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http://mathhelpforum.com/advanced-statistics/93497-test-statistic.html | 1,480,802,157,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541140.30/warc/CC-MAIN-20161202170901-00409-ip-10-31-129-80.ec2.internal.warc.gz | 171,355,307 | 10,409 | 1. ## test statistic?
X and Y stand for the size of the male and femal of a specific species of moth. Assume the distributions of X and Y are $N(\mu_{X},\sigma^{2}_{X})$ and $N(\mu_{Y},\sigma^{2}_{Y})$. $\sigma^{2}_{Y}>\sigma^{2}_{X}$. Use the modification of Z to test the hypothesis $H_{0}$: $\mu_{X}-\mu_{Y}=0$ against the alternative hypothesis $H_{1}$: $\mu_{X}-\mu_{Y}<0$.
How do you define the test statistic and a critical region that has a significance level of $\alpha=0.025$?
thank you very much
2. 1 Do you know the population variances?
If so the rejection region is $(-\infty ,-1.96)$
and the test statistic is ${\bar X-\bar Y\over \sqrt{ {\sigma^{2}_{Y}\over n_Y}+{\sigma^{2}_{X}\over n_X}}}$.
2 If you don't know $\sigma^{2}_{Y}$ and $\sigma^{2}_{X}$, then it's a t test.
BUT now I need to know if you're allowed to assume that $\sigma^{2}_{Y}=\sigma^{2}_{X}$ or not.
3 HOWEVER, if the sample sizes are large then you can approximate this via the Central Limit Theorem and use 1.
3. That is all the info I have (and part of why I am so stuck) | 336 | 1,066 | {"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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2016-50 | longest | en | 0.803377 |
http://forum.enjoysudoku.com/twin-sudoku-help-please-t31304.html | 1,534,344,654,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210133.37/warc/CC-MAIN-20180815141842-20180815161842-00660.warc.gz | 148,456,716 | 6,409 | Post the puzzle or solving technique that's causing you trouble and someone will help
I have encountered some "Twin Sudoku" puzzles, but I'm not sure what is meant by the instructions, particularly the last part.
"Twin Sudoku" consists of a pair of standard sudoku puzzles each with some starting digits. To get a complete solution for the twin puzzles, it is necessary to solve each twin puzzle, but the complete solution of the twin puzzles will be obtained by substituting the corresponding values of the digits from one twin sudoku into the other.
What exactly do they mean by "substituting the corresponding values of the digits"? What/where are the "corresponding values"?
Here are the two sets of puzzles:
#53
Code: Select all
`.............6......3.8.5.....3.2....28.9.47....8.5.....4.3.2......2..............6.4.2.3.7.......6..1.6.5..5.......9..6.3.7..2.......1..7.1.8..9.......3.5.7.3.1.`
#54
Code: Select all
`...573....3.....4....1.4...5.4.1.2.98..957..63.6.8.5.7...7.9....8.....9....845.............4..5..9...7.1.5............31.2.94............3.4.6...7..6..5..........`
enxio27
Posts: 447
Joined: 13 November 2007
### Re: Twin Sudoku? Help, please!
There must be a one-to-one correspondence between the digits of both boards, i.e., in all positions with a certain digit in the first grid, the corresponding positions in the second grid have to contain the same digit as each other (possibly a different digit from the digit in the first grid).
JasonLion
2017 Supporter
Posts: 639
Joined: 25 October 2007
Location: Silver Spring, MD, USA
### Re: Twin Sudoku? Help, please!
JasonLion wrote:There must be a one-to-one correspondence between the digits of both boards, i.e., in all positions with a certain digit in the first grid, the corresponding positions in the second grid have to contain the same digit as each other (possibly a different digit from the digit in the first grid).
So, are you saying that, for example, in the first grid of puzzle #53, R5C5 is 9 and in the second grid, R5C5 is 3, so that in every cell in the first grid where there is a 9, the corresponding cell in the second grid will be a 3? But that a 9 in the second grid will not NECESSARILY be a 3 in the first grid (in this case, a 3 in the first grid would be a 1 in the second grid, as in R3C3)?
enxio27
Posts: 447
Joined: 13 November 2007
### Re: Twin Sudoku? Help, please!
Yep.
JasonLion
2017 Supporter
Posts: 639
Joined: 25 October 2007
Location: Silver Spring, MD, USA
### Re: Twin Sudoku? Help, please!
Ahhh!! Now it makes sense! Why could they not explain all that in the first place?
Thanks for the help!
enxio27
Posts: 447
Joined: 13 November 2007
### re: "Twin Equivalent Sudokus"
the simple version
allows only re-mapping of digits
( as explained by JasonLion )
a trickier version
also allows re-mapping of lines within a chute
"Twin Equivalent Sudokus"
Pat
Posts: 3586
Joined: 18 July 2005
### Re: re: "Twin Equivalent Sudokus"
Pat wrote:
I remember that thread and discussion
tarek
Posts: 2675
Joined: 05 January 2006
### Re: re: "Twin Equivalent Sudokus"
tarek wrote:
I remember that thread and discussion | 892 | 3,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2018-34 | longest | en | 0.86478 |
https://www.convertunits.com/from/funt/to/teragram | 1,656,574,110,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103669266.42/warc/CC-MAIN-20220630062154-20220630092154-00029.warc.gz | 758,757,460 | 12,794 | ## ››Convert funt [Russia] to teragram
funt teragram
How many funt in 1 teragram? The answer is 2441899995.502.
We assume you are converting between funt [Russia] and teragram.
You can view more details on each measurement unit:
funt or teragram
The SI base unit for mass is the kilogram.
1 kilogram is equal to 2.441899995502 funt, or 1.0E-9 teragram.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between funte and teragrams.
Type in your own numbers in the form to convert the units!
## ››Want other units?
You can do the reverse unit conversion from teragram to funt, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Funt
The Russian pound (Фунт, funt) is an obsolete Russian unit of measurement of mass. In 1899 the Russian pound was the basic unit of weight and all other units of weight were formed from it.
## ››Definition: Teragram
The SI prefix "tera" represents a factor of 1012, or in exponential notation, 1E12.
So 1 teragram = 1012 grams-force.
## ››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! | 401 | 1,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2022-27 | latest | en | 0.868887 |
http://math.stackexchange.com/questions/145875/fundamental-group-of-mathbbr3-setminus-mbox2-linked-circles | 1,469,598,950,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257825366.39/warc/CC-MAIN-20160723071025-00285-ip-10-185-27-174.ec2.internal.warc.gz | 154,824,066 | 16,944 | # Fundamental group of $\mathbb{R}^{3}\setminus \{ \mbox{2 linked circles }\}$
Calculate the fundamental group of the complement in $\mathbb{R}^3$ of
$$\{ (x,y,z) \ | \ y = 0 , \ x^{2} + z^{2} = 1\} \cup \{ (x,y,z) \ | \ z = 0 , \ (x-1)^{2} + y^{2} = 1\}.$$
Note: this space is $\mathbb{R}^{3}\setminus \{ \mbox{2 linked circles }\}$.
-
I see no question above, only an order. You should never copy/paste a question from a textbook, or phrase your question as such. Moreover, you should explain what you have tried, and where you are stuck. – M Turgeon May 16 '12 at 14:50
@MTurgeon While I agree with your sentiments, it does seem like your tone is no different than OP's. We should lead by example. – Sasha May 16 '12 at 14:52
@Sasha Ok. I see no question above, only an order. I encourage you to never copy/paste a question from a textbook, or phrase your question as such. Moreover, you should try to explain what you have tried, and where you are stuck. – M Turgeon May 16 '12 at 14:55
One possible approach: $S^3\setminus \{\text{2 linked circles}\}$ is homotopy equivalent to a torus. Note that adding a point at $\infty$ to turn $\mathbb R^3$ into $S^3$ does not change $\pi_1$. – Grumpy Parsnip May 16 '12 at 14:56
As @JimConant says, you can add a point to make this $S^3-K$, where $K$ is your two linked circles. This does not change the homotopy type. Choosing another point of $S^3$ to be infinity won't change it either, so you can remove a point from $K$ to get a new (homeomorphic) space, $\mathbb{R}^3-L$, where $L$ is the union of a circle in the xy-plane, and the z-axis. Now the deformation retraction to the torus is obvious. – user641 May 16 '12 at 20:17 | 534 | 1,680 | {"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} | 3.5 | 4 | CC-MAIN-2016-30 | latest | en | 0.869403 |
http://forum.enjoysudoku.com/help-please-t34272.html | 1,539,852,961,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511761.78/warc/CC-MAIN-20181018084742-20181018110242-00434.warc.gz | 136,078,255 | 6,845 | ## Help please
Post the puzzle or solving technique that's causing you trouble and someone will help
### Help please
Hi
Does anyone understand this?
Attachments
sudoku mystery.JPG (84.84 KiB) Viewed 229 times
jakesprake
Posts: 2
Joined: 22 August 2015
### Re: Help please
Code: Select all
` 6 4 1258 | 258 9 158 | 12 7 3 7 3 128 | 268 4 168 | 126 9 5 125 25 9 | 2567 13 37 | 4 1268 18 ------------------------+----------------------+--------------------- 145 578 156 | 3 58 2 | 15679 1568 1489 9 58 135 | 46 7 46 | 135 158 2 245 2578 2356 | 1 58 9 | 3567 568 48 ------------------------+----------------------+--------------------- 25 6 4 | 579 13 37 | 8 125 19 3 1 7 | 589 2 58 | 59 4 6 8 9 25 | 45 6 145 | 125 3 7 `
Above is state after naked singles, hidden singles, box-line moves and naked/hiden sets, ie the basic set of moves. After that there are choices: a number of short chains or a single chain of length 6:
(1)r3c1 = (1-4)r4c1 = r4c9 - (4=8)r6c9 - (8=1)r3c9 - r3c1 => -5 r3c58, r4c1
However, for manual solving I rather prefer the Swordfish of 1s at r1c367, r2c367, r9c67 eliminating 1 from r45c37. The puzzle solves with singles after that.
Phil
pjb
2014 Supporter
Posts: 2025
Joined: 11 September 2011
Location: Sydney, Australia
### Re: Help please
The puzzle solved cell status in line format : 64..9..7373..4..95..9...4.....3.2...9...7...2...1.9....64...8..317.2..4689..6..37
The following diagram shows the Swordfish that solves the puzzle :
Code: Select all
`*--------------------------------------------------*| 6 4 *1258 | 258 9 *158 |*12 7 3 || 7 3 *128 | 268 4 *168 |*126 9 5 || 125 25 9 | 2567 13 37 | 4 1268 18 ||----------------+--------------+------------------|| 145 578 56-1 | 3 58 2 | 5679-1 1568 1489 || 9 58 35-1 | 46 7 46 | 35-1 158 2 || 245 2578 2356 | 1 58 9 | 3567 568 48 ||----------------+--------------+------------------|| 25 6 4 | 579 13 37 | 8 125 19 || 3 1 7 | 589 2 58 | 59 4 6 || 8 9 25 | 45 6 *145 |*125 3 7 |*--------------------------------------------------*`
Leren
<edit>
Now that I've read your original question I can go back a bit. I can't understand what the solver is saying but there is an easy way to eliminate the 1 from r7c6, and a lot more besides.
Code: Select all
`*---------------------------------------------------*| 6 4 1258 | 258 9 158 | 12 7 3 || 7 3 128 | 268 4 168 | 126 9 5 || 125 25 9 | 2567 13 *37-156 | 4 1268 18 ||---------------+-----------------+-----------------|| 145 578 156 | 3 58 2 | 15679 1568 1489 || 9 58 135 | 46 7 46 | 135 158 2 || 245 2578 2356 | 1 58 9 | 3567 568 48 ||---------------+-----------------+-----------------|| 25 6 4 | 579 13 *37-15 | 8 125 19 || 3 1 7 | 589 2 58 | 59 4 6 || 8 9 25 | 45 6 145 | 125 3 7 |*---------------------------------------------------*`
There is a hidden pair (37) in r47c6, so you can eliminate all other digits from those two cells. That's a pretty basic move, so I'm not quite sure why the solver did not suggest it. This move exposes the Swordfish.
Leren
Leren
Posts: 3312
Joined: 03 June 2012
### Re: Help please
Thanks Laren
jakesprake
Posts: 2
Joined: 22 August 2015
### Re: Help please
I don't quite understand the solver's logic, but coloring in 1 gives it's result, and is a 1-step solution.
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 6 4 1258 | 258 9 158 | 12 7 3 || 7 3 128 | 268 4 168 | 126 9 5 || 25-1 258 9 | 25678 A1358 135678 | 4 B1268 B18 |+-------------------------+-------------------------+-------------------------+| 145 578 1568 | 3 58 2 | 15679 1568 1489 || 9 58 13568 | 4568 7 4568 | 1356 1568 2 || 245 2578 23568 | 1 58 9 | 3567 568 48 |+-------------------------+-------------------------+-------------------------+| 25 6 4 | 579 135 1357 | 8 125 19 || 3 1 7 | 589 2 58 | 59 4 6 || 8 9 25 | 45 6 A145 |B125 3 7 |+-------------------------+-------------------------+-------------------------+`
1 is in r9c6 or r9c7.
If in r9c6 then also in r3c5 (box2).
If in r9c7, then also in r3c89 (box3).
So it must be in one of r3c589 and cannot be in r3c1 (or r3c6).
Thus 1 must be in r4c1 and you have singles to the end.
eleven
Posts: 1862
Joined: 10 February 2008
Return to Help with puzzles and solving techniques | 1,999 | 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} | 3.0625 | 3 | CC-MAIN-2018-43 | longest | en | 0.158572 |
https://www.roseindia.net/sql/mysql-example/mysql-random-number.shtml | 1,725,722,196,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00848.warc.gz | 938,686,520 | 12,718 | MySQL random number
You may fall in the condition when you want to create random number for your program to process some logic, here is the solution for such condition.
MySQL random number
You may fall in the condition when you want to create random number for your program to process some logic, here is the solution for such condition. You can create a random number by the use of RAND() function of MySQL. There are two versions of RAND() function : first RAND() and second is RAND(X). Random numbers are created to produce unpredictable floating-point value between 0 and 1 (including both 0 and 1). Random number generated by the use of RAND() function is not a perfect random number but it is very close to the perfect random numbers.
Here in this example we will describe, how you can create random number in MySQL. Numbers generated are in the random order which means that they lack any pattern.
```mysql> set @Number = RAND(); Query OK, 0 rows affected (0.00 sec)``` ```mysql> SELECT @Number as Number; +------------------+ | Number | +------------------+ | 0.44307944592375 | +------------------+ 1 row in set (0.00 sec) ``` ```mysql> set @Number = RAND(); Query OK, 0 rows affected (0.00 sec) mysql> SELECT @Number as Number; +------------------+ | Number | +------------------+ | 0.63046264893418 | +------------------+ 1 row in set (0.00 sec) ```
We have created a variable Number in which generated random number is being taken and thereafter we can show the result of it.
Now we are going to explain you use of RAND(X) with the simple example.
```mysql> set @Number = RAND(1); Query OK, 0 rows affected (0.00 sec)``` ```mysql> SELECT @Number as Number; +------------------+ | Number | +------------------+ | 0.40540353712198 | +------------------+ 1 row in set (0.00 sec)``` ```mysql> set @Number = RAND(1); Query OK, 0 rows affected (0.00 sec)``` ```mysql> SELECT @Number as Number; +------------------+ | Number | +------------------+ | 0.40540353712198 | +------------------+ 1 row in set (0.00 sec)```
You can see that it generates the same number since we have provided seed value in the RAND() function. | 515 | 2,136 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2024-38 | latest | en | 0.824684 |
https://talk.collegeconfidential.com/t/unsubsidized-loan-please-help/1198817 | 1,642,826,455,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320303747.41/warc/CC-MAIN-20220122043216-20220122073216-00595.warc.gz | 614,981,678 | 7,255 | <p>I understand how a subsidized loan works so don't write about that.</p>
<p>An unsubsidized loan starts to accrue interest the moment the money is disbursed and i'm responsible for the interest amount as soon as it's given to me.</p>
<p>So UF is offering my \$1,655 in unsubsidized loan at a 6.8% fixed interest rate. And I'm going to have this unsubsized loan in my file for ~3 years or so. HOW DOES THE INTEREST WORK ON THIS \$1,655. I want to defer any interest payments until after I graduate.</p>
<p>Is it every month that 6.8% is added to the 1,655 or does the interest work seperate if i defer interest payments until after I graduate. </p>
<p>What i was thinking was that every month 6.8% is added to my amount borrowed. Or is 6.8% just added to my principal balance sense day one.. Paying for the interest why in school is not an option. </p>
<p>It is 6.8% a year, not a month.</p>
<p>When you defer the interest it is capitalized. This means it is added to the principal and you pay interest on the interest.</p>
<p>great response thank you ^^. So how do you pay on interest on interest.</p>
<p>lets say \$1,655 x 6.8% = ~\$112 so that's the interest in one year so then you would multiply that by 6.8 = 112 x 6.8% = x amount?</p>
<p>Then if this is the case unsubsidized loan doesn't sound that bad. :) I could deal with 112 dollars a year or so.</p>
<p>Um I don't no if I am correct, but based on my knowledge and swimcatsmom's post. It would be something as such:</p>
<p>\$1,655 (Principal) x 6.8% (Interest Rate) = \$112.54 (Interest for a year)</p>
<p>Next, you have to add on the interest to your principal as it has accrued as such:</p>
<p>\$1,655 (Principal) + \$112.54 (Interest for a year) = \$1767.54 (Your new balance)
This above is one year. The following would be a second year.</p>
<p>\$1767.54 (Your new balance) x 6.8% (Interest Rate) = \$120.19 (Interest for the year)</p>
<p>\$1767.54 (Your old balance) + \$120.19 (Interest for the year) = \$1887.73 (Your new balance)</p>
<p>and you would basically repeat what I have done to get your final balance for the third year. </p>
<p>This is what I thought of when I read your question (although I maybe incorrect).</p>
<p>-Joe</p>
<p>thank you Joe. Sounds like it's right lol. Hope to hear from someone else to clarify or confirm.</p>
<p>Thanks,</p>
<p>Anyone else?</p>
<p>If you want to figure out how the debt will grow over time there is a mathematical formula.</p>
<p>If the interest was only calculated and capitalized annually then you would take the principal and multiply it by 1.068^n (.068 being the interest rate and n being the number of years before you start repaying the debt.) </p>
<p>So if it would be 3 years till you start paying the debt the debt would have grown 1655(1.068)^3 = \$2016 by the time you start paying it off. In reality the interest is probably calculated and added on at least a monthly basis so the amount will be a little higher (around \$2028 after 36 months)</p> | 820 | 3,003 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-05 | latest | en | 0.956669 |
http://csirnetphysics.blogspot.com/2010/12/hamiltonian-formulism-of-mechanics-part.html | 1,539,838,540,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511703.70/warc/CC-MAIN-20181018042951-20181018064451-00276.warc.gz | 88,119,294 | 16,250 | ## Thursday, December 16, 2010
### Hamiltonian formulism of mechanics (part 2)
-Following four forms of generator are possible : F1(q,Q,t) , F2(q,P,t) , F3(p,Q,t) , F4(p,P,t)
Legendre transforms are
g(u,y)=f(x,y)-ux
where u=(∂ F/∂ x)
(i) F1=F1(q,Q,t) putting it in equation 1 and finding derivative of F w.r.t. we get
pj = ∂F1(q,Q,t)/∂qj
Pj = - ∂F1(q,Q,t)/∂Qj
K=H+∂F1(q,Q,t)/∂t
(ii) putting u = -Pj , x = Qj , y = qj , g = F2(q,P,t) , f=F1 and using Legender transforms we get
F2(qj,Pj,t) = F1(qj,Qj,t)+ΣPjQj
Putting this in 1 and solving we get
pj = ∂F2(qj,Pj,t)/∂qj
Qj = ∂F2(qj,Pj,t)/∂Pj
K=H+∂F2(qj,Pj,t)/∂t
(iii) F3 = F3(Q,p,t) is the third form
Connect it to the first form using legender transforms
Since pj = ∂F1(qj,Pj,t)/∂qj and u=∂f/∂x
this implies that u=pj , x=qj , y=Qj , g=F3 , f=F1
Thus, F3 = F1(Q,q,t)-Σpjqj
or, F1(Q,q,t)=F3+Σpjqj
putting these in equation 1
qj = -∂F3(pj,Qj,t)/∂pj
Pj = -∂F3(pj,Qj,t)/∂Qj
K=H+∂F3(pj,Qj,t)/∂t
(iv) Fourth form is F4(p,P,t) , connecting F4 by F1 through legender transforms and solving we get
qj = -∂F4(pj,Pj,t)/∂pj
Qj = ∂F4(pj,Pj,t)/∂Pj
K=H+∂F4(pj,Pj,t)/∂t | 546 | 1,117 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-43 | latest | en | 0.60409 |
https://www.codespeedy.com/predicting-insurance-using-scikit-learn-in-python/ | 1,675,195,397,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499890.39/warc/CC-MAIN-20230131190543-20230131220543-00362.warc.gz | 741,798,463 | 16,634 | # Predicting insurance using Scikit-Learn in Python
Today we’ll be predicting the insurance using Scikit-Learn and Pandas in Python. We will use the Linear Regression algorithm to predict insurance. The insurance money is calculated from a Medical Cost Dataset which has various features to work with. Predicting insurance using Scikit-Learn and Pandas in Python requires a combination of Data Analytics and Machine Learning.
## Importing the .csv file using Pandas
First, download the dataset from this link. Then import the Pandas library and convert the .csv file to the Pandas dataframe. You can take any dataset of your choice. Preview your dataframe using the head() method.
```import pandas as pd
Output:
``` age sex bmi children smoker region charges
0 19 female 27.900 0 yes southwest 16884.92400
1 18 male 33.770 1 no southeast 1725.55230
2 28 male 33.000 3 no southeast 4449.46200
3 33 male 22.705 0 no northwest 21984.47061
4 32 male 28.880 0 no northwest 3866.85520```
## Predict the charge for insurance using sklearn in Python
We will store the features we are using for prediction ie. age, BMI in the X variable. And, the target value to be predicted ie. the charges in the y variable. We are only taking two features for this tutorial, you can take as many as you want. The .values() function is to convert the resulting dataframe t0 a numpy array.
```X=df[['age','bmi']].values
y=df['charges'].values
print(X)
print(y)
```
Output:
```[[19. 27.9 ]
[18. 33.77]
[28. 33. ]
...
[18. 36.85]
[21. 25.8 ]
[61. 29.07]]
[16884.924 1725.5523 4449.462 ... 1629.8335 2007.945 29141.3603]```
The next step is to import the LinearRegression package of the sklearn library to fit our regression model. Firstly, we create the regression model ‘regsr’. Then we train the model using the fit() method. We pass our features and target to our model.
```from sklearn.linear_model import LinearRegression
regsr=LinearRegression()
regsr.fit(X,y)```
Output:
`LinearRegrsesion(copy_X=True, fit_intercept=True, n_jobs=None, normalize=False)`
## Predicting our insurance
Now that we have trained our model, we can start predicting values. For example, we want to determine the insurance cost for a person of age 20 and a Body-Mass Index of 30 ie. [20,30]. We will then convert the list to a numpy array and reshape the array. This array is then passed to the predict() method.
```import numpy as np
prediction=regsr.predict(np.asarray([20,30]).reshape(-1,2))
print(prediction)```
Output:
```[8402.76367021]
```
Thus, the insurance money for this person is \$8402.76.
You can also try using other algorithms like the KNN Classification algorithm and see which one works best.
Also, check out:
### One response to “Predicting insurance using Scikit-Learn in Python”
1. Shubham Mishra says:
Dear Sir,
I’m unable to use proper libary for the Programe “SBI Life Insurance” in Jupyter Notebook,I’m Getting error,
I don’t know which libary used to load the sbi life insurance Datasets ,the algorithm is used Logistic Regression but when i’m doing..
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt | 866 | 3,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-06 | latest | en | 0.766388 |
https://web2.0calc.com/questions/algebra_21437 | 1,719,167,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862488.55/warc/CC-MAIN-20240623162925-20240623192925-00217.warc.gz | 536,603,860 | 5,440 | +0
# Algebra
0
2
1
+620
If \$s\$ is a real number, then what is the smallest possible value of \$2s^2 - 8s + 19 - 7s^2 - 8s + 20\$?
May 24, 2024
#1
+129507
+1
Simplify as
-5s^2 -16s + 39
This is a parabola that opens downward.....there is no minimum value
May 24, 2024 | 121 | 281 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2024-26 | latest | en | 0.771738 |
http://discrete.prof.ninja/501/ | 1,571,709,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987795403.76/warc/CC-MAIN-20191022004128-20191022031628-00319.warc.gz | 53,750,889 | 2,496 | # May 1st
Eulerian Circuits
## Vocab
• a WALK on a graph is a sequence of vertex, edge, vertex, edge, ... , vertex where each edge joins the vertices surrounding it.
• LENGTH OF A WALK is the number of edges.
• a CLOSED WALK has first and last vertex as the same (OPEN WALK when they are different)
• a PATH is a walk in which every vertex is distinct
• TRAIL is a walk that never repeats an edge (can repeat vertices not edges)
• CIRCUIT is a closed trail (can repeat vertices but not edges)
• CYCLE is a circuit with only the first and last vertex equal (no repeated vertices or edges other than start and stop point).
• EULERIAN CIRCUIT is a circuit that contains every vertex and every edge.
• a graph is CONNECTED when there is a walk between any two vertices
• EULERIAN TRAIL is a trail that visits every edge and every vertex (but start and stop at different points)
• HAMILTONIAN CYCLE is a cycle that hits every vertex
• a graph that contains a hamiltonian cycle is called HAMILTONIAN, a graph that contains an eulerian circuit is called EULERIAN
• for fun think about a RANDOM WALK
## Useful facts
• A graph is Eulerian if and only if it is connected and every vertex is even. (I would like to prove this to you.)
• A graph has an Eulerian trail (between two distinct vertices $$u$$ and $$v$$) if and only if it is connected and all vertices except $$u$$ and $$v$$ are even.
• Any cycle inside of a graph has no sub-cycles
• Any cycle inside of a graph uses exactly two edges from each of the vertices of that cycle.
• We all owe our lives to the brave intergalactic warriors that fight on our behalf, let us honor them.
## The problems
1. Find a connected graph with as few vertices as possible that has precisely two vertices of even degree.
2. Prove that a graph is bipartite if and only if it contains no odd cycles
3. Take two Eulerian graphs that share no vertices and connect them with a single edge. Is the resulting graph Eulerian? Does the resulting graph contain an Eulerian trail?
4. For any distinct vertices, $$u$$ and $$v$$ in a graph, prove that there is a walk between $$u$$ and $$v$$ if and only if there is a path from $$u$$ to $$v$$
5. Is the following graph Eulerian? If so find an Eulerian circuit.
6. Here is a five room house, try to draw any continuous line that crosses every wall exactly one time. What can you prove and how? Here is a sample line:
"WallsLines2". Licensed under CC BY-SA 3.0 via Wikipedia.
7. (Next daily challenge) Research question: prove that the Petersen graph is not Hamiltonian | 626 | 2,546 | {"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} | 4.125 | 4 | CC-MAIN-2019-43 | latest | en | 0.944007 |
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Topic: Help : What's the real difference between the function CPUTIME and TIC,TOC???
Replies: 0
alex Posts: 6 Registered: 12/7/04
Help : What's the real difference between the function CPUTIME and TIC,TOC???
Posted: Sep 23, 1996 4:15 PM
Hi,
I need to know what's the real difference between the function
CPUTIME and the combination between TIC,TOC?
When i run a function, it is always almost the same time
that is printed
or what MAthworks mean exactly by CPUTIME, the help given is
not very clear (that of the manual too). PArticularly the
it is not said that there's a difference between the two
function..
First i thought that we can compute with CPUTIME the effective
time needed by the CPU..
In other words, let us suppose that we have to compute
the elapsed time needed for <a_function>
> tic, <a_function>,t1 = toc
> t0=cputime;<a_function>, t2 = cputime-t;Are t1 and t2 ALWAYS equal??????
Any help will be appreciated,
S. Guetari | 311 | 1,152 | {"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} | 2.640625 | 3 | CC-MAIN-2018-17 | longest | en | 0.915596 |
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Showing results 1 to 15 of 15 Search took 0.00 seconds. Search: Posts Made By: shoff
Forum: Questions and Comments 06-23-2015, 08:25 AM Replies: 4 Views: 11,106 Posted By shoff ^^ Good advice. ^^ Good advice.
Forum: Questions and Comments 06-22-2015, 08:33 PM Replies: 0 Views: 6,206 Posted By shoff Getting loopy? I still don't really understand how to use these loops. Are there any good tutorials out there that explain the process of applying statements/functions to them? I want to say "Loop the whole program...
Forum: The Lounge 06-22-2015, 08:30 PM Replies: 1 Views: 3,646 Posted By shoff Yay Congrats! Yay Congrats!
Forum: The Lounge 06-22-2015, 08:30 PM Replies: 1 Views: 3,611 Posted By shoff I have a hard time getting my loops to do... I have a hard time getting my loops to do anything specific. What statements are you using?
Forum: The Lounge 06-19-2015, 10:34 AM Replies: 2 Views: 4,448 Posted By shoff It helps if you create a "hop" method with the... It helps if you create a "hop" method with the up/down movement housed in it. Then in your formula, the hop method is attributed to the random distance. So when a random distance is selected, it runs...
Forum: Questions and Comments 06-19-2015, 10:28 AM Replies: 0 Views: 6,206 Posted By shoff Making decisions For this assignment we did If/Else statements and I really enjoyed creating these statements in conjunction with variables. I could see how you could create a myriad of decisions to allow for future...
Forum: Questions and Comments 06-16-2015, 10:59 AM Replies: 4 Views: 8,882 Posted By shoff You will get it! It might help us to print out... You will get it! It might help us to print out the instructions and highlight the important points, that way we don't miss anything and have to back track. Other than that, it's just creating those...
Forum: Questions and Comments 06-15-2015, 05:07 PM Replies: 2 Views: 7,765 Posted By shoff Why do the lights go out when my snake talks? So I completed my fourth assignment. Methods are now getting much easier and I am figuring out how useful they are now that I am putting scenes together. I think I have got a walking method down pat,...
Forum: Questions and Comments 06-15-2015, 05:03 PM Replies: 2 Views: 8,973 Posted By shoff I don't know, that would be useful, especially... I don't know, that would be useful, especially for bipeds.
Forum: Questions and Comments 06-14-2015, 11:54 AM Replies: 0 Views: 5,524 Posted By shoff Walking mummy I worked with creating new methods in Alice, particularly for walking. So for my walk I had to use together leg movements in order. I used the "turn at speed" instead of "move at speed" because the...
Forum: The Lounge 06-11-2015, 09:50 PM Replies: 0 Views: 3,689 Posted By shoff Second assignment, the penguin and the ice. I started out wanting to do the chicken assignment, but getting the legs to move in a walking motion was too complicated to start out with. Luckily the penguin already has a "waddle" type animation...
Forum: The Lounge 06-11-2015, 09:44 PM Replies: 2 Views: 4,881 Posted By shoff Yeah, working with shapes was tricky. I couldn't... Yeah, working with shapes was tricky. I couldn't get my circle to show up unless I placed it first, then had to make sure nothing was covering it up. Very frustrating at first.
Forum: The Lounge 06-11-2015, 09:46 AM Replies: 2 Views: 3,736 Posted By shoff Sounds like you had a good first experience,... Sounds like you had a good first experience, can't wait to see more!
Forum: The Lounge 06-10-2015, 08:30 AM Replies: 2 Views: 3,953 Posted By shoff Getting started with Alice As part of my programming logic and design class, I had my first foray into Alice yesterday. Got a little tour around the program and did my first assignment moving a rabbit across the screen. It...
Forum: The Lounge 06-10-2015, 08:26 AM Replies: 2 Views: 3,645 Posted By shoff You are an overachiever I see! Congrats on your... You are an overachiever I see! Congrats on your first assignment. I will have to check out Visual Studio more in depth, it looks really cool. Thanks!
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Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Announcements Community News and Announcements Alice 3 How Do I? Works In Progress Share Your World Share Custom Classes Bugs and Trouble Shooting Suggestion Box Alice 2 How do I...? Works-In-Progress Share Worlds Share Objects Bugs and Troubleshooting Suggestion Box Educators Teaching with Alice Teaching Using Alice 3 Teaching Using Alice 2 CS Principles and Alice AP CSA and Intro to Java and Alice Workshops Alice Player and VR General Discussion Questions and Comments The Lounge | 1,242 | 4,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-33 | latest | en | 0.907595 |
https://example-a.com/answer/25460561 | 1,708,607,712,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473819.62/warc/CC-MAIN-20240222125841-20240222155841-00884.warc.gz | 248,948,037 | 8,337 | algorithmsetpartitioning
# Algorithm to produce all partitions of a list in order
I've need for a particular form of 'set' partitioning that is escaping me, as it's not quite partitioning. Or rather, it's the subset of all partitions for a particular list that maintain the original order.
I have a list of n elements `[a,b,c,...,n]` in a particular order.
I need to get all discrete variations of partitioning that maintains the order.
So, for four elements, the result will be:
``````[{a,b,c,d}]
[{a,b,c},{d}]
[{a,b},{c,d}]
[{a,b},{c},{d}]
[{a},{b,c,d}]
[{a},{b,c},{d}]
[{a},{b},{c,d}]
[{a},{b},{c},{d}]
``````
I need this for producing all possible groupings of tokens in a list that must maintain their order, for use in a broader pattern matching algorithm.
I've found only one other question that relates to this particular issue here, but it's for ruby. As I don't know the language, it looks like someone put code in a blender, and don't particularly feel like learning a language just for the sake of deciphering an algorithm, I feel I'm out of options.
I've tried to work it out mathematically so many times in so many ways it's getting painful. I thought I was getting closer by producing a list of partitions and iterating over it in different ways, but each number of elements required a different 'pattern' for iteration, and I had to tweak them in by hand.
I have no way of knowing just how many elements there could be, and I don't want to put an artificial cap on my processing to limit it just to the sizes I've tweaked together.
Solution
• You can think of the problem as follows: each of the partitions you want are characterized by a integer between 0 and 2^(n-1). Each 1 in the binary representation of such a number corresponds to a "partition break" between two consecutive numbers, e.g.
`````` a b|c|d e|f
0 1 1 0 1
``````
so the number `01101` corresponds to the partition `{a,b},{c},{d,e},{f}`. To generate the partition from a known parition number, loop through the list and slice off a new subset whenever the corresponding bit it set.
I can understand your pain reading the fashionable functional-programming-flavored Ruby example. Here's a complete example in Python if that helps.
``````array = ['a', 'b', 'c', 'd', 'e']
n = len(array)
for partition_index in range(2 ** (n-1)):
# current partition, e.g., [['a', 'b'], ['c', 'd', 'e']]
partition = []
# used to accumulate the subsets, e.g., ['a', 'b']
subset = []
for position in range(n):
subset.append(array[position])
# check whether to "break off" a new subset
if 1 << position & partition_index or position == n-1:
partition.append(subset)
subset = []
print partition
`````` | 678 | 2,684 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-10 | latest | en | 0.944027 |
https://docs.astropy.org/en/latest/api/astropy.stats.kuiper.html | 1,716,710,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058872.68/warc/CC-MAIN-20240526074314-20240526104314-00239.warc.gz | 177,915,900 | 7,836 | # kuiper#
astropy.stats.kuiper(data, cdf=<function <lambda>>, args=())[source]#
Compute the Kuiper statistic.
Use the Kuiper statistic version of the Kolmogorov-Smirnov test to find the probability that a sample like `data` was drawn from the distribution whose CDF is given as `cdf`.
Warning
This will not work correctly for distributions that are actually discrete (Poisson, for example).
Parameters:
dataarray_like
The data values.
cdf`callable()`
A callable to evaluate the CDF of the distribution being tested against. Will be called with a vector of all values at once. The default is a uniform distribution.
argslist-like, optional
Additional arguments to be supplied to cdf.
Returns:
D`float`
The raw statistic.
fpp`float`
The probability of a D this large arising with a sample drawn from the distribution whose CDF is cdf.
Notes
The Kuiper statistic resembles the Kolmogorov-Smirnov test in that it is nonparametric and invariant under reparameterizations of the data. The Kuiper statistic, in addition, is equally sensitive throughout the domain, and it is also invariant under cyclic permutations (making it particularly appropriate for analyzing circular data).
Returns (D, fpp), where D is the Kuiper D number and fpp is the probability that a value as large as D would occur if data was drawn from cdf.
Warning
The fpp is calculated only approximately, and it can be as much as 1.5 times the true value.
Stephens 1970 claims this is more effective than the KS at detecting changes in the variance of a distribution; the KS is (he claims) more sensitive at detecting changes in the mean.
If cdf was obtained from data by fitting, then fpp is not correct and it will be necessary to do Monte Carlo simulations to interpret D. D should normally be independent of the shape of CDF.
References
[1]
Stephens, M. A., “Use of the Kolmogorov-Smirnov, Cramer-Von Mises and Related Statistics Without Extensive Tables”, Journal of the Royal Statistical Society. Series B (Methodological), Vol. 32, No. 1. (1970), pp. 115-122. | 482 | 2,054 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-22 | latest | en | 0.875666 |
https://gogeometry.blogspot.com/2016/07/geometry-problem-1240-triangle-circle.html?showComment=1470070390684 | 1,670,419,126,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711162.52/warc/CC-MAIN-20221207121241-20221207151241-00758.warc.gz | 284,244,783 | 13,924 | ## Saturday, July 30, 2016
### Geometry Problem 1240: Triangle, Circle, Diameter, Circumcircle, Circumcenter, Perpendicular. Mobile Apps
Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to view more details of problem 1240.
1. If < DAC = A then < DEB = A and < DOB = 2A
Now DOB being isoceles < DBO = 90-A
But < DCA = 90-A
It follows that DBOC is concyclic and < CDA being 90, < BOC is also = 90
Sumith Peiris
Moratuwa
Sri Lanka
1. Correction to last line.......
If BO meets AC at X, it follows that DBXC is concyclic and < CDA being 90, < BXC is also = 90
2. https://goo.gl/photos/LyXa4bZgSQcpQGuR9
Draw diameter BF of circle O , lines AE and CD
Since AD⊥DC= > DC will cut circle O at F
Since AE⊥CE => AE will cut circle O at F
In triangle ABF , C is the orthocenter so AC⊥BO
3. let angle CAB = A, ABC = B and ACB = C
=> angle DEC = DEB = DAC = A => triangles ABC & EBD are similar
Extend BO to meet the circle at F and join FE
=> <FED = 90+A
Since FEDB is cyclic quadrilateral <DBF = <DBO = <ABO = 90-A
So if a triangle is formed by extending AC to intersect BO at a point P - it is right angle at P as <PAB = A & PBA = 90-A
4. Let (Q) be the circle with diameter AC, and Bt the tangent to circle (O) on B.
Circles (Q) and (O) intersect at D and E.
Line (ADB) cuts (Q) on A and (O) on B.
Line (ECB) cuts (Q) on C and (O) on B. Applying Reim’s theorem:
AC//Bt and since Bt┴BO BO┴AC. | 501 | 1,507 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2022-49 | latest | en | 0.892096 |
https://il.tradingview.com/script/Yp0mBs5b-Elder-Ray-Bear-Power-Strategy/ | 1,601,507,723,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402128649.98/warc/CC-MAIN-20200930204041-20200930234041-00393.warc.gz | 389,198,865 | 61,161 | # Elder Ray (Bear Power) Strategy
צפיות 1817
Developed by Dr Alexander Elder, the Elder-ray indicator measures buying
and selling pressure in the market. The Elder-ray is often used as part
of the Triple Screen trading system but may also be used on its own.
Dr Elder uses a 13-day exponential moving average ( EMA ) to indicate the
market consensus of value. Bear Power measures the ability of sellers to
drive prices below the consensus of value. Bear Power reflects the ability
of sellers to drive prices below the average consensus of value.
Bull Power is calculated by subtracting the 13-day EMA from the day's High.
Bear power subtracts the 13-day EMA from the day's Low.
You can use in the xPrice any series: Open, High, Low, Close, HL2 , HLC3, OHLC4 and ect...
```////////////////////////////////////////////////////////////
// Copyright by HPotter v1.0 23/07/2014
// Developed by Dr Alexander Elder, the Elder-ray indicator measures buying
// and selling pressure in the market. The Elder-ray is often used as part
// of the Triple Screen trading system but may also be used on its own.
// Dr Elder uses a 13-day exponential moving average (EMA) to indicate the
// market consensus of value. Bear Power measures the ability of sellers to
// drive prices below the consensus of value. Bear Power reflects the ability
// of sellers to drive prices below the average consensus of value.
// Bull Power is calculated by subtracting the 13-day EMA from the day's High.
// Bear power subtracts the 13-day EMA from the day's Low.
// You can use in the xPrice any series: Open, High, Low, Close, HL2, HLC3, OHLC4 and ect...
////////////////////////////////////////////////////////////
study(title="Elder Ray (Bear Power) Strategy")
Length = input(13, minval=1)
Trigger = input(0)
hline(0, color=purple, linestyle=line)
xPrice = close
xMA = ema(xPrice,Length)
DayLow = iff(dayofmonth != dayofmonth[1], low, min(low, nz(DayLow[1])))
nRes = DayLow - xMA
pos = iff(nRes > Trigger, 1,
iff(nRes < Trigger, -1, nz(pos[1], 0)))
barcolor(pos == -1 ? red: pos == 1 ? green : blue )
plot(nRes, color=blue, title="Bear Power", style = histogram)
``` | 549 | 2,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-40 | latest | en | 0.817724 |
https://physics.stackexchange.com/questions/379104/how-can-density-functional-theory-dft-be-understood-in-many-body-perturbation | 1,571,343,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986676227.57/warc/CC-MAIN-20191017200101-20191017223601-00295.warc.gz | 639,646,593 | 32,532 | # How can density functional theory (DFT) be understood in many body perturbation theory (MBPT) language?
Many body interacting fermions problems are formulated in the many body perturbation theory language using Feynman diagrams and imaginary time formalism. To the best of my knowledge the kinetic energy should be the energy of the non interacting system and any correction coming from the interaction would come as self energy corrections. In practical we rarely take the dispersion of the free electrons., it is in general a better to take the DFT dispersion. What is very confusing is that DFT already takes some of the interaction into account, so if i am to use the DFT dispersion in a many body perturbative theory, what should i take for the interaction ?
I understand how can DFT get the dispersion, and i understand how the dispersion is modified by the self energy in MBPT. What i don't understand, is what does it mean to mix both approaches.
• I assume with dispersion you really mean London Dispersion not the whole van der Waals interactions. And when you say "I understand how can DFT get the dispersion" what method are you talking about there are different ways of getting dispersion interactions into account in DFT. – physicopath Jan 10 '18 at 19:15
• by dispersion i mean the band structure calculated from DFT, the one you obtain solving KS equations. – lakehal Jan 11 '18 at 16:46
\begin{align} G^{-1} &= G_0^{-1} - \Sigma \\ &= G_{DFT}^{-1} + V_{xc} - \Sigma \end{align}
Up to now, there is no advantage of using $G_{DFT}$, because you essentially go back to $G_0$. The advantage appears when you recall that the self-energy itself depends on the Green's function.
If you use some approximation based on skeleton diagrams, you need to perform self-consistent calculations, starting from a self-energy obtained from some initial Green's function. In this case, a good initial guess for $G$ (e. g., $G_{DFT}$) would speed up convergence. An example how this works within the $GW$ approximation can be found in [B. Holm and U. von Barth, "Fully self-consistent $GW$ self-energy of the electron gas"].
If you are not going to do self-consistent calculations of some skeleton diagrams and just want to construct a perturbation theory around $G_{DFT}$, you may simply substitute $G_0 = G_{DFT} + V_{xc}$ in the usual series in $G_0$ and $V$ — then both $V$ and $V_{xc}$ will be perturbations (though it does not look particularly useful).
• I am not very familiar with self-consistent calculations of the green's function, and don't know what are exactly skeleton diagrams. So if you have a good read on that, it would be very useful. I was thinking ( or hoping ) that one can replace $G_{0}$ by $G_{DFT}$ and renormalize the interaction and use MBPT. If i substitute $G_0=G_{DFT}+V_{xc}$ then is $G_0$ the free-electron propagator ? – lakehal Mar 2 '18 at 11:26 | 706 | 2,886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2019-43 | latest | en | 0.911237 |
https://www.joannenova.com.au/2011/05/so-what-is-the-second-darn-law/print/ | 1,656,805,874,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00412.warc.gz | 876,198,848 | 9,530 | - JoNova - https://www.joannenova.com.au -
# So what is the Second Darn Law?
With nearly 500 comments on the thread on the Second Law of Thermodynamics there is obviously a need for people to discuss the basic greenhouse theory. Here’s a new thread on that theme.
From NASA:
### But there are variations
As with all these Laws of science there is no exact wording, because There Is No God Who Issues Science Decrees*. What we have are human efforts to best explain the world around us. Note that the two well known versions of the Second Law both contain the phrase “whose sole result”, meaning that heat transfer can certainly move from a colder to a warmer body if there is some other compensating movement where more heat is transferred from a hotter body to a colder one. Voila… whatever heat transfer goes from greenhouse gases to the Earth is more than countered by the heat moving from the Sun to Earth and on to space. Greenhouse gases can heat the Earth as long as the entropy of the whole system increases.
Clausius statement:
No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.
Kelvin Statement
No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.
Behind the scenes I’m still getting many emails from people wondering about this topic. I’m sorry I can’t reply to them all. Joseph Postma kindly replied to my last post on this topic in comment #97, and in reply to him I say this (his comments are in blue):
“The 2nd Law of Thermodynamics applies to net flows of heat, not to each individual photon, and it does not prevent some heat flowing from a cooler body to a warm one.”
–JoNova
JP: The laws of heat transfer DO forbid a cold body from raising the temperature of a warmer body. This is a much more physically unambiguous clarification. “Flows of heat” is ambiguous and confusing. It is obviously correct to say that a cold body does not raise the temperature of a warmer body. “Flows of heat” is ambiguous and not important, in terms of what has the ability to raise who’s temperature.
The “ability to raise temperature” is not proscribed by the 2nd law, merely by your interpretation of it. The second law does not mention “abilities” of anything.
“Net heat flows” is not ambiguous. It is the Net heat flow. Are more photons going in one direction? Yes, but the ones coming back the other way affect the NET heat flow.
JP: It is obviously correct to say that a cold body does not raise the temperature of a warmer body.
No it isn’t. It’s obviously correct to say that the heat stored in a cold object can’t raise the temp of a warmer object, but the heat is not coming from a cold object. The heat is coming from the sun. The cold body just stops it leaving the warmer object at the rate it was previously.
Why do you keep ignoring the sun? There is a continuous input of energy into the Earth?
JP: But there IS a restriction on which body can raise the temperature of its neighbours. Only the hotter body can raise the temperature of a colder neighbour.
This is preposterous. God doesn’t make rules like “photons can travel in any direction except the way it came”. Atoms radiate photons randomly and some return in the same direction.
“Technically, strictly, greenhouse gases don’t “warm” the planet (as in, they don’t supply additional heat energy), but they slow the cooling, which for all pragmatic purposes leaves the planet warmer that it would have been without them.” — Jo Nova
JP: This statement is fraught with contradiction. Is it technical or is it real? I mean this is very important! If GHG’s, as you correctly point out cannot raise the planets temperature because they are not actually sources of additional energy, then how do we arrange some logic in which they do exactly what we just said they cannot do?
The technical part is the language use, the reality part is the temperature.
The difference between whether we ascribe the raised temperature to the GHG gas or to the SUN is “technical” the reality of the warming is real. Think of the blanket analogy. Technically a blanket on your body does not heat you (if you want to be a pedant) your own body heats you. In reality, you put on a blanket and your temperature goes up.
What if we put it this way: Greenhouse gases don’t warm^ the planet but they do create a situation where the Sun can warm the world to a temperature above where it previously was without the gases?
That doesn’t break the second law, and it describes reality with a technical accuracy.
^(I don’t like this phrasing at all: the word “warm” here specifically means something other than it’s usual use — here it implies no additional heat energy is supplied that is new to the Earth-atmosphere system. Normally “to warm” means… well you know… . Life doesn’t have to be this complicated.)
In the end this boils down to the energy flows in a whole system, and people who think that the addition of greenhouse gases to our atmosphere won’t warm the planet (due to a conflict with the Second Law) are forgetting that the system includes the sun, and the extra energy flowing in continuously is what drives the system. CO2 may have little effect on Planet Earth, but that’s for other reasons. You can’t just make a comment about “the Earth-Atmosphere” part of the system and ignore the plasma ball that’s 1.4 million kilometers across and 13 million degrees in the middle.
————————————————————
### UPDATE #1 From Michael Hammer
Michael Hammer has suggested this as a line of reasoning that may help people discuss this. If you don’t agree with the end conclusion (that greenhouse gases can’t warm earth because they are not hotter than Earth), point out exactly which step in the sequence is the one you think is wrong and explain it so we can understand why.
1. Do you agree that if you stand surrounded by cold objects (say a ring of huge ice blocks) you feel cold?
2. Do you agree that the colder the object you are surrounded by the colder you feel? eg: if they were blocks of frozen CO2 (dry ice) instead of water ice would you feel colder?
3. OK now if you have been standing surrounded by extremely cold objects and then move so that you are now surrounded by merely cool objects does the move make you feel less cold than you were before?
4. In the absence of green house gases the earth is surrounded by an extremely cold object – outer space at 4K (-269C). Green house gases make the atmosphere opaque at some wavelengths. With these in place the earths surface is in effect surrounded by a merely cool atmosphere instead of the truly frigid outer space. Because the surface is now surrounded by a less cold object than it was before it is less cold.
5. Since warm and cold are opposites, less cold is the equivalent of warmer. Surrounding earth’s surface with an opaque cool atmosphere make it warmer than it would be if exposed directly to the ultra frigid outer space.
————————————————————-
*Which doesn’t mean there is No God at all. I’m staying right out of that one.
————————————————————-
### UPDATE#2 Joseph Postma in Comment #47
Thanks for posting my comments, Jo.
Firstly, I would like to re-iterate that in my paper, “Understanding the Thermodynamic Atmosphere Effect”, I in fact refer to the Laws of Heat Transfer (i.e. thermodynamics) in total, not just the 2nd Law. You may have an earlier pre-edit copy which had the reference to the 2nd Law specifically.
Secondly, the 2nd Law isn’t even the most important one when it comes to understanding how the atmosphere actually works. Conservation of energy is actually much more important, in terms of the physics. Your diagram from NASA shows just how obscure the 2nd Law can be.
Lastly, the brief mention of the Laws of heat transfer, around page 6-7 of “Understanding…”, is NOT central in any way to the paper! There are MUCH more important issues discussed within.
“heat transfer can certainly move from a colder to a warmer body if there is some other compensating movement where more heat is transferred from a hotter body to a colder one.”
This was never disputed…it is unclear why there is so much focus on this issue. What is undisputable is that more heat transfers from the warm body to the cold body and so the warm body has to cool down. That is, a cold body does not raise the temperature of a warmer body.
JN: Exactly – the greenhouse gases don’t warm the sun and the sun gets cooler (albeit slowly).
There is too much ambiguity and invitation to equivocation when using terms like “heat transfer” when what is more important is how the temperature is actually controlled. “Heat” does not equal “higher temperature”. 1 Kelvin has “heat”.
“Voila… whatever heat transfer goes from greenhouse gases to the Earth is more than countered by the heat moving from the Sun to Earth and on to space.”
And so by extension the GHG’s cannot increase the temperature of the ground. Because the net heat transfer is from the ground (warmest), outward (coolest).
“Greenhouse gases can heat the Earth as long as the entropy of the whole system increases.”
