𝑥𝐴𝑦𝜑 ↔ ∃𝑦𝑥𝐴 𝜑)
Distinct variable groups: 𝑥,𝑦 𝑦,𝐴
Allowed substitution hints: 𝜑(𝑥,𝑦) 𝐴(𝑥)
Proof of Theorem bj-rexcom4
StepHypRef Expression
1 df-rex 3056 . 2 (∃𝑥𝐴𝑦𝜑 ↔ ∃𝑥(𝑥𝐴 ∧ ∃𝑦𝜑))
2 19.42v 2030 . . . . 5 (∃𝑦(𝑥𝐴𝜑) ↔ (𝑥𝐴 ∧ ∃𝑦𝜑))
32bicomi 214 . . . 4 ((𝑥𝐴 ∧ ∃𝑦𝜑) ↔ ∃𝑦(𝑥𝐴𝜑))
43exbii 1923 . . 3 (∃𝑥(𝑥𝐴 ∧ ∃𝑦𝜑) ↔ ∃𝑥𝑦(𝑥𝐴𝜑))
5 excom 2191 . . . 4 (∃𝑥𝑦(𝑥𝐴𝜑) ↔ ∃𝑦𝑥(𝑥𝐴𝜑))
6 df-rex 3056 . . . . . 6 (∃𝑥𝐴 𝜑 ↔ ∃𝑥(𝑥𝐴𝜑))
76bicomi 214 . . . . 5 (∃𝑥(𝑥𝐴𝜑) ↔ ∃𝑥𝐴 𝜑)
87exbii 1923 . . . 4 (∃𝑦𝑥(𝑥𝐴𝜑) ↔ ∃𝑦𝑥𝐴 𝜑)
95, 8bitri 264 . . 3 (∃𝑥𝑦(𝑥𝐴𝜑) ↔ ∃𝑦𝑥𝐴 𝜑)
104, 9bitri 264 . 2 (∃𝑥(𝑥𝐴 ∧ ∃𝑦𝜑) ↔ ∃𝑦𝑥𝐴 𝜑)
111, 10bitri 264 1 (∃𝑥𝐴𝑦𝜑 ↔ ∃𝑦𝑥𝐴 𝜑)
Colors of variables: wff setvar class Syntax hints: ↔ wb 196 ∧ wa 383 ∃wex 1853 ∈ wcel 2139 ∃wrex 3051 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1871 ax-4 1886 ax-5 1988 ax-11 2183 This theorem depends on definitions: df-bi 197 df-an 385 df-ex 1854 df-rex 3056 This theorem is referenced by: bj-rexcom4a 33192
Copyright terms: Public domain W3C validator | 977 | 1,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.174695 |
https://web2.0calc.com/questions/maths_92147 | 1,582,661,390,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146127.10/warc/CC-MAIN-20200225172036-20200225202036-00238.warc.gz | 600,853,906 | 6,416 | +0
# maths
0
326
3
+701
1. Armand is planning a crawfish boil for the weekend. Crawfish are like tiny lobsters, about 3 to 4 in. long. They are especially popular in Louisiana. Crawfish are cooked in a pot of boiling water with spicy seasoning, called a “crawfish boil.” Every 10 lb of crawfish requires 24 oz of seasoning.
(a) What is the unit rate and what does this number mean?
(b) How much seasoning would Armand need for 30 lb of crawfish? Explain or show your work.
(c) What is the maximum number of pounds of crawfish that can be cooked with 36 oz of seasoning? Explain or show your work.
Oct 9, 2018
#1
+107493
+2
a) Unit rate = 24 oz / 10lb = 2.4 oz / lb
This means that we need 2.4 oz of seasoning for every pound of crawfish
b) For 30 lbs....we need 30 * 2.4 = 72 oz of seasoning
c) With 36 oz of seasoning...we can cook 36 / 2.4 = 15 lbs of crawfish
Oct 9, 2018
#2
+20309
-1
Chris can probably tell you that 'crawfish' are also known as 'crawdads' ....or 'MudBugs' in Louisiana !
Oct 9, 2018
#3
+107493
+1
HAHAHA!!!!!....I'm probably the only person in LA that doesn't eat crawfish...!!!
CPhill Oct 9, 2018 | 386 | 1,172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-10 | latest | en | 0.922057 |
https://en.wikipedia.org/wiki/Partial_autocorrelation_function | 1,722,783,899,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640404969.12/warc/CC-MAIN-20240804133418-20240804163418-00074.warc.gz | 182,572,001 | 29,297 | # Partial autocorrelation function
In time series analysis, the partial autocorrelation function (PACF) gives the partial correlation of a stationary time series with its own lagged values, regressed the values of the time series at all shorter lags. It contrasts with the autocorrelation function, which does not control for other lags.
This function plays an important role in data analysis aimed at identifying the extent of the lag in an autoregressive (AR) model. The use of this function was introduced as part of the Box–Jenkins approach to time series modelling, whereby plotting the partial autocorrelative functions one could determine the appropriate lags p in an AR (p) model or in an extended ARIMA (p,d,q) model.
## Definition
Given a time series ${\displaystyle z_{t}}$, the partial autocorrelation of lag ${\displaystyle k}$, denoted ${\displaystyle \phi _{k,k}}$, is the autocorrelation between ${\displaystyle z_{t}}$ and ${\displaystyle z_{t+k}}$ with the linear dependence of ${\displaystyle z_{t}}$ on ${\displaystyle z_{t+1}}$ through ${\displaystyle z_{t+k-1}}$ removed. Equivalently, it is the autocorrelation between ${\displaystyle z_{t}}$ and ${\displaystyle z_{t+k}}$ that is not accounted for by lags ${\displaystyle 1}$ through ${\displaystyle k-1}$, inclusive.[1]${\displaystyle \phi _{1,1}=\operatorname {corr} (z_{t+1},z_{t}),{\text{ for }}k=1,}$${\displaystyle \phi _{k,k}=\operatorname {corr} (z_{t+k}-{\hat {z}}_{t+k},\,z_{t}-{\hat {z}}_{t}),{\text{ for }}k\geq 2,}$where ${\displaystyle {\hat {z}}_{t+k}}$ and ${\displaystyle {\hat {z}}_{t}}$ are linear combinations of ${\displaystyle \{z_{t+1},z_{t+2},...,z_{t+k-1}\}}$ that minimize the mean squared error of ${\displaystyle z_{t+k}}$ and ${\displaystyle z_{t}}$ respectively. For stationary processes, the coefficients in ${\displaystyle {\hat {z}}_{t+k}}$ and ${\displaystyle {\hat {z}}_{t}}$ are the same, but reversed:[2] ${\displaystyle {\hat {z}}_{t+k}=\beta _{1}z_{t+k-1}+\cdots +\beta _{k-1}z_{t+1}\qquad {\text{and}}\qquad {\hat {z}}_{t}=\beta _{1}z_{t+1}+\cdots +\beta _{k-1}z_{t+k-1}.}$
## Calculation
The theoretical partial autocorrelation function of a stationary time series can be calculated by using the Durbin–Levinson Algorithm:${\displaystyle \phi _{n,n}={\frac {\rho (n)-\sum _{k=1}^{n-1}\phi _{n-1,k}\rho (n-k)}{1-\sum _{k=1}^{n-1}\phi _{n-1,k}\rho (k)}}}$where ${\displaystyle \phi _{n,k}=\phi _{n-1,k}-\phi _{n,n}\phi _{n-1,n-k}}$ for ${\displaystyle 1\leq k\leq n-1}$ and ${\displaystyle \rho (n)}$ is the autocorrelation function.[3][4][5]
The formula above can be used with sample autocorrelations to find the sample partial autocorrelation function of any given time series.[6][7]
## Examples
The following table summarizes the partial autocorrelation function of different models:[5][8]
Model PACF
White noise The partial autocorrelation is 0 for all lags.
Autoregressive model The partial autocorrelation for an AR(p) model is nonzero for lags less than or equal to p and 0 for lags greater than p.
Moving-average model If ${\displaystyle \phi _{1,1}>0}$, the partial autocorrelation oscillates to 0.
If ${\displaystyle \phi _{1,1}<0}$, the partial autocorrelation geometrically decays to 0.
Autoregressive–moving-average model An ARMA(p, q) model's partial autocorrelation geometrically decays to 0 but only after lags greater than p.
The behavior of the partial autocorrelation function mirrors that of the autocorrelation function for autoregressive and moving-average models. For example, the partial autocorrelation function of an AR(p) series cuts off after lag p similar to the autocorrelation function of an MA(q) series with lag q. In addition, the autocorrelation function of an AR(p) process tails off just like the partial autocorrelation function of an MA(q) process.[2]
## Autoregressive model identification
Partial autocorrelation is a commonly used tool for identifying the order of an autoregressive model.[6] As previously mentioned, the partial autocorrelation of an AR(p) process is zero at lags greater than p.[5][8] If an AR model is determined to be appropriate, then the sample partial autocorrelation plot is examined to help identify the order.
The partial autocorrelation of lags greater than p for an AR(p) time series are approximately independent and normal with a mean of 0.[9] Therefore, a confidence interval can be constructed by dividing a selected z-score by ${\displaystyle {\sqrt {n}}}$. Lags with partial autocorrelations outside of the confidence interval indicate that the AR model's order is likely greater than or equal to the lag. Plotting the partial autocorrelation function and drawing the lines of the confidence interval is a common way to analyze the order of an AR model. To evaluate the order, one examines the plot to find the lag after which the partial autocorrelations are all within the confidence interval. This lag is determined to likely be the AR model's order.[1]
## References
1. ^ a b "6.4.4.6.3. Partial Autocorrelation Plot". www.itl.nist.gov. Retrieved 2022-07-14.
2. ^ a b Shumway, Robert H.; Stoffer, David S. (2017). Time Series Analysis and Its Applications: With R Examples. Springer Texts in Statistics. Cham: Springer International Publishing. pp. 97–99. doi:10.1007/978-3-319-52452-8. ISBN 978-3-319-52451-1.
3. ^ Durbin, J. (1960). "The Fitting of Time-Series Models". Revue de l'Institut International de Statistique / Review of the International Statistical Institute. 28 (3): 233–244. doi:10.2307/1401322. ISSN 0373-1138. JSTOR 1401322.
4. ^ Shumway, Robert H.; Stoffer, David S. (2017). Time Series Analysis and Its Applications: With R Examples. Springer Texts in Statistics. Cham: Springer International Publishing. pp. 103–104. doi:10.1007/978-3-319-52452-8. ISBN 978-3-319-52451-1.
5. ^ a b c Enders, Walter (2004). Applied econometric time series (2nd ed.). Hoboken, NJ: J. Wiley. pp. 65–67. ISBN 0-471-23065-0. OCLC 52387978.
6. ^ a b Box, George E. P.; Reinsel, Gregory C.; Jenkins, Gwilym M. (2008). Time Series Analysis: Forecasting and Control (4th ed.). Hoboken, New Jersey: John Wiley. ISBN 9780470272848.
7. ^ Brockwell, Peter J.; Davis, Richard A. (1991). Time Series: Theory and Methods (2nd ed.). New York, NY: Springer. pp. 102, 243–245. ISBN 9781441903198.
8. ^ a b Das, Panchanan (2019). Econometrics in Theory and Practice : Analysis of Cross Section, Time Series and Panel Data with Stata 15. 1. Singapore: Springer. pp. 294–299. ISBN 978-981-329-019-8. OCLC 1119630068.
9. ^ Quenouille, M. H. (1949). "Approximate Tests of Correlation in Time-Series". Journal of the Royal Statistical Society, Series B (Methodological). 11 (1): 68–84. doi:10.1111/j.2517-6161.1949.tb00023.x. | 1,958 | 6,714 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 29, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-33 | latest | en | 0.787072 |
http://stackoverflow.com/questions/14710778/python-sorting-sub-strings | 1,436,246,108,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375098990.43/warc/CC-MAIN-20150627031818-00050-ip-10-179-60-89.ec2.internal.warc.gz | 256,711,375 | 17,975 | # Python sorting : sub strings
In Python
``````s= "ABCC"
n = len(s)
sorted(set([s[a:b] for a in range(n) for b in range(a+1,n+2)])
``````
gives me, alphabetically sorted sub strings with out repetitions
``````['A', 'AB', 'ABC', 'ABCC', 'B', 'BC', 'BCC', 'C', 'CC']
``````
How can I further sort it by length of sub string.
``````['A', 'B', 'C', 'AB', 'BC', 'CC', 'ABC', 'BCC', 'ABCC']
``````
-
– Ashwini Chaudhary Feb 5 '13 at 15:21
simple,
``````sorted(set(s[a:b] for a in range(n) for b in range(a+1,n+1)),
key=lambda x:(len(x),x))
``````
This creates a key by which the comparison is done. First it compares the string lengths to determine the order. If the strings have the same length, the tie-breaker is the string contents.
-
cleaned up `set([])` and `n+2` from the OP's code – georg Feb 5 '13 at 15:34
@thg435 -- Thanks. – mgilson Feb 5 '13 at 15:41
This is your solution:
``````s= "ABCC"
n = len(s)
sorted(sorted(set([s[a:b] for a in range(n) for b in range(a+1,n+2)])),key=len)
``````
-
And how is that solution wrong? Did you run it yourself? Please run it in Python and tell me if it is wrong or not. – Guddu Feb 5 '13 at 15:30
Python sorting is guaranteed to be stable, so there's nothing wrong about double `sorted`. Upvoted. – georg Feb 5 '13 at 15:33
I missed the fact that you're using two `sorted` calls(ignore my last comment(deleted)), +1. – Ashwini Chaudhary Feb 5 '13 at 15:37
mgilson....Could you prove your point? I have tested this with this string also s= "NOPQRSTUVWXYZABCDEFGHIJKLM" – Guddu Feb 5 '13 at 15:41
I am wondering why two times sorting is required, 'set' sorts the list alphabetically, then you sort the with sorted for len. – ZEN.Kamath Feb 5 '13 at 16:12 | 570 | 1,715 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2015-27 | latest | en | 0.821419 |
http://folio.furman.edu/projects/euclid/eubk1/prop22/prop22~1.htm | 1,556,269,395,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578762045.99/warc/CC-MAIN-20190426073513-20190426095513-00030.warc.gz | 62,106,691 | 1,389 | THE ELEMENTS Book I Proposition 22 Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, andB, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C. Return to Propositions Next Page | 92 | 361 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-18 | latest | en | 0.868641 |
https://thinkmad.in/puzzles/what-is-the-next-number-in-the-sequence-1-3-4-7-11-18/ | 1,643,410,341,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320306346.64/warc/CC-MAIN-20220128212503-20220129002503-00588.warc.gz | 601,598,154 | 41,611 | Select Page
# Next number sequence in 1, 3, 4, 7, 11, 18, _?
Home » Puzzles » Next number sequence in 1, 3, 4, 7, 11, 18, _?
# Question: Find the next number sequence in 1, 3, 4, 7, 11, 18,_?
Ans: 45
Explanation:
The logic used : Here we add the Answer of the first step to the given sum of the other two numbers.
5 + 2 + 5 = 12
12 + 3 + 6 = 21
21+4+7=32
32 + 5 + 8 = 45
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http://rhinocfd.com/phoenics/d_earth/d_core/inplib/q1ears/368.htm | 1,516,452,822,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889617.56/warc/CC-MAIN-20180120122736-20180120142736-00238.warc.gz | 280,622,139 | 3,117 | ```
TALK=T;RUN(1,1)
DISPLAY
HEAT-EXCHANGER SIMULATIONS BY MEANS OF PHOENICS
These examples make extensive use of the "neighbour"
technique, whereby the use of a special patch name
activates special calls in GROUND which are to be
included in GREX.
It will be useful to extend the series further, so as
to illustrate how two-phase flow can be handled on
tube and shell sides of a boiler, and to show also how
cooling-tower problems can be handled with the aid of
this technique.
ENDDIS
************************************************************
Group 1. Run Title and Number
************************************************************
************************************************************
TEXT(In-Form equivalent of library case 791 )
************************************************************
************************************************************
IRUNN = 1 ;LIBREF = 368
************************************************************
Group 2. Time dependence
* Set overall time and no. of steps
TFIRST =0. ;TLAST =20.
FSTEP = 1 ;LSTEP = 20
TFRAC(1)=0.05 ;TFRAC(2)=0.1
TFRAC(3)=0.15 ;TFRAC(4)=0.2
TFRAC(5)=0.25 ;TFRAC(6)=0.3
TFRAC(7)=0.35 ;TFRAC(8)=0.4
TFRAC(9)=0.45 ;TFRAC(10)=0.5
TFRAC(11)=0.55 ;TFRAC(12)=0.6
TFRAC(13)=0.65 ;TFRAC(14)=0.7
TFRAC(15)=0.75 ;TFRAC(16)=0.8
TFRAC(17)=0.85 ;TFRAC(18)=0.9
TFRAC(19)=0.95 ;TFRAC(20)=1.
************************************************************
Group 3. X-Direction Grid Spacing
CARTES = T
NX = 20
XULAST =10.
XFRAC(1)=0.05 ;XFRAC(5)=0.25
XFRAC(9)=0.45 ;XFRAC(13)=0.65
XFRAC(17)=0.85
************************************************************
Group 4. Y-Direction Grid Spacing
NY = 1
YVLAST =1.
YFRAC(1)=1.
************************************************************
Group 5. Z-Direction Grid Spacing
PARAB = F
NZ = 1
ZWLAST =1.
ZFRAC(1)=1.
************************************************************
Group 6. Body-Fitted Coordinates
************************************************************
Group 7. Variables: STOREd,SOLVEd,NAMEd
ONEPHS = T
NAME(14)=1STH ;NAME(15)=2NDH
NAME(149)=VPOR ;NAME(150)=EPOR
* Y in SOLUTN argument list denotes:
* 1-stored 2-solved 3-whole-field
* 4-point-by-point 5-explicit 6-harmonic averaging
SOLUTN(1STH,Y,Y,N,N,N,Y)
SOLUTN(2NDH,Y,Y,N,N,N,Y)
SOLUTN(VPOR,Y,N,N,N,N,Y)
SOLUTN(EPOR,Y,N,N,N,N,Y)
EPOR = 150 ;HPOR = 0 ;NPOR = 0 ;VPOR = 149
************************************************************
Group 8. Terms & Devices
* Y in TERMS argument list denotes:
* 1-built-in source 2-convection 3-diffusion 4-transient
* 5-first phase variable 6-interphase transport
TERMS(1STH,N,N,N,Y,Y,Y)
TERMS(2NDH,N,N,N,Y,N,Y)
DIFCUT =0.5 ;ZDIFAC =1.
GALA = F ;ADDDIF = F
HUNIT =1.
ISOLX = -1 ;ISOLY = -1 ;ISOLZ = -1
************************************************************
Group 9. Properties used if PRPS is not
stored, and where PRPS = -1.0 if it is!
RHO1 =1. ;TMP1 =0.
EL1 =0.
TSURR =0. ;TEMP0 =0.
PRESS0 =0.
DVO1DT =0. ;DRH1DP =0.
EMISS =0. ;SCATT =0.
ENUL =1.0E-05 ;ENUT =0.
CP1 =1. ;CP2 =1.
************************************************************
Group 10.Inter-Phase Transfer Processes
************************************************************
Group 11.Initial field variables (PHIs)
FIINIT(1STH)=1. ;FIINIT(2NDH)=0.
FIINIT(VPOR)=1.0E-05 ;FIINIT(EPOR)=0.5
No PATCHes yet used for this Group
FSWEEP = 1
NAMFI =CHAM
************************************************************
Group 12. Patchwise adjustment of terms
Patches for this group are printed with those
for Group 13.
Their names begin either with GP12 or &
************************************************************
Group 13. Boundary & Special Sources
PATCH(INLET1 ,CELL , 1, 1, 1, 1, 1, 1, 1, 1000)
COVAL(INLET1 ,1STH,0.1 ,1. )
PATCH(INLET2 ,CELL , 20, 20, 1, 1, 1, 1, 1, 1000)
COVAL(INLET2 ,2NDH,0.1 ,0. )
PATCH(INEW1 ,CELL , 2, 20, 1, 1, 1, 1, 1, 1000)
COVAL(INEW1 ,1STH,In-Form:source - see Grp 19)
PATCH(INEE1 ,CELL , 1, 19, 1, 1, 1, 1, 1, 1000)
COVAL(INEE1 ,2NDH,In-Form:source - see Grp 19)
PATCH(INEPLUS ,VOLUME, 1, 20, 1, 1, 1, 1, 1, 1000)
COVAL(INEPLUS ,1STH,In-Form:source - see Grp 19)
PATCH(INEMINUS,VOLUME, 1, 20, 1, 1, 1, 1, 1, 1000)
COVAL(INEMINUS,2NDH,In-Form:source - see Grp 19)
PATCH(INEOLD ,VOLUME, 1, 20, 1, 1, 1, 1, 1, 20)
COVAL(INEOLD ,1STH,In-Form:source - see Grp 19)
COVAL(INEOLD ,2NDH,In-Form:source - see Grp 19)
XCYCLE = F
EGWF = T
WALLCO = GRND2
************************************************************
Group 14. Downstream Pressure For PARAB
************************************************************
Group 15. Terminate Sweeps
LSWEEP = 25 ;ISWC1 = 1
LITHYD = 1 ;LITFLX = 1 ;LITC = 1 ;ITHC1 = 1
SELREF = T
RESFAC =1.0E-04
************************************************************
Group 16. Terminate Iterations
LITER(1STH)=20 ;LITER(2NDH)=20
ENDIT(1STH)=1.0E-03 ;ENDIT(2NDH)=1.0E-03
************************************************************
Group 17. Relaxation
RELAX(1STH,FALSDT,1.0E+09)
RELAX(2NDH,FALSDT,1.0E+09)
OVRRLX =0.
EXPERT = F ;NNORSL = F
************************************************************
Group 18. Limits
VARMAX(1STH)=1.0E+10 ;VARMIN(1STH)=-1.0E+10
VARMAX(2NDH)=1.0E+10 ;VARMIN(2NDH)=-1.0E+10
VARMAX(VPOR)=1.0E+10 ;VARMIN(VPOR)=-1.0E+10
VARMAX(EPOR)=1.0E+10 ;VARMIN(EPOR)=-1.0E+10
************************************************************
Group 19. Data transmitted to GROUND
PARSOL = F
ISG62 = 1
SPEDAT(SET,SOURCE,1STH!INEW1,C,=0.1*(1STH[-1]-1STH)!LINE)
SPEDAT(SET,SOURCE,2NDH!INEE1,C,=0.1*(2NDH[+1]-2NDH)!LINE)
SPEDAT(SET,SOURCE,1STH!INEPLUS,C,=0.05*(2NDH-1STH)!LINE)
SPEDAT(SET,SOURCE,2NDH!INEMINUS,C,=0.05*(1STH-2NDH)!LINE)
SPEDAT(SET,SOURCE,1STH!INEOLD,C,=0.5*(OLD(1STH)-1STH)!LINE)
SPEDAT(SET,SOURCE,2NDH!INEOLD,C,=0.5*(OLD(2NDH)-2NDH)!LINE)
************************************************************
Group 20. Preliminary Printout
DISTIL = T ;NULLPR = F
NDST = 0
DSTTOL =1.0E-02
EX(1STH)=0. ;EX(2NDH)=0.
EX(VPOR)=0. ;EX(EPOR)=0.
************************************************************
Group 21. Print-out of Variables
INIFLD = F ;SUBWGR = F
* Y in OUTPUT argument list denotes:
* 1-field 2-correction-eq. monitor 3-selective dumping
* 4-whole-field residual 5-spot-value table 6-residual table
OUTPUT(1STH,N,N,Y,Y,Y,Y)
OUTPUT(2NDH,N,N,Y,Y,Y,Y)
OUTPUT(VPOR,N,N,N,N,N,N)
OUTPUT(EPOR,N,N,N,N,N,N)
************************************************************
Group 22. Monitor Print-Out
IXMON = 10 ;IYMON = 1 ;IZMON = 1
NPRMON = 100000 ;NPRMNT = 1 ;TSTSWP = 5
UWATCH = T ;USTEER = T
HIGHLO = F
************************************************************
Group 23.Field Print-Out & Plot Control
NPRINT = 100000 ;NUMCLS = 5
NTPRIN = 100000 ;ISTPRF = 1 ;ISTPRL = 100000
NXPRIN = 4 ;IXPRF = 1 ;IXPRL = 10000
IPLTF = 1 ;IPLTL = 50 ;NPLT = 1
ISWPRF = 1 ;ISWPRL = 100000
ITABL = 3 ;IPROF = 3
ABSIZ =0.5 ;ORSIZ =0.4
NTZPRF = 1 ;NCOLPF = 50
ICHR = 2 ;NCOLCO = 45 ;NROWCO = 20
PATCH(PROFILES,PROFIL, 1, 20, 1, 1, 1, 1, 1, 1000)
PLOT(PROFILES,1STH,0. ,1. )
PLOT(PROFILES,2NDH,0. ,1. )
PATCH(TIMEPLOT,PROFIL, 20, 20, 1, 1, 1, 1, 1, 20)
PLOT(TIMEPLOT,1STH,0. ,1. )
PLOT(TIMEPLOT,2NDH,0. ,1. )
************************************************************
Group 24. Dumps For Restarts
SAVE = T ;NOWIPE = F
NSAVE =CHAM
STOP
``` | 2,699 | 7,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-05 | latest | en | 0.494973 |
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# Rogue AEP values and min/maxing
24 replies to this topic
### #1 Mish
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Posted 23 August 2006 - 05:29 PM
So I want to min/max my PvP and PvE gear. I took Shadowpanther.net webpage and imported it into a spreadsheet. In addition to the unmodified version, I have a "DPS" version and a "PvP" version.
In my DPS version, I take stamina and resist out of the equation.
In my PvP version, I want to add weighting to Stamina and Dodge while removing weighting for +hit. This is where I'm at.
The standard AEP values are 1 Agil = 1 Stam. I would like to give Stamina a higher value for PvP, but I'm torn about how much I should go. 1.5? 2?
I would like some comments about what each of you feel the weighting should be and why?
The same question stands for items that have +Dodge %. The current value is 1 AGI = .33% Dodge.
The full equivalencies are here for those unfamiliar with this system (unlikely):
Formula: 1 Agility = 1 Stamina = 2 Strength = .1% Crit. = 5 Defense = .33% Dodge = .2% Parry = .13% To Hit = 2 Attack Power = 4 Any Resistance = 5 Health/5 Sec. = 50 Armor
Mishana, 80 UD Rogue, Lightning's Blade (US)
### #2 exog
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Posted 23 August 2006 - 05:32 PM
why remove +hit in pvp?
it still removes white misses in pvp too afaik.
### #3 chalon
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Posted 23 August 2006 - 05:38 PM
The issue with any equivalence system is that equivalence from a DPS point of view is a sliding scale. Even a matter of buffs or not can have a huge difference. For instance, with my current gear from a DPS point-of-view, 1 crit = about 17 AP unbuffed, but with standard raid buffs 1 crit = 22.35 AP.
### #4 Vhal
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Posted 23 August 2006 - 06:32 PM
At some point, I'm going to be motivated enough to set up a page that calculates everything in relation to AP instead of Agi, which I (at least) find to be so much easier to handle conceptually -- Agi gives two benefits, thus scaling twice, and AP is directly convertable to raw weapon DPS.
But anyway, for PvP values...
I've done the rogue and the mage thing in PvP on a competitive server. I like stamina. I find it's less important on the rogue than the mage, though. I'm still reprioritizing everything post-1.12, but I tend to value stamina at 1-1.5 agi.
Hit is more important than any other stat up to +5. Of course, +5 is not terribly hard to get, and with 1.12 Precision becomes viable in more PvP builds (I'm currently running 16/12/23 daggers, taking 2 points in Precision to pick up Imp. Sprint in Combat). Past +5, its value drops tremendously -- it's only a white damage booster, and PvP is significantly biased towards yellow damage.
### #5 Maverikki
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Posted 23 August 2006 - 07:59 PM
I found a page from my bookmarks that lets you play around with the weights of different stats. Seems it lacks some equipment from AQ and onward. Maybe someone interested enough could send a message to the author and update it.
http://ccgi.brogett..../cgi-bin/AEP.pl
### #6 probiscus
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Posted 23 August 2006 - 09:07 PM
Everytime someone brings up AEP, I choke a kitten.
### #7 issei
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Posted 24 August 2006 - 12:01 AM
AEP is at first a nice general tool/guideline and then a crutch and then a handicap
### #8 Karakas
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Posted 24 August 2006 - 12:05 AM
AEP is evil. It tries to assign static values to each stat, because we all know +crit/+hit don't scale up the more damage you do per hit :rolleyes:
### #9 Vhal
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Posted 24 August 2006 - 01:34 AM
Erm, Karakas, figuring it out dynamically is kinda the point of Brogetta's calculator, linked above.
It seems to me that the two most important questions with regard to evaluating rogue gear are "How does attack power relate to crit (or hit) chance in terms of expected damage output?" and "How much expected damage output loss is justifiable to gain a certain amount of stamina?". There are, of course, many additional subtleties to it, but the actual value of the majority of pieces of gear can be determined if one has at least reasonable answers to the above questions.
Now, that said, I've never cared for quantifying the value of gear in terms of agility itself (due to it boosting both crit and AP), but find that coming up with a good relative value of AP to the other stats (crit and hit objectively, stam subjectively) is quite helpful.
### #10 Dakous
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Posted 24 August 2006 - 01:35 AM
My 50,000 AP totally saved my life this one time Vek turned and did that one off melee beat.
Everybody is your brother until the rent comes due.
### #11 Kalman
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Posted 24 August 2006 - 08:26 AM
Erm, Karakas, figuring it out dynamically is kinda the point of Brogetta's calculator, linked above.
Too bad Brogetta's has *never* done a very good job of it.
Melador> Incidentally, these last few pages are why people hate lawyers.
Viator> I really don't want to go all Kalman here.
Bury> Just imagine what the world would be like if you used your powers for good.
### #12 sp00n
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Posted 24 August 2006 - 08:31 AM
See my sig, something like Brogetta's, yet I enter the items once a month or so when something new comes up.
Still missing some sort of DPS calc though to dynamically adapt the AEP values.
Hm, sig isn't showing, here's the link: http://sp00n.pytalhost.com/wow/
Stopped Playing
### #13 Boevis
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Posted 24 August 2006 - 08:35 AM
My 50,000 AP totally saved my life this one time Vek turned and did that one off melee beat.
My 50,000 HP totally killed Razuvious this one time healers went OOM and couldn't keep the adds up.
see what I did there?
### #14 Mish
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Posted 24 August 2006 - 12:47 PM
AEP is at first a nice general tool/guideline and then a crutch and then a handicap
What do you suggest as an alternative?
Mishana, 80 UD Rogue, Lightning's Blade (US)
### #15 sp00n
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Posted 24 August 2006 - 12:55 PM
AEP is at first a nice general tool/guideline and then a crutch and then a handicap
What do you suggest as an alternative?
Stopped Playing
### #16 Mish
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Posted 24 August 2006 - 01:07 PM
I use that for doing DPS. I value stamina pretty highly for PvP though and the DPS sheet doesn't cover surviveability.
Mishana, 80 UD Rogue, Lightning's Blade (US)
### #17 Avair
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Posted 24 August 2006 - 01:46 PM
AEP is a very easy to understand rule of thumb. It works quite well enough for what it is, an inexact formula. Yes, a DPS spreadsheet will give you better results, but AEP has the virtue of not requiring a computer to calculate.
For Raiding/PvE though, don't weight stamina too heavily. AEP was a PvP formula primarily.
### #18 Maestroquark
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Posted 24 August 2006 - 02:31 PM
My 50,000 HP totally killed Razuvious this one time healers went OOM and couldn't keep the adds up.
see what I did there?
Proved the point? That equivalence measures are bad because everything taken in extremes leaves you extremely weak?
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### #19 Faytte
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Posted 24 August 2006 - 03:01 PM
More or less the superfluese damage in Naxx is either
A- easily survivable
B- will instantly kill you
As a result for the raider stamina, as a whole, isnt very important anymore. You'll notice the upgrade in stamina from t2 to t3 is fairly underwhelming (although the dps increase is very nice!).
Thats why as a raider i stick to the spreadsheets.
As far as Vek. Any rogue that far in the game should have
A- Enough HP to survive the uppercuts
B- The common sense to bandage before running back into melee. You can pop evasion or a healthstone if your that impatient.
Farewell, remorse: all good to me is lost; Evil, be thou my good.
~Paradise Lost (bk. IX, l. 171)
### #20 Dakous
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Posted 24 August 2006 - 05:52 PM
As far as Vek. Any rogue that far in the game should have
A- Enough HP to survive the uppercuts
B- The common sense to bandage before running back into melee. You can pop evasion or a healthstone if your that impatient.
The point was that if you could find low sta high impact epics that you WOULDN'T have enough HP to survive the uppercuts, so true equivelence doesn't exist. Boevis came up with an alternate example, too.
You said raider stamina isn't that important, but how much does uppercut do as opposed to how much HP does a raid geared rogue have? Forgetting, for a moment, BoK since 1/4th of the people playing WoW these days don't have that luxury.
And on your whole point about "things in Naxx oneshot"... how many things in MC oneshot back when that was the top instance and noone had epics from it? How many things in BWL oneshot? AQ40? But post gearing.. oh boy, you survive.
I'm sure there's no opportunity cost for dying every pull.
Everybody is your brother until the rent comes due.
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0 members, 0 guests, 0 anonymous users | 2,686 | 9,664 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2014-52 | longest | en | 0.918996 |
http://www.blackenterprise.com/when-pe-ratios-just-wont-cut-it/ | 1,542,635,201,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745762.76/warc/CC-MAIN-20181119130208-20181119152208-00399.warc.gz | 372,368,081 | 19,754 | When P/E ratios just won't cut it - Black Enterprise
A long time ago, in a market far, far away, price-to-earnings ratios were considered an adequate measure of a stock’s intrinsic value. But with the bull market now more than eight years old and with stock prices soaring, P/E ratios have become so astronomically high that some analysts today question whether they are an accurate gauge of a stock’s worth.
By old Wall Street standards, stocks with P/Es of 20 were overvalued and not considered good growth prospects. But that rule of thumb has less relevance in today’s market, when even cyclical stocks such as Minnesota Mining & Manufacturing Co. (NYSE: MMM) trade at a P/E of roughly 23.9.
As a result, analysts and fund managers are relying more on alternative methods of valuing stocks. The following is a glossary of such terms, as well as more basic definitions:
• Price-to-earnings ratio or P/E ratio. The P/E ratio, also called the multiple, is still the most common way to value a stock. It is the price of a stock divided by its earnings per share. It’s usually calculated using analysts’ consensus estimates of next year’s earnings, known as the forward P/E.
For example, a stock selling for \$20 that earned \$1 last year has a trailing P/E of 20. If that same stock has projected earnings of \$2 a share, it will have a forward P/E of 10.
• Book value. This is the net asset value of a company’s securities. It is calculated by figuring a company’s total net asset value, then dividing that amount by the number of shares of common stock, shares of preferred stock or bonds outstanding. Book value is one tool used to determine if a stock is underpriced and therefore a good buy.
• Dividend yield. The annual rate of return on an investment expressed as a percentage. For stocks, it is the annual dividend divided by the purchase price. It’s also called the current yield.
• EBITDA. The acronym stands for earnings before interest, taxes, depreciation and amortization. Also called cash flow, it is figured by taking a company’s net income, adding depreciation and amortization, then subtracting capital expenditures (how much money the company invests each year on plant and equipment). EBITDA is used primarily to value broadcasting and telecommunications stocks, companies with large investments in capital improvements.
• Price-to-sales ratio. A stock’s capitalization divided by its sales over the trailing 12 months. The value is the same whether the calculation is done for the whole company or on a per-share basis.
If you want to find out more about financial terms on the Web, visit www.investorwords.com. | 568 | 2,637 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-47 | latest | en | 0.96169 |
https://business-accounting.net/direct-material-variance/ | 1,600,967,961,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400219691.59/warc/CC-MAIN-20200924163714-20200924193714-00686.warc.gz | 303,497,406 | 11,746 | # Direct material variance
It measures the difference between the budgeted and the actual level of activity valued at the standard fixed cost per unit. The fixed overhead volume variance is obtained by subtracting actual units produced from budgeted units and then multiplying the result with standard fixed cost per unit. The standard fixed cost per unit is obtained by dividing the budgeted fixed overhead by the budgeted production. It can also be obtained by subtracting actual hours incurred in production from the budgeted hours and then multiplying the result with the standard fixed cost per hour.
Fixed overhead expenditure variance is calculated by subtracting the actual fixed overhead cost from the budgeted fixed overhead cost. It can be favorable when the budgeted fixed overhead is less than the actual fixed overhead or adverse when the actual costs are more than the budgeted. In budgeting (or management accounting in general), a variance is the difference between a budgeted, planned, or standard cost and the actual amount incurred/sold. Variance analysis, also described as analysis of variance or ANOVA, involves assessing the difference between two figures.
It is a tool applied to financial and operational data that aims to identify and determine the cause of the variance. In applied statistics, there are different forms of variance analysis. In project management, variance analysis helps maintain control over a project’s expenses by monitoring planned versus actual costs.
For instance, a monthly closing report might provide quantitative data about expenses, revenue and remaining inventory levels. Variances between planned and actual costs might lead to adjusting business goals, objectives or strategies. This is the difference between the actual and budgeted variable overhead costs that result from inefficient use of indirect materials and indirect labor.
Overhead variance occurs when the day-to-day costs of running a business differs from the amount budgeted and can occur with both variable and fixed overheads. Sales variance is the difference between planned or expected sales and actual sales made. Analysing sales variance helps to measure sales performance, understand market conditions and evaluate business results.
It is favorable when the actual units produced are more than the budgeted units and adverse when the number of units produced are less than the budgeted. Variance analysis is important to assist with managing budgets by controlling budgeted versus actual costs. In program and project management, for example, financial data are generally assessed at key intervals or milestones.
Variance is the difference between budgeted or planned costs or sales and actual costs incurred or sales made. Your variable components may consist of things such as indirect material, and direct labor, and supplies.
### How do you calculate spending variance?
A spending variance is the difference between the actual and expected (or budgeted) amount of an expense. The spending variance for fixed overhead is known as the fixed overhead spending variance, and is the actual expense incurred minus the budgeted expense.
Actual overhead variances are those that have been incurred and can be known at the end of a particular accounting period after the accounts have been prepared. Absorbed overheads are overheads charged to a product based on a predetermined overhead rate, which is the standard overhead absorption rate. Understanding and analysing variance helps businesses understand current outgoings and budget for future expenses. Businesses therefore carry out variance analysis – a quantitative investigation into the differences between planned and actual costs and revenues.
If such analysis is not carried out in regular intervals, it may cause a delay in the management action to control its costs. Managers acknowledge that it is impossible to exactly attain budgeted estimates, such as targeted profits. A variance is a difference between the actual figures and budgeted estimates. Overhead variance arises due to the differences between actual overhead variances and the budgeted or the absorbed variances.
When conducting variance analysis consider your actual revenue and/or costs versus your budgeted figures. Are there small, continual changes over time that are diverging from your planned budget? Analysis of these trends from month to month will help you get a better understanding of where your variance is coming from. Higher than expected expenses can also cause an unfavorable variance. For example, if your budgeted expenses were \$200,000 but your actual costs were \$250,000, your unfavorable variance would be \$50,000 or 25 percent.
• It measures the difference between the budgeted and the actual level of activity valued at the standard fixed cost per unit.
• The standard fixed cost per unit is obtained by dividing the budgeted fixed overhead by the budgeted production.
## Direct Material Variances
Variance analysis is usually associated with explaining the difference (or variance) between actual costs and the standard costs allowed for the good output. For example, the difference in materials costs can be divided into a materials price variance and a materials usage variance. The difference between the actual direct labor costs and the standard direct labor costs can be divided into a rate variance and an efficiency variance. The difference in manufacturing overhead can be divided into spending, efficiency, and volume variances. Thus, Variance Analysis is important to analyze the difference between the actual and planned behavior of an organization.
Fixed overhead may include rent, car insurance, maintenance, depreciation and more. Variance analysis for overhead is split between variances related to variable and fixed costs. This is the difference between the standard and actual cost per unit of the direct materials purchased, multiplied by the standard number of units expected to be used in the production process. This is the difference between the budgeted fixed overheads and the standard fixed overheads absorbed on actual production.
### What is the fixed overhead spending variance?
The fixed overhead spending variance is the difference between the actual amount of fixed overhead and the budgeted amount of fixed overhead. If the company spent more than it should have (according to the standard, which is set by management) on fixed overhead, then the fixed overhead spending variance is unfavorable.
## What Is A Spending Variance?
Ideally, your actual costs should match what you budgeted and your cost variance should be zero, but in practice this is fairly difficult to achieve. This is the difference between the actual and budgeted number of units sold, multiplied by the budgeted contribution margin. This measure is used to determine the impact on the overall sales margin of differences in the expected mix of units sold.
The budget is the primary tool financial analysts use to manage expenses and variances from the budget. By comparing the budget to actual numbers, analysts are able to identify any variances between budgeted and true costs. The higher the variance, the more help is needed in terms of management. The best way to manage variances is to have monthly reports and regular meetings to discuss these discrepancies with management and department heads.
Variable overhead efficiency variance is calculated by subtracting the standard budgeted hours from the actual hours incurred, and then multiplying the result with the standard variable overhead rate. A favorable variance results when the actual hours used are less than the budgeted while an adverse variance results from use of more hours than the budgeted. This is the difference between the budgeted fixed overhead expenditure and the actual fixed overhead incurred. It arises due to changes in the cost of fixed overhead during the period.
Effective variance analysis can help a company spot trends, issues, opportunities and threats to short-term or long-term success. Variance analysis helps management to understand the present costs and then to control future costs.
Unfavorable budget variances refer to the negative difference between actual revenues and what was budgeted. This usually happens when revenue is lower than expected or when expenses are higher than expected.
This also allows you to hold specific managers accountable for minimizing budget variance. Cost variance allows you to monitor the financial progression of whatever it is you are doing in your business. When cost variances are low, you know you have controlled your risks well. You also know you have retrieved and analyzed data related to operations sufficiently.
Variance calculation should always be calculated by taking the planned or budgeted amount and subtracting the actual/forecasted value. Thus a positive number is favorable and a negative number is unfavorable. | 1,606 | 8,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.919475 |
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1
Orders of variables in nested models (in R)
I am trying to use nested models to investigate the influence of 5 factors on my dependent variable. I am not interested in interactions, only the influence of each variable taken separately. My dependent variable is part4Auto, and my independent variables are:
• part1FlyingHours
• part1TypePilot
• part3SUS
• part5MWQ
So I wrote this nested model sequence which gave me the following output:
> model.baseline = lm(part4Auto ~ 1, data)
> model.1 = update(model.baseline, .~. + part1FlyingHours)
> model.2 = update(model.1, .~. + part1TypePilot)
> model.3 = update(model.2, .~. + part3SUS)
> model.4 = update(model.3, .~. + part5MWQ)
> anova(model.baseline, model.1, model.2, model.3, model.4)
Analysis of Variance Table
Model 1: part4Auto ~ 1
Model 2: part4Auto ~ part1FlyingHours
Model 3: part4Auto ~ part1FlyingHours + part1TypePilot
Model 4: part4Auto ~ part1FlyingHours + part1TypePilot + part3SUS
Model 5: part4Auto ~ part1FlyingHours + part1TypePilot + part3SUS + part5MWQ
Res.Df RSS Df Sum of Sq F Pr(>F)
1 41 22.562
2 40 21.578 1 0.9846 3.2352 0.080460 .
3 38 19.665 2 1.9125 3.1419 0.055249 .
4 37 13.418 1 6.2477 20.5279 6.241e-05 ***
5 36 10.957 1 2.4612 8.0866 0.007306 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The problem is that when I change the order of independent variables, I obtain different results which almost modify my conclusion (here I exchanged part1FlyingHours and part5MWQ):
> model.baseline = lm(part4Auto ~ 1, data)
> model.1 = update(model.baseline, .~. + part5MWQ)
> model.2 = update(model.1, .~. + part1TypePilot)
> model.3 = update(model.2, .~. + part3SUS)
> model.4 = update(model.3, .~. + part1FlyingHours)
> anova(model.baseline, model.1, model.2, model.3, model.4)
Analysis of Variance Table
Model 1: part4Auto ~ 1
Model 2: part4Auto ~ part5MWQ
Model 3: part4Auto ~ part5MWQ + part1TypePilot
Model 4: part4Auto ~ part5MWQ + part1TypePilot + part3SUS
Model 5: part4Auto ~ part5MWQ + part1TypePilot + part3SUS + part1FlyingHours
Res.Df RSS Df Sum of Sq F Pr(>F)
1 41 22.562
2 40 15.226 1 7.3367 24.1063 1.979e-05 ***
3 38 14.680 2 0.5462 0.8973 0.416588
4 37 10.978 1 3.7015 12.1619 0.001304 **
5 36 10.957 1 0.0215 0.0707 0.791882
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
On the contrary, the output given by summary() does not change.
So my question is: for nested models, does the order of introduction of variables change the results so much? Or do I have an important flaw (like unbalanced data)? And if it is the first solution, what can I do to ensure that my results are not biased or incomplete? | 1,023 | 2,779 | {"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} | 2.75 | 3 | CC-MAIN-2019-30 | latest | en | 0.560326 |
https://flashgene.com/archives/233762.html | 1,670,147,420,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710968.29/warc/CC-MAIN-20221204072040-20221204102040-00657.warc.gz | 292,588,140 | 14,276 | ## 情境引入
```N_CLIENTS = 3
num_cls, classes = 4, ['cat', 'dog', 'car', 'ship']
train_labels = [0, 3, 2, 0, 3, 2, 1, 0, 3, 3, 1, 0, 3, 2, 2] #数据集的标签列表
client_idcs = [slice(0, 4), slice(4, 11), slice(11, 15)]
# 数据集样本在client上的划分情况```
```import matplotlib.pyplot as plt
import numpy as np
plt.figure(figsize=(5,3))
plt.hist([train_labels[idc]for idc in client_idcs], stacked=False,
bins=num_cls,
label=["Client {}".format(i) for i in range(N_CLIENTS)])
plt.xticks(np.arange(num_cls), classes)
plt.legend()
plt.show()```
## bins 参数
#### bins: int or sequence or str, default: rcParams[“hist.bins”] (default: 10)
If bins is an integer, it defines the number of equal-width bins in the range.
If bins is a sequence, it defines the bin edges, including the left edge of the first bin and the right edge of the last bin; in this case, bins may be unequally spaced. All but the last (righthand-most) bin is half-open. In other words, if bins is:
`[1, 2, 3, 4]`
then the first bin is [1, 2) (including 1, but excluding 2) and the second [2, 3). The last bin, however, is [3, 4], which includes 4.
If bins is a string, it is one of the binning strategies supported by numpy.histogram_bin_edges: ‘auto’, ‘fd’, ‘doane’, ‘scott’, ‘stone’, ‘rice’, ‘sturges’, or ‘sqrt’.
`bins` 序列的刻度要参照 `hist` 函数中的x坐标刻度来设置，本任务中4个分类类别对应的x轴刻度分别为 `[0, 1, 2, 3]` 。如果我们将序列设置为 `[0, 1, 2, 3, 4]` 就表示第一个绘图区域对应的区间是 `[1, 2)` ，第2个绘图区域对应的位置是 `[1, 2)` ,第三个绘图区域对应的位置是 `[2, 3)` ，依次类推。
```plt.hist([train_labels[idc]for idc in client_idcs], stacked=False,
bins=np.arange(-0.5, 4, 1),
label=["Client {}".format(i) for i in range(N_CLIENTS)])```
## stacked参数
```plt.hist([train_labels[idc]for idc in client_idcs],stacked=True
bins=np.arange(-0.5, 4, 1),
label=["Client {}".format(i) for i in range(N_CLIENTS)])```
## rwidth 参数
#### rwidth float or None, default: None
The relative width of the bars as a fraction of the bin width. If None, automatically compute the width.
Ignored if histtype is ‘step’ or ‘stepfilled’.
```plt.hist([train_labels[idc]for idc in client_idcs],stacked=True,
bins=np.arange(-0.5, 4, 1), rwidth=0.5,
label=["Client {}".format(i) for i in range(N_CLIENTS)])``` | 788 | 2,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.476024 |
http://ccuart.org/circuit-diagram-common-emitter-amplifier | 1,537,402,486,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156314.26/warc/CC-MAIN-20180919235858-20180920015858-00504.warc.gz | 42,332,775 | 20,114 | # Circuit Diagram Common Emitter Amplifier
Circuit Diagram Common Emitter Amplifier. Thank You for visiting CCUART. Nowadays were excited to declare that we have discovered an incredibly interesting topic to be pointed out, namely Circuit Diagram Common Emitter Amplifier. Lots of people attempting to find info about Circuit Diagram Common Emitter Amplifier and certainly one of them is you, is not it?
## Explanation About Common Emitter Amplifier Working Principle And Its Applications
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## Common Emitter Amplifier Experiment
Common Emitter Amplifier Circuit Ac Equivalent Circuit For Ce Amplifier Figure 1 Npn Common Emitter Amplifier Common Emitter Amplifier Diagram Of A Practical Common Emitter Amplifier Circuit Av Vb Ve Vc Circuit Diagram Circuit Description This Is A Common Emitter Amplifier The Common Emitter Amplifier Circuit Common Emitter Configuration Img119f 3 Single Stage Common Emitter Amplifier Circuit Common Emitter Amplifier Schematic Im Expected To Design A Pnp Common Emitter Amplifier What Is The Difference Between The Common Emitter And Common Base Bias Configuration In Bjt Consider A Common Emitter Amplifier Circuit Shown In Fig 1 Schematic Diagram Common Emitter Amplifier Use A Bypassed Emitter Resistor And Set The Quiescent Current At 1ma Assume Vcc Is 15v L 1mh And Put A 62k Resistor Across Lc To Set Q 10 .
## Common Emitter Amplifier Youtube basic understanding
Regarding Image description: Image has been added by author. We thank you for your visit to CCUART. Make sure you get the information you are looking for common emitter amplifier youtube . Do not forget to share and love our reference to help further develop our website. | 463 | 2,267 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2018-39 | latest | en | 0.823192 |
https://face2ai.com/math-probability-3-3-cumulative-distribution-function/ | 1,568,673,659,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514572964.47/warc/CC-MAIN-20190916220318-20190917002318-00166.warc.gz | 481,246,802 | 27,455 | # 累积分布函数
## 累积分布函数的定义和基本性质 Definition and Basic Properties
Definition (Cumulative) Distribution Function :The distribution function or cumulative distribution function(abbreviated c.d.f) $F$ of a random variable $X$ is the function:
$$F(x)=Pr(X\leq x)\quad \text{for} \quad -\infty<x<+\infty$$
$$F(x)= \begin{cases} 0 &\text{for} &x<0\\ 1-p &\text{for}& 0\leq x<1\\ 1 &\text{for}&x\geq 1 \end{cases}$$
### 性质1：不减性 Nondecreasing
The function $F(x)$ is nondecreasing as $x$ increases;that is,if $x_1<x_2$ then $F(x_1)\leq F(x_2)$
$$\text{for }x_1\leq x_2\\ {x:x<x_1}\subset {x:x<x_2}$$
$$Pr({x:x<x_1})\leq Pr( {x:x<x_2})$$
Q.E.D.
### 性质2：有限 Limits
Limits at $\pm\infty$ $lim_{x\to -\infty}F(x)=0$ and $lim_{x\to \infty}F(x)=1$
$$\text{for }x_1\leq x_2\\ {x:x<x_1}\subset {x:x<x_2}$$
$Pr({x:x<x_1})=1 \quad \text{for } x_1=+\infty$
$Pr({x:x<x_2})=0 \quad \text{for } x_2=-\infty$
c.d.f不需要有连续性，如图：
$$F(x^{-})=lim_{y\to x,y<x}F(y)\\ F(x^{+})=lim_{y\to x,y>x}F(y)$$
### 性质3：向右连续 Continuity from the Right
Continuity from the Right.A c.d.f. is always continuous from the right: that is,$F(x)=F(x^{+})$ at every point $x$
## 从分布函数中确定概率 Determining Probabilities from the Distribution Function
### T1： $Pr(X>x) = 1 – F(x)$
Theorem: For every value $x$,
$$Pr(X>x) = 1 – F(x)$$
QED
### T2:$Pr(x_1<X\leq x_2)=F(x_2)-F(x_1)$
Theorem :For all value $x_1$ and $x_2$ such that $x_1<x_2$ :
$$Pr(x_1<X\leq x_2)=F(x_2)-F(x_1)$$
$$B=A^{c}\cap C$$
$$Pr(B)=Pr(C)-Pr(A\cap C)\\ \text{for: }A=A\cap C\\ Pr(B)=Pr(C)-Pr(A)\\ Pr(x_1<X\leq x_2)=F(x_2)-F(x_1)$$
QED
### T3:$Pr(X<x)=F(x^{-})$
Theorem For each value $x$
$$Pr(X<x)=F(x^{-})$$
### T4:$Pr(X=x)=F(x)-F(x^{-})$
Theorem For every value $x$
$$Pr(X=x)=F(x)-F(x^{-})$$
$$Pr(X=x)=Pr(X\leq x)-Pr(X<x)$$
$$F(x)=Pr(X\leq x)$$
## 离散分布的累积函数 The c.d.f. of a Discrete Distribution
1. $F(x)$ will have a jump of magnitude $f(x_i)$ at each possible valuee $x_i$ of $X$
2. $F(x)$ will be constant between every pair of successive jump
3. The distribution of a discrete random variable $X$ can be represented equally well by either the p.f. or the c.d.f. of $X$
## 连续分布的累积函数 The c.d.f. of a Continuous Distribution
Theorem Let X have a continuous distribution, and let f(x) and F(x) denote its p.d.f and the c.d.f. respectively.Then $F$ is continuous at every $x$:
$$F(x)=\int^{x}_{-\infty}f(t)dt$$
and
$$\frac{dF(x)}{dx}=f(x)$$
at all x such that f is continuous.