Not without an additional source of energy, could they do this*. GHG’s do not represent an additional source of energy. Re-emitted, or “back-IR” radiation from GHG’s, do not represent additional energy – this is rather the RESULT of the original Solar energy insolated into the system. The Laws of Thermodynamics, in total, tell you that this energy can’t be replicated for further heating. Never mind the 2nd Law all by itself. The IR and back-IR, etc, is a RESULT of what has already happened…they’re not the cause.
JN: *Never once have I said that GHGs are a source of new energy. Have I not explained at length and repeatedly that the heat comes from the sun, and the GHG’s slow the heat loss from Earth? Remember if the Earth didn’t radiate the heat away it’s final equilibrium temperature would be much higher… what would 4 billion years of input do with no heat loss?
“Why do you keep ignoring the sun? There is a continuous input of energy into the Earth?”
This is unfortunate. If my paper would actually be read it can be learned that I spend a GREAT deal of time discussing the Sun and its continuous input into the Earth system. When understood properly, as I qualitatively and partially quantitatively explained, you discover there is no need to postulate a radiative greenhouse theory in the first place because 1) there’s more than enough Solar insolation to explain the ground temperature, and 2) back-IR heat amplification is not supported experimentally – not that we should have ever expected it to be so, if one utilizes all 3 laws of thermo.
JN: This is a whole new point. If you have other reasons to think CO2’s effect is minimal, lets discuss those but the only way to get past the focus on the Second Law is to admit that as a point of reasoning, there is nothing about the second law which precludes GHGs from theoretically causing an increase in the Earths temperature. Then we can move on and discuss the other reasons. And if I can be so bold — people will find your other reasons more compelling if there is no erroneous reference to the Laws of Thermodynamics being broken at the start.
“This is preposterous. God doesn’t make rules like “photons can travel in any direction except the way it came”. Atoms radiate photons randomly and some return in the same direction.”
Obviously I did not say that. [No, but that’s the implication of a Law that “stops” this happening.–JN] I said there is a restriction on cold bodies raising the temperature of a warmer body. [Only in a closed system where there is not a third body involved like The Sun –JN] If you surrounded yourself by an entire “death star” of ice – you were placed right in the center of it – ALL of that IR energy from the ice-star, which would be tremendously more energy than you radiate at ~30C (or whatever skin temperature humans are), would not raise your temperature. It could be emitting one-billion times more thermal-IR than you do, and you could be right at the center of the sphere, but it would still not warm you. You would still cool. Even if the “death-star” you were in the center of were 29C, you would still cool.
JN: People keep warmer in Igloos, not because the heat comes from the ice, but because it comes from burning fat internally, and the ice cuts the heat loss. Why continually ignore the real heat inputs? No one is suggesting that ice or GHG’s raise the temp of a warmer body above what it would have been without another energy source…. I’m stuck in ground-hog day: the sun … the sun … the sun…. ?
“What if we put it this way: Greenhouse gases don’t warm^ the planet but they do create a situation where the Sun can warm the world to a temperature above where it previously was without the gases?”
The Sun can only warm the planet to whatever its insolation temperature is. If a planet surface is warmer it is because of additional physical contributions, such as adiabatic compression and convection, etc. Any IR energy as a result of these processes are just that – the RESULT. They cannot self-amplify the temperature or energies of the system as they are the consequence of what has already happened, not the cause of what has already happened.
Also, the discussion of the actual solar insolation and resulting expected temperature can be read about in “Understanding…”. The input insolation upon the Earth is NOT -18C…it is actually +30C on average with a maximum of around +121C, depending on local effects and extinction. It is only the “effective”, or mathematically and theoretically averaged energy OUTPUT of the entire system, as compared to a Black Body, which equates to -18C. This does not mean this is the actual kinetic temperature you should find, nor does it say where you should actually find it even if it was.
“… it describes reality with a technical accuracy.”
Well, I disagree It isn’t accurate, ENOUGH.
JN: I’m talking about a principle – not an exact number. The principle is right or wrong. Your calculations are a different thing, and accuracy matters there, but if the principles are wrong, there’s no point doing a calculation.
In regard to Hammer’s post, I will just say this:
The effect of an atmosphere on a planet is to modulate the surface temperature from the extremes it would otherwise experience, as is the case on the moon compared to Earth. ANY atmosphere, independent of GHG’s, has a thermal capacity and ability to retain heat, and so it will ALWAYS be the case that the presence of an atmosphere keeps the “bottom-of-atmosphere” average temperature higher than the effective mathematical equivalent (as just discussed). If adiabatic effects are added to this then the “bottom-of-atmosphere” temperature can rise even further – even to well above the original Solar insolation, as is the case on Venus.
So consider then what the strongest GHG, H2O vapour, does in contribution to this effect: it is the best modulator of temperature that we have in our atmosphere. Under the Sun, in the tropics with lots of H20 vapour, the temperature will raise to LESS-than whatever the Solar insolation says it should. For some reason all the H2O doesn’t amplify the temperature. Very telling.
Then at night, comparing a tropic region to a desert, the tropic region cools by only a fraction of what a desert does. They likely have roughly the same “daily-average” temperature.
So the strongest GHG actually kept it cooler under the Sun than it should have been, and cooler than a desert, even with all the extra back-IR “heating”, of which a desert experiences very little.
There’s nothing wrong with being kept warm at night…but there is a significant concern if Solar insolation heating is modulated downwards by GHG’s.
The only extra heating more CO2 could provide is that contribution to increasing the atmospheric depth and density, such as to increase the physical heating effects.
I think we agree that ideally, CO2 concentration should be above at least 1000ppmv, in order to sustain the biosphere and improve its productivity. I am not sure this can be attained via fossil fuel burning…but I DO think we should try! Even if it DID warm the planet…well that would be great. The climate has always changed and it is only the intellectually inept, and the physically lazy, who now try to use other-people’s MONEY to avoid doing the work and economic improvements to adapt to it.
But as it is, the GH postulate is borne out of an incorrect application of the Stefan-Boltzmann equation, and worse, an illogical interpretation of its result. Under such a scenario, a logician EXPECTS non-physical postulates and theories to be borne out of an illogical premise. Applying the Laws of Thermodynamics, and trying to figure out how the 2nd Law applies, is secondary to this more fundamental physical mistake. Though, again, we should expect to find an inability to agree on the correct application under such a scenario.
We (the Slayers of the Sky-Dragon) will be publishing a new paper in the coming weeks which quantitatively will put this issue to rest…if logic is allowed back into the paradigm. We will also outline (and carry out) the proper physics experiment which anyone can perform, to actually prove or disprove the thesis on either side – such is the way science actually works.
JN: Joseph, you can’t quantitatively put to a rest an argument of reason which is not based on logic. The reasoning underpins everything. Maybe CO2 doesn’t heat the Earth (for other reasons unrelated to this post), but if that’s the case — it’s not anything to do with “Laws of Thermodynamics”. Can you at least acknowledge that in the Sun-Earth-Atmosphere system it is possible that energy flows which are always going from the hot sun to cold space, could be changed by a colder entity (GHGs) which would allow a warmer entity (Earth) to get even warmer, given that a third source (the Sun) is continually adding energy?
7.5 out of 10 based on 4 ratings | 3,939 | 18,122 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2022-27 | latest | en | 0.941432 |
http://www.fixya.com/support/t26578029-solve_answer_5_3x1_26 | 1,519,191,222,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813431.5/warc/CC-MAIN-20180221044156-20180221064156-00327.warc.gz | 439,847,189 | 31,060 | Question about The Office Equipment & Supplies
# Solve the answer to 5.3x1.26= - The Office Equipment & Supplies
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Apr 11, 2009 | Nintendo Professor Layton & the Curious...
### Before i pay iwant to know if you can solve my problem
Depends on the problem and you do not have to pay for an answer, you may just have to wait.
Dec 31, 2008 | Computers & Internet
### How to use ti 89 with a problem like "-(3x+1)-(4x-7)=4-(3x+2)'
Just plug in one side of the equation and find the answer. Then plug in the other side and compare answers.
Feb 19, 2008 | Texas Instruments TI-89 Calculator
## Open Questions:
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Level 3 Expert | 1,294 | 4,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2018-09 | latest | en | 0.944712 |
https://www.jiskha.com/display.cgi?id=1332443015 | 1,529,905,634,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867493.99/warc/CC-MAIN-20180625053151-20180625073151-00281.warc.gz | 849,504,448 | 4,436 | # statistics
posted by Anonymous
Calculate the z-score for each set of data. Determine who did better on her respective test, Tonya or Lisa.
Student English Test Grade z-Score Student Math Test Grade z-Score
John 82 Jim 81
Julie 88 Jordan 85
Samuel 90 Saye 79
Tonya 86 Lisa 82
Mean 86.50 81.75
St. Dev. 3.42 2.50
Given the above information, predict the raw score for a student whose z-score in the Math test is .98.
1. PsyDAG
Didn't I answer this for you previously?
Z = (score-mean)/SD
Compare the Z scores for Tonya and Lisa.
For second question, substitute .98 for the Z score and solve for the raw score. (Are you sure that .98 = Z score and not a proportion of the scores? If so, find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (.98) and its corresponding Z score. Again, use the equation above.
2. bfdbdfb
bfsdbhdfb
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More Similar Questions | 827 | 3,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-26 | latest | en | 0.931284 |
http://electronics.stackexchange.com/questions/32158/can-the-characteristic-impedance-z0-of-a-conductor-be-measured | 1,386,588,979,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163966854/warc/CC-MAIN-20131204133246-00016-ip-10-33-133-15.ec2.internal.warc.gz | 59,544,148 | 14,292 | Can the characteristic impedance Z0 of a conductor be measured?
We have a custom made data cable at my work, is there any good way to measure $Z_0$ of a cable, or must the formulas be used?
Trying to work out what value termination resistor to use.
-
If you had it custom made, and the impedance is important, it should have been in the specification! – stevenvh May 18 '12 at 4:17
It wasn't custom made by me, all work inherited from the predecessors... as usual – fred basset May 18 '12 at 15:29
Is the cable designed to be driven with differential signals (like twisted pair) or single-ended (like coax)? – The Photon May 18 '12 at 15:32
Differential, RS-485 – fred basset May 19 '12 at 6:22
The proper basic kit is called a TDR (time-domain reflectometer). A more advanced version is called a Two Port Network Analyser, both are usually expensive pieces of specialist test kit.
However, you can measure the impedance with normal lab kit in the following way;
Build your own TDR setup; You just need a fast oscilloscope and a pulse generator. See The Wiki page for a TDR This works by sending a short pulse down the cable and measuring the amplitude of the reflected pulse.
Unless you have a very fast oscilloscope and signal generator, work with a long cable (10's of m or more) to ensure you get a decent delay (or you won't able to tell the difference between the incident and reflected pulses) However if the cable is too long the attentation will make distinguishing the reflected pulse from noise very difficult.
Depending on what construction is used, signals travel down a cable at about 0.7 * the speed of light.
Do the same with an open circuit, a known resistance and a short circuit terminating the cable. The three values should be similar, take an average.
Method
Your kit setup should be as follows, though the picture is missing the termination resistor (or short) from the tail end of the cable.
Measure the height of the pulse out and back (incident and reflected) and divide them (rho), then solve the following equations:
$$\rho = \frac{Vr}{Vi}$$
Vr is the reflected voltage
Vi is the incident voltage
The characteristic impedance is Zo
The termination impedance is Zt
$$\rho = \frac{Z_{t} - Z{o}}{Z_{t} + Z{o}}$$
-
Thanks, good answer. – fred basset May 19 '12 at 6:26
Take a length of cable, leave the end open, and measure the impedance $Z_a$. Then repeat with the cable end shorted to get $Z_b$. Both are complex impedances. Your cable's characteristic impedance is
$Z_0 = \sqrt{Z_a \times Z_b}$.
Again, this is the complex square root.
The characteristic impedance is frequency independent, but you can't measure it at DC, so not with a common multimeter. Like Telaclavo says, at low frequencies the characteristic impedance may vary; it will only be constant above 100kHz to 1MHz. Despite being frequency independent whenever possible you'll usually measure at your working frequency.
-
Just to point out that he has to measure that at the frequency of interest, using special equipment. He can't do that with an ohmmeter. – Telaclavo May 18 '12 at 13:42
@Telaclavo - obviously, but I'll add it to my answer. – stevenvh May 18 '12 at 13:44
It is (mostly) independent of frequency... above a certain frequency, which can be between 100 kHz and 1 MHz. Below that, it varies a lot. google.es/… – Telaclavo May 18 '12 at 14:18
@Telaclavo - Right again :-). BTW, feel free to edit my answer if you think it needs improving/adjusting/clarification. – stevenvh May 18 '12 at 14:38 | 882 | 3,532 | {"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} | 2.90625 | 3 | CC-MAIN-2013-48 | latest | en | 0.928759 |
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# What is an irrational number but not an integer?
Wiki User
2017-04-04 00:45:45
Best Answer
No Irrational Numbers are integers. Pi is one example.
Wiki User
2017-04-04 00:45:45
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## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
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Wiki User
2013-12-28 17:18:42
sqrt(2) and pi are two example.
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People also asked | 183 | 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} | 2.96875 | 3 | CC-MAIN-2023-06 | longest | en | 0.868731 |
https://html.rincondelvago.com/ecuacion-lineal-simple.html?url=ecuacion-lineal-simple | 1,643,431,484,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299927.25/warc/CC-MAIN-20220129032406-20220129062406-00308.warc.gz | 364,679,332 | 16,713 | # Ecuación lineal simple
3/2X + 5/2Y = 7/4
1/2X - 1/3Y = 9
3/2X + 5/2Y = 7/4 1/2X - 1/3Y = 9
3/2X = 7/4 - 5/2Y 1/2X = 9 + 1/3Y
X = 7/6 - 5/3Y X= 18 + 2/3Y
7/6 - 5/3Y = 18 + 2/3Y
7/6 - 5/3Y = 18 + 2/3Y
-5/3Y - 2/3Y = 18 - 7/6
-15/3 - 6/3 = -21/3
-7
-7Y = 18 - 7/6
-7Y = 101/6
Y = -101/42
X = 18 + 2/3Y
X = 18 + 2/3 (-101/42)
X = 18 - 101/63
X = 1134/63 - 101/63
X = 1033/63
1/2X + 7Y = 1/2x + 3
-1/9X + 5/3Y =1/2
1/2X + 7Y = 1/2x + 3 -1/9X + 5/3Y =1/2
1/2X - 1/2X = 3 -7Y -1/9X - 1/2 = 5/3Y
-3=-7Y Y = -3/45X -3/10 =Y
-3/7 = -Y
Y = 3/7
3/7 = -3/45X - 10
3/7 + 3/10 = -3/35X
30/70 + 21/70 = 51/70
51/70 = 3-/45X
X = 2295/210
7/4X + 7/2Y = -1/2Y + 9/2
1/2X + 3/9Y =1/3
7/4X + 7/2Y = -1/2Y + 9/2 1/2X + 3/9Y =1/3
7/4X = -1/2Y - 7/2Y + 9/2 1/2X = 1/3 - 3/9Y
7/4X = -8Y + 9/2 X = 2/3 - 2/3Y
X = -32/7Y +18/7
-32/7Y +18/7 = 2/3 - 2/3Y
-32/7Y + 2/3Y = 2/3 - 18/7
-82/21Y = -40/21
- Y = 20/41
Y = -20/41
X = 2/3 - 2/3Y
X = 2/3 -2/3 (-20/41)
X = 2/3 - 40/123
X = 126/369
2X +3/2Y = 7
1/2X - 1/2Y = -9
2X +3/2Y = 7 1/2X - 1/2Y = -9
2X = 7 - 3/2Y 1/2X = -9 + 1/2Y
X = 7/2 - 3/4Y X = -18 + Y
7/2 - 3/4Y = -18 + Y
-3/4Y - Y = - 18 - 7/2
-7/4Y = -43/2
Y = 86/7
X = -18 + Y
X = -18 + 86/7
X = -40/7
1/2X + 1/9Y = 3
-3/2X + 1/9Y = 1/5Y + 1
1/2X + 1/9Y = 3 -3/2X + 1/9Y = 1/5Y + 1
1/2X = 3 - 1/9Y -3/2X = 1/5Y - 1/9Y +1
X = 6 - 2/9Y -3/2X = 4/45Y +1
-X = -8/135Y - 2/3
X = 8/135Y + 2/3
6 - 2/9Y = 8/135Y + 2/3
-2/9Y - 8/135Y = 2/3 - 6
-38/135Y = -16/3
Y = 360/19
Instituto Universitario Tecnológico
De Tecnología Industrial
Rodolfo Loero Arismendi
I.U.T.I.R.L.A
Realizado Por:
Calculo Matricial
Informatica_I3NJ
Barcelona, 24 de Septiembre de 2002
Enviado por: Cadaveria Idioma: castellano País: Venezuela
Palabras clave:
Te va a interesar | 1,210 | 1,799 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2022-05 | latest | en | 0.650207 |
https://lists.mysql.com/internals/555 | 1,516,646,829,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891530.91/warc/CC-MAIN-20180122173425-20180122193425-00249.warc.gz | 732,538,615 | 4,478 | List: Internals « Previous MessageNext Message » From: Michael Widenius Date: March 4 2001 10:28am Subject: Re: Optimisation suggestion... View as plain text
```Hi!
Sorry for the delayed reply; I have been on a vacation for a week and
are slowly catching up with all old emails.
>>>>> "Antony" == Antony T Curtis <antony@stripped> writes:
Antony> Thimble Smith wrote:
>>
>> On Wed, Feb 21, 2001 at 12:59:43PM +0000, Antony T Curtis wrote:
>> > In the case where there is a SELECT on a table where no index is
>> > directly applicable, if there is a compound index where the second part
>> > (or third, etc) of the index is used in the WHERE clause, it should be
>> > possible to still use that index by being able to skip rows rather than
>> > doing a brute-force search through all the records.
>> >
>> > Any hints where I should look to see if such an optimisation could be
>> > implemented?
>>
>> It isn't clear to me how that would work. Do you mean, to scan
>> the whole index in order to find rows where the secondary index
>> part matches, instead of scanning the data file? Or do you have
>> something else in mind? I think this optimization wouldn't work,
>> but I'm not sure I understand what you're suggesting.
>>
>> Tim
Antony> The kind of thing it'd do is something like this...
Antony> If the key-part to scan is the (n)th key part of the index,
Antony> 1. Seek index to start
Antony> 2. Read key from index
Antony> 3. Check (n)th key part against end-range, if less or equal, goto 6
Antony> 4. Increment (n-1)th key part and seek to greater/equal key
Antony> 5. goto stage 2
Antony> 6. Check (n)th key part against start-range, if greater/equal, goto xx
Antony> 7. Set (n)th key part to start-range and seek to greather/equal key
Antony> 8. goto 2
Antony> 9. (the key should point to valid row in range)
Antony> 10. Seek to next key
Antony> 11. if not end of index, goto 6
The problem is that the above will only work if a major part of the
keys are identical in the first n-1 parts. (You would guess that at
least 1/3 of the keys must be identical per key block for this to
work, which is not a very common situation).
The major problem is that when it comes to scan indexes compared to
scan a fixed length data file, index scanning is at least 3 times
slower than data file scanning (for B-trees). The reason is that
there is much more seeks involved when scanning the index structure
and also because the index structure is more complicated than the data
file structure, especially when we talk about compressed indexes.
The above is at least true for MyISAM tables where the data file is
stored separately from the index.
If you have a database structure where the data is stored in the leafs
of the trees (like Berkeley DB), the above optimization could be faster
as there is as much seeks involved in scanning the index as scanning
the data.
Antony> As an aside, IBM DB2 does not allow two indexes to exist where the two
Antony> keys have the same components but are in different orders,
Antony> Something like the following will fail in DB2.
>> CREATE TABLE db.test (
>> COL1 CHAR(10) NOT NULL,
>> COL2 CHAR(10) NOT NULL,
>> INDEX (COL1, COL2),
>> INDEX (COL2, COL1)
>> );
Antony> I suggest that their product does something similar to the above.
I can't see any reason why the above should fail; For example, if you
don't allow the above, you can't optimize a wide range of queries,
like:
SELECT * from test where col1 >= "A" and col1 <= "B"
SELECT * from test where col2 >= "A" and col2 <= "B"
SELECT * from test where col1 >= "A" and col2 >= "E"
SELECT * from test where col2 >= "A" and col1 >= "E"
SELECT * from test ORDER BY col1,col2;
SELECT * from test ORDER BY col2,col1;
To optimize a major part of the above queries, you must have both
indexes! To not allow one to do this is only relevant if you are
using some kind of index type where you can't do range or order by
searching.
As MySQL keeps a note about which indexes can be used to satisfy the
WHERE and the SELECT part, it would not be hard to do this
optimization for the handlers where the data is stored in the trees
(Currently only BDB tables).
The general optimization, where you allow any operator on the key
parts is easy to do:
and add in sql_select.cc two new functions:
in this case you need to call handler::read_next() as long as the
WHERE clause was satisfied and read_next_unique() when the WHERE
clause wasn't satisfied.
The general optimization to handle cases like:
SELECT * from test where key_part2>= "A" and key_part2 <="B" or
key_part2 >="D" or key_part2 <="E"
is much harder to do; In this case we could theoretically skip to
next key_part1 key when: we have found a key (#, "F"), but to do this
one would have to do a lot of new code, in the style of opt_range.cc
Regards,
Monty
``` | 1,276 | 4,823 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2018-05 | latest | en | 0.898554 |
https://www.enotes.com/homework-help/what-radical-130017 | 1,484,720,377,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280239.54/warc/CC-MAIN-20170116095120-00374-ip-10-171-10-70.ec2.internal.warc.gz | 905,047,324 | 12,854 | pohnpei397 | College Teacher | (Level 3) Distinguished Educator
Posted on
In math, a radical is any expression that has a square root, or a cube root, or some other root like that.
In a radical, you have the radical sign (the little sign that you use in square roots - that looks like a check mark with a long horizontal line that goes over the number) and the radicand. The radicand is the number or other expression that goes under the radical sign.
Radicals can be simplified (like simplifying square root of 25 down to 5) or they can just be left as radicals (square root of 7, for example).
neela | High School Teacher | (Level 3) Valedictorian
Posted on
The roots of the numbers or algebric expressions could be expressed in index form or radical form. Radical form is another form of expressing roots
x^(1/n) is the index form or exponent form of nth root of x written like n√(x). The sign n√ stands the nth root of
is called the radical form. In mathematics, radicals could be of any order more than 2. Example: | 258 | 1,037 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-04 | longest | en | 0.89996 |
https://topclasswriter.com/pole-and-minimum-number-of-moves-math-help-needed/ | 1,701,545,436,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100448.65/warc/CC-MAIN-20231202172159-20231202202159-00674.warc.gz | 659,293,508 | 17,044 | # pole and minimum number of moves math help needed
There are three poles in front of you. One pole is stacked with 64 rings ranging in weight from one ounce (at the top) to 64 ounces (at the bottom). Your task is to move all of the rings to one of the other to poles so that they end up in the same order. The rules are that you can only move one ring at a time, you can move a ring only from one pole to another, and you cannot even temporarily place a ring on top of a lighter ring.
What is the minimum number of moves you need to make to achieve the task?
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Show a detailed proof and give a thorough explanation please. | 204 | 911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2023-50 | latest | en | 0.947311 |
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# [已答复] 一阶偏微分方程，求帮忙看下程序
zsmj12 发表于 2013-7-1 10:47:08
du/dy=-0.035d2u/d2x+0.00014(du/dx)^2 以上du/dy为偏倒数d2u/d2x为2次偏倒： 初始和边界条件为 x = 0 u = 5000K, x = L u = 2000K,L = 10m, t = 0 u = 5000K. 程序如下 function [c,f,s] = pdexvvpde(x,t,u,DuDx) c = 1; f = -0.035.*DuDx; s =0.00014.*(DuDx)^2 ; function [pl,ql,pr,qr] = pdexvvbc(xl,ul,xr,ur,t) pl = ul-5000; ql = 0; pr = ur-2000; qr = 0.00001; function u0 = pdexvvic(x) u0 = 5000; function pdexvv m = 0; x = linspace(0,0.000001,100); t = linspace(0,0.6,50000); %u = linspace(2000,5000,6); sol = pdepe(m,@pdexvvpde,@pdexvvic,@pdexvvbc,x,t); u = sol(:,:,1); figure plot(x,u(end,:),'-r') title('Numerical solution and exact solution.') xlabel('x') ylabel('u(x,end)')存在问题，希望帮忙改改
1 条回复
## 倒序浏览
math 发表于 2013-7-2 03:31:11
您需要登录后才可以回帖 登录 | 注册 本版积分规则 回帖后跳转到最后一页 | 435 | 802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | latest | en | 0.351521 |
https://arpitbhayani.me/competitive-programming-solutions/hackerrank/two-pluses | 1,585,967,160,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370519111.47/warc/CC-MAIN-20200404011558-20200404041558-00298.warc.gz | 346,069,038 | 107,610 | # TWO-PLUSES hackerrank Solution - Correct, Optimal and Working
``````/*
* Author: Arpit Bhayani
* https://arpitbhayani.me
*/
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <deque>
#include <iostream>
#include <list>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define ll long long
#define MIN(a, b) a < b ? a : b
#define MAX(a, b) a > b ? a : b
using namespace std;
int i = 0;
char ch;
while((ch = getchar()) != '\n') {
str[i++] = ch;
}
str[i] = '\0';
return i;
}
char arr[16][16];
class compar {
public:
bool operator() (int *a, int *b) {
return a[2] < b[2];
}
};
priority_queue< int *, vector<int *>, compar > q;
void make_all(int *plus, char ch) {
for(int i = 0; i < plus[2]; i++) {
arr[plus[0] + i][plus[1]] = ch;
arr[plus[0] - i][plus[1]] = ch;
}
for(int j = 0; j < plus[2]; j++) {
arr[plus[0]][plus[1] + j] = ch;
arr[plus[0]][plus[1] - j] = ch;
}
}
int process(int r, int c) {
int max_area = -1;
for(int i = 0 ; i < r; i++) {
for(int j = 0 ; j < c; j++) {
if(arr[i][j] == 'G') {
int s = 1;
int left = j - 1;
int right = j + 1;
int up = i - 1;
int down = i + 1;
while( left >= 0 && right < c && up >= 0 && down < r ) {
if(arr[i][left] != 'G' || arr[i][right] != 'G' || arr[up][j] != 'G' || arr[down][j] != 'G') {
break;
}
s++;
left--;
right++;
up --;
down++;
}
int *t = (int *) malloc(3 * sizeof(int));
t[0] = i;
t[1] = j;
t[2] = s;
q.push(t);
}
}
}
while(!q.empty()) {
int *first_plus = q.top();
priority_queue< int *, vector<int *>, compar > tq;
make_all(first_plus, 'B');
for(int i = 0 ; i < r; i++) {
for(int j = 0 ; j < c; j++) {
if(arr[i][j] == 'G') {
int s = 1;
int left = j - 1;
int right = j + 1;
int up = i - 1;
int down = i + 1;
while( left >= 0 && right < c && up >= 0 && down < r ) {
if(arr[i][left] != 'G' || arr[i][right] != 'G' || arr[up][j] != 'G' || arr[down][j] != 'G') {
break;
}
s++;
left--;
right++;
up --;
down++;
}
int *t = (int *) malloc(3 * sizeof(int));
t[0] = i;
t[1] = j;
t[2] = s;
tq.push(t);
}
}
}
int * seconds_plus = tq.top();
max_area = MAX(max_area, ( (first_plus[2] - 1)*4 + 1) * ((seconds_plus[2] - 1)*4 + 1) );
make_all(first_plus, 'G');
while(!tq.empty()) {
int *x = tq.top();
free(x);
tq.pop();
}
q.pop();
}
return max_area;
}
int main(int argc, char *argv[]) {
int r, c;
scanf("%d%d", &r, &c);
for(int i = 0 ; i < r; i++) {
scanf("%s", arr[i]);
}
printf("%d\n", process(r, c));
return 0;
}
`````` | 933 | 2,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2020-16 | longest | en | 0.285424 |
https://math.stackexchange.com/questions/4221452/issues-with-representations-of-su2-and-the-su2-algebra | 1,716,859,173,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059055.94/warc/CC-MAIN-20240528000211-20240528030211-00709.warc.gz | 322,418,555 | 36,128 | # Issues with representations of SU(2) and the su(2) algebra
I'm currently having a somewhat hard time with a rather simple task.
This is my first contact with the formalism of Lie Groups. I'm studying the SU(2) and SO(3) groups and their algebra su(2) and trying to express su(2) in different forms.
I know the algebra is defined by this commutation relation between the generators $$[J_i,J_j] = i \epsilon_{ijk}J_k$$
but I cannot prove that this relation can be expressed in other forms if we use different J definitions, for example
i)$$[J^{ij},J^{lm}]=i(\delta^{il}J^{jm}-\delta^{jl}J^{im}-\delta^{im}J^{jl}+\delta^{jm}J^{il})$$ if we use $$J^{ij} \equiv \epsilon_{ijk}J_k$$,
ii)the form the algebra takes when we use the definition $$J_i \equiv -i(x_j\partial_k-x_k\partial_j)$$
I want to be able to prove that the algebra can be represented in these different forms, but every time I apply the J definitions to the commutator I ended up with expressions that don't lead to the correct forms.
I don't know if someone else had a tough time while studying this subject for the first time, but I appreciate any help. Thank you all.
First, $$(x_j \partial_k)(x_m \partial_n) = x_j (\partial_k x_m) \partial_n + (x_j x_m) (\partial_k \partial_n) = x_j \delta_{km} \partial_n + (x_j x_m) (\partial_k \partial_n) .$$
Therefore, \begin{align} [x_j \partial_k, x_m \partial_n] &= (x_j \partial_k)(x_m \partial_n) - (x_m \partial_n)(x_j \partial_k) \\ &= (x_j \delta_{km} \partial_n + (x_j x_m) (\partial_k \partial_n)) - x_m \delta_{jn} \partial_k + (x_m x_j) (\partial_n \partial_k) \\ &= \delta_{km} x_j \partial_n - \delta_{jn} x_m \partial_k \\ \end{align} and \begin{align} [x_j \partial_k - x_k \partial_j, x_m \partial_n - x_n \partial_m] & = [x_j \partial_k, x_m \partial_n] - [x_j \partial_k, x_n \partial_m] - [x_k \partial_j, x_m \partial_n] + [x_k \partial_j, x_n \partial_m] \\& = (\delta_{km} x_j \partial_n - \delta_{jn} x_m \partial_k) - (\delta_{kn} x_j \partial_m - \delta_{jm} x_n \partial_k) \\& - (\delta_{jm} x_k \partial_n - \delta_{kn} x_m \partial_j) + (\delta_{jn} x_k \partial_m - \delta_{km} x_n \partial_j) \\& = \delta_{km} (x_j \partial_n - x_n \partial_j) - \delta_{jn} (x_m \partial_k - x_k \partial_m) \\& - \delta_{kn} (x_j \partial_m - x_m \partial_j) + \delta_{jm} (x_n \partial_k - x_k \partial_n) . \end{align} | 779 | 2,354 | {"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": 7, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-22 | latest | en | 0.781382 |
https://socialsci.libretexts.org/Courses/Cosumnes_River_College/SOC_301%3A_Social_Problems_(Ninh)/02%3A__Articulate_how_sociological_concepts_can_be_used_to_explain_social_problems./2.02%3A_Racial_and_Ethnic_Inequality/2.2.01%3A_Chapter_Introduction | 1,726,551,025,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00451.warc.gz | 484,713,643 | 30,739 | # 2.2.1: Chapter Introduction
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SOCIAL PROBLEMS IN THE NEWS
“Anger, Shock over Cross Burning in Calif. Community,” the headline said. This cross burning took place next to a black woman’s home in Arroyo Grande, California, a small, wealthy town about 170 miles northwest of Los Angeles. The eleven-foot cross had recently been stolen from a nearby church.
This hate crime shocked residents and led a group of local ministers to issue a public statement that said in part, “Burning crosses, swastikas on synagogue walls, hateful words on mosque doors are not pranks. They are hate crimes meant to frighten and intimidate.” The head of the group added, “We live in a beautiful area, but it’s only beautiful if every single person feels safe conducting their lives and living here.”
Four people were arrested four months later for allegedly burning the cross and charged with arson, hate crime, terrorism, and conspiracy. Arroyo Grande’s mayor applauded the arrests and said in a statement, “Despite the fact that our city was shaken by this crime, it did provide an opportunity for us to become better educated on matters relating to diversity.”
Sources: Jablon, 2011; Lerner, 2011; Mann, 2011Jablon, R. (2011, March 23). Anger, shock over cross burning in Calif. community. washingtonpost.com. Retrieved from http://www.washingtonpost.com/wp-dyn/content/article/2011/03/23/AR2011032300301.html; Lerner, D. (2011, July 22). Police chief says suspects wanted to “terrorize” cross burning victim. ksby.com. Retrieved from http://www.ksby.com/news/police-chief-says-suspects-wanted-to-terrorize-cross-burning-victim/; Mann, C. (2011, March 22). Cross burning in Calif. suburb brings FBI into hate crime investigation. cbsnews.com. Retrieved from http://www.cbsnews.com/.
Cross burnings like this one recall the Ku Klux Klan era between the 1880s and 1960s, when white men dressed in white sheets and white hoods terrorized African Americans in the South and elsewhere and lynched more than 3,000 black men and women. Thankfully, that era is long gone, but as this news story reminds us, racial issues continue to trouble the United States.
In the wake of the 1960s urban riots, the so-called Kerner Commission (1968, p. 1)Kerner Commission. (1968). Report of the National Advisory Commission on civil disorders. New York, NY: Bantam Books. appointed by President Lyndon Johnson to study the riots famously warned, “Our nation is moving toward two societies, one black, one white—separate and unequal.” The commission blamed white racism for the riots and urged the government to provide jobs and housing for African Americans and to take steps to end racial segregation.
More than four decades later, racial inequality in the United States continues to exist and in many ways has worsened. Despite major advances by African Americans, Latinos, and other people of color during the past few decades, they continue to lag behind non-Hispanic whites in education, income, health, and other social indicators. The faltering economy since 2008 has hit people of color especially hard, and the racial wealth gap is deeper now than it was just two decades ago.
Why does racial and ethnic inequality exist? What forms does it take? What can be done about it? This chapter addresses all these questions. We shall see that, although racial and ethnic inequality has stained the United States since its beginnings, there is hope for the future as long as our nation understands the structural sources of this inequality and makes a concerted effort to reduce it. Later chapters in this book will continue to highlight various dimensions of racial and ethnic inequality. Immigration, a very relevant issue today for Latinos and Asians and the source of much political controversy, receives special attention in Chapter 5.2 "Population and the Environment"’s discussion of population problems.
2.2.1: Chapter Introduction is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts. | 2,655 | 8,359 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-38 | latest | en | 0.193876 |
https://www.aakash.ac.in/important-concepts/physics/reflection-of-light | 1,713,722,720,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817790.98/warc/CC-MAIN-20240421163736-20240421193736-00727.warc.gz | 555,484,848 | 39,650 | • Call Now
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# Reflection of light
Let's play an interesting game. First, keep your face in front of a dull wall and observe. Now see your face on your shining floor and note the observation. Now take a bucket of water and see your face in it and observe. At last, see your face in the plane mirror at your home. If you compare the observations of all four cases you will find there is no image of your face when you see it on the wall. There is an image with very poor quality when seen on the floor. When you see the face in the water still you can see the image but not clear. You got a proper image of good quality when seen in the plane mirror. Although all four are plane surfaces, still we are getting different quality images. This phenomenon can be understood by the reflection of light. The walls of the house are rough in nature so produce diffuse reflection and regular reflection is very less when light falls on it. On the other hand, the mirror reflects almost all the light that falls on it. Let's learn more about the reflection of light!
Table of content
• Reflection of light
• Laws of Reflection
• Terminology used in reflection of light
• Object and image
• Practice problems
• FAQs
## Reflection of light
When a light ray encounters a boundary between two different media and bounces back to enter the same medium, this phenomenon is referred to as the reflection of light. The ray which falls on the surface is called the incident light ray. The ray that bounces back is known as the reflected ray. If we draw a perpendicular on the reflecting surface, at the point of incidence it is called the normal. The point where a light ray coming from an item or source appear to meet after being reflected by a surface is called an image. We can see our surroundings because light reflects off a variety of surfaces.
The reflecting surface can be plane or spherical. In the above figure, a plane reflecting surface is shown. A parallel beam of light incident on the surface after the reflection is bounced back in the same medium.
## Laws of Reflection
There are two laws that govern the reflection of light (reflected from any surface). These Laws are referred to as laws of reflection. When light from any object reflects from polished surfaces, the reflection is called regular reflection. Regular reflection produces an image. We obtain two different types of images: real and virtual, depending on the mirror (reflecting surface) and object position.
At home, we create our virtual image using plane mirrors. The image distance from a plane mirror will be equal to the object distance from the mirror. The light ray is reflected by a planar surface and is subject to the following two laws of reflection.
1. The angle of incidence and reflection is the same. i.e. (i = r).
2. At the point of incidence, the normal to the reflecting surface, the incident ray, and the reflected ray all are coplanar.
## Terminology used in reflection of light
1. Incident ray : A light beam that strikes the mirror after emerging from an object or the source.
2. Reflected rays : The light beam that bounces back after striking a reflective surface.
3. Point of incident : It is the location where the incident ray makes contact with the surface.
4. Normal : It is the line which is perpendicular to the reflecting surface at the point of incident.
5. Angle of incidence : Angle of incidence is the angle formed at the point of incidence between the incident ray and the normal to the surface. It is represented by i.
6. Angle of reflection : Angle of reflection is defined as the angle formed by the reflected ray and the normal to the surface at the point of incidence. It is represented by r.
7. Angle of deviation : Angle of deviation is the angle formed by the directions of the incident and reflected rays. Angle of deviation is given by δ=π-(i+r)=π-2i
## Object and image
Object : An object is everything which is viewed. Also it is the point of intersection of the incident rays. The object can be real and virtual. When the incident rays actually meet (or emerge from) the object is called a real object (figure 1). When the incident rays appear to meet (or emerge from) then it is called a virtual object (figure 2).
Image : The point of intersection of reflected rays is called image. It can also be real and virtual. If the reflected rays appear to meet then it is called a virtual image (figure 1). If the reflected rays actually meet at a point then it is called real image (figure 2).
### Types of reflection
The reflection of light can be of the following three types :
1. Regular/ Specular Reflection: A smooth surface, such as a mirror, a polished metal, or a calm body of water, will reflect all the light that arrives from a given direction at the same angle. This reflection is known as regular or specular reflection. The image formed is crisp and clear. All the rays of light that are reflected from a surface are parallel if a parallel light source is incident on it. Thus, a regular reflection produces an image. Hence a flat, plane mirror creates an image that is virtual, upright and the same size as the object.
1. Diffused Reflection: When light strikes rough surfaces, it reflects in a diffuse manner. When a beam of light strikes such a surface, it is reflected at different angles and in all directions. Depending on how rough their surfaces are, the light that strikes the objects in our room is diffused in all directions. This reflection does not produce any image.
1. Multiple reflection : Multiple reflections is similar to regular reflections but it uses more than one mirror. Multiple reflections occur in these situations when two mirrors are positioned at an angle. The angle formed by the two mirrors determines how many images are visible.
Application of reflection of light in our daily life
The most general application of reflection of light is given below :
• Reflection of light on the polished surface.
• Periscopes use reflection of light to let forces see oncoming enemies on the battlefield without endangering their safety.
• Medical diagnostics and optical communications are made possible through reflection of light.
• We can see the moon because the light rays coming from the sun get reflected from the surface of the moon.
• The law of reflection makes it possible to precisely determine distances between objects.
• We can see the colours of diverse objects through reflection of light.
## Practice problems
Q1. A light ray is incident on a plane surface at an angle 37o with the surface, what is the angle of reflection?
A. Given angle with the surface $\theta$s = 37o
The angle of incidence i = 90- $\theta$= 90- 37= 53o
As per the law of reflection, angle of incidence is equal to angle of reflection i = r
So the angle of reflection= 53o.
Q2. The angle of incidence for a plane reflecting surface is . What is the angle of deviation?
A. Given angle of incidence $i=\frac{\pi }{4}$
The deviation's angle is given by,
The angle of deviation is or 90o.
Q3. Write the characteristics of the image formed by the plane mirror?
A. The characteristic of the image formed by plane mirrors are as follows :
1. The image and the object from the mirror are separated by the same amount of distance.
2. The image formed is virtual and upright.
3. The image is the exact same size as the object.
4. The image produced is laterally inverted i.e. right is shown as left, and left as right.
Q4. A reflecting surface deviating the light ray at an angle 120o. What is the angle of incidence?
A. Given $\delta ={120}^{o}$
The angle of deviation is given by,
Hence the angle of incidence is 30o
## FAQs
Q1. Which of the following cases the laws of reflection hold good for?
1. plane mirror
2. concave mirror
3.convex mirror
4. all reflecting surfaces.
A. The law of reflection is valid for all the reflecting surfaces including plane mirrors, concave mirrors and convex mirrors. Concave and convex mirrors have curved surfaces but at the microscopic level it can be considered to consist of an infinite number of a plane reflecting surfaces.
Q2. How can you see the reflection of light on the moon?
A. The light coming from the sun falls on the moon and after the reflection from the moon it reaches the earth and thus we can see it. The portion on which the light falls appears bright and the rest appears dark.
Q3. Which object reflects more light?
A. Compared to objects with dull or unpolished surfaces, glossy or polished objects reflect more light. The best light reflector is made of silver metal.
Q4. Can water reflect the light?
A. Yes, water can reflect light. Also, it refracts the light. When the light falls on the water surface some of the light gets reflected and some refracted into it.
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Vacuum tube devices
1. 1. Copyright 2005 by Taylor & Francis Group
2. 2. Copyright 2005 by Taylor & Francis Group
3. 3. Copyright 2005 by Taylor & Francis Group
4. 4. Copyright 2005 by Taylor & Francis Group
5. 5. Copyright 2005 by Taylor & Francis Group
6. 6. Copyright 2005 by Taylor & Francis Group
7. 7. Copyright 2005 by Taylor & Francis Group
8. 8. Copyright 2005 by Taylor & Francis Group
9. 9. Copyright 2005 by Taylor & Francis Group
10. 10. Copyright 2005 by Taylor & Francis Group
11. 11. Copyright 2005 by Taylor & Francis Group
12. 12. | 53,291 | 30,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2017-04 | latest | en | 0.285087 |
https://devforum.roblox.com/t/understanding-cframefrommatrix-the-replacement-for-cframenew/593742 | 1,702,016,725,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100724.48/warc/CC-MAIN-20231208045320-20231208075320-00560.warc.gz | 231,765,586 | 13,263 | # Understanding CFrame.fromMatrix() - The Replacement for CFrame.new()
Hey y’all!
Happy Memorial Day to everyone! Always remember to honor those who served in the army to defend their own country.
Anyways, I’ve seen that the news of `CFrame.new(position, lookAt)` (only the one with the Look At) being deprecated has not spread very far. Many people still use this version, which works absolutely fine, but you shouldn’t use deprecated constructors in your code, should you?
The new version is `CFrame.fromMatrix()` which is a little more complicated than the deprecated version. Plus, the DevHub doesn’t clearly explain it yet. It’s more like an understand-it-yourself situation. As a result, I’ve seen a nasty stew of confusion cooking in places like the #help-and-feedback:scripting-support. But don’t worry, I’m here! I’ll show you the recipe for the tastier stew! For that, I recommend you read through this tutorial if you want to understand the concept yourself and not copy code aimlessly.
## [color=#e8434e]Disclaimer[/color]
1. You must know a little bit about vectors and the general definition of the cross product. If you don’t know check out these resources:
# The Right-Hand Rule
You may already know that the cross product returns an orthogonal (perpendicular) vector to the two given ones. But, there are two possible vectors that are perpendicular, both pointing in opposite directions. The cross product is not commutative, and therefore, it matters in which order you cross two vectors.
Luckily, something called the [color=#3296fa]Right-Hand Rule[/color] exists. For that, you’ll need your right hand (didn’t expect that did you?).
Basically, you point your [color=#2323ff]index[/color] finger in the direction of the [color=#2323ff]first[/color] vector and point your [color=#e12828]middle[/color] finger in the direction of the [color=#e12828]second[/color] vector. Point your [color=#963c96]thumb[/color] up and it will be the direction of the [color=#963c96]resulting[/color] vector from the cross product (index finger * middle finger).
View the image below for a visual explanation:
[color=#2828ff]a[/color] is the [color=#2828ff]index[/color] finger and [color=#e02626]b[/color] is the [color=#e02626]middle[/color] finger.
Note: The first two vectors don’t necessarily have to be perpendicular themselves to obtain another vector perpendicular to both of them. The angle in between can be acute or obtuse and the cross product would still return an orthogonal vector.
This is the abstract definition of the Right-Hand Rule, but we’ll be using it in action rightaway!
# Overview of CFrame.fromMatrix()
This is the API for this constructor:
## [color=#3296fa]CFrame.fromMatrix(Vector3pos, Vector3vX, Vector3vY, Vector3vZ)[/color]
Explanation of each component:
Pos: The position of the object, pretty straightforward.
vX: The direction of the right vector of a CFrame.
vY: The direction of the up vector.
vZ: [OPTIONAL] The direction of the forward/look vector.
## How do you know which 3D axis is the direction of which vector?
In Roblox Studio, have the view selector open (which displays the 3 axes).
• Select the right (or left) face of a part, which is in the direction of the X axis.
• We all know that the Y-axis is the up direction in Roblox. But still, select the top (or bottom) face of a part and you’ll see that it’s in the direction of the Y axis.
• Select the front (or back) face of a part, it’s in the direction of the Z axis.
You can remember this as the phrase “right up front” in the order of X, Y, and Z.
Also, make sure you know how these three vectors are related:
If you imagine the look vector pointing in the direction you’re facing, then the right vector is to the right and the up vector is pointing up.
## Why is vZ optional?
Well, since all faces in traditional rectangular parts’ faces are facing in the direction of vectors that are orthogonal to each other, we only need two vectors really. Roblox automatically takes the first two vectors and crosses them to find the third, so we don’t need to include that.
# Using CFrame.fromMatrix()
Now that we know about our new little friend, we can now use it in action! Since you may be used to the old `CFrame.new(pos, lookAt)` constructor, we’ll create a function to simulate that.
CFrame.new() Simulator! << oh goodness no, don’t that let that be a thing!
For that, we’ll create a function that will take the position and the direction to face and construct a CFrame using `CFrame.fromMatrix()`. And then, we’ll return that!
``````local function getCFrame(position, lookAt)
end
``````
Great, now we have a function going. What do we do first? Well, using `position` and `lookAt`, we can find the LookVector simply by subtracting the two Vector3s:
``````local lookVector = (position - lookAt).Unit
--points from the position to be at to the position to face
``````
Why make it a unit vector? Well, magnitude really doesn’t matter in this new constructor, we only need the direction component. So, making everything a magnitude of 1 will keep things consistent.
Now, before we start crossing roads vectors, we need two of them to begin with! Remember how I mentioned that the angle between the original two vectors doesn’t matter? Well, we can put that to action by creating a model vector that points straight up.
``````local modelUpVector = Vector3.new(0, 1, 0) --again, unit vectors!
``````
We can now start crossing. We have the look vector and we have the model up vector, so next, we need to find the right vector.
Let’s get the order right before crossing. But we need to rotate the forward-up-right vectors slightly:
So, we know that the forward vector is the first to come in the cross product followed by the up vector:
``````local rightVector = lookVector:Cross(modelUpVector)
--[[
Yes, the magnitude of the resulting vector is dependent on the area of the parallelogram
of the first two, but again, we only care about the direction, not magnitude!
]]
``````
Now all we need is the real up vector that’s perpendicular to the look vector. The model one we made is not orthogonal to the look vector, but we can cross the look and right vectors to find that out!
Use the right-hand rule:
So, the first vector is the right vector and the second vector is the forward vector.
``````local upVector = rightVector:Cross(lookVector)
``````
And we’re almost done, just need to return the CFrame!
``````--remember it's X, Y, Z (since Z is optional we don't need to use the look vector)
return CFrame.fromMatrix(position, rightVector, upVector)
``````
Now we can actually use this:
``````taylor.CFrame = getCFrame(taylor.Position, swift.Position)
--turn taylor to face swift!
``````
Here is the entire code (uncommented):
Expand/collapse
``````local function getCFrame(position, lookAt)
local lookVector = (position - lookAt).Unit
local modelUpVector = Vector3.new(0, 1, 0)
local rightVector = lookVector:Cross(modelUpVector)
local upVector = rightVector:Cross(lookVector)
return CFrame.fromMatrix(position, rightVector, upVector)
end
taylor.CFrame = getCFrame(taylor.Position, swift.Position)
``````
## Closing Remarks
Hopefully, after reading this, there isn’t much confusion going on in the #help-and-feedback:scripting-support category (or anywhere else) on this topic. You should now be able to cook up the tastier stew! But I still want to evaluate myself:
So, roll in them polls!
Rate this tutorial.
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Also, feel free to check out the later parts of the tutorial:
68 Likes
Your sample code using fromMatrix to substitute the supposedly deprecated CFrame eye and target constructor does not provide identical functionality in certain edge cases. This is pointed out by Egomoose in the following post: Is CFrame.new(Vector3 position, Vector3 lookAt) actually deprecated?
I discovered this discrepancy the hard way when I discovered that my math would come out completely incorrectly when using your (and the devhub’s) substitute sample code using fromMatrix, and would only return the expected output when using the legacy constructor.
I do not wish to downplay your work here on explaining how to use fromMatrix–the diagrams are great and everything is explained with diligence; however, I do wish to make a statement here (since this post will likely be seen by Roblox engineers at some point) that I do not believe the legacy CFrame.new ( Vector3 pos, Vector3 lookAt ) should be deprecated. It is compact, very easy for beginners to understand and work with, and contains behavior that is not matched 1:1 by the substitute.
6 Likes
Yeah, I honestly don’t know why it got deprecated. It was so much simpler. New additions in a programming language are supposed to be more convinient.
But, until they do that, I thought I’d explain the replacement a bit better than the DevHub does, just for the time being.
3 Likes
You should explain the solution to edge cases too since people would be discouraged but I thought the tutorial was helpful.
2 Likes
Interesting tutorial!
Barring the weirdness of the deprecation and the edge-case, the vector concepts associated with the new constructor are solid - the appreciation of matrices as groupings of vectors is seldom noted. This and your trigonometry guide are pretty good concise entries into the topics. Do keep up the good fight - technical knowledge ought to be promoted!
On a side-note though, I do agree that the deprecation of the .new(v3, v3) form is weird - it made pretty good mathematical sense, one being a position vector and the other being a relative look vector.
2 Likes
hello, thanks for the tutorial. But there are still some things I don’t understand: how would I make the CurrentCamera face the back of the player at spawn? I need this for an obby style game. I tried this code: (with your getCFrame function)
``````localPlayer.CharacterAdded:Connect(function()
local camera = workspace.CurrentCamera
camera.CFrame = getCFrame(camera.CFrame.Position,localPlayer.Character:WaitForChild("HumanoidRootPart").Position)
end)
``````
But it doesn’t work, it faces the camera at the players face and positions it near the ground. And sometimes it faces the camera to the right of the player??
Thanks for any help!