## 分位数函数 The Quantile Function
Definition Quantiles/Percentiles: Let $X$ be a random variable with c.d.f. $F$ .For each $p$ strictly between $0$ and $1$ ,define $F^{-1}(p)$ to be the smallest value $x$ such that $F(x)\geq p$ .Then $F^{-1}(p)$ is called the $p$ quantile of $X$ or the $100p$ percentile of $X$ .The function $F^{-1}$ defined here on the open interval $(0,1)$ is called the quantile function of $X$
### 中位数 Median quartiles
Definition Median/Quantiles.The $\frac{1}{2}$ quantile or the 50th percentile of a distribution is called its median.The $\frac{1}{4}$ quantile or 25th percentile is the lower quartile.The $\frac{3}{4}$ quantile or 75th percentile is called upper quartile.
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Subscribe | 1,237 | 3,129 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-39 | longest | en | 0.367206 |
https://volumeconverter.intemodino.com/conversion/gi-to-fl-oz-and-fl-oz-to-gi.html | 1,695,492,524,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506528.19/warc/CC-MAIN-20230923162848-20230923192848-00607.warc.gz | 679,672,846 | 5,784 | Convert Gills to Fluid Ounces and Fluid Ounces to Gills Online
Length Conversion Settings
Number of decimal places
Round fraction to the nearest
Gi to fl oz conversion factor and conversion formula
The gi to fl oz conversion factor is 4.8037997
1 gi = 4.8037997 fl oz
To convert X gills to fluid ounces you should use the following conversion formula:
X gills × 4.8037997 = Result in fluid ounces
Gills to Fluid Ounces Conversion Table
1 gi = 4.8037997 fl oz
2 gi = 9.6075994 fl oz
3 gi = 14.41139911 fl oz
4 gi = 19.21519881 fl oz
5 gi = 24.01899851 fl oz
6 gi = 28.82279821 fl oz
7 gi = 33.62659791 fl oz
8 gi = 38.43039762 fl oz
9 gi = 43.23419732 fl oz
10 gi = 48.03799702 fl oz
11 gi = 52.84179672 fl oz
12 gi = 57.64559642 fl oz
13 gi = 62.44939613 fl oz
14 gi = 67.25319583 fl oz
15 gi = 72.05699553 fl oz
16 gi = 76.86079523 fl oz
17 gi = 81.66459493 fl oz
18 gi = 86.46839464 fl oz
19 gi = 91.27219434 fl oz
20 gi = 96.07599404 fl oz
21 gi = 100.87979374 fl oz
22 gi = 105.68359344 fl oz
23 gi = 110.48739315 fl oz
24 gi = 115.29119285 fl oz
25 gi = 120.09499255 fl oz
26 gi = 124.89879225 fl oz
27 gi = 129.70259195 fl oz
28 gi = 134.50639166 fl oz
29 gi = 139.31019136 fl oz
30 gi = 144.11399106 fl oz
31 gi = 148.91779076 fl oz
32 gi = 153.72159046 fl oz
33 gi = 158.52539017 fl oz
34 gi = 163.32918987 fl oz
35 gi = 168.13298957 fl oz
36 gi = 172.93678927 fl oz
37 gi = 177.74058897 fl oz
38 gi = 182.54438868 fl oz
39 gi = 187.34818838 fl oz
40 gi = 192.15198808 fl oz
41 gi = 196.95578778 fl oz
42 gi = 201.75958748 fl oz
43 gi = 206.56338719 fl oz
44 gi = 211.36718689 fl oz
45 gi = 216.17098659 fl oz
46 gi = 220.97478629 fl oz
47 gi = 225.77858599 fl oz
48 gi = 230.5823857 fl oz
49 gi = 235.3861854 fl oz
50 gi = 240.1899851 fl oz
51 gi = 244.9937848 fl oz
52 gi = 249.79758451 fl oz
53 gi = 254.60138421 fl oz
54 gi = 259.40518391 fl oz
55 gi = 264.20898361 fl oz
56 gi = 269.01278331 fl oz
57 gi = 273.81658302 fl oz
58 gi = 278.62038272 fl oz
59 gi = 283.42418242 fl oz
60 gi = 288.22798212 fl oz
61 gi = 293.03178182 fl oz
62 gi = 297.83558153 fl oz
63 gi = 302.63938123 fl oz
64 gi = 307.44318093 fl oz
65 gi = 312.24698063 fl oz
66 gi = 317.05078033 fl oz
67 gi = 321.85458004 fl oz
68 gi = 326.65837974 fl oz
69 gi = 331.46217944 fl oz
70 gi = 336.26597914 fl oz
71 gi = 341.06977884 fl oz
72 gi = 345.87357855 fl oz
73 gi = 350.67737825 fl oz
74 gi = 355.48117795 fl oz
75 gi = 360.28497765 fl oz
76 gi = 365.08877735 fl oz
77 gi = 369.89257706 fl oz
78 gi = 374.69637676 fl oz
79 gi = 379.50017646 fl oz
80 gi = 384.30397616 fl oz
81 gi = 389.10777586 fl oz
82 gi = 393.91157557 fl oz
83 gi = 398.71537527 fl oz
84 gi = 403.51917497 fl oz
85 gi = 408.32297467 fl oz
86 gi = 413.12677437 fl oz
87 gi = 417.93057408 fl oz
88 gi = 422.73437378 fl oz
89 gi = 427.53817348 fl oz
90 gi = 432.34197318 fl oz
91 gi = 437.14577288 fl oz
92 gi = 441.94957259 fl oz
93 gi = 446.75337229 fl oz
94 gi = 451.55717199 fl oz
95 gi = 456.36097169 fl oz
96 gi = 461.16477139 fl oz
97 gi = 465.9685711 fl oz
98 gi = 470.7723708 fl oz
99 gi = 475.5761705 fl oz
100 gi = 480.3799702 fl oz
Fl oz to gi conversion formula and conversion factor
The fl oz to gi conversion factor is 0.20816855
1 fl oz = 0.20816855 gi
To convert X fluid ounces to gills you should use the following conversion formula:
X fluid ounces × 0.20816855 = Result in gills
Fluid Ounces to Gills Conversion Chart
1 fl oz = 0.20816855 gi
2 fl oz = 0.41633709 gi
3 fl oz = 0.62450564 gi
4 fl oz = 0.83267418 gi
5 fl oz = 1.04084273 gi
6 fl oz = 1.24901128 gi
7 fl oz = 1.45717982 gi
8 fl oz = 1.66534837 gi
9 fl oz = 1.87351692 gi
10 fl oz = 2.08168546 gi
11 fl oz = 2.28985401 gi
12 fl oz = 2.49802255 gi
13 fl oz = 2.7061911 gi
14 fl oz = 2.91435965 gi
15 fl oz = 3.12252819 gi
16 fl oz = 3.33069674 gi
17 fl oz = 3.53886528 gi
18 fl oz = 3.74703383 gi
19 fl oz = 3.95520238 gi
20 fl oz = 4.16337092 gi
21 fl oz = 4.37153947 gi
22 fl oz = 4.57970802 gi
23 fl oz = 4.78787656 gi
24 fl oz = 4.99604511 gi
25 fl oz = 5.20421365 gi
26 fl oz = 5.4123822 gi
27 fl oz = 5.62055075 gi
28 fl oz = 5.82871929 gi
29 fl oz = 6.03688784 gi
30 fl oz = 6.24505638 gi
31 fl oz = 6.45322493 gi
32 fl oz = 6.66139348 gi
33 fl oz = 6.86956202 gi
34 fl oz = 7.07773057 gi
35 fl oz = 7.28589912 gi
36 fl oz = 7.49406766 gi
37 fl oz = 7.70223621 gi
38 fl oz = 7.91040475 gi
39 fl oz = 8.1185733 gi
40 fl oz = 8.32674185 gi
41 fl oz = 8.53491039 gi
42 fl oz = 8.74307894 gi
43 fl oz = 8.95124748 gi
44 fl oz = 9.15941603 gi
45 fl oz = 9.36758458 gi
46 fl oz = 9.57575312 gi
47 fl oz = 9.78392167 gi
48 fl oz = 9.99209022 gi
49 fl oz = 10.20025876 gi
50 fl oz = 10.40842731 gi
51 fl oz = 10.61659585 gi
52 fl oz = 10.8247644 gi
53 fl oz = 11.03293295 gi
54 fl oz = 11.24110149 gi
55 fl oz = 11.44927004 gi
56 fl oz = 11.65743858 gi
57 fl oz = 11.86560713 gi
58 fl oz = 12.07377568 gi
59 fl oz = 12.28194422 gi
60 fl oz = 12.49011277 gi
61 fl oz = 12.69828132 gi
62 fl oz = 12.90644986 gi
63 fl oz = 13.11461841 gi
64 fl oz = 13.32278695 gi
65 fl oz = 13.5309555 gi
66 fl oz = 13.73912405 gi
67 fl oz = 13.94729259 gi
68 fl oz = 14.15546114 gi
69 fl oz = 14.36362968 gi
70 fl oz = 14.57179823 gi
71 fl oz = 14.77996678 gi
72 fl oz = 14.98813532 gi
73 fl oz = 15.19630387 gi
74 fl oz = 15.40447242 gi
75 fl oz = 15.61264096 gi
76 fl oz = 15.82080951 gi
77 fl oz = 16.02897805 gi
78 fl oz = 16.2371466 gi
79 fl oz = 16.44531515 gi
80 fl oz = 16.65348369 gi
81 fl oz = 16.86165224 gi
82 fl oz = 17.06982078 gi
83 fl oz = 17.27798933 gi
84 fl oz = 17.48615788 gi
85 fl oz = 17.69432642 gi
86 fl oz = 17.90249497 gi
87 fl oz = 18.11066352 gi
88 fl oz = 18.31883206 gi
89 fl oz = 18.52700061 gi
90 fl oz = 18.73516915 gi
91 fl oz = 18.9433377 gi
92 fl oz = 19.15150625 gi
93 fl oz = 19.35967479 gi
94 fl oz = 19.56784334 gi
95 fl oz = 19.77601188 gi
96 fl oz = 19.98418043 gi
97 fl oz = 20.19234898 gi
98 fl oz = 20.40051752 gi
99 fl oz = 20.60868607 gi
100 fl oz = 20.81685462 gi
Wednesday, December 16, 2020 | 2,601 | 5,967 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2023-40 | latest | en | 0.229294 |
https://datascience.stackexchange.com/questions/100334/multi-target-regression-tree-with-additional-constraint/100358#100358 | 1,638,495,243,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00454.warc.gz | 275,239,039 | 35,766 | # Multi-target regression tree with additional constraint
I have a regression problem where I need to predict three dependent variables ($$y$$) based on a set of independent variables ($$x$$): $$(y_1,y_2,y_3) = \beta_0 + \beta_1 x_1 + \beta_2 x_2 + \dots + \beta_n x_n +u.$$
To solve this problem, I would prefer to use tree-based models (i.e. gradient boosting or random forest), since the independent variables ($$x$$) are correlated and the problem is non-linear with ex-ante unknown parameterization.
I know that I could use sklearn's MultiOutputRegressor() or RegressorChain() as a meta-estimator.
However, there is an additional twist to my problem, namely that I do know that $$y_1 + y_2 - y_3 = x_1$$.
In other words, there is a fixed relation between the three $$y$$ and one of the independent variables. So essentially, the value of $$x_1$$ needs to be "distributed" in a first best manner to the (unknow) targets $$(y_1,y_2,y_3)$$ for each observation, contingent on the remaining independent variables $$x_2,\dots,x_n$$.
Of course a naive approach would be, to squeze the predicted values together somehow, so to satisfy $$\hat{y_1} + \hat{y_2} - \hat{y_3} = x_1$$. However, I wonder if there are any other options to introduce a "hard constraint" such as $$\hat{y_1} + \hat{y_2} - \hat{y_3} = x_1$$ to some (tree-based) estimator.
I noticed this post. However, I would prefer a tree-based method for reasons mentionned above.
• Is the relationship exact in the training data, or noisy? Aug 24 at 13:32
• The relation $y_1+y_2-y_3=x_1$ is almost exact (few minor residual values), while the effect of the remaining $x$ on $y$ in the sense of $y_1,y_2,y_3(x_2,...,x_n)$ is rather noisy. Aug 24 at 14:33
• I'd need to think through it some more before upgrading this to an answer, but some things to think about: (1) model just $y_1, y_2$ and then predict $\hat{y}_3 = \hat{y}_1 + \hat{y}_2 - x_1$. (2) RegressorChain, with the $y$s in order, will do essentially that but with some flexibility to change the $\hat{y}_3$. (3) Trees already do multi-output regression in a single tree, so MultiOutputRegressor shouldn't be needed. (4) If the relationship were just in the $y$s, that would be captured automatically by trees, since the leaf values are averages and the relationship is linear. Aug 25 at 14:40
• Thanks for your comment: I'm currently using RegressorChain() and I wonder if I would benefit from using option (1) in a stacking process where each of the $y_i$ is determined as "residual" for parts of the data. So first stage: Estimate two of the $\hat{y}_i$ in a chain, determine the last $\hat{\hat{y}}_i$ as residual to $x_1$. Second stage: Use the $\hat{\hat{y}}_i$ in a further modeling step to see if this information helps to reduce MSE, MAE and ensure that $\hat{y}_1+\hat{y}_2-\hat{y}_3=x_1$ is met. Do you think something like this could work? Aug 26 at 9:52 | 836 | 2,896 | {"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": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2021-49 | latest | en | 0.904502 |
https://pedagogue.app/activities-to-teach-students-to-multiply-fractions-by-whole-numbers-using-models-complete-the-equation/ | 1,719,188,545,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198864968.52/warc/CC-MAIN-20240623225845-20240624015845-00601.warc.gz | 399,332,026 | 26,284 | Activities to Teach Students to Multiply Fractions by Whole Numbers Using Models: Complete the Equation
Multiplying fractions by whole numbers using models is an important concept in mathematics. As we know, fractions are parts of a whole, and multiplying fractions with whole numbers means finding the product of a whole number and a fraction. Using models can be a useful tool to help students understand the concept of multiplying fractions by whole numbers. Here, we will discuss some activities that teachers can use to teach students to multiply fractions by whole numbers using models.
1. Using Rectangles and Grids:
One of the most popular methods for teaching multiplication of fractions by whole numbers using models is the rectangle or grid method. In this method, the rectangle or grid is divided into equal parts. For example, a rectangle or grid can be divided into four or six equal parts, and each part can be represented as a fraction. The grid or rectangle can then be shaded to represent the fraction, such as shading two of the six equal parts to represent 2/6 or 1/3.
To teach students this method, the teacher can provide the students with grids or rectangles, and ask them to shade the appropriate fraction to represent the product. For example, if the question is to find 3 x 1/2, the student would shade half of the rectangle three times. This model helps the students to understand that multiplying 3 by 1/2 is the same as adding 1/2 three times.
2. Using Fraction Circles:
Fraction circles can also be used to teach students about multiplying fractions by whole numbers. Fraction circles are circular discs that are divided into equal parts. For example, a fraction circle can be divided into four, six, or eight equal parts, and each part can be represented as a fraction. These fraction circles can be used to represent fractions as well as whole numbers.
To teach students this method, the teacher can provide the students with fraction circles and ask them to represent the fraction as a whole number, such as four circles representing the whole number four. Then, the students can use another fraction circle to represent a fraction, such as 1/3. The student would then need to find the product of 4 x 1/3, which would be represented by four fraction circles, with one-third of each circle shaded.
3. Using Word Problems:
Word problems are an excellent way to help students connect the real world to mathematical concepts. Teachers can use word problems to teach students how to multiply fractions by whole numbers using models. For example, a teacher could give students a word problem that involves baking cookies. The problem could ask how many cups of sugar are needed to make four batches of cookies, where each batch requires 2/3 cups of sugar.
To solve the problem, the students can use a model, such as a rectangle grid or fraction circles. The student would first represent the whole number four, and then multiply it by the fraction 2/3. In this way, they can visualize the problem and come up with the correct answer.
In conclusion, teaching students to multiply fractions by whole numbers is an essential concept in mathematics. By using models such as grids and fraction circles, students can visualize the concept and understand it better. Teachers can also use word problems to help students connect the mathematical concept to the real world. By incorporating these activities into their lesson plans, teachers can make learning fun and engaging for their students. | 694 | 3,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-26 | latest | en | 0.939061 |
https://www.physicsforums.com/threads/gradient-divegrance-and-curl-del-operator.118848/ | 1,511,009,869,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804881.4/warc/CC-MAIN-20171118113721-20171118133721-00252.warc.gz | 848,191,017 | 16,213 | # Gradient, divegrance and curl? del operator
1. Apr 26, 2006
### Lonley_Shepherd
Gradient, divegrance and curl??? del operator!!!
in static magnetic and electric fields, the del operator was introduced and then used to describe three different quantities.. i still cant quite figure out the physical meaning and difference between the curl,divergance and the gradient in terms of vector fields.. also the dot and cross products are still physically undefined clearly to me :S .. thnx!!
2. Apr 27, 2006
### Staff: Mentor
3. Apr 28, 2006
### Lonley_Shepherd
ohh plz do finish wut u promised!! and well the links to that other site are just stating what div and curl are... i do get wut they mean mathematically... del dot and del cross... but wen related to electric and magnetic fields.. it just isnt clear anymore.. how can a rate of change with respect to space be a scalar quantity.. and by the cross product of del to a vector the rate of change with respect to space has a direction!! just like the gradient then..
i read some texts abt the curl that refer to it as the net rotation of a field... how kan a uniform magnetic field rotate.. and when i say rotate i surely must specify an axis of rotation.,. aaaaaaaaaaaa ,.. maybe am smart enough to breath only!!!!
PS: i dunt think it was right moving my discussion here.. its physics wuts bothering me.. but thats ur call :D
Last edited: Apr 28, 2006
4. Apr 28, 2006
### nrqed
Very roughly...(I will assume you know what a scalar field and a vector field are.. A scalar field assigns a number (i.e. a scalar) to all points in space and a vector field assigns a vector to all points in space)
The gradient of a scalar roughly tells at each point the direction in which the scalar field increases the most and how large the change is at that point is (think of a scalar field as a surface, for example the surface of mountains and valleys. The gradient at a given point shows in what direction the surface goes up the steepest and how steep the surface is at that point. If part of the surface is flat, the gradient there is zero).
For the divergence of a vector field, consider a tiny (infinitesimal) volume (a cube, say). The divergence of the field at the center of the cube times the volume of the cube is the net flow of the vector field through the sides of the cube. If the divergence is positive for example, there is a net flow of the vector field out of the cube.
For the curl, consider a tiny loop (in a circle let's say). The curl of a vector field at the center of the loop is equal to the "circulation" of the vector field along the loop. In the image of the water velocity field, a nonzero curl at a point indicates that there is a net rotation of water around that point (like a tiny vortex).
This is all very crude but I just meant to give a very brief physical interpretation.
Hope this helps
Patrick
Last edited: Apr 28, 2006 | 679 | 2,917 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2017-47 | longest | en | 0.923176 |
https://web.stanford.edu/class/cs106a/handouts/lecture-1-code.html | 1,571,279,242,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986672548.33/warc/CC-MAIN-20191017022259-20191017045759-00336.warc.gz | 777,978,763 | 4,258 | CCS106A: Computer - Code - You
• Computer is powerful
• But dumb! Just runs code
• You have insight
-Think of an algorithm
-Express ideas in code
-Dumb it down so even a computer can do it
• A great marriage..
• Your insight + the computer
Story #1 - Bit's World - Simple
We have this "bit" robot, lives in a rectangular world. Use this just at the start of CS106A to get things going.
Bit can 4 things, move, left, right, paint .. show b
Bit Reference - more detailed list of Bit features
Suppose Bit starts at the upper left corner facing to the right, and then the following code runs:
``` bit.move()
bit.paint('blue')
bit.right()
bit.move()
bit.paint('green')
```
Before:
After:
The syntax `bit.move()` is a Python function call. A function hold some code. To "call" a function runs its code. This is the noun.verb function call form.
Story #2 - Running Python Code
Now look at a def that has those same steps as formal Python code:
Bit blue_green() def
Here is complete Python function that does those same steps:
```def blue_green(filename):
bit = Bit(filename)
bit.move()
bit.paint('blue')
bit.right()
bit.move()
bit.paint('green')
```
Before Running
• Run a function - runs its lines top to bottom (aka "call" that function)
• Each line does something
• In this case, for each line see bit take an action
Here is a "live" on the web version of the above def. Click "Run" on the page and watch what it does. Use the steps slider.
Story #3 - 6 Things About def Syntax
So having seen a function run, now look at the details of the syntax. Sorry, it's a bunch of small syntactic points.
Looks weird, but is basically superficial and you will get used to it quickly.
```def blue_green(filename):
bit = Bit(filename)
bit.move()
bit.paint('blue')
bit.right()
bit.move()
bit.paint('green')
```
The first code line sets up the bit-arrow as pictured above, explained in a later lecture
`bit = Bit(filename)`
• 1. A function starts with the word "def"
• 2. Followed by a name for the function, e.g. "blue_green". We'll have deeper thoughts about function naming later.
• 3. Followed by a parenthesis/colon thing, explained later
• 4. Followed by the indented lines of code of the function
• 5. line: `bit.move()` calls the move() function
When `bit.move()` is called, we see the action to the right
This is the noun.verb style of calling a function
• 6. line: `bit.paint('blue')` calls the paint() function
-passes in the parameter 'blue' as extra info the call
• The run of the function ends on its last line
A Little Talk About Syntax Errors
• Syntax - code is structured for the computer
• Very common error - type in code, with slight syntax problem
• e.g. syntax error: `bit.move[)`
• Professional programmers make that sort of "error" all the time
• Just type it in, run it, fix the errors
• Not a reflection of some flaw in the programmer
• Do not give in to the following thoughts (imposter syndrome):
-"Another syntax error .. maybe I'm not cut out for this!"
-"I bet nobody else is getting these"
• Just the nature of typing ideas into the mechanical computer language
• Exercises to desensitize: a bunch of typical syntax errors + fixing them
• Fixing these little errors is a small, normal step
• Computers are powerful for you, but you need to work their way on syntax
Code Demos - Get Errors - Fix!
• These all have errors, trying run and fix (lecture demo, or on your own)
• These are on the parlante.org experimental server, uses your Stanford login
• bit1a
• bit1b
• bit1c
• bit1d
• Common syntax errors:
-messed up "()" on function call
-ragged indentation
-missing ":"
-typo the name of a function, like "bit.mve()"
You Try One: green2 (if we have time)
Bit begins at the upper left corner facing right. Paint the square below the start square green, and the square below that green as well, ending like this:
The word pass is a placeholder for your code, remove it. Feel free to make some syntax errors getting this to work!)
• Now you write code to solve one.
• Use the Run button to try your code as you go
• Type Ctrl-Return - Like clicking the Run button
• On this problem the system is checks if the output is correct
(To be clear this is not part of your grade, just checking the output for the exercise.)
• "diff" Feature - red marks on incorrect squares
-The system knows what the world is supposed to look like, so marks differing sq | 1,087 | 4,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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 | 0.85049 |
https://pushkarrajthakur.com/how-to-double-your-money-by-investing-in-share-market-best-stocks-portfolio-pranjal-kamra/ | 1,685,932,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00357.warc.gz | 522,630,227 | 26,257 | Millions of people don’t know how to earn this much So I don’t know how interested you all must be But I am going to talk about doubling your money every 4 years don’t know how to double my money every year .
So I will talk about what I know One thing I would like to say is that why is it necessary to learn to double in 4 years And then decide whether you want to learn it or notify I play a game with you all Out of 100 people 99 will die And 1 will become a millionaire Who will play that game? You won’t play right? So when the data of the whole world says that 95-99% traders are in loss So everyone should think that I will be in that 1%
Whereas I am young, I am inexperienced, I am just starting tolerant am not going to lose my money am going to earn it Whereas I am young, I am inexperienced, I am just starting to learn Is this logical? We will not take this risk in life in COVID-19, 2% out of 100 people died Death rate was 2% mortality We were still saving ourselves from COVID-19We were saving ourselves right? So why are you behind such a COVID-19In which 99 out of 100 people die Now we know that money doubles in 4 years Which can be done in reality
How to do that? Look, there are 1% people who definitely earn a lot of money from trading But there are 1% people who win lotteries Who become millionaires from casinos We don’t do all that We do things that an average person can do Not an extremely intelligent person So first of all we have to believe that we are an average person And we need a method that an average person can earn money from Not a genius person.
Now, if most people can earn money from trading So in such a situation, machines You all want to see patterns right? So patterns identify Whatever patterns are told to you Patterns identify that a machine can do better than a human
Any pattern, I don’t know what kind of patterns are there But if money is made from patterns Then you can’t make money If money is not made from patterns, then you can make money But if money is made from patterns Then you can’t make money, then a machine will make money Which has the most expensive supercomputer Which will identify the pattern the fastest And will use that pattern to make money the fastest
If money is made from patterns, then you can’t make money Then what do you think? The world’s hedge funds, quant funds, investment banks Will let that pattern remain for you? For all of you? Why will they let it remain? They will hire the world’s best technical analysts In a package of crores Will buy the world’s best software
And will take advantage of all the patterns They won’t leave it for you So understanding these truths is very important You are young When we are young Most of us are middle classify we see a way to quickly solve the problems of our family We become over enthusiastic We become excited But that leads to a mistake, trust me make a promise here today
Do whatever you want to do And whatever I am going to tell you in the next 5 minutes You either do that If anyone wants to record this Record it For the next 5 years Same date can almost guarantee you That the way I am telling you after 5 years Its return will be equal Or it will be close Or it will be more It won’t be very less In any other way that you have learned can guarantee you one more thing
That what I am telling you In that, the effort will be 10 times less And the returns will be almost equal2%We define almost less2% up and down Either it will be equal Or 2% more or 2% less If it is more than this Trust me, I will shut down Phenology will shut down my YouTube channel Because I know Nothing else is going to work Okay
Now in the next 5 minutes We will understand what we have to dowel have to do such a simple thing And I will also tell you why these people don’t say Because I will tell you such a simple thing See, go to the doctor And you are feeling very sick from inside Very sick And you go to the doctor And the doctor tells you Kadar Veda, drink soup, go home Take Paracetamol pill What do we people say? What kind of doctor is he? He didn’t take us seriously And that same doctor Writes 4 antibiotics Spits 2 injections
So you think OH man, he is a serious doctor He listened to me properly So most experts Give you complicated advice To sound like experts If they say simple things You people don’t value them OH man, he is saying simple thing she doesn’t know anything So that’s why As an expert
If I want a lot of popularity Or as an expert If I want to sound very smart have to say complicated things Otherwise I won’t be respected So it becomes the experts’ helplessness To say complicated things Simple things don’t have respect Now what I am telling you Is very simple It is going to be the simplest thing for the day And trust me, returns Either 2% more or 2% lessor equal But the effort Is going to be 10 times less
Right now In the next 5 minutes will make your stock market portfolio Which is most probably Not guaranteed in the stock market But equal to the market Will perform better than the market or around the market Right now we will make that portfolio Let’s talk about it Throughout the day From the beginning What toothpaste do you use after waking up? How many people are from Colgate? How many people are from Peps dent? Oh man, that’s not listed, sorry Those from Colgate Buy from Colgate Those from PepsodentBuy from Holist one doneSoapWhich soap? Dettol Dettol is not listed and?
Lifebuoy Lux So buy from Lux and Lifebuoy from HUL Next Buy from Pears and Heater that Which shampoo? Clinic Plumy hair is falling off from Clinic Plus But okay, Clinic Plus So till now We have put 2 companies in our portfolio And you don’t have to buy this portfolio know, you won’t believe in easy things You can make a dummy portfolio of this The shares in Money Control Make a dummy portfolio of that Apart from that
Do whatever you want After 5 years The trading you did And the portfolio I made now We will compare the results of the blogs After 5 years, if there is a difference of more than 2%I will delete everything will go away from this space If I am teaching the wrong thing You should stop its I will stop it If it doesn’t work2 shares done
Third At home People use detergent or soap to wash utensils Vim Again HULHUL is coming again Okay Where do you buy clothes from? From which store? Yes, From the cart Pan Hussain Aditya Birla Fashion Okay don’t know this Is it listed? Jokey and Zodio, no one goes there? Westside, Zooid ZodioIt is from Westside
Trent Okay Trent, yes So the next third share Aditya Birla or Trent Fourth After thisBankWhere is the bank account? HDFCI heard it from ICICI Okay HDFC Whoever has a bank account Leave the government bank Private bank4 shares done Next T
After this Shoes Shoes and slippers Which brand? Bata Relax Relax and Bata2 more6Am I deaf? Caustics problem What are you saying? Okay. So whoever is saying whatever, that’s Shera. After that…Yes, good question. Home paint. Which paint is homemade? Asian paints? Did you paint it with Favicon? Yes. Yes. Piddleite Asian paint. Okay, stocks. Kitchen.
Cooker, etc. Which one? Kitchen. Yes. What else? What else? I can’t hear you. What else? What else? What else? What else? What else? What else? Which brand is there without a cooker? You must have never seen it. Prestige. Hawkins, Prestige.Okay? The kitchen has two stocks. Ten. Yes, Favicon should be there. Favicon is good, should be there.
Tires. Bike. Which bike do you have? Or which one do you want to buy? Pulsar. Apache. Bajaj. There are four. There are four. There are four. There are 4. And? Seven.Seven. Brownie. Brownie. Okay. There are four. There are five. There are three. Tata and others have reached 12 stocks. We have 2 minutes left. How many stocks do you need for the portfolio?15 to 20 stocks? You have reached 12. It’s not 10, it’s too concentrated. You will take loans too.
Correct? Correct? Yes.Insurance? You must be taking insurance too. Life insurance? Which brand of health insurance? Policy Bazaar? Smart? And where do you order food from? Do you pay UPI too? So, man, give me an answer. I really have a lot of problems with you all. You do all the things in the world.
Why don’t you buy the shares of the product you are giving yourself? Get the whole world to do the work from you. And I really have a lot of problems. I have been on YouTube for 5 years. I talk about every day, but you don’t understand. It’s very simple, brother. The product you are giving money to, buy the share of that product and that money will come in yourpocket.
So, the portfolio of 15 shares is made. Now we will meet after 5 years. Sure. Whatever you have learned, it will be more profitable than that. Was it simple? It’s so simple that you won’t do it, right? This is very simple. | 2,066 | 8,818 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-23 | longest | en | 0.964389 |
http://www.coursehero.com/file/3524/Solutions-to-Exam-I-Ma120White-Version-Su05/ | 1,369,555,635,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706637439/warc/CC-MAIN-20130516121717-00062-ip-10-60-113-184.ec2.internal.warc.gz | 388,170,502 | 9,304 | # Register now to access 7 million high quality study materials (What's Course Hero?) Course Hero is the premier provider of high quality online educational resources. With millions of study documents, online tutors, digital flashcards and free courseware, Course Hero is helping students learn more efficiently and effectively. Whether you're interested in exploring new subjects or mastering key topics for your next exam, Course Hero has the tools you need to achieve your goals.
7 Pages
### Solutions to Exam I Ma120(White Version), Su05
Course: MATH 120, Summer 2005
School: SDSMT
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Math 441/442: Engineering Statistics I & II, 2 Credits Each, SDSMT Spring 2004Finding Your InstructorRoger Johnson ("Roger" or "Mr. Johnson" is fine), McLaury 314B email: Roger.Johnson@sdsmt.edu, web: http:/www.mcs.sdsmt.edu/rwjohnso/ (this page av
SDSMT - MATH - 442
Math 441/442: Engineering Statistics I & II, Spring 2005, 2 Credits Each, SDSMTFinding Your InstructorRoger Johnson ("Roger" or "Mr. Johnson" is fine), McLaury 308 (inner office) email: Roger.Johnson@sdsmt.edu, web: http:/www.mcs.sdsmt.edu/rwjohnso
SDSMT - MATH - 353
Math 353: Linear Optimization Spring 2003Finding Your InstructorRoger Johnson ("Roger" or "Mr. Johnson" is fine), McLaury 314B email: Roger.Johnson@sdsmt.edu, web: http:/www.mcs.sdsmt.edu/rwjohnso/ (this page available via the link "Course Syllabi"
SDSMT - MATH - 353
SDSMT - MATH - 687
Math 687: Statistical Design & Analysis of Experiments, 3 Credits, SDSMT Spring 2004To consult the statistician after an experiment is finished is often merely to ask him [her] to conduct a post mortem examination. [S]He can perhaps say what the exp | 1,460 | 5,259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.843798 |
http://www.ck12.org/algebra/Distance-Formula/lesson/user:bHlhLnNuZWxsQGhlbnJ5LmsxMi5nYS51cw../Connecting-Algebra-and-Geometry-Through-Coordinates-MCC9-12G.GPE.7/ | 1,438,202,406,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042986625.58/warc/CC-MAIN-20150728002306-00022-ip-10-236-191-2.ec2.internal.warc.gz | 369,233,987 | 56,196 | <meta http-equiv="refresh" content="1; url=/nojavascript/">
# Distance Formula
## Using the Pythagorean Theorem to determine distances
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Practice Distance Formula
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Connecting Algebra and Geometry Through Coordinates MCC9-12G.GPE.7
### Standards:
MCC9-12.G.GPE.7 Use coordinates to compute perimeters of polygons and areas of triangles and rectangles, e.g., using the distance formula.
### DISTANCE FORUMLA
What if you were given the coordinates of two points like (6, 2) and (-3, 0). How could you determine how far apart these two points are? After completing this Concept, you'll be able to find the distance between any two points in the coordinate plane using the Distance Formula.
### Guidance
In the last section, we saw how to use the Pythagorean Theorem to find lengths. In this section, you’ll learn how to use the Pythagorean Theorem to find the distance between two coordinate points.
#### Example A
Find the distance between points A=(1,4)\begin{align*}A = (1, 4)\end{align*} and B=(5,2)\begin{align*}B = (5, 2)\end{align*} .
Solution
Plot the two points on the coordinate plane.
In order to get from point A=(1,4)\begin{align*}A = (1, 4)\end{align*} to point B=(5,2)\begin{align*}B = (5, 2)\end{align*} , we need to move 4 units to the right and 2 units down. These lines make the legs of a right triangle.
To find the distance between A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} we find the value of the hypotenuse, d\begin{align*}d\end{align*} , using the Pythagorean Theorem.
d2d=22+42=20=20=25=4.47
#### Example B
Find the distance between points C=(2,1)\begin{align*}C = (2, -1)\end{align*} and D=(3,4)\begin{align*}D = (-3, -4)\end{align*} .
Solution
We plot the two points on the graph above.
In order to get from point C\begin{align*}C\end{align*} to point D\begin{align*}D\end{align*} , we need to move 3 units down and 5 units to the left.
We find the distance from C\begin{align*}C\end{align*} to D\begin{align*}D\end{align*} by finding the length of d\begin{align*}d\end{align*} with the Pythagorean Theorem.
d2d=32+52=34=34=5.83
The Distance Formula
The procedure we just used can be generalized by using the Pythagorean Theorem to derive a formula for the distance between any two points on the coordinate plane.
Let’s find the distance between two general points A=(x1,y1)\begin{align*}A=(x_1,y_1)\end{align*} and B=(x2,y2)\begin{align*}B=(x_2,y_2)\end{align*} .
Start by plotting the points on the coordinate plane:
In order to move from point A\begin{align*}A\end{align*} to point B\begin{align*}B\end{align*} in the coordinate plane, we move x2x1\begin{align*}x_2 - x_1\end{align*} units to the right and y2y1\begin{align*}y_2 - y_1\end{align*} units up.
We can find the length d\begin{align*}d\end{align*} by using the Pythagorean Theorem:
d2=(x1x2)2+(y1y2)2
Therefore, d=(x1x2)2+(y1y2)2\begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\end{align*} . This is called the Distance Formula. More formally:
Given any two points (x1,y1)\begin{align*}(x_1, y_1)\end{align*} and (x2,y2)\begin{align*}(x_2, y_2)\end{align*} , the distance between them is d=(x1x2)2+(y1y2)2.\begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.\end{align*}
We can use this formula to find the distance between any two points on the coordinate plane. Notice that the distance is the same whether you are going from point A\begin{align*}A\end{align*} to point B\begin{align*}B\end{align*} or from point B\begin{align*}B\end{align*} to point A\begin{align*}A\end{align*} , so it does not matter which order you plug the points into the distance formula.
Let’s now apply the distance formula to the following examples.
#### Example C
Find the distance between the following points.
a) (-3, 5) and (4, -2)
b) (12, 16) and (19, 21)
c) (11.5, 2.3) and (-4.2, -3.9)
Solution
Plug the values of the two points into the distance formula. Be sure to simplify if possible.
a) d=(34)2+(5(2))2=49+49=98=72\begin{align*}d=\sqrt{(-3-4)^2+(5-(-2))^2}=\sqrt{49+49}=\sqrt{98}=7 \sqrt{2}\end{align*}
b) d=(1219)2+(1621)2=49+25=74\begin{align*}d=\sqrt{(12-19)^2+(16-21)^2}=\sqrt{49+25}=\sqrt{74}\end{align*}
c) d=(11.5+4.2)2+(2.3+3.9)2=284.93=16.88\begin{align*}d=\sqrt{(11.5+4.2)^2+(2.3+3.9)^2}=\sqrt{284.93}=16.88\end{align*}
Applications Using Distance and Midpoint Formulas
The distance and midpoint formula are useful in geometry situations where we want to find the distance between two points or the point halfway between two points.
#### Example D
Plot the points A=(4,2),B=(5,5)\begin{align*}A = ( 4, -2), B = (5, 5)\end{align*} , and C=(1,3)\begin{align*}C = (-1, 3)\end{align*} and connect them to make a triangle. Show that the triangle is isosceles.
Solution
Let’s start by plotting the three points on the coordinate plane and making a triangle:
We use the distance formula three times to find the lengths of the three sides of the triangle.
ABBCAC=(45)2+(25)2=(1)2+(7)2=50=52=(5+1)2+(53)2=(6)2+(2)2=40=210=(4+1)2+(23)2=(5)2+(5)2=50=52
Notice that AB=AC\begin{align*}AB = AC\end{align*} , therefore triangle ABC\begin{align*}ABC\end{align*} is isosceles.
#### Example E
At 8 AM one day, Amir decides to walk in a straight line on the beach. After two hours of making no turns and traveling at a steady rate, Amir is two miles east and four miles north of his starting point. How far did Amir walk and what was his walking speed?
Solution
Let’s start by plotting Amir’s route on a coordinate graph. We can place his starting point at the origin: A=(0,0)\begin{align*}A = (0, 0)\end{align*} . Then his ending point will be at B=(2,4)\begin{align*}B = (2, 4)\end{align*} .
The distance can be found with the distance formula:
dd=(20)2+(40)2=(2)2+(4)2=4+16=20=4.47 miles
Since Amir walked 4.47 miles in 2 hours, his speed is s=4.47 miles2 hours=2.24 mi/h\begin{align*}s=\frac{4.47 \ miles}{2 \ hours}=\underline{\underline{2.24 \ mi/h}}\end{align*} .
Watch this video for help with the Examples above.
### Vocabulary
• The Distance Formula states that given any two points (x1,y1)\begin{align*}(x_1, y_1)\end{align*} and (x2,y2)\begin{align*}(x_2, y_2)\end{align*} , the distance between them is
d=(x1x2)2+(y1y2)2.\begin{align*}d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.\end{align*}
### Guided Practice
Find all points on the line y=2\begin{align*}y = 2\end{align*} that are exactly 8 units away from the point (-3, 7).
Solution
Let’s make a sketch of the given situation.
Draw line segments from the point (-3, 7) to the line y=2\begin{align*}y = 2\end{align*} .
Let k\begin{align*}k\end{align*} be the missing value of x\begin{align*}x\end{align*} we are seeking.
Let's use the distance formula:Square both sides of the equation:Therefore:Use the quadratic formula:Therefore:8640kk=(3k)2+(72)2=(3k)2+25=9+6k+k239 or 0=k2+6k30=6±36+1202=6±1562=3.24 or k=9.24
The points are (-9.24, 2) and (3.24, 2).
### Practice
Find the distance between the two points.
1. (3, -4) and (6, 0)
2. (-1, 0) and (4, 2)
3. (-3, 2) and (6, 2)
4. (0.5, -2.5) and (4, -4)
5. (12, -10) and (0, -6)
6. (-5, -3) and (-2, 11)
7. (2.3, 4.5) and (-3.4, -5.2)
8. Find all points having an x\begin{align*}x-\end{align*} coordinate of -4 whose distance from the point (4, 2) is 10.
9. Find all points having a y\begin{align*}y-\end{align*} coordinate of 3 whose distance from the point (-2, 5) is 8.
10. Find three points that are each 13 units away from the point (3, 2) but do not have an x\begin{align*}x-\end{align*} coordinate of 3 or a y\begin{align*}y-\end{align*} coordinate of 2.
Find the midpoint of the line segment joining the two points.
1. Plot the points A=(1,0),B=(6,4),C=(9,2)\begin{align*}A = (1, 0), B = (6, 4), C = (9, -2)\end{align*} and D=(6,4),E=(1,0),F=(2,6)\begin{align*}D = (-6, -4), E = (-1, 0), F = (2, -6)\end{align*} . Prove that triangles ABC\begin{align*}ABC\end{align*} and DEF\begin{align*}DEF\end{align*} are congruent.
2. Plot the points A=(4,3),B=(3,4),C=(2,1),D=(1,8).\begin{align*}A = (4, -3), B = (3, 4), C = (-2, -1), D = (-1, -8).\end{align*} Show that ABCD\begin{align*}ABCD\end{align*} is a rhombus (all sides are equal)
3. Plot points A=(5,3),B=(6,0),C=(5,5).\begin{align*}A = (-5, 3), B = (6, 0), C = (5, 5).\end{align*} Find the length of each side. Show that ABC\begin{align*}ABC\end{align*} is a right triangle. Find its area.
4. Find the area of the circle with center (-5, 4) and the point on the circle (3, 2).
5. Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles and then the direction of the bike trail changes and she rides in the new direction for a while longer. When she stops Michelle is 2 miles south and 10 miles west from her starting point. Find the total distance that Michelle covered from her starting point. | 3,101 | 8,776 | {"found_math": true, "script_math_tex": 58, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2015-32 | longest | en | 0.758055 |
https://studysoup.com/tsg/calculus/268/applied-calculus/chapter/12009/4-7 | 1,563,777,567,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527531.84/warc/CC-MAIN-20190722051628-20190722073628-00301.warc.gz | 553,430,732 | 11,876 | Make \$16/hr - and build your resume - as a Marketing Coordinator!
> > > Chapter 4.7
# Solutions for Chapter 4.7: LOGISTIC GROWTH
## Full solutions for Applied Calculus | 5th Edition
ISBN: 9781118174920
Solutions for Chapter 4.7: LOGISTIC GROWTH
Solutions for Chapter 4.7
4 5 0 333 Reviews
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##### ISBN: 9781118174920
Chapter 4.7: LOGISTIC GROWTH includes 20 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. Since 20 problems in chapter 4.7: LOGISTIC GROWTH have been answered, more than 6946 students have viewed full step-by-step solutions from this chapter. Applied Calculus was written by Patricia and is associated to the ISBN: 9781118174920. This textbook survival guide was created for the textbook: Applied Calculus, edition: 5.
Key Calculus Terms and definitions covered in this textbook
• Binomial
A polynomial with exactly two terms
• Circular functions
Trigonometric functions when applied to real numbers are circular functions
• Common ratio
See Geometric sequence.
• Directed line segment
See Arrow.
• Ellipsoid of revolution
A surface generated by rotating an ellipse about its major axis
• Endpoint of an interval
A real number that represents one “end” of an interval.
• Factoring (a polynomial)
Writing a polynomial as a product of two or more polynomial factors.
• Matrix, m x n
A rectangular array of m rows and n columns of real numbers
• Nappe
See Right circular cone.
• Parametric equations for a line in space
The line through P0(x 0, y0, z 0) in the direction of the nonzero vector v = <a, b, c> has parametric equations x = x 0 + at, y = y 0 + bt, z = z0 + ct.
• Partial fraction decomposition
See Partial fractions.
• Piecewise-defined function
A function whose domain is divided into several parts with a different function rule applied to each part, p. 104.
• Rational numbers
Numbers that can be written as a/b, where a and b are integers, and b ? 0.
• Rational zeros theorem
A procedure for finding the possible rational zeros of a polynomial.
• Scalar
A real number.
• Sum of a finite arithmetic series
Sn = na a1 + a2 2 b = n 2 32a1 + 1n - 12d4,
• Synthetic division
A procedure used to divide a polynomial by a linear factor, x - a
• Vertex of a cone
See Right circular cone.
• Whole numbers
The numbers 0, 1, 2, 3, ... .
• Work
The product of a force applied to an object over a given distance W = ƒFƒ ƒAB!ƒ.
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https://byjus.com/question-answer/nine-point-charges-placed-at-the-vertices-of-an-octagon-and-centre-of-an-octagon/ | 1,696,393,826,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511351.18/warc/CC-MAIN-20231004020329-20231004050329-00261.warc.gz | 165,203,162 | 30,703 | Question
# Nine point charges placed at the vertices of an octagon and centre of an octagon as shown in the figure. Net force experienced by +q charge at the centre of the octagon is zero. If any two charges are removed simultaneously from the vertex, it observed that net force on charge +q is still zero, then the charges will be
A
+3Q and +2Q
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B
+Q and Q
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C
4Q and +3Q
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D
any pair of diagonally opposite charges
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Solution
## The correct option is D any pair of diagonally opposite chargesThe force experienced by the charge +q placed at the centre of the octagon is shown in the figure. Let a be the distance between centre and vertex of the octagon. So, Force on +q due to −Q, F02=F06=kQqa2 Similarly, F03=F07=k(3Q)qa2=3kQqa2; F04=F08=4kQqa2 and F01=F05=2kQqa2 Here, only diagonally opposite charges apply equal and opposite forces on +q. So, if we remove any pair of diagonally opposite charges from vertex of the octagon, net forces on +q will be unaffected. Hence, option (d) is the correct answer.
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Select... | 405 | 1,394 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-40 | latest | en | 0.881988 |
https://datascience.stackexchange.com/questions/74063/understanding-neural-network-probability/74080 | 1,624,431,087,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488534413.81/warc/CC-MAIN-20210623042426-20210623072426-00547.warc.gz | 183,523,161 | 38,136 | # Understanding neural network probability
I'm training a CNN model with two classes to predict. I know it gives me a probability for one class and for the other one, and I also know I can get the predicted label, but I don't the results given. Isn't the sum of the output for each evaluated input supposed to be equal 1.0? For instance:
[[0.2858745 0.85059494]
[0.2858745 0.85059494]
[0.6040499 0.5927084 ]
[0.8403308 0.291448 ]
[0.04195209 0.95504093]
[0.79433376 0.21279709]
[0.79433376 0.21279709]
[0.01326967 0.9891382 ]
[0.0153821 0.9867737 ]
[0.79433376 0.21279709]
[0.01617167 0.98520505]
[0.01351487 0.98596036]
[0.01473185 0.9846144 ]
[0.00896762 0.9899838 ]
[0.00936404 0.9893628 ]]
Is there something I didn't get?
Code:
model_05_01 = Sequential()
input_shape=(x_train.shape[1], 1)))
• Tensorflow. 3 Conv1D layers, + 2 Dense layers, the last one is model.add(Dense(2, activation='sigmoid')) – Tiago Minuzzi May 12 '20 at 22:21
While that is a valid classifier problem to solve, I'm guessing what you mean is that you have an exclusive 2-class problem: it's either positive, or negative. In that case, you want the final layer to be Dense(1, activation='sigmoid') | 410 | 1,188 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2021-25 | latest | en | 0.838817 |
https://calculatorshub.net/agriculture/mulch-calculator-map/ | 1,723,285,966,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640805409.58/warc/CC-MAIN-20240810093040-20240810123040-00732.warc.gz | 119,505,619 | 27,475 | Home » Simplify your calculations with ease. » Agriculture » Mulch Calculator Map Online
# Mulch Calculator Map Online
## Introduction
Navigating through landscaping and gardening projects can often feel like a journey through a dense forest, especially when it comes to mulch application. One of the vital tools to streamline this process is a mulch calculator map. This helpful innovation can efficiently guide you through your mulch needs, keeping your green spaces healthy and vibrant.
## Definition
A mulch calculator map is an online tool that helps determine the exact amount of mulch required for a specific area. It's a combination of a map and calculator, allowing you to visualize the designated area for mulching and estimate the mulch volume needed, based on area size and mulch depth.
## Detailed Explanation of the Calculator's Working
The mulch calculator map operates on simple principles. Users enter the area's dimensions in square meters and the desired mulch depth in centimeters. The calculator then uses these inputs to compute the total volume of mulch required in cubic meters, ensuring you have the precise mulch quantity needed for effective coverage.
## Properly Formatted Formula with Variable Description
The formula that the mulch calculator map uses is quite simple: Area (in square meters) * Depth (in centimeters) = Volume of Mulch (in cubic meters). Here, 'Area' refers to the total ground space that needs mulching, and 'Depth' refers to the thickness of the mulch layer you want to apply.
## Example
Let's say you have a garden space that is 10 square meters in area, and you wish to apply a mulch layer of 5 cm depth. Inserting these values into the calculator would give: 10 (Area) * 0.05 (Depth) = 0.5 cubic meters of mulch required.
## Applications
A. Residential Landscaping: The calculator is an invaluable tool for homeowners looking to beautify their garden spaces and improve soil health.
B. Commercial Landscaping: Professional landscapers can use this tool to accurately estimate mulch requirements for large-scale projects.
C. Agriculture: Farmers and gardeners can use the calculator to optimize mulch usage, enhancing soil fertility and crop yield.
## Most Common FAQs
What depth should mulch be?
A mulch layer of 5-7.5 cm is generally recommended for most gardens to provide effective coverage and nutrient delivery.
Can I use the mulch calculator for any mulch type?
Yes, the mulch calculator can be used for any type of mulch as it calculates the volume based on area and depth inputs.
## Conclusion
A mulch calculator map is a practical and efficient tool that transforms how we approach mulching in our gardens and landscapes. By providing precise estimates, it aids in promoting healthier landscapes, reducing waste, and saving money. Embrace this digital tool to make your green spaces thrive. | 593 | 2,873 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-33 | latest | en | 0.87988 |
https://curriculum.media.pearsoncmg.com/curriculum/math/digits/TXgrade_7/html_books/homework_helper_vol1/htmls/page_286.html | 1,606,538,716,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195069.35/warc/CC-MAIN-20201128040731-20201128070731-00080.warc.gz | 253,489,454 | 5,345 | Digits, Grade 7, Volume 1, Homework Helper
### 8-2: Initial Value
#### Key Concept
A linear relationship can be written in the form y equals m x plus b . .
You can see both the rate of change and the initial value of the relationship in the equation.
You can also see both the rate of change and the initial value of the relationship in the graph.
End ofPage 286 | 88 | 368 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2020-50 | latest | en | 0.899502 |
https://www.physicsforums.com/threads/confidence-interval.255541/ | 1,527,365,025,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867859.88/warc/CC-MAIN-20180526190648-20180526210648-00057.warc.gz | 790,361,105 | 14,062 | # Homework Help: Confidence interval
1. Sep 11, 2008
### twoflower
Let's say we know this:
$$\sqrt{n}\left(\widehat{\theta} - \theta\right) \sim \mathcal{N}\left(0, \frac{1}{F(\theta)}\right)$$
How do we get from this information to this expression of confidence interval for $\theta$?
$$\left( \widehat{\theta} \pm u_{1-\frac{\alpha}{2}}\frac{1}{\sqrt{nF\left(\widehat{\theta}\right)}}\right)$$
Where $u_{1-\frac{\alpha}{2}}$ is appropriate quantil of standard normal distribution.
Thank you.
2. Sep 11, 2008
If $$a$$ is the value from $$Z$$ (standard normal) with area $${\alpha}/2$$ to its right, you know the value of
$$\Pr\left(-u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u)$$
$$-u < \sqrt{n F(\theta)} \left(\hat \theta - \theta\right) < u$$ | 265 | 773 | {"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} | 3.140625 | 3 | CC-MAIN-2018-22 | longest | en | 0.614906 |
https://collegemathteaching.wordpress.com/tag/tent-function/ | 1,569,059,966,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574377.11/warc/CC-MAIN-20190921084226-20190921110226-00059.warc.gz | 446,279,167 | 17,024 | College Math Teaching
August 21, 2014
Calculation of the Fourier Transform of a tent map, with a calculus tip….
I’ve been following these excellent lectures by Professor Brad Osgood of Stanford. As an aside: yes, he is dynamite in the classroom, but there is probably a reason that Stanford is featuring him. 🙂
And yes, his style is good for obtaining a feeling of comradery that is absent in my classroom; at least in the lower division “service” classes.
This lecture takes us from Fourier Series to Fourier Transforms. Of course, he admits that the transition here is really a heuristic trick with symbolism; it isn’t a bad way to initiate an intuitive feel for the subject though.
However, the point of this post is to offer a “algebra of calculus trick” for dealing with the sort of calculations that one might encounter.
By the way, if you say “hey, just use a calculator” you will be BANNED from this blog!!!! (just kidding…sort of. 🙂 )
So here is the deal: let $f(x)$ represent the tent map: the support of $f$ is $[-1,1]$ and it has the following graph:
The formula is: $f(x)=\left\{\begin{array}{c} x+1,x \in [-1,0) \\ 1-x ,x\in [0,1] \\ 0 \text{ elsewhere} \end{array}\right.$
So, the Fourier Transform is $F(f) = \int^{\infty}_{-\infty} e^{-2 \pi i st}f(t)dt = \int^0_{-1} e^{-2 \pi i st}(1+t)dt + \int^1_0e^{-2 \pi i st}(1-t)dt$
Now, this is an easy integral to do, conceptually, but there is the issue of carrying constants around and being tempted to make “on the fly” simplifications along the way, thereby leading to irritating algebraic errors.
So my tip: just let $a = -2 \pi i s$ and do the integrals:
$\int^0_{-1} e^{at}(1+t)dt + \int^1_0e^{at}(1-t)dt$ and substitute and simplify later:
Now the integrals become: $\int^{1}_{-1} e^{at}dt + \int^0_{-1}te^{at}dt - \int^1_0 te^{at} dt.$
These are easy to do; the first is merely $\frac{1}{a}(e^a - e^{-a})$ and the next two have the same anti-derivative which can be obtained by a “integration by parts” calculation: $\frac{t}{a}e^{at} -\frac{1}{a^2}e^{at}$; evaluating the limits yields:
$-\frac{1}{a^2}-(\frac{-1}{a}e^{-a} -\frac{1}{a^2}e^{-a}) - (\frac{1}{a}e^{a} -\frac{1}{a^2}e^a)+ (-\frac{1}{a^2})$
Add the first integral and simplify and we get: $-\frac{1}{a^2}(2 - (e^{-a} -e^{a})$. NOW use $a = -2\pi i s$ and we have the integral is $\frac{1}{4 \pi^2 s^2}(2 -(e^{2 \pi i s} -e^{-2 \pi i s}) = \frac{1}{4 \pi^2 s^2}(2 - cos(2 \pi s))$ by Euler’s formula.
Now we need some trig to get this into a form that is “engineering/scientist” friendly; here we turn to the formula: $sin^2(x) = \frac{1}{2}(1-cos(2x))$ so $2 - cos(2 \pi s) = 4sin^2(\pi s)$ so our answer is $\frac{sin^2( \pi s)}{(\pi s)^2} = (\frac{sin(\pi s)}{\pi s})^2$ which is often denoted as $(sinc(s))^2$ as the “normalized” $sinc(x)$ function is given by $\frac{sinc(\pi x)}{\pi x}$ (as we want the function to have zeros at integers and to “equal” one at $x = 0$ (remember that famous limit!)
So, the point is that using $a$ made the algebra a whole lot easier.
Now, if you are shaking your head and muttering about how this calculation was crude that that one usually uses “convolution” instead: this post is probably too elementary for you. 🙂 | 1,052 | 3,215 | {"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": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-39 | latest | en | 0.873262 |
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## Central Tendency and Variability
by: Kaiyana Dudley
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# Central Tendency and Variability Sociology 210
Marketplace > University of Michigan > Statistics > Sociology 210 > Central Tendency and Variability
Kaiyana Dudley
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Simple intro to central tendency and variability
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Elementary Statistics
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Jennifer Barber
TYPE
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2
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Math, Statistics, variance, standard deviation
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This 2 page Class Notes was uploaded by Kaiyana Dudley on Saturday September 17, 2016. The Class Notes belongs to Sociology 210 at University of Michigan taught by Jennifer Barber in Fall 2016. Since its upload, it has received 7 views. For similar materials see Elementary Statistics in Statistics at University of Michigan.
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Date Created: 09/17/16
● Wha t is central ncy? ○ The textbook definition words it as a statistical measure that attempts to determine the single value that is most typical or most representative of the entire set of scores ○ A simpler definition: the center of most common number of a distribution ● What does central tendency look like? ○ The central tendency is usually understood as the mean, median, or mode ■ The mean is the center of gravity or the average out of all the scores ■ You can find the mean by simply adding up all of the numbers in the distribution and dividing by the number of scores ■ Algebraically, the population (or distribution) mean can be written using summations (∑) or μ, and the sample mean is represented by Xbar () ̄ ■ The median can be found by putting all of the scores in order and locating the center value ■ If there are two center values, add them together and dividing by 2 ■ The mode is the score that shows up most frequently ■ There can be no mode or more than one if certain scores are similarly common ● How do I know which measure will represent the central tendency best? ○ It's not always obvious which measure to use, but here are a few key indicators to remember ■ When there are extreme outliers, it may be best to use the median or mode ■ The mode won't explain central tendency unless it is very dominant ■ The median is less sensitive than the mean to extreme values ● What is variability? ○ Variability shows the differences between scores in a population that describes whether scores are spread out or clustered together around the center of distribution ■ This can also be referred to as the measure of dispersion ■ The range, interquartile range, or semiinterquartile range are based on the distance between scores of different rank ■ The variance and deviation are measures based on the distance of a score (or group of scores) to the center of the distribution ○ Variability basically tells us how well a score(s) represent the entire population ● Can you explain more about the measures of variability? ○ The range is the difference between the highest (upper real limit) and lowest (lower real limit) score of the entire distribution ● Where does the quartile stuff come in? ○ We mark the quartiles when the scores are ranked in ascending order ○ A distribution can be split up in four quartiles (Q1, Q2, Q3, & Q4) ■ A percentile is a value on a 1100 number scale of the distribution that tells the percentage of a score ○ Q1 is the first quartile that represents 25th percentile ■ Q2=50%, Q3=75%, and Q4=100% ○ The interquartile range is the difference between Q3 and Q1 (Q3Q1) ■ The semiinterquartile range is ((Q3Q1)/2) ● What about deviation and variance? ○ Deviation is a score's distance from the mean (Xμ) ○ The average of the deviation scores will always equal zero (∑(Xx)=0) ■ This is the population variance that is the average squared distance from the mean (but this isn't what we want because it equals zero) ■ It equals the mean squared deviation ■ For statistics, our goal is to measure the variability by finding the standard deviation from the mean, which is the square root of variance References Essentials of Statistics for the Behavioral Sciences by Frederick J Gravetter, Larry B. Wallnau, Chapter 24 Jennifer Barber Lecture 2 Slide Presentation (September 13 &15, 2016)
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# lecture03 - Lecture#3 OUTLINE Power calculations Circuit...
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Lecture 3, Slide 1 EECS40, Fall 2003 Prof. King Lecture #3 OUTLINE Power calculations Circuit elements – Voltage and current sources – Electrical resistance (Ohm’s law) • Kirchhoff’s laws Reading Chapter 2 Lecture 3, Slide 2 EECS40, Fall 2003 Prof. King If an element is absorbing power ( i.e. if p > 0), positive charge is flowing from higher potential to lower potential. p = vi if the “passive sign convention” is used: How can a circuit element absorb power? By converting electrical energy into heat (resistors in toasters), light (light bulbs), or acoustic energy (speakers); by storing energy (charging a battery). Review: Power + v _ i _ v + i or
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Lecture 3, Slide 3 EECS40, Fall 2003 Prof. King Find the power absorbed by each element: Power Calculation Example vi (W) 918 - 810 - 12 - 400 - 224 1116 p (W) Conservation of energy Î total power delivered equals total power absorbed Lecture 3, Slide 4 EECS40, Fall 2003 Prof. King Circuit Elements There are 5 ideal basic circuit elements: – voltage source – current source – resistor – inductor – capacitor Many practical systems can be modeled with just sources and resistors The basic analytical techniques for solving
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Read the riddle and answer it. In a village, there are four Mansions located at a different distance from each other. Following are the distances: The third Mansion is 60 km apart from the first Mansion. The fourth Mansion is 40 km apart from the second Mansion. The third Mansion is 10 km nearer to … Read more
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## How Can You Determine If The Barrel is More Than Half Full?
Go through the hints in the riddle and leave your answers in the comments section. There is a barrel with no lid and some wine in it. “This barrel of wine is more than half full”, says Mark. “No, it’s not”, says Steve. “It’s less than half full.” They do not have any measuring instruments … Read more
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## Find The Logic: If NIGHT→305769 MATCH→93600 Then, LIGHT → ?
Find the value of ? in the riddle below. Find The Logic…. NIGHT→305769 MATCH→93600 CANDLE→369600 ↓ LIGHT→ ? So were you able to solve the riddle? Leave your answers in the comment section below. If you get the correct answer, please share it with your friends and family on WhatsApp, Facebook and other social networking … Read more
## Riddle: How Long Will The Pills Last
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https://socratic.org/questions/how-do-you-find-points-of-inflection-and-determine-the-intervals-of-concavity-gi-10 | 1,660,154,447,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571198.57/warc/CC-MAIN-20220810161541-20220810191541-00602.warc.gz | 486,209,935 | 6,284 | # How do you find points of inflection and determine the intervals of concavity given y=4x^3+21x^2+36x-20?
Nov 26, 2017
There is an inflection point at $x = - 1.75$ and the function is concave down ($\cap$) on the interval $\left(- \infty , - 1.75\right)$, and it is concave up ($\cup$) on the interval $\left(- 1.75 , \infty\right)$.
#### Explanation:
Concavity and inflection points of a function can be determined by looking at the second derivative. If the second derivative is 0, it is an inflection point (IE where the graph changes concavity). If the second derivative is positive, you know the graph is concave up ($\cup$), and if the second derivative is negative, you know the graph is concave down ($\cap$).
So let's start by finding the second derivative for our function,
$f \left(x\right) = 4 {x}^{3} + 21 {x}^{2} + 36 x - 20$
$f ' \left(x\right) = 12 {x}^{2} + 42 x + 36$
$f ' ' \left(x\right) = 24 x + 42$
Now we can solve the following equation to find the inflection point(s):
$24 x + 42 = 0$
$24 x = - 42$
$x = - \frac{42}{24} = - 1.75$
Now, let's look at the intervals before and after our only inflection point.
If we pick a point smaller than $- 1.75$, say $- 2$, and plug it into our second derivative, we get $- 6$, which is negative, so our function must be concave down ($\cap$) on the interval $\left(- \infty , - 1.75\right)$.
If we then pick a point larger than $- 1.75$, say $0$, we get $42$, which is positive, so the function must be concave up ($\cup$) on the interval $\left(- 1.75 , \infty\right)$.
In conclusion, by analyzing the function's second derivative we figured out that it has an inflection point at $x = - 1.75$, that it is concave down ($\cap$) on the interval $\left(- \infty , - 1.75\right)$, and that it is concave up ($\cup$) on the interval $\left(- 1.75 , \infty\right)$. | 581 | 1,838 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 28, "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} | 4.78125 | 5 | CC-MAIN-2022-33 | latest | en | 0.788944 |
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# lt;In other monogamous primate species except human beings,
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<In other monogamous primate species except human beings, males and females> tend to be of the same size and to fight off predators together.
A)
B) Among species of monogamous premates other than human beings, males and females
C) The males and females of other monogamous primate species except human beings
D) Other than human beings, the males and females of monogamous primate species
E) Except for human beings, other males and females of species of monogamous primates
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10 Jan 2004, 21:43
D and E set up a false comparison.
A and C use both "other" and "except".
B makes sense, is relatively compact, and clearly conveys the meaning.
10 Jan 2004, 21:43
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Louisville, KY, United States 12/15/2011 9:54 P.M. Arrival Scan
Anchorage, AK, United States 12/15/2011 12:30 P.M. Departure Scan
12/15/2011 9:28 A.M. Arrival Scan
Narita, Japan 12/15/2011 8:40 P.M. Departure Scan
12/15/2011 7:04 P.M. Arrival Scan
Shanghai, China 12/15/2011 3:40 P.M. Departure Scan
Even with timezones, how is it possible that going from China to USA happens in 6+ hours?
0
25112
4 Solutions
Commented:
Quirks of flying with earth's rotation. It'll be a longer trip back in the opposite direction
0
Commented:
The times given in the history report are all local. The package is flying against the earth's rotation but it's passage through the international dateline at 180 degrees East or West (with local deviations) causes a 24hr jump in the "local" time. The packages starts at 1540h local in China, whose timezone is +8hr, passes through the international dateline and ends up at 2154h local in Loisville, which is in the -5hr time zone. If you were to look at the arrival and departure times from the point of view of an observer at a fixed time zone e.g. GMT or UTC (often suffixed by "Z" or zulu), then the package departure time from China would be 0740z on the 15th December. It's arrival time in Loisville would be 0254z on the 16th December, but, with the US being 5 hours behind UTC, the *local* arrival date would still be the 15th. That gives a total time between initial departure and final arrival scans of 19h14m.
0
Commented:
Sorry, with the earth's rotation!
0
Commented:
??? You say "going from china to usa" then you list itenerary from Kentucky TO China. Boy , are you going to be double screwed when you reverse your trip back home when you're going from usa to China. How long is that trip a gonna take ???
0
Commented:
^ the tracking is in reverse order: from arrival to departure
Between Japan and Alaska note the package travels backward in time
8:40pm dep --> 9:28am arr
It didn't travel via wormhole. It crossed the international dateline in the middle of the Pacific, where it jumped backward one day.
The scans are logged as local time.
0
Commented:
http://www.happyzebra.com/timezones-worldclock/difference-between-Louisville-and-Shanghai.php
The time difference is +13 hours.
So, when you leave Kentucky when it is 1:00 am (an hour after midnight), it is 2:00 pm in Shanghai (same calendar day)
The actual flight time, with all the layovers, is 13 hours greater than the 6+ hours, you see on the shedules.
So, it actually takes 19+ hours to get there.
Louisville, KY, United States 12/15/2011 9:54 P.M. Arrival Scan
Anchorage, AK, United States 12/15/2011 12:30 P.M. Departure Scan
Is the 9:54pm, the time you'd leave Louisville?
If it is, it is impossible to fly to Alaska and get off the plane and get on another plane and leave for Japan at 12:30pm
And I assume you mean 12:30 AM (??)
That is only 2.5 hours. Alaska is 3 times zones to the west = 3 hours, so the trip to Alaska is actually 3 + 2.5 = 5.5 hours.
0
Author Commented:
fascinating ...!
0
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Problem 2591. Does the coin touch the line?
Solution 2137244
Submitted on 22 Feb 2020 by Gatech AE
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
d = 1; n=2; m=2; y_correct = 0.25; assert(isequal(your_fcn_name(d,n,m),y_correct))
y = 0.2500
2 Pass
d = 1; n=1; m=1; y_correct = 0; assert(isequal(your_fcn_name(d,n,m),y_correct))
y = 0
3 Pass
d = 1; n=3; m=4; y_correct = 0.5; assert(isequal(your_fcn_name(d,n,m),y_correct))
y = 0.5000
4 Pass
d = 2; n=1; m=1; y_correct = 0; assert(isequal(your_fcn_name(d,n,m),y_correct))
y = 0
5 Pass
d = 0.5; n=2; m=3; y_correct = 0.625; assert(isequal(your_fcn_name(d,n,m),y_correct))
y = 0.6250
6 Pass
d = 2; n=1; m=4; y_correct = 0; assert(isequal(your_fcn_name(d,n,m),y_correct))
y = 0 | 348 | 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} | 3.015625 | 3 | CC-MAIN-2020-24 | latest | en | 0.624205 |
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# As far as we know, Earth is the only planet on which life
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As far as we know, Earth is the only planet on which life [#permalink]
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30 May 2009, 22:27
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12. As far as we know, Earth is the only planet on which life has evolved, and all known life forms are carbon-based. Therefore, although there might exist non-carbon-based life on planets very unlike Earth, our scientific estimates of the probability of extraterrestrial life should be generated from estimates of the number of planets like Earth and the likelihood of carbon-based life on those planets.
Which one of the following general principles most strongly supports the recommendation?
(A) There is no good reason to think that unobserved phenomena closely resemble those that have been observed.
(B) A scientific theory that explains a broad range of phenomena is preferable to a competing theory that explains only some of those phenomena.
(C) It is preferable for scientists to restrict their studies to phenomena that are observable and forego making estimates about unobservable things.
(D) A scientific theory that explains observed phenomena on the basis of a few principles that are independent of each other is preferable to a theory that explains those same phenomena on the basis of many independent principles.
(E) Estimations of probability that are more closely tied to what is known are preferable to those that are less closely tied to what is known.
OA is
[Reveal] Spoiler:
E
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30 May 2009, 23:05
Yes, only E follows the recommendation.
Kudos [?]: 3 [0], given: 1
Re: earth and life [#permalink] 30 May 2009, 23:05
Display posts from previous: Sort by | 642 | 2,654 | {"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} | 3.046875 | 3 | CC-MAIN-2017-43 | latest | en | 0.928872 |
www.rockinjlnl.com | 1,527,382,444,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00493.warc.gz | 444,932,923 | 4,006 | The Inbreeding Coefficient and Probability
from: Probability and its relationship to parts of Livestock production
by Jori (Garoutte) Butler
The inbreeding coefficient, as used in linebreeding, is a measurement of the probable chance that any genes will double when two animals are mated. This probability increases with the degree of relationship between the two individuals. By knowing the probability in such instances, a breeder can safely decide whether to try linebreeding in some cases.
Inbreeding coefficients and their related probabilities (percent chance of genes doubling) can be seen in the following table:
INBREEDING COEFFICIENTS FOR VARIOUS MATINGS
kind of mating inbreeding %
*Half first cousin (1 common grandparent) 3.12
*first cousin (2 common grandparents) 6.25
*Grandparent-Grandoffspring 12.50
*Half brother-sister 12.50
*Parent-offspring 25.00
*Full brother-sister (1 generation) 25.00
*Full brother-sister (2 generations) 37.50
*Full brother-sister (3 generations) 50.00
(Boggess)
One can see from the previous table that as relationship increases, so does the inbreeding coefficient and the probability that genes will double in each mating.
Knowing this probability is very important to breeders who incorporate linebreeding into their programs as the doubling of genes can either greatly improve or massively decline the quality of livestock as not only the good genes are doubled. The bad genes have the same probability of doubling and becoming dominant.
Source:
Boggess, Mark V. "Application of Genetic Principles" Cattle Producers Library.
(from: J. Garoutte, Mathematical Applications in Animal Sciences)
*Crossbreeding*
In crossbreeding, heterosis or hybrid vigor, is often a common calculation in determining whether or not to cross two breeds. Heterosis is the degree to which an offspring deviates from its parent's breeds. The amount of heterosis is calculated where breed A and breed B and crosses between them are raised as contemporaries. In this manner, the amount of heterosis is calculated as follows:
AB+AB - A+B
2 2
The percent of heterosis is calculated as follows:
amount of heterosis * 100
A+H/2
These formulas can be figured with many examples, but a common one is applied to weaning weights. The percentage of heterosis can also be calculated as the sum of the crossbred average minus the straightbred average multiplied by 100, all divided by the straightbred average. (Crossbreeding . . . Western Range . . .)
Annual progress for inherited traits can also be calculated. It is equal to the heritability multiplied by the selection differential, all divided by the generation interval.
The heritability is the proportion of differences between animals which are transmitted to the offspring. The selection differential is the difference between the average of a group with the selected individual, and the generation interval is the average length of the generation. (Gregory 9,10)
*Linebreeding*
The math used in linebreeding is not very difficult, but it is quite interesting. The inbreeding coefficient is a percentage of the probable chance that genes will double when related animals are mated. It is often used in determining whether or not ot mate two related animals because the higher the inbreeding coefficient is, the more likely it is that genes will double, whether they be good or bad.
The different inbreeding coefficients are related to the different kinds of matings. In a half-first cousin mating, 1 common grandparent, the inbreeding coefficient is 3.12%. First cousin matings, 2 common grandparents, have a 6.25% inbreeding coefficient. Both grandparent-grandoffspring and half-sibling matings have an inbreeding coefficient of 12.5%. Parent-offspring matings and full-sibling matings, for 1 generation, have an inbreeding coefficient of 25%. Two generation full-sibling matings have an inbreeding coefficient of 37.5% and full-sibling matings for 3 generations have an inbreeding coefficient of 50%.
By considering the inbreeding coefficient, breeder can determine whether or not it will be beneficial for them to linebreed two animals. (Boggess 1020-3)
The rate of inbreeding per generation can also be calculated considering that the herd is basically closed to outside influence and all mating is basically random where relationship is concerned. It is calculated where m is the number of males used each generation and f is the total number of females in the herd each generation by the following forumula: 1/8m+1/8F. (Gregory 14)
Sources:
Boggess, Mark V. "Application of Genetic Principles" Cattle Producer's Library
Crossbreeding Beef Cattle for Western Range Environments. Nevada Agricultural Experiment Station College of Agriculture University of Nevada-Reno, Nov. 1988
Gregory, Keith E. Beef Cattle Breeding. United States Department of Agriculture | 1,135 | 5,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2018-22 | latest | en | 0.604399 |
https://www.thecoolist.com/mystic/numerology/master-numbers/ | 1,723,002,305,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640667712.32/warc/CC-MAIN-20240807015121-20240807045121-00489.warc.gz | 779,582,110 | 64,489 | Master numbers are 11, 22, and 33 in numerology and represent high-frequency spirituality. Master numbers give insight into a person’s personality, spiritual identity, and karmic energy. Having master numbers in your numerological chart indicates that you’re close to manifesting your desired reality and achieving spiritual enlightenment.
You find out if you have a master number by calculating your numerological chart and halting your calculations if you find 11, 22, or 33. Numerology systems allow for multiple master numbers to appear in a single chart depending on the system being used. For example, you can calculate your life path master number using Pythagorean or Indian numerology. However, there are numerology practices that view 11, 22, and 33 as important but different from other belief systems. For example, Chaldean numerology includes the compound numbers 11/2, 22/4, and 33/6 as important combinations, but they aren’t classified the same as Pythagorean or Indian.
Seeing a repeating master number is a sign that the universe is trying to get your attention. For example, angel numbers feature the master numbers in their numeric sequences. The frequencies from master numbers intertwine with your angelic messages to emphasize your purpose. The master numbers influence your spiritual journey and connection with twin flames or personal growth.
Master numbers are compatible due to the high frequency at which they vibrate. The spiritual energies draw them together to help each other along their journey. The energy of each of the master numbers is powerful due to their soul’s connection with previous lives. Master numbers indicate the possession of an old soul that extends beyond the current physical reality. The appearance of master numbers is less common than 1-9 due to being larger numbers. The rarity at which they appear in numerological charts adds to their strong vibrations.
Below we examine master numbers, their meaning in your numerological chart, and why you keep seeing them.
## What are master numbers?
Master numbers are double-digit numeric combinations 11, 22, and 33 that resonate with a high vibrational frequency. Master numbers are special numbers in numerology because of their high frequency and rarity in numerological equations. People with a master number in their numerology chart should stop reducing if they have 11, 22, or 33 and analyze that number instead. Below are details about each of the three master numbers.
• Master number 11: Master number 11 is sensitive, sagacious, intuitive, and charismatic. The number 11 is the “Psychic Master” because of its knack for leadership and psychic connection to mystical elements. People with this number are often firm, intuitive, empathetic, and strongly connected to the spiritual realm.
• Master number 22: Master number 22 is practical and proactive. Individuals with a master number 22 are called “Master Builders” due to being highly ambitious and capable of manifesting their goals on a grand scale.
• Master number 33: Master number 33 is compassionate, nurturing, and dedicated to helping and healing others. Number 33 is the “Master Teacher” because it embodies unconditional love, selflessness, and spiritual understanding. Those with the master number 33 radiate a harmonious and loving energy.
### Master number 11
Master number 11 is the Psychic Master and represents the power of intuition. 11 is the first of the master numbers and embodies divine intuition and guidance. The number 11 can appear throughout your numerology chart as a life path number, expression number, or within your angel number. Master number 11 represents spiritual insight, inspirational leadership, and the pursuit of personal truths. Numerology views number 11 as a conduit of the spiritual realm. Therefore, individuals with the number 11 are highly intuitive and perceptive. Furthermore, people with master number 11 inspire others with their wisdom and guidance, leading by example. Number 11s seek deeper truths because of their interest in metaphysical and spiritual matters.
### Master number 22
Master number 22 is the Master Builder because it contains a transformative energy that manifests desires and dreams into reality. 22 is the penultimate master number and represents a shift between the physical and the metaphysical. People born under master number 22 combine their spiritual insights and practical skills to achieve material success. Master number 22 appears less frequently throughout a person’s numerology chart because large numbers are less likely to appear in mathematical equations. For example, master number 22 only shows up once for a birthday number. However, the number 3 has four chances of appearing in a birthday number through the numbers 3, 12, 21, and 30. The number 22’s rarity doubles the vibratory frequency of your spiritual message and pushes you closer to building a bridge between the physical and spiritual worlds.
### Master number 33
Master number 33 is the Master Teacher because 33 embodies compassion, empathy, and spiritual wisdom. 33 is the third and final master number that symbolizes humanitarianism and care for others. Having the master number 33 in your numerology chart is rare because it is the largest master number, which reduces the frequency of it appearing in calculations. For example, the number 33 can’t appear as a birthday number because the days of the month don’t exceed 31. Master number 33 in your chart means you’re close to reaching spiritual enlightenment and should offer guidance to others. People with master number 33 are naturally inclined towards working with others and making the world a better place. The innate gift of master number 33 is uplifting others through their guidance and shrewd perception to recoup those karmic gifts and push you through to your spiritual ascension.
## How do you find your master number?
You find your master number using mathematical calculations that reveal your numerological chart. Your numerology chart involves multiple calculations that determine your destiny number, soul urge number, and personality number. For example, add the constituent digits in your birthday together to find your destiny number. Continue adding each digit until you’re left with 1-9 or master numbers 11, 22, or 33. Master numbers are the only ones left as two digits and aren’t reduced. The energy of master numbers pertains to different areas of your life, which modulates how each is expressed. For example, your soul urge number represents your inner desires and motivations compared to your life path number, which is your life’s purpose and spiritual path. Having a master number 33 as your soul urge number means you feel compelled to offer humanitarian services or some form of therapy. Comparatively, having a master number 33 as your life path number means you are destined to follow a path that involves humanitarianism.
• Write down birthdate: Write down your full birth date in numeric form.
• Add the digits: Add all the digits in the birth date separately.
• Stop reducing: Stop reducing if the number equals a master 11, 22, or 33.
### What does it mean to have multiple master numbers in your chart?
Having multiple master numbers in your chart has three different meanings. Firstly, having multiple master numbers in your chart means the energy of these numbers influences multiple aspects of your life. Seeing multiple master numbers in your chart means the master numbers appear in different equations and impact different elements. For example, a life path master number comes from adding each digit of a birthdate and represents a person’s trajectory in life. However, an expression number is calculated using your full name and represents the legacy you leave society. Secondly, multiple master numbers in your chart indicate a strong connection with spirituality and mysticism. 11, 22, and 33 remain two-digit numbers in numerology because they carry a higher vibration and hold unique spiritual attributes. A high vibratory frequency denotes an intense spiritual connection that suggests spiritual ascension is in process. Thirdly, having multiple master numbers means you have a specific divine mission on earth. For instance, a life path calculation of 11 means you’re meant to enhance the spirituality and well-being of the collective through psychological practice, humanitarian endeavors, or charity work. A soul urge number of 33 means you’re drawn to positions of service and healing. A soul urge 33 in conjunction with life path 11 means your destiny and soul urge are closely aligned.
## What do master numbers mean in numerology?
Master numbers in numerology mean a heightened spiritual and mystical significance. Master numbers 11, 22, and 33 are imbued with mystical properties that influence an individual’s life. The importance of master numbers in numerology depends on where they manifest in a numerology chart. For example, your numerology birth number being a master number is a sign that your spirit is vibrating at a high frequency and you’re shifting into self-fulfillment. Master numbers appear throughout numerological chart calculations, which branch from the following three core numerology belief systems.
• Pythagorean: Pythagorean numerology is a modern form of numerology that assigns numbers 1-9 to the letters of the alphabet.
• Chaldean: Chaldean numerology is the oldest recorded form of numerology and creates an alphanumeric system using numbers 1-8.
• Indian: Indian numerology is a broad term that incorporates Tamil and Vedic practices to create a mystical alphanumeric system.
Below we examine three types of numerology and the practices within each branch.
### What are the Pythagorean master numbers?
Pythagorean master numbers mean your numerology chart is derived using a branch of numerology based on the teachings of the ancient Greek mathematician Pythagoras. Pythagorean numerology assigns a numerical value of 1-9 to letters of the alphabet and uses these values to analyze and interpret various aspects of a person’s life, personality, and destiny. Pythagorean numerology considers 11, 22, and 33 the master numbers and stops reducing calculations if they appear in the chart. 11, 22, and 33 have the most influence because they are endowed with profound energies. Master numbers are present throughout Pythagorean numerological charts and have multiple calculations to draw from. For example, a master soul urge number calculates the vowels of a person’s name based on the Pythagorean alphanumeric code. Additionally, a master expression number derives a numerological calculation from a person’s full birth name. Each of the three master numbers has a different meaning within Pythagorean numerology. Firstly, master 11 in Pythagorean numerology means you have a strong desire to explore your spiritual growth and reach your awakening. Number 22 is a foundational step toward enlightenment and represents the first level of master number ascension. Secondly, master 22 in Pythagorean numerology means you have the karmic energy to manifest your life’s vision. Master number 22 is a sign of shifting realities that lead you closer to actualizing your destiny. Thirdly, master number 33 means you have an innate ability to promote positivity in other people’s lives. Pythagorean numerology considers 33 a high-frequency mystical symbol of spiritual fulfillment.
### What do master numbers mean in Chaldean numerology?
Master numbers in Chaldean numerology mean your number chart contains compound numbers such as 11/2, 22/4, and 33/6. Chaldean numerology is one of the oldest forms of numerology and is attributed to the ancient Chaldeans, the ancient people of Mesopotamia. Chaldean numerology assigns a numerical value of 1-8 to letters of the alphabet based on their sound vibrations rather than their sequential order. The Chaldean system analyzes names and birthdates to gain insights into a person’s personality, life’s journey, and destiny. The compound master numbers 11, 22, and 33 have separate attributes and purposes for your numerological chart in Chaldean numerology. Firstly, master number 11 in Chaldean numerology is highly karmic or vibrational. The number 11 is associated with spiritual insight, intuition, and illumination. People with the number 11 in their numerology chart are believed to have a heightened sense of spiritual awareness and possess unique gifts of intuition. Secondly, the master number 22 in Chaldean numerology represents the potential for manifesting dreams and turning ideas into reality. Individuals with the number 22 are seen as capable of achieving great things in the material world through practical and visionary means. 22 suggests a strong sense of purpose and the ability to bring about positive change. Thirdly, master number 33 in Chaldean numerology possesses spiritual wisdom and compassion. 33 signifies a deep sense of caring for others and a strong connection to higher spiritual realms.
### What do master numbers mean in Indian numerology?
Master numbers in Indian numerology mean your numerology chart incorporates the teachings of Tamil numerology. Tamil numerology originates from the Tamil Nadu region of India. The Tamil people use a script that consists of 247 letters of 12 vowels, 18 consonants, 216 vowel and consonant combinations, and 1 ayutha ezhuthu that correlates to numbers 1-9. Master numbers in Tamil numerology mean you’re operating at a higher level of consciousness and are near a spiritual awakening. Master numbers 11, 22, and 33 have separate meanings according to Tamil numerology. Firstly, master number 11 in Tamil numerology is associated with heightened intuition and a strong connection to higher consciousness. Number 11 in your Tamil chart suggests spiritual growth and self-discovery. Secondly, master number 22 in Tamil means you have the potential to manifest your ideal desires. 22 signifies a strong sense of purpose and focus on achievements in the material and spiritual realms. Thirdly, master number 33 in Tamil numerology represents acts of healing and selfless service. 33 signifies a deep desire to uplift humanity and a strong connection to universal love and guidance.
## Are there master life path numbers?
Yes, the master life path numbers are 11, 22, and 33. The following descriptions provide insight into the life path master numbers. Firstly, master life path numbers are born with a birthdate that reduces to the number 11. Master 11 is associated with spiritual insight, intuition, and high sensitivity. Secondly, master life path number 22 people are born with a birthdate that reduces to the number 22. Master 22 signifies the ability to manifest dreams and ideals into reality on a large scale. People with life path 22 are practical visionaries who significantly contribute to the world. Thirdly, master life path number 33 stems from birthdates that reduce to 33. Master 33 represents a deep sense of compassion, healing, and a strong desire to serve others. Individuals with a 33 life path are spiritual leaders and mentors.
## What do master angel numbers mean?
Master angel numbers mean the numbers 11, 22, and 33 provide higher vibratory frequencies to angel number sequences. Angel number sequences are repeating numbers or patterns with spiritual or symbolic significance. Master numbers appear in angel number combinations and by reducing an angel number. For example, the angel number 1111 contains the master number 11. The angel number 911 reduces to 11 and contains the energy held within 11. Each master number has a high-frequency vibration that emphasizes the message of the angel number. The three master numbers have separate meanings within angel numbers. Firstly, master angel number 11 means you’re seeing the number because you need to stay on your current spiritual path and pay attention to your inner guidance. Secondly, master angel number 22 means your focus and determination lead to manifesting your dreams. You have to be patient and persevere to see your reality manifest. Thirdly, master angel number 33 means you’re receiving spiritual guidance and compassion. Continue helping others along their journey, and karma will return the support.
## What do master numbers mean for twin flames?
Master numbers for twin flames mean you’re heading toward a spiritual awakening and augmenting an unbreakable bond. Your twin flame is the other half that mirrors your soul and helps you grow throughout life. Each master number influences twin flames in three separate ways. Firstly, master number 11 for twin flames is a sign of spiritual awakening and alignment with your twin flame. Seeing master number 11 signifies that you and your twin flame are on a significant spiritual journey together. The reason you’re drawn to each other is that you must explore deeper levels of spiritual growth and awareness. Secondly, master number 22 is associated with manifesting dreams into reality. 22 indicates that your union has the potential to bring about profound positive changes not only in your lives but also in the world. You and your twin flame must work together to achieve significant goals and purposes. Thirdly, master number 33 is understood as a symbol of service and healing. Your twin flame connection is rooted in your ability to help and guide others. Your union is deeply healing, not just for you both but for those involved in your life.
## Are master numbers compatible?
Yes, all three master numbers are compatible with the others. Below are descriptions that explain the compatibility of the six master number combinations.
• 11 and 11: Two master number 11s create a deeply spiritual and harmonious relationship. However, they intensify each other’s tendencies towards anxiety or over-sensitivity. They should work to form a powerful spiritual bond with mutual understanding to support each other’s growth.
• 11 and 22: Master 11 and master 22 make a balanced pairing. The visionary drive of the 22 finds grounding with the intuitive guidance of the 11. While the 11 offers spiritual insights, the 22 brings practicality to a union of dreams and spiritual manifestation. However, effective communication is crucial to avoid misunderstandings.
• 11 and 33: The combination of the Master Psychic and Master Teacher is spiritually rich. 11 and 33 are deeply compassionate and understanding, making their bond nurturing. The 33 provides wisdom and guidance, while the 11 offers intuition and inspiration. Their shared spiritual journey is deeply fulfilling, but they must attend to practical matters distracting from their connection.
• 22 and 22: A pairing of two Master Builders is an extraordinary dynamic. Each master 22 shares ambitions and dreams that lead to significant achievements. There’s an understanding of each other’s drive and determination. However, they must be cautious about becoming overly materialistic or pushing each other too hard, leading to burnout.
• 22 and 33: The Master Builder and the Master Teacher form a relationship where dreams meet guidance. The 22 is grounded and ambitious, while the 33 brings wisdom and compassion. The combination of 22 and 33 leads to personal growth and tangible achievements. Mutual respect is essential for a 22 and 33 relationship to flourish.
• 33 and 33: Two master number 33s create a nurturing and enlightened relationship. 33 focuses on personal growth, healing, and teaching, which makes their bond deep and transformative. Each 33 uplifts and guides the other leading to a supportive and uplifting environment. However, 33s must ensure they don’t become overly wrapped in introspection and neglect the practicalities of life.
## Are master numbers rare?
Yes, master numbers are rare. Master numbers 11, 22, and 33 are rare in numerology because they are less likely to appear in numerological equations. For example, fewer birthdates reduce to master numbers as the numbers increase. The rarity of each number adds to the number’s meaning and significance. Firstly, master 11 is the least rare of the three master numbers. Seeing master 11 or having 11 in your chart indicates a rare sensitive intuition. Secondly, master 22 is the second most rare of the three master numbers. Connecting to the number 22 suggests a heightened ability to manifest your dream reality. Thirdly, master 33 is the rarest master number as birthdates seldom reduce to 33. Having 33 in your chart shows you have profound compassion, healing abilities, and a strong inclination toward selfless services.
## What is the most powerful number?
The most powerful number in numerology is the master number 33. There are three reasons why 33 is the most powerful number. Firstly, 33 is the rarest master number with a heightened spiritual vibration. Secondly, 33 is called the “Master Teacher” because it combines the passion of the number 3 with the nurturing spirit of 6. Individuals with a strong influence of 33 in their numerology charts use their creative passions to uplift those around them. Thirdly, number 33 represents selfless protective energy. People associated with the number 33 have a strong sense of devotion and determination to improve life for others. The effect 33 has on other people is a powerful example of 33’s outreach.
## Are master numbers old souls?
Yes, master numbers are old souls that have mastered qualities necessary for spiritual ascension. There are three reasons why master numbers are considered old souls. Firstly, master numbers have heightened spiritual awareness. Master numbers are associated with heightened spiritual awareness, intuition, and a strong connection to higher consciousness. Individuals who have old souls have a deep understanding of the universe and themselves. Old souls have transcended their previous lives and are continuing to work to achieve spiritual enlightenment and have a deep understanding of life’s spiritual aspects. Secondly, master numbers display the old soul characteristic of a defined life purpose. People with master numbers have a significant life purpose or mission. The life purpose aligns with the concept of old souls having a goal to fulfill in this lifetime based on their accumulated wisdom. Master numbers indicate a higher level of spiritual growth and potential in the current lifetime. Thirdly, master numbers are old souls because they have a history of unique challenges and insights. Master numbers bring unique challenges and opportunities for growth, which contribute to an individual’s wisdom and maturity. Old souls are characterized by their ability to navigate life’s challenges with grace and insight. For example, a master number frequently gives wise advice beyond their years.
## Why do I keep seeing master numbers?
You keep seeing master numbers because the universe has a specific message that needs to be brought to your attention. There are three main messages behind why you keep seeing master numbers. Firstly, your angels show you master numbers because you’re being told to focus on your spiritual awakening. Repeating master numbers symbolize spiritual awakenings and heightened awareness. Your higher self or spiritual guides are trying to get your attention to help you on your spiritual journey. Secondly, you keep seeing master numbers because you’re receiving guidance from the Divine. Master numbers are signs from the universe that you are on the right path or making the right decisions in your life. Seeing number sequences and master numbers is a way of receiving messages from a higher power, the universe, or divine beings. The numbers 11, 22, and 33 endorse positive humanitarian and entrepreneurial endeavors because you’re receiving added support from your angels who carry specific guidance or insights relating to your life. Thirdly, you keep seeing master numbers because you’re experiencing synchronicity. Synchronicity is experiencing events that feel meaningful but lack an obvious connection. For example, synchronicity is occurring if you repeatedly see the number 11. Repeating master numbers means external events align with your inner state and thoughts. Your thoughts and actions deeply connect to the spiritual realm during these serendipitous events. Master numbers remind you that your current choices greatly impact your life. | 4,804 | 24,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-33 | latest | en | 0.919201 |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/6/lesson/6.2.2/problem/6-73 | 1,716,218,541,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058291.13/warc/CC-MAIN-20240520142329-20240520172329-00292.warc.gz | 261,001,019 | 16,037 | ### Home > APCALC > Chapter 6 > Lesson 6.2.2 > Problem6-73
6-73.
The table below shows the number of gold medals the U.S. has received in the Summer Olympic Games between 1956 and 2012. Calculate the mean number (average) of gold medals the U.S. wins. Write a brief statement describing how you found the mean.
Year
# of Gold Medals
1956
$32$
1960
$34$
1964
$36$
1968
$45$
1972
$33$
1976
$34$
1984
$83$
Year
# of Gold Medals
1988
$36$
1992
$37$
1996
$44$
2000
$40$
2004
$35$
2008
$36$
2012
$46$
Divide the total number of medals by the total number of Olympic Games. | 200 | 601 | {"found_math": true, "script_math_tex": 14, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.764789 |
https://en.wikipedia.org/wiki/Wishart_distribution | 1,723,399,867,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641008125.69/warc/CC-MAIN-20240811172916-20240811202916-00315.warc.gz | 183,464,757 | 48,458 | # Wishart distribution
Notation X ~ Wp(V, n) n > p − 1 degrees of freedom (real)V > 0 scale matrix (p × p pos. def) X (p × p) positive definite matrix ${\displaystyle f_{\mathbf {X} }(\mathbf {X} )={\frac {|\mathbf {X} |^{(n-p-1)/2}e^{-\operatorname {tr} (\mathbf {V} ^{-1}\mathbf {X} )/2}}{2^{\frac {np}{2}}|{\mathbf {V} }|^{n/2}\Gamma _{p}({\frac {n}{2}})}}}$ Γp is the multivariate gamma function tr is the trace function ${\displaystyle \operatorname {E} [\mathbf {\mathbf {X} } ]=n{\mathbf {V} }}$ (n − p − 1)V for n ≥ p + 1 ${\displaystyle \operatorname {Var} (\mathbf {X} _{ij})=n\left(v_{ij}^{2}+v_{ii}v_{jj}\right)}$ see below ${\displaystyle \Theta \mapsto \left|{\mathbf {I} }-2i\,{\mathbf {\Theta } }{\mathbf {V} }\right|^{-{\frac {n}{2}}}}$
In statistics, the Wishart distribution is a generalization of the gamma distribution to multiple dimensions. It is named in honor of John Wishart, who first formulated the distribution in 1928.[1] Other names include Wishart ensemble (in random matrix theory, probability distributions over matrices are usually called "ensembles"), or Wishart–Laguerre ensemble (since its eigenvalue distribution involve Laguerre polynomials), or LOE, LUE, LSE (in analogy with GOE, GUE, GSE).[2]
It is a family of probability distributions defined over symmetric, positive-definite random matrices (i.e. matrix-valued random variables). These distributions are of great importance in the estimation of covariance matrices in multivariate statistics. In Bayesian statistics, the Wishart distribution is the conjugate prior of the inverse covariance-matrix of a multivariate-normal random-vector.[3]
## Definition
Suppose G is a p × n matrix, each column of which is independently drawn from a p-variate normal distribution with zero mean:
${\displaystyle G=(g_{i}^{1},\dots ,g_{i}^{n})\sim {\mathcal {N}}_{p}(0,V).}$
Then the Wishart distribution is the probability distribution of the p × p random matrix [4]
${\displaystyle S=GG^{T}=\sum _{i=1}^{n}g_{i}g_{i}^{T}}$
known as the scatter matrix. One indicates that S has that probability distribution by writing
${\displaystyle S\sim W_{p}(V,n).}$
The positive integer n is the number of degrees of freedom. Sometimes this is written W(V, p, n). For np the matrix S is invertible with probability 1 if V is invertible.
If p = V = 1 then this distribution is a chi-squared distribution with n degrees of freedom.
## Occurrence
The Wishart distribution arises as the distribution of the sample covariance matrix for a sample from a multivariate normal distribution. It occurs frequently in likelihood-ratio tests in multivariate statistical analysis. It also arises in the spectral theory of random matrices[citation needed] and in multidimensional Bayesian analysis.[5] It is also encountered in wireless communications, while analyzing the performance of Rayleigh fading MIMO wireless channels .[6]
## Probability density function
The Wishart distribution can be characterized by its probability density function as follows:
Let X be a p × p symmetric matrix of random variables that is positive semi-definite. Let V be a (fixed) symmetric positive definite matrix of size p × p.
Then, if np, X has a Wishart distribution with n degrees of freedom if it has the probability density function
${\displaystyle f_{\mathbf {X} }(\mathbf {X} )={\frac {1}{2^{np/2}\left|{\mathbf {V} }\right|^{n/2}\Gamma _{p}\left({\frac {n}{2}}\right)}}{\left|\mathbf {X} \right|}^{(n-p-1)/2}e^{-{\frac {1}{2}}\operatorname {tr} ({\mathbf {V} }^{-1}\mathbf {X} )}}$
where ${\displaystyle \left|{\mathbf {X} }\right|}$ is the determinant of ${\displaystyle \mathbf {X} }$ and Γp is the multivariate gamma function defined as
${\displaystyle \Gamma _{p}\left({\frac {n}{2}}\right)=\pi ^{p(p-1)/4}\prod _{j=1}^{p}\Gamma \left({\frac {n}{2}}-{\frac {j-1}{2}}\right).}$
The density above is not the joint density of all the ${\displaystyle p^{2}}$ elements of the random matrix X (such ${\displaystyle p^{2}}$-dimensional density does not exist because of the symmetry constrains ${\displaystyle X_{ij}=X_{ji}}$), it is rather the joint density of ${\displaystyle p(p+1)/2}$ elements ${\displaystyle X_{ij}}$ for ${\displaystyle i\leq j}$ (,[1] page 38). Also, the density formula above applies only to positive definite matrices ${\displaystyle \mathbf {x} ;}$ for other matrices the density is equal to zero.
### Spectral density
The joint-eigenvalue density for the eigenvalues ${\displaystyle \lambda _{1},\dots ,\lambda _{p}\geq 0}$ of a random matrix ${\displaystyle \mathbf {X} \sim W_{p}(\mathbf {I} ,n)}$ is,[8][9]
${\displaystyle c_{n,p}e^{-{\frac {1}{2}}\sum _{i}\lambda _{i}}\prod \lambda _{i}^{(n-p-1)/2}\prod _{i
where ${\displaystyle c_{n,p}}$is a constant.
In fact the above definition can be extended to any real n > p − 1. If np − 1, then the Wishart no longer has a density—instead it represents a singular distribution that takes values in a lower-dimension subspace of the space of p × p matrices.[10]
## Use in Bayesian statistics
In Bayesian statistics, in the context of the multivariate normal distribution, the Wishart distribution is the conjugate prior to the precision matrix Ω = Σ−1, where Σ is the covariance matrix.[11]: 135 [12]
### Choice of parameters
The least informative, proper Wishart prior is obtained by setting n = p.[citation needed]
The prior mean of Wp(V, n) is nV, suggesting that a reasonable choice for V would be n−1Σ0−1, where Σ0 is some prior guess for the covariance matrix.
## Properties
### Log-expectation
The following formula plays a role in variational Bayes derivations for Bayes networks involving the Wishart distribution. From equation (2.63),[13]
${\displaystyle \operatorname {E} [\,\ln \left|\mathbf {X} \right|\,]=\psi _{p}\left({\frac {n}{2}}\right)+p\,\ln(2)+\ln |\mathbf {V} |}$
where ${\displaystyle \psi _{p}}$ is the multivariate digamma function (the derivative of the log of the multivariate gamma function).
### Log-variance
The following variance computation could be of help in Bayesian statistics:
${\displaystyle \operatorname {Var} \left[\,\ln \left|\mathbf {X} \right|\,\right]=\sum _{i=1}^{p}\psi _{1}\left({\frac {n+1-i}{2}}\right)}$
where ${\displaystyle \psi _{1}}$ is the trigamma function. This comes up when computing the Fisher information of the Wishart random variable.
### Entropy
The information entropy of the distribution has the following formula:[11]: 693
${\displaystyle \operatorname {H} \left[\,\mathbf {X} \,\right]=-\ln \left(B(\mathbf {V} ,n)\right)-{\frac {n-p-1}{2}}\operatorname {E} \left[\,\ln \left|\mathbf {X} \right|\,\right]+{\frac {np}{2}}}$
where B(V, n) is the normalizing constant of the distribution:
${\displaystyle B(\mathbf {V} ,n)={\frac {1}{\left|\mathbf {V} \right|^{n/2}2^{np/2}\Gamma _{p}\left({\frac {n}{2}}\right)}}.}$
This can be expanded as follows:
{\displaystyle {\begin{aligned}\operatorname {H} \left[\,\mathbf {X} \,\right]&={\frac {n}{2}}\ln \left|\mathbf {V} \right|+{\frac {np}{2}}\ln 2+\ln \Gamma _{p}\left({\frac {n}{2}}\right)-{\frac {n-p-1}{2}}\operatorname {E} \left[\,\ln \left|\mathbf {X} \right|\,\right]+{\frac {np}{2}}\\[8pt]&={\frac {n}{2}}\ln \left|\mathbf {V} \right|+{\frac {np}{2}}\ln 2+\ln \Gamma _{p}\left({\frac {n}{2}}\right)-{\frac {n-p-1}{2}}\left(\psi _{p}\left({\frac {n}{2}}\right)+p\ln 2+\ln \left|\mathbf {V} \right|\right)+{\frac {np}{2}}\\[8pt]&={\frac {n}{2}}\ln \left|\mathbf {V} \right|+{\frac {np}{2}}\ln 2+\ln \Gamma _{p}\left({\frac {n}{2}}\right)-{\frac {n-p-1}{2}}\psi _{p}\left({\frac {n}{2}}\right)-{\frac {n-p-1}{2}}\left(p\ln 2+\ln \left|\mathbf {V} \right|\right)+{\frac {np}{2}}\\[8pt]&={\frac {p+1}{2}}\ln \left|\mathbf {V} \right|+{\frac {1}{2}}p(p+1)\ln 2+\ln \Gamma _{p}\left({\frac {n}{2}}\right)-{\frac {n-p-1}{2}}\psi _{p}\left({\frac {n}{2}}\right)+{\frac {np}{2}}\end{aligned}}}
### Cross-entropy
The cross-entropy of two Wishart distributions ${\displaystyle p_{0}}$ with parameters ${\displaystyle n_{0},V_{0}}$ and ${\displaystyle p_{1}}$ with parameters ${\displaystyle n_{1},V_{1}}$ is
{\displaystyle {\begin{aligned}H(p_{0},p_{1})&=\operatorname {E} _{p_{0}}[\,-\log p_{1}\,]\\[8pt]&=\operatorname {E} _{p_{0}}\left[\,-\log {\frac {\left|\mathbf {X} \right|^{(n_{1}-p_{1}-1)/2}e^{-\operatorname {tr} (\mathbf {V} _{1}^{-1}\mathbf {X} )/2}}{2^{n_{1}p_{1}/2}\left|\mathbf {V} _{1}\right|^{n_{1}/2}\Gamma _{p_{1}}\left({\tfrac {n_{1}}{2}}\right)}}\right]\\[8pt]&={\tfrac {n_{1}p_{1}}{2}}\log 2+{\tfrac {n_{1}}{2}}\log \left|\mathbf {V} _{1}\right|+\log \Gamma _{p_{1}}({\tfrac {n_{1}}{2}})-{\tfrac {n_{1}-p_{1}-1}{2}}\operatorname {E} _{p_{0}}\left[\,\log \left|\mathbf {X} \right|\,\right]+{\tfrac {1}{2}}\operatorname {E} _{p_{0}}\left[\,\operatorname {tr} \left(\,\mathbf {V} _{1}^{-1}\mathbf {X} \,\right)\,\right]\\[8pt]&={\tfrac {n_{1}p_{1}}{2}}\log 2+{\tfrac {n_{1}}{2}}\log \left|\mathbf {V} _{1}\right|+\log \Gamma _{p_{1}}({\tfrac {n_{1}}{2}})-{\tfrac {n_{1}-p_{1}-1}{2}}\left(\psi _{p_{0}}({\tfrac {n_{0}}{2}})+p_{0}\log 2+\log \left|\mathbf {V} _{0}\right|\right)+{\tfrac {1}{2}}\operatorname {tr} \left(\,\mathbf {V} _{1}^{-1}n_{0}\mathbf {V} _{0}\,\right)\\[8pt]&=-{\tfrac {n_{1}}{2}}\log \left|\,\mathbf {V} _{1}^{-1}\mathbf {V} _{0}\,\right|+{\tfrac {p_{1}+1}{2}}\log \left|\mathbf {V} _{0}\right|+{\tfrac {n_{0}}{2}}\operatorname {tr} \left(\,\mathbf {V} _{1}^{-1}\mathbf {V} _{0}\right)+\log \Gamma _{p_{1}}\left({\tfrac {n_{1}}{2}}\right)-{\tfrac {n_{1}-p_{1}-1}{2}}\psi _{p_{0}}({\tfrac {n_{0}}{2}})+{\tfrac {n_{1}(p_{1}-p_{0})+p_{0}(p_{1}+1)}{2}}\log 2\end{aligned}}}
Note that when ${\displaystyle p_{0}=p_{1}}$ and ${\displaystyle n_{0}=n_{1}}$we recover the entropy.