1 Like
You should probably be asking this kind of stuff in #help-and-feedback:scripting-support but see if this would help you:
Edge cases solution
``````local UNIT_X = Vector3.new(1, 0, 0)
local UNIT_Y = Vector3.new(0, 1, 0)
local UNIT_Z = Vector3.new(0, 0, 1)
local IDENTITYCF = CFrame.new()
local function getCFrame(eye, target, normalId) -- normalId would be treated as Front if you don't provide it
local cf = IDENTITYCF
local forward = target - eye
-- technically returns 0, 1, 0, NAN, NAN, NAN, NAN, NAN, NAN, NAN, NAN, NAN
if (forward:Dot(forward) == 0) then
return IDENTITYCF + eye
end
-- calculate needed rotation for adjusting face
if (normalId and normalId ~= Enum.NormalId.Front) then
local face = Vector3.FromNormalId(normalId)
local axis = math.abs(face.z) >= 1 and UNIT_Y or UNIT_Z:Cross(face)
local theta = math.acos(-face.z)
cf = CFrame.fromAxisAngle(axis, theta)
end
-- return the CFrame
forward = forward.Unit
if (math.abs(forward.y) >= 0.9999) then
return CFrame.fromMatrix(eye, -math.sign(forward.y) * UNIT_Z, UNIT_X) * cf
else
local right = forward:Cross(UNIT_Y).Unit
local up = right:Cross(forward).Unit
return CFrame.fromMatrix(eye, right, up) * cf
end
end
``````
https://devforum.roblox.com/t/is-cframe-new-vector3-position-vector3-lookat-actually-deprecated/383172/4
1 Like
I guess I’ll have to stick with the deprecated CFrame.new(). If the “replacement” doesn’t always work, then switching my existing CFrame.new()s to this new function would break a lot of things that I have already made. I don’t think Roblox will ever remove it, because it would break so many games that use it. But thanks for the tutorial, could be useful later on.
1 Like | 2,984 | 12,176 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.882469 |
http://en.cnki.com.cn/Article_en/CJFDTOTAL-DQWX201802033.htm | 1,603,163,315,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107869785.9/warc/CC-MAIN-20201020021700-20201020051700-00361.warc.gz | 39,470,622 | 9,217 | Full-Text Search：
《Chinese Journal of Geophysics》 2018-02
## Spectral element method for 3D frequency-domain marine controlled-source electromagnetic forward modeling
LIU Ling;YIN ChangChun;LIU YunHe;QIU ChangKai;HUANG Xin;ZHANG Bo;College of Geo-exploration Science and Technology,Jilin University;
Marine controlled-source electromagnetic(MCSEM)method is an important predrill reservoir appraisal method to reduce exploration risk in detecting sub-seafloor hydrocarbon reservoirs.Most 3D forward modelings for MCSEM are based on conventional numerical methods like finite-difference and finite-element method.In this paper,we introduce spectral element method(SEM)based on Gauss-Lobatto-Chebyshev(GLC)polynomials for 3D frequency-domain MCSEM.SEM is an accurate and efficient electromagnetic modeling methods due to its spectral accuracy and exponential convergence.The method combines the flexibility of finite-element method with the accuracy of spectral method by a simple application of spectral method to each element.Staring from the Maxwell′s equations,we obtain a vector Helmholtz equation of electric field.Then we use Galerkin weighted residual method to discretize the vector Helmholtz equation,in which the curl-conforming Gauss-Lobatto-Chebyshev(GLC)polynomials are chosen as basis functions.As a kind of high-order complete orthogonal polynomials,the GLC polynomials havethe characteristic of exponential convergence with the order and can derive the matrix elements analytically,which improves the modeling accuracy.Finally,we use the direct solver MUMPS to solve the final system of equations to speed up the modeling process.For numerical experiments,we first simulate responses of a 3-layer model and compare our SEM results with open-source software to check the accuracy of our SEM algorithm.After that,we analyze efficiency of SEM by comparing with finite-difference method.Finally,we apply SEM for typical 3D models and analyze the MCSEM responses.Numerical results show that SEM is an efficient and effective method for MSCEM modeling,and it can deliver very accurate results even with coarse meshes.
【Fund】： 国家重点研发计划重点专项(2016YFC0303100 2017YFC0601903);; 国家自然科学基金重点项目(41530320);国家自然科学基金面上项目(41274121);; 国家青年基金项目(41404093)联合资助
【CateGory Index】： P631.325
CAJViewer7.0 supports all the CNKI file formats; AdobeReader only supports the PDF format.
【Citations】
Chinese Journal Full-text Database 8 Hits
1 HAN Bo;HU Xiang-Yun;HUANG Yi-Fan;PENG Rong-Hua;LI Jian-Hui;CAI Jian-Chao;Hubei Subsurface Multi-scale Imaging Key Laboratory,Institute of Geophysics and Geomatics,China University of Geosciences;;3-D frequency-domain CSEM modeling using aparallel direct solver[J];地球物理学报;2015-08 2 YANG Jun;LIU Ying;WU Xiao-Ping;Laboratory of Seismology and Physics of Earth′s Interior,School of Earth and Space Sciences, University of Science and Technology of China;College of Marine Geosciences,Ocean University of China;;3D simulation of marine CSEM using vector finite element method on unstructured grids[J];地球物理学报;2015-08 3 YIN Chang-Chun;BEN Fang;LIU Yun-He;HUANG Wei;CAI Jing;Geo-Exploration Science and Technology Institute,Jilin University;;MCSEM 3D modeling for arbitrarily anisotropic media[J];地球物理学报;2014-12 4 LIU You-shan;TENG Ji-wen;XU Tao;LIU Shao-lin;SI Xiang;MA Xue-ying;Institute of Geology and Geophysics,Chinese Academy of Sciences;University of Chinese Academy of Sciences;;Numerical modeling of seismic wavefield with the SEM based on Triangles[J];地球物理学进展;2014-04 5 LI Lin;LIU Tao;HU Tian-Yue;School of Earth and Space Sciences,Peking University;Petroleum Exploration and Production Research Institute,SINOPEC;;Spectral element method with triangular mesh and its application in seismic modeling[J];地球物理学报;2014-04 6 CHEN Gui-Bo~(1,2),WANG Hong-Nian~2,YAO Jing-Jin~3,YANG Shou-Wen~21 School of Physics,Changchun University of Science and Technology,Changchun 130022,China2 School of Physics,J ilin University,Changchun 130026,China3 Institute of Geophysical and Geochemical Exploration,Geological Academe of China,Hebei Lang fang 065000,China;Modeling of electromagnetic responses of 3-D electrical anomalous body in a layered anisotropic earth using integral equations[J];地球物理学报;2009-08 7 SUN Xiang-Yang1,NIE Zai-Ping1,ZHAO Yan-Wen1,LI Ai-Yong2,LUO Xi2 1 School of Electronic Engineering,University of Electronic Science and Technology of China,Chengdu 610054,China2 China Oilfield Service Limited,Beijing 101149,China;The electromagnetic modeling of logging-while-drilling tool in tilted anisotropic formations using vector finite element method[J];地球物理学报;2008-05 8 WANG XiuMing1,2,SERIANI Geza3 & LIN WeiJun1 1 Institute of Acoustics,Chinese Academy of Sciences,Beijing 100080,China;2 CSIRO Petroleum,ARRC,PO Box 1130,Technology Park,Bentley,WA 6102,Australia;3 Instituto Nazonnale di Oceangrafia e di Geofisica Sperimentale,Borga Grotta Gigante 42/c-Sgonico,I-34016,Trieste,Italy;Some theoretical aspects of elastic wave modeling with a recently developed spectral element method[J];Science in China(Series G:Physics,Mechanics & Astronomy);2007-02
【Co-citations】
Chinese Journal Full-text Database 10 Hits
1 WANG ShuMing;DI QingYun;WANG Ruo;SU XiaoLu;Mohamed A;Hubei Subsurface Multi-scale Imaging Key Laboratory,Institute of Geophysics and Geomatics,China University of Geosciences;Institute of Geology and Geophysics,Chinese Academy of Sciences;;Removing airwave effects using electromagnetic fields decomposition for 3D MCSEM[J];地球物理学报;2018-02 2 LIU Ling;YIN ChangChun;LIU YunHe;QIU ChangKai;HUANG Xin;ZHANG Bo;College of Geo-exploration Science and Technology,Jilin University;;Spectral element method for 3D frequency-domain marine controlled-source electromagnetic forward modeling[J];地球物理学报;2018-02 3 LIN Weijun;SU Chang;SERIANI Géza;Institute of Acoustics, Chinese Academy of Sciences;University of Chinese Academy of Sciences;Istituto Nazionale di Oceanografia e di Geofisica Sperimentale;;The poly-grid spectral element method and its implementation in high performance computing[J];应用声学;2018-01 4 MA Mingxue;MAO Baohua;YUE Xizhou;LIU Baoyin;ZHANG Zhongqing;China Oilfield Services Limited;Hangzhou Xunmei Technology CO.LTD.;Zhejiang University;;Simulation and Application of a New Electromagnetic Resistivity Tool with Array Compensation Propagation While Drilling[J];测井技术;2017-06 5 Li Jianhui;Hu Xiangyun;Chen Bin;Liu Yajun;Guo Shiming;Institute of Geophysics and Geomatics,China University of Geosciences (Wuhan);State Key Laboratory for Geomechanics and Deep Underground Engineering,China University of Mining & Technology;;3D electromagnetic modeling with vector finite element for a complex-shaped loop source[J];石油地球物理勘探;2017-06 6 Minghui Zhang;Tao Xu;Zhiming Bai;Youshan Liu;Jue Hou;Guiping Yu;Key Laboratory of Earthquake Geodesy, Institute of Seismology,China Earthquake Administration;State Key Laboratory of Lithospheric Evolution, Institute of Geology and Geophysics, Chinese Academy of Science;CAS Center for Excellence in Tibetan Plateau Earth Sciences;Institute of Geophysics, China Earthquake Administration;University of Chinese Academy of Science;;Ray tracing of turning wave in elliptically anisotropic media with an irregular surface[J];Earthquake Science;2017-Z1 7 LI Gang;LI Yu-Guo;HAN Bo;DUAN Shuang-Min;College of Marine Geosciences,Ocean University of China;GEOMAR Helmholtz Centre for Ocean Research Kiel;Key Lab of Submarine Geosciences and Prospecting Techniques of Ministry of Education,Ocean University of China;Laboratory of Marine Mineral Resources,Qingdao National Laboratory for Marine Science and Technology;;2.5D marine CSEM modeling in the frequency-domain based on an improved interpolation scheme at receiver positions[J];地球物理学报;2017-12 8 LIU Ying;LI Yu-Guo;HAN Bo;Key Lab of Submarine Geosciences and Prospecting Techniques,Ministry of Education and College of Marine Geosciences,Ocean University of China;Evaluation and Detection Technology Laboratory of Marine Mineral Resources,Qingdao National Laboratory for Marine Science and Technology;;Adaptive edge finite element modeling of the 3D CSEM field on unstructured grids[J];地球物理学报;2017-12 9 DONG Xing-Peng;YANG Ding-Hui;Department of Mathematical Sciences,Tsinghua University;;Numerical modeling of the 3-D seismic wavefield with the spectral element method in spherical coordinates[J];地球物理学报;2017-12 10 Zhao Yue;Xu Feng;Li Xiu;Liu Jinpeng;Institute of Acoustics,Chinese Academy of Sciences;College of Geology Engineering and Geomatics,Chang'an University;;Characteristics of 3D TEM response on seafloor with the central loop configuration[J];石油地球物理勘探;2017-05
【Secondary Citations】
Chinese Journal Full-text Database 10 Hits | 2,243 | 8,674 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-45 | latest | en | 0.833689 |
https://www.chi-cmg.com/360-day-interest-calculator/ | 1,575,627,249,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540487789.39/warc/CC-MAIN-20191206095914-20191206123914-00234.warc.gz | 656,999,568 | 7,702 | # 360 Day Interest Calculator
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Many banks use an "Actual/360" formula to calculate payments, while Excel’s pmt function and your financial calculator use the 30/360 formula (i.e., every month earns 30 days’ interest on a 360-day year). When banks use Actual/360, it means that interest for each day is based on the nominal rate (e.g., 6.00%) divided by 360 days.
Apt Calculator "It will be a larger strategic decision which has trade implications which has geopolitical implications," telecom secretary Aruna Sundararajan told reporters on sidelines of an ITU-APT Foundation.
Calculate Accrued Interest Using the Days360 Function. For bonds that use the 30/360 day count convention, we can calculate the day count fraction using the days360 function: days360(start_date,end_date,[method]) This function will calculate the number of days between two dates using the 30/360 convention.
Want to see how fast your own savings could grow? Use the NerdWallet compound interest calculator. compound interest is a feature you can take advantage of as long as you have an account that offers a.
Commercial Business Real Estate Latest headlines for commercial real estate including development companies and large construction projects.. global Business and Financial News, Stock Quotes, and Market Data and Analysis..
365/360 Loan Calculator Definitions. Loan type. Financing start date This is the first day that interest will begin to be charged to your loan balance. This is also typically the same date that funds are distributed to the borrower. Interest rate
Us Bank Home Loan Calculator Know your monthly amortization through this home loan calculator – MANILA, Philippines – If you’re wondering how much your monthly amortization will be for a home you’re looking to buy, you’ll find a home loan calculator very helpful. A home loan calculator computes.
In the default case, an insured worker can use 180 days uninterrupted sick leave annually and no more that 360 interrupted. | 653 | 3,252 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-51 | latest | en | 0.906914 |
https://www.oercommons.org/browse?f.keyword=multiplication-property-of-equality | 1,685,268,526,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643663.27/warc/CC-MAIN-20230528083025-20230528113025-00660.warc.gz | 1,050,580,740 | 19,007 | Updating search results...
# 6 Results
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Mathematics
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10/06/2016
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Equations and Inequalities
Type of Unit: Concept
Prior Knowledge
Students should be able to:
Add, subtract, multiply, and divide with whole numbers, fractions, and decimals.
Use the symbols <, >, and =.
Evaluate expressions for specific values of their variables.
Identify when two expressions are equivalent.
Simplify expressions using the distributive property and by combining like terms.
Use ratio and rate reasoning to solve real-world problems.
Order rational numbers.
Represent rational numbers on a number line.
Lesson Flow
In the exploratory lesson, students use a balance scale to find a counterfeit coin that weighs less than the genuine coins. Then continuing with a balance scale, students write mathematical equations and inequalities, identify numbers that are, or are not, solutions to an equation or an inequality, and learn how to use the addition and multiplication properties of equality to solve equations. Students then learn how to use equations to solve word problems, including word problems that can be solved by writing a proportion. Finally, students connect inequalities and their graphs to real-world situations.
Subject:
Algebra
Mathematics
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Pearson
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Students work in pairs to critique and improve their work on the Self Check from the previous lesson.Key ConceptsTo critique and improve the task from the Self Check and to complete a similar task with a partner, students use what they know about solving equations and relating the equations to real-world situations.Goals and Learning ObjectivesSolve equations using the addition or multiplication property of equality.Write word problems that match algebraic equations.Write equations to represent a mathematical situation.
Subject:
Algebra
Material Type:
Lesson Plan
09/21/2015
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0.0 stars
Four full-year digital course, built from the ground up and fully-aligned to the Common Core State Standards, for 7th grade Mathematics. Created using research-based approaches to teaching and learning, the Open Access Common Core Course for Mathematics is designed with student-centered learning in mind, including activities for students to develop valuable 21st century skills and academic mindset.
Subject:
Mathematics
Material Type:
Full Course
Provider:
Pearson
10/06/2016
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Getting Started
Type of Unit: Introduction
Prior Knowledge
Students should be able to:
Understand ratio concepts and use ratios.
Use ratio and rate reasoning to solve real-world problems.
Identify and use the multiplication property of equality.
Lesson Flow
This unit introduces students to the routines that build a successful classroom math community, and it introduces the basic features of the digital course that students will use throughout the year.
An introductory card sort activity matches students with their partner for the week. Then over the course of the week, students learn about the routines of Opening, Work Time, Ways of Thinking, Apply the Learning (some lessons), Summary of the Math, Reflection, and Exercises. Students learn how to present their work to the class, the importance of students’ taking responsibility for their own learning, and how to effectively participate in the classroom math community.
Students then work on Gallery problems, to further explore the resources and tools and to learn how to organize their work.
The mathematical work of the unit focuses on ratios and rates, including card sort activities in which students identify equivalent ratios and match different representations of an equivalent ratio. Students use the multiplication property of equality to justify solutions to real-world ratio problems.
Subject:
Mathematics
Material Type:
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Provider:
Pearson
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Review the multiplication property of equality. Demonstrate the use of “ask myself” questions to understand a problem before solving it. Have students discuss the strategies that they can use when they feel stuck on a problem. Direct partners to solve a problem using a ratio table and equations, and then justify their solution in a presentation using the multiplication property of equality. Have each student write a Summary of the Mathematics in the lesson and work together to create a classroom summary.Key ConceptsStudents use the multiplication property of equality to justify their solution to a ratio problem.Goals and Learning ObjectivesBefore starting to work on a problem, make sense of the problem by using “ask myself” questions.Persevere in solving a problem even when feeling stuck.Solve a ratio problem using two different strategies.Link arithmetic and algebraic methods to solve a ratio problem.Use the multiplication property of equality to solve ratio problems
Subject:
Ratios and Proportions
Material Type:
Lesson Plan | 1,065 | 5,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-23 | latest | en | 0.880098 |
https://smart-answers.com/mathematics/question18872425 | 1,660,044,797,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570921.9/warc/CC-MAIN-20220809094531-20220809124531-00133.warc.gz | 489,767,072 | 34,981 | , 03.11.2020 02:20, ajime
# What is 15 plus 2000
### Other questions on the subject: Mathematics
Follow these steps using the algebra tiles to solve the equation −5x + (−2) = −2x + 4. 1. add 5 positive x-tiles to both sides and create zero pairs. 2. add 4 negative unit tiles to both sides and create zero pairs. 3. divide the unit tiles evenly among the x-tiles. x =
Mathematics, 21.06.2019 20:00, ismailear18
Anyone? 15m is what percent of 60m; 3m; 30m; 1.5 km?
Mathematics, 21.06.2019 22:50, sonyav732
Which of the following is closest to 32.9 x 7.5? a: 232 b: 259 c: 220 d: 265 | 215 | 584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2022-33 | latest | en | 0.785981 |
https://edukite.org/course/artificial-intelligence-for-robotics-su/ | 1,628,148,785,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046155458.35/warc/CC-MAIN-20210805063730-20210805093730-00014.warc.gz | 234,659,379 | 24,716 | • No products in the cart.
Have you ever dreamt of understanding how an autonomous car works?
In this course you will learn how to program all the major systems of a robotic car. This class will teach you basic methods in Artificial Intelligence, including: probabilistic inference, planning and search, localization, tracking and control, all with a focus on robotics. At the end of the course, you will leverage what you learned by solving the problem of a runaway robot.
### Assessment
This course does not involve any written exams. Students need to answer 5 assignment questions to complete the course, the answers will be in the form of written work in pdf or word. Students can write the answers in their own time. Each answer needs to be 200 words (1 Page). Once the answers are submitted, the tutor will check and assess the work.
### Certification
Edukite courses are free to study. To successfully complete a course you must submit all the assignment of the course as part of the assessment. Upon successful completion of a course, you can choose to make your achievement formal by obtaining your Certificate at a cost of £49.
Having an Official Edukite Certification is a great way to celebrate and share your success. You can:
• Show it to prove your success
Course Credit: Stanford University
### Course Curriculum
Localization Overview Introduction 00:04:00 Nanodegree Program 00:01:00 Localization 00:02:00 Total Probability 00:05:00 Uniform Probability Quiz 00:01:00 Uniform Distribution 00:01:00 Generalized Uniform Distribution 00:01:00 Probability After Sense 00:02:00 Compute Sum 00:01:00 Normalize Distribution 00:01:00 pHit and pMiss 00:00:00 Sum of Probabilities 00:01:00 Sense Function 00:01:00 Normalized Sense Function 00:01:00 Test Sense Function 00:01:00 Multiple Measurements 00:01:00 Exact Motion 00:01:00 Move Function 00:01:00 Inexact Motion 1 00:02:00 Inexact Motion 2 00:01:00 Inexact Motion 3 00:01:00 Inexact Move Function 00:01:00 Limit Distribution Quiz 00:01:00 Move Twice 00:01:00 Move 1000 00:01:00 Sense and Move 00:02:00 Sense and Move 2 00:01:00 Localization Summary 00:01:00 Formal Definition of Probability 1 00:01:00 Formal Definition of Probability 2 00:01:00 Formal Definition of Probability 3 00:01:00 Bayes’ Rule 00:03:00 Cancer Test 00:01:00 Theorem of Total Probability 00:02:00 Coin Flip Quiz 00:01:00 Two Coin Quiz 00:01:00 Assessment Submit Your Assignment 00:00:00 Certification 00:00:00
## 4.7
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1367 STUDENTS | 764 | 2,828 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-31 | longest | en | 0.841156 |
http://newtoyslist.com/rubik/rubix-building-products-how-to-build-up-a-rubiks-cube.html | 1,566,123,888,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027313747.38/warc/CC-MAIN-20190818083417-20190818105417-00362.warc.gz | 138,202,993 | 6,080 | 5 After all the labeling is completed, the cubes are put in their final packaging. This can be a small box that has an instruction booklet included or a plastic blister pack with a cardboard backing. The package serves the dual purpose of protecting the Rubik's cube from damage caused by shipping and advertising the product. The Rubik's cube packages are put into cases and moved to a pallet. The pallets are then loaded on trucks and the products are shipped all over the world.
### When you get round to building the Rubik’s Cube, you will find it is not as hard as it appears. The instructions are quite easy to follow and it will probably take you about fifteen minutes. When you get round to placing the colored tiles, pay attention to where they are supposed to go. Because once you snap them into place. you will not be able to remove them. Having said that. you can still use the Rubik’s Cube. What you will not be able to do is follow the instruction guide and solve the puzzle.
If you have 2 adjacent well permuted corners- turn the upper face once clockwise (U). That move will reposition the corners into a situation which only one well permuted corner will remain while the other three corners needed to be rotated counter-clockwise. Now just execute the algorithm above, and by this single execution you actually completed this step (remember to execute this algorithm from the correct angle – when the well permuted corner is on the back right. see algorithm image above).
The Rubik's cube appears to be made up of 26 smaller cubes. In its solved state, it has six faces, each made up of nine small square faces of the same color. While it appears that all of the small faces can be moved, only the corners and edges can actually move. The center cubes are each fixed and only rotate in place. When the cube is taken apart it can be seen that the center cubes are each connected by axles to an inner core. The corners and edges are not fixed to anything. This allows them to move around the center cubes. The cube maintains its shape because the corners and edges hold each other in place and are retained by the center cubes. Each piece has an internal tab that is retained by the center cubes and trapped by the surrounding pieces. These tabs are shaped to fit along a curved track that is created by the backs of the other pieces. The central cubes are fixed with a spring and rivet and retain all the surrounding pieces. The spring exerts just the right pressure to hold all the pieces in place while giving enough flexibility for a smooth and forgiving function. Rubiks Build It Solve It
Maybe all it takes to solve a Rubik's Cube is to see how one is made, and that's what kids get to do with the Rubik's Build It, Solve It kit. It comes with all the pieces, tools, and instructions kids need to build their very own Rubik's Cube. Once built, there is a 10-page instruction booklet that guides kids through solving a Rubik's Cube. From identifying the parts of a Rubik's Cube to solving basic Rubik's puzzles, kids are given a slow introduction on how to use their Rubik's Cube and progress to harder and harder challenges.
Español: hacer patrones con el cubo de Rubik, Français: faire des formes originales avec votre Rubik’s Cube, Português: Fazer Padrões Incríveis Usando um Cubo Mágico, Deutsch: Mit einem Zauberwürfel beeindruckende Muster machen, Italiano: Creare Fantastiche Composizioni Sul Cubo Di Rubik, Русский: сделать замысловатый узор кубика Рубика, Bahasa Indonesia: Membuat Pola Kubus Rubik yang Keren, Nederlands: Gave patronen maken op een Rubiks kubus
Rubik’s Build It Solve It is very similar to the traditional Rubik’s cube, but with a slight twist. With this cube you get all of the tools and bit’s and pieces along with the instructions. This is all you will need in order you you to build a Rubik’s Cube of your own. Once you have fitted the cube together, an instruction booklet has been included, and it is 10-pages in length. Finally, you will learn after all these years the process of solving Rubik’s Cube. You will be shown everything from identifying the parts of the cube along with solving basic puzzles. When you buy this toy, your children will be shown a slow and steady way of using the cube. They will then progress further with the toy to learn even harder challenges.
The Rubik's cube appears to be made up of 26 smaller cubes. In its solved state, it has six faces, each made up of nine small square faces of the same color. While it appears that all of the small faces can be moved, only the corners and edges can actually move. The center cubes are each fixed and only rotate in place. When the cube is taken apart it can be seen that the center cubes are each connected by axles to an inner core. The corners and edges are not fixed to anything. This allows them to move around the center cubes. The cube maintains its shape because the corners and edges hold each other in place and are retained by the center cubes. Each piece has an internal tab that is retained by the center cubes and trapped by the surrounding pieces. These tabs are shaped to fit along a curved track that is created by the backs of the other pieces. The central cubes are fixed with a spring and rivet and retain all the surrounding pieces. The spring exerts just the right pressure to hold all the pieces in place while giving enough flexibility for a smooth and forgiving function. Rubix Building Solutions
Do you remember those complicated little Rubik’s block that we would sit there trying to figure out for what seems like hours? Did any of you guys/girls ever solve them? Maybe all it takes for us to solve the “cube” would be for us to see what it’s all about. While there are many mesmerizing toys that are about to emerge into our world, today, we would like to take a close look at the Rubik’s Build It, Solve It, because we believe this is the one-way ticket to finally solving the cube! Rubix Building Solutions
Whether you complete all 6 stages or 1, be sure to tell your teacher about this program so all your classmates can solve with you! Teachers from all over the country use our program, at no cost, to teach their classes not only to solve, but math, art, science, and more. Hundreds of schools compete at solving cubes as a group and classes create really cool mosaic designs too. We even have ongoing mosaic contests each year. So check out our site and learn how you can do even more with a Rubik's® Cube!
```You can find assembly instructions for the BrickPi3 here. We will need to assemble the case, attach the BrickPi3, the Raspberry Pi, the Raspberry Pi Camera, add an SD Card, and add batteries. To make the software easier to setup, Raspbian for Robots comes with most of the software you will need already setup. You will need at least an 8 GB SD Card, and you will want to expand the disk to fit the full size of the SD Card.
```
Alright, let us be clear here. Do you remember how aggravated you used to get when trying to do the traditional Rubik’s Cube. Did you ever try to solve it, remember how frustrating it could be. Well, let me tell you that the Rubik’s Build It Solve building kit offers you a behind-the-scenes look. Winning Moves even give you a 10-page instruction manual. But this does not mean you could be solving the puzzle within seconds like a professional.
If you have 2 adjacent well permuted corners- turn the upper face once clockwise (U). That move will reposition the corners into a situation which only one well permuted corner will remain while the other three corners needed to be rotated counter-clockwise. Now just execute the algorithm above, and by this single execution you actually completed this step (remember to execute this algorithm from the correct angle – when the well permuted corner is on the back right. see algorithm image above). Rubiks Build It Solve It Instructions
This Rubik’s Build It, Solve It kit is for one player – it is recommended for ages 8 and up. As we said before, it is great for children and adults that don’t mind trying to figure out how the cube works – it’s great for those that enjoy putting puzzles together. This kit right here is going to give an inside look on how the cube works and how it’s put together. Plus, you’ll receive some tips in the instruction manual on how to solve it.
Hi, your aluminum cube is beautiful. My 6 yr old son has asked me to help him make a cube. So I went to the Home Depot today and looked around the plumbing and hardware sections. I only found some PVC pipe attachments but nothing with 6 knobs. I also found some nuts and bolts but I don't know which kind to get. Could you give me some idea of what to buy? I don't have very many tools at home but I do have a drill if I needed to drill through the PVC pipe and copper pipe cutters if I needed to use metal piping. Please help.
Keeping white on top, turn the cube so that a different colour face is toward you. Follow the above instructions again. Repeat with the other two faces until the white cross is complete. This step is quite intuitive; you can do it for sure but it does take a little practice. Just move the white edges to their places not messing up the ones already fixed. Rubiks Build It Solve It Review
Puzzle makers have been creating problems for people to solve for centuries. Some of the earliest puzzles date back to the time of the ancient Greeks and Romans. The Chinese have a ring puzzle that is thought to have been developed during the second century A.D. This was first described by Italian mathematician Girolamo Carolano (Cardan) in 1550. When the printing press was invented, complete books of mathematical and mechanical problems designed specifically for recreation were circulated. Rubix Build | 2,165 | 9,766 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-35 | latest | en | 0.962849 |
cloanc.com | 1,726,553,576,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00230.warc.gz | 143,491,510 | 18,977 | # California Proposition 19 Benefit Calculator
There are really just a couple options when it comes time to make a decision on the disposition of real estate at the time of distribution. Sell the property and distribute the proceeds or one (or more) of the beneficiaries can keep the property.
If the person(s) that keeps the property is the son or daughter of the deceased, Proposition 19 allows for a transfer of the property taxes paid by the parents up to the property’s full base year value plus \$1,022,600 (adjusted up from \$1,000,000 on 2/16/2023) if the following conditions are met:
1. The property was the primary residence of the surviving spouse at the time of death.
2. The property will be the primary residence of one of the children taking the property at distribution.
3. The other beneficiaries are also children of the deceased (though one may be able to get all of their siblings share even if there are non-children beneficiaries).
Proposition 19 also allows for anyone over the age of 55 to transfer their current primary residence property tax base to a replacement primary residence in California. This allows a child beneficiary to take their parents’ property at distribution and sell the property. One has 12 months to move into the property. The parent’s property tax base can then be applied to a new property as long as it is owner occupied. If the last surviving spouse died before February 15, 2021, non-owner occupied properties may qualify under Proposition 58.
In order to qualify for the Proposition 19 exclusion for reassessment of property taxes and take advantage of your parents’ low Base Year Value, the distribution of the trust or estate must be equal. Each sibling must get the same amount whether it is cash, equity or other assets. As long as each child beneficiary gets the same value, the child taking the property almost always can get full benefit of the parents’ low Base Year Value. Many times the trust or estate will not have enough cash or other assets to make an equal distribution. In these situations the trust or estate can borrow the money from a third party (not the person taking the property) and use the loan proceeds to pay off the other siblings share. This may or may not be a great solution. The calculator below will show you how long it will take for the property tax savings to cover the costs of a loan from a third party. One needs to be fairly certain they will live in the home longer than the time it takes to break-even on third party loan costs or have a plan to transfer the low tax base to another property after the sale of the parents’ home.
# PROPOSITION 19 BENEFIT CALCULATOR
PLEASE MODIFY THE FIELDS SHOWN IN RED TO CALCULATE YOUR ESTIMATED BENEFIT
This calculator is for estimation purposes only. The trust loan benefit calculator should be used as a tool to determine if it makes sense to move forward. Exact property tax savings should be discussed with your attorney. Another factor to consider is the interest paid on the outstanding loan balance. Most of our borrowers payoff or pay down our loan immediately keeping the interest costs to a minimum. Borrowers paying off our loan with a refinance may pay interest on a larger balance which could affect the break-even point calculation. If you are interested in taking advantage of the California Proposition 19 Exclusion from property tax reassessment on an inherited home, please call us at 877-464-1066. We are always available to answer any questions you have and to help you begin the financing process. | 730 | 3,561 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.932449 |
https://software.intel.com/content/www/us/en/develop/documentation/mkl-developer-reference-fortran/top/appendix-b-routine-and-function-arguments/matrix-arguments.html | 1,596,745,497,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00543.warc.gz | 460,521,053 | 44,065 | ## Developer Reference
• 2020.2
• 07/15/2020
• Public Content
Contents
# Matrix Arguments
Matrix arguments of the
Intel®
Math Kernel Library
routines can be stored in
either one- or two-dimensional
arrays, using the following storage schemes:
Full storage
is the simplest scheme.
A matrix
A
is stored in a two-dimensional array
a
, with the matrix element
a
ij
stored in the array element
a
(
i
,
j
)
. , where
lda
is the leading dimension of array
a
.
If a matrix is triangular (upper or lower, as specified by the argument
uplo
), only the elements of the relevant triangle are stored; the remaining elements of the array need not be set.
Routines that handle symmetric or Hermitian matrices allow for either the upper or lower triangle of the matrix to be stored in the corresponding elements of the array:
if
uplo
='U'
,
a
ij
is stored as described for
i
j
, other elements of
a
need not be set.
if
uplo
='L',
a
ij
is stored as described for
j
i
, other elements of
a
need not be set.
Packed storage
allows you to store symmetric, Hermitian, or triangular matrices more compactly: the relevant triangle (again, as specified by the argument
uplo
) is packed by columns in a one-dimensional array
ap
:
if
uplo
='U',
a
ij
is stored in
ap
(
i
+
j
(
j
- 1)/2)
for
i
j
if
uplo
='L'
,
a
ij
is stored in
ap
(
i
+ (2*
n
-
j
)*(
j
- 1)/2)
for
j
i
.
In descriptions of LAPACK routines, arrays with packed matrices have names ending in
p
.
Band storage
is as follows: an
m
-by-
n
band matrix with
kl
non-zero sub-diagonals and
ku
non-zero super-diagonals is stored compactly in
a two-dimensional array
ab
with
kl
+
ku
+ 1 rows and
n
columns. Columns of the matrix are stored in the corresponding columns of the array, and diagonals of the matrix are stored in rows of the array
. Thus,
a
ij
is stored in
ab
(
ku
+1+
i
-
j
,
j
) for max(1,
j
-
ku
i
min(
n
,
j
+
kl
).
Use the band storage scheme only when
kl
and
ku
are much less than the matrix size
n
. Although the routines work correctly for all values of
kl
and
ku
, using the band storage is inefficient if your matrices are not really banded.
The band storage scheme is illustrated by the following example, when
m
=
n
= 6,
kl
= 2,
ku
= 1
Array elements marked * are not used by the routines:
When a general band matrix is supplied for LU factorization , space must be allowed to store
kl
additional super-diagonals generated by fill-in as a result of row interchanges. This means that the matrix is stored according to the above scheme, but with
kl
+
ku
super-diagonals. Thus,
a
ij
is stored in
ab
(
kl
+
ku
+1+
i
-
j
,
j
)
for
max(1,
j
-
ku
i
min(
n
,
j
+
kl
)
.
The band storage scheme for LU factorization is illustrated by th | 806 | 2,689 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2020-34 | latest | en | 0.832668 |
http://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq?hide_answers=1&beginning=16380 | 1,369,514,690,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368706298270/warc/CC-MAIN-20130516121138-00056-ip-10-60-113-184.ec2.internal.warc.gz | 312,780,117 | 13,426 | # Questions on Algebra: Linear Equations, Graphs, Slope answered by real tutors!
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Algebra: Linear Equations, Graphs, Slope Solvers Lessons Answers archive Quiz In Depth
Question 443123: Solve the following systems of equation 2x + y =11 3x + 4y =4 Click here to see answer by swincher4391(726)
Question 443187: find the slope and y intercept of the line in the given equation. y=-9/13x-7 Click here to see answer by swincher4391(726)
Question 443184: Find x and y intercepts and graph the equation plotting the intercepts, show me how you do this step by step please. Create graph. 5x -6y = -45 Click here to see answer by ankor@dixie-net.com(15661)
Question 443486: I need to know how to find the slope of an equation like 9x-12y=8? Click here to see answer by rfer(12670)
Question 443483: i need help learning how to work this problem: y=-4/3x+5 Click here to see answer by MathLover1(6638)
Question 443523: Find the slope if it exists of the straight line containing the pair of points (4,5)and(9,-9) Click here to see answer by chriswen(106)
Question 443637: the general form of a quadratic equation is y= ax +bx+c 1 .draw the graph with equation y= x -3x-4 for value of x from -2 to +5. 2. write the down the minmum value of y and the value of x for this point 3. label the line of symmetry. Click here to see answer by rwm(914)
Question 443630: What’s the equation of a line that passes through points (0, -1) and (2, 3)? A. y = 2x − 1 B. y = 2x + 1 C. x = 2y − 1 D. x = 2y + 1 Click here to see answer by rwm(914)
Question 440276: Find the general equation of the line: y = (2)/(3) x - 7 a)2x-3y=21 b)2x+3y=21 c)2x-3y=-21 d)2x+3y=-21 Click here to see answer by rwm(914)
Question 443712: The cost y in dollars, for renting a car from a leasing firm is given by y = 0.6x + 45, where x is the distance driven, in miles. Graph this equation. Click here to see answer by josmiceli(9697)
Question 443740: Hi. I'm stuck on a problem: Find the slope of each line whose equation is given. If the slope is undefined, state this. x = −7 Thank you for your help! Click here to see answer by scott8148(6628)
Question 443763: Complete the table equation value of m value of b slope y intercept y=1 Click here to see answer by rwm(914)
Question 443781: I have 3 questions: 1) I= nE/(nr+R) (solve for n) 2) 1/a + 1/b = c (solve for a). I think the LCD is ab. To isolate a, I need to divide both sides by bc. Then I get lost! 3) Using the quadratic formula, solve 5x^2+8x+2=0. a=5, b=8, c=2; plug them into the quadratic formula -b plus or minus the square root of(b^2-4ac)divided by 2a. I am having a hard time with the steps for the answer because I looked up the answer but can't seem to figure the steps correctly. I beleive the answer to be (-4 plus or minus the square root of 6)divided by 5. But I keep coming up with (-8 plus or minus the square root of 24)divided by 10. I would really appreciate your help! Thanks so much. Click here to see answer by edjones(7569)
Question 443872: Which is an equation for the line that contains the points (-2, 3) and (2, -1)? Click here to see answer by josmiceli(9697)
Question 443870: one number is 15 more than another. the sum of twice the larger number and 3 times the smaller number is 180. whats the numbers? Click here to see answer by chriswen(106)
Question 444168: write the equation -4x + 2y = 1 in slope-intercept form. give the slope and the y-intercept Click here to see answer by mananth(12270)
Question 444167: graph the equation 4x - 3y = 12 Click here to see answer by mananth(12270)
Question 444158: find the x- and y- intercepts for the graph of of the equation 5x-3y=15 Click here to see answer by mananth(12270)
Question 444242: I am not sure how to resolved this math problem: Find points of intersection for 2x-4y>6 3x+2y<6 (Solve by graphing and state 1 solution and 1 non-solution) Thank you very much for your help Elik Click here to see answer by benni1013(197)
Question 444242: I am not sure how to resolved this math problem: Find points of intersection for 2x-4y>6 3x+2y<6 (Solve by graphing and state 1 solution and 1 non-solution) Thank you very much for your help Elik Click here to see answer by stanbon(57407)
Question 444242: I am not sure how to resolved this math problem: Find points of intersection for 2x-4y>6 3x+2y<6 (Solve by graphing and state 1 solution and 1 non-solution) Thank you very much for your help Elik Click here to see answer by MathLover1(6638)
Question 444510: y=-x+7, y-x=3 find out if the equation lines are perpendicular Click here to see answer by chriswen(106)
Question 444520: Given that f(x)=x^2-3x+1 and g(x)=sqrt(4-x), find the following if it exist. (f+g)(3) Click here to see answer by solver91311(16897)
Question 444602: trying to understand how to do this? my teacher isnt giving us notes and i dont know how to do it. give 3 solutionsfor the following equations. y = x + -8 Click here to see answer by stanbon(57407)
Question 444698: Write the equation of the line which passes through (2, 1) and is perpendicular to x = Click here to see answer by oberobic(2304)
Question 444723: Sketch a line passing through the point (2, 1) and having slope -1/2 Click here to see answer by stanbon(57407)
Question 444719: Find the slope–intercept form for the line that satisfies the given condition: Slope -4/3, y-intercept -5 Click here to see answer by oberobic(2304)
Question 444779: WRITE Y-3 = 1/2 (X + 4)IN SLOPE INTERCEPT FORM Click here to see answer by rfer(12670)
Question 444813: I need to find the equation of the line containg (-2, -7) and (-5, -8) Click here to see answer by mananth(12270)
Question 444834: given the linear equation y=2x-4 find the y corrdinates of the points (0) (-1)and (2) plot these points and graph on linear equation Click here to see answer by stanbon(57407)
Question 444882: find the slope of the line containing the given pair of points (1,-9)and (-2,-3) if line is undefined state so. Click here to see answer by stanbon(57407)
Question 445012: Solve for y: -3x + 2Y = -12 Seems too easy, just move everything except the y to the right side of the equation. What am I missing? Thanks! Click here to see answer by Alan3354(30993)
Question 445007: Find the slope of the line containing the given pair of points. If the line is undefined, state so. (-4, -5) and (-4, 5) I'm needing a little help with this particular problem. Also, would you mind showing how you arrived to your answer, so that I'll understand it. Thank you so much. Click here to see answer by rwm(914)
Question 444883: A home's back yard drops 9 feet over a horizontal distance of 12ft. find slope of this back yard. Round answer to two decimals Click here to see answer by rwm(914)
Question 444828: Find the value of x so the line that passes through (x, 5) and (12, 2) is perpendicular to the line that passes through (0, 1) and (2, 7). Click here to see answer by rwm(914) | 2,131 | 7,271 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2013-20 | latest | en | 0.85636 |
https://codegolf.stackexchange.com/questions/195535/currency-exchange | 1,585,870,167,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370508367.57/warc/CC-MAIN-20200402204908-20200402234908-00044.warc.gz | 401,472,744 | 29,936 | # Currency exchange [duplicate]
You are starting up a cryptocurrency exchange website which supports conversion from USD to two currencies, foo-coin and bar-coin. Write a program that takes the exchange rate for each coin to USD as arguments and outputs the maximum value of USD that cannot be completely divided into the two currencies (assuming these crypto-coins can only be purchased as full coins):
e.g. If the two exchange rates are $23 and$18, then 41 dollars could be split into 1 foo-coin and 1 bar-coin, but there is no way to completely split 42 dollars between the two coins. The output of the program for these two values is 373, as that is the largest dollar amount that CANNOT be divided into whole values of foo-coin and bar-coin.
Winner is the program with the fewest characters.
edit: You may assume the GCD of the two rates is 1
• Can you add more test cases? I am not sure I even understand the question. What does "cannot be completely divided into the two currencies" mean in this case? How is 373 "completely divided" into foo-coin and bar-coin? – Wisław Nov 8 '19 at 19:48
• @wislaw does the updated description clarify things well enough? – Ryan Burrow Nov 8 '19 at 19:55
• This is a simpler version of this challenge. – Arnauld Nov 8 '19 at 19:57
• What's the expected output if $\gcd(x,y)\neq 1$? Or is it guaranteed that it won't happen? – Arnauld Nov 8 '19 at 20:04
• The title of exchange rates is confusing. It made me think the challenge would be about converting an amount in one currency to another given the exchange rate. – xnor Nov 8 '19 at 20:09
# 05AB1E, 3 bytes
<P<
Try it online!
This is of course only assuming that the greatest common denominator of the inputs (exchange rates) is 1, because otherwise this number is not well-defined.
Explanation:
The number we want to compute is called the Frobenius Number and can be calculated with the formula $$\\operatorname{Frob}(a,b) = (a-1)(b-1) - 1\$$.
< | Subtract 1 from each input
P | Take their product
< | Subtract 1
# Swift, 42 bytes
func d(f:Int,b:Int)->Int{return f*b-(f+b)}
Try it online!
• Wouldn't f*b-f-b be 2 bytes shorter? – Arnauld Nov 8 '19 at 22:37
# Pyth, 7 bytes
t*thQte
Try it online!
As @Wisław pointed out, this is simply the Frobenius Number. Also uses (a-1)(b-1)-1, as the alternate ab-a-b is a bit longer.
## How it works
*thQ - Multiply the first number minus one
te - By the second number (the Q is implicit) minus one
t - Minus one | 678 | 2,487 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.28125 | 3 | CC-MAIN-2020-16 | latest | en | 0.932565 |
http://math4finance.com/general/99-pointsthree-basketball-teams-measured-the-height-of-each-player-on-their-team-the-table-shows-the-mean-and-the-mean-absolute-deviation-of-the-heights-for-each-team-which-statements-are-correct-select-each-correct-answer-a-the-heights-of | 1,718,504,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00843.warc.gz | 19,698,090 | 7,512 | Q:
# 99 POINTSThree basketball teams measured the height of each player on their team. The table shows the mean and the mean absolute deviation of the heights for each team. Which statements are correct? Select each correct answer. A.) The heights of the Panthers’ players and those of the Warriors vary about the same amount. B.) The heights of the Buffaloes’ players and those of the Panthers vary about the same amount. C.) The heights of the Warriors’ players vary more than do the Panthers’ heights. D.)The heights of the Buffaloes’ players vary more than do the Warriors’ heights.
Accepted Solution
A:
The answers are B and D hope this helps. | 144 | 651 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-26 | latest | en | 0.934235 |
https://www.wtamu.edu/~cbaird/sq/mobile/2013/04/09/can-you-make-a-shock-wave-of-light-by-breaking-the-light-barrier-just-like-supersonic-airplanes-break-the-sound-barrier/ | 1,726,207,575,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00214.warc.gz | 995,671,760 | 3,684 | # Can you make a shock wave of light by breaking the light barrier just like supersonic airplanes break the sound barrier?
Category: Physics
Published: April 9, 2013
By: Christopher S. Baird, author of The Top 50 Science Questions with Surprising Answers and physics professor at West Texas A&M University
Yes and no. It depends on the material you are in. In order to keep things from traveling into the past, and thereby preserve local conservation of mass-energy, and thereby prohibit the universe from exploding in an instant, nothing can travel faster than the speed of light in vacuum. Supersonic airplanes break the sound barrier by flying faster than the speed of sound. This is possible because sound is just a traveling vibration of air molecules. As airplanes approach the speed of sound, their sound waves pile up into a wall of air pressure that shatters apart weak airplanes. Airplanes that are strong enough can poke through this wall of air pressure and create a shock wave that trails behind them. When this sonic shock wave passes ground observers, we hear it as a sonic boom. By analogy, if a space ship traveled faster than the speed of light, it would create a shock wave made entirely of light. The problem is that nothing can go faster than the speed of light in vacuum, so a space ship can never go fast enough to break the light barrier. It's not a question of engineering, but of fundamental physics. As an object approaches the speed of light, it takes an increasing amount of energy to accelerate. It would take literally an infinite amount of energy for a space ship to exactly reach the speed of light in vacuum. This fact is verified every day in particle accelerators. Using the electrical power consumption of a small town, particle accelerators channel that energy into getting a handful of tiny particles like protons and electrons traveling very close to the speed of light in vacuum. Every year, particle accelerators are being improved to reach ever higher speeds. For instance, one decade the record was that the particles were traveling at 99.99% the speed of light in vacuum, and then the next decade an improved machine reached 99.999% the speed of light in vacuum, and the following decade saw a record of 99.9999% (these numbers are for illustration purposes only and are not exact). The current record is held by the LHC which has accelerated protons to 99.999997% the speed of light in vacuum. A ship in space can't ever reach the speed of light in vacuum, and therefore can't ever break the light barrier and create an optical shock wave.
Everything discussed so far applies only to vacuum. The situation gets interesting when you are not in vacuum. When traveling through a material such as water or glass, the speed of light in that material is significantly slower than the speed of light in vacuum. An object can go faster than the speed of light in a material without breaking any fundamental laws. And as you would expect, an object traveling through a material at a speed that is faster than the light in that material does indeed create an optical shock wave. This shock wave of light is known as Cherenkov radiation. In order for the effect to be noticeable to the naked eye, the object has to be traveling very quickly through a fairly dense material. For instance, nuclear reactors spit out electrons at very high speeds as by-products of the nuclear reaction. These high-speed electrons travel through the cooling water faster than the speed of light in the water, and therefore create shock waves of light. This Cerenkov radiation is observed as an eerie blue glow.
Even though air is very close to being a vacuum, it is not exactly a vacuum. The speed of light in air is slightly less than the speed of light in vacuum, so objects can actually break the light barrier in air without going faster than the fundamental limit of the speed of light in vacuum. It is just much harder to get Cherenkov radiation in air than in water because higher speeds are required. But it does happen. Supernovae shoot out particles at amazingly high speeds (but always less than the speed of light in vacuum). When these particles hit earth's atmosphere, they are known as cosmic rays. Some of these cosmic rays and their by-products are indeed fast enough to create optical shock waves as they travel through the air. Although the atmospheric Cherenkov radiation from cosmic rays is too faint to see with the naked eye, it is detected by cameras and serves as an important tool for scientists studying cosmic rays. | 918 | 4,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2024-38 | latest | en | 0.934102 |
https://bestmaths.net/online/index.php/year-levels/year-9/year-9-topics/probab/faq/ | 1,657,124,890,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104675818.94/warc/CC-MAIN-20220706151618-20220706181618-00571.warc.gz | 160,744,364 | 3,874 | ## Probability FAQ
1. What is the difference between an experiment and a trial?
An experiment could be rolling a die 10 times, whereas each of the 10 rolls would be a trial.
2. Can a probability be 0?
Yes. An event with a probability of 0 cannot happen.
3. What is the highest probability?
The highest probability is 1.
If the probability is 1, an event is certain to happen | 98 | 381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2022-27 | latest | en | 0.82287 |
https://ksusentinel.com/2021/05/26/the-magic-of-benfords-law-05-26-2021-marcelo-viana/ | 1,719,132,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00190.warc.gz | 307,918,072 | 30,767 | # The Magic of Benford’s Law – 05/26/2021 – Marcelo Viana
The astronomer Simon Newcomb discovered in 1881 that it is much more common in astronomical data that the leading digit (the left one) is small (1, 2, …) than large (9, 8, “…). that the number begins with a digit d corresponds roughly to the decimal logarithm of (1 + 1 / d): For d = 1 this is 30.1%, for d = 9, however, only 4.6%.
This surprising fact was rediscovered in 1937 by the physicist Frank Benford, who pointed out that this is also the case with the most diverse types of data: city population, numbers in this edition of Folha, cases of Covid in different countries and states, sizes of volcanoes, time intervals between Heartbeats, points in basketball games, etc.
There are exceptions when the data is artificial (cell counts in Rio de Janeiro always start with 9) or vary within a limited range (adult height almost always starts with 5 or 6). However, it is widely confirmed that the vast majority of natural data follows Benford’s Law, as I outlined in the last issue of this column. This means that real data can be distinguished from false or fraudulent data.
One request concerns the examination of tax returns: if the claim is genuine, the amounts must comply with Benford law, so any discrepancy is an indication that the claim falls into the fine net and is carefully analyzed. The IRS and its counterparts in other countries do not disclose their methods, so we ignore how the law is actually applied. But it’s a free tool and it’s very easy to use.
I know, dear reader, that must seem very naive to you: surely a good cheater is smart enough to “boil” his values according to Benford’s Law, right? Well, it’s not that easy because the law has a property called scale invariance. In practice this means that regardless of the unit used, the law must always apply!
Even if the declaration data is well “made” in reais, the IRS can convert it, for example, into Japanese yen, Swiss francs or Indian rupees. The numbers are completely different but still have to comply with Benford’s Law: discrepancies in any of these currencies are suspicious evidence. It’s complicated now, isn’t it?
And Benford’s Law has other notable properties that make it even more difficult to deceive. Stay for next week.
PRESENT LINK: Did you like this column? The subscriber can release five free hits from each link per day. Just click the blue F below. | 557 | 2,424 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.94378 |
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# Le Monde Company's Allocation of joint production costs
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
Le Monde Company is a manufacturer of chemicals for various purposes. One of the processes used by Le Monde produces HTP-3, a chemical used in hot tubs and swimming pools; PST-4, a chemical used in pesticides; and RJ-5, a product that is sold to fertilizer manufacturers. Le Monde uses the net realized value method to allocate joint production costs. The ratio of output quantities to input quantities of direct material used in the joint process remains consistent from month to month. Le Monde Company uses FIFO (first in, first out) in valuing its finished goods inventory.
Data regarding operations for the month of October are as follows. During this month, Le Monde incurred joint production costs of \$1,360,000 in the manufacture of HTP-3, PST-4 and RJ-5.