### KL-divergence
The Kullback–Leibler divergence of ${\displaystyle p_{1}}$ from ${\displaystyle p_{0}}$ is
{\displaystyle {\begin{aligned}D_{KL}(p_{0}\|p_{1})&=H(p_{0},p_{1})-H(p_{0})\\[6pt]&=-{\frac {n_{1}}{2}}\log |\mathbf {V} _{1}^{-1}\mathbf {V} _{0}|+{\frac {n_{0}}{2}}(\operatorname {tr} (\mathbf {V} _{1}^{-1}\mathbf {V} _{0})-p)+\log {\frac {\Gamma _{p}\left({\frac {n_{1}}{2}}\right)}{\Gamma _{p}\left({\frac {n_{0}}{2}}\right)}}+{\tfrac {n_{0}-n_{1}}{2}}\psi _{p}\left({\frac {n_{0}}{2}}\right)\end{aligned}}}
### Characteristic function
The characteristic function of the Wishart distribution is
${\displaystyle \Theta \mapsto \operatorname {E} \left[\,\exp \left(\,i\operatorname {tr} \left(\,\mathbf {X} {\mathbf {\Theta } }\,\right)\,\right)\,\right]=\left|\,1-2i\,{\mathbf {\Theta } }\,{\mathbf {V} }\,\right|^{-n/2}}$
where E[⋅] denotes expectation. (Here Θ is any matrix with the same dimensions as V, 1 indicates the identity matrix, and i is a square root of −1).[9] Properly interpreting this formula requires a little care, because noninteger complex powers are multivalued; when n is noninteger, the correct branch must be determined via analytic continuation.[14]
## Theorem
If a p × p random matrix X has a Wishart distribution with m degrees of freedom and variance matrix V — write ${\displaystyle \mathbf {X} \sim {\mathcal {W}}_{p}({\mathbf {V} },m)}$ — and C is a q × p matrix of rank q, then [15]
${\displaystyle \mathbf {C} \mathbf {X} {\mathbf {C} }^{T}\sim {\mathcal {W}}_{q}\left({\mathbf {C} }{\mathbf {V} }{\mathbf {C} }^{T},m\right).}$
### Corollary 1
If z is a nonzero p × 1 constant vector, then:[15]
${\displaystyle \sigma _{z}^{-2}\,{\mathbf {z} }^{T}\mathbf {X} {\mathbf {z} }\sim \chi _{m}^{2}.}$
In this case, ${\displaystyle \chi _{m}^{2}}$ is the chi-squared distribution and ${\displaystyle \sigma _{z}^{2}={\mathbf {z} }^{T}{\mathbf {V} }{\mathbf {z} }}$ (note that ${\displaystyle \sigma _{z}^{2}}$ is a constant; it is positive because V is positive definite).
### Corollary 2
Consider the case where zT = (0, ..., 0, 1, 0, ..., 0) (that is, the j-th element is one and all others zero). Then corollary 1 above shows that
${\displaystyle \sigma _{jj}^{-1}\,w_{jj}\sim \chi _{m}^{2}}$
gives the marginal distribution of each of the elements on the matrix's diagonal.
George Seber points out that the Wishart distribution is not called the “multivariate chi-squared distribution” because the marginal distribution of the off-diagonal elements is not chi-squared. Seber prefers to reserve the term multivariate for the case when all univariate marginals belong to the same family.[16]
## Estimator of the multivariate normal distribution
The Wishart distribution is the sampling distribution of the maximum-likelihood estimator (MLE) of the covariance matrix of a multivariate normal distribution.[17] A derivation of the MLE uses the spectral theorem.
## Bartlett decomposition
The Bartlett decomposition of a matrix X from a p-variate Wishart distribution with scale matrix V and n degrees of freedom is the factorization:
${\displaystyle \mathbf {X} ={\textbf {L}}{\textbf {A}}{\textbf {A}}^{T}{\textbf {L}}^{T},}$
where L is the Cholesky factor of V, and:
${\displaystyle \mathbf {A} ={\begin{pmatrix}c_{1}&0&0&\cdots &0\\n_{21}&c_{2}&0&\cdots &0\\n_{31}&n_{32}&c_{3}&\cdots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\n_{p1}&n_{p2}&n_{p3}&\cdots &c_{p}\end{pmatrix}}}$
where ${\displaystyle c_{i}^{2}\sim \chi _{n-i+1}^{2}}$ and nij ~ N(0, 1) independently.[18] This provides a useful method for obtaining random samples from a Wishart distribution.[19]
## Marginal distribution of matrix elements
Let V be a 2 × 2 variance matrix characterized by correlation coefficient −1 < ρ < 1 and L its lower Cholesky factor:
${\displaystyle \mathbf {V} ={\begin{pmatrix}\sigma _{1}^{2}&\rho \sigma _{1}\sigma _{2}\\\rho \sigma _{1}\sigma _{2}&\sigma _{2}^{2}\end{pmatrix}},\qquad \mathbf {L} ={\begin{pmatrix}\sigma _{1}&0\\\rho \sigma _{2}&{\sqrt {1-\rho ^{2}}}\sigma _{2}\end{pmatrix}}}$
Multiplying through the Bartlett decomposition above, we find that a random sample from the 2 × 2 Wishart distribution is
${\displaystyle \mathbf {X} ={\begin{pmatrix}\sigma _{1}^{2}c_{1}^{2}&\sigma _{1}\sigma _{2}\left(\rho c_{1}^{2}+{\sqrt {1-\rho ^{2}}}c_{1}n_{21}\right)\\\sigma _{1}\sigma _{2}\left(\rho c_{1}^{2}+{\sqrt {1-\rho ^{2}}}c_{1}n_{21}\right)&\sigma _{2}^{2}\left(\left(1-\rho ^{2}\right)c_{2}^{2}+\left({\sqrt {1-\rho ^{2}}}n_{21}+\rho c_{1}\right)^{2}\right)\end{pmatrix}}}$
The diagonal elements, most evidently in the first element, follow the χ2 distribution with n degrees of freedom (scaled by σ2) as expected. The off-diagonal element is less familiar but can be identified as a normal variance-mean mixture where the mixing density is a χ2 distribution. The corresponding marginal probability density for the off-diagonal element is therefore the variance-gamma distribution
${\displaystyle f(x_{12})={\frac {\left|x_{12}\right|^{\frac {n-1}{2}}}{\Gamma \left({\frac {n}{2}}\right){\sqrt {2^{n-1}\pi \left(1-\rho ^{2}\right)\left(\sigma _{1}\sigma _{2}\right)^{n+1}}}}}\cdot K_{\frac {n-1}{2}}\left({\frac {\left|x_{12}\right|}{\sigma _{1}\sigma _{2}\left(1-\rho ^{2}\right)}}\right)\exp {\left({\frac {\rho x_{12}}{\sigma _{1}\sigma _{2}(1-\rho ^{2})}}\right)}}$
where Kν(z) is the modified Bessel function of the second kind.[20] Similar results may be found for higher dimensions. In general, if ${\displaystyle X}$ follows a Wishart distribution with parameters, ${\displaystyle \Sigma ,n}$, then for ${\displaystyle i\neq j}$, the off-diagonal elements
${\displaystyle X_{ij}\sim {\text{VG}}(n,\Sigma _{ij},(\Sigma _{ii}\Sigma _{jj}-\Sigma _{ij}^{2})^{1/2},0)}$. [21]
It is also possible to write down the moment-generating function even in the noncentral case (essentially the nth power of Craig (1936)[22] equation 10) although the probability density becomes an infinite sum of Bessel functions.
## The range of the shape parameter
It can be shown [23] that the Wishart distribution can be defined if and only if the shape parameter n belongs to the set
${\displaystyle \Lambda _{p}:=\{0,\ldots ,p-1\}\cup \left(p-1,\infty \right).}$
This set is named after Gindikin, who introduced it[24] in the 1970s in the context of gamma distributions on homogeneous cones. However, for the new parameters in the discrete spectrum of the Gindikin ensemble, namely,
${\displaystyle \Lambda _{p}^{*}:=\{0,\ldots ,p-1\},}$
the corresponding Wishart distribution has no Lebesgue density.
## References
1. ^ a b Wishart, J. (1928). "The generalised product moment distribution in samples from a normal multivariate population". Biometrika. 20A (1–2): 32–52. doi:10.1093/biomet/20A.1-2.32. JFM 54.0565.02. JSTOR 2331939.
2. ^ Livan, Giacomo; Novaes, Marcel; Vivo, Pierpaolo (2018), Livan, Giacomo; Novaes, Marcel; Vivo, Pierpaolo (eds.), "Classical Ensembles: Wishart-Laguerre", Introduction to Random Matrices: Theory and Practice, SpringerBriefs in Mathematical Physics, Cham: Springer International Publishing, pp. 89–95, doi:10.1007/978-3-319-70885-0_13, ISBN 978-3-319-70885-0, retrieved 2023-05-17
3. ^ Koop, Gary; Korobilis, Dimitris (2010). "Bayesian Multivariate Time Series Methods for Empirical Macroeconomics". Foundations and Trends in Econometrics. 3 (4): 267–358. doi:10.1561/0800000013.
4. ^ Gupta, A. K.; Nagar, D. K. (2000). Matrix Variate Distributions. Chapman & Hall /CRC. ISBN 1584880465.
5. ^ Gelman, Andrew (2003). Bayesian Data Analysis (2nd ed.). Boca Raton, Fla.: Chapman & Hall. p. 582. ISBN 158488388X. Retrieved 3 June 2015.
6. ^ Zanella, A.; Chiani, M.; Win, M.Z. (April 2009). "On the marginal distribution of the eigenvalues of wishart matrices" (PDF). IEEE Transactions on Communications. 57 (4): 1050–1060. doi:10.1109/TCOMM.2009.04.070143. hdl:1721.1/66900. S2CID 12437386.
7. ^ Livan, Giacomo; Vivo, Pierpaolo (2011). "Moments of Wishart-Laguerre and Jacobi ensembles of random matrices: application to the quantum transport problem in chaotic cavities". Acta Physica Polonica B. 42 (5): 1081. arXiv:1103.2638. doi:10.5506/APhysPolB.42.1081. ISSN 0587-4254. S2CID 119599157.
8. ^ Muirhead, Robb J. (2005). Aspects of Multivariate Statistical Theory (2nd ed.). Wiley Interscience. ISBN 0471769851.
9. ^ a b Anderson, T. W. (2003). An Introduction to Multivariate Statistical Analysis (3rd ed.). Hoboken, N. J.: Wiley Interscience. p. 259. ISBN 0-471-36091-0.
10. ^ Uhlig, H. (1994). "On Singular Wishart and Singular Multivariate Beta Distributions". The Annals of Statistics. 22: 395–405. doi:10.1214/aos/1176325375.
11. ^ a b c Bishop, C. M. (2006). Pattern Recognition and Machine Learning. Springer.
12. ^ Hoff, Peter D. (2009). A First Course in Bayesian Statistical Methods. New York: Springer. pp. 109–111. ISBN 978-0-387-92299-7.
13. ^ Nguyen, Duy. "AN IN DEPTH INTRODUCTION TO VARIATIONAL BAYES NOTE". Retrieved 15 August 2023.
14. ^ Mayerhofer, Eberhard (2019-01-27). "Reforming the Wishart characteristic function". arXiv:1901.09347 [math.PR].
15. ^ a b Rao, C. R. (1965). Linear Statistical Inference and its Applications. Wiley. p. 535.
16. ^ Seber, George A. F. (2004). Multivariate Observations. Wiley. ISBN 978-0471691211.
17. ^ Chatfield, C.; Collins, A. J. (1980). Introduction to Multivariate Analysis. London: Chapman and Hall. pp. 103–108. ISBN 0-412-16030-7.
18. ^ Anderson, T. W. (2003). An Introduction to Multivariate Statistical Analysis (3rd ed.). Hoboken, N. J.: Wiley Interscience. p. 257. ISBN 0-471-36091-0.
19. ^ Smith, W. B.; Hocking, R. R. (1972). "Algorithm AS 53: Wishart Variate Generator". Journal of the Royal Statistical Society, Series C. 21 (3): 341–345. JSTOR 2346290.
20. ^ Pearson, Karl; Jeffery, G. B.; Elderton, Ethel M. (December 1929). "On the Distribution of the First Product Moment-Coefficient, in Samples Drawn from an Indefinitely Large Normal Population". Biometrika. 21 (1/4). Biometrika Trust: 164–201. doi:10.2307/2332556. JSTOR 2332556.
21. ^ Fischer, Adrian; Gaunt, Robert E.; Andrey, Sarantsev. "The Variance-Gamma Distribution: A Review". ArXiv. Retrieved 28 June 2024.
22. ^ Craig, Cecil C. (1936). "On the Frequency Function of xy". Ann. Math. Statist. 7: 1–15. doi:10.1214/aoms/1177732541.
23. ^ Peddada and Richards, Shyamal Das; Richards, Donald St. P. (1991). "Proof of a Conjecture of M. L. Eaton on the Characteristic Function of the Wishart Distribution". Annals of Probability. 19 (2): 868–874. doi:10.1214/aop/1176990455.
24. ^ Gindikin, S.G. (1975). "Invariant generalized functions in homogeneous domains". Funct. Anal. Appl. 9 (1): 50–52. doi:10.1007/BF01078179. S2CID 123288172.
25. ^ Dwyer, Paul S. (1967). "Some Applications of Matrix Derivatives in Multivariate Analysis". J. Amer. Statist. Assoc. 62 (318): 607–625. doi:10.1080/01621459.1967.10482934. JSTOR 2283988. | 6,915 | 20,575 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 61, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.696879 |
https://www.convertunits.com/from/pace+%5Bgreat%5D/to/kilometer | 1,601,080,405,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00630.warc.gz | 779,547,307 | 11,827 | ## ››Convert pace [great] to kilometre
pace [great] kilometer
Did you mean to convert pace [great] pace [Roman] to kilometer
## ››More information from the unit converter
How many pace [great] in 1 kilometer? The answer is 656.16797900262.
We assume you are converting between pace [great] and kilometre.
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pace [great] or kilometer
The SI base unit for length is the metre.
1 metre is equal to 0.65616797900262 pace [great], or 0.001 kilometer.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between pace [great] and kilometres.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of pace [great] to kilometer
1 pace [great] to kilometer = 0.00152 kilometer
10 pace [great] to kilometer = 0.01524 kilometer
50 pace [great] to kilometer = 0.0762 kilometer
100 pace [great] to kilometer = 0.1524 kilometer
200 pace [great] to kilometer = 0.3048 kilometer
500 pace [great] to kilometer = 0.762 kilometer
1000 pace [great] to kilometer = 1.524 kilometer
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## ››Definition: Kilometer
A kilometre (American spelling: kilometer, symbol: km) is a unit of length equal to 1000 metres (from the Greek words khilia = thousand and metro = count/measure). It is approximately equal to 0.621 miles, 1094 yards or 3281 feet.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more! | 537 | 2,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-40 | latest | en | 0.790884 |
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# P32_040 - . 75 B max /B max = ( r 2 /R ) 1 , or r 2 = R/ ....
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40. (a) From Sample Problem 32-3 we know that B r for r R and B r 1 for r R .S ot h e maximum value of B occurs at r = R , and there are two possible values of r at which the magnetic ±eld is 75% of B max . W edenotethesetwova luesas r 1 and r 2 ,where r 1 <R and r 2 >R .Th en 0 . 75 B max /B max = r 1 /R ,or r
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Unformatted text preview: . 75 B max /B max = ( r 2 /R ) 1 , or r 2 = R/ . 75 = 1 . 3 R . (b) From Eqs. 32-39 and 32-41, B max = i d 2 R = i 2 R = ( 4 10 7 T m / A ) (6 . 0 A) 2 (0 . 040 m) = 3 . 10 5 T ....
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## Resistors in Parallel Page 1
This guide builds on our Resistors in Series guide. If you have not read that guide already, we suggest that you take a moment and look at it first. You can view our Resistors in Series guide here.
Resistors in Parallel
When you have two resistors in a circuit and they are tied together so that they receive the same voltage, they are said to be in parallel.
When you have resistors in parallel, they allow more current to pass through the circuit than if you have just one resistor. Think about cars on the freeway. If you have a narrow section of freeway, that is similar to resistance and less cars will be able to make it through that section. This is the same as a single resistor in a circuit.
Circuit with Single Path Traffic with Single Path
However, if you have more than one narrow section in parallel with each other the cars have a choice of which route they can take. Overall more cars are able to make it though the narrow sections. This is the same thing as resistors in parallel.
Circuit with Multiple Paths Traffic with Multiple Paths
Here you can see that more traffic is able to make it through when you have parallel roads. Like wise, more current can flow through the circuit when you have multiple resistors in parallel.
In the above schematic I1 represents the current through R1, while I2 represents the current through R2. This is to show that the current has more than one path to take.
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#### Written by Jason Bauer
Jason Bauer is an owner and programmer for Portforward.com. He's allergic to twitter and facebook, but you can find more of his articles in the Guides section.
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# Solar Week Tuesday: Guess the Celebrity
This is an online game associated with activities during Solar Week, a twice-yearly event in March and October during which classrooms are able to interact with scientists studying the Sun. Outside of Solar Week, information, activities, and... (View More)
# Quaky-Shaky Design Challenge
In this engineering challenge, student teams are introduced to the engineering design process, and then construct and test an earthquake-resistant structure. The lesson plan includes teacher support, student worksheets, multimedia assets, and links... (View More)
# Sounds of the Sun
This is an activity about the Doppler effect. Learners begin by simulating the noise made by a passing siren. After learning that the change in pitch results from movement, they investigate the definition of frequency, calculate change in frequency,... (View More)
# The Scale of Things
This is an activity about size and scale. Learners will create and walk through a distance scale model of the size of the Solar System. This activity requires a straight line distance of approximately 295 meters (300 yards).
Audience: Elementary school
# The Reasons for the Seasons
This is an activity about seasons. Learners begin by brainstorming a list of activities and events that occur in each season. Next, learners perform an experiment by comparing the temperature on thermometers left under a lamp for different lengths... (View More)
# Light, Directly
This is an activity about the concept of direct versus indirect sunlight. Learners construct and use a sun angle analyzer to investigate the effect of angle on area illuminated. The fraction of light on each square of the analyzer is then calculated... (View More)
# Adding the Moon: Using a Classroom Model to Explore the Movement of the Sun, Earth, and Moon
This is an activity about the motion of the Sun, Earth and Moon, specifically rotation and revolution. After identifying what they already know about the Sun, Earth and Moon, learners will observe and manipulate a styrofoam ball model of the Sun,... (View More)
Keywords: NSTA2013
Audience: Elementary school
# How Can the Little Moon Hide the Giant Sun? Exploring Size and Distance
This is an activity exploring the concept that distance affects how we perceive an object's size, specifically pertaining to the size of the Sun and the Moon as seen from Earth. Learners will complete a hands on activity where two balls of differing... (View More)
Keywords: NSTA2013
Audience: Elementary school
# Motion of the Sun and Earth: Using a Playground Model to Explore Rotation and Revolution
This is an activity about the rotation of the Earth and Sun, and the Earth's revolution around the Sun. In chalk, learners will draw the Sun-Earth system, complete with Earth's orbit, and then act out the rotation and revolution of a yearly cycle.... (View More)
Keywords: NSTA2013
Audience: Elementary school
# Motion of the Sun and Earth: Using a Classroom Model to Explore Rotation and Revolution
This is an activity about the motion of the Sun and the Earth, specifically rotation and revolution. After identifying what they already know about the Sun, learners will observe and manipulate a styrofoam ball model of the Sun and Earth. This... (View More)
Keywords: NSTA2013
Audience: Elementary school
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https://garagedoorrepairaurora.co/best-garage-door-maintenance-centennial-colorado-80155-call-now.html | 1,544,692,991,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824601.32/warc/CC-MAIN-20181213080138-20181213101638-00155.warc.gz | 622,240,348 | 12,787 | “Especially on houses where the garage is front and center, the garage door absolutely has to look good,” says Casey McGrath, a real estate practitioner in Kitsap County, Wash. And it has to operate smoothly: Americans use the garage more than any other entry to the house, including the front door, according to a survey commissioned by window and door manufacturer JELD-WEN.
In 1992 the United States Consumer Product Safety Commission released new rules for automatic garage door openers. Anything manufactured after 1993 was required to include either an electric eye (a pair of sensors that detect an object obstructing the doorway) or a wall-mounted control button that users hold down in order to close the door entirely. Most manufacturers opted for the electric eye method, sometimes referred to as safety sensors.
A spring design manual, also called a rate book, gives tables that relate the torque constant ("rate") and maximum turns for springs of given wire size, diameter, and length. For example, a typical page in a rate book would show a table for a given wire size and inside diameter, the maximum inch-pounds (MIP) of torque available for a standard lifetime of 10,000 cycles in that size, the weight of the spring per linear inch, and the rates of the spring (as IPPT, inch-pounds per turn) for each of various lengths. From these figures one can calculate the lifting capacity, substitutions, conversions, and cycle life upgrades for a door of given weight and drum geometry. The weight-lifting capacity of a given spring is calculated based on its torque constant (IPPT, or inch-pounds per turn), which is the rotational version of the spring constant that characterizes the spring. The IPPT constant is found from tables giving IPPT for given spring dimensions (wire-size/diameter/length). The same tables may indicate the maximum number of turns for various expected lifetimes in cycles. The torque required to balance a given door can be calculated from the weight of the door times the moment arm of the drums (as we do below under "Calculating the Forces We Will Be Handling"). The ultimate torque of the spring in the fully-wound condition is the number of turns (when fully-wound) times the IPPT constant. Choosing a spring to balance the door then simply requires matching the ultimate torque of the spring to the balancing torque.
Measurements: With the door in the down position, I measure a wire size of 0.273 inches, outside diameter of 2.0 inches, and overall length of 41.5 inches. Relaxing the spring shortens the length by about 7.5 coils of wire, so to estimate the relaxed length, we deduct the wire diamter of 0.273 inches times 7.5 from the 41.5 inch wound length, yielding an estimated relaxed length of 39.5 inches. The mean coil diameter is 2.0 - 0.273, or 1.73 inches. Perhaps this was actually a 40-inch-long spring with a 1.5 ID, 1.75 mean diameter, and 2.0 OD, but let's continue on calculating with the actually observed sizes. The number of coils in the relaxed spring is the relaxed length of 39.5 inches divided by the wire size of 0.273 inches, or about 145 coils. Deducting about 5 dead coils at the ends yields 140 active coils.
```The Genie Chain Drive garage door opener combines The Genie Chain Drive garage door opener combines powerful reliable performance with smooth operation to create a long-lasting and easy-to-install garage door opener. The unit's electronic push-button programming makes installation and setup fast and easy. Complete with accessories the system features a reliable DC motor that delivers quiet smooth operation ... More + Product Details Close
```
The open-ended work-order trick: You may be very surprised if you allow work to proceed without signing a work order with a specified price. Or, you may sign a work order, and think you're protected against open-ended wallet-reaching, only to find a much higher price due at the finish than you expected, because you signed a "parts as needed" order that got loaded up with a long list of parts (that likely were still in serviceable condition). You might have been quoted a price, but then get a bill for that price plus a lot more added for the "service call" and the "parts", and be told the quote was just for the labor. While this is the normal way of abusing your finances down at the hospital, you shouldn't agree to it for a garage door service call. These guys are not doctors.
Garage doors provide one of the major entryways into your home. Therefore, it is essential to ensure that they are secure and in good working order. If you are having problems with your current door, then you need to have a professional determine if it can be repaired or if it will need to be replaced. Here are some of the most common factors that will affect the cost of a garage door repair.
Screw - Screw-driven lift systems attach a long screw to the motor to move the garage door. These systems are powerful, fast, relatively quiet and require less maintenance than typical chain- or belt-driven systems. Genie and Overhead Door are two popular brands that make screw-driven systems. Based on our research of Genie’s line, you can expect to pay \$244 on average for a screw-driven garage door opener.
Measurements: With the door in the down position, I measure a wire size of 0.273 inches, outside diameter of 2.0 inches, and overall length of 41.5 inches. Relaxing the spring shortens the length by about 7.5 coils of wire, so to estimate the relaxed length, we deduct the wire diamter of 0.273 inches times 7.5 from the 41.5 inch wound length, yielding an estimated relaxed length of 39.5 inches. The mean coil diameter is 2.0 - 0.273, or 1.73 inches. Perhaps this was actually a 40-inch-long spring with a 1.5 ID, 1.75 mean diameter, and 2.0 OD, but let's continue on calculating with the actually observed sizes. The number of coils in the relaxed spring is the relaxed length of 39.5 inches divided by the wire size of 0.273 inches, or about 145 coils. Deducting about 5 dead coils at the ends yields 140 active coils.
We specialize in all varieties of garage door repair work. We have the tools and knowledge to get the job done safely. It can be somewhat dangerous to take on a garage door repair on your own, and most garage door companies will advise against attempting a DIY fix. There are some hazards to watch out for when working around these heavy and high tension doors. We recommend you give the experts at Girard's a call to address the issue in a safe and timely manner.
Since their invention in the 1920s, electric garage door openers have come a long way. Garage door openers work by using a trolley connected to an arm that attaches to the top of the garage door and slides back and forth on a track, which opens and closes the garage door. When operating the motor, a chain or belt turns and pulls the trolley along the track. A good garage door opener will have a horsepower of 1/2 HP, 3/4 HP, or 1-1/4 HP. Garage door openers have the ability to open and close a limited number of times in power outage emergencies. Security is something else to consider when purchasing an opener. It's helpful to have sensors that will stop the operation of the garage door when a person, vehicle, or other obstacle is in the way.
aluminium frames B&D B and D doors and openers BnD Colorbond Custom garage doors cyclone Danmar Doors Garage door insulation garage door insurance garage door motors garage door remote garage door repairs garage door seals garage doors perth garage door springs and cables garage door struts insulated garage doors keyless entry Knotwood lifting cables Merlin garage door motors Merlin garage door openers Perth perth garage door repairs perth garage doors perth garage door servicing replacement remote reprogram garage door remote roller doors sectional garage doors Steel-Line tension springs tracks and rollers wind struts
```If you have paired springs, you can take a shortcut here instead of using locking pliers. Simply apply a slight torsion to the bar by clamping one of the springs with an easy half-turn or so applied. This will hold the lift cables in slight tension while you wind the other spring. If you have a single spring design, you can't use this trick, and have to use the locking pliers.
```
We’ve been reviewing the best garage door openers for more than half a decade. This year we chose the Chamberlain Premium as our top pick. It’s an efficient, reliable garage door opener, and it comes with a backup battery system that works in a power outage. This model also has a preinstalled timer that closes your garage door automatically so you don't have to worry about whether you left the garage door open.
Direct Drive - Unlike chain- and belt-driven systems, where the motor remains stationary while moving a chain, a direct drive motor moves while the chain stays in place. Because of this, the only moving part is the gear the motor uses to move while lifting the door. To our knowledge, only one company makes direct drive garage door openers: Sommer. The company claims the motors are more efficient and quieter than any other type of opener. However, this is the most expensive type of opener we’ve found, costing around \$286 on average.
The door and tracks at this stage of the repair are in a minimum-energy condition. This is a good opportunity to work on any hinges, bearings, rollers, cables, or tracks that need tightening, repair, lubrication, or replacement. Again, these parts should be available from the spring source, and should be ordered based on a pre-inspection. Home-improvement stores carry some of these parts, but the type and quality may not be the best.
Call Girard's Garage Door Service, and one of our technicians will walk you through all the options to find the perfect garage door for your taste and budget. We work with the top manufacturers across the country to ensure the utmost safety and security for your family. We’re a volume dealer, so we’re able to beat the prices that most local companies charge. We offer excellent service at an affordable price. Check out our garage door options here.
I obtained my parts, as described on this page, from American Garage Door Supply Inc. (http://www.americandoorsupply.com/) They may have raised their small-order prices or imposed a minimum order since I ordered from them in 2002 at the prices in my essay above. The Web site offers a free catalog by PDF download or mail. Springs in 2005 were priced at \$2.25/lb.
Whatever problem you are having with your garage door or garage door opener, you can count on Aaron’s Garage Door Service to provide you with the best service in town. We are an honest company that has built the reputation of our quality service. We don’t deceive people by offering a free service call or discount coupons to get the job then jack up the prices at the end. You can always count on Aaron’s Garage Door Service when you have any type of garage door problem.
We were very pleased with the service we received. The receptionist that took our information over the phone was courteous and efficient. The young man, Nate, who came to the house was on time, respectful of our wants and needs, and very knowledgeable about the job. He was able to fix our door easily. Finally we have a garage door that works with a push of the button! We highly recommend this company for your garage door needs.read more
A torsion spring counterbalance system consists of one or two tightly wound up springs on a steel shaft with cable drums at both ends. The entire apparatus mounts on the header wall above the garage door and has three supports: a center bearing plate with a steel or nylon bearing and two end bearing plates at both ends. The springs themselves consist of the steel wire with a stationary cone at one end and a winding cone at the other end. The stationary cone is attached to the center bearing plate. The winding cone consists of holes every 90 degrees for winding the springs and two set screws to secure the springs to the shaft. Steel counterbalance cables run from the roller brackets at the bottom corners of the door to a notch in the cable drums. When the door is raised, the springs unwind and the stored tension lifts the door by turning the shaft, thus turning the cable drums, wrapping the cables around the grooves on the cable drums. When the door is lowered, the cables unwrap from the drums and the springs are rewound to full tension.[7]
```Horsepower: The horsepower measurement, often shortened to HP, describes the power the garage door opener motor has. A motor with a greater horsepower measurement will open and close the door more quickly, while also being able to handle larger and heavier doors. Motors between 1/2 HP and 1 HP are the most common for residential garages, FeldCo says.
```
Speed of a thrown winding bar:: The springs, being in balance with the door, effectively are able to launch a typical 150 lb door at 10.6 mph speed. An 18-inch long by 1/2-inch diameter steel winding bar happens to weigh about 1 pound. Since momentum is conserved, this 150:1 ratio in weight of the door to the winding bar means the fully-wound springs could potentially throw a winding bar at 10.6 mph * 150 = 1590 mph = 2332 ft/sec, assuming the energy were perfectly coupled and transferred. If the energy transfer were only 1/3 efficient, this would still be the 800 ft/sec speed of a typical pistol bullet. Except it is a foot-and-a-half metal spear, not a bullet.
Resetting the drums, if needed: If the drums were incorrectly set in their old positions, one must reset both drums in new positions on the shaft. This is complicated by the presence of old dimples in the torsion shaft from previous setting(s), which must be avoided lest they improperly influence the new setting of the drums. To begin this process of resetting the drums, the door must first be lowered and resting level on the floor, the spring(s) must be in the unwound condition with their set-screws loosened, and the lift cables wrapped around the drums. If for some reason the door does not rest level on the floor, such as the floor being uneven, then insert temporary shims between the door bottom and the floor to bring the door up to level. Loosen the set-screws on the drums, and turn the torsion shaft to avoid the old dimples from the set-screws in the old drum position. Tighten the set-screw on the left drum (that is, on your left as you face the door from in the garage), creating a new dimple, and apply tension to its cable with the locking-pliers technique, enough tension to keep the cable taut but not enough to start to move the door up. Attach and wind the cable on the opposite (right) drum by hand until the cable is similarly taut, and set the screw, remembering that tightening the screw will tend to add a bit of extra tension to the cable. Both drums should now be fixed on the torsion shaft, with the cables about equally taut (listen to the sound when you pluck them like a guitar string) and the door still level on the ground. Setting the left drum first, and the right drum second, will allow you to take up any slack in the cable introduced by the left drum rotating slightly with respect to the torsion shaft as you tighten the set screws. This alignment and balance of the cables, drums, and door is critical to smooth operation and proper closing. If you have a single-spring assembly, the distance along the torsion tube from the spring cone to one drum is longer than to the other drum, which allows a bit more twist to one side than the other, and you may have to compensate with the setting of the drums.
Horsepower You’ll also want to consider the horsepower of your opener. Typically, 1/2 horsepower will work for most doors in newer homes. If you have a heavier solid wood door, though, you may want to opt for 3/4 horsepower so it can lift it with ease. You can also get 1 and 1 1/2 horsepower models, but those are a better fit for commercial or special situations.
Drive type Garage door openers work in a variety of ways. A chain-drive model raises and lowers doors with a metal chain — these tend to be inexpensive and able to handle heavier doors, but they’re loud. A belt-drive opener uses a belt and operates more smoothly and quietly. Screw-drive openers, which lift doors with a threaded steel rod, are also quiet and don’t require much maintenance. Finally, a direct-drive opener operates as a single unit — the whole contraption moves to lift the door. These tend to be the quietest, smoothest mechanisms available.
The optician's trick: The serviceman looks over your door with lots of scowling, chin-scratching, and tsk-tsking. You ask, "how much?" He replies with the fair price. If you don't flinch at that price, he says, "for the parts", while quoting a large additional cost for the labor. If you still don't flinch, he adds, "each," while pointing back and forth to your pair of springs. (I hope none of you service people are reading this!) I call this the "optician's trick" after the old vaudeville joke about lenses, frames, and left/right.
When you’re thinking about replacing your garage door, there are many options to consider. Picking out materials and styles is sometimes stressful for homeowners. Neighborhood Garage Door Repair has some helpful tips on choosing the right material for your door. Garage Door Materials Manufacturers offer different materials when it comes to designing a garage door. […]
Drive type Garage door openers work in a variety of ways. A chain-drive model raises and lowers doors with a metal chain — these tend to be inexpensive and able to handle heavier doors, but they’re loud. A belt-drive opener uses a belt and operates more smoothly and quietly. Screw-drive openers, which lift doors with a threaded steel rod, are also quiet and don’t require much maintenance. Finally, a direct-drive opener operates as a single unit — the whole contraption moves to lift the door. These tend to be the quietest, smoothest mechanisms available.
At this point I weighed the unlifted door to confirm and fine-tune my calculations. This is not strictly necessary, but it makes the adjustments easier to perform, if you happen to have a scale with the requisite capacity. With some helpers, we first lifted the door a few inches and rested it on blocks of wood to provide clearance underneath. Then I slid a 400-pound-capacity freight scale under the center of the door, we lifted again to remove the blocks, and lowered the door gently onto the scale. This door weighed in at 238 pounds, which is very heavy for a single-car door. Since the outside of the door carries the 3/4-inch plywood paneling to match the house, and that plywood weighs about 2 lbs/sq-ft, I estimate the door weight to be about 7 x 10 x 2 = 140 lbs of paneling with the rest 238 - 140 = 98 lbs the interior panels, hardware, and cobwebs. Knowing this total weight will help later in adjusting the torsion on the springs. After weighing, we removed the scale and blocks, leaving the door fully lowered again. Had I not had a high-capacity freight scale, I might have improvised a crude weighing device from levers and smaller weights of known mass, or a lever arm pressing a reduced proportion of the full weight onto a lower-capacity scale. Another factor to remember is that The weight of a wood door can vary with humidity.
A knowledgeable installer with good inventory can offer you upgraded spring lifetimes by using longer, heavier springs than were originally installed. For example, you may be offered more expensive springs with expected lifetimes of 15, 25, or even 100 thousand cycles, instead of the standard 10 thousand. The difference in labor to substitute this upgrade is nil. Since the dealer's cost of springs is proportional to weight, and typically a small part of the job price anyway, the dealer's cost for this upgrade is slight. This would seem to be a excellent option to offer every customer, and if correctly calculated and reasonably priced, one that you should take as cost-effective. Yet you may not be offered such an upgrade, if the installer is not adept at making the rather simple calculations, or if the optional springs are not on his truck, or if you're not around to be asked, or if the installer just doesn't like selling or taking time to discuss such details.
The technician that came out, Todd Noel, was very friendly and professional. He fixed my problem quickly and did not overcharge me for something that I did not need. He also made some great recommendations for future work. Because of his level of service and my customer experience, I highly recommend Precision and will use them again in the future.
The garage door opener traditionally has been a simple device. And it still is for the most part. But as with a lot of household items, technological advances are finding their way into garage door opener hardware. There are quite a few cool features in modern garage door openers, some of which are things you never knew you needed to automate the process of opening and closing a garage door.
Smartphone Control: Many newer garage door openers allow you to connect the device to your home's Wi-Fi network, according to RW Garage Doors. You then can open and close the door through a smartphone app. In fact, many of these apps will give you an alert on the smartphone when the garage door is open for a certain period. Some opener models even can connect to your existing Smart Home system, incorporating all your appliances on one system.
Thanks for considering our wide array of Overhead Door™ residential, commercial and industrial doors, openers, operators and accessories to complete your home, business or remodeling project. We are confident that you will find a door and opener product to meet your needs. After all, we lead the industry with our comprehensive selection of residential and commercial door systems. Overhead Door Corporation is a proud member of DASMA.
To estimate the maximum physical force required to wind these springs, consider that they are balancing the weight of the door with a torque applied to a lift drum on each end of the torsion shaft. The lift drums have a 2-inch radius, which is the standard residential size, and corresponds conveniently to about a 1-foot circumference. If we pessimistically assume the 10-by-7-foot door has a weight of 350 pounds, this implies a torque of 350 pounds on a 2-inch radius, that is, 700 inch-pounds, or 58 foot-pounds. Each of the two springs should be exerting slightly less than half of the balancing torque, or 29 foot-pounds. Compare this to, say, the bolts in an automobile, which are typically torqued to values of about 50 foot-pounds, or tire lug nuts, which may be torqued to well over 100 foot-pounds.
Chris was very professional & thorough. He arrived on time & was able to answer all questions that I had. I was extremely satisfied with the quality of the work he performed. I spoke several times on the phone with Mrs. Carol who keep me informed prior to, during, and after the installation of my garage door. She was very professional, easy to talk with and resolved any misunderstanding that I had.
## We lead busy and sometimes hectic lives. It's easy for someone low on sleep and high on stress to accidently bump into their garage door leaving a little damage. If you're lucky it's just an individual garage door panel that's been damaged. This will leave you with the option of garage door panel replacment. There's a chance it might cost you upwards of \$150-200 to spot repair damaged garage door panel. If it happens to be an in-production model, they probably will be able to repair or replace the entire panel (if needed) for \$250-400. Unfortunately if you've got an older model of garage door the panels may no longer be in production. This might seem like a disadvantage because you'll end up having to just replace the entire door. Many times it's actually easier and more cost effective to just replace the garage door.
```Together with our alliances, we proudly serve all of Australia, including Adelaide, Brisbane, Gold Coast, Melbourne, Perth and Sydney, with top quality sectional garage doors, roller doors, steel garage doors, remote garage doors, tilt garage doors, timber garage doors as well as commercial garage doors, commercial shutters and industrial shutters. If you need custom garage doors, we’re the ones you need to call. Our manufacturers have been selected for their capacity to meet and exceed our requirements for superior craftsmanship, reliability and a single-minded focus on excellence in customer service.
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Whether you're planning on moving in the future or you're fixing up your forever home, replacing a garage door can provide an excellent return on investment and boost your home's resale value. According to Remodeling magazine's 2016 Cost vs. Value report, garage door upgrades provide a nationwide average return on investment of over 90%, making it one of the top five home improvement projects.
```The chain-driven Chamberlain C410's .5-horsepower motor has enough power to handle most residential garage doors. Owners say it lifts even heavy doors easily. While chain-driven openers generally are louder than belt-driven models, most say that, like its predecessor model, it's actually pretty good in that regard, and we saw fewer complaints about installation challenges, too. The opener is also well-equipped in the feature department, though you'll need to spend a bit extra if you want to use its smart functions.
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Garage door repair is a specialized job, which is typically handled by a garage door repair service. Professional garage door repair technicians can test or repair a garage door system and fix cosmetic blemishes on doors. Common requests include help with jammed or inoperable doors, slow or erratic doors, unusual sounds, dents or scrapes on the door, and general system testing. Garage door repair professionals can work on single-car, double-car and RV-size garage doors.
With all the styles offered by Clopay®, many homeowners fall in love with multiple doors. To help you choose the best garage door for your home, Clopay® has developed a comprehensive buying guide, outlining styles, construction materials, insulation, wind resistance, automatic garage door openers and more. We also allow you to view different door styles on your home with our Door Imagination System. Simply upload your home or choose from a variety of sample homes with one-car, two-car, and three-car garages to design the best garage door for your home. | 5,707 | 26,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.93557 |
http://www.google.com/patents/US6234482?dq=patent:7076806 | 1,467,322,521,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783399385.17/warc/CC-MAIN-20160624154959-00111-ip-10-164-35-72.ec2.internal.warc.gz | 586,552,355 | 17,568 | Patents
Publication number US6234482 B1 Publication type Grant Application number US 09/354,940 Publication date May 22, 2001 Filing date Jul 15, 1999 Priority date Jul 15, 1999 Fee status Lapsed Publication number 09354940, 354940, US 6234482 B1, US 6234482B1, US-B1-6234482, US6234482 B1, US6234482B1 Inventors Thomas S. Henderson Original Assignee Thomas S. Henderson Export Citation Patent Citations (11), Referenced by (45), Classifications (8), Legal Events (5) External Links:
Method for playing a dice game
US 6234482 B1
Abstract
A dice game utilizing three dice begins with players making at least one of a single roll wager or a multiple roll wager. Single roll wagers are resolved after each roll of the dice. The single roll wagers include group wagers in which players wager on mutually exclusive groups of sums, each group having substantially the same probability of winning. Multiple roll wagers may require several rolls of the dice to be resolved because, if the number rolled is neither a predetermined losing outcome nor the outcome wagered upon, the wager is neither collected nor paid and an additional roll is required to resolve the wager. A shooter shoots the dice and the numbers facing upward on the dice are noted. Additionally, the numbers are summed. The wagers are then resolved. A jackpot wager is available to the shooter without the shooter making any additional wager. The shooter wins the jackpot by rolling a predetermined combination of numbers in a predetermined sequence of at least two consecutive rolls of the dice.
Images(3)
Claims(7)
I claim:
1. A method for playing a dice game utilizing three dice, or electronic representations thereof, having numerical indicia on their faces comprising:
the players making at least one of the following wagers:
(1) a single roll wager, said single roll wager including group wagers wherein players wager on a plurality of mutually exclusive groups of sums having 4, 8, 12, 15; 6, 9, 13, 17; 3, 5, 10, 14, 16, 18, said group wagers being paid when a sum in the group wagered upon is rolled and collected when any other sum is rolled or (2) a multiple roll wager;
designating a shooter from among the wagering players to roll said dice;
the shooter rolling said dice;
determining the numbers showing on the faces of said dice and summing the numbers;
for each player making a single roll wager, paying the player based on the amount wagered if the single roll outcome wagered upon is rolled, otherwise collecting the player's single roll wager;
for each player making a multiple roll wager, paying the player based on the amount wagered if the multiple roll outcome wagered upon is rolled, collecting the multiple roll wager if a predetermined losing outcome is rolled, and neither collecting nor paying the multiple roll wager if any other outcome is rolled; and
paying a jackpot to said shooter, without said shooter making any wager in addition to the shooter's multiple roll or single roll wager, when the shooter rolls a predetermined combination of numbers in a predetermined sequence, said sequence comprising at least two consecutive rolls of the dice.
2. The method of claim 1 wherein said jackpot comprises:
a first sum paid to the shooter when the dice show a predetermined jackpot combination on a first roll and a different predetermined jackpot combination on a second roll.
3. The method of claim 1 wherein said jackpot comprises:
a second sum, said second sum paid to the shooter when the dice show the same predetermined jackpot combination on at least two consecutive rolls.
4. The method of claim 1 wherein said jackpot comprises:
a third sum, said third sum paid to the shooter when the dice show the same predetermined jackpot combination on a first roll and a second roll, and a different predetermined jackpot combination on a third roll.
5. The method of claim 1 wherein said single roll wagers further comprise:
straight wagers wherein straight wagers are paid when dice show consecutive numbers on a single roll and collected when the dice show non-consecutive numbers.
6. The method of claim 1 wherein said multiple roll wagers comprise
totals wagers wherein players wager on the sum of the dice, excluding at least one predetermined losing sum, the players being able to increase, decrease, or remove said totals wagers after each roll of the dice.
7. A method for playing a dice game utilizing three dice or an electrical equivalent thereof, each die having six sides numbered one through six, comprising the steps of:
players placing at least one of the following wagers:
(1) a multiple roll wager or
(2) a single roll wager;
designating a shooter to roll said dice, said shooter being selected from the wagering players;
rolling said dice;
determining the numbers showing on the faces of said dice;
calculating the sum of the numbers showing on the faces of said dice;
paying or collecting single roll wagers, said single roll wagers comprising:
over wagers wherein players wagering over are paid when the sum rolled exceeds eleven, said over wagers being collected when the sum rolled is less than twelve;
under wagers wherein players wagering under are paid when the sum rolled is less than ten, said under wagers being collected when the sum rolled is more than nine;
group wagers wherein the possible sums, excluding at least one predetermined losing sum, are divided among three mutually exclusive groups having 4, 8, 12, 15; 6, 9, 13, 17; 3, 5, 10, 14, 16, 18, said group wagers being paid when a sum in the group wagered upon is rolled, otherwise said group wagers being collected;
any pairs wagers wherein players wagering any pairs are paid when exactly two of the dice show the same number, said any pairs wagers being collected when fewer than two or more than two dice show the same number;
called pair wagers wherein a player wagering a called pair is paid when exactly two of the dice show a number specified by said player, said called pairs wagers being collected when fewer than two or more than two dice show the specified number;
any trips wagers wherein players wagering any trips are paid when three dice show the same number, said any trips wagers being collected when fewer than three dice show the same number;
called trips wagers wherein a player wagering called trips is paid when three dice show a number specified by said player, said called trips wagers being collected when fewer than three dice show the specified number;
straight wagers wherein players wagering straight are paid when dice show consecutive numbers, said straight wagers being collected when the dice show non-consecutive numbers; paying or collecting multiple roll wagers said multiple roll wagers comprising:
totals wagers wherein players may wager on the sum of the dice, excluding at least one predetermined losing sum, the totals wagers being paid when the sum rolled equals the sum wagered upon, all of said totals wagers being collected when the sum of the dice is a predetermined losing sum, the totals wagers being neither collected nor paid when any other sum is rolled, said players with totals wagers remaining being able to increase, decrease, or remove said totals wagers before the next roll of the dice; and
field wagers wherein players may wager on a group of sums of the dice excluding at least one predetermined losing sum, the field wagers being paid when the sum rolled is in the group of sums wagered upon, all of said field wagers being collected when the sum of the dice is a predetermined losing sum, the field wagers being neither collected nor paid when any other sum is rolled, said players with field wagers remaining being able to increase, decrease, or remove said field wagers before the next roll of the dice; and
paying a jackpot to the shooter without said shooter making any wager in addition to the shooter's multiple roll or single roll wager, said jackpot comprising:
a first sum being paid to the shooter when the dice show three threes on a first roll and two of any other number on a second roll;
a second sum being larger than said first sum, said second sum being paid to the shooter when the dice show three threes on a first roll and three of any other number on a second roll;
a third sum being larger than said second sum, said third sum being paid to the shooter when the dice show three threes on a first roll and a second roll, and two of any other number on a third roll;
a fourth sum being larger than said third sum, said fourth sum being paid to the shooter when the dice show three threes on a first roll and a second roll and three of any other number on a third roll;
a fifth sum being larger than said fourth sum, said fifth sum being paid to the shooter when the dice show three threes on a first roll, a second roll, and a third roll.
Description
FIELD OF THE INVENTION
The present invention relates to a dice game. Specifically, the present game is a dice game with a jackpot payout in which players may wager on a single roll of the dice or multiple rolls of the dice.
BACKGROUND OF THE INVENTION
Dice games are well known in the art. Two of the most well known dice games are Craps and Sic Bo. In Craps, players place multiple roll wagers on the sum of the dice, single roll wagers on the sum of the dice, and multiple roll wagers on the combination of numbers shown on the face of the dice. For example, a pass line wager is paid when a natural, i.e. a seven or eleven, is rolled on the initial, or come out, roll or a point number, i.e. four, five, six, eight, nine, or ten, is rolled on the come out roll and then repeated before rolling a seven. The pass line wager is collected if a craps, i.e. two, three, or twelve, is rolled on the come out roll or a point number is rolled and a seven is rolled before the point number is repeated.
Conversely, a don't pass line wager is paid when a craps is rolled on the come out roll or a point number is rolled and a seven is rolled before the point number is repeated. The don't pass line wager is collected if a natural is rolled on the come out roll or a point number is rolled and then repeated before rolling a seven. A wager related to the pass line wager and the don't pass line wager is the free odds wager. Should a shooter roll a point number on the come out roll, any player with a pass line wager may place an additional free odds wager which will be paid at actual odds if the point number is rolled before a seven. For example, if the point number is six, a player with a free odds wager will be paid at 6:5 if the shooter rolls a six before rolling a seven. Likewise, any player with a don't pass line wager may place an additional wager which will be paid at actual odds if a seven is rolled before the point number. Thus, if the point number is six, a player with a free odds wager will be paid at 5:6 if the shooter rolls a seven before rolling a six.
A come wager and a don't come wager are treated similarly to the pass wager and don't pass wager, respectively, with the roll immediately following the placing of the wager treated as the come out roll for the come and don't come wagers.
Other possible wagers include place, or box, wagers in which a player wagers that a selected number will be rolled before a seven is rolled. The player may also choose to bet the converse, that is, that a seven will be rolled before a selected number is rolled. Yet another wager is the hardway wager in which a player wagers that the next roll of a selected sum will include double numbers. For example, the ten hardway wager is paid if the next ten rolled includes two fives and collected if the next ten rolled includes a four and a six. A hardway wager is also collected if a seven is rolled before a winning roll is rolled. Finally, a player may wager on a group of single roll sums. These single roll wagers include the field wager on the numbers two, three, five, nine, ten, eleven, and twelve and wagers on various other sums and dice combinations such as any craps, any sevens, two aces, two sixes, and the like.
Sic Bo is a single roll dice game in which players may wager on the sum of the dice. For example, the player may wager on the sum of fifteen without regard to the combination of dice required to form a fifteen, i.e. the wager would win if the dice show 5-5-5, 4-5-6, or 3-6-6. Alternatively or additionally, the player may wager on the combination of dice, for example, wagering on the combination of 5-5-5. Three dice are rolled and all wagers are immediately resolved; that is, there are no multiple roll wagers in Sic Bo.
The drawback of these games is that there is no opportunity to receive a large bonus. Moreover, there is no wager in Craps or Sic Bo on a subgroup of sums wherein each group is mutually exclusive and each group has substantially the same probability of winning.
Thus, there is a need in the art for a dice game which includes a jackpot payout and wagers on a single roll of the dice or on multiple rolls of the dice wherein the wagers are paid or collected based on the sum of the dice and the combination of numbers showing on the dice.
SUMMARY OF THE INVENTION
Three dice, or electronic representations thereof, are provided for the game of the present invention. The dice are preferably cubical with numerical indicia on each face.
The players place single roll wagers and multiple roll wagers on a table layout. Single roll wagers, as the name implies, are resolved after each roll of the dice. In the game of the present invention, as few as one single roll wager, the group wager, could be provided. However, the preferred embodiment contemplates a plurality of single roll wagers to retain player interest. In a group wager, players wager on one of a plurality of mutually exclusive groups of sums having substantially the same probability of containing the number rolled. In other words, the groups are constructed such that there is substantially the same probability that any of the groups will be a winner on each roll. The group wagers are paid when a sum in the group wagered upon is rolled and collected when any other sum is rolled. Other single roll wagers could include wagers on consecutive numbers on the three dice, the size of the sum of the numbers rolled, and whether any of the dice include matching numbers.
Additionally, the game of the present invention includes at least one multiple roll wager. Multiple roll wagers, as the name implies, may require multiple rolls of the dice to be resolved. That is, if the outcome wagered upon occurs, the multiple roll wager is paid. Conversely, if a designated losing outcome occurs, the multiple roll wager is collected. If neither a winning nor losing outcome is rolled, the multiple roll wager is neither paid nor collected. When this occurs, the player may leave the wager for the next roll, remove the wager from the layout, or increase or decrease the amount wagered. In a preferred embodiment, multiple roll wagers are placed on the sum of the numbers. That is, a player making a multiple roll wager selects a sum from all possible sums excluding at least one predetermined losing sum.
After the wagers are placed, a shooter is designated from among the players who placed a wager. The shooter rolls the dice and the numbers showing on the faces of the dice are noted. Additionally, the numbers are summed. The single roll wagers and multiple roll wagers are resolved as described above.
Additionally, the present game provides a jackpot. Only the shooter is eligible for the jackpot. However, the shooter need not make a wager in addition to the shooter's single roll or multiple roll wager to be eligible for the jackpot. The shooter wins the jackpot by rolling a predetermined combination of numbers in a predetermined sequence of at least two consecutive rolls of the dice. In a preferred embodiment, the jackpot includes a plurality of predetermined sequences, each sequence having a different payoff depending on the probability of the sequence occurring.
It is an object of the present invention to provide a dice game in which a jackpot payout is available to the shooter without requiring the shooter to make an additional jackpot wager. Another object of the present invention is to provide a dice game in which both single roll wagers and multiple roll wagers are available to the players. Also, it is an object of the present invention is to provide a dice game in which a player may wager on the sum of the dice or the combination of numbers on the dice.
BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is a top view of the layout of according to an embodiment of the present invention;
FIG. 2 is a flow chart of the jackpot payout according to an embodiment of the present invention.
DESCRIPTION
Reference is now made to the figures wherein like parts are referred to by like numerals throughout. Referring to FIG. 1, the layout 10 of the present invention has reflection-inversion symmetry. That is, one side of the layout 10 is an inverted reflection of the opposite side of the layout 10. This allows players on either side of the table to easily place wagers on the layout 10. A plurality of boxes 12 are marked on the layout 10 where players may place gaming chips to designate their wagers. The layout 10 includes a dealer area 14 from which the dealer may pay winning wagers, collect losing wagers, distribute gaming chips, and the like.
These boxes 12 include at least one single roll wager 16 and at least one multiple roll wager 18. Although the game requires only one single roll wager 16, the group wager 20, in a preferred embodiment a plurality of single roll wagers 16 are provided to maintain player interest. That is, a number of different single roll wagers 16 may be provided so that the player does not become bored with the game. As the name indicates, single roll wagers 16 are wagers which are placed before rolling dice and are resolved after each roll. In other words, single roll wagers 16 are immediately paid or collected after each roll.
In the group wager 20, the possible sums, excluding predetermined losing sums, are divided into a plurality of groups 20 such that the probability of rolling a sum in any one group 20 is substantially the same. The purpose of this is to insure that players are equally likely to win regardless of which group 20 is selected, thus providing the casino banking the game with predictability.
For example, in a game using three cubical dice with sides numbered one through six, the possible sums are three through eighteen inclusive. If the designated losing outcomes are seven and eleven, the sums could be divided into three groups 20—four, eight, twelve, fifteen; six, nine, thirteen, seventeen; and three, five, ten, fourteen, sixteen, eighteen. In a single roll, the probability that a sum will be contained in one of these three groups 20 is substantially the same because there are fifty-six ways to roll a total in the first group 22, fifty-nine ways to roll a total in the second group 24, and fifty-nine ways to roll a total in the third group 26. It can be seen that an infinite number of groupings is possible given that the only requirements are that there be at least one designated losing sum excluded from the groupings and that the remaining sums be divided into at least two groups 20 having substantially the same probability of containing a sum rolled. In a preferred embodiment, each group 20 is designated with a different color 22, 24, 26 on the layout 10.
It is contemplated that many other single roll wagers 16 on the sum of the dice or the combination of numbers showing on the dice may be possible. For example, a player may wager on the sum of the dice, whether the sum is odd or even, and the like. One possible single roll wager 16 is a straight wager 28. In a straight wager 28, the player is rewarded if the three dice show consecutive numbers. For example, if the numbers rolled are 2-3-4, the straight wagers 28 are paid. If the numbers rolled are non-consecutive, the straight wagers 28 are collected.
Another single roll wager 16 which could be provided is a wager on whether two or more of the dice include matching numbers. For example a pairs wager 30 could be provided in which a wagering player wins if exactly two of the dice have matching numbers such as a roll of 2-2-5. A variation on this wager is a called pairs wager 32 in which the player must designate, or call, which number will appear on both dice. Likewise, a trips wager 34 and a called trips wager 36 may be provided for players to wager that all three dice will have matching numbers.
Yet another single roll wager 16 which could be provided is a wager on the relative size of the sum. For example, an over wager 38 could be provided in which the player wagers that the sum will exceed eleven. Likewise, an under wager 40 could be provided in which the player wagers that the sum will be less than ten.
In a preferred embodiment, the single roll wagers 16 are paid according to Table 1.
TABLE 1 Wager Payout Over 11 or Under 10 \$2 wager wins \$2 \$5 wager wins \$7 Group wager with three groups \$2 wager wins \$4 \$5 wager wins \$12 Straight 7 to 1 Any pair 1 to 1 Called pair 11 to 1 Any trip 30 to 1 Called trip 185 to 1
The game of the present invention also contemplates at least one multiple roll wager 18. As the name indicates, a multiple roll wager 18 is a wager which may require more than one roll to resolve the wager. In other words, some outcomes, designated as losing outcomes, result in the loss of the wager; the outcome wagered upon is a winning outcome; and any other outcome does not affect the wager.
While the multiple roll wager 18 could be a wager on the combination of numbers rolled, in a preferred embodiment, the multiple roll wager 18 is a wager on the sum of the dice and is referred to as a totals wager 42. Thus, in a preferred embodiment, the sums seven and eleven are designated losing outcomes. Any other sum may be wagered upon. If the shooter rolls a seven or eleven, all multiple roll wagers 18 are collected by the casino. If the shooter rolls any other sum, players wagering upon the sum rolled are rewarded. The other wagers are neither rewarded nor collected. In a preferred embodiment, after each roll, a player with an unaffected wager may leave the wager on the layout 10 or may increase, decrease, or remove the wager. For example, if the designated losing sums are seven and eleven, and a player wagers on nine, his wager is unaffected until the shooter rolls a seven, nine, or eleven. When the shooter rolls a sum other than seven, nine, or eleven, the player wagering on nine may leave the wager on nine, or increase, decrease, or remove the wager.
Another multiple roll wager 18 which may be offered is a field wager in which a player may wager on a group of sums rather than a single sum. In a preferred embodiment, two field wagers are available: a large field wager 44 for the sums fifteen, sixteen, seventeen, and eighteen and a small field wager 46 for the sums three, four, five, and six.
In a preferred embodiment, multiple roll wagers 18 on the sum of the dice are paid according to the probability that the sum will be rolled. That is, the less likely sums are paid at a higher rate than the more likely sums as shown in Table 2.
TABLE 2 Sum wagered Payoff for \$2 Payoff for \$5 Payoff for \$10 upon wager wager wager 3 or 18 \$72 \$180 \$360 4 or 17 \$24 \$60 \$120 5 or 16 \$12 \$30 \$60 6 or 15 \$7 \$18 \$36 14 \$4 \$12 \$24 8 or 13 \$3 \$8 \$16 9 or 12 \$2 \$7 \$14 10 \$2 \$6 \$13 Small and large \$9 field
After the players place their wagers, a shooter is designated from among the wagering players and the shooter rolls three dice. While the preferred embodiment contemplates cubical, six-sided dice, dice with any number of sides could be used. Moreover, electronic representations of dice, such as a processor which randomly selects numbers from a data structure storing the numbers one through six, could substitute for physical dice. Thus, the game could be played on an electronic machine, a personal computer, a computer network, such as the Internet, or the like. After three numbers are obtained through random selection, the numbers are noted and the sum of the numbers is calculated. The single roll wagers 16 and multiple roll wagers 18 are resolved as described above.
The game of the present invention also includes a jackpot payout. The jackpot payout is available to the shooter only. No other player shares in the jackpot payout. However, the jackpot is available to the shooter without the shooter making any wager other than a single roll wager 16 or a multiple roll wager 18. In other words, the shooter does not need to make a jackpot wager to be eligible for the jackpot payout; the shooter is eligible merely by being the shooter.
The jackpot is won by rolling a predetermined combination of numbers in a predetermined sequence of at least two rolls. The predetermined combination could be any of a variety of possible combinations. Likewise, any pay table could be used to reward jackpot rolls. However, FIG. 2 shows a schematic of the preferred embodiment of the jackpot payout. A player becomes eligible for the jackpot by rolling 50 three threes 52. If the shooter rolls a non-triple 56 on the second roll 54, the shooter wins a first payout 58 and the shooter's jackpot play ends. If the shooter rolls a triple 56 other than threes on the second roll 54, the shooter wins a second payout 62 greater than the first payout 58 and the shooter's jackpot play ends. If the shooter rolls three threes 60 on the second roll 54, the shooter is eligible for a jackpot payout on the third roll 64. If the shooter rolls a non-triple 66 on the third roll 64, the shooter wins a third payout 68 greater than the second payout 62 and the shooter's jackpot play ends. If the shooter rolls a triple 66 other than threes on the third roll 64, the shooter wins a fourth payout 72 greater than the third payout 68 and the shooter's jackpot play ends. If the player rolls three threes 70 for a third time, the player wins a fifth payout 74—the jackpot.
An advantage of the present dice game is that a jackpot payout is available to the shooter without requiring the shooter to make an additional jackpot wager. Another advantage of the present game is that both single roll wagers 16 and multiple roll wagers 18 are available to the players. Yet another advantage of the present invention is players may wager on the sum of the dice or the combination of numbers on the dice.
Patent Citations
Cited PatentFiling datePublication dateApplicantTitle
US4312508Aug 4, 1980Jan 26, 1982Wood Michael WDice game and teaching aid
US5133559Feb 19, 1991Jul 28, 1992Page Robert ACasino dice game
US5265881Jun 30, 1992Nov 30, 1993Vincent DohertyMethod of playing a dice or card game
US5350175Jan 7, 1994Sep 27, 1994Dean DiLulloBetting game method of play
US5364101Apr 22, 1993Nov 15, 1994Spooner James JCasino dice game method
US5405145Aug 10, 1994Apr 11, 1995Jones; Albert C.Dice game with wild die
US5456467Feb 9, 1995Oct 10, 1995Hoover; Betty L.Method of playing a poker dice game
US5542671Mar 2, 1994Aug 6, 1996Stewart; Walter M.Method for playing game of dice
US5806847 *Jul 10, 1997Sep 15, 1998White; Roger L.Wagering game employing dice
US5879006 *Dec 17, 1996Mar 9, 1999Bowling; James P.Method of playing a three dice game
US5997000 *Aug 22, 1995Dec 7, 1999Nakano; YoshioCombined dice and card game
Non-Patent Citations
Reference
1John Scarne, Scarne on Dice, pp. 336-341, Wilshire Book Company.
2Pyramid Dice, Pamphlet distributed at Circus Circus, Las Vegas Nevada.
Referenced by
Citing PatentFiling datePublication dateApplicantTitle
US6601848May 22, 2002Aug 5, 2003William P. Timmons, Sr.Dice game
US6805352Oct 3, 2003Oct 19, 2004Enlil-Enki Enterprises, S.A.Craps game with progressive jackpot
US6854732 *Nov 25, 2002Feb 15, 2005Ernest W. MoodySix dice game
US6893019Mar 13, 2003May 17, 2005Daniel E. GaygenBetting game using one die of one color and two die of another color and giving special status to a roll of one on the single die
US6896264Aug 27, 2003May 24, 2005Jose Cherem HaberMethod of playing a dice wagering game
US6899330 *Apr 5, 2004May 31, 2005Fredrick I. ZinkBowling dice game
US6974132Mar 18, 2003Dec 13, 2005Nicholas SorgeMethod of play and game surface for a dice game having a progressive jackpot
US7152863Jan 23, 2004Dec 26, 2006Scheb Jr PaulGame of chance
US7185889 *Sep 22, 2004Mar 6, 2007Vanzanten David SCasino table wagering game and method therefor
US7255350 *Jun 16, 2003Aug 14, 2007Timmons Sr William PDice game
US7278636 *Dec 20, 2004Oct 9, 2007Zajac John DMethod and apparatus for playing a dice game
US7434808Aug 26, 2005Oct 14, 2008Nicholas SorgeMethod of play and game surface for a dice game
US7874905 *Jan 25, 2011Universal Entertainment CorporationDice game method and dice game machine
US8011663 *Sep 6, 2011Scheb Jr PaulGame of chance
US8074991 *Sep 7, 2004Dec 13, 2011Hui Chuan ChenLet me roll sic bo
US8221219 *Jul 17, 2012Dorr Robert CGaming machine displaying one wagered-on game symbol and method of play
US8545305Jun 28, 2010Oct 1, 2013Wms Gaming Inc.Devices, systems, and methods for dynamically simulating a component of a wagering game
US8900047 *Oct 9, 2013Dec 2, 2014Nathaniel FerrellDice-based gaming system
US9098981Oct 1, 2012Aug 4, 2015Fresh Idea Global LimitedParadise box gaming system
US9171418Nov 21, 2012Oct 27, 2015Bally Gaming, Inc.Gaming devices and gaming systems with multiple display device arrangement
US9214060Oct 28, 2014Dec 15, 2015Fresh Idea Global LimitedGaming center allowing switching between games based upon historical results
US9240095Feb 23, 2011Jan 19, 2016Solution Champion LimitedMethods and systems for playing baccarat jackpot with an option for insurance betting
US20030073485 *Oct 15, 2002Apr 17, 2003Friedman Stacy A.Method of playing wagering games
US20030107172 *Nov 25, 2002Jun 12, 2003Moody Ernest W.Six dice game
US20030218300 *Jun 16, 2003Nov 27, 2003Timmons William P.Dice game
US20040130094 *Oct 3, 2003Jul 8, 2004Enlil-Enki Enterprises, S.A.Craps game with progressive jackpot
US20040178575 *Mar 13, 2003Sep 16, 2004Gaygen Daniel E.Betting game using one die of one color and two matching color dice of a different color
US20050040593 *Mar 18, 2003Feb 24, 2005Nicholas SorgeMethod of play and game surface for a dice game having a progressive jackpot
US20050151317 *Dec 20, 2004Jul 14, 2005Zajac John D.Method and apparatus for playing a dice game
US20050161879 *Jan 23, 2004Jul 28, 2005Scheb Paul Jr.Game of chance
US20050184456 *Feb 19, 2004Aug 25, 2005Lowery Jeffery R.Casino dice game
US20050280208 *Aug 26, 2005Dec 22, 2005Nicholas SorgeMethod of play and game surface for a dice game
US20060049578 *Sep 7, 2004Mar 9, 2006Chen Hui CLet me roll Sic Bo
US20060061036 *Sep 22, 2004Mar 23, 2006Vanzanten David SCasino table wagering game and method therefor
US20060202416 *Mar 14, 2005Sep 14, 2006Madden Robert BMethod of a progressive payout dice game
US20080096641 *Oct 19, 2007Apr 24, 2008Aruze Corp.Dice game method and dice game machine
US20080096642 *Oct 19, 2007Apr 24, 2008Aruze Corp.Dice game method and dice game machine
US20080182640 *Jan 22, 2008Jul 31, 2008Aruze Corp.Multiple participants type gaming device and playing method thereof
US20090250870 *Aug 28, 2008Oct 8, 2009Krumenacker Thomas CVariant Games of Chance Using Three Dice
US20090250872 *Mar 18, 2008Oct 8, 2009Scheb Jr PaulGame of Chance
US20090256310 *Apr 9, 2009Oct 15, 2009Alfred SanchezPoker cube game apparatus and method of play
US20110241294 *Oct 6, 2011Marco GuldenaarCasino game and a set of six-face cubic colored dice
US20120289300 *Apr 20, 2012Nov 15, 2012Jay ChunMethods and systems for playing sic bo jackpot
US20140265124 *Feb 28, 2014Sep 18, 2014Madson Products, LlcGames of chance and methods for playing such games
WO2007024738A2 *Aug 24, 2006Mar 1, 2007Nicholas SorgeA dice game
Classifications
U.S. Classification273/274, 273/309, 273/139
International ClassificationA63F3/00, A63F9/04
Cooperative ClassificationA63F9/04, A63F3/00157
European ClassificationA63F9/04
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Effective date: 20130522 | 7,834 | 32,532 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-26 | latest | en | 0.906212 |
https://www.kylesconverter.com/mass-flow/milligrams-per-day-to-ounces-troy-per-year | 1,670,094,589,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00682.warc.gz | 915,762,768 | 5,722 | # Convert Milligrams Per Day to Ounces Troy Per Year
### Kyle's Converter > Mass Flow > Milligrams Per Day > Milligrams Per Day to Ounces Troy Per Year
Milligrams Per Day (mg/d) Ounces Troy Per Year (oz t/yr)* Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
Reverse conversion?
Ounces Troy Per Year to Milligrams Per Day
(or just enter a value in the "to" field)
#### Please share if you found this tool useful:
Unit Descriptions
1 Milligram per Day:
Mass flow of milligrams across a threshold per unit time of a day. 1 milligram per day = 0.000001/86400 kilograms per second (SI base unit). 1 mg/d ≈ 1.1574074 x 10-11 kg/s.
1 Ounce Troy per Year:
Mass flow of troy ounces across a threshold per unit time of a year. A 365 day civil year. 1 Troy ounce per year = 0.0311034768/31536000 kilograms per second (SI base unit). 1 oz t/yr ≈ 9.862 847 79 x 10-10 kg/s.
Conversions Table
1 Milligrams Per Day to Ounces Troy Per Year = 0.011770 Milligrams Per Day to Ounces Troy Per Year = 0.8215
2 Milligrams Per Day to Ounces Troy Per Year = 0.023580 Milligrams Per Day to Ounces Troy Per Year = 0.9388
3 Milligrams Per Day to Ounces Troy Per Year = 0.035290 Milligrams Per Day to Ounces Troy Per Year = 1.0562
4 Milligrams Per Day to Ounces Troy Per Year = 0.0469100 Milligrams Per Day to Ounces Troy Per Year = 1.1735
5 Milligrams Per Day to Ounces Troy Per Year = 0.0587200 Milligrams Per Day to Ounces Troy Per Year = 2.347
6 Milligrams Per Day to Ounces Troy Per Year = 0.0704300 Milligrams Per Day to Ounces Troy Per Year = 3.5205
7 Milligrams Per Day to Ounces Troy Per Year = 0.0821400 Milligrams Per Day to Ounces Troy Per Year = 4.694
8 Milligrams Per Day to Ounces Troy Per Year = 0.0939500 Milligrams Per Day to Ounces Troy Per Year = 5.8675
9 Milligrams Per Day to Ounces Troy Per Year = 0.1056600 Milligrams Per Day to Ounces Troy Per Year = 7.041
10 Milligrams Per Day to Ounces Troy Per Year = 0.1174800 Milligrams Per Day to Ounces Troy Per Year = 9.388
20 Milligrams Per Day to Ounces Troy Per Year = 0.2347900 Milligrams Per Day to Ounces Troy Per Year = 10.5615
30 Milligrams Per Day to Ounces Troy Per Year = 0.35211,000 Milligrams Per Day to Ounces Troy Per Year = 11.735
40 Milligrams Per Day to Ounces Troy Per Year = 0.469410,000 Milligrams Per Day to Ounces Troy Per Year = 117.3502
50 Milligrams Per Day to Ounces Troy Per Year = 0.5868100,000 Milligrams Per Day to Ounces Troy Per Year = 1173.5022
60 Milligrams Per Day to Ounces Troy Per Year = 0.70411,000,000 Milligrams Per Day to Ounces Troy Per Year = 11735.0225 | 831 | 2,535 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-49 | latest | en | 0.579672 |
https://www.geeksforgeeks.org/check-whether-arithmetic-progression-can-formed-given-array/?ref=leftbar-rightbar | 1,618,358,913,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038075074.29/warc/CC-MAIN-20210413213655-20210414003655-00507.warc.gz | 889,983,789 | 24,222 | Related Articles
Check whether Arithmetic Progression can be formed from the given array
• Difficulty Level : Medium
• Last Updated : 13 Apr, 2021
Given an array of n integers. The task is to check whether an arithmetic progression can be formed using all the given elements. If possible print “Yes”, else print “No”.
Examples:
```Input : arr[] = {0, 12, 4, 8}
Output : Yes
Rearrange given array as {0, 4, 8, 12}
which forms an arithmetic progression.
Input : arr[] = {12, 40, 11, 20}
Output : No```
Method 1 (Simple)
A simple solution is to first find the smallest element, then find second smallest element and find the difference between these two. Let this difference be d. After finding the difference, find third smallest, fourth smallest and so on. After finding every i-th smallest smallest (from third onward), find the difference between value of current element and value of previous element. If difference is not same as d, return false. If all elements have same difference, return true. Time complexity of this solution is O(n2)
Method 2(Use Sorting)
The idea is to sort the given array. After sorting, check if differences between consecutive elements are same or not. If all differences are same, Arithmetic Progression is possible.
Below is the implementation of this approach:
C++
`// C++ program to check if a given array``// can form arithmetic progression``#include``using` `namespace` `std;` `// Returns true if a permutation of arr[0..n-1]``// can form arithmetic progression``bool` `checkIsAP(``int` `arr[], ``int` `n)``{`` ``if` `(n == 1)`` ``return` `true``;` ` ``// Sort array`` ``sort(arr, arr + n);` ` ``// After sorting, difference between`` ``// consecutive elements must be same.`` ``int` `d = arr[1] - arr[0];`` ``for` `(``int` `i=2; i
Java
`// Java program to check if a given array``// can form arithmetic progression``import` `java.util.Arrays;` `class` `GFG {`` ` ` ``// Returns true if a permutation of`` ``// arr[0..n-1] can form arithmetic`` ``// progression`` ``static` `boolean` `checkIsAP(``int` `arr[], ``int` `n)`` ``{`` ``if` `(n == ``1``)`` ``return` `true``;`` ` ` ``// Sort array`` ``Arrays.sort(arr);`` ` ` ``// After sorting, difference between`` ``// consecutive elements must be same.`` ``int` `d = arr[``1``] - arr[``0``];`` ``for` `(``int` `i = ``2``; i < n; i++)`` ``if` `(arr[i] - arr[i-``1``] != d)`` ``return` `false``;`` ` ` ``return` `true``;`` ``}`` ` ` ``//driver code`` ``public` `static` `void` `main (String[] args)`` ``{`` ``int` `arr[] = { ``20``, ``15``, ``5``, ``0``, ``10` `};`` ``int` `n = arr.length;`` ` ` ``if``(checkIsAP(arr, n))`` ``System.out.println(``"Yes"``);`` ``else`` ``System.out.println(``"No"``);`` ``}``}` `// This code is contributed by Anant Agarwal.`
Python3
`# Python3 program to check if a given``# array can form arithmetic progression` `# Returns true if a permutation of arr[0..n-1]``# can form arithmetic progression``def` `checkIsAP(arr, n):`` ``if` `(n ``=``=` `1``): ``return` `True` ` ``# Sort array`` ``arr.sort()` ` ``# After sorting, difference between`` ``# consecutive elements must be same.`` ``d ``=` `arr[``1``] ``-` `arr[``0``]`` ``for` `i ``in` `range``(``2``, n):`` ``if` `(arr[i] ``-` `arr[i``-``1``] !``=` `d):`` ``return` `False` ` ``return` `True` `# Driver code``arr ``=` `[ ``20``, ``15``, ``5``, ``0``, ``10` `]``n ``=` `len``(arr)``print``(``"Yes"``) ``if``(checkIsAP(arr, n)) ``else` `print``(``"No"``)` `# This code is contributed by Anant Agarwal.`
C#
`// C# program to check if a given array``// can form arithmetic progression``using` `System;` `class` `GFG {`` ` ` ``// Returns true if a permutation of`` ``// arr[0..n-1] can form arithmetic`` ``// progression`` ``static` `bool` `checkIsAP(``int` `[]arr, ``int` `n)`` ``{`` ``if` `(n == 1)`` ``return` `true``;`` ` ` ``// Sort array`` ``Array.Sort(arr);`` ` ` ``// After sorting, difference between`` ``// consecutive elements must be same.`` ``int` `d = arr[1] - arr[0];`` ``for` `(``int` `i = 2; i < n; i++)`` ``if` `(arr[i] - arr[i - 1] != d)`` ``return` `false``;`` ` ` ``return` `true``;`` ``}`` ` ` ``//Driver Code`` ``public` `static` `void` `Main ()`` ``{`` ``int` `[]arr = {20, 15, 5, 0, 10};`` ``int` `n = arr.Length;`` ` ` ``if``(checkIsAP(arr, n))`` ``Console.WriteLine(``"Yes"``);`` ``else`` ``Console.WriteLine(``"No"``);`` ``}``}` `// This code is contributed by vt_m.`
PHP
``
Javascript
``
Output
`Yes`
Time Complexity: O(n Log n).
Method 3(Use Hashing)
1. Find out the smallest and second smallest elements
2. Find different between the two elements. d = second_smallest – smallest
3. Store all elements in a hashmap and return “NO” if duplicate element found (can be done together with step 1).
4. Now start from “second smallest element + d” and one by one check n-2 terms of Arithmetic Progression in hashmap. If any value of progression is missing, return false.
5. Return “YES” after end of the loop.
Below is the implementation of this method.
C++
`// C++ program to check if a given array``// can form arithmetic progression``#include ``using` `namespace` `std;` `// Returns true if a permutation of arr[0..n-1]``// can form arithmetic progression``bool` `checkIsAP(``int` `arr[], ``int` `n)``{`` ``unordered_map<``int``, ``int``> hm;`` ``int` `smallest = INT_MAX, second_smallest = INT_MAX;`` ``for` `(``int` `i = 0; i < n; i++) {`` ` ` ``// Find the smallest and and`` ``// update second smallest`` ``if` `(arr[i] < smallest) {`` ``second_smallest = smallest;`` ``smallest = arr[i];`` ``}`` ` ` ``// Find second smallest`` ``else` `if` `(arr[i] != smallest`` ``&& arr[i] < second_smallest)`` ``second_smallest = arr[i];`` ` ` ``// Check if the duplicate element found or not`` ``if` `(hm.find(arr[i]) == hm.end())`` ``hm[arr[i]]++;`` ` ` ``// If duplicate found then return false`` ``else`` ``return` `false``;`` ``}`` ` ` ``// Find the difference between smallest and second`` ``// smallest`` ` ` ``int` `diff = second_smallest - smallest;`` ` ` ``// As we have used smallest and`` ``// second smallest,so we`` ``// should now only check for n-2 elements`` ``for` `(``int` `i = 0; i < n - 1; i++) {`` ``if` `(hm.find(second_smallest) == hm.end())`` ``return` `false``;`` ``second_smallest += diff;`` ``}`` ``return` `true``;``}` `// Driven Program``int` `main()``{`` ``int` `arr[] = { 20, 15, 5, 0, 10 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``(checkIsAP(arr, n)) ? (cout << ``"Yes"` `<< endl)`` ``: (cout << ``"No"` `<< endl);` ` ``return` `0;` ` ``// This code is contributed by Raman Jha``}`
Python3
`# Python3 program to check if a given array``# can form arithmetic progression` `# Returns true if a permutation of arr[0..n-1]``# can form arithmetic progression``def` `checkIsAP(arr, n):`` ` ` ``hm ``=` `{}`` ``smallest ``=` `float``(``'inf'``)`` ``second_smallest ``=` `float``(``'inf'``)`` ` ` ``for` `i ``in` `range``(n):`` ` ` ``# Find the smallest and and`` ``# update second smallest`` ``if` `(arr[i] < smallest):`` ``second_smallest ``=` `smallest`` ``smallest ``=` `arr[i]`` ` ` ``# Find second smallest`` ``elif` `(arr[i] !``=` `smallest ``and`` ``arr[i] < second_smallest):`` ``second_smallest ``=` `arr[i]`` ` ` ``# Check if the duplicate element found or not`` ``if` `arr[i] ``not` `in` `hm:`` ``hm[arr[i]] ``=` `1`` ` ` ``# If duplicate found then return false`` ``else``:`` ``return` `False`` ` ` ``# Find the difference between smallest`` ``# and second smallest`` ``diff ``=` `second_smallest ``-` `smallest`` ` ` ``# As we have used smallest and`` ``# second smallest,so we`` ``# should now only check for n-2 elements`` ``for` `i ``in` `range``(n``-``1``):`` ``if` `(second_smallest) ``not` `in` `hm:`` ``return` `False`` ` ` ``second_smallest ``+``=` `diff`` ` ` ``return` `True` `# Driver code``arr ``=` `[ ``20``, ``15``, ``5``, ``0``, ``10` `]``n ``=` `len``(arr)` `if` `(checkIsAP(arr, n)):`` ``print``(``"Yes"``)``else``:`` ``print``(``"No"``)`` ` `# This code is contributed by rohitsingh07052`
Output
`Yes`
Time Complexity: O(n)
Auxiliary Space: O(n)
Thanks to Chenna Rao for suggesting this method,
Method 4(Using counting sort)
We can reduce space required in method 3 if given array can be modified.
1. Find smallest and second smallest elements.
2. Find d = second_smallest – smallest
3. Subtract smallest element from all elements.
4. Now if given array represent AP, all elements should be of form i*d where i varies from 0 to n-1.
5. One by one divide all reduced elements with d. If any element is not divisible by d, return false.
6. Now if array represents AP, it must be a permutation of numbers from 0 to n-1. We can easily check this using counting sort.
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. | 3,185 | 9,943 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.710391 |
https://www.physicsforums.com/threads/how-do-you-find-the-angle-of-a-slope-if-you-are-only-given-coefficient-of-friction.472600/ | 1,590,751,778,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00563.warc.gz | 871,902,359 | 16,737 | # How do you find the angle of a slope if you are only given coefficient of friction
A skier is traveling at a constant speed of 4 m/s on a ski slope. The kinetic coefficient of frictionbetween her skis and the slope is 0.2. Find the angle of the slope.
attempt:
sin feda = f/w but i have no weight
f=mgsinfeda and N=mg cos feda
all of this ends up being mu=tan feda but dont understand where the Velocity in the problem factors in or if since its a constant force and the net sum of a constant moving anything is zero if it does not factor into the equasion at all. i have drawn a vector sum diagram with net sum =0. We worked out one of these in class but he gave us the wieght of the person and the angle. i cant figure out how to do it without those variables.
Last edited:
Related Introductory Physics Homework Help News on Phys.org
I will gladly assist you with the problem, but you must show your attempt at a solution before I can do so as per PF rules.
Last edited:
gneill
Mentor
A skier is traveling at a constant speed of 4 m/s on a ski slope. The kinetic coefficient of frictionbetween her skis and the slope is 0.2. Find the angle of the slope.
attempt:
sin feda = f/w but i have no weight
f=mgsinfeda and N=mg cos feda
all of this ends up being mu=tan feda but dont understand where the Velocity in the problem factors in or if since its a constant force and the net sum of a constant moving anything is zero if it does not factor into the equasion at all. i have drawn a vector sum diagram with net sum =0. We worked out one of these in class but he gave us the wieght of the person and the angle. i cant figure out how to do it without those variables.
Generally you will find with these sorts of problems that information that is not given (such as weight or mass) will eventually cancel out in the equations involved. So unless you're "addicted" to plugging in numbers as soon as possible for every case, there shouldn't be a problem.
In this case the skier is said to be traveling at a constant speed downhill. That implies no acceleration, and hence no force imbalance. So you need to find expressions for the forces involved. If you assume that the skier has a mass "m", what are expressions for the downhill force due to gravity and the frictional force?
Oh, and here's a ---> θ <--- for cutting and pasting
i posted the solution but knowing the rules had to delete it:
giving you a hint:
if velocity is constant; net force = 0
just equate all forces in it | 592 | 2,491 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2020-24 | longest | en | 0.940894 |
https://nonsmooth.gricad-pages.univ-grenoble-alpes.fr/siconos/getting_started/tutorial_python/BulletIO.html | 1,723,369,871,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640983659.65/warc/CC-MAIN-20240811075334-20240811105334-00009.warc.gz | 344,271,144 | 8,249 | # Collections of rigid bodies with Bullet based contact detection (Siconos/Mechanics)#
Author: Maurice Bremond, Vincent Acary 2013–2016
You may refer to the source code of this set of examples, found here.
## Description of the physical problems : rigid bodies collection with contact and Coulomb friction#
In this set of examples (cubes.py, n_cubes.py, …), we model collections of rigid bodies associated with shapes (primitive (sphere, cube capsule, etc), convex hull, mesh) that interact with contact and friction (see fig-n_cubes).
The example use extensively the mechanics_run python code that enable to create and manage the geometrical data and the results of the simulation in hdf5 file.
The results cab be viewed with the siconos_vview and the siconos_vexport python scripts
A brief description of examples is as follows :
• cube.py is the simplest example with few cubes that fall down on a rigid fixed plane. The scene is built using calls to mechanics_run methods such as
• addPrimitiveShape for creating a primitive shape
```with Hdf5(mode='r+') as io:
io.addPrimitiveShape('Ground', 'Box', (100, 100, .5))
```
```with Hdf5(mode='r+') as io:
(-1.0, 1.0, -1.0),
(-1.0, -1.0, -1.0),
(-1.0, -1.0, 1.0),
(-1.0, 1.0, 1.0),
(1.0, 1.0, 1.0),
(1.0, 1.0, -1.0),
(1.0, -1.0, -1.0),
(1.0, -1.0, 1.0)])
```
• addObject for associating Newton Euler Dynamical System to a shape
```with Hdf5(mode='r+') as io:
io.addObject('cube', [Contactor('Cube')], translation=[0, 0, 2],
velocity=[10, 0, 0, 1, 1, 1],
mass=1)
```
The computation is launched using the method run() with default arguments
```with Hdf5(mode='r+') as io:
io.run(with_timer=False,
time_stepping=None,
space_filter=None,
body_class=None,
shape_class=None,
face_class=None,
edge_class=None,
gravity_scale=1,
t0=0,
T=10,
h=0.0005,
multipoints_iterations=True,
theta=0.50001,
Newton_max_iter=20,
set_external_forces=None,
solver=Numerics.SICONOS_FRICTION_3D_NSGS,
itermax=100000,
tolerance=1e-8,
numerics_verbose=False,
output_frequency=None)
```
• n_cubes.py . This is an extension of cubes.py where it is possible to build a reactangular pile of cubes
• cube_scene.py, cube_simulation.py n_cubes_scene.py n_cubes_simulation.py . These examples are a different treatment of the same example but the construction of the scene and the simulation are separated into differents. In that way, it is possible to produce a scene file and then to perfoms several simulations. Furthermore, if the result of the simulation is stored in the hdf5 file, we restart from the last state of the system to continue the simulation. it is alo possible to concatenate in time several simulations in that way.
• MultipleContactorsAndNSlaws.py . This is an example where two shapes are fixed to the same mechanical system. It allows to build clusters and to get some complex contact shapes.
• bar.py, bar_contact.py This simple example provides one with a simple simulation where the performance of Lie integrator are important for long term simulation of rotating bodie in a stable way. It allows also to test large inertia ratio in the solver.
• convexhull.py . This test shos how to build a collection of convexhull shapes
• PairWise_test_bullet.py . This test file performs the test of contacting pairs of various nature (primitive, convex hull). Meshes have to be added.
• chute.py. This example is a little more complex one where the a collection of polyhedra falls down into an hopper. | 912 | 3,451 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-33 | latest | en | 0.730626 |
https://www.teacherspayteachers.com/Product/Geometry-Task-Cards-4th-Grade-Math-Centers-2678435 | 1,524,356,357,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945459.17/warc/CC-MAIN-20180421223015-20180422003015-00458.warc.gz | 893,145,714 | 21,415 | ## Main Categories
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Geometry - 4th Grade Math Flip and Go Cards
30 math task cards that reinforce 4th grade common core math standards - Identify lines and angles and classify shapes by properties of their lines and angles.
Do you want a quick and easy way for students to practice and review lines and angles within different shapes? I know that every minute of the instructional day is precious. There is never enough time to do it all. I designed Flip and Go Math Cards to maximize instructional time and efficiently reinforce math skills throughout all parts of the school day. They are easy to assemble. Simply print, cut and go! Oh, and don’t forget to put a ring on it!
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Teachers Pay Teachers is an online marketplace where teachers buy and sell original educational materials. | 998 | 5,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2018-17 | latest | en | 0.909901 |
https://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP?From=short+ton | 1,696,130,155,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510734.55/warc/CC-MAIN-20231001005750-20231001035750-00654.warc.gz | 776,112,830 | 3,897 | Partner with ConvertIt.com
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http://www.coderanch.com/t/261904/java-programmer-OCPJP/certification/preceedence | 1,464,386,726,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049277091.36/warc/CC-MAIN-20160524002117-00186-ip-10-185-217-139.ec2.internal.warc.gz | 421,864,951 | 7,721 | Win a copy of Design for the Mind this week in the Design forum!
regarding preceedence
shreya prabhu
Ranch Hand
Posts: 31
b1=true,b2=true,b3=false
b3&=b1|b2->this evaluates to false
b3=b3&b1|b2->this evaluates to true
why does it give different result?
Keith Lynn
Ranch Hand
Posts: 2409
Originally posted by archana prabhu:
b1=true,b2=true,b3=false
b3&=b1|b2->this evaluates to false
b3=b3&b1|b2->this evaluates to true
why does it give different result?
| has a higher precedence than &=
so b3 &= b1|b2 is the same as b3 &= (b1|b2) which is b3 &= true which is false since b3 is false.
& has a higher precedence than |
so b3 = b3 & b1 | b2 is the same as b3 = (b3 & b1) | b2 = false | true = true. | 245 | 705 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-22 | latest | en | 0.850816 |
http://stackoverflow.com/questions/9869524/how-to-convert-list-of-intable-strings-to-int | 1,438,079,997,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042981856.5/warc/CC-MAIN-20150728002301-00126-ip-10-236-191-2.ec2.internal.warc.gz | 224,129,694 | 21,027 | # How to convert list of intable strings to int
In Python, I want to convert a list of strings:
``````l = ['sam','1','dad','21']
``````
and convert the integers to integer types like this:
``````t = ['sam',1,'dad',21]
``````
I tried:
``````t = [map(int, x) for x in l]
``````
but is showing an error.
How could I convert all intable strings in a list to int, leaving other elements as strings?
My list might be multi-dimensional. A method which works for a generic list would be preferable:
`l=[['aa','2'],['bb','3']]`
-
Note that `[map(int, x) for x in l]` will try to turn each string into a list of integers, character by character. You probably meant either `map(int, l)` or `[int(x) for x in l]`. – Thomas Wouters Mar 26 '12 at 9:31
`[int(x) for x in l]` will throw `ValueError`s for non-numeric strings. – Secator Mar 26 '12 at 9:53
I'd use a custom function:
``````def try_int(x):
try:
return int(x)
except ValueError:
return x
``````
Example:
``````>>> [try_int(x) for x in ['sam', '1', 'dad', '21']]
``````
Edit: If you need to apply the above to a list of lists, why didn't you converted those strings to int while building the nested list?
Anyway, if you need to, it's just a matter of choice on how to iterate over such nested list and apply the method above.
One way for doing that, might be:
``````>>> list_of_lists = [['aa', '2'], ['bb', '3']]
>>> [[try_int(x) for x in lst] for lst in list_of_lists]
[['aa', 2], ['bb', 3]]
``````
You can obviusly reassign that to `list_of_lists`:
``````>>> list_of_lists = [[try_int(x) for x in lst] for lst in list_of_lists]
``````
-
+1, but why `list` + `map` instead of a list comprehension? – larsmans Mar 26 '12 at 9:36
@larsmans: Emh... I don't know... I was probably dazzled by the OP `map()`. I'll update the answer in a moment :) – Rik Poggi Mar 26 '12 at 9:39
I've been so bold as to do it for you. – larsmans Mar 26 '12 at 9:41
@larsmans: You sir are fast! Thanks :) – Rik Poggi Mar 26 '12 at 9:42
@sum2000: You should've stated that from the beginning :) Anyway just iterate over your list_of_list and apply the method above on each sub list. – Rik Poggi Mar 26 '12 at 9:51
I would create a generator to do it:
``````def intify(lst):
for i in lst:
try:
i = int(i)
except ValueError:
pass
yield i
intified_list = list(intify(lst))
# or if you want to modify an existing list
# lst[:] = intify(lst)
``````
If you want this to work on a list of lists, just:
``````new_list_of_lists = map(list, map(intify, list_of_lists))
``````
-
The idea is right, but the implementation is not very elegant; the `str`-to-`int` conversion and iteration should really be separated as in Rik Poggi's answer. – larsmans Mar 26 '12 at 9:35
@larsmans I disagree. I assumed based on the question that this is only going to be used on sequences, so why require the extra step of `map`? Not everything has to be written in functional programming style. (Also, purely as a bonus, this gives the same result on both Python 2 and Python 3). – agf Mar 26 '12 at 9:40
`map` is not needed, but separating iteration from purely per-element conversions, in my experience, leads to much cleaner and more readable code. It's a matter of separating concerns, not my FP fetish ;) – larsmans Mar 26 '12 at 9:44
@agf please look at the edited part. – sum2000 Mar 26 '12 at 9:47
@sum2000 I already updated mine to match. The same basic idea could be used for the other answers. – agf Mar 26 '12 at 9:49
For multidimenson lists, use recursive technique may help.
``````from collections import Iterable
def intify(maybeLst):
try:
return int(maybeLst)
except:
if isinstance(maybeLst, Iterable) and not isinstance(lst, str):
return [intify(i) for i in maybeLst] # here we call intify itself!
else:
return maybeLst
print intify(maybeLst)
``````
-
How about using map and lambda
``````>>> map(lambda x:int(x) if x.isdigit() else x,['sam','1','dad','21'])
``````
or with List comprehension
``````>>> [int(x) if x.isdigit() else x for x in ['sam','1','dad','21']]
>>>
``````
As mentioned in the comment, as isdigit may not capture negative numbers, here is a refined condition to handle it notable a string is a number if its alphanumeric and not a alphabet :-)
``````>>> [int(x) if x.isalnum() and not x.isalpha() else x for x in ['sam','1','dad','21']]
``````
-
`str.isdigit()` isn't the right test to see if `int()` will work, specifically when you can have negative values. Calling `int()` is the right test. – Thomas Wouters Mar 26 '12 at 9:29
i am doing `l1=[int(x) if x.isdigit() else x for x in l]`, it is showing error `AttributeError: 'list' object has no attribute 'isdigit'` – sum2000 Mar 26 '12 at 9:33
@sum2000 This method only works if all the items in the list are strings. It sounds like some of the items in your list aren't. – agf Mar 26 '12 at 9:34
How is l defined? can you tell me the result of `[type(x) for x in l]` – Abhijit Mar 26 '12 at 9:35
Your updated version is still wrong for negative numbers. `'-21'.isalnum()` is `False`. Calling `int` and handling the error is the only simple solution that always works. Not everything needs to be a one-liner. – agf Mar 26 '12 at 9:36
Use `isdigit()` to check each character in the string to see if it is a digit.
Example:
``````mylist = ['foo', '3', 'bar', '9']
t = [ int(item) if item.isdigit() else item for item in mylist ]
print(t)
``````
-
See Thomas' comment on the other equivalent answer about why this can be wrong. – agf Mar 26 '12 at 9:32
Shouldn't `"-1"` be able to convert to int? – John La Rooy Mar 26 '12 at 9:41 | 1,672 | 5,570 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2015-32 | latest | en | 0.87979 |
https://compilers.iecc.com/comparch/article/92-12-071 | 1,501,094,934,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426372.41/warc/CC-MAIN-20170726182141-20170726202141-00581.warc.gz | 613,527,939 | 3,085 | # Re: question on control dependence
## Paul Havlak <paco@cs.rice.edu>Tue, 15 Dec 1992 19:13:56 GMT
From comp.compilers
Related articles
Extension Languages marks@iris.mincom.oz.au (1992-12-14)
question on control dependence mcconnel@newta1.cs.uiuc.edu (1992-12-14)
Re: question on control dependence cliffc@rice.edu (1992-12-15)
Re: question on control dependence preston@dawn.cs.rice.edu (1992-12-15)
Re: question on control dependence preston@dawn.cs.rice.edu (1992-12-15)
Re: question on control dependence paco@cs.rice.edu (Paul Havlak) (1992-12-15)
Re: question on control dependence bwilson@shasta.stanford.edu (1992-12-15)
Re: question on control dependence paco@cs.rice.edu (1992-12-16)
Re: question on control dependence dkchen@sp91.csrd.uiuc.edu (1992-12-16)
| List of all articles for this month |
Newsgroups: comp.compilers From: Paul Havlak Organization: Center for Research on Parallel Computation Date: Tue, 15 Dec 1992 19:13:56 GMT Keywords: optimize, design, bibliography References: 92-12-056 92-12-065
mcconnel@newta1.cs.uiuc.edu writes:
I've come up with a simple algorithm for computing control dependences, ...
Looks to me like the algorithm should work. I dislike it anyway
for two reasons:
* It's not very general; many languages allow more than two
branches out of a node.
* It uses bit vectors. Sure, you only require O(N) bit-vector
steps for N nodes, but the bit vectors are O(N) long,
so you're always paying O(N^2) time and space.
My favorite algorithm for control dependence calculation is in [3].
It requires no data structures besides the control-flow graph and the
postdominator tree, as input, and the control dependence graph, as output.
The algorithm takes time linear in the number of control dependences,
which is usually linear in the number of nodes.
First, build the postdominator tree using the method of [4]
(almost-linear) or [5] (linear). Then to enumerate the set CD(X, L) [the
control dependence successors of a node X with label L], do the following:
* CD(X, L) = {}
* Assign Y = the sink of X's control-flow outedge with label L
* While (Y != immediate postdominator of X)
CD(X, L) += Y
Assign Y = immediate postdominator of Y
(where += denotes addition of member to set)
Note that this method can also be used in the construction of SSA form;
see [3] for details.
References:
[3]
@inproceedings{CFS:Compact,
Author={R. Cytron and J. Ferrante and V. Sarkar},
Title={Compact representations for control dependence},
BookTitle=SIGPLAN90,
Pages={337--351},
Month=Jun,
Year={1990}}
[4]
@Article{LeTa:Dom,
Author = {T. Lengauer and R. E. Tarjan},
Title = {A fast algorithm for Finding Dominators in a flowgraph},
Journal = TOPLAS,
Volume = 1,
Pages = {121--141},
Year = 1979}
Actually, [3] cites another method for building the postdominator
tree which, I just realize, I've never looked up:
[5]
@inproceedings{Harel:Dom,
Author = {Dov Harel},
Title = {A linear-time algorithm for finding dominators in
flow graphs and related problems},
BookTitle = {Symposium on Theory of Computing},
Month = May,
Year = {1985}}
--
Paul Havlak Dept. of Computer Science
Graduate Student Rice University, Houston TX 77251-1892
PFC/ParaScope projects (713) 527-8101 x2738 paco@cs.rice.edu
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Post a followup to this message | 907 | 3,273 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-30 | longest | en | 0.800971 |
https://de.mathworks.com/matlabcentral/cody/solutions/147112 | 1,563,732,840,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00395.warc.gz | 364,856,952 | 15,676 | Cody
# Problem 167. Pizza!
Solution 147112
Submitted on 11 Oct 2012 by Greg
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
%% z = 1; a = 1; v_correct = pi; assert(isequal(pizza(z,a),v_correct))
2 Pass
%% z = 2; a = 1; v_correct = 4*pi; assert(isequal(pizza(z,a),v_correct))
3 Pass
%% z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct))
4 Pass
%% z = 1; a = 2; v_correct = 2*pi; assert(isequal(pizza(z,a),v_correct)) | 195 | 566 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-30 | latest | en | 0.57815 |
https://www.transtutors.com/questions/saggy-butz-corporation-issued-40-million-of-12-10-year-bonds-on-march-1-2011-the-mar-1084507.htm | 1,542,590,192,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744803.68/warc/CC-MAIN-20181119002432-20181119024432-00370.warc.gz | 991,951,845 | 18,495 | # Saggy Butz Corporation issued $40 million of 12%, 10 year bonds on March,1 2011. The market rate of. 1 answer below » Saggy Butz Corporation issued$40 million of 12%, 10 year bonds on March,1 2011. The market rate of interest at the date of issue was 14%. Interest is payable on September 1 and March 1. The fiscal year of the corporation ends on November 30. Prepare all journal entries for the first year on the bonds. The company uses the straight line method on amortization.
Par Value $40,000,000.00 Future Value$ 40,000,000.00 Coupon Rate 6.00% Rate 7.00% Nper 20.00 PMT = Par Value * Coupon Rate PMT = $40,000,000 * 6% PMT$ 2,400,000.00 Using excel formula, Present Value $35,762,394.30 Journal Entries in the books of Saggy Butz Corporation Date Particulars Debit Credit 1-Mar-11 Cash$35,762,394.30 Discount on issue of bonds $4,237,605.70 Bonds Payable$ 40,000,000.00 (To record issue of... | 272 | 902 | {"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} | 2.703125 | 3 | CC-MAIN-2018-47 | longest | en | 0.797071 |
https://electronics.stackexchange.com/questions/203403/how-to-simulate-adjustable-negative-resistance-with-active-components | 1,726,148,560,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651457.35/warc/CC-MAIN-20240912110742-20240912140742-00739.warc.gz | 215,398,643 | 41,384 | # How to simulate adjustable 'negative resistance' with active components?
I'm building a decade resistance box using 2W resistors, from 10x1M ohm to 10x0.1 ohm. However, due to some contact resistance, the 0.1 to 1 ohm resistance will likely turn out to be ~1.1 ohm to 2 ohms (the rest of which will be in a tolerable margin of error). Is there a way I can 'offset' the ~1 ohm of wiper and contact resistance with some active components, with an adjustable potentiometer for fine tuning?
I'm thinking -1 ohm +/- 0.5 ohms ought to do, and ideally it would handle up to 2W just like the other resistors. Is such a circuit possible?
• Find lower contact resistances.
– user16324
Commented Nov 29, 2015 at 19:53
• I can't, really. Even passive units costing thousands of dollars have residual resistances of around 1 ohm and report their accuracy after having subtracted their 'zero resistance'. I'm looking for a way to do this internally, if such a thing is possible. Commented Nov 29, 2015 at 20:01
• @Ehryk: My 1953 Siemens resistance decade has a total contact resistance of 80mOhm. Most modern mid price range resistance boxes I know of have contact resistances around 300mOhm. Even my cheap 1% quickly slapped together resistance decade has a resistance of 800mOhm. Maybe you were not looking at the right place? Commented Nov 29, 2015 at 20:16
• Perhaps, but I'm building my own and I'd like to be able to trim out whatever is there so that I get a 'true' 0.1, 0.2, 0.3 ohm with an acceptable percentage of error (the resistors are all 1%). Is this possible? Commented Nov 29, 2015 at 21:58
• I.E. even with 300mOhm, your '0.1 ohm' setting would resultin 0.4 ohms, no? 400% error? Commented Nov 29, 2015 at 21:59 | 496 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.932992 |
https://www.hackmath.net/en/math-problem/27501 | 1,721,688,935,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517927.60/warc/CC-MAIN-20240722220957-20240723010957-00578.warc.gz | 674,330,508 | 8,572 | # Between 27501
Dan and Honza live 4 km from each other. They agreed to meet on the way between the two homes. Dan went out at 2 p.m. at a speed of 5 km/h. Honza rode towards him on a bicycle at 15 km/h. What time did they meet, and how far did John travel?
Result
t = 14:12 hh:mm
s2 = 3 km
## Step-by-step explanation:
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Do you want to convert velocity (speed) units?
Do you want to convert time units like minutes to seconds? | 174 | 663 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2024-30 | latest | en | 0.95874 |
http://physics.stackexchange.com/questions/70495/how-much-energy-is-consumed-by-a-aircraft-to-maintain-a-speed-of-sound | 1,469,570,127,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825124.55/warc/CC-MAIN-20160723071025-00236-ip-10-185-27-174.ec2.internal.warc.gz | 184,245,753 | 16,507 | # How much energy is consumed by a aircraft to maintain a speed of sound? [closed]
Let's assume that the aircraft is 1000kg and it is flying in a air density of 1.225kg/m^3 at the speed of sound in air.
Just how much energy does it require per second to maintain flying at this speed.(Please make other necessary assumptions.)
Thank you.
-
## closed as off-topic by Chris White, akhmeteli, Nathaniel, Dilaton, Qmechanic♦Jul 10 '13 at 12:40
This question appears to be off-topic. The users who voted to close gave this specific reason:
If this question can be reworded to fit the rules in the help center, please edit the question.
Questions that basically come down to "solve this numerical problem for me" usually aren't well received here... did you try to solve this yourself? What did you try? Where did you get stuck? – Kyle Oman Jul 9 '13 at 23:31
Sorry, and thanks for letting me know. – user26863 Jul 10 '13 at 8:44
"Other necessary assumptions?" Okay, the air craft is a long, thin rod about the diameter of a pencil, with tiny wings. No wait, it's a 500 kg jet engine, dragging behind it a large diameter open parachute made of 400 kg of nylon. – Kaz Jul 10 '13 at 22:37
## 1 Answer
Along the lines of what Kyle is saying, nobody is going to do the calculation for you, but I think the answer is conceptually more comlpicated that the other two clear regimes (supersonic and subsonic).
Supersonic flight and subsonic flight incur two different kind of airflow. Trying to travel at exactly the speed of sound would present a lot of difficulties in the first place, because the speed of sound is sensitive to air composition and temperature, which will change as you traverse the sky.
Regardless, this speed of travel, near Mach 1, is called "transonic" and encompasses a range where both supersonic and subsonic flows co-exist in complicated and chaotic ways: it can cause a lot of instabilities and put undesirable sheer forces on the aircraft, so the aerodynamic design of the aircraft is going to have a significant effect. Between the increased drag force associated with transonic speeds, the instabilities causing control problems, and possibly having to endure several shock waves, I speculate that it would take a lot more energy than traveling at either supersonic or subsonic speeds.