HTP-3 PST-4 RJ-5
Finished goods inventory in gallons (oct 1) 18,000 52,000 3,000
October sales in gallons 650,000 325,000 150,000
October production in gallons 700,000 350,000 170,000
Additional processing costs \$699,200 \$652,800 \$48,000
Final sales value per gallon \$3.20 4.80 \$4.00
1. Determine Le Monde Company's allocation of joint production costs for the month of October. (Carry calculation of relative proportions to four decimal places)
2. Determine the dollar values of the finished goods inventories for HTP-3, PST-4 and RJ-5 as of October 31. (Round the cost per gallon to the nearest cent)
3. Suppose Le Monde Company has a new opportunity to sell PST-4 at the split off point for \$3.04 per gallon. Prepare an analysis showing whether the company should sell PST-4 at the split off point or continue to process this product further. | 441 | 1,815 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2019-09 | longest | en | 0.899695 |
http://www.pgrocer.net/CIS122/asgn/JSifasgn/JScodeasgn.html | 1,670,271,168,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711045.18/warc/CC-MAIN-20221205200634-20221205230634-00753.warc.gz | 84,832,837 | 1,110 | On all flowcharts, assume that as you look at the flowchart the Y goes to the right and the N goes to the left. All code should be written in JavaScript.
## Flowchart #1:
Problem #1: Write the code shown in flowchart #1 using simple if statements.
Problem #2: Write the code shown in flowchart #1 using compound if statements. Note that sometimes they are called complex instead of compound.
## Flowchart #2:
Problem #3: Write the code shown in flowchart #2 using simple if statements.
Problem #4: Could the code shown in flowchart #2 be written using a compound AND statement. Explain your answer.
## Flowchart #3:
Problem #5: Write the code shown in flowchart #3 using simple if statements.
Problem #6: Could the code shown in flowchart #3 be written using a compound AND statement? Explain your answer
## Flowchart #4:
Problem #7: Write the code shown in flowchart #4 using simple if statements.
Problem #8: Write the code shown in flowchart #4 using a compound if statements.
## Flowchart #5:
Problem #9: Write the code shown in flowchart #5 using simple if statements.
Problem #10: Write the code shown in flowchart #5 using compound if statements.
## Flowchart #6:
Problem #11: Write the code shown in flowchart #6 using simple if statements.
Problem #12: Write the code shown in flowchart #6 using compound if statements.
## Flowchart #7:
Problem #13: Write the code shown in flowchart #7 using simple if statements.
Problem #14: Could the code shown in flowchart #7 be written using a compound OR statement? Explain your answer.
## Flowchart #8:
Problem #15: Write the code shown in flowchart #8 using simple if statements.
Problem #16: Write the code shown in flowchart #8 using compound if statements.
## Flowchart #9:
Problem #17: Write the code shown in flowchart #9 using simple if statements.
Problem #18: Could the code shown in flowchart #9 be written using a compound statement? Explain your answer. | 454 | 1,943 | {"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} | 2.953125 | 3 | CC-MAIN-2022-49 | latest | en | 0.736998 |
https://www.clutchprep.com/chemistry/practice-problems/115982/a-voltaic-cell-with-ni-ni2-and-co-co2-half-cells-has-the-following-initial-conce | 1,611,687,482,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704803308.89/warc/CC-MAIN-20210126170854-20210126200854-00001.warc.gz | 711,715,468 | 39,012 | # Problem: A voltaic cell with Ni/Ni 2+ and Co/Co2+ half-cells has the following initial concentrations: [Ni2+] = 0.80 M; [Co 2+] = 0.20 M.(a) What is the initial Ecell?
###### FREE Expert Solution
Step 1. Write the two half-cell reactions and determine the half-cell potentials (refer to the Standard Reduction Potential Table)
Ni2+(aq) + 2 e- Ni(s) E° = -0.230 V
Co2+(aq) + 2 e- Co(s) E° = -0.277 V
Step 2. Identify the reduction half-reaction (cathode) and the anode half-reaction (anode)
Co2+(aq) + 2 e- Co(s) E° = -0.277 V ↓ E° → oxidation → anode
Ni2+(aq) + 2 e- Ni(s) E° = -0.230 V ↑ E° → reduction → cathode
Step 3. Get the overall reaction by balancing the number of electrons transferred then adding the oxidation half-reaction and reduction half-reaction.
lose electrons oxidation → anode
gain electrons reduction → cathode
82% (82 ratings)
###### Problem Details
A voltaic cell with Ni/Ni 2+ and Co/Co2+ half-cells has the following initial concentrations: [Ni2+] = 0.80 M; [Co 2+] = 0.20 M.
(a) What is the initial Ecell? | 349 | 1,162 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-04 | latest | en | 0.694356 |
https://docs.3di.live/b_infiltration.html | 1,632,849,080,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060877.21/warc/CC-MAIN-20210928153533-20210928183533-00623.warc.gz | 263,886,173 | 4,980 | # 3.2. Infiltration¶
3Di supports to methods for computing infiltration. The so-called simple infiltration and the Horton based infiltration. The simple infiltration uses an infiltration rate that is constant in time, while the Horton based infiltration is based on the concept of infiltration that reaches in time an equilibrium rate. In 3Di, the infiltration computed with the first method removes the water from the model. For the second method, a ground water layer is defined from which water can also exfiltrate in case the groundwater level reaches the surface level.
## 3.2.1. Simple Infiltration¶
Infiltration is the process of water slowly sinking into the soil. The infiltration rate depends on the type of soil and the land coverage. The infiltration rate is constant in time throughout the simulation. There are two exception for this: 1) the availability of water and when a user defines a maximum infiltration capacity.
In 3Di, the infiltration rate is defined for subgrid cell and defined in mm/day. The infiltration is computed per computational cell. Therefore, the infiltration rate per pixel is translated to values per computational cell and can depend on the water level.
When using the subgrid method, water start filling a cell from the lowest subgrid cell. When simulating rainfall run-off scenarios, the overall water depths are relatively small, therefore often only a small part of a computational cell is wet. If in this part of the cell the infiltration rate is low, the infiltration will be limited. However, in reality rain falls over a full spatial domain, and reaches the lowest areas only after a while, but might be infiltrated before it reaches that ares. Therefore, the total infiltration rate per computational can be made dependent of the rain. Besides defining the infiltration rates, a user defines also the infiltration_surface_option that can be found in the global settings table. In the Figure below, an overview is given of the various types.
Infiltration during rainfall acts on the whole surface within one computational cell (left). When rain stops, infiltration only acts on those pixels below the water level.
The first option, defines the infiltration rate in a cell based on the assumption that the full cell is wet. In other words, the infiltration rate in a cell is independent of the water level (left panel). The second option, defines the infiltration rate on the cumulative rates in the wet areas (middle panel). In this case the infiltration rate is dependent of the water level. The third option is the default option and is a combination of the two previous options. In this case the infiltration rate of the full cell domain is used when it is raining and only the infiltration rate of the wet area is used when it is dry (right panel).
In 3Di, many raster input variables are stored in tables. This yields as well for the infiltration rate, as this allows the infiltration rate to be dependent of the water level and for an efficient storage of the information. The table increment for the infiltration tables equals the 2D increment.
### 3.2.1.1. Maximum infiltration capacity¶
De maximum infiltration capacity is the maximum volume of water that can infiltrate during one simulation. The user defines a layer with a thickness per pixel in meters. The maximum infiltration per computational cell is the sum of within all pixels. It will stop infiltrating when the maximum infiltration is reached.
### 3.2.1.2. Input¶
The user can define all the aspects for simple infiltration in the simple_infiltration_table.
## 3.2.2. Horton based infiltration¶
The second possibility to add infiltration to a system, is infiltration based on the Horton equation. This describes a infiltration rate that is initially higher and decays to an equilibrium infiltration. This type of infiltration is always used in combination with a groundwater level. In such, the infiltration is always limited. More information about the Horton based infiltration can be found here Horton based infiltration. | 803 | 4,057 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-39 | longest | en | 0.910427 |
https://topic.alibabacloud.com/zqpop/commons-math_188489.html | 1,568,786,942,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573184.25/warc/CC-MAIN-20190918044831-20190918070831-00396.warc.gz | 698,057,426 | 10,274 | # commons math
Alibabacloud.com offers a wide variety of articles about commons math, easily find your commons math information here online.
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### Introduction and use of Apache Commonsmath
Apache Commons Math is a set of functions biased toward scientific calculation, mainly for linear algebra, mathematical analysis, probability and statistics.Although I graduated from mathematics, I was also holding the "mathematical analysis" gnawing, but for a long time, these concepts are beginning unfamiliar, write a little example, only for reference.Packagetest.ffm83.commons.math;Importorg.apache.commo
### Java implementation of the normal distribution function (commons-math-3.3)
1. The calculation process is: Extract the report raw data--> Calculate the financial ratio value by the above formula--> Group the ratio value according to the grouping rules above--> Assign corresponding values to each group--> based on each variable grouping assignment and weight, calculate the quantitative model score S1 by Formula (the formula is as follows)whichThe above-mentioned calculus formula is a normal distribution and is the standard normal distribution of the integral formula, cu
### Apache CommonsMath Library simple and practical
Packagetest.ffm83.commons.math;importorg.apache.commons.math3.linear.array2drowrealmatrix; Import org.apache.commons.math3.linear.LUDecomposition; Importorg.apache.commons.math3.linear.RealMatrix; Importorg.apache.commons.math3.stat.descriptive.moment.GeometricMean; Importorg.apache.commons.math3.stat.descriptive.moment.Kurtosis; Importorg.apache.commons.math3.stat.descriptive.moment.Mean; importorg.apache.commons.math3.stat.descriptive.moment.Skewness; Importorg.apache.commons.math3.stat.descri
### Math component usage example
Import org. Apache. commons. Math. Stat. descriptive. Moment. geometricmean;Import org. Apache. commons. Math. Stat. descriptive. Moment. kurtosis;Import org. Apache. commons. Math. Stat. descriptive. Moment. Mean;Import org. Apac
### Statistical tools in APACHE common
Package com. njs. math; Import org. apache. commons. math. stat. descriptive. moment. GeometricMean;Import org. apache. commons. math. stat. descriptive. moment. Kurtosis;Import org. apache. commons. math. stat. descriptive. momen
### Java Finance and mathematics
1, Commons-math/commons-lang-math The above two packets are under Apache, the former is more powerful than the latter, the latter have the function before, the latter mainly to solve some of the usual procedures in the basic mathematical calculations, mainly in the range of Judgments (*range), random number generation (jvmrandom,randomutils), Fractional processing (fraction), digital conversion, size judgm
### ) Overview of Apache commons
(). getzipcode ();If (zipcode. Equals ("90210 ")){Address = location. getaddress ();Break;}} 11. commons Lang Http://jakarta.apache.org/commons/lang/ Note: This Toolkit can be seen as an extension of Java. Lang. Provides tool classes such as stringutils, stringescapeutils, randomstringutils, tokenizer, and wordutils. 12. commons Logging Http://jakarta.apache.org/commons/logging/ Note: Do you know log4j? 13. commons
### Introduction to Apache commons
equivalentAddress = NULL;Collection locations = vendor. getlocations ();Iterator it = locations. iterator ();While (it. hasnext ()){Location = (location) it. Next ();String zipcode = location. getaddress (). getzipcode ();If (zipcode. Equals ("90210 ")){Address = location. getaddress ();Break;}} 11. commons Lang Http://jakarta.apache.org/commons/lang/ Note: This Toolkit can be seen as an extension of Java. Lang. Provides tool classes such as stringutils, stringescapeutils, randomstringutils, to
### Spark's Workcount
\commons-cli\commons-cli\1.2\commons-cli-1.2.jar; F:\repo\org\apache\commons\commons-math\2.1\commons-math-2.1.jar; F:\repo\xmlenc\xmlenc\0.52\xmlenc-0.52.jar; F:\repo\commons-configuration\commons-configuration\1.6\commons-configuration-1.6.jar; F:\repo\commons-collections\commons-collections\3.2.1\commons-collections
### Java EE Servlet several path
,tools.jar,commons-beanutils*.jar,commons-codec*.jar, commons-collections*.jar,commons-dbcp*.jar,commons-digester*.jar,commons-fileupload*.jar,commons-httpclient*. Jar,commons-io*.jar,commons-lang*.jar,commons-logging*.jar,commons-math*.jar,commons-pool*.jar,jstl.jar, Taglibs-standard-spec-*.jar,geronimo-spec-jaxrpc*.jar,wsdl4j*.jar,ant.jar,ant-junit*.jar,aspectj*.jar,jmx.jar, H2*.jar,hibernate*.jar,httpcli
### Apache Jackrabbit Oak 1.5.9 released
[OAK-4680]-Unify The usage of Commons Math 3 across the project[OAK-4691]-use utility backends from Oak-segment-tar in Oak-run[OAK-4697]-Optimize read of old node state[OAK-4703]-Update jackrabbit version to 2.13.2[OAK-4715]-Reduce Documentstore reads for local changes[OAK-4716]-Upgrade dependency to Oak-segment-tar to version0.0.10Task[OAK-4686]-Documentnodestoretest#compareonbranch fails forupdate.limit=1
### Open-source tools included in Apache commons
(). getzipcode ();If (zipcode. Equals ("90210 ")){Address = location. getaddress ();Break;}} Commons LangHttp://jakarta.apache.org/commons/lang/Note: This Toolkit can be seen as an extension of Java. Lang. Provides tool classes such as stringutils, stringescapeutils, randomstringutils, tokenizer, and wordutils. Commons LoggingHttp://jakarta.apache.org/commons/logging/Note: Do you know log4j? Commons mathHttp://jakarta.apache.org/commons/
### GroupId and Artifactid Explained
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(); if (Zipcode.equals ("90210")) { address = location.getaddress (); break; }}Xi. Commons Langhttp://jakarta.apache.org/commons/lang/Description: This toolkit can be seen as an extension to Java.lang. Tools such as StringUtils, Stringescapeutils, Randomstringutils, Tokenizer, wordutils are provided.12, Commons Logginghttp://jakarta.apache.org/commons/logging/Description: Do you know log4j? 13. Commons Mathhttp://jakarta.apache.org/common
### Get started with the HBase programming API put
-1.0.13.jar;d:\software\Hbase-1.2.3\lib\commons-digester-1.8.jar;d:\software\hbase-1.2.3\lib\commons-el-1.0.jar;d:\software\ Hbase-1.2.3\lib\commons-httpclient-3.1.jar;d:\software\hbase-1.2.3\lib\commons-io-2.4.jar;d:\software\ Hbase-1.2.3\lib\commons-lang-2.6.jar;d:\software\hbase-1.2.3\lib\commons-logging-1.2.jar;d:\software\ Hbase-1.2.3\lib\commons-math-2.2.jar;d:\software\hbase-1.2.3\lib\commons-math3-3
### Go About the Apache Commons toolset
(It.hasnext ()) {Location location = (location) it.next ();String ZipCode = location.getaddress (). Getzipcode ();if (Zipcode.equals ("90210")) {Address = location.getaddress ();Break}}Xi. Commons Langhttp://jakarta.apache.org/commons/lang/Description: This toolkit can be seen as an extension to Java.lang. Tools such as StringUtils, Stringescapeutils, Randomstringutils, Tokenizer, wordutils are provided.12, Commons Logginghttp://jakarta.apache.org/commons/logging/Description: Do you know log4j?
### Jar packet minimum static dependency implementation
; Slf4j_api_1_4_3; Slf4j_api_1_4_3-> Slf4j_log4j12_1_4_3; Slf4j_log4j12_1_4_3-> Slf4j_api_1_4_3; Slf4j_log4j12_1_4_3-> log4j_1_2_15; } Import Java.io.BufferedReader; Import Java.io.FileReader; Import Java.util.HashMap; Import Java.util.LinkedHashMap; Import Java.util.LinkedHashSet; Import Java.util.Map; Grph file analysis based on Jaranalyzer generated public class Getminjar {private static hashmap Only the least static-dependent jar pack is obtained, and the dynamically loaded progr
### SVN View the revision number of the specified release _SVN
Assuming that each release is tagged, you can run SVN Info/path/to/svn/tag and return the following information: [Plain] View plain copy Path: [Path] URL: [URL] Revision: [Current Revision] Node kind:directory last Changed Auth Or: [author] last Changed Rev: [Specify the revision number of the final path change] For example, suppose you want to get the revision number of Release 2.0 of Apache Commons Math
### Getting Started with Java Web programming--spring Boot project Build
tomcat.util.scan.StandardJarScanFilter.jarsToSkip ", * "As follows:#-Test JARs (JUnit, Cobertura and dependencies) Tomcat.util.scan.standardjarscanfilter.jarstoskip=bootstrap.jar, Commons-daemon.jar,tomcat-juli.jar,annotations-api.jar,el-api.jar,jsp-api.jar,servlet-api.jar, Websocket-api.jar,jaspic-api.jar,catalina.jar,catalina-ant.jar,catalina-ha.jar,catalina-storeconfig.jar, Catalina-tribes.jar,jasper.jar,jasper-el.jar,ecj-*.jar,tomcat-api.jar,tomcat-util.jar,tomcat-util-scan.jar, Tomcat-coyo
### Hadoop 2.5 HDFs Namenode–format error Usage:java namenode [-backup] |
/commons-beanutils-1.7.0.jar:/usr/hadoop-2.2.0/share/hadoop/common/lib/ Log4j-1.2.17.jar:/usr/hadoop-2.2.0/share/hadoop/common/lib/jackson-xc-1.8.8.jar:/usr/hadoop-2.2.0/share/hadoop /common/lib/commons-math-2.1.jar:/usr/hadoop-2.2.0/share/hadoop/common/lib/commons-logging-1.1.1.jar:/usr/ Hadoop-2.2.0/share/hadoop/common/lib/commons-httpclient-3.1.jar:/usr/hadoop-2.2.0/share/hadoop/common/lib/jaxb-impl-2.2.
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If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days. | 2,721 | 10,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2019-39 | latest | en | 0.665171 |
http://julesandjames.blogspot.com/2011/03/pigeons-vs-sciencebloggers-round-two.html | 1,469,929,064,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258948335.92/warc/CC-MAIN-20160723072908-00105-ip-10-185-27-174.ec2.internal.warc.gz | 149,185,434 | 25,280 | ## Friday, March 04, 2011
### Pigeons vs sciencebloggers, round two
And now for the follow-up.
The question was, if someone has two children, and tells you that (at least) one of them is a boy born on a Tuesday, what is the probability that the other child is also a boy? And the intended answer, of course, is 13/27. Why? Well, including the birth day (of week) as a variable, at the outset for a two-child family there are 142 = 196 possible and equiprobable pairs such as (B-Mon, G-Wed) where each child has a gender and birthday (of week) and the ordering denotes birth order. The vast majority of these pairs don't have a Tuesday boy and can be ignored. Of those that do, 7 of them look like (B-Tue, G-any) and another 7 are (G-any, B-Tue). Similarly, there are 7 pairs (B-Tue, B-any) and 7 (B-any, B-Tue). But wait! The case (B-Tue, B-Tue) has been counted twice! In fact there are only 13 cases where one child is a Tuesday boy and the other is also a boy. So that makes the probability 13/(13+14) = 13/27 where one child is a Tuesday boy and the other is also a boy.
However, there's a sting in the tail. Let's say you take a bunch of people with two children, and ask them if one of their children is a boy. Pick such a parent at random. We have already seen that in 1/3 of cases where the answer was yes, the other child will also be a boy. Now ask this parent for the day of the week that their son was born on (if they have a choice, they can flip a coin and pick one son at random, preferably out of sight so as not to give the game away). You get the reply..."Tuesday". Or some other day. Maybe they will say Saturday. In this case, hearing the day of the week on which a son was born does *not* change the probability that the other child is also a boy. So anyone who interpreted the original situation as meaning "a parent of two children, at least one of whom is a boy, tells you the day of the week on which their son (or a random son if applicable) was born, and their answer was `Tuesday'" would be completely justified in concluding that the probability of the other child being a boy was still 1/3. This solution was mentioned in More or Less on 11 June, and it was (re)listening to this old podcast (probably no longer available, but here's a related web page) that prompted me to blog it at last. What Tim Harford could have gone on to point out, but didn't (IIRC), is that even the original "two children" problem is typically under-determined in the way that it is presented. If a parent of two children is asked for the gender of one randomly-selected child, then irrespective of their reply, the probability of their other child being a boy (or alternatively, being the same gender as the one they gave) is...50%. So the solution to this problem also depends on how this "one child is a boy" parent is found.
Similar weaknesses can be found in most statements of the Monty Hall problem too. If all that is reported is the observed actions of the game show host on one occasion, then we don't really have enough information to generalise to a rigorous frequency-based calculation. Maybe Monty only opens another door on the occasions that the player originally picked the car, in which case swapping will lose. Maybe Monte opens a random door (and might expose the car)...in which case the player should still swap, and in fact the probabilities are unchanged from the original problem, in fact. Wikipedia discusses several alternative interpretations in some detail.
Contrary to Gary Foshee's statement on this web page, I don't think this sort of ambiguity is particularly controversial, it's just the result of trying to dress up a mathematical problem in natural-sounding (but slightly imprecise) English. When the problem is stated unambiguously, it's not particularly difficult. At least for pigeons.
And to any pigeons still reading, all I can say is: coo.
Ron Broberg said...
David B. Benson said...
Coo to you too.
Rattus Norvegicus said...
But Ron, are you the walrus?
Dallas said...
I love this problem! In the US, if the first child is a boy, that actual data shows that 52.2% of the time the second child is a girl.
http://www.in-gender.com/XYU/Odds/Gender_Odds.aspx
Even though there is a 51/49 Boy/Girl birth ratio. :)
James Annan said...
Dallas, think you might have misread that, the page seems to say the probability of the second child being a girl if the first is a boy is actually 50%, though I agree the ratios aren't usually so even...except in maths puzzles!
Dallas said...
Yes, the probability is 50% according to the link, the actual results, last table I think, showed the results of their survey as 52.2% chance of mixed (i.e. boy + girl) in a two child situation.
So my "first child boy" was misleading, though, there is a slightly greater probability of a first child boy (51/49 birth ratio).
Their actual results were very close to the normal birth ratio rather than the rounded 50/50 estimate.
So, I my opinion, the 13/27 prediction should actually be, 50/50, if you consider the historic birth ration :)
David B. Benson said...
So far the pigeons are batting 1000.
Dallas said...
Exactly :)
Steve Easterbrook said...
Shorter explanation: the problem isn't a probability problem at all, it's a sampling problem. And as the sampling method is never stated (as least not in any formulation I've ever seen), the answer cannot be determined.
Anonymous said...
You can take it one step further. Pick one parent from the same bunch of people with two children, and ask them to tell you the gender of one of their children. You get the reply "boy." The probability that this parent has two boys in indeed 1/2, not 1/3, for the exact same reason hearing "Tuesday" did not change the answer in your variation. If there are 100 parents in this group, 75 will *have* a boy in this group, but only 50 will *say* that they have a boy; 25 of those that could will say "girl." In either subgroup, 25 will have two boys, so the probability is different in the two cases.
Most versions of this question are ambiguous because they do not distinguish between those two cases. 1/3 (or 13/27) is correct only if you were looking for a boy (or a boy born on Tuesday). If it was a random fact, and the fact appropriate for either child could have been stated, the answer is always 1/2. And if you aren't told which is true, you can assume the latter cases (because it represents no bias), but not the first. | 1,548 | 6,474 | {"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} | 4.09375 | 4 | CC-MAIN-2016-30 | longest | en | 0.982964 |
http://www.chegg.com/homework-help/questions-and-answers/automobile-battery-charged-constant-current-2a-hours-terminal-voltage-battery-v-11-05t-t-0-q167926 | 1,394,605,420,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394021400132/warc/CC-MAIN-20140305121000-00007-ip-10-183-142-35.ec2.internal.warc.gz | 271,394,880 | 7,062 | energy and cost
0 pts ended
An automobile battery is charged with a constant current of 2A for five hours. The terminal voltage of the battery is v=11 +0.5t for t>0.
a. Find the energy delivered to the battery during the fivehours
b. If electrical energy costs 10 cents/k Wh, find the cost ofcharging the battery for five hours.
Answer to b. is 1.23 cents | 93 | 356 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2014-10 | latest | en | 0.917014 |
https://www.teachoo.com/10309/3063/NCERT-Question-19/category/NCERT-Questions/ | 1,679,778,487,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945372.38/warc/CC-MAIN-20230325191930-20230325221930-00126.warc.gz | 1,121,190,506 | 31,839 | NCERT Questions
Class 9
Chapter 11 Class 9 - Work and Energy
## Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
If there are several forces acting on the object, but net force is 0
Then acceleration will be 0
Since
Force = Mass × Acceleration
If Net Force = 0
Then, Acceleration = 0
Get live Maths 1-on-1 Classs - Class 6 to 12 | 115 | 418 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2023-14 | longest | en | 0.944687 |
https://ru.scribd.com/document/345307702/matlab3 | 1,610,948,164,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514121.8/warc/CC-MAIN-20210118030549-20210118060549-00622.warc.gz | 542,997,157 | 118,135 | Вы находитесь на странице: 1из 26
# x=[0:0.
01:10];
y=[x.^3-15*x.^2+66*x-80];
plot(x,y,':k')
grid on
hold off
x=[1 7 2 4 6 1];
y=[5 5 2 7 2 5];
plot(x,y)
grid on
x=[3 2 2 3 5 6 6 5];
y=[2 3 5 6 6 5 3 2];}
??? y=[2 3 5 6 6 5 3 2];}
|
{Error: Unbalanced or unexpected parenthesis or bracket.
}
x=[3 2 2 3 5 6 6 5];
y=[2 3 5 6 6 5 3 2];
plot(x,y)
grid on
x=[3 2 2 3 5 5 6 6 5 3];
y=[6 5 3 2 6 2 3 5 6 6];
plot(x,y)
grid on
x=[3 2 2 3 5 5 6 6 5 3];
y=[6 5 3 2 6 2 3 5 6 6];
plot(x,'r',y 'y')
??? plot(x,'r',y 'y')
|
{Error: Unexpected MATLAB expression.
}
x=[3 2 2 3 5 5 6 6 5 3];
y=[6 5 3 2 6 2 3 5 6 6];
plot(x,'r',y ,'y')
{??? Error using ==> plot
Not enough input arguments.
}
ejemplo2
x =
Columns 1 through 8
0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700
Columns 9 through 16
0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500
Columns 17 through 24
0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300
Columns 25 through 32
0.2400 0.2500 0.2600 0.2700 0.2800 0.2900 0.3000 0.3100
Columns 33 through 40
0.3200 0.3300 0.3400 0.3500 0.3600 0.3700 0.3800 0.3900
Columns 41 through 48
0.4000 0.4100 0.4200 0.4300 0.4400 0.4500 0.4600 0.4700
Columns 49 through 56
0.4800 0.4900 0.5000 0.5100 0.5200 0.5300 0.5400 0.5500
Columns 57 through 64
0.5600 0.5700 0.5800 0.5900 0.6000 0.6100 0.6200 0.6300
Columns 65 through 72
0.6400 0.6500 0.6600 0.6700 0.6800 0.6900 0.7000 0.7100
Columns 73 through 80
0.7200 0.7300 0.7400 0.7500 0.7600 0.7700 0.7800 0.7900
Columns 81 through 88
0.8000 0.8100 0.8200 0.8300 0.8400 0.8500 0.8600 0.8700
Columns 89 through 96
0.8800 0.8900 0.9000 0.9100 0.9200 0.9300 0.9400 0.9500
Columns 97 through 104
0.9600 0.9700 0.9800 0.9900 1.0000 1.0100 1.0200 1.0300
Columns 105 through 112
1.0400 1.0500 1.0600 1.0700 1.0800 1.0900 1.1000 1.1100
Columns 113 through 120
1.1200 1.1300 1.1400 1.1500 1.1600 1.1700 1.1800 1.1900
Columns 121 through 128
1.2000 1.2100 1.2200 1.2300 1.2400 1.2500 1.2600 1.2700
Columns 129 through 136
1.2800 1.2900 1.3000 1.3100 1.3200 1.3300 1.3400 1.3500
Columns 137 through 144
1.3600 1.3700 1.3800 1.3900 1.4000 1.4100 1.4200 1.4300
Columns 145 through 152
1.4400 1.4500 1.4600 1.4700 1.4800 1.4900 1.5000 1.5100
Columns 153 through 160
1.5200 1.5300 1.5400 1.5500 1.5600 1.5700 1.5800 1.5900
Columns 161 through 168
1.6000 1.6100 1.6200 1.6300 1.6400 1.6500 1.6600 1.6700
Columns 169 through 176
1.6800 1.6900 1.7000 1.7100 1.7200 1.7300 1.7400 1.7500
Columns 177 through 184
1.7600 1.7700 1.7800 1.7900 1.8000 1.8100 1.8200 1.8300
Columns 185 through 192
1.8400 1.8500 1.8600 1.8700 1.8800 1.8900 1.9000 1.9100
Columns 193 through 200
1.9200 1.9300 1.9400 1.9500 1.9600 1.9700 1.9800 1.9900
Columns 201 through 208
2.0000 2.0100 2.0200 2.0300 2.0400 2.0500 2.0600 2.0700
Columns 209 through 216
2.0800 2.0900 2.1000 2.1100 2.1200 2.1300 2.1400 2.1500
Columns 217 through 224
2.1600 2.1700 2.1800 2.1900 2.2000 2.2100 2.2200 2.2300
Columns 225 through 232
2.2400 2.2500 2.2600 2.2700 2.2800 2.2900 2.3000 2.3100
Columns 233 through 240
2.3200 2.3300 2.3400 2.3500 2.3600 2.3700 2.3800 2.3900
Columns 241 through 248
2.4000 2.4100 2.4200 2.4300 2.4400 2.4500 2.4600 2.4700
Columns 249 through 256
2.4800 2.4900 2.5000 2.5100 2.5200 2.5300 2.5400 2.5500
Columns 257 through 264
2.5600 2.5700 2.5800 2.5900 2.6000 2.6100 2.6200 2.6300
Columns 265 through 272
2.6400 2.6500 2.6600 2.6700 2.6800 2.6900 2.7000 2.7100
Columns 273 through 280
2.7200 2.7300 2.7400 2.7500 2.7600 2.7700 2.7800 2.7900
Columns 281 through 288
2.8000 2.8100 2.8200 2.8300 2.8400 2.8500 2.8600 2.8700
Columns 289 through 296
2.8800 2.8900 2.9000 2.9100 2.9200 2.9300 2.9400 2.9500
Columns 297 through 304
2.9600 2.9700 2.9800 2.9900 3.0000 3.0100 3.0200 3.0300
Columns 305 through 312
3.0400 3.0500 3.0600 3.0700 3.0800 3.0900 3.1000 3.1100
Columns 313 through 320
3.1200 3.1300 3.1400 3.1500 3.1600 3.1700 3.1800 3.1900
Columns 321 through 328
3.2000 3.2100 3.2200 3.2300 3.2400 3.2500 3.2600 3.2700
Columns 329 through 336
3.2800 3.2900 3.3000 3.3100 3.3200 3.3300 3.3400 3.3500
Columns 337 through 344
3.3600 3.3700 3.3800 3.3900 3.4000 3.4100 3.4200 3.4300
Columns 345 through 352
3.4400 3.4500 3.4600 3.4700 3.4800 3.4900 3.5000 3.5100
Columns 353 through 360
3.5200 3.5300 3.5400 3.5500 3.5600 3.5700 3.5800 3.5900
Columns 361 through 368
3.6000 3.6100 3.6200 3.6300 3.6400 3.6500 3.6600 3.6700
Columns 369 through 376
3.6800 3.6900 3.7000 3.7100 3.7200 3.7300 3.7400 3.7500
Columns 377 through 384
3.7600 3.7700 3.7800 3.7900 3.8000 3.8100 3.8200 3.8300
Columns 385 through 392
3.8400 3.8500 3.8600 3.8700 3.8800 3.8900 3.9000 3.9100
Columns 393 through 400
3.9200 3.9300 3.9400 3.9500 3.9600 3.9700 3.9800 3.9900
Columns 401 through 408
4.0000 4.0100 4.0200 4.0300 4.0400 4.0500 4.0600 4.0700
Columns 409 through 416
4.0800 4.0900 4.1000 4.1100 4.1200 4.1300 4.1400 4.1500
Columns 417 through 424
4.1600 4.1700 4.1800 4.1900 4.2000 4.2100 4.2200 4.2300
Columns 425 through 432
4.2400 4.2500 4.2600 4.2700 4.2800 4.2900 4.3000 4.3100
Columns 433 through 440
4.3200 4.3300 4.3400 4.3500 4.3600 4.3700 4.3800 4.3900
Columns 441 through 448
4.4000 4.4100 4.4200 4.4300 4.4400 4.4500 4.4600 4.4700
Columns 449 through 456
4.4800 4.4900 4.5000 4.5100 4.5200 4.5300 4.5400 4.5500
Columns 457 through 464
4.5600 4.5700 4.5800 4.5900 4.6000 4.6100 4.6200 4.6300
Columns 465 through 472
4.6400 4.6500 4.6600 4.6700 4.6800 4.6900 4.7000 4.7100
Columns 473 through 480
4.7200 4.7300 4.7400 4.7500 4.7600 4.7700 4.7800 4.7900
Columns 481 through 488
4.8000 4.8100 4.8200 4.8300 4.8400 4.8500 4.8600 4.8700
Columns 489 through 496
4.8800 4.8900 4.9000 4.9100 4.9200 4.9300 4.9400 4.9500
Columns 497 through 504
4.9600 4.9700 4.9800 4.9900 5.0000 5.0100 5.0200 5.0300
Columns 505 through 512
5.0400 5.0500 5.0600 5.0700 5.0800 5.0900 5.1000 5.1100
Columns 513 through 520
5.1200 5.1300 5.1400 5.1500 5.1600 5.1700 5.1800 5.1900
Columns 521 through 528
5.2000 5.2100 5.2200 5.2300 5.2400 5.2500 5.2600 5.2700
Columns 529 through 536
5.2800 5.2900 5.3000 5.3100 5.3200 5.3300 5.3400 5.3500
Columns 537 through 544
5.3600 5.3700 5.3800 5.3900 5.4000 5.4100 5.4200 5.4300
Columns 545 through 552
5.4400 5.4500 5.4600 5.4700 5.4800 5.4900 5.5000 5.5100
Columns 553 through 560
5.5200 5.5300 5.5400 5.5500 5.5600 5.5700 5.5800 5.5900
Columns 561 through 568
5.6000 5.6100 5.6200 5.6300 5.6400 5.6500 5.6600 5.6700
Columns 569 through 576
5.6800 5.6900 5.7000 5.7100 5.7200 5.7300 5.7400 5.7500
Columns 577 through 584
5.7600 5.7700 5.7800 5.7900 5.8000 5.8100 5.8200 5.8300
Columns 585 through 592
5.8400 5.8500 5.8600 5.8700 5.8800 5.8900 5.9000 5.9100
Columns 593 through 600
5.9200 5.9300 5.9400 5.9500 5.9600 5.9700 5.9800 5.9900
Columns 601 through 608
6.0000 6.0100 6.0200 6.0300 6.0400 6.0500 6.0600 6.0700
Columns 609 through 616
6.0800 6.0900 6.1000 6.1100 6.1200 6.1300 6.1400 6.1500
Columns 617 through 624
6.1600 6.1700 6.1800 6.1900 6.2000 6.2100 6.2200 6.2300
Columns 625 through 632
6.2400 6.2500 6.2600 6.2700 6.2800 6.2900 6.3000 6.3100
Columns 633 through 640
6.3200 6.3300 6.3400 6.3500 6.3600 6.3700 6.3800 6.3900
Columns 641 through 648
6.4000 6.4100 6.4200 6.4300 6.4400 6.4500 6.4600 6.4700
Columns 649 through 656
6.4800 6.4900 6.5000 6.5100 6.5200 6.5300 6.5400 6.5500
Columns 657 through 664
6.5600 6.5700 6.5800 6.5900 6.6000 6.6100 6.6200 6.6300
Columns 665 through 672
6.6400 6.6500 6.6600 6.6700 6.6800 6.6900 6.7000 6.7100
Columns 673 through 680
6.7200 6.7300 6.7400 6.7500 6.7600 6.7700 6.7800 6.7900
Columns 681 through 688
6.8000 6.8100 6.8200 6.8300 6.8400 6.8500 6.8600 6.8700
Columns 689 through 696
6.8800 6.8900 6.9000 6.9100 6.9200 6.9300 6.9400 6.9500
Columns 697 through 704
6.9600 6.9700 6.9800 6.9900 7.0000 7.0100 7.0200 7.0300
Columns 705 through 712
7.0400 7.0500 7.0600 7.0700 7.0800 7.0900 7.1000 7.1100
Columns 713 through 720
7.1200 7.1300 7.1400 7.1500 7.1600 7.1700 7.1800 7.1900
Columns 721 through 728
7.2000 7.2100 7.2200 7.2300 7.2400 7.2500 7.2600 7.2700
Columns 729 through 736
7.2800 7.2900 7.3000 7.3100 7.3200 7.3300 7.3400 7.3500
Columns 737 through 744
7.3600 7.3700 7.3800 7.3900 7.4000 7.4100 7.4200 7.4300
Columns 745 through 752
7.4400 7.4500 7.4600 7.4700 7.4800 7.4900 7.5000 7.5100
Columns 753 through 760
7.5200 7.5300 7.5400 7.5500 7.5600 7.5700 7.5800 7.5900
Columns 761 through 768
7.6000 7.6100 7.6200 7.6300 7.6400 7.6500 7.6600 7.6700
Columns 769 through 776
7.6800 7.6900 7.7000 7.7100 7.7200 7.7300 7.7400 7.7500
Columns 777 through 784
7.7600 7.7700 7.7800 7.7900 7.8000 7.8100 7.8200 7.8300
Columns 785 through 792
7.8400 7.8500 7.8600 7.8700 7.8800 7.8900 7.9000 7.9100
Columns 793 through 800
7.9200 7.9300 7.9400 7.9500 7.9600 7.9700 7.9800 7.9900
Columns 801 through 808
8.0000 8.0100 8.0200 8.0300 8.0400 8.0500 8.0600 8.0700
Columns 809 through 816
8.0800 8.0900 8.1000 8.1100 8.1200 8.1300 8.1400 8.1500
Columns 817 through 824
8.1600 8.1700 8.1800 8.1900 8.2000 8.2100 8.2200 8.2300
Columns 825 through 832
8.2400 8.2500 8.2600 8.2700 8.2800 8.2900 8.3000 8.3100
Columns 833 through 840
8.3200 8.3300 8.3400 8.3500 8.3600 8.3700 8.3800 8.3900
Columns 841 through 848
8.4000 8.4100 8.4200 8.4300 8.4400 8.4500 8.4600 8.4700
Columns 849 through 856
8.4800 8.4900 8.5000 8.5100 8.5200 8.5300 8.5400 8.5500
Columns 857 through 864
8.5600 8.5700 8.5800 8.5900 8.6000 8.6100 8.6200 8.6300
Columns 865 through 872
8.6400 8.6500 8.6600 8.6700 8.6800 8.6900 8.7000 8.7100
Columns 873 through 880
8.7200 8.7300 8.7400 8.7500 8.7600 8.7700 8.7800 8.7900
Columns 881 through 888
8.8000 8.8100 8.8200 8.8300 8.8400 8.8500 8.8600 8.8700
Columns 889 through 896
8.8800 8.8900 8.9000 8.9100 8.9200 8.9300 8.9400 8.9500
Columns 897 through 904
8.9600 8.9700 8.9800 8.9900 9.0000 9.0100 9.0200 9.0300
Columns 905 through 912
9.0400 9.0500 9.0600 9.0700 9.0800 9.0900 9.1000 9.1100
Columns 913 through 920
9.1200 9.1300 9.1400 9.1500 9.1600 9.1700 9.1800 9.1900
Columns 921 through 928
9.2000 9.2100 9.2200 9.2300 9.2400 9.2500 9.2600 9.2700
Columns 929 through 936
9.2800 9.2900 9.3000 9.3100 9.3200 9.3300 9.3400 9.3500
Columns 937 through 944
9.3600 9.3700 9.3800 9.3900 9.4000 9.4100 9.4200 9.4300
Columns 945 through 952
9.4400 9.4500 9.4600 9.4700 9.4800 9.4900 9.5000 9.5100
Columns 953 through 960
9.5200 9.5300 9.5400 9.5500 9.5600 9.5700 9.5800 9.5900
Columns 961 through 968
9.6000 9.6100 9.6200 9.6300 9.6400 9.6500 9.6600 9.6700
Columns 969 through 976
9.6800 9.6900 9.7000 9.7100 9.7200 9.7300 9.7400 9.7500
Columns 977 through 984
9.7600 9.7700 9.7800 9.7900 9.8000 9.8100 9.8200 9.8300
Columns 985 through 992
9.8400 9.8500 9.8600 9.8700 9.8800 9.8900 9.9000 9.9100
Columns 993 through 1000
9.9200 9.9300 9.9400 9.9500 9.9600 9.9700 9.9800 9.9900
Columns 1001 through 1008
10.0000 10.0100 10.0200 10.0300 10.0400 10.0500 10.0600 10.0700
Columns 1009 through 1016
10.0800 10.0900 10.1000 10.1100 10.1200 10.1300 10.1400 10.1500
Columns 1017 through 1024
10.1600 10.1700 10.1800 10.1900 10.2000 10.2100 10.2200 10.2300
Columns 1025 through 1032
10.2400 10.2500 10.2600 10.2700 10.2800 10.2900 10.3000 10.3100
Columns 1033 through 1040
10.3200 10.3300 10.3400 10.3500 10.3600 10.3700 10.3800 10.3900
Columns 1041 through 1048
10.4000 10.4100 10.4200 10.4300 10.4400 10.4500 10.4600 10.4700
Columns 1049 through 1056
10.4800 10.4900 10.5000 10.5100 10.5200 10.5300 10.5400 10.5500
Columns 1057 through 1064
10.5600 10.5700 10.5800 10.5900 10.6000 10.6100 10.6200 10.6300
Columns 1065 through 1072
10.6400 10.6500 10.6600 10.6700 10.6800 10.6900 10.7000 10.7100
Columns 1073 through 1080
10.7200 10.7300 10.7400 10.7500 10.7600 10.7700 10.7800 10.7900
Columns 1081 through 1088
10.8000 10.8100 10.8200 10.8300 10.8400 10.8500 10.8600 10.8700
Columns 1089 through 1096
10.8800 10.8900 10.9000 10.9100 10.9200 10.9300 10.9400 10.9500
Columns 1097 through 1104
10.9600 10.9700 10.9800 10.9900 11.0000 11.0100 11.0200 11.0300
Columns 1105 through 1112
11.0400 11.0500 11.0600 11.0700 11.0800 11.0900 11.1000 11.1100
Columns 1113 through 1120
11.1200 11.1300 11.1400 11.1500 11.1600 11.1700 11.1800 11.1900
Columns 1121 through 1128
11.2000 11.2100 11.2200 11.2300 11.2400 11.2500 11.2600 11.2700
Columns 1129 through 1136
11.2800 11.2900 11.3000 11.3100 11.3200 11.3300 11.3400 11.3500
Columns 1137 through 1144
11.3600 11.3700 11.3800 11.3900 11.4000 11.4100 11.4200 11.4300
Columns 1145 through 1152
11.4400 11.4500 11.4600 11.4700 11.4800 11.4900 11.5000 11.5100
Columns 1153 through 1160
11.5200 11.5300 11.5400 11.5500 11.5600 11.5700 11.5800 11.5900
Columns 1161 through 1168
11.6000 11.6100 11.6200 11.6300 11.6400 11.6500 11.6600 11.6700
Columns 1169 through 1176
11.6800 11.6900 11.7000 11.7100 11.7200 11.7300 11.7400 11.7500
Columns 1177 through 1184
11.7600 11.7700 11.7800 11.7900 11.8000 11.8100 11.8200 11.8300
Columns 1185 through 1192
11.8400 11.8500 11.8600 11.8700 11.8800 11.8900 11.9000 11.9100
Columns 1193 through 1200
11.9200 11.9300 11.9400 11.9500 11.9600 11.9700 11.9800 11.9900
Columns 1201 through 1208
12.0000 12.0100 12.0200 12.0300 12.0400 12.0500 12.0600 12.0700
Columns 1209 through 1216
12.0800 12.0900 12.1000 12.1100 12.1200 12.1300 12.1400 12.1500
Columns 1217 through 1224
12.1600 12.1700 12.1800 12.1900 12.2000 12.2100 12.2200 12.2300
Columns 1225 through 1232
12.2400 12.2500 12.2600 12.2700 12.2800 12.2900 12.3000 12.3100
Columns 1233 through 1240
12.3200 12.3300 12.3400 12.3500 12.3600 12.3700 12.3800 12.3900
Columns 1241 through 1248
12.4000 12.4100 12.4200 12.4300 12.4400 12.4500 12.4600 12.4700
Columns 1249 through 1256
12.4800 12.4900 12.5000 12.5100 12.5200 12.5300 12.5400 12.5500
Columns 1257 through 1264
12.5600 12.5700 12.5800 12.5900 12.6000 12.6100 12.6200 12.6300
Columns 1265 through 1272
12.6400 12.6500 12.6600 12.6700 12.6800 12.6900 12.7000 12.7100
Columns 1273 through 1280
12.7200 12.7300 12.7400 12.7500 12.7600 12.7700 12.7800 12.7900
Columns 1281 through 1288
12.8000 12.8100 12.8200 12.8300 12.8400 12.8500 12.8600 12.8700
Columns 1289 through 1296
12.8800 12.8900 12.9000 12.9100 12.9200 12.9300 12.9400 12.9500
Columns 1297 through 1304
12.9600 12.9700 12.9800 12.9900 13.0000 13.0100 13.0200 13.0300
Columns 1305 through 1312
13.0400 13.0500 13.0600 13.0700 13.0800 13.0900 13.1000 13.1100
Columns 1313 through 1320
13.1200 13.1300 13.1400 13.1500 13.1600 13.1700 13.1800 13.1900
Columns 1321 through 1328
13.2000 13.2100 13.2200 13.2300 13.2400 13.2500 13.2600 13.2700
Columns 1329 through 1336
13.2800 13.2900 13.3000 13.3100 13.3200 13.3300 13.3400 13.3500
Columns 1337 through 1344
13.3600 13.3700 13.3800 13.3900 13.4000 13.4100 13.4200 13.4300
Columns 1345 through 1352
13.4400 13.4500 13.4600 13.4700 13.4800 13.4900 13.5000 13.5100
Columns 1353 through 1360
13.5200 13.5300 13.5400 13.5500 13.5600 13.5700 13.5800 13.5900
Columns 1361 through 1368
13.6000 13.6100 13.6200 13.6300 13.6400 13.6500 13.6600 13.6700
Columns 1369 through 1376
13.6800 13.6900 13.7000 13.7100 13.7200 13.7300 13.7400 13.7500
Columns 1377 through 1384
13.7600 13.7700 13.7800 13.7900 13.8000 13.8100 13.8200 13.8300
Columns 1385 through 1392
13.8400 13.8500 13.8600 13.8700 13.8800 13.8900 13.9000 13.9100
Columns 1393 through 1400
13.9200 13.9300 13.9400 13.9500 13.9600 13.9700 13.9800 13.9900
Columns 1401 through 1408
14.0000 14.0100 14.0200 14.0300 14.0400 14.0500 14.0600 14.0700
Columns 1409 through 1416
14.0800 14.0900 14.1000 14.1100 14.1200 14.1300 14.1400 14.1500
Columns 1417 through 1424
14.1600 14.1700 14.1800 14.1900 14.2000 14.2100 14.2200 14.2300
Columns 1425 through 1432
14.2400 14.2500 14.2600 14.2700 14.2800 14.2900 14.3000 14.3100
Columns 1433 through 1440
14.3200 14.3300 14.3400 14.3500 14.3600 14.3700 14.3800 14.3900
Columns 1441 through 1448
14.4000 14.4100 14.4200 14.4300 14.4400 14.4500 14.4600 14.4700
Columns 1449 through 1456
14.4800 14.4900 14.5000 14.5100 14.5200 14.5300 14.5400 14.5500
Columns 1457 through 1464
14.5600 14.5700 14.5800 14.5900 14.6000 14.6100 14.6200 14.6300
Columns 1465 through 1472
14.6400 14.6500 14.6600 14.6700 14.6800 14.6900 14.7000 14.7100
Columns 1473 through 1480
14.7200 14.7300 14.7400 14.7500 14.7600 14.7700 14.7800 14.7900
Columns 1481 through 1488
14.8000 14.8100 14.8200 14.8300 14.8400 14.8500 14.8600 14.8700
Columns 1489 through 1496
14.8800 14.8900 14.9000 14.9100 14.9200 14.9300 14.9400 14.9500
Columns 1497 through 1504
14.9600 14.9700 14.9800 14.9900 15.0000 15.0100 15.0200 15.0300
Columns 1505 through 1512
15.0400 15.0500 15.0600 15.0700 15.0800 15.0900 15.1000 15.1100
Columns 1513 through 1520
15.1200 15.1300 15.1400 15.1500 15.1600 15.1700 15.1800 15.1900
Columns 1521 through 1528
15.2000 15.2100 15.2200 15.2300 15.2400 15.2500 15.2600 15.2700
Columns 1529 through 1536
15.2800 15.2900 15.3000 15.3100 15.3200 15.3300 15.3400 15.3500
Columns 1537 through 1544
15.3600 15.3700 15.3800 15.3900 15.4000 15.4100 15.4200 15.4300
Columns 1545 through 1552
15.4400 15.4500 15.4600 15.4700 15.4800 15.4900 15.5000 15.5100
Columns 1553 through 1560
15.5200 15.5300 15.5400 15.5500 15.5600 15.5700 15.5800 15.5900
Columns 1561 through 1568
15.6000 15.6100 15.6200 15.6300 15.6400 15.6500 15.6600 15.6700
Columns 1569 through 1576
15.6800 15.6900 15.7000 15.7100 15.7200 15.7300 15.7400 15.7500
Columns 1577 through 1584
15.7600 15.7700 15.7800 15.7900 15.8000 15.8100 15.8200 15.8300
Columns 1585 through 1592
15.8400 15.8500 15.8600 15.8700 15.8800 15.8900 15.9000 15.9100
Columns 1593 through 1600
15.9200 15.9300 15.9400 15.9500 15.9600 15.9700 15.9800 15.9900
Columns 1601 through 1608
16.0000 16.0100 16.0200 16.0300 16.0400 16.0500 16.0600 16.0700
Columns 1609 through 1616
16.0800 16.0900 16.1000 16.1100 16.1200 16.1300 16.1400 16.1500
Columns 1617 through 1624
16.1600 16.1700 16.1800 16.1900 16.2000 16.2100 16.2200 16.2300
Columns 1625 through 1632
16.2400 16.2500 16.2600 16.2700 16.2800 16.2900 16.3000 16.3100
Columns 1633 through 1640
16.3200 16.3300 16.3400 16.3500 16.3600 16.3700 16.3800 16.3900
Columns 1641 through 1648
16.4000 16.4100 16.4200 16.4300 16.4400 16.4500 16.4600 16.4700
Columns 1649 through 1656
16.4800 16.4900 16.5000 16.5100 16.5200 16.5300 16.5400 16.5500
Columns 1657 through 1664
16.5600 16.5700 16.5800 16.5900 16.6000 16.6100 16.6200 16.6300
Columns 1665 through 1672
16.6400 16.6500 16.6600 16.6700 16.6800 16.6900 16.7000 16.7100
Columns 1673 through 1680
16.7200 16.7300 16.7400 16.7500 16.7600 16.7700 16.7800 16.7900
Columns 1681 through 1688
16.8000 16.8100 16.8200 16.8300 16.8400 16.8500 16.8600 16.8700
Columns 1689 through 1696
16.8800 16.8900 16.9000 16.9100 16.9200 16.9300 16.9400 16.9500
Columns 1697 through 1704
16.9600 16.9700 16.9800 16.9900 17.0000 17.0100 17.0200 17.0300
Columns 1705 through 1712
17.0400 17.0500 17.0600 17.0700 17.0800 17.0900 17.1000 17.1100
Columns 1713 through 1720
17.1200 17.1300 17.1400 17.1500 17.1600 17.1700 17.1800 17.1900
Columns 1721 through 1728
17.2000 17.2100 17.2200 17.2300 17.2400 17.2500 17.2600 17.2700
Columns 1729 through 1736
17.2800 17.2900 17.3000 17.3100 17.3200 17.3300 17.3400 17.3500
Columns 1737 through 1744
17.3600 17.3700 17.3800 17.3900 17.4000 17.4100 17.4200 17.4300
Columns 1745 through 1752
17.4400 17.4500 17.4600 17.4700 17.4800 17.4900 17.5000 17.5100
Columns 1753 through 1760