-
Hello, I do agree that at transonic speed there are a lot more that need to be taking into account but if it was supersonic, would you say that considering the work done against the drag is a good approximation of the energy consumption? Thanks. – user26863 Jul 10 '13 at 8:47 | 605 | 2,594 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-30 | latest | en | 0.965345 |
http://razhayesheitanparastan.com/xzq0tg/1j13cd74msztf1t9/ | 1,580,186,951,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00112.warc.gz | 132,148,693 | 10,080 | # Graphing Parabolas Worksheet Algebra 1 Answers Sec 7 Homework Ppt Download
Thursday, September 19th 2019. | Worksheets Excel
## Animal Farm Worksheets
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If you're interested in figuring out how to Quote with Excel, Open a and construct your workbook when . Excel just gives you the capability to unhide one too. Fortunately, it includes a feature which do this very processthe . When entering dates, it still defaults to the current year if the year portion of the date isn't entered. | 348 | 1,661 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2020-05 | latest | en | 0.867064 |
https://www.sanfoundry.com/mathematics-questions-answers-geometry-distance-formula/ | 1,726,783,271,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00257.warc.gz | 905,279,536 | 19,961 | # Class 10 Maths MCQ – Coordinate Geometry – Distance Formula
This set of Class 10 Maths Chapter 7 Multiple Choice Questions & Answers (MCQs) focuses on “Coordinate Geometry – Distance Formula”.
1. The distance between the points (5, 7) and (8, -5) is ________
a) √153
b) √154
c) √13
d) √53
Explanation: Using distance formula,
Distance between (5, 7) and (8, -5) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(8-5)^2 + (-5-7)^2}$$
= $$\sqrt {(3)^2 + (-12)^2}$$
= $$\sqrt {9 + 144}$$
= √153
2. The distance of the point (9, -12) from origin will be ___________
a) 13
b) 15
c) 14
d) 17
Explanation: Distance between (9, -12) and (0, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(0-9)^2 + (0 + 12)^2}$$
= $$\sqrt {(9)^2 + (-12)^2}$$
= $$\sqrt {81 + 144}$$
= √225 = 15
3. What will be the value of x, if the distance between the points (5, 11) and (2, x) is 10?
a) -11 + √91, -11 – √91
b) 11 + √91, 11 – √91
c) 11 + √91, 11 + √91
d) -11 + √91, 11 – √91
Explanation: Distance between (5, 11) and (2, x) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(2-5)^2 + (x-11)^2}$$
= $$\sqrt {x^2-22x + 121 + (-3)^2}$$
= $$\sqrt {x^2-22x + 121 + 9}$$
= $$\sqrt {x^2-22x + 130}$$
The distance between (5, 11) and (2, x) is 10
∴ $$\sqrt {x^2-22x + 130}$$ = 10
Squaring on both sides we get,
x2 – 22x + 130 = 100
x2 – 22x + 130 – 100 = 0
x2 – 22x + 30 = 0
x = 11 + √91, 11 – √91
4. What will be the point of x-axis which will be equidistant from the points (9, 8) and (3, 2)?
a) (10, 0)
b) (13, 0)
c) (11, 0)
d) (12, 0)
Explanation: Let the point on x-axis be (x, 0)
Distance between (9, 8) and (x, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(x-9)^2 + (0-8)^2}$$
= $$\sqrt {x^2-18x + 81 + (-8)^2}$$
= $$\sqrt {x^2-18x + 81 + 64}$$
= $$\sqrt {x^2-18x + 145}$$
Distance between (3, 2) and (x, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(x-3)^2 + (0-2)^2}$$
= $$\sqrt {x^2-6x + 9 + (-2)^2}$$
= $$\sqrt {x^2-6x + 9 + 4}$$
= $$\sqrt {x^2-6x + 13}$$
Since, the point ( x, 0) is equidistant to (3, 2) and (9, 8)
The distances will be equal
∴ $$\sqrt {x^2-18x + 145} = \sqrt {x^2-6x + 13}$$
Squaring on both sides we get,
x2 – 18x + 145 = x2 – 6x + 13
-18x + 145 = -6x + 13
-18x + 6x = -145 + 13
-12x = -132
x = $$\frac {132}{12}$$ = 11
The point is (11, 0)
5. What will be the point of y-axis which will be equidistant from the points (-1, 0) and (3, 9)?
a) (5, $$\frac {89}{18}$$)
b) (1, $$\frac {89}{18}$$)
c) (16, $$\frac {89}{18}$$)
d) (0, $$\frac {89}{18}$$)
Explanation: Let the point on y-axis be (0, y)
Distance between (-1, 0) and (0, y) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(0 + 1)^2 + (y-0)^2}$$
= $$\sqrt {y^2 + (1)^2}$$
= $$\sqrt {y^2 + 1}$$
Distance between (3, 9) and (0, y) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(0-3)^2 + (y-9)^2}$$
= $$\sqrt {y^2-18y + 81 + (-3)^2}$$
= $$\sqrt {y^2-18y + 81 + 9}$$
= $$\sqrt {y^2-18y + 90}$$
Since, the point (0, y) is equidistant from (-1, 0) and (3, 9)
The distances will be equal
∴ $$\sqrt {y^2 + 1} = \sqrt {y^2-18y + 90}$$
Squaring on both sides we get,
y2 + 1 = y2 – 18y + 90
1 – 90 = -18y
-89 = -18y
y = $$\frac {89}{18}$$
The point is (0, $$\frac {89}{18}$$)
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6. If the point P(a, b) is equidistant from the points (3, 1) and (2, 0) then ____________
a) a + b = -3
b) a – b = -3
c) a + b = 3
d) a – b = 3
Explanation: The point is (a, b)
Distance between (3, 1) and (a, b) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(a-3)^2 + (b-1)^2}$$
= $$\sqrt {a^2-6a + 9 + b^2-2b + 1}$$
= $$\sqrt {a^2-6a + 10 + b^2-2b}$$
Distance between (2, 0) and (a, b) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(a-2)^2 + (b-0)^2}$$
= $$\sqrt {a^2-4a + 4 + b^2 }$$
Since, the point (a, b) is equidistant from (-1, 0) and (3, 9)
The distances will be equal
∴ $$\sqrt {a^2-6a + 10 + b^2-2b} = \sqrt {a^2-4a + 4 + b^2 }$$
Squaring on both sides we get,
a2 – 6a + 10 + b2 – 2b = a2 – 4a + 4 + b2
-6a + 10 – 2b = -4a + 4
-2a – 6 = 2b
-a – b = 3
a + b = -3
7. The point on y-axis which is at a distance 5 unit from the point (-5, 7) is ___________
a) (7, 0)
b) (0, 7)
c) (1, 7)
d) (7, 7)
Explanation: Let the point on y-axis be (0, y)
Distance between (-5, 7) and (0, y) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(0 + 5)^2 + (y-7)^2}$$
= $$\sqrt {y^2-14y + 49 + 25}$$
= $$\sqrt {y^2-14y + 74}$$
The distance between (-5, 7) and (0, y) is 5
∴ $$\sqrt {y^2-14y + 74}$$ = 5
Squaring on both sides, we get,
y2 – 14y + 74 = 25
y2 – 14y + 49 = 0
y = 7, 7
Hence, the point is (0, 7)
8. The point on x-axis which is at a distance 12 unit from the point (4, 6) is ___________
a) (-4 + $$\sqrt {11i}$$, 0), (-4 – $$\sqrt {11i}$$, 0)
b) (-4 – $$\sqrt {11i}$$, 0), (4 – $$\sqrt {11i}$$, 0)
c) (4 – $$\sqrt {11i}$$, 0), (4 – $$\sqrt {11i}$$, 0)
d) (4 + $$\sqrt {11i}$$, 0), (4 – $$\sqrt {11i}$$, 0)
Explanation: Let the point on x-axis be (x, 0)
Distance between (4, 6) and (x, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(x-4)^2 + (0-6)^2}$$
= $$\sqrt {x^2-8x + 16 + 36}$$
= $$\sqrt {x^2-8x + 52}$$
The distance between (4, 6) and (x, 0) is 12
∴ $$\sqrt {x^2-8x + 52}$$ = 12
Squaring on both sides, we get,
x2 – 8x + 52 = 25
x2 – 8x + 27 = 0
x = 4 + $$\sqrt {11i}$$, 4 – $$\sqrt {11i}$$
9. If A(0, 3), B(5, 0) and C(-5, 0) are the vertices of ∆ABC, then the triangle is __________
a) Right-angled
b) Isosceles
c) Scalene
d) Equilateral
Explanation: Distance between (0, 3) and (5, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(5-0)^2 + (0-3)^2}$$
= $$\sqrt {5^2 + -3^2 }$$
= $$\sqrt {25 + 9}$$
= √34
Distance between (5, 0) and (-5, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(-5-5)^2 + (0-0)^2}$$
= $$\sqrt {-10^2}$$
= 10
Distance between (0, 3) and (-5, 0) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(-5-0)^2 + (0-3)^2}$$
= $$\sqrt {-5^2 + (-3)^2}$$
= $$\sqrt {25 + 9}$$
= √34
Since, the two sides of the triangle are equal.
Hence, the triangle will be isosceles triangle.
10. The area of the triangle if A (-1, -1), B(-1, 3) and C (2, -1) are the vertices of the triangle is ____________
a) 8 units
b) 4 units
c) 6 units
d) 5 units
Explanation: Distance between A (-1, -1) and B (-1, 3) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(-1 + 1)^2 + (3 + 1)^2}$$
= $$\sqrt {4^2}$$
= √16
= 4
Distance between B (-1, 3) and C(2, -1) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(2 + 1)^2 + (-1-3)^2}$$
= $$\sqrt {3^2 + (-4)^2}$$
= $$\sqrt {9 + 16}$$
= 5
Distance between A (-1, -1) and C (2, -1) = $$\sqrt {(x_2-x_1)^2 + (y_2-y_1)^2}$$
= $$\sqrt {(2 + 1)^2 + (-1 + 1)^2}$$
= $$\sqrt {3^2 + 0^2}$$
= 3
Now, AC2 + AB2 = 42 + 32 = 16 + 9 = 25
BC2 = 52 = 25
Hence, it is a right-angled triangle, right-angled at A.
Area of triangle = $$\frac {1}{2}$$ × base × height = $$\frac {1}{2}$$ × 4 × 3 = 6 units
Sanfoundry Global Education & Learning Series – Mathematics – Class 10.
To practice all chapters and topics of class 10 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected] | 3,487 | 7,151 | {"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} | 4.46875 | 4 | CC-MAIN-2024-38 | latest | en | 0.720122 |
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## Example Questions
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### Example Question #1 : Parallel Reasoning
It has been scientifically established that all dogs bark. As a result, any animal that barks is a dog. So, if a person hears an animal bark, that person can safely conclude that the animal is a dog.
Which of the following arguments most closely parallels the flawed reasoning above?
If all high interest debt should be avoided, and if some debt is high interest, then some debt should be avoided.
All high interest debt should be avoided. Debt that is not high interest need not be avoided. So, people should prefer low interest debt.
High interest debt should sometimes be avoided. As a result, some debt that should be avoided is high interest debt. So, a person can safely conclude that high interest debt should be avoided.
Only high interest debt is debt that should be avoided. Debt that is not high interest should not be avoided.
All debt that should be avoided is high interest debt because all high interest debt should be avoided. Debt that should be avoided must be high interest debt.
All debt that should be avoided is high interest debt because all high interest debt should be avoided. Debt that should be avoided must be high interest debt.
Explanation:
The argument assumes that if all A are B, then all B must be A. That is the central flaw of the argument because it may be true that all dogs bark and that seals also bark. Similarly, although it may be true that all high interest debt should be avoided, there may be other debt that should be avoided as well. It is important to note that the reasoning can be parallell even when the order of the argument may be slightly different.
### Example Question #1 : Determining Which Answer Uses Parallel Reasoning To The Argument Provided
Everyone who thinks the Rams would win the championship thought that Jones would receive the award for Most Valuable Player. But Jones did not receive the award for Most Valuable Player. Therefore, anyone who believes the Rams will win the championship is wrong.
Which one of the following arguments contains flawed reasoning most similar to that in the argument above?
Anyone who thinks that eating before exercising is a good idea has never taken a health class. But Jim has never taken a health class and knows that he should not eat before exercising. Therefore, taking a health class is not necessary for you to know eating before exercising is not a good idea.
Anyone who thinks chickens are ugly thinks ducks are ugly. Since there is no reason to think ducks are ugly there is no reason to think chickens are ugly.
Anyone who believes animals deserve better treatment believes that animals are capable of moral judgment. You do not believe that animals deserve better treatment so you do not think they are capable of moral judgment.
Anyone that believes that seagulls migrate based on advanced spatial recognition patterns believes that most bird species have highly developed frontal cortexes. But is has been conclusively proven that most bird species do not have highly developed frontal cortexes. Thus, the belief that seagulls migrated based on advanced spatial recognition patterns is false.
If you believe in fairies then you do not believe in vampires. Since Cindy believes in vampires, she cannot believe in fairies.
Anyone that believes that seagulls migrate based on advanced spatial recognition patterns believes that most bird species have highly developed frontal cortexes. But is has been conclusively proven that most bird species do not have highly developed frontal cortexes. Thus, the belief that seagulls migrated based on advanced spatial recognition patterns is false.
Explanation:
The correct answer parallels the flawed reasoning of the stimulus. Both give a situation where the first argument is dismissed on the basis that the second argument presented is false.
### Example Question #2 : Determining Which Answer Uses Parallel Reasoning To The Argument Provided
While there are certain dog-training techniques that generally tend to be effective for most dogs, it is important to keep in mind that each dog will respond particularly well to certain techniques that other dogs might not be receptive to. Therefore, the best practice is to keep general principles in mind when training dogs, while tailoring their educations to their particular traits.
Which of the following propositions does this reasoning above most closely conform to?
Unique individuals and unique circumstances call for specialized training to meet their needs.
Even though general principles are broad enough to encompass most individuals or circumstances, there will always be outliers that require special attention.
Although dogs are different, they are similar enough that a basic set of training principles can be applied to them universally, with slight room for variation.
While a general set of principles exist for accomplishing a goal, it is important to acknowledge the unique circumstances or qualities that may exist, and to take them into consideration.
It is dangerous to apply a broad set of principles to a set of unique individuals because they will react to the same set of circumstances differently.
While a general set of principles exist for accomplishing a goal, it is important to acknowledge the unique circumstances or qualities that may exist, and to take them into consideration.
Explanation:
The correct answer most closely conforms with the excerpt.
The incorrect answers are wrong for the following reasons:
1. Saying that dogs should be treated the same with "slight room for variation" emphasizes continuity over individualization, and therefore is wrong.
2. The statement "It is dangerous to apply a broad set of principles. . . " is not stated in the excerpt.
3. "Unique individuals and unique circumstances call for specialized training to meet their needs" exaggerates the need for individualization.
4. The concept of "outliers" is not present in the excerpt, a red flag that this is an incorrect answer choice.
### Example Question #4 : Parallel Reasoning
Economist: All hamburger joints must offer fries and drinks to maximize their revenue. The Burger Shack is a hamburger joint. Because it offers fries and drinks, its revenue is clearly being maximized.
The flawed reasoning in which one of the following is most similar to that in the economist’s argument?
Every piano teacher needs to have a clear course of instruction and a patient personality to succeed. Beverly, a piano teacher, has both of these, so she is undoubtedly successful.
Every traffic jam is caused by two factors: impatient driving and lack of available roads. In the city of San Calistranus, there are plenty of available roads and few impatient drivers. So, clearly, there are few traffic jams in San Calistranus.
For a video game to sell well, it must include both a memorable character and lots of action. Arkham’s Revenge is a video game that has been selling well. Therefore, it must have a memorable character and lots of action.
It will never snow unless the temperature is below 40 degrees and there is sufficient humidity. On Thursday there was sufficient humidity, but the temperature was 45 degrees, so it didn’t snow.
Successful campaigning relies on two factors: powerful action committees and sufficient exposure. Sarah Strong was a congressional candidate in the last election. She must have had a powerful action committee and sufficient exposure, because she ran a successful campaign.
Every piano teacher needs to have a clear course of instruction and a patient personality to succeed. Beverly, a piano teacher, has both of these, so she is undoubtedly successful.
Explanation:
The correct answer choice is the only one which commits the fallacy of affirming the consequent. In other words, both it and the original argument improperly assume the converse of their conditional statements. We know that the Burger Shack needs to offer fries and drinks to maximize its revenue, but we do not know that its revenue is necessarily maximized if it offers them. Likewise, we do not know whether Beverly is a successful piano teacher simply because she has a clear course of instruction and a patient personality.
### Example Question #5 : Parallel Reasoning
All limes are green. Therefore, any fruit that is green is a lime. If a person sees a fruit that is green, that person may assume that the fruit is a lime.
Which of the following most closely parallels the flawed reasoning above?
Liberal views are expressed in immigration courts. Therefore, immigration courts are liberal. If a person encounters a liberal view, that person can assume that the view was expressed in an immigration court.
Immigration courts tend to adopt more liberal views. Therefore, if a person encounters a conservative view, it is unlikely that the view was expressed in an immigration court.
Immigration courts adopt more liberal views. Therefore, any court that adopts a more liberal view is an immigration court. If a person encounters a court that has a more liberal view, that person may assume that the court is an immigration court.
Immigration courts have more liberal views. Therefore, liberal views are only expressed by immigration courts. Therefore, if a person encounters an immigration court, that person can assume that the court will have a liberal view.
Liberal views are often expressed in immigration courts. Therefore, is a person encounters a liberal view, it is likely that the view was expressed in an immigration court.
Immigration courts adopt more liberal views. Therefore, any court that adopts a more liberal view is an immigration court. If a person encounters a court that has a more liberal view, that person may assume that the court is an immigration court.
Explanation:
The flawed reasoning in the text is as follows:
X has trait Y. Therefore, anything with trait Y must be X. If a person encounters something with trait Y, then it must be X.
The correct answer properly reflects this reasoning. All the other answer choices do not properly follow this pattern: they may leave out a link of the causal reasoning or misconstruct it.
### Example Question #3 : Determining Which Answer Uses Parallel Reasoning To The Argument Provided
If a wine receives a high score from the National Sommelier Association, it is more likely to be sold in fine-dining restaurants. A new wine from Oregon just received a high score from the National Sommelier Association. Therefore, it is more likely to be sold in fine-dining restaurants.
Which of the following choices most closely reflects the reasoning in the argument above?
Brighter colors attract more attention. The professor attracts a lot of attention. Therefore, he probably tends to wear more bright colors.
Older coins are more valuable. Anita has a coin believed to have originated from the colonial era, but lacks any proof. Therefore, it is questionable whether her coin is valuable.
Using recycled materials is beneficial to the environment. Some newspapers only use recycled paper in their products. Therefore, a newspaper is likely to be beneficial to the environment.
Pearls that are larger than average are more likely to retail for a higher price. A pearl diver found a dozen pearls that are much larger than average. Therefore, the pearls are more likely to retail for a higher price.
Famous actors tend to appear in popular movies. Popular movies often feature famous actors. Therefore, it is unlikely to find a popular movie without famous actors.
Pearls that are larger than average are more likely to retail for a higher price. A pearl diver found a dozen pearls that are much larger than average. Therefore, the pearls are more likely to retail for a higher price.
Explanation:
The flow of reasoning in the argument is that if X has Y quality, then Z is likely to happen.
In the text,
X = wine
Y = receives a high score from the National Sommelier Association
Z = likely to be sold at fine dining restaurants
The correct answer follows this reasoning:
Pearls that are larger than average are more likely to retail for a higher price. A pearl diver found a dozen pearls that are much larger than average. Therefore, the pearls are more likely to retail for a higher price.
X = pearls
Y = larger than average
Z = more likely to retail at a higher price
### Example Question #4 : Determining Which Answer Uses Parallel Reasoning To The Argument Provided
All English Springer Spaniels have long hair. All Rottweilers have short hair. Each of Tina's dogs is a cross between an English Springer Spaniel and a Rottweiler. Therefore, Tina's dogs have medium-length hair.
Which one of the following uses flawed reasoning that most closely resembles the flawed reasoning used in the argument above?
All players on the Wildcats have brown hair. All players on the Razorbacks have red hair. Members of the Moye family are on both the Wildcats and the Razorbacks. Therefore, some members of the Moye family have brown hair and others have red hair.
All typists who practice at least one hour per day can type one hundred words per minute. But some typists who do not practice can also type one hundred words per minute. Mike, a typist, practices thirty minutes per day. Therefore, Mike types fifty words per minute.
All halogen gases are toxic to humans. All non-radioactive noble gases are non-toxic to humans. "Nobagen" gas is a mixture of a halogen gas and a noble gas. Therefore, "nobagen" gas is moderately toxic to humans.
All cars made by Chord are very well made. All cars made by Fysler are very poorly made. Half of the cars on Jim's lot are very well made and the other half are very poorly made. Therefore, half of the cars on Jim's lot are Chords and half are Fyslers.
All economists know linear algebra. All physicists know relativistic mechanics. Wilma is both an economist and a physicist. Therefore, Wilma knows both linear algebra and relativistic mechanics.
All halogen gases are toxic to humans. All non-radioactive noble gases are non-toxic to humans. "Nobagen" gas is a mixture of a halogen gas and a noble gas. Therefore, "nobagen" gas is moderately toxic to humans.
Explanation:
The flawed reasoning used in the passage is that a combination of two "parent" items with different attributes necessarily yields a "child" product having attributes that are averages of its parents' attributes. The correct answer uses parallel reasoning inasmuch as the argument uses the fact that halogen and noble gases differ with respect to toxicity to conclude that a combination of such gases would yield a gas having toxicity that is the average of the toxicity of its "parent" gases.
### Example Question #5 : Determining Which Answer Uses Parallel Reasoning To The Argument Provided
A high school football coach has made public comments criticizing the decision by the football coaching staff of the local university to not play their star quarterback. However, we should not listen to the high school coach’s criticism. His high school football team has not won a game in several seasons.
The flawed reasoning above most closely resembles which of the following arguments?
We should not listen to the neurobiologist’s predictions about the future state of the economy because he has no formal training in economics.
We should not heed the weather channel's warnings. They have failed to correctly predict the past twenty rainy days.
It is likely that the scholar plagiarized this paper because she has been known to plagiarize in the past.
We should not listen to the car salesperson because she has an incentive to ignore negative features of the car in order to make commission off of a sale.
We should not listen to this art critic's negative comments because it is well known that the art critic is a mediocre artist.
We should not listen to this art critic's negative comments because it is well known that the art critic is a mediocre artist.
Explanation:
The stimulus holds that an opinion is false on the basis that the person with that opinion has made past mistakes in the same area. The correct answer is similar because it follows the same pattern of critiquing a position based on the person with the opinion not the opinion itself.
### Example Question #9 : Parallel Reasoning
The Forestry Service has issued new warnings about forest fires, which hikers in the National Forest are encouraged to read thoroughly. The Forestry Service believes that more than two-thirds of forest fires last year were down to human error. Their hope is to seriously curtail the amount of fires for the coming year.
Which statement uses reasoning parallel to the reasoning used in the above statement?
A school having a class that teaches teenagers safer driving practices in order to lower the number of car wrecks
A company creating a training video to explain new procedures in their warehouses
A non-profit organization providing relief to victims of natural disasters
A football coach instituting a new formation to gain an advantage on opponents
A small business owner adding new pieces to his inventory to increase sales
A school having a class that teaches teenagers safer driving practices in order to lower the number of car wrecks
Explanation:
The Forestry Service's main goal is to reduce the number of forest fires from year to year. Their reasoning is that educating people on how fires are started is the most effective way to achieve this goal. Similarly, the school wishes to reduce car wrecks among teenage drivers by educating teenagers about safe driving practices.
### Example Question #10 : Parallel Reasoning
Few people have ever seen the initiation rituals of the secret society, as its leadership believes that if anyone reports these deep mysteries then the society will cease to have an attraction.
The reasoning used by the secret society’s leadership is most closely paralleled by __________.
a company enforcing a non-disclosure agreement about its new products
a criminal organization threatening any members who reveal information to law enforcement
a magician not revealing how tricks are performed in order to keep attracting an audience
a production company not committing itself to any specific future projects
a government agency designating specific documents as classified material | 3,591 | 18,538 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-04 | latest | en | 0.909652 |
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# Ramnarayan Krishnamurthy
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Dan Doel dan.doel at gmail.com
Sat Mar 26 20:11:20 UTC 2016
```The extra allocation/time isn't caused by allocating pairs (neither
version does this). It is caused by allocating Doubles, because GHC
doesn't figure out that the second part of your pairs is always
demanded (I think), so the corresponding argument is left boxed. sumh
essentially turns into:
sumh (D# d#) = sumh1 d#
sumh1 d# = case loop1 1# 0# of
(# _, ans #) -> ans
where
loop1 :: Double# -> Double# -> (# Double, Double #)
loop1 i# s#
| i# >= d# = (# D# i#, D# s# #)
| otherwise = loop2 (i# + 2#) (D# ...)
loop2 :: Double# -> Double -> (# Double, Double #)
loop2 i# s
| i# >= d# = (# D# i#, s #)
| otherwise = loop2 (i# + 2#) (...)
So, it starts as a loop on unboxed arguments for one step, but then
switches over to boxing the second argument after the first iteration.
If you consider loop2 in isolation (and the hypothetical loop that
both of these came from), it is not strict in s. So there is no reason
to keep s evaluated except by considering the calling context, which I
guess GHC is not doing.
One way to fix this is by modifying the predicate to make is stricter.
Instead of `((>=n) . fst)` you can use:
p (i, s) = s `seq` (i >= n)
Then it will be obvious to GHC that it shouldn't delay computing `s`
at each step.
This may be a similar failure to the old 'constructed product result
inside unboxed tuples' bug, where a function with an `ST s Int` result
could internally be optimized to return an unboxed integer, but it
isn't. Your example isn't exactly the same, but it has a similar
flavor. A call site is strict in part of the result of a function, and
generating code for that strict case would enable better optimization
of the callee, but the optimization would have to act on individual
parts of an unboxed tuple (which in your case comes from optimizing a
tuple), and that isn't done.
-- Dan
On Sat, Mar 26, 2016 at 1:50 PM, George Colpitts
<george.colpitts at gmail.com> wrote:
> The following higher order function, sumh, seems to be 3 to 14 times slower
> than the equivalent recursive function, sumr:
>
> sumh :: Double -> Double
> sumh n =
> snd \$ until ((>= n) . fst) as' (1, 0)
> where
> as' (i,s) =
> (i + 2, s + (-1) / i + 1 / (i + 1))
>
> sumr :: Double -> Double
> sumr n =
> as' 1 0
> where
> as' i s
> | i >= n = s
> | otherwise = as' (i + 2) (s + (-1) / i + 1 / (i + 1))
>
> This is true in 7.10.3 as well as 8.0.1 so this is not a regression. From
> the size usage my guess is that this is due to the allocation of tuples in
> sumh. Maybe there is a straightforward way to optimize sumh but I couldn't
> find it. Adding a Strict pragma didn't help nor did use of
> -funbox-strict-fields -flate-dmd-anal. Have I missed something or should I
> file a bug?
>
> Timings from 8.0.1 rc2:
>
> ghc --version
> The Glorious Glasgow Haskell Compilation System, version 8.0.0.20160204
> bash-3.2\$ ghc -O2 -dynamic sum.hs
> ghc -O2 -dynamic sum.hs
> [1 of 1] Compiling Main ( sum.hs, sum.o )
> bash-3.2\$ ghci
> (0.05 secs,)
> Prelude Main> sumh (10^6)
> -0.6931466805602525
> it :: Double
> (0.14 secs, 40,708,016 bytes)
> Prelude Main> sumr (10^6)
> -0.6931466805602525
> it :: Double
> (0.01 secs, 92,000 bytes)
>
> Thanks
> George
>
> _______________________________________________
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May 2, 2016
# Homework Help: geometry
Posted by omar on Wednesday, June 29, 2011 at 7:55am.
abcd is a rectangle such that ab=-3 radical sign 48+2 radical sign 75+4radical sign108 and Bc=4 radical sign 27 +radical sign 300 prove that abcd is a square | 85 | 259 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2016-18 | longest | en | 0.879414 |
https://www.cfd-online.com/Forums/openfoam-programming-development/86164-fvc-div-surfacevectorfield.html | 1,506,042,625,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688103.0/warc/CC-MAIN-20170922003402-20170922023402-00046.warc.gz | 753,683,667 | 18,770 | # fvc::div for surfaceVectorField
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March 16, 2011, 05:18 fvc::div for surfaceVectorField #1 Member Artem Shaklein Join Date: Feb 2010 Location: Russia, Izhevsk Posts: 43 Rep Power: 9 Greetings ! I want to calculate divergence of velocity. Firstly, I have volVectorField U. After some manipulations with U, I obtain surfaceScalarField Usurface. Next, i try to calculate divergence: fvc::div(Usurface), but this action leads to unexpected result. For example, when i compute div of mesh.Sf() (dimension [m^2]), i recieve volVectorField (dimension [m^-1]) instead of volScalarField (dimension [m^1]). But computation grad of mag(mesg.Sf()) (surfaceScalarField, dimension [m^2]) leads to volVectorField with dimension [m^1]. This is right, There are my questions: -Why fvc::grad(surfaceScalarField) gives correct result, but fvc::div(surfaceVectorField) - incorrect? May be i make mistake in calculations? -Can i calculate divergence of surfaceVectorField using fvc::div? Thank you! Best regards! Last edited by ARTem; March 16, 2011 at 11:10.
January 9, 2017, 08:21
#2
New Member
toboto
Join Date: Jun 2016
Posts: 13
Rep Power: 3
I've exactly the same problem. When I interpolate velocity field to the surface I got a surfaceVectorField. When I try to calculate the divergence of this field I got a volVectorField while I expect a volScalarField.
According to the programmers' manual:
Quote:
The fvc::div function can take as its argument either a surfaceField, in which case φf is specified directly, or a volField which is interpolated to the face by central differencing
Update:
I found in the source code of fvcDiv.H that if the fvc::div is applied to a volField it will return a field of one rank lower
Code:
GeometricField <typename innerProduct<vector, Type>::type, fvPatchField, volMesh> > div
(
const GeometricField<Type, fvPatchField, volMesh>&,
const word& name
);
While if fvc::div is applied to a scalarField it will return a field of the same rank!!!
Code:
tmp<GeometricField<Type, fvPatchField, volMesh>> div
(
const GeometricField<Type, fvsPatchField, surfaceMesh>&
);
Any explanation is highly appreciated.
Best regards.
Last edited by toboto; January 9, 2017 at 09:52. Reason: Add additional information
January 14, 2017, 17:47 #3 Senior Member Sergei Join Date: Dec 2009 Posts: 195 Rep Power: 10 If you check source code you'll see there're 14 overloaded functions fvc:div() for several combinations of input parameters. This set of functions can be broken down into three groups with one "primary" function within each group being called by other functions in the same group. Those primary functions are declared as: Code: template tmp< GeometricField< typename innerProduct< vector, Type >::type, fvPatchField, volMesh > > div ( const GeometricField< Type, fvPatchField, volMesh > & vf, const word &name ) Code: template tmp< GeometricField< Type, fvPatchField, volMesh > > div ( const surfaceScalarField &flux, const GeometricField< Type, fvPatchField, volMesh > &vf, const word &name ) Code: template tmp< GeometricField< Type, fvPatchField, volMesh > > div ( const GeometricField< Type, fvsPatchField, surfaceMesh > &ssf ) The first function on the list is the most general out of three: Code: 75 template 76 tmp 77 < 78 GeometricField 79 < 80 typename innerProduct::type, fvPatchField, volMesh 81 > 82 > 83 div 84 ( 85 const GeometricField& vf, 86 const word& name 87 ) 88 { 89 return fv::divScheme::New 90 ( 91 vf.mesh(), vf.mesh().divScheme(name) 92 )().fvcDiv(vf); 93 } It selects (through the Runtime Selection Mechanism) one of the available div schemes with name "name" and calls its member function fvcDiv. Currently the only one available div scheme in OpenFOAM is Gauss scheme (see Divergence theorem). Its fvcDiv is: Code: 42 template 43 tmp 44 < 45 GeometricField 46 ::type, fvPatchField, volMesh> 47 > 48 gaussDivScheme::fvcDiv 49 ( 50 const GeometricField& vf 51 ) 52 { 53 tmp 54 < 55 GeometricField 56 ::type, fvPatchField, volMesh> 57 > tDiv 58 ( 59 fvc::surfaceIntegrate 60 ( 61 this->mesh_.Sf() & this->tinterpScheme_().interpolate(vf) 62 ) 63 ); 64 65 tDiv().rename("div(" + vf.name() + ')'); 66 67 return tDiv; 68 } Firstly, it interpolates your volume field (defined in cell centres) onto a surface field (defined in cell face centres) : Code: this->tinterpScheme_().interpolate(vf) Secondly, it makes a dot (inner) product (&) of the mesh face vectors and the interpolated field. It is that operation which reduces the rank! : Code: this->mesh_.Sf() & this->tinterpScheme_().interpolate(vf) This dot product can be thought of as an operation of getting surface flux of the field through the cell faces. Lastly, it loops over the cells and for each cell it sums the fluxes through the cell's faces. This total sum of fluxes for each cell is then divided by a cell's volume Code: 42 template 43 void surfaceIntegrate 44 ( 45 Field& ivf, 46 const GeometricField& ssf 47 ) 48 { 49 const fvMesh& mesh = ssf.mesh(); 50 51 const labelUList& owner = mesh.owner(); 52 const labelUList& neighbour = mesh.neighbour(); 53 54 const Field& issf = ssf; 55 56 forAll(owner, facei) 57 { 58 ivf[owner[facei]] += issf[facei]; 59 ivf[neighbour[facei]] -= issf[facei]; 60 } 61 62 forAll(mesh.boundary(), patchi) 63 { 64 const labelUList& pFaceCells = 65 mesh.boundary()[patchi].faceCells(); 66 67 const fvsPatchField& pssf = ssf.boundaryField()[patchi]; 68 69 forAll(mesh.boundary()[patchi], facei) 70 { 71 ivf[pFaceCells[facei]] += pssf[facei]; 72 } 73 } 74 75 ivf /= mesh.Vsc(); 76 } The type of an object returned from fvcDiv is Code: 45 GeometricField 46 ::type, fvPatchField, volMesh> It is a GeometricField with elements of type innerProduct::type : Code: 88 template 89 class innerProduct 90 { 91 public: 92 93 typedef typename typeOfRank 94 < 95 typename pTraits::cmptType, 96 int(pTraits::rank) + int(pTraits::rank) - 2 97 >::type type; 98 }; innerProduct is a class template. type nested in innerProduct defines the type of an object which inner product (basically, dot operation) of two objects (of classes arg1 and arg2) results in. typeOfRank is an example of a well known design pattern called "Trait". It can be instantiated with a specific value parameter rank: Code: 44 template 45 class typeOfRank 46 {}; You just provide it with a rank and it tells you of what type an object of this rank is. OpenFOAM defines three partial specialisations of class template typeOfRank with template parameters (ranks) 0, 1, 2 for types scalar, vector and tensor, accordingly Code: 41 template 42 class typeOfRank 43 { 44 public: 45 46 typedef Cmpt type; 47 }; Code: 129 template 130 class typeOfRank 131 { 132 public: 133 134 typedef Vector type; 135 }; Code: 178 template 179 class typeOfRank 180 { 181 public: 182 183 typedef Tensor type; 184 }; pTraits is ... well another example of design pattern "Trait". If you specialise it for a specific type it can tell you the rank of any object of that type. Of course, in OpenFOMA it is allready specialised for scalar, vector and tensor. The other two "primary" fvc:div functions mentioned in the very beginning of this very lengthy post can be dissected in the similar way. If something is unclear please feel free to ask. qjh888 and mengweilm425 like this. Last edited by Zeppo; January 15, 2017 at 06:21.
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All times are GMT -4. The time now is 21:10. | 2,225 | 7,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-39 | latest | en | 0.777958 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/294/4/a/o/ | 1,610,947,439,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00480.warc.gz | 905,691,151 | 59,053 | # Properties
Label 294.4.a.o Level $294$ Weight $4$ Character orbit 294.a Self dual yes Analytic conductor $17.347$ Analytic rank $0$ Dimension $2$ CM no Inner twists $1$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$294 = 2 \cdot 3 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$4$$ Character orbit: $$[\chi]$$ $$=$$ 294.a (trivial)
## Newform invariants
Self dual: yes Analytic conductor: $$17.3465615417$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{2})$$ Defining polynomial: $$x^{2} - 2$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$1$$ Twist minimal: yes Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \sqrt{2}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + 2 q^{2} + 3 q^{3} + 4 q^{4} + ( 6 + \beta ) q^{5} + 6 q^{6} + 8 q^{8} + 9 q^{9} +O(q^{10})$$ $$q + 2 q^{2} + 3 q^{3} + 4 q^{4} + ( 6 + \beta ) q^{5} + 6 q^{6} + 8 q^{8} + 9 q^{9} + ( 12 + 2 \beta ) q^{10} + ( 2 + 6 \beta ) q^{11} + 12 q^{12} + ( 24 - 15 \beta ) q^{13} + ( 18 + 3 \beta ) q^{15} + 16 q^{16} + ( 66 - 11 \beta ) q^{17} + 18 q^{18} + ( 60 + 46 \beta ) q^{19} + ( 24 + 4 \beta ) q^{20} + ( 4 + 12 \beta ) q^{22} + ( -38 - 102 \beta ) q^{23} + 24 q^{24} + ( -87 + 12 \beta ) q^{25} + ( 48 - 30 \beta ) q^{26} + 27 q^{27} + ( -56 + 150 \beta ) q^{29} + ( 36 + 6 \beta ) q^{30} + ( 216 - 54 \beta ) q^{31} + 32 q^{32} + ( 6 + 18 \beta ) q^{33} + ( 132 - 22 \beta ) q^{34} + 36 q^{36} + ( -140 - 180 \beta ) q^{37} + ( 120 + 92 \beta ) q^{38} + ( 72 - 45 \beta ) q^{39} + ( 48 + 8 \beta ) q^{40} + ( 18 + 127 \beta ) q^{41} + ( -64 + 288 \beta ) q^{43} + ( 8 + 24 \beta ) q^{44} + ( 54 + 9 \beta ) q^{45} + ( -76 - 204 \beta ) q^{46} + ( -132 - 338 \beta ) q^{47} + 48 q^{48} + ( -174 + 24 \beta ) q^{50} + ( 198 - 33 \beta ) q^{51} + ( 96 - 60 \beta ) q^{52} + ( 134 - 192 \beta ) q^{53} + 54 q^{54} + ( 24 + 38 \beta ) q^{55} + ( 180 + 138 \beta ) q^{57} + ( -112 + 300 \beta ) q^{58} + ( 168 + 298 \beta ) q^{59} + ( 72 + 12 \beta ) q^{60} + ( -252 + 353 \beta ) q^{61} + ( 432 - 108 \beta ) q^{62} + 64 q^{64} + ( 114 - 66 \beta ) q^{65} + ( 12 + 36 \beta ) q^{66} + ( -192 - 144 \beta ) q^{67} + ( 264 - 44 \beta ) q^{68} + ( -114 - 306 \beta ) q^{69} + ( -198 + 342 \beta ) q^{71} + 72 q^{72} + ( -156 - 595 \beta ) q^{73} + ( -280 - 360 \beta ) q^{74} + ( -261 + 36 \beta ) q^{75} + ( 240 + 184 \beta ) q^{76} + ( 144 - 90 \beta ) q^{78} + ( -424 - 300 \beta ) q^{79} + ( 96 + 16 \beta ) q^{80} + 81 q^{81} + ( 36 + 254 \beta ) q^{82} + ( -324 + 80 \beta ) q^{83} + 374 q^{85} + ( -128 + 576 \beta ) q^{86} + ( -168 + 450 \beta ) q^{87} + ( 16 + 48 \beta ) q^{88} + ( -306 - 175 \beta ) q^{89} + ( 108 + 18 \beta ) q^{90} + ( -152 - 408 \beta ) q^{92} + ( 648 - 162 \beta ) q^{93} + ( -264 - 676 \beta ) q^{94} + ( 452 + 336 \beta ) q^{95} + 96 q^{96} + ( -1092 + 133 \beta ) q^{97} + ( 18 + 54 \beta ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q + 4q^{2} + 6q^{3} + 8q^{4} + 12q^{5} + 12q^{6} + 16q^{8} + 18q^{9} + O(q^{10})$$ $$2q + 4q^{2} + 6q^{3} + 8q^{4} + 12q^{5} + 12q^{6} + 16q^{8} + 18q^{9} + 24q^{10} + 4q^{11} + 24q^{12} + 48q^{13} + 36q^{15} + 32q^{16} + 132q^{17} + 36q^{18} + 120q^{19} + 48q^{20} + 8q^{22} - 76q^{23} + 48q^{24} - 174q^{25} + 96q^{26} + 54q^{27} - 112q^{29} + 72q^{30} + 432q^{31} + 64q^{32} + 12q^{33} + 264q^{34} + 72q^{36} - 280q^{37} + 240q^{38} + 144q^{39} + 96q^{40} + 36q^{41} - 128q^{43} + 16q^{44} + 108q^{45} - 152q^{46} - 264q^{47} + 96q^{48} - 348q^{50} + 396q^{51} + 192q^{52} + 268q^{53} + 108q^{54} + 48q^{55} + 360q^{57} - 224q^{58} + 336q^{59} + 144q^{60} - 504q^{61} + 864q^{62} + 128q^{64} + 228q^{65} + 24q^{66} - 384q^{67} + 528q^{68} - 228q^{69} - 396q^{71} + 144q^{72} - 312q^{73} - 560q^{74} - 522q^{75} + 480q^{76} + 288q^{78} - 848q^{79} + 192q^{80} + 162q^{81} + 72q^{82} - 648q^{83} + 748q^{85} - 256q^{86} - 336q^{87} + 32q^{88} - 612q^{89} + 216q^{90} - 304q^{92} + 1296q^{93} - 528q^{94} + 904q^{95} + 192q^{96} - 2184q^{97} + 36q^{99} + O(q^{100})$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
−1.41421 1.41421
2.00000 3.00000 4.00000 4.58579 6.00000 0 8.00000 9.00000 9.17157
1.2 2.00000 3.00000 4.00000 7.41421 6.00000 0 8.00000 9.00000 14.8284
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Atkin-Lehner signs
$$p$$ Sign
$$2$$ $$-1$$
$$3$$ $$-1$$
$$7$$ $$1$$
## Inner twists
This newform does not admit any (nontrivial) inner twists.
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 294.4.a.o yes 2
3.b odd 2 1 882.4.a.t 2
4.b odd 2 1 2352.4.a.bu 2
7.b odd 2 1 294.4.a.l 2
7.c even 3 2 294.4.e.k 4
7.d odd 6 2 294.4.e.m 4
21.c even 2 1 882.4.a.bb 2
21.g even 6 2 882.4.g.be 4
21.h odd 6 2 882.4.g.bk 4
28.d even 2 1 2352.4.a.bw 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
294.4.a.l 2 7.b odd 2 1
294.4.a.o yes 2 1.a even 1 1 trivial
294.4.e.k 4 7.c even 3 2
294.4.e.m 4 7.d odd 6 2
882.4.a.t 2 3.b odd 2 1
882.4.a.bb 2 21.c even 2 1
882.4.g.be 4 21.g even 6 2
882.4.g.bk 4 21.h odd 6 2
2352.4.a.bu 2 4.b odd 2 1
2352.4.a.bw 2 28.d even 2 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{4}^{\mathrm{new}}(\Gamma_0(294))$$:
$$T_{5}^{2} - 12 T_{5} + 34$$ $$T_{11}^{2} - 4 T_{11} - 68$$
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$( -2 + T )^{2}$$
$3$ $$( -3 + T )^{2}$$
$5$ $$34 - 12 T + T^{2}$$
$7$ $$T^{2}$$
$11$ $$-68 - 4 T + T^{2}$$
$13$ $$126 - 48 T + T^{2}$$
$17$ $$4114 - 132 T + T^{2}$$
$19$ $$-632 - 120 T + T^{2}$$
$23$ $$-19364 + 76 T + T^{2}$$
$29$ $$-41864 + 112 T + T^{2}$$
$31$ $$40824 - 432 T + T^{2}$$
$37$ $$-45200 + 280 T + T^{2}$$
$41$ $$-31934 - 36 T + T^{2}$$
$43$ $$-161792 + 128 T + T^{2}$$
$47$ $$-211064 + 264 T + T^{2}$$
$53$ $$-55772 - 268 T + T^{2}$$
$59$ $$-149384 - 336 T + T^{2}$$
$61$ $$-185714 + 504 T + T^{2}$$
$67$ $$-4608 + 384 T + T^{2}$$
$71$ $$-194724 + 396 T + T^{2}$$
$73$ $$-683714 + 312 T + T^{2}$$
$79$ $$-224 + 848 T + T^{2}$$
$83$ $$92176 + 648 T + T^{2}$$
$89$ $$32386 + 612 T + T^{2}$$
$97$ $$1157086 + 2184 T + T^{2}$$ | 3,270 | 6,581 | {"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} | 2.75 | 3 | CC-MAIN-2021-04 | latest | en | 0.310083 |
http://www.physicsforums.com/showthread.php?s=36e2a839de8619d94f7285224dc929bd&p=4027179 | 1,371,713,458,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368710963930/warc/CC-MAIN-20130516132923-00054-ip-10-60-113-184.ec2.internal.warc.gz | 626,189,298 | 10,059 | ## GPE question
This is probably child's play for you guys but can you show me how to figure the gpe of a 500lb weight lifted straight up 40'? In joules?
I assume the work done is equal to the gpe?
Thank guys
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Quote by RobVG This is probably child's play for you guys but can you show me how to figure the gpe of a 500lb weight lifted straight up 40'? In joules? I assume the work done is equal to the gpe? Thank guys
I conjecture that "gpe" is an acronym for "gravitational potential energy". If this is true, then the increase in "gpe" is equal to the work.
This is a a badly posed question. Not only is it being asked using an obscure acronym. In addition, your are not asked about the change in gpe. You are being asked as though there is an absolute quantity called gpe.
There isn't enough information given in the problem to calculate the absolute gpe. To answer the question quantitatively, one has to make a guess as to what the examiner was asking. One has to hypothesize that what is being asked for is the change in gpe. Only than one can say that the answer is the work.
A badly posed question can throw experts. Especially experts! So you have no reason to be embarrassed.
Mentor I think the question is posed fine. GPE is used as an acronym here all the time and 40' is a clear change in height.
## GPE question
The question is fine, but it's a trick question in the sense that the units are mixed.
The Joule is the SI, or metric, unit of energy. This is equivalent to work done.
The question is posed in Imperial "British" units of inches and pounds.
Convert inches to metres (one inch=2.54 centimetres) and pounds to Kg (1 lb = .454 Kg) and use your wits to find the gravitational field strength of Earth as I don't want to do your homework for you and then multiply the mass, the vertical displacement and the gravitational field strength together to get the work done.
Mentor Granted - as an American HVAC engineer, I switch back and forth so often I barely notice anymore.
Allow me to simplify. How do I figure out how many Newtons are required to lift a 500lb weight a distance of 40' in the air. 'GPE' is was it was called 30 years ago in high school physics. We wrote it as such. Thanks for any help
Mentor Simply google the lb:N conversion factor and multiply. Height is now irrelevant of course.
Quote by RobVG Allow me to simplify. How do I figure out how many Newtons are required to lift a 500lb weight a distance of 40' in the air. 'GPE' is was it was called 30 years ago in high school physics. We wrote it as such. Thanks for any help
You just made a mistake. "Newtons" is not a unit of energy.
Newtons is a unit of force, just like pounds (lbs).
The energy unit in the English system is foot-pounds (ft-lbs). The change in GPE could be measured in joules or ft-lbs.
MY GAWD! Who are you guys? Are you just a bunch of snotty children? russ_watters- "Height is now irrelevant of course." W=F*D Work is measured in Joules. How the hell can D be irrelevant? And Darwin123, yours is the most rude response I have ever gotten to an inquiry. And then you follow up with this- "You just made a mistake. "Newtons" is not a unit of energy. Newtons is a unit of force, just like pounds (lbs)." I was asking what "force" is required to lift the weight." Try reading my second question again if it's within your limited ability to comprehend- you troll. I will happily Google what I need and find my own answers. You can be damn sure I won't search for answers again on this forum.
Mentor
Quote by RobVG W=F*D Work is measured in Joules. How the hell can D be irrelevant? "You just made a mistake. "Newtons" is not a unit of energy. Newtons is a unit of force, just like pounds (lbs)." I was asking what "force" is required to lift the weight."
Well, the thing is (and I think this might be what caused the confusion), 500 lbs IS the force. The pound is a unit of force, not a unit of mass. In particular, 500 lbs is the weight of the object, and its "weight" is defined as the force with which Earth's gravity pulls down on it. So...you already know the force.
You know the force, and you know the distance, so all you have to do is multiply these two numbers, and you'll arrive at your answer for the increase in potential energy. Now, since you want the answer in joules, which are newton-metres, a metric unit, it's probably best if you first convert both the force and the distance to metric units. So, as russ suggested, all you have to do is look up the conversion factors between pounds and newtons, and between metres and feet.
With his comment about the height being irrelevant, what russ was trying to get at was that if all you wanted know was the force and not the GPE (which is what you implied in post #6), then the height was no longer required as a value in solving the problem. To get the force in newtons, all you have to do is convert the 500 lbs of force to newtons. I think that's all he was saying, and no rudeness was intended.
I strongly advise you to give this forum a chance. You'll find no better place for accurate physics advice and help online. Please let me know if you need any further clarification!
I understand the question and we use gpe all the time. We do not use lb and feet though. I don't know why some people have to be rude
Mentor
Quote by RobVG MY GAWD! Who are you guys? Are you just a bunch of snotty children?
Calm down. We're trying to help, but you are not making it very easy to help you.
russ_watters- "Height is now irrelevant of course." W=F*D Work is measured in Joules. How the hell can D be irrelevant?
You changed the question in the post I was quoting, saying you were no longer interested in work (Joules) and now were only interested in force (Newtons).
Darwin123 pointed that error out -- which is most certainly not rude.
I will happily Google what I need and find my own answers. You can be damn sure I won't search for answers again on this forum.
If the help we've given you so far isn't good enough to get you what you need, then you may well be beyond our ability to help. Good luck. Thread closed.
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https://www.calculatoratoz.com/en/focal-parameter-of-hyperbola-given-latus-rectum-and-semi-conjugate-axis-calculator/Calc-38990 | 1,685,289,301,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644309.7/warc/CC-MAIN-20230528150639-20230528180639-00505.warc.gz | 776,394,678 | 38,524 | ## Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis Solution
STEP 0: Pre-Calculation Summary
Formula Used
Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2)
p = b^2/sqrt(((2*b^2)/L)^2+b^2)
This formula uses 1 Functions, 3 Variables
Functions Used
sqrt - Square root function, sqrt(Number)
Variables Used
Focal Parameter of Hyperbola - (Measured in Meter) - Focal Parameter of Hyperbola is the shortest distance between any of the foci and directrix of the corresponding wing of the Hyperbola.
Semi Conjugate Axis of Hyperbola - (Measured in Meter) - Semi Conjugate Axis of Hyperbola is half of the tangent from any of the vertices of the Hyperbola and chord to the circle passing through the foci and centered at the center of the Hyperbola.
Latus Rectum of Hyperbola - (Measured in Meter) - Latus Rectum of Hyperbola is the line segment passing through any of the foci and perpendicular to the transverse axis whose ends are on the Hyperbola.
STEP 1: Convert Input(s) to Base Unit
Semi Conjugate Axis of Hyperbola: 12 Meter --> 12 Meter No Conversion Required
Latus Rectum of Hyperbola: 60 Meter --> 60 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
p = b^2/sqrt(((2*b^2)/L)^2+b^2) --> 12^2/sqrt(((2*12^2)/60)^2+12^2)
Evaluating ... ...
p = 11.1417202906231
STEP 3: Convert Result to Output's Unit
11.1417202906231 Meter --> No Conversion Required
11.1417202906231 11.14172 Meter <-- Focal Parameter of Hyperbola
(Calculation completed in 00.004 seconds)
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## < 8 Focal Parameter of Hyperbola Calculators
Focal Parameter of Hyperbola given Latus Rectum and Semi Transverse Axis
Focal Parameter of Hyperbola = ((Semi Transverse Axis of Hyperbola*Latus Rectum of Hyperbola)/2)/sqrt(Semi Transverse Axis of Hyperbola^2+((Semi Transverse Axis of Hyperbola*Latus Rectum of Hyperbola)/2)^2)
Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis
Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2)
Focal Parameter of Hyperbola
Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/sqrt(Semi Transverse Axis of Hyperbola^2+Semi Conjugate Axis of Hyperbola^2)
Focal Parameter of Hyperbola given Eccentricity and Semi Conjugate Axis
Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola/(Eccentricity of Hyperbola/sqrt(Eccentricity of Hyperbola^2-1))
Focal Parameter of Hyperbola given Linear Eccentricity and Semi Transverse Axis
Focal Parameter of Hyperbola = (Linear Eccentricity of Hyperbola^2-Semi Transverse Axis of Hyperbola^2)/Linear Eccentricity of Hyperbola
Focal Parameter of Hyperbola given Eccentricity
Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/(Semi Transverse Axis of Hyperbola*Eccentricity of Hyperbola)
Focal Parameter of Hyperbola given Eccentricity and Semi Transverse Axis
Focal Parameter of Hyperbola = Semi Transverse Axis of Hyperbola/Eccentricity of Hyperbola*(Eccentricity of Hyperbola^2-1)
Focal Parameter of Hyperbola given Linear Eccentricity and Semi Conjugate Axis
Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/Linear Eccentricity of Hyperbola
## < 4 Focal Parameter of Hyperbola Calculators
Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis
Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2)
Focal Parameter of Hyperbola
Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/sqrt(Semi Transverse Axis of Hyperbola^2+Semi Conjugate Axis of Hyperbola^2)
Focal Parameter of Hyperbola given Eccentricity and Semi Transverse Axis
Focal Parameter of Hyperbola = Semi Transverse Axis of Hyperbola/Eccentricity of Hyperbola*(Eccentricity of Hyperbola^2-1)
Focal Parameter of Hyperbola given Linear Eccentricity and Semi Conjugate Axis
Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/Linear Eccentricity of Hyperbola
## Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis Formula
Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2)
p = b^2/sqrt(((2*b^2)/L)^2+b^2)
## What is Hyperbola?
A Hyperbola is a type of conic section, which is a geometric figure that results from intersecting a cone with a plane. A Hyperbola is defined as the set of all points in a plane, the difference of whose distances from two fixed points (called the foci) is constant. In other words, a Hyperbola is the locus of points where the difference between the distances to two fixed points is a constant value. The standard form of the equation for a Hyperbola is: (x - h)²/a² - (y - k)²/b² = 1
## What is Focal Parameter of a Hyperbola and how is it calculated?
The focal parameter of the Hyperbola is the shortest distance from a focus to the corresponding directrix. It is calculated by the formula p= b2/√(a2+b2) where p is the focal parameter of the Hyperbola, b is the semi conjugate axis of the Hyperbola and a is the semi transverse axis of the Hyperbola.
## How to Calculate Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis?
Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis calculator uses Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2) to calculate the Focal Parameter of Hyperbola, The Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis formula is defined as the shortest distance between any of the foci and directrix of the corresponding wing of the Hyperbola and is calculated using the latus rectum and semi-conjugate axis of the Hyperbola. Focal Parameter of Hyperbola is denoted by p symbol.
How to calculate Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis using this online calculator? To use this online calculator for Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis, enter Semi Conjugate Axis of Hyperbola (b) & Latus Rectum of Hyperbola (L) and hit the calculate button. Here is how the Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis calculation can be explained with given input values -> 11.14172 = 12^2/sqrt(((2*12^2)/60)^2+12^2).
### FAQ
What is Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis?
The Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis formula is defined as the shortest distance between any of the foci and directrix of the corresponding wing of the Hyperbola and is calculated using the latus rectum and semi-conjugate axis of the Hyperbola and is represented as p = b^2/sqrt(((2*b^2)/L)^2+b^2) or Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2). Semi Conjugate Axis of Hyperbola is half of the tangent from any of the vertices of the Hyperbola and chord to the circle passing through the foci and centered at the center of the Hyperbola & Latus Rectum of Hyperbola is the line segment passing through any of the foci and perpendicular to the transverse axis whose ends are on the Hyperbola.
How to calculate Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis?
The Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis formula is defined as the shortest distance between any of the foci and directrix of the corresponding wing of the Hyperbola and is calculated using the latus rectum and semi-conjugate axis of the Hyperbola is calculated using Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/sqrt(((2*Semi Conjugate Axis of Hyperbola^2)/Latus Rectum of Hyperbola)^2+Semi Conjugate Axis of Hyperbola^2). To calculate Focal Parameter of Hyperbola given Latus Rectum and Semi Conjugate Axis, you need Semi Conjugate Axis of Hyperbola (b) & Latus Rectum of Hyperbola (L). With our tool, you need to enter the respective value for Semi Conjugate Axis of Hyperbola & Latus Rectum of Hyperbola and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Focal Parameter of Hyperbola?
In this formula, Focal Parameter of Hyperbola uses Semi Conjugate Axis of Hyperbola & Latus Rectum of Hyperbola. We can use 10 other way(s) to calculate the same, which is/are as follows -
• Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/sqrt(Semi Transverse Axis of Hyperbola^2+Semi Conjugate Axis of Hyperbola^2)
• Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/Linear Eccentricity of Hyperbola
• Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola/(Eccentricity of Hyperbola/sqrt(Eccentricity of Hyperbola^2-1))
• Focal Parameter of Hyperbola = (Linear Eccentricity of Hyperbola^2-Semi Transverse Axis of Hyperbola^2)/Linear Eccentricity of Hyperbola
• Focal Parameter of Hyperbola = Semi Transverse Axis of Hyperbola/Eccentricity of Hyperbola*(Eccentricity of Hyperbola^2-1)
• Focal Parameter of Hyperbola = Semi Conjugate Axis of Hyperbola^2/(Semi Transverse Axis of Hyperbola*Eccentricity of Hyperbola)
• Focal Parameter of Hyperbola = ((Semi Transverse Axis of Hyperbola*Latus Rectum of Hyperbola)/2)/sqrt(Semi Transverse Axis of Hyperbola^2+((Semi Transverse Axis of Hyperbola*Latus Rectum of Hyperbola)/2)^2)
• Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/sqrt(Semi Transverse Axis of Hyperbola^2+Semi Conjugate Axis of Hyperbola^2)
• Focal Parameter of Hyperbola = Semi Transverse Axis of Hyperbola/Eccentricity of Hyperbola*(Eccentricity of Hyperbola^2-1)
• Focal Parameter of Hyperbola = (Semi Conjugate Axis of Hyperbola^2)/Linear Eccentricity of Hyperbola
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http://isabelle.in.tum.de/repos/isabelle/annotate/f4da791d4850/doc-src/TutorialI/Types/Pairs.thy | 1,582,482,103,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00056.warc.gz | 81,908,604 | 7,751 | doc-src/TutorialI/Types/Pairs.thy
author nipkow Fri Dec 01 12:15:47 2000 +0100 (2000-12-01) changeset 10560 f4da791d4850 child 10608 620647438780 permissions -rw-r--r--
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nipkow@10560 1 (*<*)theory Pairs = Main:(*>*) nipkow@10560 2 nipkow@10560 3 section{*Pairs*} nipkow@10560 4 nipkow@10560 5 text{*\label{sec:products} nipkow@10560 6 Pairs were already introduced in \S\ref{sec:pairs}, but only with a minimal nipkow@10560 7 repertoire of operations: pairing and the two projections @{term fst} and nipkow@10560 8 @{term snd}. In any nontrivial application of pairs you will find that this nipkow@10560 9 quickly leads to unreadable formulae involvings nests of projections. This nipkow@10560 10 section is concerned with introducing some syntactic sugar to overcome this nipkow@10560 11 problem: pattern matching with tuples. nipkow@10560 12 *} nipkow@10560 13 nipkow@10560 14 subsection{*Notation*} nipkow@10560 15 nipkow@10560 16 text{* nipkow@10560 17 It is possible to use (nested) tuples as patterns in $\lambda$-abstractions, nipkow@10560 18 for example @{text"\(x,y,z).x+y+z"} and @{text"\((x,y),z).x+y+z"}. In fact, nipkow@10560 19 tuple patterns can be used in most variable binding constructs. Here are nipkow@10560 20 some typical examples: nipkow@10560 21 \begin{quote} nipkow@10560 22 @{term"let (x,y) = f z in (y,x)"}\\ nipkow@10560 23 @{term"case xs of [] => 0 | (x,y)#zs => x+y"}\\ nipkow@10560 24 @{text"\(x,y)\A. x=y"}\\ nipkow@10560 25 @{text"{(x,y). x=y}"}\\ nipkow@10560 26 @{term"\(x,y)\A. {x+y}"} nipkow@10560 27 \end{quote} nipkow@10560 28 *} nipkow@10560 29 nipkow@10560 30 text{* nipkow@10560 31 The intuitive meaning of this notations should be pretty obvious. nipkow@10560 32 Unfortunately, we need to know in more detail what the notation really stands nipkow@10560 33 for once we have to reason about it. The fact of the matter is that abstraction nipkow@10560 34 over pairs and tuples is merely a convenient shorthand for a more complex nipkow@10560 35 internal representation. Thus the internal and external form of a term may nipkow@10560 36 differ, which can affect proofs. If you want to avoid this complication, nipkow@10560 37 stick to @{term fst} and @{term snd} and write @{term"%p. fst p + snd p"} nipkow@10560 38 instead of @{text"\(x,y). x+y"} (which denote the same function but are quite nipkow@10560 39 different terms). nipkow@10560 40 nipkow@10560 41 Internally, @{term"%(x,y). t"} becomes @{text"split (\x y. t)"}, where nipkow@10560 42 @{term split} is the uncurrying function of type @{text"('a \ 'b nipkow@10560 43 \ 'c) \ 'a \ 'b \ 'c"} defined as nipkow@10560 44 \begin{center} nipkow@10560 45 @{thm split_def} nipkow@10560 46 \hfill(@{thm[source]split_def}) nipkow@10560 47 \end{center} nipkow@10560 48 Pattern matching in nipkow@10560 49 other variable binding constructs is translated similarly. Thus we need to nipkow@10560 50 understand how to reason about such constructs. nipkow@10560 51 *} nipkow@10560 52 nipkow@10560 53 subsection{*Theorem proving*} nipkow@10560 54 nipkow@10560 55 text{* nipkow@10560 56 The most obvious approach is the brute force expansion of @{term split}: nipkow@10560 57 *} nipkow@10560 58 nipkow@10560 59 lemma "(\(x,y).x) p = fst p" nipkow@10560 60 by(simp add:split_def) nipkow@10560 61 nipkow@10560 62 text{* This works well if rewriting with @{thm[source]split_def} finishes the nipkow@10560 63 proof, as in the above lemma. But if it doesn't, you end up with exactly what nipkow@10560 64 we are trying to avoid: nests of @{term fst} and @{term snd}. Thus this nipkow@10560 65 approach is neither elegant nor very practical in large examples, although it nipkow@10560 66 can be effective in small ones. nipkow@10560 67 nipkow@10560 68 If we step back and ponder why the above lemma presented a problem in the nipkow@10560 69 first place, we quickly realize that what we would like is to replace @{term nipkow@10560 70 p} with some concrete pair @{term"(a,b)"}, in which case both sides of the nipkow@10560 71 equation would simplify to @{term a} because of the simplification rules nipkow@10560 72 @{thm Product_Type.split[no_vars]} and @{thm fst_conv[no_vars]}. This is the nipkow@10560 73 key problem one faces when reasoning about pattern matching with pairs: how to nipkow@10560 74 convert some atomic term into a pair. nipkow@10560 75 nipkow@10560 76 In case of a subterm of the form @{term"split f p"} this is easy: the split nipkow@10560 77 rule @{thm[source]split_split} replaces @{term p} by a pair: nipkow@10560 78 *} nipkow@10560 79 nipkow@10560 80 lemma "(\(x,y).y) p = snd p" nipkow@10560 81 apply(simp only: split:split_split); nipkow@10560 82 nipkow@10560 83 txt{* nipkow@10560 84 @{subgoals[display,indent=0]} nipkow@10560 85 This subgoal is easily proved by simplification. The @{text"only:"} above nipkow@10560 86 merely serves to show the effect of splitting and to avoid solving the goal nipkow@10560 87 outright. nipkow@10560 88 nipkow@10560 89 Let us look at a second example: nipkow@10560 90 *} nipkow@10560 91 nipkow@10560 92 (*<*)by simp(*>*) nipkow@10560 93 lemma "let (x,y) = p in fst p = x"; nipkow@10560 94 apply(simp only:Let_def) nipkow@10560 95 nipkow@10560 96 txt{* nipkow@10560 97 @{subgoals[display,indent=0]} nipkow@10560 98 A paired @{text let} reduces to a paired $\lambda$-abstraction, which nipkow@10560 99 can be split as above. The same is true for paired set comprehension: nipkow@10560 100 *} nipkow@10560 101 nipkow@10560 102 (*<*)by(simp split:split_split)(*>*) nipkow@10560 103 lemma "p \ {(x,y). x=y} \ fst p = snd p" nipkow@10560 104 apply simp nipkow@10560 105 nipkow@10560 106 txt{* nipkow@10560 107 @{subgoals[display,indent=0]} nipkow@10560 108 Again, simplification produces a term suitable for @{thm[source]split_split} nipkow@10560 109 as above. If you are worried about the funny form of the premise: nipkow@10560 110 @{term"split (op =)"} is the same as @{text"\(x,y). x=y"}. nipkow@10560 111 The same procedure works for nipkow@10560 112 *} nipkow@10560 113 nipkow@10560 114 (*<*)by(simp split:split_split)(*>*) nipkow@10560 115 lemma "p \ {(x,y). x=y} \ fst p = snd p" nipkow@10560 116 nipkow@10560 117 txt{*\noindent nipkow@10560 118 except that we now have to use @{thm[source]split_split_asm}, because nipkow@10560 119 @{term split} occurs in the assumptions. nipkow@10560 120 nipkow@10560 121 However, splitting @{term split} is not always a solution, as no @{term split} nipkow@10560 122 may be present in the goal. Consider the following function: nipkow@10560 123 *} nipkow@10560 124 nipkow@10560 125 (*<*)by(simp split:split_split_asm)(*>*) nipkow@10560 126 consts swap :: "'a \ 'b \ 'b \ 'a" nipkow@10560 127 primrec nipkow@10560 128 "swap (x,y) = (y,x)" nipkow@10560 129 nipkow@10560 130 text{*\noindent nipkow@10560 131 Note that the above \isacommand{primrec} definition is admissible nipkow@10560 132 because @{text"\"} is a datatype. When we now try to prove nipkow@10560 133 *} nipkow@10560 134 nipkow@10560 135 lemma "swap(swap p) = p" nipkow@10560 136 nipkow@10560 137 txt{*\noindent nipkow@10560 138 simplification will do nothing, because the defining equation for @{term swap} nipkow@10560 139 expects a pair. Again, we need to turn @{term p} into a pair first, but this nipkow@10560 140 time there is no @{term split} in sight. In this case the only thing we can do nipkow@10560 141 is to split the term by hand: nipkow@10560 142 *} nipkow@10560 143 apply(case_tac p) nipkow@10560 144 nipkow@10560 145 txt{*\noindent nipkow@10560 146 @{subgoals[display,indent=0]} nipkow@10560 147 Again, @{text case_tac} is applicable because @{text"\"} is a datatype. nipkow@10560 148 The subgoal is easily proved by @{text simp}. nipkow@10560 149 nipkow@10560 150 In case the term to be split is a quantified variable, there are more options. nipkow@10560 151 You can split \emph{all} @{text"\"}-quantified variables in a goal nipkow@10560 152 with the rewrite rule @{thm[source]split_paired_all}: nipkow@10560 153 *} nipkow@10560 154 nipkow@10560 155 (*<*)by simp(*>*) nipkow@10560 156 lemma "\p q. swap(swap p) = q \ p = q" nipkow@10560 157 apply(simp only:split_paired_all) nipkow@10560 158 nipkow@10560 159 txt{*\noindent nipkow@10560 160 @{subgoals[display,indent=0]} nipkow@10560 161 *} nipkow@10560 162 nipkow@10560 163 apply simp nipkow@10560 164 done nipkow@10560 165 nipkow@10560 166 text{*\noindent nipkow@10560 167 Note that we have intentionally included only @{thm[source]split_paired_all} nipkow@10560 168 in the first simplification step. This time the reason was not merely nipkow@10560 169 pedagogical: nipkow@10560 170 @{thm[source]split_paired_all} may interfere with certain congruence nipkow@10560 171 rules of the simplifier, i.e. nipkow@10560 172 *} nipkow@10560 173 nipkow@10560 174 (*<*) nipkow@10560 175 lemma "\p q. swap(swap p) = q \ p = q" nipkow@10560 176 (*>*) nipkow@10560 177 apply(simp add:split_paired_all) nipkow@10560 178 (*<*)done(*>*) nipkow@10560 179 text{*\noindent nipkow@10560 180 may fail (here it does not) where the above two stages succeed. nipkow@10560 181 nipkow@10560 182 Finally, all @{text"\"} and @{text"\"}-quantified variables are split nipkow@10560 183 automatically by the simplifier: nipkow@10560 184 *} nipkow@10560 185 nipkow@10560 186 lemma "\p. \q. swap p = swap q" nipkow@10560 187 apply simp; nipkow@10560 188 done nipkow@10560 189 nipkow@10560 190 text{*\noindent nipkow@10560 191 In case you would like to turn off this automatic splitting, just disable the nipkow@10560 192 responsible simplification rules: nipkow@10560 193 \begin{center} nipkow@10560 194 @{thm split_paired_All} nipkow@10560 195 \hfill nipkow@10560 196 (@{thm[source]split_paired_All})\\ nipkow@10560 197 @{thm split_paired_Ex} nipkow@10560 198 \hfill nipkow@10560 199 (@{thm[source]split_paired_Ex}) nipkow@10560 200 \end{center} nipkow@10560 201 *} nipkow@10560 202 (*<*) nipkow@10560 203 end nipkow@10560 204 (*>*) | 3,799 | 10,313 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-10 | latest | en | 0.675972 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A064626 Football tournament numbers: the number of possible point series for a tournament of n teams playing each other once where 3 points are awarded to the winning team and 1 to each in the case of a tie. 7
%I
%S 1,2,7,40,355,3678,37263,361058,3403613,31653377,292547199,2696619716
%N Football tournament numbers: the number of possible point series for a tournament of n teams playing each other once where 3 points are awarded to the winning team and 1 to each in the case of a tie.
%C This sequence reflects the now common 3-point rule of international football where the sum of total points awarded depends on the outcome of each match. The classical 2-point rule is equivalent to that for chess tournaments (A007747).
%H A. Ivanyi, L. Lucz, T. Matuszka, and S. Pirzada, <a href="http://www.acta.sapientia.ro/acta-info/C4-2/info42-7.pdf">Parallel enumeration of degree sequences of simple graphs</a>, Acta Univ. Sapientiae, Informatica, 4, 2 (2012) 260-288. - From _N. J. A. Sloane_, Feb 15 2013
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Three_points_for_a_win">Three points for a win</a>
%e For 2 teams there are 2 possible outcomes: [0, 3] and [1, 1], so a(2) = 2.
%e For 3 teams the outcomes are [0, 3, 6], [1, 3, 4], [3, 3, 3], [1, 1, 6], [1, 2, 4], [0, 4, 4] and [2, 2, 2], so a(3) is 7. Note that the outcome [3, 3, 3] can be obtained in two ways: (A beats B, B beats C, C beats A) or (B beats A, A beats C, C beats B).
%Y Cf. A007747, A047730, A064422, A152789.
%K nonn,nice,more,hard,changed
%O 1,2
%A Thomas Schulze (jazariel(AT)tiscalenet.it), Sep 30 2001
%E a(8) and a(9) from _Jon E. Schoenfield_, May 05 2007
%E a(10) from Ming Li (dawnli(AT)ustc.edu), Jun 20 2008
%E a(11) from _Jon E. Schoenfield_, Sep 04 2008
%E a(12) from _Jon E. Schoenfield_, Dec 12 2008
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Last modified August 2 02:40 EDT 2021. Contains 346409 sequences. (Running on oeis4.) | 763 | 2,323 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2021-31 | latest | en | 0.791332 |
https://cs.stackexchange.com/questions/37956/why-is-the-optimal-cut-off-for-switching-from-quicksort-to-insertion-sort-machin | 1,586,515,308,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371893683.94/warc/CC-MAIN-20200410075105-20200410105605-00395.warc.gz | 387,766,631 | 37,219 | # Why is the optimal cut-off for switching from Quicksort to Insertion sort machine dependent?
I fail to understand why cut off value would be system dependent, and not a constant.
Cutoff to insertion sort. As with mergesort, it pays to switch to insertion sort for tiny arrays. The optimum value of the cutoff is system-dependent, but any value between 5 and 15 is likely to work well in most situations.
Would it be safe to assume that optimal cut off value is equal to optimal set size for Insert sort?
• Because computers are not identical. You are switching from a stack intensive torrent of recursive calls to a memory intensive set of loops. The latter will be most efficient if you can stay inside the size of the memory caches on the CPU. These vary from model to model, hence it is system dependent. – Thorbjørn Ravn Andersen Feb 5 '15 at 0:54
• think of it as probably closely related to more general metrics eg the speed of the insertion sort vs speed of quicksort on $n$ items. these differences are amplified for low values of $n$ because there is less "law of averages" working out. – vzn Feb 5 '15 at 5:52
Because the actual running time (in seconds) of real code on a real computer depends on how fast that computer runs the instructions and how fast it retrieves the relevant data from memory, how well it caches it and so on. Insertion sort and quicksort use different instructions and hava different memory access patterns. So the running time of quicksort versus insertion sort for any particular dataset on any particular system will depend both on the instructions used to implement those two sorting routines and the memory access patterns of the data. Given the different mixes of instructions, it's perfectly possible that insertion sort is faster for lists of up to ten items on one system, but only for lists up to six items on some other system.
The relative costs of various operations are different on different machines, and compilers have varying degrees of ability to optimize various constructs. David Richerby goes into somewhat more detail on that, but the last half-sentence of the highlighted quote is perhaps the most important. In many cases where one algorithm is more efficient than another for small data sets, and another is more efficient for large data sets, the performance differences between the two algorithms are apt to be rather small for data sets near the "break-even" point. As a simple example, assume for numerical simplicity that the cost of insertion sort is precisely $C_1 · N·N$ with some constant $C_1$, and the cost of quick sort is precisely $C_2 · N·\lg(N)$ for some other constant $C_2$. Then consider the relative behaviors for two sets of constants.
First of all, suppose that the constants are $C_1 =1$ and $C_2 = 2$. Then for $N=4$ both sorts will take time 16; for $N=16$, insertion sort will take time 256 and quick sort will take time 128. For $N=8$, insertion sort will take time 64 and quick sort will take time 48.
Now suppose the constants are $C_1 = 1$ and $C_2 = 4$. Then for $N=4$, insertion sort will take time 16 while quick sort will take time 32; for $N=16$, both sorts will take time 256. For $N=8$, insertion sort will take time 64 and quick sort will take time 96.
In the former scenario, the break-even point was 4, and in the latter it was 16, but despite the $4:1$ difference in break-even point, the time required for insertion sort and quick sort will be within 50% of each other for $N=8$. If one knows that the real situation is somewhere between those two scenarios, determining the exact break-even point may allow some performance improvement versus simply using a value of 8, but using 8 will be at most 50% worse than using the optimal value. Note further that because a program is unlikely to spend most of its time sorting regions of size 8, the 50% difference in time to sort regions of that size will generally have only a small effect on overall sorting performance.
Hand-write an insertion sort and a merge sort for a list with 3 items, down to assembly code. Pay attention to:
• # of comparisons performed
• other lower level or higher level considerations: # of assembly instructions, variance introduced by the randomness in quicksort, etc.
Now see if you can give different weights to the operations that might represent two different systems - for instance, one that's comparison-optimized and one that's register loading-optimized, and compute total . If you can get different results, then it's systems-dependent. If you always get the same result, the one algorithm is always faster.
This should give you a rigorous way to think about these problems.
• There seems to be something missing. – Raphael Feb 4 '15 at 18:54
• Clearly, anything missing is already covered in "other computer science things". – corsiKa Feb 5 '15 at 0:00
• Where would I find an exhaustive list of "other computer science things"? – Antonios Hadjigeorgalis May 29 '16 at 11:35
• @AntoniosHadjigeorgalis didn't mean to imply there is one. Perhaps you're trying to save energy on a particular processor or perhaps you're performing a sort on pancakes and want to minimize the number of times the bottom of a pancake is touched. Should pancake sort go from a toy interview problem to a serious area of research you'd probably see new concepts to model this sort of question as new questions arise. The field is growing of course. – djechlin May 29 '16 at 15:50
• Anyway I replaced the phrase that was causing offense. – djechlin May 29 '16 at 15:51
Imagine an implementation where a function call is very, very expensive. Say so expensive that in the time it takes to do one function call, you can sort an array of 1000 elements with insertion sort.
Do you think that would influence the optimal switching point?
When a partition reaches the size that can fit in a machine dependent sized cache, then it will be faster . It's more interesting to look at the mathematics behind the machine independent answer of 10 being the cut off point. C_N is the number of comparisons to sort N elements; in quick sort, each element is compared with the partitioning element until the left and right pointers cross ( so the last comparison happens once again ), so there are N + 1 comparisons for the current partitioning call. Then there are recursive calls on the left and right partitions either side of the partitioning element. In randomized quicksort, where all the elements are shuffled first, then the average number of comparisons in the two partitions is given by ${ \sum_{j=0}^N \big( C_{j-1} + C_{N-j-1} \big) \over N }$.
The goal is then to simplify the solution of the quicksort average comparison count equation , and then incorporate the given average comparison cost of insertion sort ${j(j-1) \over 4 }$ at a given cutoff $M$. The sum is simplified, because there is symmetry of C terms around half $N$. ${2 \over N} \sum_{j=0}^N C_{j-1}$ Eliminate the sum by subtracting the equation for $N-1$ $$C_N = N + 1 + {2 \over N} \sum_{j=0}^N C_{j-1}$$ minus $$C_{N-1} = (N-1) + 1 + { 2 \over N-1} \sum_{j=0}^{N-1} C_{j-1}$$ Multiply by N and N-1 respectively the equations to make the subtraction easier. $$N C_N - (N-1) C_{N-1} = N(N+1) - (N-1)(N) + 2 \bigg(\sum_{j=0}^N C_{j-1} -\sum_{j=0}^{N-1} C_{j-1} \bigg)$$ Collecting terms and subtracting the sums, $$N C_N = (N-1)C_{N-1} + 2N + 2 \bigg( C_{N-1} \bigg) = (N+1)C_{N-1} + 2N$$ Dividing by N(N+1) , the suitable summation factor, which simplifies to a recurrence equation that can be telescoped. $${C_N \over N+1} = {C_{N-1} \over N} + { 2 \over N+1}$$ Telescope, whilst maintaining that $C_{N-1} <> C_{M}$ $${C_N \over N+1} = {C_{M+1} \over M+2} + \bigg(2 \sum_{j=M+3}^{N+1} {1 \over j} = 2 \big( \sum_{j=1}^{N+1} {1\over j} - \sum_{j=1}^{M+2} {1\over j} \big) \bigg)$$ $C_{M+1}$ follows the original recurrence equation, but with the insertion cost inserted instead of recursive partitioning. $${C_{M+1} \over M + 2} = (M+1) + 1 + 2 \sum_{j=1}^{M+1} {j(j-1) \over 4}$$ Since $\sum_{j=1}^{P} { P \choose Q} = { P+1 \choose Q+1}$ $${C_{M+1} \over M + 2 } = M + 2 + 2/2 { M+2 \choose 3} = M + 2 + (M+2)(M+1)(M) / 6$$ $$C_{M+1} = 1 + { M(M+1) \over 6 }$$
Since $\sum_{j=1}^N {1\over j} = H_N \approx ln N + \gamma$ , the whole equation looks like $$C_N = (N + 1) \bigg(1+ {M(M+1) \over 6(M+2)} + 2\big(ln (N+1) - ln (M+2) \big) \bigg)$$
So for N = 1000, C_N is minimum when M = 10, independent of the machine.
This is a student answer (mine) to an Exercise from the AofA website, Introduction Analysis of Algorithms Sedgewick The answer is a middle school algebra expansion of some answers found on a discussion forum, mainly this thread : Flores, Yao and Spreng , and a it's a good idea to try to independently reach their answers. I got to the first one, and found it came up with 6 which was the wrong answer, and re-read the thread several times before being able to understand the abbreviated answer by Yao, having to dive into the book to get his assumptions , which are explained above in a middle school algebra way ( that's about my level, after 30+ years of not doing maths ). | 2,449 | 9,145 | {"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} | 2.546875 | 3 | CC-MAIN-2020-16 | latest | en | 0.892981 |
http://www.physics.udel.edu/~watson/phys345/fall1998/class/15-decimal.html | 1,545,075,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829115.83/warc/CC-MAIN-20181217183905-20181217205905-00325.warc.gz | 446,685,918 | 1,257 | ## PHYS345 Electricity and Electronics
Review of Decimal Scheme
Most familiar!
"Place holders" are powers of 10
Power of 10: Weight: 3 2 1 0 1000 100 10 1 103 102 101 100
Ten symbols or characters needed:
0, 1, 2, 3, 4, 5, 6, 7, 8, and 9
Example:
1998 = (1 x 103) + (9 x 102) + (9 x 10) + 8 = 1000 + 900 + 90 + 8 | 149 | 319 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-51 | latest | en | 0.495373 |
https://community.fabric.microsoft.com/t5/Desktop/Calculating-Hourly-Usage-Data-based-on-last-entry/m-p/115115 | 1,722,684,582,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00267.warc.gz | 140,661,490 | 51,900 | cancel
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Frequent Visitor
## Calculating Hourly Usage Data based on last entry
Seems like a simple problem however I cannot find a solution. Can anyone assist?
Operators use a piece of equipment and then enter a series of checks for mainteance Items. Within that they enter the meter reading. To tabulate how much our entire fleet is being used and be able to slice it with any given day I need the usage field to populate with the amount of hours used based upon the last meter reading. I could set up a separate table with starts/ends of each week to get the max and min but this seems like a workaround. Matt
DateTimeStamp Vehicle Reading Usage 1/1/2017 08:00:00 a 1 0 1/1/2017 08:00:01 b 2 0 1/1/2017 08:00:02 c 3 0 1/1/2017 16:00:03 a 4 3 1/1/2017 16:00:04 b 5 3 1/1/2017 16:00:05 c 6 3 1/2/2017 08:00:06 a 10 6 1/2/2017 08:00:07 b 11 6 1/2/2017 08:00:08 c 12 6 1/2/2017 16:00:09 a 15 5 1/2/2017 16:00:10 b 16 5 1/2/2017 16:00:11 c 17 5
1 ACCEPTED SOLUTION
Frequent Visitor
Sorry, none of the responses below I was able to get working. I even bought the Russo and Ferrari book to try to understand the Earlier Function.
I solved it within the source data.
=IF(ISBLANK(D2),"0",Iferror(D2-large(QUERY(\$C\$2:D2,"Select D where C matches '"&C2&"' order by D desc limit 2"),2),0))
Line C is my vehicle ID and Line D is the hour reading the operators input each day.
3 REPLIES 3
Frequent Visitor
A response I received on email. Couldn't get it to work.
Matt
Column = CALCULATE(MAX('T2'[Type]),FILTER(ALL('T2'),'T2'[ID]=EARLIER('T1'[ID]) &&'T2'[Date]<=RELATED(T1[Date])))
Frequent Visitor
Sorry, none of the responses below I was able to get working. I even bought the Russo and Ferrari book to try to understand the Earlier Function.
I solved it within the source data.
=IF(ISBLANK(D2),"0",Iferror(D2-large(QUERY(\$C\$2:D2,"Select D where C matches '"&C2&"' order by D desc limit 2"),2),0))
Line C is my vehicle ID and Line D is the hour reading the operators input each day.
Microsoft Employee
Hi @mshodge,
According to your description above, you should be able to use the formula below to create a new measure to calculate the usage with the amount of hours used based upon the last meter reading and show it on the report with other fields.
```NewMeasure =
RETURN
Regards
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Join the community in Stockholm for expert Microsoft Fabric learning including a very exciting keynote from Arun Ulag, Corporate Vice President, Azure Data. | 806 | 2,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2024-33 | latest | en | 0.880992 |
https://13dipty.medium.com/recursion-ca6baab74aaf | 1,701,896,063,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100603.33/warc/CC-MAIN-20231206194439-20231206224439-00507.warc.gz | 104,042,095 | 39,051 | # Recursion
--
Most of us know what recursion is and can write recursive code if we are asked to. But when we try to solve advance problems(i.e binary tree, binary search tree etc.) and see recursive solutions, we get confused about how that code is working, so in this post, I will try to explain how a recursive code works behind the scene with a simple factorial finding recursive code, and we will be able to trace more complex recursive code the same way.
Now let’s see the c++ code to return the factorial of a number :
How does this code work?
At first main() function calls the factorial(5), so in the stack at first main() is added and after that at the top of the stack factorial(5) is added.
In the function initially n = 5 then on line 5 it checks if n = 0, as it is not program goes to line 8 and return 5 * factorial(4). So, factorial(4) is added to the stack. Here, 5 * factorial(4) is called from factorial(5), which means whenever factorial(4) will return anything it will return to factorial(5).
Now n = 4 and again on line 5 it checks if n = 0, as it is not program goes to line 8 and return 4* factorial(3). So, factorial(3) is added to the stack.
For n = 3 and again on line 5 it checks if n = 0, as it is not program goes to line 8 and return 3 * factorial(2). So, factorial(2) is added to the stack.
For n = 2 and again on line 5 it checks if n = 0, as it is not program goes to line 8 and return 2 * factorial(1). So, factorial(1) is added to the stack.
For n = 1 and again on line 5 it checks if n = 0, as it is not program goes to line 8 and return 1* factorial(0). So, factorial(0) is added to the stack.
For factorial(0) now on the line 5 it check if n == 0, and this time n is equal to 0, so 1 will be returned to factorial(0). By the way, I didn’t talk about why we have this if(n == 0) statement, this is our base case and every recursive code needs to have a base case so that it is guaranteed the code will stop executing eventually without running out of memory. Now, let's see the stack after 1 is returned to factorial(0).
As 1 is returned to factorial(0), now we have
1 * 1(the value of factorial(0)) -> 1 at the top of the stack. Remember 1 * factorial (0) was called by factorial(1), so the result of 1 * factorial(0)->1 will be returned to factorial(1) and 1 * facctorial(0) will be poped out of stack.
1 is returned to factorial(1), now we have 2 * 1(value of factorial(1)) ->2 at the top of the stack and 2 will be returned to factorial(2) as 2 * factorial(1) was called by factorial(2) and 2 * factorial(1) will be popped out of the stack.
2 is returned to factorial(2), now we have 3 * 2(value of factorial(2)) ->6 at the top of the stack and 6 will be returned to factorial(3) as 3 * factorial(2) was called by factorial(3) and 3 * factorial(2) will be popped out of the stack.
6 is returned to factorial(3), now we have 4 * 6(value of factorial(3)) -> 24 at the top of the stack and 24 will be returned to factorial(4) as 4 * factorial(3) was called by factorial(4) and 4 * factorial(3) will be popped out of the stack.
As 24 is returned to factorial(4), now we have 5* 24(value of factorial(4)) -> 120 at the top of the stack and 120 will be returned to factorial(5) as 5 * factorial(4) was called by factorial(5) and 5 * factorial(4) will be popped out of the stack.
120 will be returned to factorial(5), so we have the result of factorial(5)->120 and this will be at the top of the stack. As factorial(5) was called by main() function, its value 120 will be returned to main() function and factorial(5) will be popped out of the stack.
So, this is how recursion works. If you have found this post helpful, leave a clap, and if there are any mistakes feel free to let me know. | 1,033 | 3,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-50 | longest | en | 0.934768 |
http://mathhelpforum.com/discrete-math/188213-cubic-graphs-bridge.html | 1,481,274,714,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542687.37/warc/CC-MAIN-20161202170902-00237-ip-10-31-129-80.ec2.internal.warc.gz | 182,743,573 | 9,534 | ## Cubic graphs with a bridge
How can you prove that a cubic graph with a bridge cannot be 3-edge colored?
I guess one could try a proof by contradiction, so we assume a 3 edge coloring is possible for such a graph. By than I am not sure in which direction to continue.
Or maybe a proof by contrapositive, so let say we have a cubic graph with a 3-edge coloring. How do we show this graph is bridgeless? | 98 | 405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2016-50 | longest | en | 0.967809 |
https://aakashdigitalsrv1.meritnation.com/ask-answer/question/prove-that-angles-in-the-same-segment-of-a-circle-are-equal/circles/513519 | 1,670,363,493,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00873.warc.gz | 108,850,155 | 8,117 | # Prove that angles in the same segment of a circle are equal
Hi!
Here is the proof of the given statement.
Theorem: Angles is the same segment of a circle are equal
By the mentioned theorem
Reflex ∠AOB = 2∠APB and reflex ∠AOB = 2∠AQB
∴ 2∠APB = 2∠AQB
⇒ ∠APB = ∠AQB
Hence, ∠APB = ∠AQ B
Cheers!
• 124
Thank you very much sir
• 29
What are you looking for? | 133 | 358 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-49 | latest | en | 0.891926 |
https://rebab.net/what-is-5-percent-of-200000/ | 1,653,418,294,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00213.warc.gz | 548,869,729 | 4,659 | What is % of ?
What is out of ?
### X is Y Percent of What Calculator
is % of what?
Using this tool you deserve to find any percentage in three ways. So, we think you got to us searching for answers like:1) What is 5 percent (%) that 200000?2) 5 is what percent the 200000?Or might be: just how much is 5 percent the 200000?
See the solutions to these difficulties below.
If friend are trying to find a
### 1) What is 5% of 200000?
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% / 100 = part / totality replace the provided values:
5 / 100 = component / 200000
Cross multiply:
5 x 200000 = 100 x Part, or
1000000 = 100 x component
Now, divide by 100 and also get the answer:
Part = 1000000 / 100 = 10000
### 2) What is 5 the end of 200000?
This concern is equivalent to: "5 is what percent of 200000?" Or What percent 5 is the end of 200000?
Use again the same portion formula:
% / 100 = part / entirety replace the provided values:
% / 100 = 5 / 200000
Cross multiply:
% x 200000 = 5 x 100
Divide by 200000 to acquire the percentage:
% = (5 x 100) / 200000 = 0.0025%
A shorter means to calculate x the end of y
You can easily discover 5 is the end of 200000, in one step, by simply splitting 5 by 200000, climate multiplying the an outcome by 100. So,
5 is out of 200000 = 5/200000 x 100 = 0.0025%
To find an ext examples, just pick one in ~ the bottom the this page. | 418 | 1,403 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-21 | latest | en | 0.897259 |
http://www.knowledgiate.com/what-are-cross-rates-and-chain-rule/ | 1,550,261,615,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247479101.30/warc/CC-MAIN-20190215183319-20190215205319-00620.warc.gz | 382,648,792 | 11,694 | # What are Cross Rates and Chain Rule
India, buying rates are calculated on the assumption that the foreign exchange acquired is disposed of abroad in the international market and the proceeds realized in US dollars. The US dollars thus acquired would be sold in the local inter-bank market to realize the rupee.
For example, if the bank purchased a DEM 10,000 bill it is assumed that it will sell the DEM at the Singapore market and acquire US dollars there. The US dollars are then sold in the inter-bank market against Indian rupee.
The bank would get the rate for US dollars in terms of Indian rupees in India. This would be the inter-bank rate for US dollars. It would also get the rate for US dollars in terms of Deutsche marks at the Singapore market. The bank has to quote the rats to the customer for Deutsche mark in terms of Indian rupees.
The fixing of rate of exchange between the foreign currency and Indian rupee through the medium of some other currency is done by what is known as ‘Chaos Rule’.
The rate thus obtained is the ‘cross rate’ between these currencies. For example, let us assume that in the inter-bank market dollar is quoted at Rs. 34.50 and at Singapore market the dollar is quoted at DEM 1.6000. With this information, the rate of exchange of Deutsche mark in terms of rupees may be calculate’ as follows:
?Rs. = DEM 1 …(1)
if DEM 1.6000 = USD 1 …(2)
and USD 1 = Rs. 34.50 …(3)
It should be noted that the currency which appears as the second item (right hand side) in the first equation appears as first item (left-hand side) in the next equation. Thus Mark appears on the right-hand side in the first equation and left-hand side in the second equation. US dollar which appears on the right-hand side in the second equation appears on the left-hand side in the third equation.
The rate of exchange between Indian rupee and Deutsche mark can be calculated by dividing the product of the right-hand side by the product of the left-hand side.
(34.50x 1x 1)/(1.6000 x 1 )= Rs.21.5625.
It would be immediately seen that the above calculation involves simply dividing the rupee-dollar rate by the dollar-mark rate.
Note: If rupee-sterling rate is arrived at taking inter-bank rupee-dollar rate and sterling-dollar rate abroad, the rupee-sterling rate is arrived at by multiplying the two rates and not by dividing. This is because the sterling-dollar rate is quoted as GBP 1 = USD ? and not USD 1 = GBP ? as in other currencies.
Close | 569 | 2,474 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2019-09 | latest | en | 0.958634 |
http://www.ima.org.uk/viewItem.cfm-cit_id=383373.html | 1,488,227,100,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501173761.96/warc/CC-MAIN-20170219104613-00374-ip-10-171-10-108.ec2.internal.warc.gz | 447,110,122 | 5,393 | A A A+
# Structure and Randomness: Pages from Year One of a Mathematical Blog
Terence Tao
AMERICAN MATHEMATICAL SOCIETY 2009, 298 PAGES
PRICE (PAPERBACK) £27.50 ISBN 978-0-8218-4695-7
Terence Tao is a distinguished mathematician, perhaps best known for his work in combinatorics and number theory, linked especially to the theory of arithmetic progressions of prime numbers. In 2007, he turned his homepage into a weblog, and this book collects some of his online writings which first appeared there. In the book’s collection of some of these blogs, it sketches out unusual proofs for classical theorems, the texts of three of his invited lectures, a selection of discussions of open problems, and a few number curiosities.
The book follows themes. One of these is given by the title, and treated most fully in the chapter called "structure and randomness". The idea is to divide mathematics into three parts: the highly structured or regular objects, the effectively random objects, and the "hybrid" objects built from both structured and random components. Tao then goes on to describe how mathematics tries to deal with these different cases using algebra, geometry, analysis and probability, going into detail on the problems faced when working, in his area, with prime numbers.
Prime numbers he gives as an example of a hybrid set and explains how you need a large variety of mathematical tools to attempt to work in this area, such as algebra, geometry, analysis, probability, decompositions, algorithms and evolution equations. The style of the book gives sketches of mathematical ideas with extensive bibliography and at times I would have liked more mathematical detail to allow my journey to become continuous through the book rather than in discrete steps.
Another interesting theme running through the book is tricks and patterns of establishing results which can be used across many different types of problems. Parts of most chapters in this book are put aside to this theme. For example in Chapter 1 Tao gives a family of tricks for improving inequalities, a technique he calls amplification.
These collections of blogs, are for mathematicians with a good understanding of abstract algebra, algebraic geometry, functional analysis, graph theory, harmonic analysis, Lie algebras, mathematical logic, measure theory, number theory, partial differential equations and real analysis. My knowledge in some of these areas is patchy, forgotten and sometimes a little vague, but the book still gave me a wonderful insight into the world of a mathematician working at the edge of our understanding. The book will be of interest to anyone who is, or who has studied mathematics, and also those interested in mathematics coming from a physics, statistics, economics and computer science background.
Steve Humble FIMA NCETM
Mathematics Today December 2010
Structure and Randomness: Pages from Year One of a Mathematical Blog can be purchased at Amazon.co.uk | 597 | 2,969 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-09 | latest | en | 0.957983 |
http://openstudy.com/updates/4fb2d386e4b055653428bccc | 1,448,553,614,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398447758.91/warc/CC-MAIN-20151124205407-00335-ip-10-71-132-137.ec2.internal.warc.gz | 178,740,989 | 14,172 | ## Math4Life 3 years ago Evaluate the infinte geometric series.
1. Math4Life
$\sum_{n=1}^{\infty} (-1/2 )^{n-2}$
2. Ket-kat
Okay, this is an infinite solution since the top is your upper limit and the bottom is the lower limit
3. Ket-kat
so, -1/2^1-2 and so on - all the way to infinity - is going to be your answer
4. Math4Life
$\sum_{n=1}^{\infty} 3(0.4)^{n-1}$
5. Math4Life
that also and thanks keep explaining
6. Math4Life
1. -1/2^1-2 all the way to infinity (my answer for number 1?)
7. Ket-kat
Same thing for this equation. Replace the n with the lower limit (1) all the way to infinity. You don't have to show the work up to infinity, just lead up to it
8. Math4Life
@Ket-kat so what would that be for number 2?
9. Ket-kat
?
10. Math4Life
@Ket-kat whats the answer for 1 and 2, i'm confused.
11. 15Trev
hate that stuff
12. Math4Life
@15Trev can you still help me though?
13. Ket-kat
Okay Like I said. In both equations on the left side of the stigma, replace the n with: 1,2,3,4,5,6,7,8,9,10.......... infinity And solve the equations :)
14. 15Trev
i think so
15. Math4Life
Thanks
16. Ket-kat
<3 No worries
17. 15Trev
we did this last chapter
18. Ket-kat
FLVS? I'm doing it now xD
19. Math4Life
Nice so its fresh for you
20. 15Trev
k
21. 15Trev
so first u have to know that whats in parenths is ur rate
22. 15Trev
multiply parenths by # next to them
23. Math4Life
1. $\sum_{n=1}^{\infty} (-1/2)^{n-1}$ 2. $\sum_{n=1}^{\infty} 3(0.4)^{n-1}$
24. 15Trev
like 3x.4
25. Math4Life
=12
26. 15Trev
no
27. 15Trev
.4
28. Math4Life
29. 15Trev
ya
30. 15Trev
now take 1 and subtract by # in ( )
31. 15Trev
like 1-.4
32. Math4Life
.6
33. 15Trev
forget everythng i said
34. Math4Life
*easily erases everything you said from brain*
35. Math4Life
@15Trev soooooo...?
36. 15Trev
wait
37. 15Trev
k lets do #2
38. Math4Life
39. 15Trev
later
40. Math4Life
ok
41. 15Trev
remember everything i siad erlier actualy
42. Math4Life
WTF.. ok lol :)
43. saifoo.khan
wtf = wow, that's fantastic.
44. Math4Life
@saifoo.khan exactly
45. saifoo.khan
Totally.
46. 15Trev
u can leave now
47. 15Trev
wats 1-.4
48. Math4Life
.6
49. Math4Life
@15Trev its .6
50. 15Trev
just get someone else to do it
51. 15Trev
i forget
52. Math4Life
@15trev nobody else is helping
53. 15Trev
well if i brought home my book i would know
54. 15Trev
do u hav the formula
55. Math4Life
No :/
56. Math4Life
@15Trev i just turned it in without those answers :/
57. 15Trev
what do u mean
58. 15Trev
i goota take a poop brb
59. Math4Life
@15Trev nvm just ill send u a link to my new question
60. 15Trev
k
61. Math4Life
@15Trev can u oovoo?
62. 15Trev
not now
63. Math4Life
@15Trev why?
64. 15Trev
the person im babysitting is right here
65. 15Trev
66. Math4Life
@15Trev ur not babysitting ur bro? and im in the middle of posting it
67. 15Trev
no
68. 15Trev
nabor
69. 15Trev
she likes minecraft
70. 15Trev
lol
71. Math4Life
@15Trev lol to bad both servers we like r down
72. 15Trev
ya
73. Math4Life
@15Trev so when r u done babysittin?
74. 15Trev
730
75. Math4Life
@15Trev I gotta mow the lawn, so later we can oovoo
76. 15Trev
k just finished babysittin | 1,228 | 3,247 | {"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} | 4.03125 | 4 | CC-MAIN-2015-48 | longest | en | 0.777282 |
https://achieverstudent.com/hotel-margin-regression-report/ | 1,660,733,215,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00163.warc.gz | 101,782,207 | 14,881 | # Hotel Margin Regression Report
Please submit your EXCEL file with summary here – MUST BE DONE IN EXCEL – I have to show the teacher how I got the answers so please use formulas in excel or use a pivot table if needed
Location analysis is one function of operations management. Deciding where to locate a plant, warehouse, or retail outlet is a critical decision for any organization. A large number of variables must be considered in this decision problem. For example, a production facility must be located close to suppliers of raw resources and supplies, skilled labor and transportation to customers. Retail outlet must consider the type and number of potential customers. In the next example, we describe an application of regression analysis to find profitable locations for a motel chain.
La Quinata Motor Inns is a moderately priced chain of motor inns located across the United States. Its market is the frequent business traveler. The chain recently launched a campaign to increase market share by building new inns. The management of the chain is aware of the difficulty in choosing locations for new motels Moveover, making decisions without adequate information often results in poor decisions. Consequently, the chain’s management acquired data on 100 randomly selected inns belonging to La Quinta. The objects were to predict which sites are likely to be profitable.
To measure profitability, La Qinata used operation margin, which is the ratio of the sum of profit, depreciation and interest expenses divided by total revenue. (Although occupancy is often used as a measure of a motels success, the company statistician concluded that occupancy was too unstable, especially during economic turbulence.) The higher the operation margin, the greater the success of the inn. La Quinta defines profitable inns as those with a operating margin in excess of 50%. unprofitable inns are those with margins of less than 30%. After a discussion with a number of experienced managers, La Quinta decided to select one or two independent variables from each of the following categories, competition, market awareness, demand generators, demographics, and physical location. To measure the degree of competition, they determined the total number of motels and hotel rooms within 3 miles of each La Qunita inn. Market awareness was measured by the number of miles to the closest competing motel. Two variables that represent sources of customers were chosen. The amount of office space and college and university enrollment in the surrounding community are demand generators. Both of these are measures of economic activity. A demographic variable that describes the community is the median household income. Finally, as a measure of the physical qualities of the location La Quinta choose the distance to the downtown core. These data are stored using the following format
Colum 1: y = operating margin, in percent
Colum 2: x1 = Total number of motel and hotel rooms with 3 miles of La Quinta inn
Colum 3:X2 = Number of miles to closet competition
Colum 4: X3= office space in thousands of square feet in surrounding community
Colum 5: X4 = College and university enrollment (in thousands) in nearby university or college
Colum 6: X5 = median household income (in thousands) in surrounding community
colum 7 : X6= Distance (in miles) to the downtown core
Develop a regress analysis
b. Test to determine where there is enough evidence to infer that the model is valid
Test each of the slop coefficients
Interpret the coefficients
Predict with 95% confidence the operating margin of a site with the following characteristics
Don't use plagiarized sources. Get Your Custom Essay on
Hotel Margin Regression Report
Just from \$13/Page
There are 3,815 rooms within 3 miles of the site, the closest other hotel or motel is 9 miles away, the amount of office space is 476,000 square fee , there is one college and one university with a total enrollment of 24,500 students, the median income in the area is \$35,000, and the distance to the downtown core is 11.2 miles.
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Order your essay today and save 20% with the discount code SEARCHGO | 877 | 4,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2022-33 | latest | en | 0.947229 |
http://studentsmerit.com/paper-detail/?paper_id=31507 | 1,480,717,959,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540698.78/warc/CC-MAIN-20161202170900-00113-ip-10-31-129-80.ec2.internal.warc.gz | 257,905,387 | 7,244 | Details of this Paper
There was no dominated strategy using pure strategies.
Description
solution
Question
1. There was no dominated strategy using pure strategies.;a. Use mixed strategies to find a dominated strategy to eliminate. Can you now find an IEDS;solution to the game? (Hint: You may be able to eliminate a strategy for only one player to start;IEDS.);b. Is the solution in (a) the same as the NE solution?;U;Player 1;D;L;2, 0;M;1, 1;Player 2;C;R;1, 2;3, 1;0, 2;3, 3;3, 0;1, 3;2, 1;2. There was no NEsolution in pure strategies.;a. Find the mixed strategy NE solution. Show the best response functions for each player.;b. What are the expected payoffs to each player from the mixed strategy solution?;Player 1;D;Player 2;L;R;U;1, 1;0, 2;2, 1;0, 2;3. Recall question 1 from Problem Set 1. Two students, Alex and Carrie pick study locations;simultaneously. The payoffs are the utilities of the grades from studying. The payoffs are;vonNeumann Morgenstern utilities and are given below in the normal form game.;a. Is there a dominant strategy solution?;b. What are all the NE solutions? Be sure to include both pure and mixed strategies. Show the;best response functions for each player.;Carrie;Library;Union;Library;10.0, 6.0;Union;4.0, 4.0;4.0, 4.0;Alex;8.0, 8.0
Paper#31507 | Written in 18-Jul-2015
Price : \$27 | 413 | 1,328 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2016-50 | longest | en | 0.864111 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_INN/TRS/secret2005/aprove2.trs.Thm26:POLO_7172_DP:NO.html.lzma | 1,719,285,297,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00392.warc.gz | 89,096,866 | 2,208 | Term Rewriting System R:
[x, y]
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
GE(s(x), s(y)) -> GE(x, y)
MINUS(s(x), s(y)) -> MINUS(x, y)
DIV(x, y) -> IFY(ge(y, s(0)), x, y)
DIV(x, y) -> GE(y, s(0))
IFY(true, x, y) -> IF(ge(x, y), x, y)
IFY(true, x, y) -> GE(x, y)
IF(true, x, y) -> DIV(minus(x, y), y)
IF(true, x, y) -> MINUS(x, y)
Furthermore, R contains three SCCs.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polynomial Ordering`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Remaining`
Dependency Pair:
GE(s(x), s(y)) -> GE(x, y)
Rules:
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))
Strategy:
innermost
The following dependency pair can be strictly oriented:
GE(s(x), s(y)) -> GE(x, y)
There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(GE(x1, x2)) = x1 POL(s(x1)) = 1 + x1
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 4`
` ↳Dependency Graph`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Remaining`
Dependency Pair:
Rules:
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polynomial Ordering`
` →DP Problem 3`
` ↳Remaining`
Dependency Pair:
MINUS(s(x), s(y)) -> MINUS(x, y)
Rules:
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))
Strategy:
innermost
The following dependency pair can be strictly oriented:
MINUS(s(x), s(y)) -> MINUS(x, y)
There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(MINUS(x1, x2)) = x1 POL(s(x1)) = 1 + x1
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 5`
` ↳Dependency Graph`
` →DP Problem 3`
` ↳Remaining`
Dependency Pair:
Rules:
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Polo`
` →DP Problem 2`
` ↳Polo`
` →DP Problem 3`
` ↳Remaining Obligation(s)`
The following remains to be proven:
Dependency Pairs:
IF(true, x, y) -> DIV(minus(x, y), y)
IFY(true, x, y) -> IF(ge(x, y), x, y)
DIV(x, y) -> IFY(ge(y, s(0)), x, y)
Rules:
ge(x, 0) -> true
ge(0, s(x)) -> false
ge(s(x), s(y)) -> ge(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
div(x, y) -> ify(ge(y, s(0)), x, y)
ify(false, x, y) -> divByZeroError
ify(true, x, y) -> if(ge(x, y), x, y)
if(false, x, y) -> 0
if(true, x, y) -> s(div(minus(x, y), y))
Strategy:
innermost
Innermost Termination of R could not be shown.