17.5200 17.5300 17.5400 17.5500 17.5600 17.5700 17.5800 17.5900
Columns 1761 through 1768
17.6000 17.6100 17.6200 17.6300 17.6400 17.6500 17.6600 17.6700
Columns 1769 through 1776
17.6800 17.6900 17.7000 17.7100 17.7200 17.7300 17.7400 17.7500
Columns 1777 through 1784
17.7600 17.7700 17.7800 17.7900 17.8000 17.8100 17.8200 17.8300
Columns 1785 through 1792
17.8400 17.8500 17.8600 17.8700 17.8800 17.8900 17.9000 17.9100
Columns 1793 through 1800
17.9200 17.9300 17.9400 17.9500 17.9600 17.9700 17.9800 17.9900
Columns 1801 through 1808
18.0000 18.0100 18.0200 18.0300 18.0400 18.0500 18.0600 18.0700
Columns 1809 through 1816
18.0800 18.0900 18.1000 18.1100 18.1200 18.1300 18.1400 18.1500
Columns 1817 through 1824
18.1600 18.1700 18.1800 18.1900 18.2000 18.2100 18.2200 18.2300
Columns 1825 through 1832
18.2400 18.2500 18.2600 18.2700 18.2800 18.2900 18.3000 18.3100
Columns 1833 through 1840
18.3200 18.3300 18.3400 18.3500 18.3600 18.3700 18.3800 18.3900
Columns 1841 through 1848
18.4000 18.4100 18.4200 18.4300 18.4400 18.4500 18.4600 18.4700
Columns 1849 through 1856
18.4800 18.4900 18.5000 18.5100 18.5200 18.5300 18.5400 18.5500
Columns 1857 through 1864
18.5600 18.5700 18.5800 18.5900 18.6000 18.6100 18.6200 18.6300
Columns 1865 through 1872
18.6400 18.6500 18.6600 18.6700 18.6800 18.6900 18.7000 18.7100
Columns 1873 through 1880
18.7200 18.7300 18.7400 18.7500 18.7600 18.7700 18.7800 18.7900
Columns 1881 through 1888
18.8000 18.8100 18.8200 18.8300 18.8400 18.8500 18.8600 18.8700
Columns 1889 through 1896
18.8800 18.8900 18.9000 18.9100 18.9200 18.9300 18.9400 18.9500
Columns 1897 through 1904
18.9600 18.9700 18.9800 18.9900 19.0000 19.0100 19.0200 19.0300
Columns 1905 through 1912
19.0400 19.0500 19.0600 19.0700 19.0800 19.0900 19.1000 19.1100
Columns 1913 through 1920
19.1200 19.1300 19.1400 19.1500 19.1600 19.1700 19.1800 19.1900
Columns 1921 through 1928
19.2000 19.2100 19.2200 19.2300 19.2400 19.2500 19.2600 19.2700
Columns 1929 through 1936
19.2800 19.2900 19.3000 19.3100 19.3200 19.3300 19.3400 19.3500
Columns 1937 through 1944
19.3600 19.3700 19.3800 19.3900 19.4000 19.4100 19.4200 19.4300
Columns 1945 through 1952
19.4400 19.4500 19.4600 19.4700 19.4800 19.4900 19.5000 19.5100
Columns 1953 through 1960
19.5200 19.5300 19.5400 19.5500 19.5600 19.5700 19.5800 19.5900
Columns 1961 through 1968
19.6000 19.6100 19.6200 19.6300 19.6400 19.6500 19.6600 19.6700
Columns 1969 through 1976
19.6800 19.6900 19.7000 19.7100 19.7200 19.7300 19.7400 19.7500
Columns 1977 through 1984
19.7600 19.7700 19.7800 19.7900 19.8000 19.8100 19.8200 19.8300
Columns 1985 through 1992
19.8400 19.8500 19.8600 19.8700 19.8800 19.8900 19.9000 19.9100
Columns 1993 through 2000
19.9200 19.9300 19.9400 19.9500 19.9600 19.9700 19.9800 19.9900
Column 2001
20.0000
grid on
uiopen('C:\Documents and Settings\Laboratorio\Mis documentos\MATLAB\pilar.txt',1
)
uiopen('C:\Documents and Settings\Laboratorio\Mis documentos\MATLAB\ejemplo2.m',
true);
ejemplo2
x =
Columns 1 through 8
0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700
Columns 9 through 16
0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500
Columns 17 through 24
0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300
Columns 25 through 32
0.2400 0.2500 0.2600 0.2700 0.2800 0.2900 0.3000 0.3100
Columns 33 through 40
0.3200 0.3300 0.3400 0.3500 0.3600 0.3700 0.3800 0.3900
Columns 41 through 48
0.4000 0.4100 0.4200 0.4300 0.4400 0.4500 0.4600 0.4700
Columns 49 through 56
0.4800 0.4900 0.5000 0.5100 0.5200 0.5300 0.5400 0.5500
Columns 57 through 64
0.5600 0.5700 0.5800 0.5900 0.6000 0.6100 0.6200 0.6300
Columns 65 through 72
0.6400 0.6500 0.6600 0.6700 0.6800 0.6900 0.7000 0.7100
Columns 73 through 80
0.7200 0.7300 0.7400 0.7500 0.7600 0.7700 0.7800 0.7900
Columns 81 through 88
0.8000 0.8100 0.8200 0.8300 0.8400 0.8500 0.8600 0.8700
Columns 89 through 96
0.8800 0.8900 0.9000 0.9100 0.9200 0.9300 0.9400 0.9500
Columns 97 through 104
0.9600 0.9700 0.9800 0.9900 1.0000 1.0100 1.0200 1.0300
Columns 105 through 112
1.0400 1.0500 1.0600 1.0700 1.0800 1.0900 1.1000 1.1100
Columns 113 through 120
1.1200 1.1300 1.1400 1.1500 1.1600 1.1700 1.1800 1.1900
Columns 121 through 128
1.2000 1.2100 1.2200 1.2300 1.2400 1.2500 1.2600 1.2700
Columns 129 through 136
1.2800 1.2900 1.3000 1.3100 1.3200 1.3300 1.3400 1.3500
Columns 137 through 144
1.3600 1.3700 1.3800 1.3900 1.4000 1.4100 1.4200 1.4300
Columns 145 through 152
1.4400 1.4500 1.4600 1.4700 1.4800 1.4900 1.5000 1.5100
Columns 153 through 160
1.5200 1.5300 1.5400 1.5500 1.5600 1.5700 1.5800 1.5900
Columns 161 through 168
1.6000 1.6100 1.6200 1.6300 1.6400 1.6500 1.6600 1.6700
Columns 169 through 176
1.6800 1.6900 1.7000 1.7100 1.7200 1.7300 1.7400 1.7500
Columns 177 through 184
1.7600 1.7700 1.7800 1.7900 1.8000 1.8100 1.8200 1.8300
Columns 185 through 192
1.8400 1.8500 1.8600 1.8700 1.8800 1.8900 1.9000 1.9100
Columns 193 through 200
1.9200 1.9300 1.9400 1.9500 1.9600 1.9700 1.9800 1.9900
Columns 201 through 208
2.0000 2.0100 2.0200 2.0300 2.0400 2.0500 2.0600 2.0700
Columns 209 through 216
2.0800 2.0900 2.1000 2.1100 2.1200 2.1300 2.1400 2.1500
Columns 217 through 224
2.1600 2.1700 2.1800 2.1900 2.2000 2.2100 2.2200 2.2300
Columns 225 through 232
2.2400 2.2500 2.2600 2.2700 2.2800 2.2900 2.3000 2.3100
Columns 233 through 240
2.3200 2.3300 2.3400 2.3500 2.3600 2.3700 2.3800 2.3900
Columns 241 through 248
2.4000 2.4100 2.4200 2.4300 2.4400 2.4500 2.4600 2.4700
Columns 249 through 256
2.4800 2.4900 2.5000 2.5100 2.5200 2.5300 2.5400 2.5500
Columns 257 through 264
2.5600 2.5700 2.5800 2.5900 2.6000 2.6100 2.6200 2.6300
Columns 265 through 272
2.6400 2.6500 2.6600 2.6700 2.6800 2.6900 2.7000 2.7100
Columns 273 through 280
2.7200 2.7300 2.7400 2.7500 2.7600 2.7700 2.7800 2.7900
Columns 281 through 288
2.8000 2.8100 2.8200 2.8300 2.8400 2.8500 2.8600 2.8700
Columns 289 through 296
2.8800 2.8900 2.9000 2.9100 2.9200 2.9300 2.9400 2.9500
Columns 297 through 304
2.9600 2.9700 2.9800 2.9900 3.0000 3.0100 3.0200 3.0300
Columns 305 through 312
3.0400 3.0500 3.0600 3.0700 3.0800 3.0900 3.1000 3.1100
Columns 313 through 320
3.1200 3.1300 3.1400 3.1500 3.1600 3.1700 3.1800 3.1900
Columns 321 through 328
3.2000 3.2100 3.2200 3.2300 3.2400 3.2500 3.2600 3.2700
Columns 329 through 336
3.2800 3.2900 3.3000 3.3100 3.3200 3.3300 3.3400 3.3500
Columns 337 through 344
3.3600 3.3700 3.3800 3.3900 3.4000 3.4100 3.4200 3.4300
Columns 345 through 352
3.4400 3.4500 3.4600 3.4700 3.4800 3.4900 3.5000 3.5100
Columns 353 through 360
3.5200 3.5300 3.5400 3.5500 3.5600 3.5700 3.5800 3.5900
Columns 361 through 368
3.6000 3.6100 3.6200 3.6300 3.6400 3.6500 3.6600 3.6700
Columns 369 through 376
3.6800 3.6900 3.7000 3.7100 3.7200 3.7300 3.7400 3.7500
Columns 377 through 384
3.7600 3.7700 3.7800 3.7900 3.8000 3.8100 3.8200 3.8300
Columns 385 through 392
3.8400 3.8500 3.8600 3.8700 3.8800 3.8900 3.9000 3.9100
Columns 393 through 400
3.9200 3.9300 3.9400 3.9500 3.9600 3.9700 3.9800 3.9900
Columns 401 through 408
4.0000 4.0100 4.0200 4.0300 4.0400 4.0500 4.0600 4.0700
Columns 409 through 416
4.0800 4.0900 4.1000 4.1100 4.1200 4.1300 4.1400 4.1500
Columns 417 through 424
4.1600 4.1700 4.1800 4.1900 4.2000 4.2100 4.2200 4.2300
Columns 425 through 432
4.2400 4.2500 4.2600 4.2700 4.2800 4.2900 4.3000 4.3100
Columns 433 through 440
4.3200 4.3300 4.3400 4.3500 4.3600 4.3700 4.3800 4.3900
Columns 441 through 448
4.4000 4.4100 4.4200 4.4300 4.4400 4.4500 4.4600 4.4700
Columns 449 through 456
4.4800 4.4900 4.5000 4.5100 4.5200 4.5300 4.5400 4.5500
Columns 457 through 464
4.5600 4.5700 4.5800 4.5900 4.6000 4.6100 4.6200 4.6300
Columns 465 through 472
4.6400 4.6500 4.6600 4.6700 4.6800 4.6900 4.7000 4.7100
Columns 473 through 480
4.7200 4.7300 4.7400 4.7500 4.7600 4.7700 4.7800 4.7900
Columns 481 through 488
4.8000 4.8100 4.8200 4.8300 4.8400 4.8500 4.8600 4.8700
Columns 489 through 496
4.8800 4.8900 4.9000 4.9100 4.9200 4.9300 4.9400 4.9500
Columns 497 through 504
4.9600 4.9700 4.9800 4.9900 5.0000 5.0100 5.0200 5.0300
Columns 505 through 512
5.0400 5.0500 5.0600 5.0700 5.0800 5.0900 5.1000 5.1100
Columns 513 through 520
5.1200 5.1300 5.1400 5.1500 5.1600 5.1700 5.1800 5.1900
Columns 521 through 528
5.2000 5.2100 5.2200 5.2300 5.2400 5.2500 5.2600 5.2700
Columns 529 through 536
5.2800 5.2900 5.3000 5.3100 5.3200 5.3300 5.3400 5.3500
Columns 537 through 544
5.3600 5.3700 5.3800 5.3900 5.4000 5.4100 5.4200 5.4300
Columns 545 through 552
5.4400 5.4500 5.4600 5.4700 5.4800 5.4900 5.5000 5.5100
Columns 553 through 560
5.5200 5.5300 5.5400 5.5500 5.5600 5.5700 5.5800 5.5900
Columns 561 through 568
5.6000 5.6100 5.6200 5.6300 5.6400 5.6500 5.6600 5.6700
Columns 569 through 576
5.6800 5.6900 5.7000 5.7100 5.7200 5.7300 5.7400 5.7500
Columns 577 through 584
5.7600 5.7700 5.7800 5.7900 5.8000 5.8100 5.8200 5.8300
Columns 585 through 592
5.8400 5.8500 5.8600 5.8700 5.8800 5.8900 5.9000 5.9100
Columns 593 through 600
5.9200 5.9300 5.9400 5.9500 5.9600 5.9700 5.9800 5.9900
Columns 601 through 608
6.0000 6.0100 6.0200 6.0300 6.0400 6.0500 6.0600 6.0700
Columns 609 through 616
6.0800 6.0900 6.1000 6.1100 6.1200 6.1300 6.1400 6.1500
Columns 617 through 624
6.1600 6.1700 6.1800 6.1900 6.2000 6.2100 6.2200 6.2300
Columns 625 through 632
6.2400 6.2500 6.2600 6.2700 6.2800 6.2900 6.3000 6.3100
Columns 633 through 640
6.3200 6.3300 6.3400 6.3500 6.3600 6.3700 6.3800 6.3900
Columns 641 through 648
6.4000 6.4100 6.4200 6.4300 6.4400 6.4500 6.4600 6.4700
Columns 649 through 656
6.4800 6.4900 6.5000 6.5100 6.5200 6.5300 6.5400 6.5500
Columns 657 through 664
6.5600 6.5700 6.5800 6.5900 6.6000 6.6100 6.6200 6.6300
Columns 665 through 672
6.6400 6.6500 6.6600 6.6700 6.6800 6.6900 6.7000 6.7100
Columns 673 through 680
6.7200 6.7300 6.7400 6.7500 6.7600 6.7700 6.7800 6.7900
Columns 681 through 688
6.8000 6.8100 6.8200 6.8300 6.8400 6.8500 6.8600 6.8700
Columns 689 through 696
6.8800 6.8900 6.9000 6.9100 6.9200 6.9300 6.9400 6.9500
Columns 697 through 704
6.9600 6.9700 6.9800 6.9900 7.0000 7.0100 7.0200 7.0300
Columns 705 through 712
7.0400 7.0500 7.0600 7.0700 7.0800 7.0900 7.1000 7.1100
Columns 713 through 720
7.1200 7.1300 7.1400 7.1500 7.1600 7.1700 7.1800 7.1900
Columns 721 through 728
7.2000 7.2100 7.2200 7.2300 7.2400 7.2500 7.2600 7.2700
Columns 729 through 736
7.2800 7.2900 7.3000 7.3100 7.3200 7.3300 7.3400 7.3500
Columns 737 through 744
7.3600 7.3700 7.3800 7.3900 7.4000 7.4100 7.4200 7.4300
Columns 745 through 752
7.4400 7.4500 7.4600 7.4700 7.4800 7.4900 7.5000 7.5100
Columns 753 through 760
7.5200 7.5300 7.5400 7.5500 7.5600 7.5700 7.5800 7.5900
Columns 761 through 768
7.6000 7.6100 7.6200 7.6300 7.6400 7.6500 7.6600 7.6700
Columns 769 through 776
7.6800 7.6900 7.7000 7.7100 7.7200 7.7300 7.7400 7.7500
Columns 777 through 784
7.7600 7.7700 7.7800 7.7900 7.8000 7.8100 7.8200 7.8300
Columns 785 through 792
7.8400 7.8500 7.8600 7.8700 7.8800 7.8900 7.9000 7.9100
Columns 793 through 800
7.9200 7.9300 7.9400 7.9500 7.9600 7.9700 7.9800 7.9900
Columns 801 through 808
8.0000 8.0100 8.0200 8.0300 8.0400 8.0500 8.0600 8.0700
Columns 809 through 816
8.0800 8.0900 8.1000 8.1100 8.1200 8.1300 8.1400 8.1500
Columns 817 through 824
8.1600 8.1700 8.1800 8.1900 8.2000 8.2100 8.2200 8.2300
Columns 825 through 832
8.2400 8.2500 8.2600 8.2700 8.2800 8.2900 8.3000 8.3100
Columns 833 through 840
8.3200 8.3300 8.3400 8.3500 8.3600 8.3700 8.3800 8.3900
Columns 841 through 848
8.4000 8.4100 8.4200 8.4300 8.4400 8.4500 8.4600 8.4700
Columns 849 through 856
8.4800 8.4900 8.5000 8.5100 8.5200 8.5300 8.5400 8.5500
Columns 857 through 864
8.5600 8.5700 8.5800 8.5900 8.6000 8.6100 8.6200 8.6300
Columns 865 through 872
8.6400 8.6500 8.6600 8.6700 8.6800 8.6900 8.7000 8.7100
Columns 873 through 880
8.7200 8.7300 8.7400 8.7500 8.7600 8.7700 8.7800 8.7900
Columns 881 through 888
8.8000 8.8100 8.8200 8.8300 8.8400 8.8500 8.8600 8.8700
Columns 889 through 896
8.8800 8.8900 8.9000 8.9100 8.9200 8.9300 8.9400 8.9500
Columns 897 through 904
8.9600 8.9700 8.9800 8.9900 9.0000 9.0100 9.0200 9.0300
Columns 905 through 912
9.0400 9.0500 9.0600 9.0700 9.0800 9.0900 9.1000 9.1100
Columns 913 through 920
9.1200 9.1300 9.1400 9.1500 9.1600 9.1700 9.1800 9.1900
Columns 921 through 928
9.2000 9.2100 9.2200 9.2300 9.2400 9.2500 9.2600 9.2700
Columns 929 through 936
9.2800 9.2900 9.3000 9.3100 9.3200 9.3300 9.3400 9.3500
Columns 937 through 944
9.3600 9.3700 9.3800 9.3900 9.4000 9.4100 9.4200 9.4300
Columns 945 through 952
9.4400 9.4500 9.4600 9.4700 9.4800 9.4900 9.5000 9.5100
Columns 953 through 960
9.5200 9.5300 9.5400 9.5500 9.5600 9.5700 9.5800 9.5900
Columns 961 through 968
9.6000 9.6100 9.6200 9.6300 9.6400 9.6500 9.6600 9.6700
Columns 969 through 976
9.6800 9.6900 9.7000 9.7100 9.7200 9.7300 9.7400 9.7500
Columns 977 through 984
9.7600 9.7700 9.7800 9.7900 9.8000 9.8100 9.8200 9.8300
Columns 985 through 992
9.8400 9.8500 9.8600 9.8700 9.8800 9.8900 9.9000 9.9100
Columns 993 through 1000
9.9200 9.9300 9.9400 9.9500 9.9600 9.9700 9.9800 9.9900
Columns 1001 through 1008
10.0000 10.0100 10.0200 10.0300 10.0400 10.0500 10.0600 10.0700
Columns 1009 through 1016
10.0800 10.0900 10.1000 10.1100 10.1200 10.1300 10.1400 10.1500
Columns 1017 through 1024
10.1600 10.1700 10.1800 10.1900 10.2000 10.2100 10.2200 10.2300
Columns 1025 through 1032
10.2400 10.2500 10.2600 10.2700 10.2800 10.2900 10.3000 10.3100
Columns 1033 through 1040
10.3200 10.3300 10.3400 10.3500 10.3600 10.3700 10.3800 10.3900
Columns 1041 through 1048
10.4000 10.4100 10.4200 10.4300 10.4400 10.4500 10.4600 10.4700
Columns 1049 through 1056
10.4800 10.4900 10.5000 10.5100 10.5200 10.5300 10.5400 10.5500
Columns 1057 through 1064
10.5600 10.5700 10.5800 10.5900 10.6000 10.6100 10.6200 10.6300
Columns 1065 through 1072
10.6400 10.6500 10.6600 10.6700 10.6800 10.6900 10.7000 10.7100
Columns 1073 through 1080
10.7200 10.7300 10.7400 10.7500 10.7600 10.7700 10.7800 10.7900
Columns 1081 through 1088
10.8000 10.8100 10.8200 10.8300 10.8400 10.8500 10.8600 10.8700
Columns 1089 through 1096
10.8800 10.8900 10.9000 10.9100 10.9200 10.9300 10.9400 10.9500
Columns 1097 through 1104
10.9600 10.9700 10.9800 10.9900 11.0000 11.0100 11.0200 11.0300
Columns 1105 through 1112
11.0400 11.0500 11.0600 11.0700 11.0800 11.0900 11.1000 11.1100
Columns 1113 through 1120
11.1200 11.1300 11.1400 11.1500 11.1600 11.1700 11.1800 11.1900
Columns 1121 through 1128
11.2000 11.2100 11.2200 11.2300 11.2400 11.2500 11.2600 11.2700
Columns 1129 through 1136
11.2800 11.2900 11.3000 11.3100 11.3200 11.3300 11.3400 11.3500
Columns 1137 through 1144
11.3600 11.3700 11.3800 11.3900 11.4000 11.4100 11.4200 11.4300
Columns 1145 through 1152
11.4400 11.4500 11.4600 11.4700 11.4800 11.4900 11.5000 11.5100
Columns 1153 through 1160
11.5200 11.5300 11.5400 11.5500 11.5600 11.5700 11.5800 11.5900
Columns 1161 through 1168
11.6000 11.6100 11.6200 11.6300 11.6400 11.6500 11.6600 11.6700
Columns 1169 through 1176
11.6800 11.6900 11.7000 11.7100 11.7200 11.7300 11.7400 11.7500
Columns 1177 through 1184
11.7600 11.7700 11.7800 11.7900 11.8000 11.8100 11.8200 11.8300
Columns 1185 through 1192
11.8400 11.8500 11.8600 11.8700 11.8800 11.8900 11.9000 11.9100
Columns 1193 through 1200
11.9200 11.9300 11.9400 11.9500 11.9600 11.9700 11.9800 11.9900
Columns 1201 through 1208
12.0000 12.0100 12.0200 12.0300 12.0400 12.0500 12.0600 12.0700
Columns 1209 through 1216
12.0800 12.0900 12.1000 12.1100 12.1200 12.1300 12.1400 12.1500
Columns 1217 through 1224
12.1600 12.1700 12.1800 12.1900 12.2000 12.2100 12.2200 12.2300
Columns 1225 through 1232
12.2400 12.2500 12.2600 12.2700 12.2800 12.2900 12.3000 12.3100
Columns 1233 through 1240
12.3200 12.3300 12.3400 12.3500 12.3600 12.3700 12.3800 12.3900
Columns 1241 through 1248
12.4000 12.4100 12.4200 12.4300 12.4400 12.4500 12.4600 12.4700
Columns 1249 through 1256
12.4800 12.4900 12.5000 12.5100 12.5200 12.5300 12.5400 12.5500
Columns 1257 through 1264
12.5600 12.5700 12.5800 12.5900 12.6000 12.6100 12.6200 12.6300
Columns 1265 through 1272
12.6400 12.6500 12.6600 12.6700 12.6800 12.6900 12.7000 12.7100
Columns 1273 through 1280
12.7200 12.7300 12.7400 12.7500 12.7600 12.7700 12.7800 12.7900
Columns 1281 through 1288
12.8000 12.8100 12.8200 12.8300 12.8400 12.8500 12.8600 12.8700
Columns 1289 through 1296
12.8800 12.8900 12.9000 12.9100 12.9200 12.9300 12.9400 12.9500
Columns 1297 through 1304
12.9600 12.9700 12.9800 12.9900 13.0000 13.0100 13.0200 13.0300
Columns 1305 through 1312
13.0400 13.0500 13.0600 13.0700 13.0800 13.0900 13.1000 13.1100
Columns 1313 through 1320
13.1200 13.1300 13.1400 13.1500 13.1600 13.1700 13.1800 13.1900
Columns 1321 through 1328
13.2000 13.2100 13.2200 13.2300 13.2400 13.2500 13.2600 13.2700
Columns 1329 through 1336
13.2800 13.2900 13.3000 13.3100 13.3200 13.3300 13.3400 13.3500
Columns 1337 through 1344
13.3600 13.3700 13.3800 13.3900 13.4000 13.4100 13.4200 13.4300
Columns 1345 through 1352
13.4400 13.4500 13.4600 13.4700 13.4800 13.4900 13.5000 13.5100
Columns 1353 through 1360
13.5200 13.5300 13.5400 13.5500 13.5600 13.5700 13.5800 13.5900
Columns 1361 through 1368
13.6000 13.6100 13.6200 13.6300 13.6400 13.6500 13.6600 13.6700
Columns 1369 through 1376
13.6800 13.6900 13.7000 13.7100 13.7200 13.7300 13.7400 13.7500
Columns 1377 through 1384
13.7600 13.7700 13.7800 13.7900 13.8000 13.8100 13.8200 13.8300
Columns 1385 through 1392
13.8400 13.8500 13.8600 13.8700 13.8800 13.8900 13.9000 13.9100
Columns 1393 through 1400
13.9200 13.9300 13.9400 13.9500 13.9600 13.9700 13.9800 13.9900
Columns 1401 through 1408
14.0000 14.0100 14.0200 14.0300 14.0400 14.0500 14.0600 14.0700
Columns 1409 through 1416
14.0800 14.0900 14.1000 14.1100 14.1200 14.1300 14.1400 14.1500
Columns 1417 through 1424
14.1600 14.1700 14.1800 14.1900 14.2000 14.2100 14.2200 14.2300
Columns 1425 through 1432
14.2400 14.2500 14.2600 14.2700 14.2800 14.2900 14.3000 14.3100
Columns 1433 through 1440
14.3200 14.3300 14.3400 14.3500 14.3600 14.3700 14.3800 14.3900
Columns 1441 through 1448
14.4000 14.4100 14.4200 14.4300 14.4400 14.4500 14.4600 14.4700
Columns 1449 through 1456
14.4800 14.4900 14.5000 14.5100 14.5200 14.5300 14.5400 14.5500
Columns 1457 through 1464
14.5600 14.5700 14.5800 14.5900 14.6000 14.6100 14.6200 14.6300
Columns 1465 through 1472
14.6400 14.6500 14.6600 14.6700 14.6800 14.6900 14.7000 14.7100
Columns 1473 through 1480
14.7200 14.7300 14.7400 14.7500 14.7600 14.7700 14.7800 14.7900
Columns 1481 through 1488
14.8000 14.8100 14.8200 14.8300 14.8400 14.8500 14.8600 14.8700
Columns 1489 through 1496
14.8800 14.8900 14.9000 14.9100 14.9200 14.9300 14.9400 14.9500
Columns 1497 through 1504
14.9600 14.9700 14.9800 14.9900 15.0000 15.0100 15.0200 15.0300
Columns 1505 through 1512
15.0400 15.0500 15.0600 15.0700 15.0800 15.0900 15.1000 15.1100
Columns 1513 through 1520
15.1200 15.1300 15.1400 15.1500 15.1600 15.1700 15.1800 15.1900
Columns 1521 through 1528
15.2000 15.2100 15.2200 15.2300 15.2400 15.2500 15.2600 15.2700
Columns 1529 through 1536
15.2800 15.2900 15.3000 15.3100 15.3200 15.3300 15.3400 15.3500
Columns 1537 through 1544
15.3600 15.3700 15.3800 15.3900 15.4000 15.4100 15.4200 15.4300
Columns 1545 through 1552
15.4400 15.4500 15.4600 15.4700 15.4800 15.4900 15.5000 15.5100
Columns 1553 through 1560
15.5200 15.5300 15.5400 15.5500 15.5600 15.5700 15.5800 15.5900
Columns 1561 through 1568
15.6000 15.6100 15.6200 15.6300 15.6400 15.6500 15.6600 15.6700
Columns 1569 through 1576
15.6800 15.6900 15.7000 15.7100 15.7200 15.7300 15.7400 15.7500
Columns 1577 through 1584
15.7600 15.7700 15.7800 15.7900 15.8000 15.8100 15.8200 15.8300
Columns 1585 through 1592
15.8400 15.8500 15.8600 15.8700 15.8800 15.8900 15.9000 15.9100
Columns 1593 through 1600
15.9200 15.9300 15.9400 15.9500 15.9600 15.9700 15.9800 15.9900
Columns 1601 through 1608
16.0000 16.0100 16.0200 16.0300 16.0400 16.0500 16.0600 16.0700
Columns 1609 through 1616
16.0800 16.0900 16.1000 16.1100 16.1200 16.1300 16.1400 16.1500
Columns 1617 through 1624
16.1600 16.1700 16.1800 16.1900 16.2000 16.2100 16.2200 16.2300
Columns 1625 through 1632
16.2400 16.2500 16.2600 16.2700 16.2800 16.2900 16.3000 16.3100
Columns 1633 through 1640
16.3200 16.3300 16.3400 16.3500 16.3600 16.3700 16.3800 16.3900
Columns 1641 through 1648
16.4000 16.4100 16.4200 16.4300 16.4400 16.4500 16.4600 16.4700
Columns 1649 through 1656
16.4800 16.4900 16.5000 16.5100 16.5200 16.5300 16.5400 16.5500
Columns 1657 through 1664
16.5600 16.5700 16.5800 16.5900 16.6000 16.6100 16.6200 16.6300
Columns 1665 through 1672
16.6400 16.6500 16.6600 16.6700 16.6800 16.6900 16.7000 16.7100
Columns 1673 through 1680
16.7200 16.7300 16.7400 16.7500 16.7600 16.7700 16.7800 16.7900
Columns 1681 through 1688
16.8000 16.8100 16.8200 16.8300 16.8400 16.8500 16.8600 16.8700
Columns 1689 through 1696
16.8800 16.8900 16.9000 16.9100 16.9200 16.9300 16.9400 16.9500
Columns 1697 through 1704
16.9600 16.9700 16.9800 16.9900 17.0000 17.0100 17.0200 17.0300
Columns 1705 through 1712
17.0400 17.0500 17.0600 17.0700 17.0800 17.0900 17.1000 17.1100
Columns 1713 through 1720
17.1200 17.1300 17.1400 17.1500 17.1600 17.1700 17.1800 17.1900
Columns 1721 through 1728
17.2000 17.2100 17.2200 17.2300 17.2400 17.2500 17.2600 17.2700
Columns 1729 through 1736
17.2800 17.2900 17.3000 17.3100 17.3200 17.3300 17.3400 17.3500
Columns 1737 through 1744
17.3600 17.3700 17.3800 17.3900 17.4000 17.4100 17.4200 17.4300
Columns 1745 through 1752
17.4400 17.4500 17.4600 17.4700 17.4800 17.4900 17.5000 17.5100
Columns 1753 through 1760
17.5200 17.5300 17.5400 17.5500 17.5600 17.5700 17.5800 17.5900
Columns 1761 through 1768
17.6000 17.6100 17.6200 17.6300 17.6400 17.6500 17.6600 17.6700
Columns 1769 through 1776
17.6800 17.6900 17.7000 17.7100 17.7200 17.7300 17.7400 17.7500
Columns 1777 through 1784
17.7600 17.7700 17.7800 17.7900 17.8000 17.8100 17.8200 17.8300
Columns 1785 through 1792
17.8400 17.8500 17.8600 17.8700 17.8800 17.8900 17.9000 17.9100
Columns 1793 through 1800
17.9200 17.9300 17.9400 17.9500 17.9600 17.9700 17.9800 17.9900
Columns 1801 through 1808
18.0000 18.0100 18.0200 18.0300 18.0400 18.0500 18.0600 18.0700
Columns 1809 through 1816
18.0800 18.0900 18.1000 18.1100 18.1200 18.1300 18.1400 18.1500
Columns 1817 through 1824
18.1600 18.1700 18.1800 18.1900 18.2000 18.2100 18.2200 18.2300
Columns 1825 through 1832
18.2400 18.2500 18.2600 18.2700 18.2800 18.2900 18.3000 18.3100
Columns 1833 through 1840
18.3200 18.3300 18.3400 18.3500 18.3600 18.3700 18.3800 18.3900
Columns 1841 through 1848
18.4000 18.4100 18.4200 18.4300 18.4400 18.4500 18.4600 18.4700
Columns 1849 through 1856
18.4800 18.4900 18.5000 18.5100 18.5200 18.5300 18.5400 18.5500
Columns 1857 through 1864
18.5600 18.5700 18.5800 18.5900 18.6000 18.6100 18.6200 18.6300
Columns 1865 through 1872
18.6400 18.6500 18.6600 18.6700 18.6800 18.6900 18.7000 18.7100
Columns 1873 through 1880
18.7200 18.7300 18.7400 18.7500 18.7600 18.7700 18.7800 18.7900
Columns 1881 through 1888
18.8000 18.8100 18.8200 18.8300 18.8400 18.8500 18.8600 18.8700
Columns 1889 through 1896
18.8800 18.8900 18.9000 18.9100 18.9200 18.9300 18.9400 18.9500
Columns 1897 through 1904
18.9600 18.9700 18.9800 18.9900 19.0000 19.0100 19.0200 19.0300
Columns 1905 through 1912
19.0400 19.0500 19.0600 19.0700 19.0800 19.0900 19.1000 19.1100
Columns 1913 through 1920
19.1200 19.1300 19.1400 19.1500 19.1600 19.1700 19.1800 19.1900
Columns 1921 through 1928
19.2000 19.2100 19.2200 19.2300 19.2400 19.2500 19.2600 19.2700
Columns 1929 through 1936
19.2800 19.2900 19.3000 19.3100 19.3200 19.3300 19.3400 19.3500
Columns 1937 through 1944
19.3600 19.3700 19.3800 19.3900 19.4000 19.4100 19.4200 19.4300
Columns 1945 through 1952
19.4400 19.4500 19.4600 19.4700 19.4800 19.4900 19.5000 19.5100
Columns 1953 through 1960
19.5200 19.5300 19.5400 19.5500 19.5600 19.5700 19.5800 19.5900
Columns 1961 through 1968
19.6000 19.6100 19.6200 19.6300 19.6400 19.6500 19.6600 19.6700
Columns 1969 through 1976
19.6800 19.6900 19.7000 19.7100 19.7200 19.7300 19.7400 19.7500
Columns 1977 through 1984
19.7600 19.7700 19.7800 19.7900 19.8000 19.8100 19.8200 19.8300
Columns 1985 through 1992
19.8400 19.8500 19.8600 19.8700 19.8800 19.8900 19.9000 19.9100
Columns 1993 through 2000
19.9200 19.9300 19.9400 19.9500 19.9600 19.9700 19.9800 19.9900
Column 2001
20.0000
ejemplo3
x =
Columns 1 through 8
0 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700
Columns 9 through 16
0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500
Columns 17 through 24
0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300
Columns 25 through 32
0.2400 0.2500 0.2600 0.2700 0.2800 0.2900 0.3000 0.3100
Columns 33 through 40
0.3200 0.3300 0.3400 0.3500 0.3600 0.3700 0.3800 0.3900
Columns 41 through 48
0.4000 0.4100 0.4200 0.4300 0.4400 0.4500 0.4600 0.4700
Columns 49 through 56
0.4800 0.4900 0.5000 0.5100 0.5200 0.5300 0.5400 0.5500
Columns 57 through 64
0.5600 0.5700 0.5800 0.5900 0.6000 0.6100 0.6200 0.6300
Columns 65 through 72
0.6400 0.6500 0.6600 0.6700 0.6800 0.6900 0.7000 0.7100
Columns 73 through 80
0.7200 0.7300 0.7400 0.7500 0.7600 0.7700 0.7800 0.7900
Columns 81 through 88
0.8000 0.8100 0.8200 0.8300 0.8400 0.8500 0.8600 0.8700
Columns 89 through 96
0.8800 0.8900 0.9000 0.9100 0.9200 0.9300 0.9400 0.9500
Columns 97 through 104
0.9600 0.9700 0.9800 0.9900 1.0000 1.0100 1.0200 1.0300
Columns 105 through 112
1.0400 1.0500 1.0600 1.0700 1.0800 1.0900 1.1000 1.1100
Columns 113 through 120
1.1200 1.1300 1.1400 1.1500 1.1600 1.1700 1.1800 1.1900
Columns 121 through 128
1.2000 1.2100 1.2200 1.2300 1.2400 1.2500 1.2600 1.2700
Columns 129 through 136
1.2800 1.2900 1.3000 1.3100 1.3200 1.3300 1.3400 1.3500
Columns 137 through 144
1.3600 1.3700 1.3800 1.3900 1.4000 1.4100 1.4200 1.4300
Columns 145 through 152
1.4400 1.4500 1.4600 1.4700 1.4800 1.4900 1.5000 1.5100
Columns 153 through 160
1.5200 1.5300 1.5400 1.5500 1.5600 1.5700 1.5800 1.5900
Columns 161 through 168
1.6000 1.6100 1.6200 1.6300 1.6400 1.6500 1.6600 1.6700
Columns 169 through 176
1.6800 1.6900 1.7000 1.7100 1.7200 1.7300 1.7400 1.7500
Columns 177 through 184
1.7600 1.7700 1.7800 1.7900 1.8000 1.8100 1.8200 1.8300
Columns 185 through 192
1.8400 1.8500 1.8600 1.8700 1.8800 1.8900 1.9000 1.9100
Columns 193 through 200
1.9200 1.9300 1.9400 1.9500 1.9600 1.9700 1.9800 1.9900
Columns 201 through 208
2.0000 2.0100 2.0200 2.0300 2.0400 2.0500 2.0600 2.0700
Columns 209 through 216
2.0800 2.0900 2.1000 2.1100 2.1200 2.1300 2.1400 2.1500
Columns 217 through 224
2.1600 2.1700 2.1800 2.1900 2.2000 2.2100 2.2200 2.2300
Columns 225 through 232
2.2400 2.2500 2.2600 2.2700 2.2800 2.2900 2.3000 2.3100
Columns 233 through 240
2.3200 2.3300 2.3400 2.3500 2.3600 2.3700 2.3800 2.3900
Columns 241 through 248
2.4000 2.4100 2.4200 2.4300 2.4400 2.4500 2.4600 2.4700
Columns 249 through 256
2.4800 2.4900 2.5000 2.5100 2.5200 2.5300 2.5400 2.5500
Columns 257 through 264
2.5600 2.5700 2.5800 2.5900 2.6000 2.6100 2.6200 2.6300
Columns 265 through 272
2.6400 2.6500 2.6600 2.6700 2.6800 2.6900 2.7000 2.7100
Columns 273 through 280
2.7200 2.7300 2.7400 2.7500 2.7600 2.7700 2.7800 2.7900
Columns 281 through 288
2.8000 2.8100 2.8200 2.8300 2.8400 2.8500 2.8600 2.8700
Columns 289 through 296
2.8800 2.8900 2.9000 2.9100 2.9200 2.9300 2.9400 2.9500
Columns 297 through 304
2.9600 2.9700 2.9800 2.9900 3.0000 3.0100 3.0200 3.0300
Columns 305 through 312
3.0400 3.0500 3.0600 3.0700 3.0800 3.0900 3.1000 3.1100
Columns 313 through 320
3.1200 3.1300 3.1400 3.1500 3.1600 3.1700 3.1800 3.1900
Columns 321 through 328
3.2000 3.2100 3.2200 3.2300 3.2400 3.2500 3.2600 3.2700
Columns 329 through 336
3.2800 3.2900 3.3000 3.3100 3.3200 3.3300 3.3400 3.3500
Columns 337 through 344
3.3600 3.3700 3.3800 3.3900 3.4000 3.4100 3.4200 3.4300
Columns 345 through 352
3.4400 3.4500 3.4600 3.4700 3.4800 3.4900 3.5000 3.5100
Columns 353 through 360
3.5200 3.5300 3.5400 3.5500 3.5600 3.5700 3.5800 3.5900
Columns 361 through 368
3.6000 3.6100 3.6200 3.6300 3.6400 3.6500 3.6600 3.6700
Columns 369 through 376
3.6800 3.6900 3.7000 3.7100 3.7200 3.7300 3.7400 3.7500
Columns 377 through 384
3.7600 3.7700 3.7800 3.7900 3.8000 3.8100 3.8200 3.8300
Columns 385 through 392
3.8400 3.8500 3.8600 3.8700 3.8800 3.8900 3.9000 3.9100
Columns 393 through 400
3.9200 3.9300 3.9400 3.9500 3.9600 3.9700 3.9800 3.9900
Columns 401 through 408
4.0000 4.0100 4.0200 4.0300 4.0400 4.0500 4.0600 4.0700
Columns 409 through 416
4.0800 4.0900 4.1000 4.1100 4.1200 4.1300 4.1400 4.1500
Columns 417 through 424
4.1600 4.1700 4.1800 4.1900 4.2000 4.2100 4.2200 4.2300
Columns 425 through 432
4.2400 4.2500 4.2600 4.2700 4.2800 4.2900 4.3000 4.3100
Columns 433 through 440
4.3200 4.3300 4.3400 4.3500 4.3600 4.3700 4.3800 4.3900
Columns 441 through 448
4.4000 4.4100 4.4200 4.4300 4.4400 4.4500 4.4600 4.4700
Columns 449 through 456
4.4800 4.4900 4.5000 4.5100 4.5200 4.5300 4.5400 4.5500
Columns 457 through 464
4.5600 4.5700 4.5800 4.5900 4.6000 4.6100 4.6200 4.6300
Columns 465 through 472
4.6400 4.6500 4.6600 4.6700 4.6800 4.6900 4.7000 4.7100
Columns 473 through 480
4.7200 4.7300 4.7400 4.7500 4.7600 4.7700 4.7800 4.7900
Columns 481 through 488
4.8000 4.8100 4.8200 4.8300 4.8400 4.8500 4.8600 4.8700
Columns 489 through 496
4.8800 4.8900 4.9000 4.9100 4.9200 4.9300 4.9400 4.9500
Columns 497 through 504
4.9600 4.9700 4.9800 4.9900 5.0000 5.0100 5.0200 5.0300
Columns 505 through 512
5.0400 5.0500 5.0600 5.0700 5.0800 5.0900 5.1000 5.1100
Columns 513 through 520
5.1200 5.1300 5.1400 5.1500 5.1600 5.1700 5.1800 5.1900
Columns 521 through 528
5.2000 5.2100 5.2200 5.2300 5.2400 5.2500 5.2600 5.2700
Columns 529 through 536
5.2800 5.2900 5.3000 5.3100 5.3200 5.3300 5.3400 5.3500
Columns 537 through 544
5.3600 5.3700 5.3800 5.3900 5.4000 5.4100 5.4200 5.4300
Columns 545 through 552
5.4400 5.4500 5.4600 5.4700 5.4800 5.4900 5.5000 5.5100
Columns 553 through 560
5.5200 5.5300 5.5400 5.5500 5.5600 5.5700 5.5800 5.5900
Columns 561 through 568
5.6000 5.6100 5.6200 5.6300 5.6400 5.6500 5.6600 5.6700
Columns 569 through 576
5.6800 5.6900 5.7000 5.7100 5.7200 5.7300 5.7400 5.7500
Columns 577 through 584
5.7600 5.7700 5.7800 5.7900 5.8000 5.8100 5.8200 5.8300
Columns 585 through 592
5.8400 5.8500 5.8600 5.8700 5.8800 5.8900 5.9000 5.9100
Columns 593 through 600
5.9200 5.9300 5.9400 5.9500 5.9600 5.9700 5.9800 5.9900
Columns 601 through 608
6.0000 6.0100 6.0200 6.0300 6.0400 6.0500 6.0600 6.0700
Columns 609 through 616
6.0800 6.0900 6.1000 6.1100 6.1200 6.1300 6.1400 6.1500
Columns 617 through 624
6.1600 6.1700 6.1800 6.1900 6.2000 6.2100 6.2200 6.2300
Columns 625 through 632
6.2400 6.2500 6.2600 6.2700 6.2800 6.2900 6.3000 6.3100
Columns 633 through 640
6.3200 6.3300 6.3400 6.3500 6.3600 6.3700 6.3800 6.3900
Columns 641 through 648
6.4000 6.4100 6.4200 6.4300 6.4400 6.4500 6.4600 6.4700
Columns 649 through 656
6.4800 6.4900 6.5000 6.5100 6.5200 6.5300 6.5400 6.5500
Columns 657 through 664
6.5600 6.5700 6.5800 6.5900 6.6000 6.6100 6.6200 6.6300
Columns 665 through 672
6.6400 6.6500 6.6600 6.6700 6.6800 6.6900 6.7000 6.7100
Columns 673 through 680
6.7200 6.7300 6.7400 6.7500 6.7600 6.7700 6.7800 6.7900
Columns 681 through 688
6.8000 6.8100 6.8200 6.8300 6.8400 6.8500 6.8600 6.8700
Columns 689 through 696
6.8800 6.8900 6.9000 6.9100 6.9200 6.9300 6.9400 6.9500
Columns 697 through 704
6.9600 6.9700 6.9800 6.9900 7.0000 7.0100 7.0200 7.0300
Columns 705 through 712
7.0400 7.0500 7.0600 7.0700 7.0800 7.0900 7.1000 7.1100
Columns 713 through 720
7.1200 7.1300 7.1400 7.1500 7.1600 7.1700 7.1800 7.1900
Columns 721 through 728
7.2000 7.2100 7.2200 7.2300 7.2400 7.2500 7.2600 7.2700
Columns 729 through 736
7.2800 7.2900 7.3000 7.3100 7.3200 7.3300 7.3400 7.3500
Columns 737 through 744
7.3600 7.3700 7.3800 7.3900 7.4000 7.4100 7.4200 7.4300
Columns 745 through 752
7.4400 7.4500 7.4600 7.4700 7.4800 7.4900 7.5000 7.5100
Columns 753 through 760
7.5200 7.5300 7.5400 7.5500 7.5600 7.5700 7.5800 7.5900
Columns 761 through 768
7.6000 7.6100 7.6200 7.6300 7.6400 7.6500 7.6600 7.6700
Columns 769 through 776
7.6800 7.6900 7.7000 7.7100 7.7200 7.7300 7.7400 7.7500
Columns 777 through 784
7.7600 7.7700 7.7800 7.7900 7.8000 7.8100 7.8200 7.8300
Columns 785 through 792
7.8400 7.8500 7.8600 7.8700 7.8800 7.8900 7.9000 7.9100
Columns 793 through 800
7.9200 7.9300 7.9400 7.9500 7.9600 7.9700 7.9800 7.9900
Columns 801 through 808
8.0000 8.0100 8.0200 8.0300 8.0400 8.0500 8.0600 8.0700
Columns 809 through 816
8.0800 8.0900 8.1000 8.1100 8.1200 8.1300 8.1400 8.1500
Columns 817 through 824
8.1600 8.1700 8.1800 8.1900 8.2000 8.2100 8.2200 8.2300
Columns 825 through 832
8.2400 8.2500 8.2600 8.2700 8.2800 8.2900 8.3000 8.3100
Columns 833 through 840
8.3200 8.3300 8.3400 8.3500 8.3600 8.3700 8.3800 8.3900
Columns 841 through 848
8.4000 8.4100 8.4200 8.4300 8.4400 8.4500 8.4600 8.4700
Columns 849 through 856
8.4800 8.4900 8.5000 8.5100 8.5200 8.5300 8.5400 8.5500
Columns 857 through 864
8.5600 8.5700 8.5800 8.5900 8.6000 8.6100 8.6200 8.6300
Columns 865 through 872
8.6400 8.6500 8.6600 8.6700 8.6800 8.6900 8.7000 8.7100
Columns 873 through 880
8.7200 8.7300 8.7400 8.7500 8.7600 8.7700 8.7800 8.7900
Columns 881 through 888
8.8000 8.8100 8.8200 8.8300 8.8400 8.8500 8.8600 8.8700
Columns 889 through 896
8.8800 8.8900 8.9000 8.9100 8.9200 8.9300 8.9400 8.9500
Columns 897 through 904
8.9600 8.9700 8.9800 8.9900 9.0000 9.0100 9.0200 9.0300
Columns 905 through 912
9.0400 9.0500 9.0600 9.0700 9.0800 9.0900 9.1000 9.1100
Columns 913 through 920
9.1200 9.1300 9.1400 9.1500 9.1600 9.1700 9.1800 9.1900
Columns 921 through 928
9.2000 9.2100 9.2200 9.2300 9.2400 9.2500 9.2600 9.2700
Columns 929 through 936
9.2800 9.2900 9.3000 9.3100 9.3200 9.3300 9.3400 9.3500
Columns 937 through 944
9.3600 9.3700 9.3800 9.3900 9.4000 9.4100 9.4200 9.4300
Columns 945 through 952
9.4400 9.4500 9.4600 9.4700 9.4800 9.4900 9.5000 9.5100
Columns 953 through 960
9.5200 9.5300 9.5400 9.5500 9.5600 9.5700 9.5800 9.5900
Columns 961 through 968
9.6000 9.6100 9.6200 9.6300 9.6400 9.6500 9.6600 9.6700
Columns 969 through 976
9.6800 9.6900 9.7000 9.7100 9.7200 9.7300 9.7400 9.7500
Columns 977 through 984
9.7600 9.7700 9.7800 9.7900 9.8000 9.8100 9.8200 9.8300
Columns 985 through 992
9.8400 9.8500 9.8600 9.8700 9.8800 9.8900 9.9000 9.9100
Columns 993 through 1000
9.9200 9.9300 9.9400 9.9500 9.9600 9.9700 9.9800 9.9900
Columns 1001 through 1008
10.0000 10.0100 10.0200 10.0300 10.0400 10.0500 10.0600 10.0700
Columns 1009 through 1016
10.0800 10.0900 10.1000 10.1100 10.1200 10.1300 10.1400 10.1500
Columns 1017 through 1024
10.1600 10.1700 10.1800 10.1900 10.2000 10.2100 10.2200 10.2300
Columns 1025 through 1032
10.2400 10.2500 10.2600 10.2700 10.2800 10.2900 10.3000 10.3100
Columns 1033 through 1040
10.3200 10.3300 10.3400 10.3500 10.3600 10.3700 10.3800 10.3900
Columns 1041 through 1048
10.4000 10.4100 10.4200 10.4300 10.4400 10.4500 10.4600 10.4700
Columns 1049 through 1056
10.4800 10.4900 10.5000 10.5100 10.5200 10.5300 10.5400 10.5500
Columns 1057 through 1064
10.5600 10.5700 10.5800 10.5900 10.6000 10.6100 10.6200 10.6300
Columns 1065 through 1072
10.6400 10.6500 10.6600 10.6700 10.6800 10.6900 10.7000 10.7100
Columns 1073 through 1080
10.7200 10.7300 10.7400 10.7500 10.7600 10.7700 10.7800 10.7900
Columns 1081 through 1088
10.8000 10.8100 10.8200 10.8300 10.8400 10.8500 10.8600 10.8700
Columns 1089 through 1096
10.8800 10.8900 10.9000 10.9100 10.9200 10.9300 10.9400 10.9500
Columns 1097 through 1104
10.9600 10.9700 10.9800 10.9900 11.0000 11.0100 11.0200 11.0300
Columns 1105 through 1112
11.0400 11.0500 11.0600 11.0700 11.0800 11.0900 11.1000 11.1100
Columns 1113 through 1120
11.1200 11.1300 11.1400 11.1500 11.1600 11.1700 11.1800 11.1900
Columns 1121 through 1128
11.2000 11.2100 11.2200 11.2300 11.2400 11.2500 11.2600 11.2700
Columns 1129 through 1136
11.2800 11.2900 11.3000 11.3100 11.3200 11.3300 11.3400 11.3500
Columns 1137 through 1144
11.3600 11.3700 11.3800 11.3900 11.4000 11.4100 11.4200 11.4300
Columns 1145 through 1152
11.4400 11.4500 11.4600 11.4700 11.4800 11.4900 11.5000 11.5100
Columns 1153 through 1160
11.5200 11.5300 11.5400 11.5500 11.5600 11.5700 11.5800 11.5900
Columns 1161 through 1168
11.6000 11.6100 11.6200 11.6300 11.6400 11.6500 11.6600 11.6700
Columns 1169 through 1176
11.6800 11.6900 11.7000 11.7100 11.7200 11.7300 11.7400 11.7500
Columns 1177 through 1184
11.7600 11.7700 11.7800 11.7900 11.8000 11.8100 11.8200 11.8300
Columns 1185 through 1192
11.8400 11.8500 11.8600 11.8700 11.8800 11.8900 11.9000 11.9100
Columns 1193 through 1200
11.9200 11.9300 11.9400 11.9500 11.9600 11.9700 11.9800 11.9900
Columns 1201 through 1208
12.0000 12.0100 12.0200 12.0300 12.0400 12.0500 12.0600 12.0700
Columns 1209 through 1216
12.0800 12.0900 12.1000 12.1100 12.1200 12.1300 12.1400 12.1500
Columns 1217 through 1224
12.1600 12.1700 12.1800 12.1900 12.2000 12.2100 12.2200 12.2300
Columns 1225 through 1232
12.2400 12.2500 12.2600 12.2700 12.2800 12.2900 12.3000 12.3100
Columns 1233 through 1240
12.3200 12.3300 12.3400 12.3500 12.3600 12.3700 12.3800 12.3900
Columns 1241 through 1248
12.4000 12.4100 12.4200 12.4300 12.4400 12.4500 12.4600 12.4700
Columns 1249 through 1256
12.4800 12.4900 12.5000 12.5100 12.5200 12.5300 12.5400 12.5500
Columns 1257 through 1264
12.5600 12.5700 12.5800 12.5900 12.6000 12.6100 12.6200 12.6300
Columns 1265 through 1272
12.6400 12.6500 12.6600 12.6700 12.6800 12.6900 12.7000 12.7100
Columns 1273 through 1280
12.7200 12.7300 12.7400 12.7500 12.7600 12.7700 12.7800 12.7900
Columns 1281 through 1288
12.8000 12.8100 12.8200 12.8300 12.8400 12.8500 12.8600 12.8700
Columns 1289 through 1296
12.8800 12.8900 12.9000 12.9100 12.9200 12.9300 12.9400 12.9500
Columns 1297 through 1304
12.9600 12.9700 12.9800 12.9900 13.0000 13.0100 13.0200 13.0300
Columns 1305 through 1312
13.0400 13.0500 13.0600 13.0700 13.0800 13.0900 13.1000 13.1100
Columns 1313 through 1320
13.1200 13.1300 13.1400 13.1500 13.1600 13.1700 13.1800 13.1900
Columns 1321 through 1328
13.2000 13.2100 13.2200 13.2300 13.2400 13.2500 13.2600 13.2700
Columns 1329 through 1336
13.2800 13.2900 13.3000 13.3100 13.3200 13.3300 13.3400 13.3500
Columns 1337 through 1344
13.3600 13.3700 13.3800 13.3900 13.4000 13.4100 13.4200 13.4300
Columns 1345 through 1352
13.4400 13.4500 13.4600 13.4700 13.4800 13.4900 13.5000 13.5100
Columns 1353 through 1360
13.5200 13.5300 13.5400 13.5500 13.5600 13.5700 13.5800 13.5900
Columns 1361 through 1368
13.6000 13.6100 13.6200 13.6300 13.6400 13.6500 13.6600 13.6700
Columns 1369 through 1376
13.6800 13.6900 13.7000 13.7100 13.7200 13.7300 13.7400 13.7500
Columns 1377 through 1384
13.7600 13.7700 13.7800 13.7900 13.8000 13.8100 13.8200 13.8300
Columns 1385 through 1392
13.8400 13.8500 13.8600 13.8700 13.8800 13.8900 13.9000 13.9100
Columns 1393 through 1400
13.9200 13.9300 13.9400 13.9500 13.9600 13.9700 13.9800 13.9900
Columns 1401 through 1408
14.0000 14.0100 14.0200 14.0300 14.0400 14.0500 14.0600 14.0700
Columns 1409 through 1416
14.0800 14.0900 14.1000 14.1100 14.1200 14.1300 14.1400 14.1500
Columns 1417 through 1424
14.1600 14.1700 14.1800 14.1900 14.2000 14.2100 14.2200 14.2300
Columns 1425 through 1432
14.2400 14.2500 14.2600 14.2700 14.2800 14.2900 14.3000 14.3100
Columns 1433 through 1440
14.3200 14.3300 14.3400 14.3500 14.3600 14.3700 14.3800 14.3900
Columns 1441 through 1448
14.4000 14.4100 14.4200 14.4300 14.4400 14.4500 14.4600 14.4700
Columns 1449 through 1456
14.4800 14.4900 14.5000 14.5100 14.5200 14.5300 14.5400 14.5500
Columns 1457 through 1464
14.5600 14.5700 14.5800 14.5900 14.6000 14.6100 14.6200 14.6300
Columns 1465 through 1472
14.6400 14.6500 14.6600 14.6700 14.6800 14.6900 14.7000 14.7100
Columns 1473 through 1480
14.7200 14.7300 14.7400 14.7500 14.7600 14.7700 14.7800 14.7900
Columns 1481 through 1488
14.8000 14.8100 14.8200 14.8300 14.8400 14.8500 14.8600 14.8700
Columns 1489 through 1496
14.8800 14.8900 14.9000 14.9100 14.9200 14.9300 14.9400 14.9500
Columns 1497 through 1504
14.9600 14.9700 14.9800 14.9900 15.0000 15.0100 15.0200 15.0300
Columns 1505 through 1512
15.0400 15.0500 15.0600 15.0700 15.0800 15.0900 15.1000 15.1100
Columns 1513 through 1520
15.1200 15.1300 15.1400 15.1500 15.1600 15.1700 15.1800 15.1900
Columns 1521 through 1528
15.2000 15.2100 15.2200 15.2300 15.2400 15.2500 15.2600 15.2700
Columns 1529 through 1536
15.2800 15.2900 15.3000 15.3100 15.3200 15.3300 15.3400 15.3500
Columns 1537 through 1544
15.3600 15.3700 15.3800 15.3900 15.4000 15.4100 15.4200 15.4300
Columns 1545 through 1552
15.4400 15.4500 15.4600 15.4700 15.4800 15.4900 15.5000 15.5100
Columns 1553 through 1560
15.5200 15.5300 15.5400 15.5500 15.5600 15.5700 15.5800 15.5900
Columns 1561 through 1568
15.6000 15.6100 15.6200 15.6300 15.6400 15.6500 15.6600 15.6700
Columns 1569 through 1576
15.6800 15.6900 15.7000 15.7100 15.7200 15.7300 15.7400 15.7500
Columns 1577 through 1584
15.7600 15.7700 15.7800 15.7900 15.8000 15.8100 15.8200 15.8300
Columns 1585 through 1592
15.8400 15.8500 15.8600 15.8700 15.8800 15.8900 15.9000 15.9100
Columns 1593 through 1600
15.9200 15.9300 15.9400 15.9500 15.9600 15.9700 15.9800 15.9900
Columns 1601 through 1608
16.0000 16.0100 16.0200 16.0300 16.0400 16.0500 16.0600 16.0700
Columns 1609 through 1616
16.0800 16.0900 16.1000 16.1100 16.1200 16.1300 16.1400 16.1500
Columns 1617 through 1624
16.1600 16.1700 16.1800 16.1900 16.2000 16.2100 16.2200 16.2300
Columns 1625 through 1632
16.2400 16.2500 16.2600 16.2700 16.2800 16.2900 16.3000 16.3100
Columns 1633 through 1640
16.3200 16.3300 16.3400 16.3500 16.3600 16.3700 16.3800 16.3900
Columns 1641 through 1648
16.4000 16.4100 16.4200 16.4300 16.4400 16.4500 16.4600 16.4700
Columns 1649 through 1656
16.4800 16.4900 16.5000 16.5100 16.5200 16.5300 16.5400 16.5500
Columns 1657 through 1664
16.5600 16.5700 16.5800 16.5900 16.6000 16.6100 16.6200 16.6300
Columns 1665 through 1672
16.6400 16.6500 16.6600 16.6700 16.6800 16.6900 16.7000 16.7100
Columns 1673 through 1680
16.7200 16.7300 16.7400 16.7500 16.7600 16.7700 16.7800 16.7900
Columns 1681 through 1688
16.8000 16.8100 16.8200 16.8300 16.8400 16.8500 16.8600 16.8700
Columns 1689 through 1696
16.8800 16.8900 16.9000 16.9100 16.9200 16.9300 16.9400 16.9500
Columns 1697 through 1704
16.9600 16.9700 16.9800 16.9900 17.0000 17.0100 17.0200 17.0300
Columns 1705 through 1712
17.0400 17.0500 17.0600 17.0700 17.0800 17.0900 17.1000 17.1100
Columns 1713 through 1720
17.1200 17.1300 17.1400 17.1500 17.1600 17.1700 17.1800 17.1900
Columns 1721 through 1728
17.2000 17.2100 17.2200 17.2300 17.2400 17.2500 17.2600 17.2700
Columns 1729 through 1736
17.2800 17.2900 17.3000 17.3100 17.3200 17.3300 17.3400 17.3500
Columns 1737 through 1744
17.3600 17.3700 17.3800 17.3900 17.4000 17.4100 17.4200 17.4300
Columns 1745 through 1752
17.4400 17.4500 17.4600 17.4700 17.4800 17.4900 17.5000 17.5100
Columns 1753 through 1760
17.5200 17.5300 17.5400 17.5500 17.5600 17.5700 17.5800 17.5900
Columns 1761 through 1768
17.6000 17.6100 17.6200 17.6300 17.6400 17.6500 17.6600 17.6700
Columns 1769 through 1776
17.6800 17.6900 17.7000 17.7100 17.7200 17.7300 17.7400 17.7500
Columns 1777 through 1784
17.7600 17.7700 17.7800 17.7900 17.8000 17.8100 17.8200 17.8300
Columns 1785 through 1792
17.8400 17.8500 17.8600 17.8700 17.8800 17.8900 17.9000 17.9100
Columns 1793 through 1800
17.9200 17.9300 17.9400 17.9500 17.9600 17.9700 17.9800 17.9900
Columns 1801 through 1808
18.0000 18.0100 18.0200 18.0300 18.0400 18.0500 18.0600 18.0700
Columns 1809 through 1816
18.0800 18.0900 18.1000 18.1100 18.1200 18.1300 18.1400 18.1500
Columns 1817 through 1824
18.1600 18.1700 18.1800 18.1900 18.2000 18.2100 18.2200 18.2300
Columns 1825 through 1832
18.2400 18.2500 18.2600 18.2700 18.2800 18.2900 18.3000 18.3100
Columns 1833 through 1840
18.3200 18.3300 18.3400 18.3500 18.3600 18.3700 18.3800 18.3900
Columns 1841 through 1848
18.4000 18.4100 18.4200 18.4300 18.4400 18.4500 18.4600 18.4700
Columns 1849 through 1856
18.4800 18.4900 18.5000 18.5100 18.5200 18.5300 18.5400 18.5500
Columns 1857 through 1864
18.5600 18.5700 18.5800 18.5900 18.6000 18.6100 18.6200 18.6300
Columns 1865 through 1872
18.6400 18.6500 18.6600 18.6700 18.6800 18.6900 18.7000 18.7100
Columns 1873 through 1880
18.7200 18.7300 18.7400 18.7500 18.7600 18.7700 18.7800 18.7900
Columns 1881 through 1888
18.8000 18.8100 18.8200 18.8300 18.8400 18.8500 18.8600 18.8700
Columns 1889 through 1896
18.8800 18.8900 18.9000 18.9100 18.9200 18.9300 18.9400 18.9500
Columns 1897 through 1904
18.9600 18.9700 18.9800 18.9900 19.0000 19.0100 19.0200 19.0300
Columns 1905 through 1912
19.0400 19.0500 19.0600 19.0700 19.0800 19.0900 19.1000 19.1100
Columns 1913 through 1920
19.1200 19.1300 19.1400 19.1500 19.1600 19.1700 19.1800 19.1900
Columns 1921 through 1928
19.2000 19.2100 19.2200 19.2300 19.2400 19.2500 19.2600 19.2700
Columns 1929 through 1936
19.2800 19.2900 19.3000 19.3100 19.3200 19.3300 19.3400 19.3500
Columns 1937 through 1944
19.3600 19.3700 19.3800 19.3900 19.4000 19.4100 19.4200 19.4300
Columns 1945 through 1952
19.4400 19.4500 19.4600 19.4700 19.4800 19.4900 19.5000 19.5100
Columns 1953 through 1960
19.5200 19.5300 19.5400 19.5500 19.5600 19.5700 19.5800 19.5900
Columns 1961 through 1968
19.6000 19.6100 19.6200 19.6300 19.6400 19.6500 19.6600 19.6700
Columns 1969 through 1976
19.6800 19.6900 19.7000 19.7100 19.7200 19.7300 19.7400 19.7500
Columns 1977 through 1984
19.7600 19.7700 19.7800 19.7900 19.8000 19.8100 19.8200 19.8300
Columns 1985 through 1992
19.8400 19.8500 19.8600 19.8700 19.8800 19.8900 19.9000 19.9100
Columns 1993 through 2000
19.9200 19.9300 19.9400 19.9500 19.9600 19.9700 19.9800 19.9900
Column 2001
20.0000
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http://oeis.org/A321501/internal | 1,586,172,942,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371624083.66/warc/CC-MAIN-20200406102322-20200406132822-00511.warc.gz | 131,458,546 | 3,417 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A321501 Numbers not of the form (x - y)(x^2 - y^2) with x > y > 0; complement of A321499. 2
%I
%S 0,1,2,4,6,8,10,12,14,18,20,22,26,28,30,34,36,38,42,44,46,50,52,54,58,
%T 60,62,66,68,70,74,76,78,82,84,86,90,92,94,98,100,102,106,108,110,114,
%U 116,118,122,124,126,130,132,134,138,140,142,146,148,150,154,156,158,162,164,166,170,172,174,178
%N Numbers not of the form (x - y)(x^2 - y^2) with x > y > 0; complement of A321499.
%C Equivalently, numbers not of the form (x - y)^2*(x + y) or d^2*(2m + d), for (x, y) = (m+d, m). This shows that excluded are all squares d^2 > 0 times any number of the same parity and larger than d. In particular, for d=1, all odd numbers > 1, and for d=2, 4*(even numbers > 4) = 8*(odd numbers > 2). For larger d, no further (neither odd nor even) numbers are excluded.
%C So apart from 0, 1 and 8, this consists of even numbers not multiple of 8. All these numbers occur, since for larger (odd or even) d, no additional term is excluded.
%F Asymptotic density is 3/8.
%F a(n) = round((n-2)*9/8)*2 for all n > 6.
%e a(1) = 0, a(2) = 1 and a(3) = 2 obviously can't be of the form (x - y)(x^2 - y^2) with x > y > 0, which is necessarily greater than 1*3 = 3.
%e See A321499 for examples of the terms that are not in the sequence.
%o (PARI) is(n)={!n||!fordiv(n,d, d^2*(d+2)>n && break; n%d^2&&next; bittest(n\d^2-d,0)||return)} \\ Uses the initial definition. More efficient variant below:
%o (PARI) select( is_A321501(n)=!bittest(n,0)&&(n%8||n<9)||n<3, [0..99]) \\ Defines the function is_A321501(). The select() command is an illustration and a check.
%o (PARI) A321501_list(M)={setunion([1],setminus([0..M\2]*2,[2..M\8]*8))} \\ Return all terms up to M; more efficient than to use select(...,[0..M]) as above.
%o (PARI) A321501(n)=if(n>6,(n-2)*9\/8*2,n>3,n*2-4,n-1)
%Y See A321499 for the complement: numbers of the form (x-y)(x^2-y^2).
%Y See A321491 for numbers of the form (x+y)(x^2+y^2).
%K nonn,easy
%O 1,3
%A _M. F. Hasler_, Nov 22 2018
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Last modified April 6 07:02 EDT 2020. Contains 333267 sequences. (Running on oeis4.) | 936 | 2,508 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-16 | latest | en | 0.743677 |
https://math.stackexchange.com/questions/2861045/hyperbola-asymptotes-from-conic-general-equation?noredirect=1 | 1,563,773,743,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527531.84/warc/CC-MAIN-20190722051628-20190722073628-00308.warc.gz | 469,289,791 | 34,296 | # Hyperbola asymptotes from conic general equation [duplicate]
This question already has an answer here:
If I have the coefficients of the following equation:
$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$
And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F?
i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas.
## marked as duplicate by Ng Chung Tak, Arnaud Mortier, Brahadeesh, Rhys Steele, Xander HendersonAug 6 '18 at 1:05
• What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward. – amd Jul 24 '18 at 5:59
First find the centre of the conic. This is the point $(u,v)$ such that the equation of the conic can be rewritten as $$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$ For the conic to be a hyperbola, the quadratic part has to factor into distinct linear factors over $\Bbb R$: $$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$ Then the asymptotes are $$R_i (X-u)+S_i(Y-v)=0$$ ($i=1$, $2$).
• Looks good. I just need to figure out the degenerate cases and how to handle them... – Justin Jul 24 '18 at 18:52
The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as: $$(ax+by+c)(dx+ey+f)=g.$$ You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $A\dots F$ to obtain $a\dots g$.
In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $A\ne0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.
In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2\times2$ symmetric.
It can be centered with
$$(x+A^{-1}b)^TA(x+A^{-1}b)-b^TA^{-T}A^Tb+c=y^TAy+c'=0.$$
Then, by diagonalizing $A$,
$$y^TP\Lambda P^{-1}y+c'=z^T\Lambda z+c'=0.$$
For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is
$$\lambda_uu^2-\lambda_vv^2=(\sqrt{\lambda_u}u+\sqrt{\lambda_v}v)(\sqrt{\lambda_u}u-\sqrt{\lambda_v}v)=-c'.$$
The two factors are the asymptotes, and in the original coordinates
$$x=P^{-1}z-A^{-1}b.$$
The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = \begin{bmatrix}A&\frac B2&\frac D2\\\frac B2&C&\frac E2\\\frac D2&\frac E2&F\end{bmatrix}$$ and the line at infinity is $\mathscr l = [0,0,1]^T$. Let $\mathscr l_\times$ be the “cross product matrix” of $\mathscr l$. We can compute the hyperbola-line intersection by finding a value of $\alpha$ that makes $$\mathscr l_\times^TQ\mathscr l_\times+\alpha\mathscr l_\times = \begin{bmatrix}C&-\alpha-\frac B2&0\\\alpha-\frac B2&A&0\\0&0&0\end{bmatrix}$$ a rank-one matrix. This occurs when the principal $2\times2$ minor vanishes, which leads to a quadratic equation in $\alpha$ with solutions $\pm\frac12\sqrt{B^2-4AC}$. Taking the positive root gives the matrix $$\begin{bmatrix}C&-\frac12\left(B+\sqrt{B^2-4AC}\right)&0 \\ -\frac12\left(B^2-\sqrt{B^2-4AC}\right)&A&0\\0&0&0\end{bmatrix}.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $C\ne0$, then the two points are $\mathbf p_1 = \left[C,-\frac12\left(B+\sqrt{B^2-4AC}\right),0\right]^T$ and $\mathbf p_2 = \left[C,-\frac12\left(B-\sqrt{B^2-4AC}\right),0\right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $\mathscr m_i = Q\mathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]\mathscr m_i=0$, or $[x,y,1]Q\mathbf p_i=0$. | 1,210 | 3,739 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-30 | latest | en | 0.863203 |
https://www.shaalaa.com/question-bank-solutions/fill-in-the-blank-the-empty-set-is-a-____-of-every-set-concept-cardinal-property-sets_120592 | 1,618,607,411,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00365.warc.gz | 1,108,131,515 | 8,651 | # Fill in the Blank: the Empty Set is a ____ of Every Set. - Mathematics
Fill in the Blanks
Fill in the blank:
The empty set is a ____ of every set.
#### Solution
The empty set is a subset of every set.
Concept: Concept for Cardinal Property of Sets
Is there an error in this question or solution?
#### APPEARS IN
Selina Concise Mathematics Class 7 ICSE
Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)
Exercise 13 (C) | Q 1.3 | 117 | 449 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-17 | latest | en | 0.808609 |
https://dhfasthealth.com/qa/question-how-many-30-minutes-is-5-hours.html | 1,600,525,570,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400192778.51/warc/CC-MAIN-20200919142021-20200919172021-00256.warc.gz | 351,046,615 | 7,807 | # Question: How Many 30 Minutes Is 5 Hours?
## How many hours is 5 hours a week?
Weeks to Hours Conversion TableWeeksHours2 Weeks336 Hours3 Weeks504 Hours4 Weeks672 Hours5 Weeks840 Hours22 more rows.
## What is 2 hours and 45 minutes expressed in just minutes?
What is 2 hours and 45 minutes expressed in just minutes? 2 × 60 = 120. 120 + 45 = 165. The answer is 165 minutes.
## What is 6.75 hours in hours and minutes?
6.75 hours with the decimal point is 6.75 hours in terms of hours. 6:75 with the colon is 6 hours and 75 minutes. . 75 = fractional hours.
## What is .50 of an hour?
For example, 50 percent of an hour equals 30 minutes, because 0.50 * 60 equals 30.
## How many minutes does a year have?
An average Gregorian year is 365.2425 days (52.1775 weeks, 8765.82 hours, 525949.2 minutes or 31556952 seconds). For this calendar, a common year is 365 days (8760 hours, 525600 minutes or 31536000 seconds), and a leap year is 366 days (8784 hours, 527040 minutes or 31622400 seconds).
## Whats .75 of an hour?
Decimal Hours-to-Minutes Conversion ChartMinutesTenths of an HourHundredths of an Hour44.7.7445.7.7546.7.7647.7.7855 more rows
## What is .8 of an hour?
Billing Increment Chart—Minutes to Tenths of an HourMinutesTime31-36.637-42.743-48.849-54.96 more rows
## What is 5.5 in hours and minutes?
Minutes to Hours Conversion TableMinutesHours330 Minutes5.5 Hours345 Minutes5.75 Hours360 Minutes6 Hours375 Minutes6.25 Hours55 more rows
## What is .25 of an hour?
Conversion Chart – Minutes to Hundredths of an Hour Enter time in Oracle Self Service as hundredths of an hour. For example 15 minutes (¼ hour) equals . 25, 30 minutes (½ hour) equals . 5, etc.
## How many 30 minutes is 6 hours?
Hours to Minutes Conversion TableHoursMinutes3 Hours180 Minutes4 Hours240 Minutes5 Hours300 Minutes6 Hours360 Minutes20 more rows
## What decimal is 5 hours 30 minutes?
Time to Decimal Conversion TableTimeHoursSeconds05:30:005.51980005:35:005.5833332010005:40:005.6666672040005:45:005.7520700115 more rows
## How many seconds does 5 hours have?
Which is the same to say that 5 hours is 18000 seconds.
## How much is 0.01 hours?
This conversion of 0.01 hours to minutes has been calculated by multiplying 0.01 hours by 60 and the result is 0.6 minutes.
## What is the decimal for 20 minutes?
0.33Minute Conversion ChartMinutesDecimal Conversion180.30190.32200.33210.3526 more rows
## How do you calculate minutes?
First, divide the number of seconds by 60 to get it into minutes form. Add this to your number of minutes, then divide again by 60 to get hours. | 741 | 2,592 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2020-40 | latest | en | 0.760184 |
http://projecteuler.net/problem=202 | 1,409,640,322,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535921869.7/warc/CC-MAIN-20140901014521-00319-ip-10-180-136-8.ec2.internal.warc.gz | 461,892,948 | 2,223 | ## Laserbeam
### Problem 202
Published on Saturday, 5th July 2008, 02:00 pm; Solved by 1477
Three mirrors are arranged in the shape of an equilateral triangle, with their reflective surfaces pointing inwards. There is an infinitesimal gap at each vertex of the triangle through which a laser beam may pass.
Label the vertices A, B and C. There are 2 ways in which a laser beam may enter vertex C, bounce off 11 surfaces, then exit through the same vertex: one way is shown below; the other is the reverse of that.
There are 80840 ways in which a laser beam may enter vertex C, bounce off 1000001 surfaces, then exit through the same vertex.
In how many ways can a laser beam enter at vertex C, bounce off 12017639147 surfaces, then exit through the same vertex? | 187 | 767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2014-35 | latest | en | 0.931038 |
http://www.jiskha.com/display.cgi?id=1352858101 | 1,498,487,496,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320763.95/warc/CC-MAIN-20170626133830-20170626153830-00366.warc.gz | 547,276,650 | 3,890 | # math
posted by .
What is 3^-2 (it is 3 to the -2 squared)
I thought the answer was 1/9
and my friend thought it is .1 repeated
.1 with the line over it
can you tell us which one of us is correct?
Thank you
• math -
You are both correct since
.1111.... = 1/9 | 84 | 266 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2017-26 | latest | en | 0.987946 |
https://homeworkdave.com/product/algorithms-homework-9/ | 1,660,118,837,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00762.warc.gz | 301,811,803 | 30,374 | Sale!
# Algorithms Homework 9
\$25.00
Category:
Homework 9
1. Describe a divide-and-conquer algorithm that accepts a positive integer
n and computes blg nc (that is, the largest integer x such that 2x ≤ n).
Your algorithm should take O(lg(lg n)) time.
Hint: you may wish to base your approach on one-sided binary search,
which starts at 1, doubles the value until it becomes too large, then performs binary search between the last value that worked and the first value
that failed. (Divide-and-conquer algorithms aren’t required to be recursive.) You may assume that the square root function takes Θ(1) time,
though there is an O(lg lg n) algorithm that does not use the square root.
1
## Reviews
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https://www.slideshare.net/MeeranBanday/magic-square-23397544 | 1,516,168,052,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886815.20/warc/CC-MAIN-20180117043259-20180117063259-00368.warc.gz | 1,011,492,998 | 35,952 | Successfully reported this slideshow.
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# Magic square
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### Magic square
1. 1. Magic SquareMeeran Ali AhmadClass IXth –AR.no- 08
2. 2. This presentation aims to explore the possibleapplications of magic squares in everyday life.In doing this, different types of magic squareswill be investigated and the methods used toconstruct them. How magic squares haveevolved and where they originally came fromshould also be considered. Also, variations onthe basic magic square will be looked at to seeif these have any practical applications.
3. 3. In recreational mathematics, a magic square of order n is an arrangementof n2 numbers, usually distinct integers, in a square, such that the nnumbers in all rows, all columns, and both diagonals sum to the sameconstant. A normal magic square contains the integers from 1 to n2. Theterm "magic square" is also sometimes used to refer to any of varioustypes of word square. Normal magic squares exist for all orders n ≥ 1 except n = 2, although thecase n = 1 is trivial, consisting of a single cell containing the number 1.The smallest nontrivial case, shown below, is of order 3. The constant sum in every row, column and diagonal is called the magicconstant or magic sum, M. The magic constant of a normal magic squaredepends only on n and has the value. In the following magic square , you mayobserve that the 5 numbers in all rows,all columns, and both diagonals sum to 205.9 77 43 75 137 57 63 21 2723 3 41 79 5955 61 19 25 4581 7 39 5 73
4. 4. Magic squares were known to Chinesemathematicians, as early as 650 BCE and Arabmathematicians, possibly as early as the 7th century,when the Arabs conquered northwestern parts of theIndian subcontinent and learned Indian mathematicsand astronomy, including other aspects ofcombinatorial mathematics. The first magic squares oforder 5 and 6 appear in an encyclopedia from Baghdadcirca 983 CE, the Encyclopedia of the Brethren of Purity(Rasail Ihkwan al-Safa); simpler magic squares wereknown to several earlier Arab mathematicians. Some ofthese squares were later used in conjunction withmagic letters as in (Shams Al-maarif) to assist Arabillusionists and magicians.
5. 5. There are many ways to construct magic squares, but the standard(and most simple) way is to follow certainconfigurations/formulas which generate regular patterns. Magicsquares exist for all values of n, with only one exception: it isimpossible to construct a magic square of order 2. Magic squarescan be classified into three types: odd, doubly even (n divisible byfour) and singly even (n even, but not divisible by four). Odd anddoubly even magic squares are easy to generate; the constructionof singly even magic squares is more difficult but several methodsexist, including the LUX method for magic squares (due to JohnHorton Conway) and the Strachey method for magic squares. Group theory was also used for constructing new magic squaresof a given order from one of them, please see. The number of different n×n magic squares for n from 1 to 5, notcounting rotations and reflections: 1, 0, 1, 880, 275305224 (sequence A006052).The number for n = 6has been estimated to 1.7745×1019.
6. 6. Magic squares were used in a number of ways in the ancientperiod such as- MusicThe main area of the application of magic squares to music is in rhythm,rather than notes. Indian musicians seem to have applied them to theirmusic and they seem to be useful in time cycles and additive rhythm. .This is because for rhythm, consecutive numbers 1 to are not used to fillthe cells of the magic square. SudokuSudoku was first introduced in 1979 and became popular in Japan during the1980’s (Pegg & Weisstein, 2006). It has recently become a very popularpuzzle in Europe, but it is actually a form of Latin square. A Sudokusquare is a 9x9 grid, split into 9 3x3 sub-squares. Each sub-square is filledin with the numbers 1 to where , so that the 9x9 grid becomes a Latinsquare. This means each row and column contain the numbers 1 to 9only once. Therefore each row, column and sub-square will sum to thesame amount.
7. 7. Mathematicians today do not need tospeculate and attach meaning to magic squaresto make them important, as has been done inthe past with Chinese and other myths. Thesquares were thought to be mysterious andmagic, although now it is clear that they arejust ways of arranging numbers and symbolsusing certain rules. They can be applied tomusic and Sudoku as has been discussed butare mainly of interest in mathematics for their“magic” properties rather than their practicalapplications. | 1,156 | 4,710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2018-05 | latest | en | 0.896859 |
https://gaming.stackexchange.com/questions/252320/how-much-cdr-is-required-for-permanent-vengeance-uptime/252321 | 1,624,462,593,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488539480.67/warc/CC-MAIN-20210623134306-20210623164306-00359.warc.gz | 276,931,821 | 39,357 | How much CDR is required for permanent Vengeance uptime?
In Diablo 3, Demon Hunters have access to the Vengeance skill, which among other things makes the Demon Hunter immune to crowd control and increases his/her damage by 40%.
With patch 2.4 adding a legendary power to the Dawn hand crossbow, which reduces the cooldown of Vengeance by up to 65%, how much CDR (cooldown reduction) is required to achieve permanent uptime on Vengeance?
• More than %26 if you use Dawn. – Ave Jan 18 '16 at 12:46
• @ardaozkal Technically true, since 36.5 is indeed more than 26. – Svj0hn Jan 19 '16 at 9:37
TL;DR
Without Dawn, you need ~77.8% CDR.
With Dawn, you need ~36.5% CDR.
(This is the number you see in your inventory, not the sum of CDR on your gear. See below for explanation and how to achieve this).
Maths
How CDR stacks
As briefly explained in this answer, Cooldown Reduction (CDR) stacks multiplicatively. This often leads to confusion among newer players, as the final number is smaller than the sum of CDR on your equipment.
For example:
Say you have two items that reduce the cooldown of a skill by 50 % and 20 % respectively. A naive interpretation is that the resulting cooldown will be reduced by 50 % + 20 % = 70 %. But the cooldown is only reduced by 60%. Why?
The formula for cooldown reduction goes like this:
`[CD] = [BCD] * (1 - CDR1) * (1 - CDR2) * (1 - CDR3) * ...`,
where `CDR1, CDR2, ...` are the CDR numbers for each individual contributor, expressed as a fraction (so 20 % = 0.2) and `[BCD]` is the base cooldown of the skill.
So for the example above (50 % and 20 %), we get:
`[CD] = [BCD] * (1 - 0.5) * (1 - 0.2) = [BCD] * 0.5 * 0.8 = [BCD] * 0.4`
So the resulting cooldown is 40 % of the base cooldown.
Is this 'diminishing returns'?
No. There is no reduced effectiveness of CDR as more is added. Adding 20 % CDR will increase the amount of casts possible in a given time by `1 / (1 - 0.2) = 1.25`, or 25 %. The apparent diminishing is just an effect of using percentage-based cooldown reduction (fixed time reduction would be a different manner).
You can think of it as follows:
Any subsequent CDR effect reduces the remaining cooldown by exactly how much it says. Since earlier CDR reduces the cooldown, there is less cooldown to be reduced, and thus the effect of subsequent CDR appears smaller (as measured in seconds).
Permanent Vengeance
Vengeance has a 90 second cooldown and a 20 second duration. To maintain permanent uptime, we need to reduce the cooldown below 20 seconds.
We then have the following relation:
`20 > 90 * (1 - CDR)`,
which when solved for `CDR` becomes (given `CDR < 1`):
`CDR > 1 - 20/90` which approximates to `CDR > 0.778`.
We thus need at least 77.8% CDR for permanent uptime.
Keep in mind that `CDR` here represents the resulting combined cooldown reduction from all your gear, subject to the stacking detailed above. This is in most cases unrealistic to achieve with CDR from gear, paragons and gem. Instead, we can look to Dawn.
Permanent Vengeance with Dawn
Now we have Dawn, which reduces the cooldown of Vengeance by 50-65 %. We can plug this into the CDR formula as one of the factors:
`20 > 90 * (1 - 0.65) * (1 - CDR)`,
which when solved for `CDR` gives (again, given `CRD < 1`):
`CDR > 1 - 20 / (90 * (1 - 0.65))` or approximately `CDR > 0.365`.
Which tells us we need at least 36.5 % CDR from gear, paragon and/or gems.
This assumes that your Dawn rolled a perfect 65 %. This is a fair assumption, considering most people (as of 2.4) use Dawn in Kanai's Cube rather than actually equip it (it automatically gets the best possible roll when in the cube).
Ways of getting enough CDR for permanent uptime
We have the following commonly obtainable sources of CDR. I will assume Dawn is being used with a 65 % roll, as is the most common case:
• Paragon - up to 10 %
• Weapons - up to 10 %
• Armor, Quiver, Jewelry - up to 8 %
• Diamond in head slot socket - up to 12.5 % (25 % with Leoric's Crown)
• Gogoc of Swiftness - 15 % (but not fully consistent)
The most common suggestion is to use Paragon (10 %), Diamond (12.5 %) and three pieces of equipment (8 %). This brings it to a total of:
`[CD] = [BCD] * (1 - 0.1) * (1 - 0.125) * (1 - 0.08)^3`,
which becomes:
`[CD] ~ [BCD] * 0.613`, which equates to CDR ~ 38.7 %.
This is slightly over what is needed, but that is a good thing. Not only does it give you margin of error during gameplay (overlap), but it also allows you to keep permanent uptime even if your gear doesn't roll perfectly.
You can mix and match any sources of CDR to arrive at the 36.5 % required, but in my personal opinion, I suggest using Paragon, Diamond in head, Shoulders, and then choosing between Quiver, Gloves and your Rings, depending on your preferred balance of toughness and damage (and what you happen to find).
• I know I'm really late here, but this works great if you also use Obsidian Ring of the Zodiac. I normally use that and Dawn, extracted or equipped depending on the quality I find. – Jerome Viveiros Mar 19 '20 at 13:57 | 1,408 | 5,067 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-25 | longest | en | 0.917122 |
http://mathhelpforum.com/algebra/16455-graph-problem.html | 1,526,875,844,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863923.6/warc/CC-MAIN-20180521023747-20180521043747-00243.warc.gz | 179,640,378 | 10,699 | 1. ## Graph Problem
Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.
Could somone please provide some pointers on how to tackle these sorts of questions. I find them a bit scary.
For the first question.
the graph given is $\displaystyle y = 2x^2 + x - 4$
the x-intercepts give the x-values for which y = 0
so for this graph, the x-intercepts are the x-values such that:
$\displaystyle 2x^2 + x - 4 = 0$
or in other words, where $\displaystyle 2x^2 + x = 4$
so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph
For the second question:
we want the solutions to $\displaystyle 2x^2 - x - 5 = 0$
this is where $\displaystyle 2x^2 + x - 4 = 2x + 1$
So draw the line $\displaystyle y = 2x + 1$, the x-values of the points of intersection are your solutions
3. Originally Posted by Jhevon
For the first question.
the graph given is $\displaystyle y = 2x^2 + x - 4$
the x-intercepts give the x-values for which y = 0
so for this graph, the x-intercepts are the x-values such that:
$\displaystyle 2x^2 + x - 4 = 0$
or in other words, where $\displaystyle 2x^2 + x = 4$
so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph
For the second question:
we want the solutions to $\displaystyle 2x^2 - x - 5 = 0$
this is where $\displaystyle 2x^2 + x - 4 = 2x + 1$
So draw the line $\displaystyle y = 2x + 1$, the x-values of the points of intersection are your solutions
I can't see where the 2x + 1 from 2x^2 + x - 4 = 2x + 1 came from. Thanks.
I can't see where the 2x + 1 from 2x^2 + x - 4 = 2x + 1 came from. Thanks.
we find solutions from graphs based on intersections. therefore we need to look for those desired intersections, or sometimes create them.
in the following problem we were asked to draw a line to find solutions to the equation $\displaystyle 2x^2 - x - 5 = 0$, but i know nothing about that equation from the graph, all i know about from the graph is the equation $\displaystyle y = 2x^2 + x - 4$, so i have to somehow create this graph from the new graph given so i can get the required intersections.
so we begin with the new graph given:
$\displaystyle 2x^2 - x - 5 = 0$ (remember, my objective is to extrapolate $\displaystyle 2x^2 + x - 4 = \mbox { something }$ from this graph
since i need a +x i have to add 2 x's to -x, but since this is an equation, i have to add it to both sides, so i get:
$\displaystyle 2x^2 + x - 5 = 2x$
now i need to get a -4 instead of a -5, so i add 1 to both sides, we get:
$\displaystyle 2x^2 + x - 4 = 2x + 1$
I changed the graph of $\displaystyle 2x^2 - x - 5 = 0$ to get this, so my solutions to the new equation are exactly the same solutions to the one asked. since this new equation has the graph that i have originally, i simply need to find the intersection points between my old graph and the line 2x + 1
Originally Posted by Jhevon
For the first question.
the graph given is $\displaystyle y = 2x^2 + x - 4$
the x-intercepts give the x-values for which y = 0
so for this graph, the x-intercepts are the x-values such that:
$\displaystyle 2x^2 + x - 4 = 0$
or in other words, where $\displaystyle 2x^2 + x = 4$
so the solutions to the first question is simply the x-intercepts, which you can tell by looking on the graph
For the second question:
we want the solutions to $\displaystyle 2x^2 - x - 5 = 0$
this is where $\displaystyle 2x^2 + x - 4 = 2x + 1$
So draw the line $\displaystyle y = 2x + 1$, the x-values of the points of intersection are your solutions | 1,082 | 3,632 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.931256 |
https://openpress.usask.ca/introtoappliedstatsforpsych/chapter/12-9-between-and-within-factors/ | 1,716,791,310,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00469.warc.gz | 361,997,633 | 23,463 | 12. ANOVA
# 12.9 Between and Within Factors
So far, all of our factors have been between subject, or independent, factors. But it is possible to have any or all of the factors as within subject, or dependent, factors in a so-called repeated measures design. In a repeated measures design you will have more than one DV, more than one measurement from each subject. When you have more than one DV per subject, you have measured a vector[1] from each subject not just a number. When you measure a vector instead of a number you need multivariate statistics. The repeated measures approach is an approach that is between univariate and multivariate statistics. With repeated measures you set up your ANOVA as if it were a univariate design and use modified sums of squares and corresponding test statistics. Certain assumptions need to be satisfied before you can do repeated measures ANOVA with the most important criteria being one known as “sphericity”[2]. If sphericity fails then you need to use the full-blown multivariate approach known as MANOVA (Multivariate ANOVA). If you use SPSS to do a within subjects ANOVA then you can use the sphericity hypothesis test output in the same way that you used the Levine’s test output when deciding to use the homoscedastic or heteroscedastic -test result from SPSS. Sphericity is so if SPSS fails to reject then you can use the repeated measures results. If for the sphericity test then you reject and you will need to set up a MANOVA.
When you have a two-way (or higher factorial) ANOVA then mixed designs are possible where one factor is a between subjects factor and the other is a within subjects factor.
# 12.9.1 *One-way ANOVA with between factors
To be completed in a later edition of this text.
1. A vector is a collection of numbers. We will have more to say about vectors in Chapter 17.
2. Sphericity will be covered in a later edition of this text in a MANOVA chapter. | 440 | 1,930 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-22 | latest | en | 0.940054 |
https://www.webenotes.com/classification-of-power-system-stability/ | 1,695,997,543,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510516.56/warc/CC-MAIN-20230929122500-20230929152500-00722.warc.gz | 1,160,479,951 | 26,683 | Contents
## CLASSIFICATION OF POWER SYSTEM STABILITY
A typical modern power system is a high-order multivariable process whose dynamic response is influenced by a wide array of devices with different characteristics and response rates. Stability is a condition of equilibrium between opposing forces. Depending on the network topology, system operating condition, and the form of disturbance, different sets of opposing forces may experience sustained imbalance leading to different forms of instability. A systematic basis for the classification of power system stability is given below.
#### Need for Classification
Power system stability is essentially a single problem; however, the various forms of instabilities that a power system may undergo cannot be properly understood and effectively dealt with by treating it as such. Because of the high dimensionality and complexity of stability problems, it helps to make simplifying assumptions to analyze specific types of problems using an appropriate degree of detail of system representation and appropriate analytical techniques. Analysis of stability, including identifying key factors that contribute to instability and devising methods of improving stable operation, is greatly facilitated by the classification of stability into appropriate categories. Classification, therefore, is essential for meaningful practical analysis and resolution of power system stability problems.
#### Categories of Stability
The classification of power system stability proposed here is based on the following considerations:
• The physical nature of the resulting mode of instability as indicated by the main system variable in which instability can be observed.
• The size of the disturbance considered, influences the method of calculation and prediction of stability.
• The devices, processes, and time span that must be taken into consideration in order to assess stability.
The figure gives the overall picture of the power system stability problem, identifying its categories and subcategories. The following are descriptions of the corresponding forms of stability phenomena.
1) ROTOR ANGLE STABILITY:Â
Rotor angle stability refers to the ability of synchronous machines of an interconnected power system to remain in synchronism after being subjected to a disturbance. It depends on the ability to maintain/restore equilibrium between electromagnetic torque and mechanical torque of each synchronous machine in the system. The instability that may result occurs in the form of increasing angular swings of some generators leading to their loss of synchronism with other generators.
The rotor angle stability problem involves the study of the electromechanical oscillations inherent in power systems. A fundamental factor in this problem is the manner in which the power outputs of synchronous machines vary as their rotor angles change. Under steady-state conditions, there is an equilibrium between the input mechanical torque and the output electromagnetic torque of each generator, and the speed remains constant. If the system is perturbed, this equilibrium is upset, resulting in acceleration or deceleration of the rotors of the machines according to the laws of motion of a rotating body. If one generator temporarily runs faster than another, the angular position of its rotor relative to that of the slower machine will advance. The resulting angular difference transfers part of the load from the slow machine to the fast machine, depending on the power-angle relationship. This tends to reduce the speed difference and hence the angular separation.
The power-angle relationship is highly nonlinear. Beyond a certain limit, an increase in angular separation is accompanied by a decrease in power transfer such that the angular separation is increased further. Instability results if the system cannot absorb the kinetic energy corresponding to these rotor speed differences. For any given situation, the stability of the system depends on whether or not the deviations in angular positions of the rotors result in sufficient restoring torques. Loss of synchronism can occur between one machine and the rest of the system, or between groups of machines, with synchronism maintained within each group after separating from each other.
The change in electromagnetic torque of a synchronous machine following a perturbation can be resolved into two components:
• Synchronizing torque component, in phase with rotor angle deviation.
• Damping torque component, in phase with the speed deviation.
System stability depends on the existence of both components of torque for each of the synchronous machines. Lack of sufficient synchronizing torque results in aperiodic or non-oscillatory instability, whereas lack of damping torque results in oscillatory instability.
For convenience in analysis and for gaining useful insight into the nature of stability problems, it is useful to characterize rotor angle stability in terms of the following two subcategories:
Small-disturbance (or small-signal) rotor angle stability:Â is concerned with the ability of the power system to maintain synchronism under small disturbances. The disturbances are considered to be sufficiently small that linearization of system equations is permissible for purposes of analysis.
Small-disturbance stability depends on the initial operating state of the system. The instability that may result can be of two forms:
I) Increase in rotor angle through a non-oscillatory or aperiodic mode due to lack of synchronizing torque, or
II) Rotor oscillations of increasing amplitude due to lack of sufficient damping torque.
In today’s power systems, small-disturbance rotor angle stability problem is usually associated with insufficient damping of oscillations. The aperiodic instability problem has been largely eliminated by use of continuously acting generator voltage regulators; however, this problem can still occur when generators operate with constant excitation when subjected to the actions of excitation limiters (field current limiters). Small-disturbance rotor angle stability problems may be either local or global in nature.
LOCAL PLANT MODE OSCILLATIONS:
Local problems involve a small part of the power system and are usually associated with rotor angle oscillations of a single power plant against the rest of the power system. Such oscillations are called local plant mode oscillations.
The stability (damping) of these oscillations depends on the strength of the transmission system as seen by the power plant, generator excitation control systems, and plant output.
INTER-AREA MODE OSCILLATIONS:
Global problems are caused by interactions among large groups of generators and have widespread effects. They involve oscillations of a group of generators in one area swinging against a group of generators in another area. Such oscillations are called inter-area mode oscillations. Their characteristics are very complex and significantly different from those of local plant mode oscillations. Load characteristics, in particular, have a major effect on the stability of inter-area modes.
The time frame of interest in small-disturbance stability studies is on the order of 10 to 20 seconds following a disturbance.
Large-disturbance rotor angle stability or transient stability:Â as it is commonly referred to, is concerned with the ability of the power system to maintain synchronism when subjected to a severe disturbance, such as a short circuit on a transmission line. The resulting system response involves large excursions of generator rotor angles and is influenced by the nonlinear power-angle relationship.
Transient stability depends on both the initial operating state of the system and the severity of the disturbance.
Instability is usually in the form of aperiodic angular separation due to insufficient synchronizing torque, manifesting as first-swing instability. However, in large power systems, transient instability may not always occur as first swing instability associated with a single mode; it could be a result of the superposition of a slow inter-area swing mode and a local-plant swing mode causing a large excursion of rotor angle beyond the first swing. It could also be a result of nonlinear effects affecting a single mode causing instability beyond the first swing.
The time frame of interest in transient stability studies is usually 3 to 5 seconds following the disturbance. It may extend to 10–20 seconds for very large systems with dominant inter-area swings.
As identified in Figure, small-disturbance rotor angle stability as well as transient stability are categorized as short-term phenomena.
2) VOLTAGE STABILITY:
Voltage stability refers to the ability of a power system to maintain steady voltages at all buses in the system after being subjected to a disturbance from a given initial operating condition.
It depends on the ability to maintain/restore equilibrium between load demand and load supply from the power system. The instability that may result occurs in the form of a progressive fall or rise of voltages of some buses. A possible outcome of voltage instability is loss of load in an area or tripping of transmission lines and other elements by their protective systems leading to cascading outages. Loss of synchronism of some generators may result from these outages or from operating conditions that violate the field current limit.
Progressive drop in bus voltages can also be associated with rotor angle instability. For example, the loss of synchronism of machines as rotor angles between two groups of machines approach 180 causes a rapid drop in voltages at intermediate points in the network close to the electrical center. Normally, protective systems operate to separate the two groups of machines and the voltages recover to levels depending on the post-separation conditions. If, however, the system is not so separated, the voltages near the electrical center rapidly oscillate between high and low values as a result of repeated “pole slips” between the two groups of machines. In contrast, the type of sustained fall of voltage that is related to voltage instability involves loads and may occur where rotor angle stability is not an issue.
The term voltage collapse is also often used. It is the process by which the sequence of events accompanying voltage instability leads to a blackout or abnormally low voltages in a significant part of the power system. Stable (steady) operation at low-voltage may continue after transformer tap changers reach their boost limit, with intentional and/or unintentional tripping of some load. The remaining load tends to be voltage sensitive, and the connected demand at normal voltage is not met.
The driving force for voltage instability is usually the loads; in response to a disturbance, power consumed by the loads tends to be restored by the action of motor slip adjustment, distribution voltage regulators, tap-changing transformers, and thermostats.
Restored loads increase the stress on the high-voltage network by increasing the reactive power consumption and causing further voltage reduction. A run-down situation causing voltage instability occurs when load dynamics attempt to restore power consumption beyond the capability of the transmission network and the connected generation.
A major factor contributing to voltage instability is the voltage drop that occurs when active and reactive power flows through inductive reactances of the transmission network; this limits the capability of the transmission network for power transfer and voltage support. The power transfer and voltage support are further limited when some of the generators hit their field or armature current time-overload capability limits.
Voltage stability is threatened when a disturbance increases the reactive power demand beyond the sustainable capacity of the available reactive power resources.
While the most common form of voltage instability is the progressive drop of bus voltages, the risk of overvoltage instability also exists and has been experienced at least on one system. It is caused by a capacitive behavior of the network (EHV transmission lines operating below surge impedance loading) as well as by under-excitation limiters preventing generators and/or synchronous compensators from absorbing the excess reactive power. In this case, the instability is associated with the inability of the combined generation and transmission system to operate below some load level. In their attempt to restore this loaded power, transformer tap changers cause long-term voltage instability.
Voltage stability problems may also be experienced at the terminals of HVDC links used for either long-distance or back-to-back applications. They are usually associated with HVDC links connected to weak ac systems and may occur at rectifier or inverter stations, and are associated with the unfavorable reactive power “load” characteristics of the converters. The HVDC link control strategies have a very significant influence on such problems since the active and reactive power at the ac/dc junction is determined by the controls. If the resulting loading on the ac transmission stresses it beyond its capability, voltage instability occurs. Such a phenomenon is relatively fast with the time frame of interest being in the order of one second or less. Voltage instability may also be associated with converter transformer tap-changer controls, which is a considerably slower phenomenon.
Recent developments in HVDC technology (voltage source converters and capacitor commutated converters) have significantly increased the limits for stable operation of HVDC links in weak systems as compared with the limits for line commutated converters.
One form of voltage stability problem that results in uncontrolled over-voltages is the self-excitation of synchronous machines.
This can arise if the capacitive load of a synchronous machine is too large. Examples of excessive capacitive loads that can initiate self-excitation are open-ended high-voltage lines and shunt capacitors and filter banks from HVDC stations.
The over-voltages that result when generator load changes to capacitive are characterized by an instantaneous rise at the instant of change followed by a more gradual rise. This latter rise depends on the relation between the capacitive load component and machine reactances together with the excitation system of the synchronous machine. The negative field current capability of the exciter is a feature that has a positive influence on the limits for self-excitation.