Duration:
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https://www.sailnet.com/threads/heavier-better-motion-comfort.144210/?u=127367 | 1,674,770,665,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00440.warc.gz | 964,900,932 | 22,239 | 1 - 1 of 1 Posts
#### kwaltersmi
Joined
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2,038 Posts
Weight/displacement is just one factor related to the motion of a sailboat. But don't take my word for it. Here's what Ted Brewer says about his "Comfort Ratio":
"This is a ratio that I dreamed up, tongue-in-cheek, as a measure of motion comfort but it has been widely accepted and, indeed, does provide a reasonable comparison between yachts of similar type. It is based on the fact that the faster the motion the more upsetting it is to the average person. Given a wave of X height, the speed of the upward motion depends on the displacement of the yacht and the amount of waterline area that is acted upon. Greater displacement, or lesser WL area, gives a slower motion and more comfort for any given sea state.
Beam does enter into it as wider beam increases stability, increases WL area, and generates a faster reaction. The formula takes into account the displacement, the WL area, and adds a beam factor. The intention is to provide a means to compare motion comfort of vessels of similar type and size, not to compare that of a Lightning class sloop with that of a husky 50 foot ketch."
The comfort ratio formula is as follows: Displacement in pounds / (.65 x (0.7 LWL + 0.3 LOA) x B^1.333). Brewer says ratios vary from 5.0 for a light displacement daysailer to the high 60.0's for a super heavy ocean cruiser.
More here if you're interested: Comfort, Capsizing, and SailCalc
And hopefully Bob will chime in too...
1 - 1 of 1 Posts | 358 | 1,502 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2023-06 | latest | en | 0.940249 |
https://affairscloud.com/reasoning-questions-syllogism-set-23/ | 1,686,405,738,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657720.82/warc/CC-MAIN-20230610131939-20230610161939-00715.warc.gz | 107,978,702 | 38,050 | # Reasoning Questions: Syllogism Set 23
Hello Aspirants. Welcome to Online Reasoning Section with explanation in AffairsCloud.com. Here we are creating Best question samples from Syllogism with explanation, which is common for competitive exams. We have included Some questions that are asked in previous exams !!!
Questions penned by Yogit
1. Direction:Q(1 – 2)
Statements: All bags are novels. No novel is a pen. All pens are pencils.
Conclusions: -1) No bag is a pen 2) all bag being pencils is a possibility.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 5) if both conclusions I and II follow
Explanation :
2. Statements: All bags are novels. No novel is a pen. All pens are pencils.Conclusions: 1) No pencil is a bag 2) All novels being pencils is a possibility.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 2) if only conclusion II follows.
Explanation :
3. Direction:Q(3 – 4)
Statements: -No boy is a graduate. Some graduates are girls. All girls are intelligent.
Conclusions: -1) All boys being intelligent is a possibility 2) No boy is a girl.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 1) if only conclusion I follows.
Explanation :
4. Conclusions: 1) No intelligent is a graduate 2) At least some intelligent are graduates.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 2) if only conclusion II follows.
Explanation :
5. Statements: – No river is white. All whites are blacks. Some blacks are huts.
Conclusions: – 1) No river is black 2) At least some whites are huts
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 4) if neither conclusion I nor II follows.
Explanation :
6. Statements: No virus is hardware. Some hardware are antivirus. All applications are hardware.
Conclusions: 2) No application is a virus 2) some virus are applications.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 1) if only conclusion I follows.
Explanation :
7. Statements: All wings are plastic. No sheet is a wing. All sheets are fibres.
Conclusions: 1) Some fibres are sheets 2) Some fibres if they are wings must be plastic.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 5) if both conclusions I and II follow
Explanation :
8. Statements: Some desks are circular. Some rectangles are desk.
Conclusions: 1) Some desk being circular is a possibility 2) some rectangles are circular
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 4) if neither conclusion I nor II follows.
Explanation :
9. Direction:Q(9 –10)
Statements: All cows are birds. All birds are bat. Some birds are sparrows.
Conclusions: 1) All cows are bats 2) All bats being sparrows is a possibility.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 5) if both conclusions I and II follow
Explanation :
10. Conclusions: 1) Some bats are not birds 2) All sparrows are birds.
1) if only conclusion I follows.
2) if only conclusion II follows.
3) if either conclusion I or II follows.
4) if neither conclusion I nor II follows.
5) if both conclusions I and II follow
Answer – 4) if neither conclusion I nor II follows.
Explanation : only 2 follows (when all birds overlap all sparrow)
Note: Dear Readers if you have any doubt in any chapter in Quants you can ask here. We will clear your doubts
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AffairsCloud Ebook - Support Us to Grow | 1,138 | 4,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2023-23 | longest | en | 0.889088 |
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MIT Course , Spring 2005 , Prof. Gilbert Strang
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# Overview
The Geometry of Linear Equations - Elimination with Matrices-Multiplication and Inverse Matrices - Factorization into A = LU - Transposes, Permutations, Spaces R^n-Column Space and Nullspace -Solving Ax = 0: Pivot Variables, Special Solutions - Solving Ax = b: Row Reduced Form R - Independence, Basis, and Dimension - The Four Fundamental Subspaces - Matrix Spaces; Rank 1; Small World Graphs-Graphs, Networks, Incidence Matrices - Orthogonal Vectors and Subspaces - Projections onto Subspaces - Projection Matrices and Least Squares - Orthogonal Matrices and Gram - Schmidt
Properties of Determinants - Determinant Formulas and Cofactors - Cramers Rule, Inverse Matrix, and Volume - Eigenvalues and Eigenvectors - Diagonalization and Powers of A - Differential Equations and exp(At) - Markov Matrices - Fourier Series - Symmetric Matrices and Positive Definiteness - Complex Matrices - Fast Fourier Transform - Positive Definite Matrices and Minima - Similar Matrices and Jordan Form - Singular Value Decomposition - Linear Transformations and Their Matrices - Change of Basis - Image Compression - Left and Right Inverses - Pseudoinverse - Final Course Review
### Lecture 20: Cramers Rule, Inverse Matrix, and Volume
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https://de.maplesoft.com/support/help/maple/view.aspx?path=dsolve/rkf45 | 1,716,164,487,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00804.warc.gz | 177,037,942 | 30,828 | rkf45 - Maple Help
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# Online Help
###### All Products Maple MapleSim
dsolve/numeric/rkf45
find numerical solution of ordinary differential equations
Calling Sequence dsolve(odesys, numeric, method=rkf45, vars, options) dsolve(numeric, method=rkf45, procopts, options)
Parameters
odesys - set or list; ordinary differential equation(s) and initial conditions numeric - literal name; instruct dsolve to find a numerical solution method=rkf45 - literal equation; numerical method to use vars - (optional) any indeterminate function of one variable, or a set or list of them, representing the unknowns of the ODE problem options - (optional) equations of the form keyword = value procopts - options used to specify the ODE system using a procedure (procedure, initial, start, number, and procvars). For more information, see dsolve/numeric/IVP.
Description
• The dsolve command with the options numeric and method=rkf45 finds a numerical solution using a Fehlberg fourth-fifth order Runge-Kutta method with degree four interpolant. This is the default method of the type=numeric solution for initial value problems when the stiff argument is not used. The other non-stiff method is a Runge-Kutta method with the Cash-Karp coefficients, ck45.
Modes of Operation
• The rkf45 method has two distinct modes of operation (for procedure-type outputs).
• With the range option
When used with the range option, the method computes the solution for the IVP over the specified range, storing the solution information internally, and uses that information to rapidly interpolate the desired solution value for any call to the returned procedure(s).
Though possible, it is not recommended that the returned procedure be called for points outside the specified range.
This method can be used in combination with the refine option of odeplot to produce an adaptive plot (that is, odeplot uses the precomputed points to produce the plot when refine is specified.)
It is not recommended that this method be used for problems in which the solution can become singular, as each step is stored, and many steps may be taken when near a singularity, so memory usage can become a significant issue.
The storage of the interpolant in use by this method can be disabled by using the interpolation=false option described below. This is recommended for high accuracy solutions where storage of the interpolant (in addition to the discrete solution) requires too much memory. Disabling the interpolant is not generally recommended because the solution values are obtained from an interpolation of the 5 closest points, and does not necessarily provide an interpolant with order 4 error.
• Without the range option
When used without the range option, the IVP solution values are not stored, but rather computed when requested.
Because not all solution values are stored, computation must restart at the initial values whenever a point is requested between the initial point and the most recently computed point (to avoid reversal of the integration direction), so it is advisable to collect solution values moving away from the initial value.
Options
• The following options are available for the rkf45 method.
'output' = keyword or array 'known' = name or list of names 'abserr' = numeric 'relerr' = numeric 'initstep' = numeric 'interr' = boolean 'maxfun' = integer 'number' = integer 'procedure' = procedure 'start' = numeric 'initial' = array 'procvars' = list 'startinit' = boolean 'implicit' = boolean 'optimize' = boolean 'compile' = boolean or auto 'range' = numeric..numeric 'events' = list 'event_pre' = keyword 'event_maxiter' = integer 'event_iterate' = keyword 'event_initial' = boolean 'complex' = boolean
output
Specifies the desired output from dsolve. The keywords procedurelist, listprocedure, or operator provide procedure-type output, the keyword piecewise provides output in the form of piecewise functions over a specified range of independent variable values, and a 1-D array or Array provide output at fixed values of the independent variable. For more information, see dsolve/numeric.
known
Specifies user-defined known functions. For more information, see dsolve/numeric.
abserr, relerr, and initstep
Specify the desired accuracy of the solution, and the starting step size for the method. For more information, see dsolve/Error_Control. The default values for rkf45 are abserr=1e-7 and relerr=1e-6. The value for initstep, if not specified, is determined by the method, taking into account the local behavior of the ODE system.
interr
By default this is set to true, and controls whether the solution interpolant error (including the interpolant on index-1 variables for DAE problems) is integrated into the error control. When set to $\mathrm{false}$, areas where the solutions is varying rapidly (e.g. a discontinuity in a derivative due to a piecewise) may have a much larger solution error than dictated by the specified error tolerances. When set to $\mathrm{true}$, the step size is reduced to minimize error in these regions, but for problems where there is a jump discontinuity in the variables, the integration may fail with an error indicating that a singularity is present. In the latter case where an error is thrown, it may be advantageous to model the discontinuities using events (see dsolve/Events).
maxfun
Specifies a maximum on the number of evaluations of the right-hand side of the first order ODE system. This option is disabled by specifying $\mathrm{maxfun}=0$. The default value for rkf45 is $30000$.
number, procedure, start, initial, and procvars
These options are used to specify the IVP using procedures. For more information, see dsolve/numeric/IVP.
startinit, implicit, and optimize
These options control the method and behavior of the computation. For more information on startinit and implicit, see dsolve/numeric/IVP. For more information on optimize, see dsolve/numeric.
compile
This option specifies that the internally generated procedures that are used to compute the numeric solution be compiled for efficiency. Note that this option will only work if $\mathrm{Digits}$ is set within the hardware precision range, and the input function contains only evalhf capable functions (e.g. only elementary mathematical functions like exp, sin, and ln). See dsolve/Efficiency. By default this value is set to false. If set to true and a compile is not possible, an error will be thrown. If set to auto and a compile is not possible, the uncompiled procedures will be used directly.
range
Determines the range of values of the independent variable for which solution values are required. Use of this option significantly changes the behavior of the method for the procedure-style output types discussed in dsolve/numeric (see description of with/without range in Modes of Operation section of this help page).
events, event_pre, event_maxiter, event_iterate, event_initial
These options are used to specify and control the behavior of events for the numerical solution. These options are discussed in detail in dsolve/Events.
complex
Accepts a boolean value indicating if the problem is (or will become) complex valued. By default this is detected based on the input system and initial data, but in cases where the input system is procedure defined, or the system is initially real, it may be necessary to specify complex=true to obtain the solution. It is assumed that for an initially real system that becomes complex, the point at which this transition occurs is considered to be a singularity, so if complex=true is not specified, the integration will halt at that point.
Additional Information
• By setting infolevel[dsolve] to $2$, information on the last mesh point evaluation is provided in the event of an error.
• The rkf45 method is capable of computing high-accuracy solutions for IVPs because the precision of the computation can be increased by changing the Digits environment variable. However, it is recommended that the higher order dverk78 or gear methods be used instead.
• Results can be plotted by using the function odeplot in the plots package.
Examples
Solution with a range - solution over $0..1$ is stored:
> $\mathrm{dsys1}≔\left\{\frac{ⅆ}{ⅆt}x\left(t\right)=y\left(t\right),\frac{ⅆ}{ⅆt}y\left(t\right)=x\left(t\right)+y\left(t\right),x\left(0\right)=2,y\left(0\right)=1\right\}$
${\mathrm{dsys1}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{x}{}\left({t}\right){=}{y}{}\left({t}\right){,}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{x}{}\left({t}\right){+}{y}{}\left({t}\right){,}{x}{}\left({0}\right){=}{2}{,}{y}{}\left({0}\right){=}{1}\right\}$ (1)
> $\mathrm{dsol1}≔\mathrm{dsolve}\left(\mathrm{dsys1},\mathrm{numeric},\mathrm{output}=\mathrm{listprocedure},\mathrm{range}=0..1\right):$
> $\mathrm{dsol1x}≔\mathrm{subs}\left(\mathrm{dsol1},x\left(t\right)\right):$$\mathrm{dsol1y}≔\mathrm{subs}\left(\mathrm{dsol1},y\left(t\right)\right):$
> $\left[\mathrm{dsol1x}\left(0\right),\mathrm{dsol1y}\left(0\right)\right]$
$\left[{2.}{,}{1.}\right]$ (2)
> $\left[\mathrm{dsol1x}\left(0.4\right),\mathrm{dsol1y}\left(0.4\right)\right]$
$\left[{2.69118458493332}{,}{2.60811748317261}\right]$ (3)
> $\left[\mathrm{dsol1x}\left(1.0\right),\mathrm{dsol1y}\left(1.0\right)\right]$
$\left[{5.58216753140384}{,}{7.82688926499912}\right]$ (4)
Solution without a range - each point is computed when requested:
> $\mathrm{dsol2}≔\mathrm{dsolve}\left(\mathrm{dsys1},\mathrm{numeric},\mathrm{method}=\mathrm{rkf45},\mathrm{output}=\mathrm{procedurelist}\right):$
> $\mathrm{dsol2}\left(0\right)$
$\left[{t}{=}{0.}{,}{x}{}\left({t}\right){=}{2.}{,}{y}{}\left({t}\right){=}{1.}\right]$ (5)
> $\mathrm{dsol2}\left(0.4\right)$
$\left[{t}{=}{0.4}{,}{x}{}\left({t}\right){=}{2.69118458493332}{,}{y}{}\left({t}\right){=}{2.60811748317261}\right]$ (6)
> $\mathrm{dsol2}\left(1.0\right)$
$\left[{t}{=}{1.0}{,}{x}{}\left({t}\right){=}{5.58216753140384}{,}{y}{}\left({t}\right){=}{7.82688926499911}\right]$ (7)
array solution:
> $\mathrm{dsys3}≔\left\{{\mathrm{D}}^{\left(2\right)}\left(x\right)\left(t\right)=-y\left(t\right),{\mathrm{D}}^{\left(2\right)}\left(y\right)\left(t\right)=\mathrm{D}\left(x\right)\left(t\right)+y\left(t\right)\right\}$
${\mathrm{dsys3}}{≔}\left\{{{\mathrm{D}}}^{\left({2}\right)}{}\left({x}\right){}\left({t}\right){=}{-}{y}{}\left({t}\right){,}{{\mathrm{D}}}^{\left({2}\right)}{}\left({y}\right){}\left({t}\right){=}{\mathrm{D}}{}\left({x}\right){}\left({t}\right){+}{y}{}\left({t}\right)\right\}$ (8)
> $\mathrm{init3}≔\left\{x\left(0\right)=1,\mathrm{D}\left(x\right)\left(0\right)=0,y\left(0\right)=0,\mathrm{D}\left(y\right)\left(0\right)=1\right\}$
${\mathrm{init3}}{≔}\left\{{x}{}\left({0}\right){=}{1}{,}{y}{}\left({0}\right){=}{0}{,}{\mathrm{D}}{}\left({x}\right){}\left({0}\right){=}{0}{,}{\mathrm{D}}{}\left({y}\right){}\left({0}\right){=}{1}\right\}$ (9)
> $\mathrm{dsol3}≔\mathrm{dsolve}\left(\mathrm{dsys3}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∪\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{init3},\mathrm{numeric},\mathrm{method}=\mathrm{rkf45},\mathrm{output}=\mathrm{Array}\left(\left[0,0.6,1.1,1.5,2.3,2.5\right]\right)\right)$
${\mathrm{dsol3}}{≔}\left[\begin{array}{c}{\mathrm{Typesetting}}{:-}{\mathrm{_Hold}}{}\left(\left[\left[\begin{array}{ccccc}t& x{}\left(t\right)& \frac{ⅆ}{ⅆt}{}x{}\left(t\right)& y{}\left(t\right)& \frac{ⅆ}{ⅆt}{}y{}\left(t\right)\end{array}\right]\right]\right)\\ {\mathrm{Typesetting}}{:-}{\mathrm{_Hold}}{}\left(\left[\left[\begin{array}{ccccc}0.0& 1.0& 0.0& 0.0& 1.0\\ 0.6& 0.9634120119365587& -0.1848064408121733& 0.6311283877215605& 1.1482184527487318\\ 1.1& 0.7669147709231557& -0.6543598518932257& 1.269965819734427& 1.4212746228163817\\ 1.5& 0.38774842823764183& -1.2827043365714699& 1.8884338106657934& 1.6704527648091125\\ 2.3& -1.3945766019956574& -3.372723596313235& 3.379309930393405& 1.9781469943175773\\ 2.5& -2.139341661159309& -4.088058562317105& 3.773067611974174& 1.9487169011577972\end{array}\right]\right]\right)\end{array}\right]$ (10)
procedurelist output:
> $\mathrm{deqn4}≔\left\{\frac{{ⅆ}^{3}}{ⅆ{t}^{3}}y\left(t\right)-2\left(\frac{{ⅆ}^{2}}{ⅆ{t}^{2}}y\left(t\right)\right)+2y\left(t\right)\right\}$
${\mathrm{deqn4}}{≔}\left\{\frac{{{ⅆ}}^{{3}}}{{ⅆ}{{t}}^{{3}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){-}{2}{}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){+}{2}{}{y}{}\left({t}\right)\right\}$ (11)
> $\mathrm{init4}≔\left\{y\left(0\right)=1,\mathrm{D}\left(y\right)\left(0\right)=1,{\mathrm{D}}^{\left(2\right)}\left(y\right)\left(0\right)=1\right\}$
${\mathrm{init4}}{≔}\left\{{y}{}\left({0}\right){=}{1}{,}{\mathrm{D}}{}\left({y}\right){}\left({0}\right){=}{1}{,}{{\mathrm{D}}}^{\left({2}\right)}{}\left({y}\right){}\left({0}\right){=}{1}\right\}$ (12)
> $\mathrm{dsol4}≔\mathrm{dsolve}\left(\mathrm{deqn4}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∪\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{init4},\mathrm{numeric},\mathrm{method}=\mathrm{rkf45},\mathrm{relerr}=1.{10}^{-8},\mathrm{abserr}=1.{10}^{-8},\mathrm{maxfun}=0,\mathrm{output}=\mathrm{procedurelist}\right)$
${\mathrm{dsol4}}{≔}{\mathbf{proc}}\left({\mathrm{x_rkf45}}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{...}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}{\mathbf{end proc}}$ (13)
> $\mathrm{dsol4}\left(0.05\right)$
$\left[{t}{=}{0.05}{,}{y}{}\left({t}\right){=}{1.05124946305338}{,}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{1.04995673867843}{,}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{0.997371821503983}\right]$ (14)
> $\mathrm{dsol4}\left(-0.45\right)$
$\left[{t}{=}{-0.45}{,}{y}{}\left({t}\right){=}{0.648630782805870}{,}\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{0.571757535785405}{,}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{t}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{0.870897136753006}\right]$ (15)
Piecewise output for complex-valued problem:
> $\mathrm{deqn5}≔\left\{\frac{ⅆ}{ⅆt}y\left(t\right)=Iy\left(t\right)\right\}$
${\mathrm{deqn5}}{≔}\left\{\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({t}\right){=}{I}{}{y}{}\left({t}\right)\right\}$ (16)
> $\mathrm{init5}≔\left\{y\left(0\right)=1\right\}$
${\mathrm{init5}}{≔}\left\{{y}{}\left({0}\right){=}{1}\right\}$ (17)
> $\mathrm{dsol5}≔\mathrm{dsolve}\left(\mathrm{deqn5}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}∪\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{init5},\mathrm{numeric},\mathrm{method}=\mathrm{rkf45},\mathrm{output}=\mathrm{piecewise},\mathrm{complex}=\mathrm{true},\mathrm{range}=0..\mathrm{Pi}\right):$
> $\mathrm{length}\left(\mathrm{dsol5}\right)$
${7753}$ (18)
> $\genfrac{}{}{0}{}{{\mathrm{dsol5}}_{2}}{\phantom{t=0}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{{\mathrm{dsol5}}_{2}}}{t=0}$
${y}{}\left({0}\right){=}{0.9999999997}{+}{4.9066668}{×}{{10}}^{{-12}}{}{I}$ (19)
> $\genfrac{}{}{0}{}{{\mathrm{dsol5}}_{2}}{\phantom{t={\mathrm{evalf}}_{15}\left(\mathrm{Pi}\right)}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{{\mathrm{dsol5}}_{2}}}{t={\mathrm{evalf}}_{15}\left(\mathrm{Pi}\right)}$
${y}{}\left({3.14159265358979}\right){=}{-1.000000643}{+}{2.430759107}{×}{{10}}^{{-8}}{}{I}$ (20)
References
Enright, W.H.; Jackson, K.R.; Norsett, S.P.; and Thomsen, P.G. "Interpolants for Runge-Kutta Formulas." ACM TOMS, Vol. 12, (1986): 193-218.
Fehlberg, E. "Klassische Runge-Kutta-Formeln vierter und niedrigerer Ordnung mit Schrittweiten-Kontrolle und ihre Anwendung auf Waermeleitungsprobleme". Computing, Vol. 6, (1970): 61-71.
Forsythe, G.E.; Malcolm, M.A.; and Moler, C.B. Computer Methods for Mathematical Computations. New Jersey: Prentice Hall, 1977.
Shampine, L.F. and Corless, R.M. "Initial Value Problems for ODEs in Problem Solving Environments". J. Comp. Appl. Math, Vol. 125(1-2), (2000): 31-40.
See Also | 5,392 | 16,190 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 52, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-22 | latest | en | 0.642211 |
http://math.stackexchange.com/questions/183656/i-want-to-to-know-how-to-show-that-this-function-is-bijective-between-these-sets | 1,469,386,050,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824133.26/warc/CC-MAIN-20160723071024-00014-ip-10-185-27-174.ec2.internal.warc.gz | 161,693,012 | 17,325 | # I want to to know how to show that this function is bijective between these sets [duplicate]
Possible Duplicate:
Is this function injective and surjective?
Let $f(x)=x^2$. In each of the following cases, is this function injective and/or surjective?
1. $f: \mathbb{R} \longrightarrow [0,\infty)$, I know this is surjective but not injective.
2. $f: \mathbb{C} \longrightarrow \mathbb{C}$. This one, I dont know how to see. I mean the function evaluated in the imaginary unity is -1, so I don't know how to deal with this.
3. $f: \mathbb{R} \longrightarrow \mathbb{R}$. This is neither surjective nor injective.
4. $f: \mathbb{R} \cup \{x \in \mathbb{C} : \mathrm{Re}(x) = 0\} \longrightarrow \mathbb{R}$. I dont really know how to deal with complex function definitions.
5. $f: \{z=x+iy: i^2=-1, y>0\} \cup \{z=x+iy: i^2=-1, y=0 \text{ and } x \ge 0\} \longrightarrow \mathbb{C}$..
Thanks a lot. I do know what surjective and bijective means, but I don't know how to prove it over complex subsets.
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## marked as duplicate by Gerry Myerson, t.b., Jennifer Dylan, LVK, J. M.Aug 20 '12 at 19:31
Better to edit your original question than to repost it. Better yet to accept some answers to your other questions. Please see meta.math.stackexchange.com/questions/3399/… – Gerry Myerson Aug 18 '12 at 10:48
$\Bbb R\subseteq\Bbb C$, so if $f$ is not injective on $\Bbb R$, it can’t be injective on $\Bbb C$, either. That is, if there are two different real numbers $x$ and $y$ such that $f(x)=f(y)$, those two real numbers are also complex numbers showing that $f$ is not injective on $\Bbb C$. Thus, for (2) all that’s left is to decide whether $f:\Bbb C\to\Bbb C$ is surjective. In other words, if $z\in\Bbb C$, is there always a complex number $w$ such that $w^2=z$? This is most easily answered if you know the exponential representation of complex numbers: every complex number can be written in the form $re^{i\theta}$, where $r,\theta\in\Bbb R$ and $r\ge 0$. Given such a number $re^{i\theta}$, can you find real numbers $s$ and $\varphi$ such that $\left(se^{i\varphi}\right)^2=re^{i\theta}$?
In (4) the domain of $f$ is the two coordinate axes in the complex plane. Alternatively, it’s the set of all $x+iy$ such that at least one of $x$ and $y$ is $0$, so it’s the set of all real numbers together with all purely imaginary numbers, i.e., numbers of the form $yi$. You already know that $f$ is not injective on the domain $\Bbb R$, which is included in this set, so you can conclude right away that $f$ is not injective on this set either. Thus, once again it comes down to deciding whether $f$ is surjective. Suppose that $r\in\Bbb R$. If $r\ge 0$ there is certainly an $x$ in the domain of $f$ such that $x^2=r$. What if $r<0$? For instance, is there a number of the form $yi$ whose square is $-4$?
Try (2) and (4), and if you get them, see whether the experience helps you with (5); if not, let me know, and I’ll add some hints for (5).
Added: For (5) let $D=\{x+iy:y>0\}\cup\{x:x\ge 0\}$, and consider the function $$f:D\to\Bbb C:z\mapsto z^2\;.$$ Pictorially speaking, $D$ is everything above the real axis in the complex plane together with all of the non-negative real numbers. In terms of the exponential representation of complex numbers, $$D=\left\{re^{i\theta}:r\ge 0\text{ and }0\le\theta<\pi\right\}\;.$$
Suppose that $se^{i\varphi}$ is any complex number. Show that you can always choose $s$ and $\varphi$ so that $s\ge0$ and $0\le\varphi<2\pi$, just as you can when you choose the polar coordinates of a point in the plane. Then find an $re^{i\theta}\in D$ such that $\left(re^{i\theta}\right)^2=se^{i\varphi}$; this will show that $f$ is surjective.
To decide whether $f$ is injective, suppose that $re^{i\theta},se^{i\varphi}\in D$ and $\left(re^{i\theta}\right)^2=\left(se^{i\varphi}\right)^2$, i.e., that $r^2e^{2i\theta}=s^2e^{2i\varphi}$. Knowing that $r,s\ge 0$ and $0\le\theta,\varphi<\pi$, can you show from this that $r=s$ and $\theta=\varphi$? If so, you’ll have proved that $f$ is injective.
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Hi Brian, i really appreciate your help, what i can look now is that the third one is neither injetive nor surjective, the fourth is not injective but indeed is surjective, the last one i really dont have any idea. so am i ok? can you help me with the last one? – Sebastian Griotberg Aug 17 '12 at 15:20
@Sebastian: You’re right about (2) and (4). Give me a few minutes to add something about (5) to my answer. – Brian M. Scott Aug 17 '12 at 15:25
Thanks a lot. i really appreciate your help. – Sebastian Griotberg Aug 17 '12 at 15:28 | 1,449 | 4,578 | {"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} | 3.6875 | 4 | CC-MAIN-2016-30 | latest | en | 0.829919 |
http://slideplayer.com/slide/4971536/ | 1,580,179,678,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00495.warc.gz | 148,594,364 | 22,748 | # 1 Polynomial Church-Turing thesis A decision problem can be solved in polynomial time by using a reasonable sequential model of computation if and only.
## Presentation on theme: "1 Polynomial Church-Turing thesis A decision problem can be solved in polynomial time by using a reasonable sequential model of computation if and only."— Presentation transcript:
1 Polynomial Church-Turing thesis A decision problem can be solved in polynomial time by using a reasonable sequential model of computation if and only if it can be solved in polynomial time by a Turing machine. P is the class of these decision problems
2 Search Problems: NP L is in NP iff there is a language L’ in P and a polynomial p so that:
3 Intuition The y-strings are the possible solutions to the instance x. We require that solutions are not too long and that it can be checked efficiently if a given y is indeed a solution or witness (we have a “simple” search problem)
4 Reductions A reduction r of L 1 to L 2 is a polynomial time computable map so that 8 x: x 2 L 1 iff r(x) 2 L 2 We write L 1 · L 2 if L 1 reduces to L 2. Intuition: Efficient software for L 2 can also be used to efficiently solve L 1.
5 NP-hardness A language L is called NP-hard iff 8 L’ 2 NP: L’ · L Intuition: Software for L is strong enough to be used to solve any simple search problem. Proposition: If some NP-hard language is in P, then P=NP.
6 NPC A language L 2 NP that is NP-hard is called NP-complete. NPC := the class of NP-complete problems. Proposition: L 2 NPC ) [L 2 P iff P=NP].
7 L is in NP means: There is a language L’ in P and a polynomial p so that L 1 · L 2 means: For some polynomial time computable map r : 8 x: x 2 L 1 iff r(x) 2 L 2 L is NP-hard means: 8 L’ 2 NP: L’ · L L is in NPC means: L 2 NP and L is NP-hard
8 How to establish NP-hardness Thousands of natural problems are NP-complete: Empirical fact: Most natural problems in NP are either in P or NP-hard. Lemma: If L 1 is NP-hard and L 1 · L 2 then L 2 is NP-hard. We need to establish one problem to be NP-hard, the rest follows using chains of reductions. Cook (1972) established SAT to be NP-hard.
9 SAT ILP MILP MAX INDEPENDENT SET MIN VERTEX COLORING HAMILTONIAN CYCLE TSP TRIPARTITE MATCHING SET COVER KNAPSACK BINPACKING
10 Boolean functions f: {false, true} n ! {false,true} Example: XOR(x 1, x 2 ) = x 1 © x 2 XOR(true, true) = false In this course 0=false, 1=true. XOR(1,1)=0
11 How to represent Boolean functions on a computer Tables. Formulae.
12 Tables More compact representation as Boolean String: “0110”. A function f: {0,1} n ! {0,1} can be represented as a table using … ……..2^n bits
13 Boolean Formulae X 1, X 2, …, X n are formulae. If f is a formula then : f is a formula. If f 1 and f 2 are formulae then (f 1 ) Æ (f 2 ) and (f 1 ) Ç (f 2 ) are formulae. Sometimes we leave out parentheses….
14 Formulae represent Boolean functions “(x 1 Æ : x 2 ) Ç (x 2 Æ : x 1 )” represents the function XOR. Sometimes formula-representation is much more compact than table representation. Sometimes not. It’s never much less compact.
15 Two special classes of Boolean formulae f(x 1, x 2 ) = x 1 © x 2 DNF: f(x 1, x 2 ) = (x 1 Æ : x 2 ) Ç (x 2 Æ : x 1 ) CNF: f(x 1, x 2 ) = ( : x 1 Ç : x 2 ) Æ (x 1 Ç x 2 ) A CNF is any conjunction of clauses (disjunctions of literals). A DNF is any disjunction of terms (conjunctions of literals). Any function on n variables can be described by a CNF (DNF) formula containing at most 2 n clauses (terms), each containing at most n literals.
16 SAT SAT: Given a Boolean function in CNF representation, is there a way to assign truth values to the variables so that the function evaluates to true? SAT: Given a CNF, is it true that it does not represent the constant-0 function? Input: ( : x 1 Ç : x 2 ) Æ (x 1 Ç x 2 ) Output: Yes. Input: ( : x 1 Ç : x 2 ) Æ (x 1 Ç x 2 ) Æ (x 1 Ç : x 2 ) Æ ( : x 1 Ç x 2 ) Output: No.
17 SAT SAT is in NP. Cook’s theorem (1972): SAT is NP-hard. Hence, SAT is NP-complete. It is a universal search problem and we do not think it has a polynomial time algorithm. If we find one, we get \$1.000.000.
18 DNF? Suppose we ask the same question of DNF formulas, rather than CNF formulas. Can you get a million dollars for solving this problem?
19 ⋀ Boolean Circuits ⋁ ⌐ X1X1 X2X2 X3X3 ⋀
20 Circuits A circuit C is a directed acyclic graph. Nodes in C are called gates. Types of gates: –Variable (or input) gates of indegree 0, labeled X 1, X 2, …X n –Constant gates of indegree 0, labeled 0,1 –AND-, OR-gates of indegree 2, –NOT-, COPY-gates of indegree 1. –m distinguished gates are output gates. The circuit C computes a Boolean function C:{0,1} n ! {0,1} m
21 ⋀ A formula can be viewed as a special kind of circuit ⋁ ⌐ X1X1 X2X2 X3X3 ⋀ f(X 1, X 2, X 3 ) = : X 1 Ç (X 1 Æ X 2 Æ X 3 )
22 Are formulas and circuits in fact the same thing? Not quite! Given a circuit, we can write down an equivalent formula, but in may become much bigger. A circuit is allowed to reuse the result of a sub-computation without doing the computation again!
23 Multi-output circuits Any function f: {0,1} n ! {0,1} m is represented by some circuit. Proof: Any function f: {0,1} n ! {0,1} can be described by a Boolean formula. A formula can also be viewed as a circuit. Combine m such circuits.
24 Tables – Formulae - Circuits For a function f: {0,1} n ! {0,1}, let s t = 2 n be the size (in bits) of its representation as a table. Let s f be the size (in bits) of its smallest representation as a formula. Then, – |s f | · 10 |s t | 2 (formulae are not much less compact than tables). –For all n, there is a function so that |s f | · 10 (log |s t |) 2 (for some functions, formulae are much more compact) Let s c be the size (in bits) of its smallest representation as a circuit. Then, –|s c | · 10 |s f | 2 (circuits are not much less compact than formulae). –Is it true than there for all n is a function so that circuits are much more compact (e.g., |s c | · 10 (log |s f |) 2 )? This is open! You won’t get \$1.000.000 for solving this, but you’ll still get pretty famous ….
25 Circuits vs. Turing Machines Circuits can be given as inputs to algorithms but they can also be seen as computational devices themselves! Like Turing Machines, circuits C: {0,1} n ! {0,1} solve decision problems on {0,1} n. Unlike Turing machines, circuits takes inputs of a fixed input length n only.
26 Theorem Given Turing Machine M running in time at most p(n) on inputs of length n, where p is a polynomial. For every n, there is a circuit C n with at most O(p(n) 2 ) gates so that 8 x 2 {0,1} n : C n (x)=1 iff M accepts x. The map 1 n ! C n is polynomial time computable.
27 Intuition behind proof
28 Problem: Cycles! Flip-Flop, stores one bit.
29 The Tableau Method Time 0 Time 1 Time t … 12 s Can be replaced by acyclic Boolean circuit of size ≈ s
30 CIRCUIT SAT CIRCUIT SAT: Given a Boolean circuit, is there a way to assign truth values to the input gates variables, so that the output gate evaluates to true? Generalizes SAT, as CNFs are formulas and formulas are circuits.
31 Example ⋀ ⋁ ⌐ X1X1 X2X2 X3X3 ⋀ Input: Output: Yes
32 Example ⋀ ⌐ X1X1 X2X2 X3X3 ⋀ Input: Output: No ⋀
33 CIRCUIT SAT CIRCUIT SAT is in NP. Proof of Cook’s theorem: – CIRCUIT SAT reduces to SAT. – CIRCUIT SAT is NP-hard. – Hence, SAT is NP-hard.
34 CIRCUIT SATSAT
35 CIRCUIT SAT is NP-hard Given an arbitrary language L in NP we must show that L reduces to CIRCUIT SAT. This means: We must construct a polynomial time computable map r mapping instances of L to circuits, so that 8 x: x 2 L, r(x) 2 CIRCUIT SAT The only thing we know about L is that there is a language L’ in P and a polynomial p, so that:
36 Theorem Given Turing Machine M running in time at most p(n) on inputs of length n, where p is a polynomial. For every n, there is a circuit C n of size at most O(p(n) 2 ) so that 8 x 2 {0,1} n : C n (x)=1 iff M accepts x. The map 1 n ! C n is polynomial time computable.
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Similar presentations | 2,409 | 8,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2020-05 | latest | en | 0.902598 |
https://physics.stackexchange.com/questions/545215/question-about-a-2d-harmonic-oscillator-with-incommensurate-frequencies-and-inte | 1,719,116,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862430.93/warc/CC-MAIN-20240623033236-20240623063236-00083.warc.gz | 386,804,488 | 40,607 | # Question about a 2D Harmonic Oscillator with incommensurate frequencies and Integrability
In Classical Dynamics by José & Saletan [section 4.2.2] they give the example of a 2D Harmonic Oscillator whose equations of motion are
$$$$\ddot{x}_i+\omega_i^2x_i=0 \ \ \ \ \ \text{for} \ \ \ \ i=1,2\tag{3.38}$$$$
This system has two obvious conserved quantities
$$$$E_i = \frac{1}{2}\dot{x}_i^2+\frac{1}{2}\omega_i^2x^2 \tag{3.39}$$$$ which are just the energies of each independent oscillator. The motion is obviously integrable and everything works out. However, in their explanation on section 4.2.2 they use this example to show that if the two frequencies are incommensurate
$$$$\frac{\omega_1} {\omega_2 } \notin \mathbb{Q}$$$$
then the motion is not periodic since the trajectory $$(x_1(t),x_2(t))$$ will never return to its initial position again. Because of this, solutions densely populate the phase space of the system and any conserved quantity defined as
$$$$\Gamma (x_1,x_2,\dot{x}_1,\dot{x}_2)=C$$$$
will be pathological discontinous. This is because for any initial condition $$\chi_0=(x_1,x_2,\dot{x}_1,\dot{x}_2)$$ there's another point arbitrarily close that belongs to a trajectory with an arbitrary different value of $$\Gamma$$. I think I understand the explanation. However, he claims that when we have this pathological we can't define conserved quantities other than $$E_1$$ and $$E_2$$. This, to me, sounds like it implies the system is not integrable, due to a lack of constants of motion. But I already know the system is fully integrable given it's just two copies of an harmonic oscillator. So my main questions are:
1. Why are they saying that we can't define conserved quantities other than $$E_1$$ and $$E_2$$? What's special about those? They are also constants of motion defined as functions of $$x_i$$ and $$\dot{x}_i$$.
2. What is the relation between incommensurate frequencies, the lack of conserved quantities and integrability?
• You have a system with 4 degrees of freedom and a set of 2 independent first integrals. That's all you need to conclude that your system is integrable. Why would you want more first integrals? Commented Apr 20, 2020 at 13:23
• I know that two independent first integrals are enough. What I don't understand is then why is he giving this example to show systems where there's a lack of constants of motion. Commented Apr 20, 2020 at 17:05
1. OP has already noted that the 2D harmonic oscillator is completely Liouville-integrable with 2 globally defined, Poisson-commuting, real integrals of motion $$H_1$$ and $$H_2$$.
2. Since the phase space has 4 real dimensions, there can at most be 3 independent real integrals of motion, and 4 independent real constants of motion. By definition an integral of motion cannot depend explicitly on time $$t$$ while a constant of motion can, cf. e.g. this Phys.SE post.
3. We can rewrite the 2D harmonic oscillator \begin{align}H~:=~&H_1+H_2, \cr H_j~:=~&\frac{p_j^2}{2}+\frac{\omega_j^2q_j^2}{2}~=~\omega_jz_j^{\ast}z_j,\qquad j~\in~\{1,2\},\end{align}\tag{A} in complex notation \begin{align}z_j~:=~&\sqrt{\frac{\omega_j}{2}}q_j + \frac{ip_j}{\sqrt{2\omega_j}}, \cr \{z^{\ast}_j, z_k\}_{PB}~=~&i\delta_{j,k},\qquad j,k~\in~\{1,2\}.\end{align}\tag{B} For technical reasons we exclude the singular zero-leaf, i.e. the phase-space becomes $$M=(\mathbb{C}^{\times})^2$$, where $$\mathbb{C}^{\times}:=\mathbb{C}\backslash\{0\}$$. The phase-space $$M$$ has 2 complex dimensions. We can easily find 2 independent, globally defined, complex constants of motion $$F_j~:=~z_je^{i\omega_j t}, \qquad j~\in~\{1,2\},\tag{C}$$ which is the maximal number. The two Hamiltonians $$H_j=\omega_j|F_j|^2$$ depend on their absolute values.
4. On one hand, if $$\frac{\omega_1}{\omega_2}~=~\frac{n_1}{n_2}~\in~\mathbb{Q}\tag{D}$$ are commensurate frequencies, then we can construct a globally defined, complex integral of motion $$\frac{z_1^{n_2}}{z_2^{n_1}}.\tag{E}$$ Its argument is independent of $$H_1$$ and $$H_2$$, which shows that the system is maximally superintegrable.
5. On the other hand, if the frequencies are incommensurate, then we can only define a 3rd independent integral of motion $${\rm Im}\left(\frac{{\rm Ln}(z_1)}{\omega_1}-\frac{{\rm Ln}(z_2)}{\omega_2}\right)\tag{F}$$ locally, because of the branch-cut of the complex logarithm $${\rm Ln}$$.
References:
1. J.V. Jose & E.J. Saletan, Classical Dynamics: A Contemporary Approach, 1998; Subsection 4.2.2 p. 183-185. | 1,361 | 4,491 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 29, "wp-katex-eq": 0, "align": 2, "equation": 4, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-26 | latest | en | 0.916885 |
https://www.onlinemathlearning.com/subtraction-word-problems.html | 1,718,929,812,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862032.71/warc/CC-MAIN-20240620235751-20240621025751-00579.warc.gz | 832,680,062 | 10,247 | # Subtraction Word Problems (1-step word problems)
These lessons look at some examples of subtraction word problems that can be solved in one step, illustrating the use of bar models or block diagrams in the solution process.
We will illustrate how block diagrams can be used to help you to visualize the subtraction word problems in terms of the information given and the data that needs to be found. The block diagrams or block modeling method is used in Singapore Math.
Example:
Jessica has 1135 beads. 604 beads are red and the rest are blue. How many blue beads does she have?
Solution:
1135 – 604 = 531
Example:
James and Ken donated \$2300 to a charitable organization. Ken donated \$658. How much did James donate?
Solution:
2300 – 658 = 1642
James donated \$1642.
Example:
The price of a car is \$2795 and the price of a motorbike is \$1063. What is the difference between the prices of the 2 vehicles?
Solution:
2795 – 1063 = 1732
The difference between the prices of the 2 vehicles is \$1732.
Example:
There are 967 chairs in a hall. During an event, 761 chairs were occupied. How many chairs were not occupied?
Solution:
967 – 761 = 206
206 chairs were not occupied.
Examples of subtraction word problems
Examples:
1. 134 girls and 119 boys took part in an art competition. How many more girls than boys were there?
2. Mei Lin saved \$184. She saved \$63 more than Betty. How much did Betty saved?
3. John read 32 pages in the morning. He read 14 pages less in the afternoon.
a) How many pages did he read in the afternoon?
b) How many pages did they read altogether?
A visual way to solve world problems using bar modeling
This type of word problem uses the part-whole model.
Example:
Mr. Oliver bought 88 pencils. he sold 26 of them. How many pencils did he have left?
A visual way to solve world problems using bar modeling
This type of word problem uses the part-whole model. Because the part is missing, this is a subtraction problem.
Example:
There are 98 hats, 20 of them are pink and the rest are yellow. How many yellow hats are there?
Example:
Cayla did 88 sit-ups in the morning. Nekira did 32 sit-ups at night. How many more sit-ups did Cayla do than Nekira?
How to use bar modeling in Singapore math to solve word problems that deal with comparing?
Example:
Adam has 11 fewer lollipops than Hope. If Adam has 16 lollipops, how many lollipop does Hope have?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 613 | 2,608 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.75 | 5 | CC-MAIN-2024-26 | latest | en | 0.951224 |
https://encyclopediaofmath.org/index.php?title=Elementary_theory&oldid=46803 | 1,652,682,443,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00574.warc.gz | 305,721,333 | 8,341 | # Elementary theory
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A collection of closed formulas of first-order predicate logic. The elementary theory $\mathop{\rm Th} ( K)$ of a class $K$ of algebraic systems (cf. Algebraic system) of signature $\Omega$ is defined to be the collection of all closed formulas of the first-order predicate logic of signature $\Omega$ that are true in all systems of $K$. If $K$ consists of a single system $A$, then the elementary theory of the class $K$ is the elementary theory of the system $A$. Two algebraic systems of the same signature are said to be elementarily equivalent if their elementary theories are the same. An algebraic system $A$ of signature $\Omega$ is called a model of an elementary theory $T$ of signature $\Omega$ if all formulas of $T$ are true in $A$. An elementary theory is called consistent if it has models. A consistent elementary theory is called complete if any two models of it are elementarily equivalent. The class of all models of an elementary theory $T$ is denoted by $\mathop{\rm Mod} ( T)$. An elementary theory $T$ is called solvable (or decidable) if the set of formulas $\textrm{ Th Mod } ( T)$( that is, the set of all logical consequences of $T$) is recursive. A class $K$ of algebraic systems of signature $\Omega$ is called axiomatizable if there exists an elementary theory $T$ of signature $\Omega$ such that $K = \mathop{\rm Mod} ( T)$. In this case $T$ is called a collection of axioms for $K$. A class $K$ is axiomatizable if and only if $K = \textrm{ Mod Th } ( K)$. For example, the class of dense linear orders without a smallest or largest element is axiomatizable, its elementary theory is solvable and any two systems of this class are elementarily equivalent, since the elementary theory of this class is complete; moreover, its elementary theory is finitely axiomatizable. The class of finite cyclic groups is not axiomatizable; however, its elementary theory is solvable, and hence recursively axiomatizable. There are examples of finitely-axiomatizable unsolvable elementary theories. They include those of groups, rings, fields, and others. However, a complete recursively-axiomatizable elementary theory is necessarily solvable. Therefore, to prove the solvability of a recursively-axiomatizable elementary theory it is sufficient to observe that it is complete.
Several methods for proving completeness are known. An elementary theory is called categorical in cardinality $\alpha$( cf. Categoricity in cardinality) if all its models of cardinality $\alpha$ are isomorphic. An elementary theory that is categorical in some infinite cardinality and has no finite models is necessarily complete. For example, the elementary theory of algebraically closed fields of a given characteristic is recursively axiomatizable and categorical in every uncountable cardinality; it has no finite models, and therefore it is complete and solvable. In particular, the elementary theory of the field of complex numbers is solvable. Two formulas in the same signature as that of a theory $T$ are equivalent in the theory $T$ if they contain the same variables and if, for any model $A$ of $T$ and any assignment of elements of $A$ to their free variables, the formulas are either both true or both false. A complete elementary theory $T$ of finite or countable signature is countably categorical if and only if for every $n$ there are finitely many formulas with $n$ free variables $v _ {1} \dots v _ {n}$ such that every formula of the appropriate signature with $v _ {1} \dots v _ {n}$ as free variables is equivalent in $T$ to one of those formulas. A complete theory of finite or countable signature that is categorical in one uncountable cardinality is also categorical in every other uncountable cardinality. A system $A$ of signature $\Omega$ is called an elementary subsystem of a system $B$ of the same signature if $A$ is a subsystem of $B$ and if for every formula $\Phi ( v _ {1} \dots v _ {n} )$ of the first-order predicate logic of $\Omega$ with free variables $v _ {1} \dots v _ {n}$ and all $a _ {1} \dots a _ {n} \in A$, the truth of $\Phi ( a _ {1} \dots a _ {n} )$ in $A$ implies its truth in $B$. An elementary theory $T$ is called model complete if for any two models $A$ and $B$ of it the fact that $A$ is a subsystem of $B$ implies that it is an elementary subsystem. It turns out that a model-complete theory having a model that can be isomorphically imbedded in every model of the theory is complete. Two systems of the same signature which satisfy the same prenex formulas without existential quantifiers are called universally equivalent. A model-complete elementary theory all models of which are universally equivalent is complete. Using the technique of model completeness one can prove that real-closed fields, in particular, the field of real numbers, have a complete and solvable elementary theory. Among the other solvable elementary theories are those of addition of natural numbers and integers, of Abelian groups, of $p$- adic number fields, of finite fields, of residue class fields, of ordered Abelian groups, and of Boolean algebras.
The general study of unsolvable elementary theories was initiated by A. Tarski in the 1940s, but even earlier, in 1936, A. Church had proved the unsolvability of first-order predicate logic and J. Rosser, also in 1936, had proved the unsolvability of the arithmetic of the natural numbers. The elementary theory $\mathop{\rm Th} ( K)$ of a class $K$ of algebraic systems of the same signature $\Omega$ is said to be inseparable if there is no recursive set of formulas containing $\mathop{\rm Th} ( K)$ and not containing any closed formula which is false in all systems in $K$. The elementary theory of a class $K _ {1}$ of systems of signature $\langle P ^ {(} 2) \rangle$ consisting of a single two-place predicate is called relatively definable in the elementary theory of a class $K _ {2}$ of systems of signature $\Omega _ {2}$ if there exist formulas $\Phi ( v _ {0} ; u _ {1} \dots u _ {s} )$ and $\Psi ( v _ {1} , v _ {2} ; u _ {1} \dots u _ {s} )$ of $\Omega _ {2}$ such that for every system $A _ {1}$ in $K _ {1}$ one can find a system $A _ {2}$ in $K _ {2}$ and elements $b _ {1} \dots b _ {s}$ in $A _ {2}$ for which the set $X = \{ {x \in A _ {2} } : {\Phi ( x ; b _ {1} \dots b _ {s} ) \textrm{ is true in } A _ {2} } \}$ together with the predicate $P ^ {(} 2)$, defined on $X$ so that $P ^ {(} 2) ( x , y )$ is true if and only if $\Psi ( x , y ; b _ {1} \dots b _ {s} )$ is true in $A _ {2}$, forms an algebraic system isomorphic to $A _ {1}$. This definition extends naturally to the theory of classes $K _ {1}$ of arbitrary signature. If the elementary theory of a class $K _ {1}$ is inseparable and relatively definable in the elementary theory of a class $K _ {2}$, then that of $K _ {2}$ is also inseparable. This makes it possible to prove that the elementary theories of many classes of algebraic systems are inseparable. Here it is convenient to take as $\mathop{\rm Th} ( K _ {1} )$ the elementary theory of all finite binary relations or that of all finite symmetric relations, or similar elementary theories. Inseparable elementary theories are unsolvable. So are those of the field of rational numbers and of many classes of rings and fields. The unsolvability of the elementary theory of finite groups is an important result of A.I. Mal'tsev.
#### References
[1] Yu.L. Ershov, I.A. Lavrov, A.D. Taimanov, M.A. Taitslin, "Elementary theories" Uspekhi Mat. Nauk , 20 : 4 (1965) pp. 37–108 (In Russian) [2] Yu.L. Ershov, "Decision problems and constructivizable models" , Moscow (1980) (In Russian)
Somewhat more generally one defines recursive separability and inseparability for pairs of theories rather than for a single theory. Thus, two disjoint sets of natural numbers $A$ and $B$ are said to be recursively separable if there exists a recursive set $A _ {1}$ containing $A$ which is disjoint from $B$. This is a symmetric notion. The sets $A$ and $B$ are (recursively) inseparable if they are not recursively separable. The single theory definition of inseparability results if this is applied to the sets $R$ and $T$ of refutable and provable formulas, or rather their associated sets of Gödel numbers $R ^ { \star }$ and $T ^ { \star }$. | 2,106 | 8,369 | {"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} | 3.21875 | 3 | CC-MAIN-2022-21 | latest | en | 0.908952 |
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