As in the case of rotor angle stability, it is useful to classify voltage stability into the following subcategories:
Large-disturbance voltage stability: refers to the system’s ability to maintain steady voltages following large disturbances such as system faults, loss of generation, or circuit contingencies. This ability is determined by the system and load characteristics, and the interactions of both continuous and discrete controls and protections. Determination of large-disturbance voltage stability requires the examination of the nonlinear response of the power system over a period of time sufficient to capture the performance and interactions of such devices as motors, under-load transformer tap changers, and generator field-current limiters. The study period of interest may extend from a few seconds to tens of minutes.
Small-disturbance voltage stability: refers to the system’s ability to maintain steady voltages when subjected to small perturbations such as incremental changes in system load.
This form of stability is influenced by the characteristics of loads, continuous controls, and discrete controls at a given instant of time. This concept is useful in determining, at any instant, how the system voltages will respond to small system changes. With appropriate assumptions, system equations can be linearized for analysis thereby allowing the computation of valuable sensitivity information useful in identifying factors influencing stability. This linearization, however, cannot account for nonlinear effects such as tap changer controls (dead bands, discrete tap steps, and time delays). Therefore, a combination of linear and nonlinear analyzes is used in a complementary manner.
As noted above, the time frame of interest for voltage stability problems may vary from a few seconds to tens of minutes. Therefore, voltage stability may be either a short-term or a long-term phenomenon as identified in Figure 1.
Short-term voltage stability involves the dynamics of fast-acting load components such as induction motors, electronically controlled loads, and HVDC converters.
The study period of interest is in the order of several seconds, and analysis requires the solution of appropriate system differential equations; this is similar to the analysis of rotor angle stability. Dynamic modeling of loads is often essential. In contrast to angle stability, short circuits near loads are important. It is recommended that the term transient voltage stability not be used.
Long-term voltage stability involves slower-acting equipment such as tap-changing transformers, thermostatically controlled loads, and generator current limiters. The study period of interest may extend to several or many minutes, and long-term simulations are required for the analysis of system dynamic performance. Stability is usually determined by the resulting outage of equipment, rather than the severity of the initial disturbance.
Instability is due to the loss of long-term equilibrium (e.g., when loads try to restore their power beyond the capability of the transmission network and connected generation), the post-disturbance steady-state operating point being small-disturbance unstable, or a lack of attraction toward the stable post-disturbance equilibrium (e.g., when remedial action is applied too late). The disturbance could also be a sustained load buildup (e.g., morning load increase). In many cases, static analysis can be used to estimate stability margins, identify factors influencing stability, and screen a wide range of system conditions and a large number of scenarios. Where the timing of control actions is important, this should be complemented by quasi-steady-state time-domain simulations.
The distinction between Voltage and Rotor Angle Stability:
It is important to recognize that the distinction between rotor angle stability and voltage stability is not based on weak coupling between variations in active power/angle and reactive power/voltage magnitude. In fact, the coupling is strong for stressed conditions and both rotor angle stability and voltage stability are affected by pre-disturbance active power as well as reactive power flows. Instead, the distinction is based on the specific set of opposing forces that experience sustained imbalance and the principal system variable in which the consequent instability is apparent.
FREQUENCY STABILITY:
Frequency stability refers to the ability of a power system to maintain steady frequency following a severe system upset resulting in a significant imbalance between generation and load.
It depends on the ability to maintain/restore equilibrium between system generation and load, with minimum unintentional loss of load. The instability that may result occurs in the form of sustained frequency swings leading to the tripping of generating units and/or loads.
Severe system upsets generally result in large excursions of frequency, power flows, voltage, and other system variables, thereby invoking the actions of processes, controls, and protections that are not modeled in conventional transient stability or voltage stability studies. These processes may be very slow, such as boiler dynamics, or only triggered for extreme system conditions, such as volts/Hertz protection tripping generators.
In large interconnected power systems, this type of situation is most commonly associated with conditions following the splitting of systems into islands. Stability in this case is a question of whether or not each island will reach a state of operating equilibrium with minimal unintentional loss of load. It is determined by the overall response of the island as evidenced by its mean frequency, rather than the relative motion of machines. Generally, frequency stability problems are associated with inadequacies in equipment responses, poor coordination of control and protection equipment, or insufficient generation reserve. Examples of such problems are reported in the references. In isolated island systems, frequency stability could be of concern for any disturbance causing a relatively significant loss of load or generation.
During frequency excursions, the characteristic times of the processes and devices that are activated will range from a fraction of seconds, corresponding to the response of devices such as under-frequency load shedding and generator controls and protections, to several minutes, corresponding to the response of devices such as prime mover energy supply systems and load voltage regulators. Therefore, as identified in Figure, frequency stability may be a short-term phenomenon or a long-term phenomenon.
An example of short-term frequency instability is the formation of an under-generated island with insufficient under-frequency load shedding such that frequency decays rapidly causing a blackout of the island within a few seconds. On the other hand, more complex situations in which frequency instability is caused by steam turbine over-speed controls or boiler/reactor protection and controls are longer-term phenomena with the time frame of interest ranging from tens of seconds to several minutes.
During frequency excursions, voltage magnitudes may change significantly, especially for islanding conditions with under-frequency load shedding that unloads the system. Voltage magnitude changes, which may be higher in percentage than frequency changes, affect the load-generation imbalance.
High voltage may cause undesirable generator tripping by poorly designed or coordinated loss of excitation relays or volts/Hertz relays. In an overloaded system, the low voltage may cause undesirable operation of impedance relays. | 3,975 | 22,389 | {"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} | 2.890625 | 3 | CC-MAIN-2023-40 | longest | en | 0.939574 |
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# What is the maximum number of sheep that Ruben's pen will
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What is the maximum number of sheep that Ruben's pen will [#permalink]
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02 Aug 2011, 19:54
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What is the maximum number of sheep that Ruben's pen will hold?
1. If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4.
2. Currently, there are 12 sheep in the pen.
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02 Aug 2011, 20:08
Let the maximum capacity of pen = x and the number of sheep present as of now = 'y'.
so you cannot directly solve for 'x' & hence '1' is not sufficient. However, you will get following equation:
2/3(x) - 8 = y - 1/4(y) = 3/4(y)
Statement (2) alone is not sufficient (obviously).
Now, together the equations seem to be sufficient, but let us just solve the equations:
2/3(x) - 8 = 3/4(12) = 9, giving x = 25.5. We cannot have 25.5 sheep (sheep could be either 25 or 26), hence I will answer 'E'.
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02 Aug 2011, 20:28
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KUDOS
Using statement (1): If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4.
Let the capacity of the pen be 'a' and the number of sheep currently in the pen be 'b'.
Then (2/3)*a - 8 = b - (b/4)
and b = (2/3)*a
Solving these equations simultaneously gives us:
a = 48 and b=32. Sufficient.
Using statement (2): Just knowing the number of sheep currently in the pen is insufficient to tell us the capacity of the pen. Insufficient.
Therefore (A) it is.
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Last edited by GyanOne on 03 Aug 2011, 01:01, edited 1 time in total.
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02 Aug 2011, 21:01
Asher wrote:
What is the maximum number of sheep that Ruben's pen will hold?
1. If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4.
2. Currently, there are 12 sheep in the pen.
1.
Let "x" be the number of sheep when the pen is full.
$$\frac{1}{4}*\frac{2}{3}*x=8$$
$$x=48$$
Sufficient.
Ans: "A"
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02 Aug 2011, 22:02
Gyanone, fluke, could you please explain in detail. I am not able to follow the explanation.
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02 Aug 2011, 22:03
.....................................
Let the capacity of the pen be 'a' and the number of sheep currently in the pen be 'b'.
Then (2/3)*a - 8 = b - (b/4)
and x = (2/3)*y ???
Solving these equations simultaneously gives us:
a = 48 and b=32. Sufficient.
.....................................
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02 Aug 2011, 22:07
@ fluke:
--------------------------------------
Let "x" be the number of sheep when the pen is full.
\frac{1}{4}*\frac{2}{3}*x=8
--------------------------------------
How could you arrive at equation in 1 variable? The stem says "the number of sheep in the pen will decrease by 1/4". We don't know how many sheep are in the pen (it does not say that the max. capacity of the pen will decrease by 1/4)? Both statements are different. So, you have to have two variables.
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02 Aug 2011, 22:10
Also, to add on, 48 does not satisfy the stem "If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4. "
2/3x-8 gives = 24
and decrease by 1/4 gives 36
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02 Aug 2011, 22:13
1
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puneet478 wrote:
Gyanone, fluke, could you please explain in detail. I am not able to follow the explanation.
1. If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4.
Let's take it one by one:
If 8 sheep are removed from the pen when the pen is 2/3 full.
(2/3)x-> This is 2/3 full because "x" is the number of sheep when the pen is full
We got to remove 8 sheep from it
(2/3)x-8
The number of sheep in the pen is decreased by 1/4.
This statement means:
If there were 100 sheep in the pen, it decreases by 25.
If there were 20 sheep in the pen, it decreases by 5.
So,
(2/3)x-8=(2/3)x-(1/4)(2/3)x
Or in other words:
(1/4)(2/3)x=8. Because, you removed 8 sheep and the number decreased by (1/4)(2/3)x
Thus, we know x, which is the total capacity. The question asks us to find just that.
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02 Aug 2011, 22:32
@ fluke...
(2/3)x-8=(2/3)x-(1/4)(2/3)x, how did you arrive at part highlighted in boldface?
Stem says number of sheep (in the pen at that particular time) decreases to 1/4. How can we assume that it was 2/3x at that time? We simply do not know!
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03 Aug 2011, 01:05
puneet478,
apologies - my second statement was supposed to be:
b = (2/3)*a (edited now in the original solution).
This is valid because statement (1) says:
If 8 sheep are removed from the pen when it is 2/3 full, the number of sheep in the pen will decrease by 1/4.
Note the highlighted part. This means that the pen is 2/3 full when we remove the 8 sheep from it.
This means the number of sheep currently in the pen is 2/3 of the pen's capacity.
=> b = (2/3)*a
You can (rather must) take the number of sheep currently in the pen to be 2/3 of the pen's capacity because that is exactly what statement (1) says.
Happy to explain further if you still have doubts.
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03 Aug 2011, 01:20
what i understand is that: 2/3x = total no. of sheep in the pen (assuming that sheep is the only animal in the pen)
Am i right?
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03 Aug 2011, 03:04
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Yes, assuming the x in your post is the capacity of the pen.
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03 Aug 2011, 05:36
@gyanone, I think you & fluke are right. I was thinking on different lines. Thanks for the clarifications.
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03 Aug 2011, 05:53
Ans is A.
from option 1.
lets assume max numer of sheeps = x
so current no of sheeps = 2/3 x
now by reducing it by number 8 the total number of sheep reduces by 1/4 .
that means it reduces by 2/3 x * 1/4
now we can put this in equation
2/3 x * 1/4 = 8
and we can find the value of X
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03 Aug 2011, 07:51
thanks fluke and GyanOne.. very detailed and clear explanation.. +1
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Re: Knewton DS [#permalink] 03 Aug 2011, 07:51
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$y'(x)=\frac {2 x y(x)^2+y(x)^2+4 x y(x) \log (2 x+1)+2 y(x) \log (2 x+1)+2 x \log ^2(2 x+1)+\log ^2(2 x+1)-2}{2 x+1}$ Mathematica : cpu = 0.228767 (sec), leaf count = 22
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When boiled down to its basic parts, OSPF’s LSDB is much more simple than it appears. In reality, the LSDB is merely a listing of nodes and a description of the relationship between those nodes described using LSAs. Type-1 Router LSAs describe the router nodes in the network and their relationship to other nodes. Type-2 Network LSAs describe segments on the network where multiple routers share a relationship. Type-3 Summary LSAs are used to provide routing to prefixes between areas. Type-4 ASBR Summary LSAs inject topology information into an area about an ASBR from another area that is flooding Type-5 External LSAs into the domain.
Each LSA type acts as a single piece of the puzzle that, when put together, creates a graph of the full network topology. OSPF uses this graph to run SPF and calculate an SPT for each router in the topology.
This blog will attempt to explain these concepts as simply as possible. As such, some examples have been contrived or overly simplified in order to convey the overall idea.
LSDB as a Graph
To understand the LSDB better, it helps to think of it as a graph. Graphs are mathematical structures made up of a set of objects that are in some way related to each other. Each object is called a vertex or node. A relationship between two nodes is called an edge. They are usually diagrammatically depicted as follows:
In the above we have nodes A through E. The relationship between nodes A and B, A and C, and so on forms an edge. It is possible by following the edges between nodes to plot a path between any two nodes without passing through any node twice. For instance, to reach Node D from Node A, one could follow the A-B-C-D path. Notice there are several other paths through which A could reach D. The challenge is deciding on a criteria that makes one path preferred over another.
OSPF uses Dijkstra’s shortest path first algorithm to find the shortest paths between points on the graph. In order for the algorithm to work, there needs to be a weight or cost assigned to each edge on the graph as follows:
The SPF algorithm adds up the total weight of each available path, the path with the lowest total weight becomes the best path. In the example above, the path A-C-D is considered the best path because the collective weights of the edges is lowest among all other available paths.
Networks can easily be expressed in the same terms. Each network segment or networking device is a node. The relationship between those nodes form an edge, or link. These links have different characteristics, such as bandwidth, delay, monetary cost, or reliability. These characteristics can be used as weights on the graph. Routers describe these relationships using Link State Advertisements (LSAs), creating a different type of LSA for each type of relationship.
The Link State Database is a collection of these LSAs, containing a list of all of the nodes in the network and the relationship between those nodes. The SPF algorithm is run against the LSDB to find the shortest path between related nodes to reach all nodes in the topology.
When represented in such a way, it is a simple task to use this graph as input for the SPF algorithm to determine the shortest paths. Each router does this by first setting itself as the root of a tree and finding the shortest paths from it to every other node in the network. The result is a Shortest Path Tree (SPT) that spans from the calculating router to every other node.
As stated earlier, LSAs are the building blocks of the LSDB each describing a relationship and therefore a different piece of the overall network topology. There are several different types of LSAs. This blog will explain LSA Types 1 through 5 as they are the most common LSA types encountered when first learning about OSPF.
All LSAs regardless of type have an LSA header, which identifies important pieces of information about the LSA. For example, the Type-1 Router LSA Header is as follows:
LS age: 36
Options: (No TOS-capability, DC)
LS Seq Number: 80000004
Checksum: 0x2B0B
From top to bottom:
1. The LSA’s age in seconds. Once an LSA reaches Max Age or 3600s, it is removed from the database.
2. LSA Options that control how OSPF routers treat the LSA
3. The LSA Type. In this case it is a Type-1 Router LSA
4. An Identifier for the LSA called the Link State ID. In this case, 1.1.1.1
5. The router that originally advertised the LSA. Only this router can modify or prematurely remove the LSA from the LSDB. For this LSA, 1.1.1.1 is the advertising router and is the only router that can modify or prematurely age out the LSA.
6. Sequence number which is used to identify newer copies of an LSA
7. Checksum that is used to detect corruption in the LSA.
This header can also be referenced in the show ip ospf database output. The output however refers to the Link State ID as Link ID which is inaccurate. A Link ID identifies a single link whereas a Link State ID identifies an LSA that may contain many links.
R1#sh ip ospf database
OSPF Router with ID (1.1.1.1) (Process ID 1)
1.1.1.1 1.1.1.1 858 0x80000009 0x00949E 4
A key topic for understanding is that the Link State ID (or Link ID from the show ip ospf database output) is not an IP address. In the above, the presence of 1.1.1.1 as a Link ID does not mean there is network connectivity to ip address 1.1.1.1/32. Instead, it signals that there is a router node called 1.1.1.1 on the graph in the LSDB.
The next sections describe how LSAs are used to describe the relationships between nodes in the LSDB graph. To better understand the function of each LSA, a reference topology will be built piece-by-piece as new LSAs are introduced.The first LSA discussed is the Router LSA.
Type-1 Router LSAs
The most important LSA is the Router LSA or Type-1 LSA. It is the LSA to which all other LSAs attach. This LSA describes a router node itself and all of its relationships or link adjacencies to other nodes. It is originated by every router in the topology and flooded to all routers in a single area. OSPF identifies four types of link adjacencies using Type-1 Router LSAs:
1. Stub Network
3. Link to a multiaccess network
NOTE: There is a distinction between an “interface” and a “link”. An interface describes a physical or logical interface on the router (i.e. loopback interfaces, ethernet ports). A link describes a relationship on the graph between two nodes. Virtual links are good examples of this concept. A virtual link describes a logical connection to another OSPF router, however, the physical interfaces used to make up the path may change dynamically depending on network conditions.
Let’s examine how the fields in the Type-1 Router LSAs are populated to describe R1’s adjacencies from the reference diagram.
R1 has three links in this network: one to R2 , one to another network segment, and one on the network segment 192.168.1.0/24. R1’s job is now to describe these links in a Type-1 Router LSA and advertise them to R2. R1 will supply three pieces of information to R2 and any other router in the OSPF domain:
1. An ID to identify itself as a node on the graph
3. The relationship to other nodes connected to those links (if applicable)
In order to identify a node on the graph, OSPF uses a 32-bit number as the Node Identifier in the LSDB, called the router ID (RID). By default, OSPF chooses the highest IP address on an up loopback interface first and then the highest IP address on any up interface second as its RID. The RID can also be statically defined with the router-id command in OSPF configuration mode.
When R1 creates its Type-1 Router LSA, it will use its RID (1.1.1.1) as the Link State ID and advertising router field of the LSA header as follows:
R1#show ip ospf database router self-originate
OSPF Router with ID (1.1.1.1) (Process ID 1)
LS age: 835
Options: (No TOS-capability, DC)
The next section details on one of the four types of link adjacencies carried in the Type-1 Router LSA, Stub Networks.
Describing Stub Networks
R1 is the only router on the network segment 192.168.1.0/24 in the above topology. Such links are classified as stub networks by OSPF because the segment does not include another OSPF router. Since there is only one router connected, traffic will be routed to or from the segment, but never through it. In other words, the segment 192.168.1.0/24 will not be used as a transit segment.
R1 models this link in its Type-1 Router LSA as if it were a link to another node using the network prefix as the Link ID and the subnet mask as the Link Data. In this way, the network appears as a node directly connected to R1. Out of all of the link adjacencies described by a Type-1 Router LSA, the stub network link description is the only one that carries network layer addressing information. The following is R1’s link description for the stub network 192.168.1.0/24.
R1#show ip ospf database router self-originate
OSPF Router with ID (1.1.1.1) (Process ID 1)
LS age: 835
Options: (No TOS-capability, DC)
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 1
This concludes R1’s description of its stub network. Next, R1 needs to model the link between itself and R2.
In OSPF, a point-to-point link adjacency is a link that connects to only one other OSPF router. In the above, R1’s 12.1.1.1 link only connects to R2 and is expressed as a point-to-point link in OSPF. OSPF exhibits what may seem to be odd behavior when it comes to point-to-point links. When modeling a point-to-point link, OSPF will assume the link is unnumbered.
An unnumbered point-to-point link, is a point-to-point link to which no network layer addressing information is applied. Network layer addressing information is not needed on a point-to-point link because there are only two devices on the link (i.e. between R1 and R2 in the above). Traffic for such links will always be routed through the link and never destined to the link. Nothing is gained from assigning an address to each interface in this scenario.
However, on Cisco IOS an interface cannot forward IP traffic unless an IP address is assigned to it. For unnumbered point-to-point links, the ip unnumbered command allows a point-to-point interface to borrow an IP address from another interface on the router. This enables the IP protocol stack to run on the interface, allowing the forwarding of control and transit IP traffic over the point-to-point interface.
On R1, OSPF models this point-to-point link as follows:
Link connected to: another Router (point-to-point)
(Link ID) Neighboring Router ID: 2.2.2.2
Number of MTID metrics: 0
TOS 0 Metrics: 10
“Node 1.1.1.1 has a link called 12.1.1.1 that is connected to node 2.2.2.2”.
NOTE: Some versions of Cisco IOS will set the Link Data for ip unnumbered interfaces to a random internal value. The engineer must configure the interface-id snmp-if-index command under OSPF configuration mode to enable the use of the SNMP interface ID for Link Data in the Type-1 Router LSA.
Virtual Links, links between two nodes on the graph used to connect to the backbone, are expressed as unnumbered point-to-point links as well. Because virtual-link adjacencies are also described by the Type-1 Router LSA in the same manner, they are not going to be expounded upon in this blog.
If addressing information is assigned to to the point-to-point link, OSPF will model this separately from the point-to-point link description as if it were a stub network attached to the router. The link between R1 and R2 has been assigned the 12.1.1.0/24 subnet. OSPF attaches this subnet off R1 using a second stub link description:
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 1
With this link description, R1 is identifying that it is also connected to a network node 12.1.1.0/24.
The OSPF Lie
It may appear as if OSPF is lying when it advertises a stub network along with the point-to-point link, especially given our previous understanding of stub networks. A stub network is not supposed to connect to another router and it is not supposed to be used for transit traffic. However, in the point-to-point link to R2 above, this is not true. It appears as if transit traffic will cross the stub network 12.1.1.0/24.
In order to reconcile this inconsistency, one must understand that logically R1 considers R2 and the stub network 12.1.1.0/24 as two different nodes in the network. This distinction is key. Each link to these nodes is used as independent relationships in the graph when computing shortest paths.
When R1 describes a point-to-point link, it first models the link to node R2 as described in the section above. Then it describes the link to the stub network node 12.1.1.0/24, which is assigned to the point-to-point link. Each link is considered separately when performing SPF computations. From a graphical point of view, this makes sense. The graphical relationship between R1, R2, and the stub network node is as follows:
From OSPF’s perspective, these relationships are independent. There is a point-to-point relationship between R1 and R2 and both routers have a network node 12.1.1.0/24 attached. The SPF calculation behaves in the following way regarding these two descriptions:
• When calculating the path between Nodes 1.1.1.1. and 2.2.2.2, OSPF uses the point-to-point link description.
• When calculating the path to the 12.1.1.0/24 network, OSPF uses the stub network link description.
Understanding this, it is clear OSPF has not lied. It is not using the 12.1.1.0/24 network node to transit traffic. It is using the point-to-point link description. Only traffic destined to 12.1.1.0/24 will be routed using the stub network link description. Traffic destined to other remote network nodes will be routed across the point-to-point link which has no addressing information attached to it.
Addressing information for the point-to-point link does not have to be modeled in the Type-1 Router LSA. Advertising this information facilitates routing directly to those interfaces. If direct connectivity to the interfaces is not required, this addressing information can be suppressed using the ip ospf prefix-suppression command in interface configuration mode. With this configuration, the addressing information for the point-to-point link (the 12.1.1.0/24 subnet) will not be modeled in the Type-1 Router LSA, eliminating it from the LSDB.
The basics of Type-1 Router LSAs as far as stub and point-to-point links are concerned have been covered. To explain the next type of link described by Type-1 Router LSAs, we need to add a new section to the topology and introduce a new LSA type.
Type-2 Network LSAs
The below shows an example of a multiaccess network. R1 has an interface on a multiaccess network shared with R3 and R4. Multiaccess networks allow multiple devices to connect to the same network segment. This is achieved by transparently bridging traffic between each device connected to the segment. For OSPF, this transparent bridging creates a situation where a single OSPF router can have several OSPF routers connected to one link.
How would R1 model the relationship with R3 and R4 on the multiaccess segment using a Type-1 Router LSA alone? A point-to-point link description only describes a link to a single router. For this reason, R1 can not describe its relationship to both R3 and R4 in a single point-to-point link description. R1 would have to use multiple point-to-point link descriptions to describe these relationships. R3 and R4 would have to do the same, leading to the following graphical topology:
If the graph above were described using words, the following would be stated:
“Node 1.1.1.1 is connected to Node 3.3.3.3 and Node 4.4.4.4. Node 3.3.3.3 is connected to Node 1.1.1.1 and Node 4.4.4.4. Node 4.4.4.4 is connected to Node 1.1.1.1 and Node 3.3.3.3.”
Stating the relationships in this way is cumbersome, not only when spoken but also in the LSDB. It takes up a lot of space. In OSPF terms, this “space” translates to the overall size of the LSDB. A Large LSDB increases how long it takes to calculate shortest paths during an SPF run and consumes more router memory.
Another way of expressing this information is to create a node that represents the multiaccess segment and have each router on the segment advertise their relationship to the new node. The result is the following graphical representation:
If the graph above were described using words, the following would be stated:
“Nodes 1.1.1.1, 3.3.3.3, and 4.4.4.4 are all connected to Transit Network Node 134.1.1.4.”
This is a simpler statement that takes up less space. OSPF assumes this network node provides connectivity to the routers to which it is attached, treating the network node as if it were another OSPF router on the graph. By solving the path to the network node, OSPF also solves the path to each router connected to the network node. This node is typically referred to as a pseudo node, because it does not physically exist in the network. It is a logical construct to help define the relationship between nodes on a multiaccess segment.
One router from the segment, called the Designated Router, is elected to create the pseudo node using a Type-2 Network LSA. In the example topology, R4 is acting as the DR on the segment and generates the Type-2 Network LSA with the following information:
1. R4 uses its interface IP address on the segment as the Link State ID of the pseudo node in the Type-2 Network LSA header.
2. The Type-2 Network LSA generated by R4 contains a list of all router nodes attached to the pseudo node.
The following is the Type-2 Network LSA generated by R4 identifying the pseudo node 134.1.1.4 and its attached nodes 4.4.4.4, 1.1.1.1, and 3.3.3.3.
R1# show ip ospf database network
--- omitted ---
Options: (No TOS-capability, DC)
LS Seq Number: 80000001
Checksum: 0x9CDC
Length: 36
Attached Router: 4.4.4.4
Attached Router: 1.1.1.1
Attached Router: 3.3.3.3
R1, as a router connected to the segment, uses a transit network link description in its Type-1 Router LSA to describe its link to the pseudo node. The transit link description has the Link ID of 134.1.1.4, which is both the DR’s interface address on the segment and the Link State ID of the pseudo node. Next, the Link data for the transit link description is set to R1’s interface address on the segment.
R1# show ip ospf database router self-originate
--- omitted ---
Options: (No TOS-capability, DC)
--- omitted ---
Link connected to: a Transit Network
Number of MTID metrics: 0
TOS 0 Metrics: 10
Compared to a point-to-point link description, the Link ID for the transit link description is a pseudo node identifier, while in the point-to-point link description, the Link ID is the RID of a neighboring router.
The transit network link description in R1’s Type-1 Router LSA does not include network layer information. In order to derive network layer information, the Type-2 Network LSA contains the netmask for the segment:
R1# show ip ospf database network
--- omitted ---
Options: (No TOS-capability, DC)
LS Seq Number: 80000001
Checksum: 0x9CDC
Length: 36
Because the Link State ID of the Type-2 Network LSA is the address of the DR, other routers in the topology can find the network address by applying the mask to the Link State ID of the Type-2 Network LSA. In the above, the network address of the network node is 134.1.1.0/24.
The combination of these two LSAs spells out R1’s link as follows
“Node 1.1.1.1 has a transit link to pseudo node 134.1.1.4. Pseudo node 134.1.1.4 has netmask /24 and connects nodes 1.1.1.1, 3.3.3.3 and 4.4.4.4.”
At this point, Type-1 Router and Type-2 Network LSAs have been fully covered. The entire network topology can be modeled with these two LSAs alone. Doing so, however, can cause problems as the network grows larger. To deal with this, OSPF uses the concept of Areas.
LSDB and Areas
Left as it is, the LSDB will continue to grow in size as the network grows. Eventually, it can become so large that it increases the time it takes to run the SPF algorithm and the memory demand on the router. Another negative side effect of a large LSDB is that routing updates affect all routers and force a full recalculation of SPF. If a link goes down in one section of the network, the entire network will have to adjust their LSDBs and recalculate the shortest paths.
OSPF avoids these issues by allowing the network engineer to segment the network into Areas. Most of the computational overhead of SPF is spent calculating the shortest paths between Type-1 Router LSAs and Type-2 Network LSAs as they contain the most dense topology information. When a topology is split into different areas, these LSAs are not permitted across area boundaries, decreasing the size of the LSDB. Instead, this topology information is hidden by routers that sit between each area, termed Area Border Routers (ABRs).
With such an implementation, all OSPF routers in the topology will no longer have a synchronized LSDB, and therefore cannot arrive at the same outcome when running the SPF algorithm. Only routers in the same area will have synchronized LSDBs and can agree on the internal topology of the area. Therefore, full SPF computations are run separately within each area and not over the entire topology. Topology changes in one area will not cause a full SPF calculation in other areas.
In the topology below, routers in Area 1 will maintain a synchronized LSDB separate of routers in Area 0. SPF is run in Area 1 to determine the shortest paths within that area. A separate SPF calculation is performed in Area 0 to determine the shortest paths within that area.
NOTE: ABRs in the topology contain a separate LSDB for each area to which they are attached and will run SPF for each LSDB. In the above, ABR1 performs an SPF calculation for Area 1 and a separate SPF calculation for Area 0.
ABRs rely on the Type-3 Summary LSA to advertise condensed topology information between areas.
Type-3 Summary LSAs
Instead of describing the details of what nodes exist in an area and how they are interconnected, Type-3 Summary LSAs only advertise prefixes. In a multi-area OSPF domain, there exists two different kinds of prefixes:
• intra-area: prefixes from within a single area
• inter-area: prefixes from outside a single area
An ABR will generate a Type-3 Summary LSA for all intra-area and inter-area prefixes from its routing table and flood them into the other areas to which it is attached. From the topology above, ABR1 would advertise all intra-area prefixes from Area 1 to Area 0 as Type-3 LSAs. Likewise it will advertise all intra-area and inter-area prefixes from Area 0 to Area 1 as Type-3 LSAs. The same happens between Area 0 and Area 2.
In order to make the topology more resistant to loops, OSPF enforces a two-layer hierarchy. All areas must have a connection to the top layer of the hierarchy which is Area 0, also known as the backbone. Type-3 Summary LSAs are exchanged through the backbone down to the lower layer of the hierarchy where all other areas exist. This creates a hub-and-spoke topology like the below:
This organization also changes the ABR’s Type-3 Summary LSA flooding algorithm as follows:
• Type-3 LSAs are generated for intra-area and inter-area prefixes from the backbone area and flooded to non-backbone areas.
• Type-3 LSAs are generated only for intra-area routes in non-backbone areas and flooded into the backbone area.
• An ABR only uses Type-3 LSAs from the Backbone area in the SPF calculation
For example, in the above, ABR1 will generate a Type-3 Summary LSA for all intra-area prefixes in Area 1 and flood them into Area 0. ABR1 will also generate a Type-3 Summary LSA for intra-area and inter-area prefixes from Area 0 and flood them into Area 1.
NOTE: The above algorithm implies that an ABR can only generate Type-3 Summary LSAs if it is connected to Area 0. This may differ among vendor implementations, however, this is how ABRs function in Cisco IOS.
To demonstrate these concepts, the reference topology is further built upon to create a multiarea OSPF domain.
The topology shows R1, R2, R3, and R4 belong to the backbone Area 0. R2, R3, and R4 are ABRs with links to other areas. R2 has a link in Area 256, R3 has a link in Area 37, and R4 has a link in area 489.
Looking at the LSDB on R2 for router LSAs yields this result:
R2#show ip ospf database
OSPF Router with ID (2.2.2.2) (Process ID 1)
1.1.1.1 1.1.1.1 1576 0x80000004 0x0035FF 4
2.2.2.2 2.2.2.2 1329 0x80000003 0x006C6F 2
3.3.3.3 3.3.3.3 1359 0x80000003 0x009A5C 1
4.4.4.4 4.4.4.4 1255 0x80000004 0x00608A 1
--- some output has been omitted ---
2.2.2.2 2.2.2.2 1319 0x80000002 0x007E34 2
5.5.5.5 5.5.5.5 1195 0x80000003 0x005688 4
6.6.6.6 6.6.6.6 743 0x80000004 0x0018B4 3
--- some output has been omitted ---
Notice, R2 has separate Type-1 Router LSAs for the backbone area and Area 256. Also notice, the nodes 5.5.5.5 (R5) and 6.6.6.6 (R6) that belong to Area 256 are not listed in the Type-1 Router LSAs in Area 0. Likewise, nodes 1.1.1.1 (R1), 3.3.3.3 (R3), or 4.4.4.4 (R4) that belong to Area 0 are not listed in Type-1 Router LSAs for Area 256. The details of each area are kept hidden. This is evidenced by examining the router LSAs on R1 and R6:
R1#show ip ospf database
OSPF Router with ID (1.1.1.1) (Process ID 1)
1.1.1.1 1.1.1.1 12 0x80000006 0x003102 4
2.2.2.2 2.2.2.2 53 0x80000004 0x006A70 2
3.3.3.3 3.3.3.3 14 0x80000005 0x00965E 1
4.4.4.4 4.4.4.4 13 0x80000006 0x005C8C 1
R6#sh ip ospf database
OSPF Router with ID (6.6.6.6) (Process ID 1)
2.2.2.2 2.2.2.2 582 0x80000002 0x007E34 2
5.5.5.5 5.5.5.5 457 0x80000003 0x005688 4
6.6.6.6 6.6.6.6 3 0x80000004 0x0018B4 3
R1, R6, R7, R8, and R9 only have interfaces in their respective areas, making them internal routers. As such, they only have detailed information about the topology within their area. R2, R3, and R4, as ABRs, have interfaces in multiple areas. They have detailed information about the topology of all areas to which they are connected.
To provide connectivity between areas, R2, as an ABR, performs the following:
1. Looks in its RIB for all OSPF intra-area routes from Area 256.
2. Advertises them as Type-3 Summary LSAs into the backbone Area 0.
3. Looks into its RIB for all intra-area and inter-area prefixes from the backbone Area 0.
4. Floods these prefixes into its non-backbone Area 256 as Type-3 Summary LSAs, excluding those routes that belong to Area 256.
Here is an example of R2’s routing table for the prefix 192.168.5.0/24:
R2#show ip route ospf | inc 192.168.5.0
O 192.168.5.0/24 [110/11] via 25.1.1.5, 00:11:53, Ethernet0/2
Notice it has an “O” next to it. This means it is a route to an intra-area prefix. R2 will take this prefix and advertise it in the following Type-3 Summary LSA into Area 0:
R2#show ip ospf database summary 192.168.5.0
OSPF Router with ID (2.2.2.2) (Process ID 1)
Summary Net Link States (Area 0)
LS age: 830
Options: (No TOS-capability, DC, Upward)
Link State ID: 192.168.5.0 (summary Network Number)
LS Seq Number: 80000001
Checksum: 0xD2E8
Length: 28
MTID: 0 Metric: 11
R2 sets the Link State ID to the network number of the summary address and inserts its own RID as the advertising router. The netmask and metric are all encoded in this LSA as well. Finally, we can see this LSA has been flooded in Area 0.
Routers in Area 0 will perform the following:
• Add this prefix to their routing tables
• Solve the shortest path to the advertising router 2.2.2.2
• Add the metric to their final route computation
This process results in the following inter-area prefix, denoted by “O IA” in the routing table:
R1#show ip route ospf | i 192.168.5.0
O IA 192.168.5.0/24 [110/21] via 12.1.1.2, 01:06:23, Ethernet0/1
Through this process, OSPF behaves no differently than a Distance Vector protocol. It simply advertises the network prefix, a direction (towards 2.2.2.2), and a cost (in this case Metric: 11). Doing so, full SPF calculations are not required when prefixes are added or removed between areas. The ABRs simply age out the Type-3 Summary LSAs and all other routers remove them from their respective LSDBs.
Dealing with External Information
External routing information is any information that has been injected into the OSPF process with the redistribute command. This information could originate from a different routing protocol or a different OSPF process. The routers configured to inject this external information into the OSPF domain are called the Autonomous System Border Router or ASBR and use Type-5 external LSAs to perform the flooding.
NOTE: Type-7 LSAs can also be used to inject external information into a not-so-stubby area. However the details behind this are outside the scope of this blog.
Type-5 External LSAs
In the below, the reference topology has been extended to include external prefixes. The external prefixes 172.16.56.0/24 and 172.16.65.0/23 are being injected into the OSPF domain by the ASBR R6 in Area 256. R6 generates an individual Type-5 External LSA for both of these prefixes. Type-5 External LSAs have a domain-wide flooding scope and as such do not belong to any specific area. As the Type-5 External LSA is flooded throughout the domain, no other router can modify the contents or regenerate the LSA.
The Type-5 External LSAs generated by R6 below contains three important pieces of information:
• The Link State ID is always set the to the external network being flooded into the OSPF domain
• The Advertising Router field is set to the RID of the ASBR, which is 6.6.6.6 in this case
• The Forward Address is set to 0.0.0.0. This value indicates all OSPF routers should solve the path to the advertising router, R6, to reach the external prefix.
LS age: 685
Options: (No TOS-capability, DC, Upward)
Link State ID: 172.16.56.0 (External Network Number )
LS Seq Number: 80000001
Checksum: 0x4350
Length: 36
Metric Type: 2 (Larger than any link state path)
MTID: 0
Metric: 20
External Route Tag: 0
LS age: 685
Options: (No TOS-capability, DC, Upward)
Link State ID: 172.16.65.0 (External Network Number )
LS Seq Number: 80000001
Checksum: 0xDFAA
Length: 36
Metric Type: 2 (Larger than any link state path)
MTID: 0
Metric: 20
External Route Tag: 0
The above can be translated as such:
“The external networks x.x.x.x can be reached via the node 6.6.6.6.”
Internal routers in Area 256, such as R5 or R2, can solve the path to the ASBR because they have the Type-1 Router LSA for the ASBR (6.6.6.6 or R6) in their LSDBs:
R5#sh ip ospf database
OSPF Router with ID (5.5.5.5) (Process ID 1)
2.2.2.2 2.2.2.2 1527 0x80000002 0x007E34 2
5.5.5.5 5.5.5.5 1526 0x80000003 0x00FB64 5
6.6.6.6 6.6.6.6 1526 0x80000003 0x001AB3 3
R6’s Type-1 Router LSA contains sufficient information to determine it has a point-to-point link to node 5.5.5.5 which is R5. R5 will have this information and, as a result, can solve the path to R6 in the topology.
R5#show ip ospf database router 6.6.6.6
OSPF Router with ID (5.5.5.5) (Process ID 1)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 682
Options: (No TOS-capability, DC)
LS Seq Number: 80000003
Checksum: 0x1AB3
Length: 60
AS Boundary Router
[--omitted--]
Link connected to: another Router (point-to-point)
(Link ID) Neighboring Router ID: 5.5.5.5
Number of MTID metrics: 0
TOS 0 Metrics: 10
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 10
Note: Here it is clearer that the OSPF RID is just a 32-bit identification of a node, and not an IP address. If the RID of 6.6.6.6 in the advertising field of the Type-5 LSA truly represented an IP address, R5 wouldn’t need to refer to a Type-1 Router LSA to solve reachability to the external prefixes. R5 could just perform a normal destination-based IP lookup for 6.6.6.6 in its routing table. This also stresses the point that just because a router has an LSA for a prefix or node, that does not mean it can reach that particular prefix or node.
We understand from above that, within the area the ASBR is located, a Type-1 Router LSA plays an important role in locating the ASBR on R5. However, for routers in other areas, such as R1 in Area 0, there is no entry in the LSDB for node 6.6.6.6:
R1#show ip ospf database
OSPF Router with ID (1.1.1.1) (Process ID 1)
1.1.1.1 1.1.1.1 1247 0x80000005 0x00290C 4
2.2.2.2 2.2.2.2 1196 0x80000003 0x006C6F 2
3.3.3.3 3.3.3.3 1213 0x80000004 0x00411C 2
4.4.4.4 4.4.4.4 1217 0x80000004 0x006FD8 2
This is because topology information is lost between areas in favor of smaller LSDBs. R1 cannot solve the path to this node using a Type-1 Router LSA because it doesn’t have one in its LSDB to begin with. R1 will have to find summary information for another router in Area 0 that has a tree built to node 6.6.6.6.
This is where Type-4 ASBR Summary LSAs are added to the mix. Using Type-4 Summary LSAs, routers in a different area than the ASBR can solve the SPT to another router that has sufficient information to build a tree to the ASBR. These routers are the ABRs.
Type-4 ASBR Summary LSAs
Type-4 ASBR Summary LSAs are generated by ABRs and flooded into all of their attached areas. It serves two important purposes for areas in which the ASBR does not exist:
1. Injects topology information about the ASBR back into the LSDBs of other areas
2. Allows routers in other areas to make the best choice of entry point into the area the ASBR resides.
For the first point, Type-4 ASBR Summary LSAs simply alert all routers in other areas which router in their area has sufficient information to reach the ASBR. This allows the routers in other areas to perform an inter-area lookup to solve the path to the ASBR.
A case for the second point is made when there are multiple ABRs in an area with differing costs to reach the ASBR. When generating the Type-4 ASBR Summary LSA, the ABR lists its own cost to reach the ASBR. Without the Type-4 ASBR Summary LSA, the routers in other areas would not know which ABR to use as the best entry point into the area containing the ASBR.
In the reference topology, when R6 redistributes external prefixes into the OSPF routing domain, it sets a bit in its Type-1 Router LSA indicating that it is an ASBR
LSA-type 1 (Router-LSA), len 60
--- omitted ---
Sequence Number: 0x80000004
Checksum: 0x18b4
Length: 60
Flags: 0x02 ((E) AS boundary router)
.... .0.. = (V) Virtual link endpoint: No
.... ..1. = (E) AS boundary router: Yes
.... ...0 = (B) Area border router: No
This bit is called the External bit or the E-bit. Once set, it signals to the ABR (R2) that R6 is injecting external information into the OSPF domain. The same can be observed using the show ip ospf database router [RID of asbr] command.
R5#show ip ospf database router 6.6.6.6
OSPF Router with ID (6.6.6.6) (Process ID 1)
LS age: 481
Options: (No TOS-capability, DC)
LS Seq Number: 80000004
Checksum: 0x18B4
Length: 60
AS Boundary Router
The ABR will generate a Type-4 ASBR Summary LSA for the ASBR and flood the Type-4 Summary LSA into all other areas to which it is connected.
The following is the Type-4 ASBR Summary LSA generated by R2 for the ASBR R6 that is flooded into Area 0:
R2#show ip ospf database asbr-summary
OSPF Router with ID (2.2.2.2) (Process ID 1)
Summary ASB Link States (Area 0)
LS age: 0
Options: (No TOS-capability, DC, Upward)
LS Type: Summary Links(AS Boundary Router)
LS Seq Number: 80000003
Checksum: 0xEE17
Length: 28
MTID: 0 Metric: 20
Note: It is not necessary to flood a Type-4 ASBR Summary LSA in the area the ASBR is located. As previously stated, the area the ASBR is located has sufficient information in its LSDB (specifically a Type-1 Router LSA) to recurse to the ASBR.
Notice the advertising router field is set to 2.2.2.2 (R2) in the Type-4 ASBR Summary LSA. This indicates R1, R3, and R4 will solve the shortest path to R2 in order to reach the ASBR R6. Because the ABR sets itself as the advertising router for the Type-4 ASBR Summary LSA, these LSAs have an area-wide flooding scope and will not be flooded across the entire OSPF domain like a Type-5 External LSA.
Using the Type-4 ASBR Summary LSA, R2 is saying:
“I can be used to reach the ASBR 6.6.6.6 with a metric of 20.”
Other ABRs in the topology (R3 and R4) will generate a new Type-4 ASBR Summary LSA, setting themselves as the advertising router for the LSA. For example in Area 489, the Type-4 ASBR Summary LSA for the node 6.6.6.6 is as follows:
R4#sh ip ospf 1 489 data asbr-summary
OSPF Router with ID (4.4.4.4) (Process ID 1)
Summary ASB Link States (Area 489)
LS age: 31
Options: (No TOS-capability, DC, Upward)
LS Type: Summary Links(AS Boundary Router)
LS Seq Number: 80000001
Checksum: 0x7F6C
Length: 28
MTID: 0 Metric: 40
Type-4 Summary LSAs contain only the following:
• RID of the ASBR
• The RID of the advertising ABR
• Advertising ABR’s metric to reach the ASBR.
Type-4 ASBR Summary LSAs do not contain IP addressing information. They merely serve as pointers with information about how to reach an ASBR in the OSPF domain.
Connecting the LSAs
At this point it may seem like the LSDB is just a random array of LSAs with no real structure. In actuality, all of the LSAs in the database are linked by keys. These keys are Link State IDs, and Link IDs that are populated with RIDs, interface address, or other identifiers. Using these keys, the LSDB contains all of the information necessary to allow the SPF algorithm to discover the shortest path between any two nodes.
In order to get a good idea of how the different types of LSAs are connected in the LSDB, it helps to have an example or two to follow.
This section examines how the LSAs in the LSDB are used in R1’s calculation of the shortest path to reach the 192.168.8.0/24 network and to the external network 172.16.65.0/24. At each step, R1 uses a combination of Link State IDs and Link IDs as keys to discover its ultimate next hop along the path to reach the destination node.
To solve the path to the 192.168.8.0/24 network, R1 uses 192.168.8.0 as a key to look up information about the node:
R1#show ip ospf database summary 192.168.8.0
OSPF Router with ID (1.1.1.1) (Process ID 1)
Summary Net Link States (Area 0)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 668
Options: (No TOS-capability, DC, Upward)
Link State ID: 192.168.8.0 (summary Network Number)
LS Seq Number: 80000002
Checksum: 0x733C
Length: 28
MTID: 0 Metric: 11
R1 knows node 4.4.4.4 (R4) advertised the link to this network with a metric of 11. R1 uses 4.4.4.4 as a key to look up the next LSA with details of how to reach node 4.4.4.4. It matches 4.4.4.4’s Type-1 Router LSA:
R1#show ip ospf database router 4.4.4.4
OSPF Router with ID (1.1.1.1) (Process ID 1)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 796
Options: (No TOS-capability, DC)
LS Seq Number: 80000004
Checksum: 0x6FD8
Length: 48
Area Border Router
AS Boundary Router
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 1
Link connected to: a Transit Network
Number of MTID metrics: 0
TOS 0 Metrics: 10
R4 has two links. The first link is a link to a stub network that will not transit R1’s traffic. The second link is a link to a transit network pseudo node 134.1.1.4 with a metric of 10. R1 uses 134.1.1.4 as a key to find out how to reach the pseudo network node in the LSDB. It finds the following Type-2 Network LSA:
R1#show ip ospf database network 134.1.1.4
OSPF Router with ID (1.1.1.1) (Process ID 1)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 918
Options: (No TOS-capability, DC)
LS Seq Number: 80000002
Length: 36
Attached Router: 4.4.4.4
Attached Router: 1.1.1.1
Attached Router: 3.3.3.3
R1 discovers pseudo node 134.1.1.4 also has a link to node 1.1.1.1, which is R1 itself. R1 uses its own RID as the key to find its own Router LSA and discover which link it should use to reach the pseudo node:
R1#show ip ospf database router self-originate
OSPF Router with ID (1.1.1.1) (Process ID 1)
LS age: 1079
Options: (No TOS-capability, DC)
LS Seq Number: 80000005
Checksum: 0x290C
Length: 72
--- omitted ---
Link connected to: a Transit Network
Number of MTID metrics: 0
TOS 0 Metrics: 10
R1 has verified it has also advertised a connection to this pseudo node. This verification finishes R1’s search. It can use the pseudo node to reach node 4.4.4.4, which can reach the network node 192.168.8.0/24 on R8. R1 uses R4’s interface address 134.1.1.4 as next hop, adds the cost up along the path to equal 21, and writes this in its RIB as the inter-area route below:
R1#show ip route ospf | i 192.168.8.0
O IA 192.168.8.0/24 [110/21] via 134.1.1.4, 00:53:42, Ethernet0/0
NOTE: R1 will also calculate R3’s path to reach the pseudo-node, but the cost value would be 31 vs the 21 of going directly to R4. This is because R1 would forward to R3, which would forward to R4. R4’s cost to the pseudo-node is not used in R1’s calculation, because it is the incoming interface and not the outgoing interface for the traffic. This leads to the cost of 21 instead of 31. In R3’s case, R3’s interface would be used to both receive R1’s packet and forward the packet to R4, which is why it is added in the total cost of the path.
To solve the path to the external network, R1 examines the Type-5 External LSA for 172.16.65.0/24:
R1#show ip ospf database external 172.16.65.0
OSPF Router with ID (1.1.1.1) (Process ID 1)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 1624
Options: (No TOS-capability, DC, Upward)
Link State ID: 172.16.65.0 (External Network Number )
LS Seq Number: 80000002
Checksum: 0xDDAB
Length: 36
Metric Type: 2 (Larger than any link state path)
MTID: 0
Metric: 20
External Route Tag: 0
R1 notes the Metric of 20 to reach the external prefix. Since the redistributing router R6 belongs to an area that R1 does not belong to, R1 needs to build an SPT to someone in its area that already has an SPT built to the ASBR, R6. The listed forward address of 0.0.0.0 forces R1 to use the RID of the advertising router as a key to find the appropriate next hop for the prefix. It matches this Type-4 ASBR Summary LSA:
R1#show ip ospf database asbr-summary
OSPF Router with ID (1.1.1.1) (Process ID 1)
Summary ASB Link States (Area 0)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 1754
Options: (No TOS-capability, DC, Upward)
LS Type: Summary Links(AS Boundary Router)
LS Seq Number: 80000002
Checksum: 0xF016
Length: 28
MTID: 0 Metric: 20
Using this LSA, R1 knows node 2.2.2.2 (R2) can be used to reach the ASBR node 6.6.6.6. It notes the metric of 20 and uses 2.2.2.2 as a key to find out how to reach that node discovering node 2.2.2.2’s Type-1 Router LSA:
R1#show ip ospf database router 2.2.2.2
OSPF Router with ID (1.1.1.1) (Process ID 1)
Routing Bit Set on this LSA in topology Base with MTID 0
LS age: 1942
Options: (No TOS-capability, DC)
LS Seq Number: 80000003
Checksum: 0x6C6F
Length: 48
Area Border Router
Link connected to: another Router (point-to-point)
(Link ID) Neighboring Router ID: 1.1.1.1
Number of MTID metrics: 0
TOS 0 Metrics: 10
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 10
R1 finds a Type-1 Router LSA for node 2.2.2.2. Node 2.2.2.2 has a link to a stub network, which cannot transit R1’s traffic. It also has a link to node 1.1.1.1, which is R1 itself. R1 uses its own RID as the key to find its own Router LSA to find the appropriate link:
R1#show ip ospf database router self-originate
OSPF Router with ID (1.1.1.1) (Process ID 1)
LS age: 62
Options: (No TOS-capability, DC)
LS Seq Number: 80000006
Checksum: 0x270D
Length: 72
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 1
Link connected to: another Router (point-to-point)
(Link ID) Neighboring Router ID: 2.2.2.2
Number of MTID metrics: 0
TOS 0 Metrics: 10
Link connected to: a Stub Network
Number of MTID metrics: 0
TOS 0 Metrics: 10
R1 verifies it is advertising a point-to-point link to 2.2.2.2 as well. It has now completed the path to reach the external network. In this case R1 does not add up the costs to reach the external prefix; it lists only the cost associated with the Type-5 External LSA of 20 and adds R2 as the next hop:
R1#show ip route ospf | i 172.16.65.0
O E2 172.16.65.0 [110/20] via 12.1.1.2, 01:10:10, Ethernet0/1
NOTE: R1 does record the total cost of the path in what’s known as a forward metric. The details of why R1 does this is outside the scope of this blog.
This example is a bit contrived, but is used to demonstrate how the different LSAs are linked in the LSDB. In reality, R1 uses a much more thorough approach to build its SPT, however, the details of this are outside the scope of this blog.
Summary
When first learning about OSPF’s LSDB, it helps to think of it as a graph. Graphs are collections of nodes connected by edges. For OSPF, these nodes are routers, transit networks, stub networks, and external networks described by LSAs. LSAs are linked through Link State IDs and Link IDs forming edges.
This graph creates a complete picture of the network topology from which OSPF can calculate the shortest path between these nodes. | 12,534 | 46,004 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.961087 |
https://www.spreadsheet123.com/ExcelTemplates/breakeven-analysis.html | 1,695,307,841,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00215.warc.gz | 1,138,831,002 | 20,686 | # Break Even Analysis
Calculate the Breal-Even Point using this Break-Even Analysis Calculator - by Alex Bejanishvili
Posted in category FINANCIAL STATEMENTS
Break-even analysis is used by businesses to determine the time it will take to become profitable and usually is a part of a Business Plan. Whether you are a startup business, or are thinking of adding a product or service to your existing business venture, it is useful to take the time to run a break-even analysis to determine whether it is a financially viable venture, in terms of how many units you will need to sell to break even, and how long it will take to do so.
The actual calculations required to perform a break-even analysis are relatively simple. The calculations however are the easy part, determining your optimum sales unit price and accompanying variable costs can be much more difficult. Even fixed costs can be less than straightforward, as you may, for example, find yourself with a choice of premises with different rates of rent. While a break-even calculator cannot make these decisions for you, it can make the decision making process significantly easier for you as a business owner, by enabling you to easily look at different scenarios for break-even. Our Break Even Calculator will help you run the reports and analysis you need for your business.
## Break Even Analysis
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### Break-Even Point Calculator for Multiple Products
This version of Break-Even calculator allows to calculate break-even units and break-even price and break-even period for multiple products using additional Sales Forecast feature. This feature helps to define the proportion of the product in the total sale as a constant over the period of time.
View Screenshot
## Using Break Even Calculator
There are a number of ways in which break-even can be viewed. While they essentially all use the same information to perform a calculation, there are differences between them, and you may favour one method over the others. Our break-even calculator allows you to work out your break-even number of units, break-even price and break-even payback period. The information you enter can be easily changed, allowing you to easily look at factors such as the impact of a small increase in cost or sale price.
The break-even number of units is the number of units that you would have to sell in order to break-even, or put another way, the point at which your investment will start to make a positive return.
The break-even price is the dollar amount of sales you would have to make in order to cover your costs, i.e. the point at which your total revenue equals your total costs.
The break-even payback period is the amount of time it will take you to break-even.
Our break-even calculator also allows you to calculate a net income before taxes (NIBT) using the monthly number of units sold. This will enable you to see how your business will perform over a period of time, allowing you to plan for the future.
## Break Even Chart
A breakeven chart is a useful way of displaying your total revenue, total cost and profit (or loss), along with you're a break-even point. There are two ways of calculating this break-even point:
TC = TRorFC + VC = P * X
• TC - Total Costs
• TR - Total Revenue
• FC - Total Fixed Costs
• TR - Total Variable Costs
• P - Sale Price per Unit
• U - Number of Unites Sold
You will find a version of this chart on each of the spreadsheets, with the number of units running along the X axis, and cost/revenue/profit running along the Y axis.
## Solving Break Even Analysis Problems
For all break-even calculations, you will need to know the fixed costs, variable costs and sales price per unit.
### Solving for number of units
There is a dedicated sheet on the spreadsheet for calculating the break-even number of units. The formula used for this calculation is:
X = FC / ( P - VC )
The definition of the variables used in this equation are listed below:
• X - Number of Units to Break-Even
• P - Sale Price per Unit
• FC - Total Fixed Cost
• VC - Total Variable Cost Per Unit
Enter your unit sale price above the chart, and use the areas below it to enter your fixed and variable costs. The fixed costs section should be used for entering costs that do not vary depending on the number of units sold or manufactured, for example rent. The variable costs is split into two sections, one for costs equating to a dollar amount per unit, for example, manufacturing costs, and another for percentage costs per unit, e.g. a licencing fee or commission.
Once you have entered all of this information, the chart will show you your total revenue, total cost and profit (or loss) based on the number of units sold. Below the chart, you will find the number break-even units and break-even sales.
### Solving for the Break-Even Price
The next sheet on the spreadsheet calculates the break-even price per unit. This can be useful if you want to try and calculate the price you need to sell a set number of units at in order to break-even. The calculations used for this are:
P = 1 / ( 1 - VC-P ) * ( VC-D + ( FC / X ) )
• P - Sale Price per Unit
• X - Number of Units to Break-Even
• FC - Total Fixed Cost
• VC-P - Total Variable Cost Per Unit Percentage-Based
• VC-D - Total Variable Cost Per Unit Dollar-Based
At the top of the sheet, enter the number of units you are looking to sell. As with the break-even units sheet, enter your fixed and variable costs below the chart in the appropriate sections. Again, the chart will show you your total revenue, total costs and profit (or loss) based on the number of units sold. This time however, the focus is on the unit sale price information. The chart shows the total revenue at which you will break-even, and below the chart, this information is given, along with the break-even sales and number of units (as entered at the top of the sheet).
### Solving for the payback period
Unlike the other two break-even calculations, the payback period requires fixed costs and start-up costs to be split, for example, a piece of manufacturing equipment may only need to be purchased once, whereas rent is an ongoing fixed cost.
T = SC / ( ( P - VC ) * x ) - RFC
• T - Payback Period in months
• SC - Total Start-up Costs
• P - Sale Price per Unit
• VC - Total Variable Cost Per Unit
• RFC - Total Recurring Fixed Cost
• x - Number of Units sold per month
Enter your unit sale price and number of monthly number of units at the top of the page. As before, there is space to fill out your recurring fixed and variable costs, and additionally for this spreadsheet, there is a section for entering your start-up costs.
To the right of the data entry section, a chart will again show you the total revenue, total cost, and profit (or loss). Below the chart, you will find all of the information you have input, along with the break-even time and break-even units.
### Solving for break-even sales
There are two methods of calculating the break-even sales as a dollar amount.
S = P * UorS = FC / (1 - VC / P)
• S - Break-even sales
• P - Sale Price per Unit
• U - Number of units
• FC - Total Fixed Cost
• VC - Total Variable Cost Per Unit
The second part of this equation (1 - VC / P) is also known at the contribution margin ratio CMR, so the formula can be more simply expressed as:
S = FC * CMR
• S - Break-even sales
• FC - Total Fixed Cost
• CMR - Contribution margin ratio
The break-even sales amount can also be altered to take the payback period into account:
S = P * MU * T
• S - Break-even sales
• P - Sale Price per Unit
• MU - Number of units per month
• T - Payback period in month
Should you require a certain profit level, adding this into your fixed costs will allow you to calculate your break-even sales after taking this into consideration. There are a number of reasons why this could be helpful, for example if you are relying on the profits for your personal income, rather than taking a salary that is already accounted for within the fixed costs.
### Net Profit Before Tax
The Break-even (Payback Period) sheet contains a chart and table for the net profit before tax (NPBT). The NPBT is a commonly used measure of profitability, and is sometimes also referred to as earnings before tax, or pre-tax profit. As the name suggests, it is calculated by subtracting expenses, with the exception of corporate income tax, from income
On the spreadsheet itself, the net profit before tax can be calculated by selecting the period (number of months), and number of units you want to calculate for. You may wish to use your solved payback period, or number of units for this calculation, or might prefer to look over a set period of time, for example six months or a year. Using this information, along with the costs and revenue information entered elsewhere on the Break-even (Payback Period) sheet, the NPBT calculator will tell you the total number of units sold, total cost and total revenue, along with a value for the NPBT. | 2,057 | 9,331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.872559 |
https://www.jobduniya.com/Placement-Papers/ABB/ABB-Placement-Paper-2-May-2007-Vadodra.html | 1,582,668,462,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00112.warc.gz | 772,319,533 | 9,632 | # Placement Papers: ABB Placement Paper 2 May 2007 Vadodra
Examrace Placement Series prepares you for the toughest placement exams to top companies.
TECHNICAL PAPER-40 QUESTION, 45-MINUTES
1. In a ckt, We r giving voltage of 50 Hz as well as 60. Then what will be the resultant frequency.
1. less than 50
2. more than 60
3. in between 50 & 60
4. none according to our conclusion answer will be none because if we apply two frequency component resultant frequency we can not say with such an ease. U should confirm the answer
2. In a ckt a single resistor is connected across a dc source, what will be the effect on current in first resistor if we connect one more resistance in parallel with earlier one. Answer. No change since it is a parallel combination.
3. why we don't like flashover in transmission line (t-line) -
1. it may create earth fault
2. it reduces the life of insulator.
4. total no of strands in a arcs conductor is 81, then what is the no. Of conductor in its outer layer (a). 36 (b) 18 (c) 24. Also read some more on acsr.
5. Two questions based on p. u. Calculation like, p. u. Calculation is given with respect to some old base and u have to calculate it with reference to new base (new resistance/old). = (mva new/mva old) * (old voltage/new voltage) 2. Other question is based upon transfer of p. u calculation in transformer i.e., how base changes when we we move from primary to secondary or like wise. Read some more on p. u calculation.
6. which table is referred for sag calculation-stringing chart answer
7. in a R-L ckt a ac voltage is applied, such that instantaneous power is negative for 2ms, then what will be the power factor.
1. 9 deg
2. 18 deg
3. 36 Deg. (I don't know the correct ans).
8. In an incandescent lamp
1. luminous intensity is more than non-luminous intensity
2. , less,
Ans: Since efficiency is less than 100%, hence ans is (b), u should confirm it further.
1. series motor
2. shunt motor
3. Compound motor
Ans: (b)
10. In a 60Hz induction motor full load speed is 850 rpm then what is the Synchronous speed (a). 900 rpm (b), 950 rpm (c), 1600 rpm. Ans: (a)
11. A sync. Motor is running at synch. Speed, if al of sudden D. C. Excitation is removed, then
1. it will rotate at slip speed
2. it will stop
3. it will continue to rotate at sync. Speed
Ans: (a)
12. A transmission line is designed for 50Hz, 440KV. If we want to transfer power at 60Hz, 440 KV, then the power transfer capability will
1. decrease
2. Increase
3. None
Ans: Decrease as P = (| Vt | Ef | sin (delta) )/X, where (delta) is torque angle.
13. Increased rotor resistance in rotor ckt of induction motor is related with
1. high starting torque
2. more speed variation, Ans: (a)
14. In the formulae E = 4.44 f N? is
1. Avg value
2. Rms value
3. Maximum value
15. Voltage & current in a ckt is given by V = V1 + j V2 and I = I1 + j I2, then rms power is (refer book by Edministrator on NETWORK. ).
16. Input impedence of MOSFET is more than BJT. (Ans).
17. Remember truth table of AND, NOR, NAND, OR, EX-OR ETC
18. Conversion of Binary number into Equivalent decimal No.
19. Megger is used for the measurement of (a) Insulation resistance (b), Conductor resistance. Ans: (a)
20. Form factor for sinusoidal as well as DC
21. Formulae of Regulation (Vs-Vr) * 100/Vr, then transmission line is
1. short transmission line
2. long
3. medium. Ans: (a)
22. Improvement in power factor reduces
1. power consumed by consumer
2. power generation
3. both a & b
24. No-load test for Synchronous motor, the graph is drawn-stator open ckt emf Vs field current (Ans: a)
25. An AC voltage of 50Hz is impressed in a resistive ckt, the oscillating power has a frequency (a) 50 Hz (b), 100 (c), no oscillating power is there in resistive ckt Ans: (a)
26. Insulation used in transformer ___________leakage flux.
1. increases
2. decreases Ans: (b)
27. After rain what happens to Insulator (a) break-down strength of Insulator decreases (b), Arch length reduces, Ans: (b)? [Confirm it]
28. Diversity factor helps to (what?) [Read diversity factor, load factor, Reserve capacity factor in depth, with calculation]
29. Why capacitance is shown as a Shunt element in analysis of transmission line
1. it is between Conductor & earth
2. because Admittance is used for calculation of capacitive reactance
Ans: (a)
30. B-R-Y sequence is followed in three phase system, if phase voltage in B-phase is Vm sin 100, then the phase voltage in R-phase would be (a) Vm sin (-20) Ans: (a)
31. In a particular ckt I = Im Sin (wt-270) and V = Vm Sin wt, then type of ckt is (a) pure resistive ckt. [Ans]
32. In a L-R ckt energy lost = 2000 W, energy conserved = 500W, then what is the time constant. Ans: Time constant = L/R = 0.5
33. In electro-dynamometer A'meter & wattmeter the type of scale is Ans: Non-uniform
34. For the same current carrying capacity corona loss of ACSR will be ________than copper conductor (a). more (b), less (c), equal Ans: (b)
35. A R-C ckt, supplied with DC, a bulb is connected across the Capacitor, then what happens to the illumination, if we change the capacitance. Ans: No change at all
## Quantitative Paper
45 Questions____45 minutes
1. About 10 quanti questions (based on Mixture, Work etc. Of very easy type)
2. What is GDP?
3. Vector algebra, codition for Co-planer vector etc.
4. Gravitation, geo-synchronous satellite (it's hight, orbit, radius etc.), escape velocity, how g (gravitational accln) varies, about gravitational potential.
5. Basic electricity and Magnetism____Biot-savart law, current carrying conductor properties.
6. Nuclear physics, Bohr's constant, and Other theories related.
7. Problem based on VIBGYOR, how wave length and frequency is varying.
8. Questions based on Plank's Theory, E = hv
9. V = u + at, V2 = u2 + 2as and W = mgh questions based on above theory
10. Faraday's laws of electrolysis, m = Zit
11. Heat conduction problem.
12. Co lour-coding of resistor (BBROYGBVGW)
13. How velocity of light changes in different medium while frequency remain unchanged.
14. statistics, calculation of mode, co-efficient regression (3 − 4 Questions)
15. f (x) = Sin x + Cos x, find the maximum value of the function. Ans: Sqrt (2)
16. Formulae for parallel plate capacitor and force between plates. | 1,735 | 6,330 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2020-10 | latest | en | 0.895384 |
https://en.wikiversity.org/wiki/Fundamental_Physics/Electronics/RL_Circuit/RL_Series | 1,623,523,551,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586239.2/warc/CC-MAIN-20210612162957-20210612192957-00490.warc.gz | 232,953,017 | 8,860 | # Fundamental Physics/Electronics/RL Circuit/RL Series
${\displaystyle V_{L}+V_{R}=0}$
${\displaystyle L{\frac {d}{dt}}i(t)+Ri(t)=0}$
${\displaystyle {\frac {d}{dt}}i(t)=-{\frac {R}{L}}i(t)}$
${\displaystyle \int {\frac {di(t)}{i(t)}}=-\int {\frac {R}{L}}dt}$
${\displaystyle Lni(t)=-{\frac {R}{L}}t+c}$
${\displaystyle i(t)=e^{-{\frac {R}{L}}t+c}}$
${\displaystyle i(t)=Ae^{-{\frac {R}{L}}t}}$ | 167 | 395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-25 | latest | en | 0.356376 |
https://kunduz.com/questions-and-answers/anormal-distribution-has-a-mean-of-20-and-a-standard-deviation-of-5-use-the-68-95-99-7-rule-to-find-the-percentage-of-values-in-the-distribution-between-20-and-35-what-is-the-percentage-of-values-in-318835/ | 1,721,739,041,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518029.87/warc/CC-MAIN-20240723102757-20240723132757-00606.warc.gz | 317,214,413 | 32,104 | Question:
# Anormal distribution has a mean of 20 and a standard
Last updated: 11/18/2023
Anormal distribution has a mean of 20 and a standard deviation of 5 Use the 68 95 99 7 rule to find the percentage of values in the distribution between 20 and 35 What is the percentage of values in the distribution between 20 and 35 Save | 86 | 331 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.909201 |
https://gmatclub.com/forum/an-empty-fuel-tank-with-a-capacity-of-200-gallons-was-fill-133301.html?fl=similar | 1,495,952,851,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609598.5/warc/CC-MAIN-20170528043426-20170528063426-00064.warc.gz | 948,011,899 | 65,720 | Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
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# An empty fuel tank with a capacity of 200 gallons was fill
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An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 04:59
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An empty fuel tank with a capacity of 200 gallons was filled partially with fuel A and then to capacity with fuel B. Fuel A contains 12% ethanol by volume and fuel B contains 16% ethanol by volume. If the full fuel tank contains 30 gallons of ethanol, how many gallons of fuel A were added?
A 160
B 150
C 100
D 80
E 50
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:06
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Expert's post
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macjas wrote:
An empty fuel tank with a capacity of 200 gallons was filled partially with fuel A and then to capacity with fuel B. Fuel A contains 12% ethanol by volume and fuel B contains 16% ethanol by volume. If the full fuel tank contains 30 gallons of ethanol, how many gallons of fuel A were added?
A 160
B 150
C 100
D 80
E 50
Say there are A gallons of fuel A in the tank, then there would be 200-A gallons of fuel B.
The amount of ethanol in A gallons of fuel A is 0.12A;
The amount of ethanol in 200-A gallons of fuel B is 0.16(200-A);
Since the total amount of ethanol is 30 gallons then 0.12A+0.16(200-A)=30 --> A=50.
Hope it's clear.
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:13
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marked E. Solved it this way:
total ethanol = 30 lts in 200lts therefore % = 30/200 *100 = 15%
now creating an equation in
Let qty of fuel A= x lts and Fuel b = 200-x lts
there fore 0.12x +0.16(200-x) = 30, solving we get x = 50lts.
Hope its clear.
Last edited by vibhav on 27 May 2012, 05:16, edited 1 time in total.
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:16
additionally as be see the mixture's overall % is closer to 16% ethanol mixture, this should show you that the 16% ethanol mixture's quantity is much higher than 12% mixture.
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:18
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Another way to do it (weighted averages):
30 of 200
12% = 15% 16%
A-------------Mixture--------------B
3 1
So 1 part of B and 3 parts of A make 200
So, A is 25% of the total 200 i.e 50
Last edited by macjas on 27 May 2012, 05:23, edited 2 times in total.
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:19
To solve this problem using mixtures is also possible;
qty of A/qty of b = % of B (16%) - final %(15) / final %(15) - % of A(12%)
i.e. a/b = 1/4 and we know a+B = 200 hence a= 200/4 = 50
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:20
@macjas thats a quick method as well! nice.
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 05:24
sorry... was having trouble formatting my post correctly
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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27 May 2012, 06:15
4
KUDOS
Hi,
The Allegation method is very useful in this case.
The percentage of ethanol in two mixtures = 12% & 16%
Final percentage in mixture = 30/200 * 100 = 15%
Following is the diagrammatic representation of the method:
Attachments
Allegations.jpg [ 11.3 KiB | Viewed 4068 times ]
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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08 Jul 2013, 01:10
Bumping for review and further discussion*. Get a kudos point for an alternative solution!
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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18 May 2014, 15:14
We can solve this problem using weighted averages method.
First we need to find 30 is what % of 200 => 15%;
12%(A) ------------ 15%Final Mix ------------------ 16%(B)
3 1
w1/w2 = 1/3 ;
gallons of fuel A added = 1/4(200) =50;
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink]
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22 May 2014, 02:16
1
KUDOS
Let's take Fuel of type A = A litre
Fuel of type B = B litre
Ethanol in type A = 12% => 0.12A
Ethanol in type B = 16% => 0.16B
Now,
A + B = 200
0.12A + 0.16B = 30
Multiplying first with 0.16 and subtracting second
0.16A - 0.12A = 32-30
0.04A = 2
A = 2/0.04=> A = 50
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Re: An empty fuel tank with a capacity of 200 gallons was fill [#permalink] 22 May 2014, 02:16
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• Weight- a measure of the force of gravity on an object. The weight of an object is the gravitational force between the object and the Earth. The more mass the object has the greater its weight will be.Weight is a force, so it is measured in newtons. On the surface of the Earth an object with a mass of 1 kg has a weight of about 10 N.
• Mass- a measure of how much matter an object contains. The more matter an object contains, the greater the mass
• Gravity- the attraction between two objects that have mass. Even us, as humans, attract other objects, although our mass is too small for that force to be very strong. Gravitational forces increase when masses are bigger and the objects are closer.
• The mass of an object stays the same wherever it is, but it’s weight can change. The moon has less mass than the earth, so it’s gravity is less than the earth’s gravity. This means that objects weigh less on the moon than on the earth.
• The Moon has less mass than the Earth, so the moon's gravity is less than the Earth's. This means that objects weigh less on the Moon than they do on the Earth. The moons gravity it roughly around one sixth of the earth's. A 120kg astronaut weighing 1200 N on earth would only weigh 200 N on the moon. The mass remains the same.
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https://mathexamination.com/lab/2-%C3%97-2-real-matrices.php | 1,611,433,668,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703538431.77/warc/CC-MAIN-20210123191721-20210123221721-00313.warc.gz | 440,422,535 | 8,664 | ## Take My 2 × 2 Real Matrices Lab
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Visualisation is a really effective technique, and is something that you ought to not miss over. Imagine what it would certainly feel like to actually affix the component and also be able to go from there, without the measurement.
Currently, allow's check out the second example. Allow's take the very same procedure as previously, now the student needs to keep in mind that they are going to move back one step. If you tell them that they need to move back one action, but after that you eliminate the idea of needing to move back one step, then they won't recognize just how to wage the problem, they won't understand where to search for that step, as well as the procedure will certainly be a mess.
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An aid service can make it less complicated for you to handle your 2 × 2 Real Matrices class. Furthermore, you can learn more concerning yourself as well as help you succeed. Discover the most effective tutoring solution as well as you will certainly have the ability to take your research study skills to the next level. | 2,596 | 12,681 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2021-04 | latest | en | 0.977526 |
https://www.gentsetrekvogels.be/2016_06/why-the-45-degree-line-of-aggregate-supply.html | 1,606,385,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141187753.32/warc/CC-MAIN-20201126084625-20201126114625-00319.warc.gz | 668,206,944 | 5,935 | # Why The 45 Degree Line Of Aggregate Supply
Aggregate demand ad is the total demand for goods and services produced within the economy over a period of time.Aggregate demand ad is composed of various components.Ad cig x-m c consumer expenditure on goods and services.I gross capital investment i.E.Investment spending on capital goods e.G.Factories and machines.
• ## Aggregate Demand S Cool The Revision Website
Aggregate demand and supply analysis is very similar to the analysis in the supply and demand topic.The big difference is that aggregate demand and supply refer to the aggregates of the whole economy.The supply and demand analysis in the first topic is used in microeconomics to look at the behaviour of individual consumers, producers and industries.
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• ## Simple Keynesian Model Of Income Determination
Simple keynesian model of income determination 1.Of national income is determined in keyness two sector model.Ci curve represents the aggregate expenditure and 45 degree oz line represents aggregate supply of output.The goods and services are provided by firms when they think they can sell them in themarket.There will be equilibrium in.
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• ## Aggregate Supply Definition Investopediam
Aggregate supply, also known as total output, is the total supply of goods and services produced within an economy at a given overall price in a given period.
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• ## Inflationary And Deflationary Gapsrecessionary Gap
Inflationary and deflationary gaps.Aggregate demand and as aggregate supply is not equal to the level of full employment, then two situations can arise.Let us assume initially that the aggregate expenditure curves ae interests the 45 degree line at point e to the left of full employment line or potential income.
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• ## Keynesian Cross Model Conspecte Com
There is only one level of y where aggregate demand is equal to y, thepoint where ad cutts the 45-degree line.This level is called the equilibrium level of gdp and it is denoted by y.Formally, y is defined implicitly by ydy y.Justification.Note that we have not said anything about the aggregate supply.
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• ## What Is The Slope Of A 45 Degree Line Bartleby
The slope of 45-degree line is always positive.In a 45-degree line, any intersecting point of values in.Maria can read 20 pages of economics in an hour.She can also read 50 pages of sociology in an hour.She spends.Principles of microeconomics mindtap course list tax rates suppose taxes.
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• ## The Expenditure Output Model Social Sci Libretexts
The intersection of the aggregate expenditure line with the 45-degree lineat point e 0 in figure 1will show the equilibrium for the economy, because it is the point where aggregate expenditure is equal to output or real gdp.After developing an understanding of what the aggregate expenditures schedule means, we will return to this.
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• ## The Keynesian Aggregate Expenditure Model
Aggregate consumption function the keynesian model assumes that there is a positive relationship be-tween consumption and income.However, as in-come increases, consump-tion expands by a smaller amount.Thus, the slope of the consumption func-tion line is less than 1 c less than the slope of the 45-degree line.Exhibit 1 6 9 12 15 45.
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• ## Why The 45 Degree Line Of Aggregate Supply
Why the 45 degree line of aggregate supply -, why the 45 degree line of aggregate supply as a leading global manufacturer of crushing and milling equipment, we.Contact supplier.Aggregate supply as curve - cliffsnotes study guides.The aggregate supply curve depicts the quantity of real.Is a vertical line,.
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• ## Reading Equilibrium And The Expenditure Output
To understand why the point of intersection between the aggregate expenditure function and the 45-degree line is a macroeconomic equilibrium, consider what would happen if an economy found itself to the right of the equilibrium point e, say point h in figure b.8, where output is higher than the equilibrium.
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• ## What Is The Slope Of The Aggregate Demand Curve
Therefore, the increase in consumer saving results in an increase in the supply of loanable funds, which decreases the real interest rate and increases the level of investment in the economy.Since investment is a category of gdp and therefore a component of aggregate demand, a decrease in the price level leads to an increase in aggregate demand.
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• ## Decrease Decrease On The 45 Degree Line Diagram The
Decrease decrease on the 45-degree line diagram, the 45-degree line shows points where real aggregate expenditure equals real gdp on the 45-degree line diagram, for points that lie below the 45-degree line planned aggregate expenditure is less than gdp if planned aggregate expenditure is less than total production, firms will experience an unplanned increase in inventories how does a decrease.
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• ## What Is Significance Of 45 Degree Homework Help
Aggregate supply is the total quantity of the commodity that is offered for sale in the market at any particular point of time.It is the total quantity of commodities and services that the firms are willing and able to sell in the market.It is represented by the upward sloping 45-degree line.
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• ## Notes On Aggregate Supply And Its Component
Advertisements notes on aggregate supply and its component aggregate supply is the money value of total output available in the economy for purchase during a given period.When expressed.In physical terms, aggregate supply refers to the total production of goods and services in an economy.It is assumed that in short run, prices of.
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• ## Reading The Expenditure Output Model
The 45-degree line shows all points where aggregate expenditures and output are equal.The aggregate expenditure schedule shows how total spending or aggregate expenditure increases as output or real gdp rises.The intersection of the aggregate expenditure schedule and the 45-degree line.
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• ## An Angle Of How Many Degrees Is Found In The
Answer the 45-degree line represents an aggregate supply curve which embodies the idea that, as long as the economy is operating at less than full employment, anything demanded will.
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• ## Aggregate Demand And Aggrate Supply
Aggregate demand and aggrate supply - economic krugman vip vip.The aggregate demand curve and the income-expenditure model planned aggregate spending 45-degree line e ae planned 2 2 ae.
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• ## Economics Review Chapter 25 Flashcards Quizlet
Keynesian economics focuses on explaining why recessions and depressions occur, as well as offering a for minimizing their effects.Policy prescription.Aggregate demand is more likely to than aggregate supply in the short run.Shift substantially.45- degree line.Refer to.
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• ## Amosweb Is Economics Encyclonomic Webpedia
The 45-degree line assists in the keynesian economics evaluation and analysis of the macroeconomy.In general, a 45-degree line is so named because it forms a 45-degree angle with both the vertical or horizontal axes of a typical right-angle diagram.
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• ## The Aggregate Expenditure Model Course Hero
One is the aggregate expenditure function, which shows how aggregate expenditures increase as real gdp increases.The aggregate expenditure function component looks like a line on the graph with a positive slope.The second component to the aggregate expenditure model is a 45-degree line that shows where aggregate expenditures ae equals real gdp.
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• ## Why Is The Consumption Line And 45degree Lines
The question is as below.The answer is a.I understand why it is a, but do not understand why it isnt c.Doesnt the point where the consumption line and 45 degrees line intersect represents the equilibrium income level.
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• ## Why The Degree Line Of Aggregate Supply
Why is aggregate supply curve, a 45 degree line from , the aggregate supply curve is represented by the 45 line throughout this line the planned expenditure is equal to the planned output that is as y expenditure.Unit iv study guide aggregate expenditure, , analyze shifts of the aggregate supply curve , aggregate expenditure line 2 , the.
More | 1,646 | 8,106 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-50 | latest | en | 0.923267 |
https://nl.mathworks.com/matlabcentral/answers/1921535-how-to-run-operation-row-by-row-ex-row-n-divide-row-n-1 | 1,718,415,374,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00438.warc.gz | 390,640,881 | 27,253 | # how to run operation row by row. Ex row(n) divide row(n+1)
3 views (last 30 days)
Noriham B on 2 Mar 2023
Commented: DUY Nguyen on 2 Mar 2023
clc
clear
syms f(x)
f(x) = exp(-((x).^2));
a =0; % lower limit
b = 1; %upper limit
for j=1:6
n = 2.^j;
h = (b-a)/n;
kamir = vpa(int(f,x,a,b));
s = 0.5*(f(a)+f(b));
for i=1:n-1
s = s+f(a+i*h);
end
I =vpa(h*s);
fprintf(' \t %6d \t %f.4 \n ',n,I)
end
Above is my code. after run, we will get 2 column...
for the second column...what I want to do is...I want to compute row(n)/row(n+1)
Hopefully dear all professor/Dr/expert can help me with this problem..thank you in advance
DUY Nguyen on 2 Mar 2023
Hi, in this modified code, I added a prev_I variable to store the previous value of I. After computing I, we compute the ratio as I / prev_I and print it in the third column!
clear
syms f(x)
prev_I=1;
f(x) = exp(-((x).^2));
a =0; % lower limit
b = 1; %upper limit
fprintf('n\t\tIntegral\tRatio\n');
n Integral Ratio
for j= 1:6
n = 2.^j;
h = (b-a)/n;
kamir = vpa(int(f,x,a,b));
s = 0.5*(f(a)+f(b));
for i=1:n-1
s = s+f(a+i*h);
end
I =vpa(h*s);
if i==1
ratio =1;
else
ratio = I / prev_I;
end
fprintf('%d\t\t%.4f\t\t%.4f\n', n, I, ratio)
prev_I = I;
end
2 0.7314 1.0000 4 0.7430 1.0159 8 0.7459 1.0039 16 0.7466 1.0010 32 0.7468 1.0002 64 0.7468 1.0001
##### 3 CommentsShow 1 older commentHide 1 older comment
Noriham B on 2 Mar 2023
its working actually..juz that
should be
else
ratio = prev_I/I ;
the rest is ok...tq so much sir for helping me...
now its working..may God bless u
DUY Nguyen on 2 Mar 2023
Oh, yes! I see.
No problem :)
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Start Hunting! | 680 | 1,816 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2024-26 | latest | en | 0.716908 |
https://www.physicsforums.com/threads/capacitor-very-basic-questions.601126/ | 1,531,809,004,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589573.27/warc/CC-MAIN-20180717051134-20180717071134-00565.warc.gz | 959,767,981 | 15,001 | # Capacitor-very basic questions
1. Apr 28, 2012
### vig
Capacitor---very basic questions
Ive already gone through many posts on working of capacitors...but if i consider a very simple circuit...say i connect a function generator (which generates a sawtooth wave) across a capacitor...(no resistance in series)...and i measure voltage accross the capacitor, what will be the waveform like?..one theory says that since the capacitor is connected in parallel to the function generator, i must get the same sawtooth waveform..however, voltage across capacitor can not change abruptly and so i should be getting a smooth waveform across it..it would be great if someone can clear things out...
2. Apr 28, 2012
### phyzguy
Re: Capacitor---very basic questions
If it is an ideal voltage generator, and an ideal capacitor, the voltage generator will supply the infinite current necessary to get an abrupt change in the voltage across the capacitor. In practice, both the capacitor and the voltage generator have some resistance and some inductance, and the voltage generator is limited in the amount of current it can supply. These non-ideal elements will determine exactly what the waveform looks like.
3. Apr 28, 2012
### vig
Re: Capacitor---very basic questions
'voltage generator will supply the infinite current necessary to get an abrupt change in the voltage across the capacitor'...could u pls elaborate on this?..and considering the most ideal situation, the waveform across a capacitor would show sharp peaks??...and if a resistor is connected in series, and the function generator can still source infinite current (say), the waveform would remain the same?
4. Apr 28, 2012
### Redbelly98
Staff Emeritus
Re: Capacitor---very basic questions
The current into a capacitor is proportional to the rate of change of the voltage across the capacitor. An instantaneously abrupt change in voltage represents an infinitely large rate of change for the voltage, therefore the current would have to be infinite to produce such a voltage change across a capacitor.
Yes, in this idealized and very impossible scenario physguy described, that is exactly what would happen. (Ignoring that pesky detail about it being impossible.)
No, a resistor limits the current that an ideal voltage source would produce.
5. May 2, 2012
### vig
Re: Capacitor---very basic questions
so very crudely speaking...its not a capacitor's fault that it does not allow a sharp change in voltage..it is essentially the inability of the source to provide an infinite current?...
6. May 2, 2012
### phyzguy
Re: Capacitor---very basic questions
Suppose I'm trying to fill a swimming pool. Would you say, "It's not the swimming pool's fault that it doesn't fill up immediately, it's my inability to find a big enough hose!"?
7. May 2, 2012
### Redbelly98
Staff Emeritus
Re: Capacitor---very basic questions
Not quite. A real capacitor has some resistance, as do the connecting wires. And any closed circuit has a nonzero inductance too. So there are several factors at play that limit the current.
If we spent more time thinking about how real devices behave, we might come up with other factors too.
8. May 2, 2012
### vig
Re: Capacitor---very basic questions
Is'nt it true?..if u could get a big hose then the swimming pool would fill up immediately..:uhh: | 756 | 3,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2018-30 | latest | en | 0.866421 |
http://math.stackexchange.com/questions/46478/the-discrepancy-of-certain-sequences | 1,469,495,498,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824499.16/warc/CC-MAIN-20160723071024-00240-ip-10-185-27-174.ec2.internal.warc.gz | 159,902,716 | 16,989 | # The discrepancy of certain sequences
To my knowledge, the best upper bound for the discrepancy of sequences of the type $(n\alpha) (\mod 1), n=1,2,...$ is $$\frac{ND_N(\alpha)}{\log N\log\log N}\to \frac{2}{\pi^2}$$ in measure.
My first question is this: Given a particular double sequence $a_{ij}=i\alpha+j\beta (\mod 1)$, what is the optimal bound on the discrepancy in terms of $\alpha$ and $\beta$? For $1\le i,j\le N$ it is straight forward to show that the sequence of $N^2$ terms has a discrepancy no larger than that of its "best direction", i.e. $$D_{N^2}(a_{ij})\le \min\{D_N(i\alpha),D_N(j\beta)\}.$$ I would very much like to know if this could be improved.
Secondly, I'm interested in the discrepancy of polynomial sequences. Given a polynomial of order $d$, what is the discrepancy of $(p(n)) (\mod 1)$, $n=1,2,...$, depending on $d$ and the coefficients?
I would be very grateful for any enlightenment or references you could give.
- | 277 | 955 | {"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} | 2.921875 | 3 | CC-MAIN-2016-30 | latest | en | 0.872645 |
https://www.stata.com/statalist/archive/2010-04/msg00433.html | 1,563,486,178,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525829.33/warc/CC-MAIN-20190718211312-20190718233312-00308.warc.gz | 851,018,931 | 4,477 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# st: RE: Odds ratio
From "Visintainer, Paul" To "'statalist@hsphsun2.harvard.edu'" Subject st: RE: Odds ratio Date Thu, 8 Apr 2010 16:05:27 -0400
```Rosie,
Many health professionals find the ORs difficult to interpret, especially if the base proportion is common. It doesn't sound like the reviewer is questioning your analysis, just the way you present the results -- and, depending on your study, the ORs may look rather strange. For example, if you did a survey, and 60% of one group answered a question positively, while 90% of another group answered it positively, your OR would be 6.0. This can be quite confusing for someone trying to interpret this OR like a relative risk.
One approach might be to try a program by Joseph Hilbe, called -oddsrisk-. It will convert ORs from a logistic regression to a relative risk with 95%CIs. In the example above, the OR was 6, but the conversion gave 1.5 (e.g., .90/.60 = 1.5). This probably will make more sense to certain readers.
Note that using the term "relative risk" depends on your study. If your logistic model was developed on a survey (e.g., cross-sectional), then your ORs are prevalence ORs. If you convert them using the -oddsrisk- program, you'll have "prevalence ratios", not relative risks.
Best,
-p
________________________________________________
Paul F. Visintainer, PhD
Baystate Medical Center
Division of Academic Affairs - 3rd Floor
Springfield, MA 01199
-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu [mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of Rosie Chen
Sent: Thursday, April 08, 2010 2:48 PM
To: statalist@hsphsun2.harvard.edu
Subject: st: Odds ratio
Hello, dear all,
I have a question regarding a reviewer's comment on my use of odds ratio in interpreting the results of a logistic regression, and would appreciate it very much if you can provide any insight or any references for responding to the comment.
The reviewer commented that all results are expressed in terms of odds ratios which makes it very difficult to assess the magnitude of the effect. Probabilities and changes in probabilities would be much easier to interpret. My impression is that, although it is true that predicted probabilities might be easier to understand, odds ratios have been used extensively in research when we interpret results from logit models.
Do you have any suggestions regarding how to respond to this comment, or do you have any statistics textbooks in your mind that recommend odds ratio as a standard approach reporting results from logistic models?
Thank you very much in advance!
Rosie
*
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* http://www.ats.ucla.edu/stat/stata/
----------------------------------------------------------------------
CONFIDENTIALITY NOTICE: This email communication and any attachments may contain confidential and privileged information for the use of the designated recipients named above. If you are not the intended recipient, you are hereby notified that you have received this communication in error and that any review, disclosure, dissemination, distribution or copying of it or its contents is prohibited. If you have received this communication in error, please reply to the sender immediately or by telephone at (413) 794-0000 and destroy all copies of this communication and any attachments. For further information regarding Baystate Health's privacy policy, please visit our Internet web site at http://www.baystatehealth.com.
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```
• References: | 885 | 3,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-30 | longest | en | 0.885922 |
http://www.averest.org/examples/ModelsOfComputation/Synchrony/WriteConflicts/top.html | 1,632,783,194,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058552.54/warc/CC-MAIN-20210927211955-20210928001955-00149.warc.gz | 71,627,059 | 1,884 | // ************************************************************************** //
// //
// eses eses //
// eses eses //
// eses eseses esesese eses Embedded Systems Group //
// ese ese ese ese ese //
// ese eseseses eseseses ese Department of Computer Science //
// eses eses ese eses //
// eses eseses eseseses eses University of Kaiserslautern //
// eses eses //
// //
// ************************************************************************** //
// This program shows that we can not simply stop the fixpoint computation as //
// soon as all values are known values. In the first iteration of the macro //
// step, y will obtain the value true (due to the must-assignment), and x //
// receives the value false due to the reaction to absence. Hence, all values //
// are known after this first iteration. However, a second iteration reveals //
// that we also have to execute the then branch of the if-statement which //
// results in a write conflict that changes the value of y to TOP. //
// The program also demonstrates that we need not only the additional value //
// BOT, but also TOP to obtain a full lattice, so that also write conflicts //
// are properly modeled. //
// ************************************************************************** //
module top(bool ?x,!y) {
y=true;
if(!x) y=false;
} drivenby drv0 {
x = false;
} drivenby drv1 {
x = true;
}
// ***********************************
// macro step number 0
// ***********************************
// active locations: __start
// ***********************************
// --------------------------------
// iteration 0 of macro step 0
// --------------------------------
// before executing must actions:
// x: BOT
// y: BOT
// before reaction to absence:
// x: BOT
// y: 1
// after reaction to absence:
// x: 0
// y: 1
// --------------------------------
// iteration 1 of macro step 0
// --------------------------------
// before executing must actions:
// x: 0
// y: 1
// before reaction to absence:
// x: 0
// y: TOP
// after reaction to absence:
// x: 0
// y: TOP
// --------------------------------
// iteration 2 of macro step 0
// --------------------------------
// before executing must actions:
// x: 0
// y: TOP
// before reaction to absence:
// x: 0
// y: TOP
// after reaction to absence:
// x: 0
// y: TOP
// --------------------------------
// -> fixpoint found (causality problem)
// -> final environment (selected latest reincarnations)
// x: 0
// y: TOP
// -> no delayed assignments
// --------------------------------
// --------------------------------
// encountered causality problem
// -------------------------------- | 689 | 3,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2021-39 | latest | en | 0.22877 |
http://www.emouseatlas.org/emap/analysis_tools_resources/software/woolz/html_Core/group__AlgFit.html | 1,539,965,996,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512411.13/warc/CC-MAIN-20181019145850-20181019171350-00260.warc.gz | 462,060,401 | 4,055 | Woolz Image Processing Version 1.7.5
AlgFit
## Files
file AlgLinearFit.c
Provides functions for fitting linear models to data, ie linear regression.
file AlgPolyLSQ.c
Provides functions for fitting a polynomial using least squares.
## Functions
AlgError AlgLinearFit1D (int datSz, double *datXA, double *datYA, double *dstA, double *dstB, double *dstSigA, double *dstSigB, double *dstQ)
Computes the least squares best fit straight line (y = a + bx) through the given data, ie linear regression. This function is based on the function fit(): Press W. H., Teukolsky S. A., Vetterling W. T. and Flannery B. P, Numerical Recipies in C, 1992, CUP. More...
AlgError AlgLinearFitIdx1D (double *datXA, double *datYA, int *idxXA, int *idxYA, int idxASz, double *dstA, double *dstB, double *dstSigA, double *dstSigB, double *dstQ)
Computes the least squares best fit straight line (y = a + bx) through the given data, ie linear regression. More...
AlgError AlgPolynomialLSq (double *xVec, double *yVec, int vecSz, int polyDeg, double *cVec)
Attempts to fit a polynomial to the given data using a least squares approach. More...
## Function Documentation
AlgError AlgLinearFit1D ( int datSz, double * datXA, double * datYA, double * dstA, double * dstB, double * dstSigA, double * dstSigB, double * dstQ )
Computes the least squares best fit straight line (y = a + bx) through the given data, ie linear regression. This function is based on the function fit(): Press W. H., Teukolsky S. A., Vetterling W. T. and Flannery B. P, Numerical Recipies in C, 1992, CUP.
Returns
Error code.
Parameters
datSz Number of elements in given data arrays array. datXA Data array with 'x' values. datYA Data array with 'y' values. dstA Destination ptr for intercept 'a', may be NULL. dstB Destination ptr for gradient 'b', may be NULL. dstSigA Destination ptr for std dev of 'a', may be NULL. dstSigB Destination ptr for std dev of 'b', may be NULL. dstQ Destination ptr for goodness of fit, may be NULL.
References ALG_ERR_FUNC, ALG_ERR_NONE, and AlgGammaP().
AlgError AlgLinearFitIdx1D ( double * datXA, double * datYA, int * idxXA, int * idxYA, int idxASz, double * dstA, double * dstB, double * dstSigA, double * dstSigB, double * dstQ )
Computes the least squares best fit straight line (y = a + bx) through the given data, ie linear regression.
Returns
Error code.
Parameters
datXA Data array with 'x' values. datYA Data array with 'y' values. idxXA Index array with indicies into the 'x' data buffer. for the values use examine. idxYA Index array with indicies into the 'y' data buffer. for the values use examine. idxASz Number of elements in each of the given index arrays. dstA Destination ptr for intercept 'a', may be NULL. dstB Destination ptr for gradient 'b', may be NULL. dstSigA Destination ptr for std dev of 'a', may be NULL. dstSigB Destination ptr for std dev of 'b', may be NULL. dstQ Destination ptr for goodness of fit, may be NULL.
References ALG_ERR_FUNC, ALG_ERR_NONE, and AlgGammaP().
AlgError AlgPolynomialLSq ( double * xVec, double * yVec, int vecSz, int polyDeg, double * cVec )
Attempts to fit a polynomial to the given data using a least squares approach.
Returns
Error code.
Parameters
xVec Data vector x of size vecSz. yVec Data vector y of size vecSz. vecSz Size of data vectors. polyDeg Degree of ploynomial. cVec Destination vector for the polynomial coefficients, which must have at least polyDeg + 1 elements. | 953 | 3,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2018-43 | latest | en | 0.489424 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/112/2/i/b/ | 1,653,564,516,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604794.68/warc/CC-MAIN-20220526100301-20220526130301-00002.warc.gz | 1,001,433,664 | 58,090 | # Properties
Label 112.2.i.b Level $112$ Weight $2$ Character orbit 112.i Analytic conductor $0.894$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$112 = 2^{4} \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 112.i (of order $$3$$, degree $$2$$, not minimal)
## Newform invariants
Self dual: no Analytic conductor: $$0.894324502638$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-3})$$ Defining polynomial: $$x^{2} - x + 1$$ x^2 - x + 1 Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 28) Sato-Tate group: $\mathrm{SU}(2)[C_{3}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a primitive root of unity $$\zeta_{6}$$. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - \zeta_{6} + 1) q^{3} - 3 \zeta_{6} q^{5} + ( - 2 \zeta_{6} + 3) q^{7} + 2 \zeta_{6} q^{9} +O(q^{10})$$ q + (-z + 1) * q^3 - 3*z * q^5 + (-2*z + 3) * q^7 + 2*z * q^9 $$q + ( - \zeta_{6} + 1) q^{3} - 3 \zeta_{6} q^{5} + ( - 2 \zeta_{6} + 3) q^{7} + 2 \zeta_{6} q^{9} + (3 \zeta_{6} - 3) q^{11} + 2 q^{13} - 3 q^{15} + (3 \zeta_{6} - 3) q^{17} - \zeta_{6} q^{19} + ( - 3 \zeta_{6} + 1) q^{21} + 3 \zeta_{6} q^{23} + (4 \zeta_{6} - 4) q^{25} + 5 q^{27} - 6 q^{29} + (7 \zeta_{6} - 7) q^{31} + 3 \zeta_{6} q^{33} + ( - 3 \zeta_{6} - 6) q^{35} + \zeta_{6} q^{37} + ( - 2 \zeta_{6} + 2) q^{39} + 6 q^{41} + 4 q^{43} + ( - 6 \zeta_{6} + 6) q^{45} - 9 \zeta_{6} q^{47} + ( - 8 \zeta_{6} + 5) q^{49} + 3 \zeta_{6} q^{51} + (3 \zeta_{6} - 3) q^{53} + 9 q^{55} - q^{57} + ( - 9 \zeta_{6} + 9) q^{59} + \zeta_{6} q^{61} + (2 \zeta_{6} + 4) q^{63} - 6 \zeta_{6} q^{65} + (7 \zeta_{6} - 7) q^{67} + 3 q^{69} + ( - \zeta_{6} + 1) q^{73} + 4 \zeta_{6} q^{75} + (9 \zeta_{6} - 3) q^{77} - 13 \zeta_{6} q^{79} + (\zeta_{6} - 1) q^{81} - 12 q^{83} + 9 q^{85} + (6 \zeta_{6} - 6) q^{87} - 15 \zeta_{6} q^{89} + ( - 4 \zeta_{6} + 6) q^{91} + 7 \zeta_{6} q^{93} + (3 \zeta_{6} - 3) q^{95} - 10 q^{97} - 6 q^{99} +O(q^{100})$$ q + (-z + 1) * q^3 - 3*z * q^5 + (-2*z + 3) * q^7 + 2*z * q^9 + (3*z - 3) * q^11 + 2 * q^13 - 3 * q^15 + (3*z - 3) * q^17 - z * q^19 + (-3*z + 1) * q^21 + 3*z * q^23 + (4*z - 4) * q^25 + 5 * q^27 - 6 * q^29 + (7*z - 7) * q^31 + 3*z * q^33 + (-3*z - 6) * q^35 + z * q^37 + (-2*z + 2) * q^39 + 6 * q^41 + 4 * q^43 + (-6*z + 6) * q^45 - 9*z * q^47 + (-8*z + 5) * q^49 + 3*z * q^51 + (3*z - 3) * q^53 + 9 * q^55 - q^57 + (-9*z + 9) * q^59 + z * q^61 + (2*z + 4) * q^63 - 6*z * q^65 + (7*z - 7) * q^67 + 3 * q^69 + (-z + 1) * q^73 + 4*z * q^75 + (9*z - 3) * q^77 - 13*z * q^79 + (z - 1) * q^81 - 12 * q^83 + 9 * q^85 + (6*z - 6) * q^87 - 15*z * q^89 + (-4*z + 6) * q^91 + 7*z * q^93 + (3*z - 3) * q^95 - 10 * q^97 - 6 * q^99 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + q^{3} - 3 q^{5} + 4 q^{7} + 2 q^{9}+O(q^{10})$$ 2 * q + q^3 - 3 * q^5 + 4 * q^7 + 2 * q^9 $$2 q + q^{3} - 3 q^{5} + 4 q^{7} + 2 q^{9} - 3 q^{11} + 4 q^{13} - 6 q^{15} - 3 q^{17} - q^{19} - q^{21} + 3 q^{23} - 4 q^{25} + 10 q^{27} - 12 q^{29} - 7 q^{31} + 3 q^{33} - 15 q^{35} + q^{37} + 2 q^{39} + 12 q^{41} + 8 q^{43} + 6 q^{45} - 9 q^{47} + 2 q^{49} + 3 q^{51} - 3 q^{53} + 18 q^{55} - 2 q^{57} + 9 q^{59} + q^{61} + 10 q^{63} - 6 q^{65} - 7 q^{67} + 6 q^{69} + q^{73} + 4 q^{75} + 3 q^{77} - 13 q^{79} - q^{81} - 24 q^{83} + 18 q^{85} - 6 q^{87} - 15 q^{89} + 8 q^{91} + 7 q^{93} - 3 q^{95} - 20 q^{97} - 12 q^{99}+O(q^{100})$$ 2 * q + q^3 - 3 * q^5 + 4 * q^7 + 2 * q^9 - 3 * q^11 + 4 * q^13 - 6 * q^15 - 3 * q^17 - q^19 - q^21 + 3 * q^23 - 4 * q^25 + 10 * q^27 - 12 * q^29 - 7 * q^31 + 3 * q^33 - 15 * q^35 + q^37 + 2 * q^39 + 12 * q^41 + 8 * q^43 + 6 * q^45 - 9 * q^47 + 2 * q^49 + 3 * q^51 - 3 * q^53 + 18 * q^55 - 2 * q^57 + 9 * q^59 + q^61 + 10 * q^63 - 6 * q^65 - 7 * q^67 + 6 * q^69 + q^73 + 4 * q^75 + 3 * q^77 - 13 * q^79 - q^81 - 24 * q^83 + 18 * q^85 - 6 * q^87 - 15 * q^89 + 8 * q^91 + 7 * q^93 - 3 * q^95 - 20 * q^97 - 12 * q^99
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/112\mathbb{Z}\right)^\times$$.
$$n$$ $$15$$ $$17$$ $$85$$ $$\chi(n)$$ $$1$$ $$-\zeta_{6}$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
65.1
0.5 − 0.866025i 0.5 + 0.866025i
0 0.500000 + 0.866025i 0 −1.50000 + 2.59808i 0 2.00000 + 1.73205i 0 1.00000 1.73205i 0
81.1 0 0.500000 0.866025i 0 −1.50000 2.59808i 0 2.00000 1.73205i 0 1.00000 + 1.73205i 0
$$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
7.c even 3 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 112.2.i.b 2
3.b odd 2 1 1008.2.s.p 2
4.b odd 2 1 28.2.e.a 2
7.b odd 2 1 784.2.i.d 2
7.c even 3 1 inner 112.2.i.b 2
7.c even 3 1 784.2.a.d 1
7.d odd 6 1 784.2.a.g 1
7.d odd 6 1 784.2.i.d 2
8.b even 2 1 448.2.i.c 2
8.d odd 2 1 448.2.i.e 2
12.b even 2 1 252.2.k.c 2
20.d odd 2 1 700.2.i.c 2
20.e even 4 2 700.2.r.b 4
21.g even 6 1 7056.2.a.bw 1
21.h odd 6 1 1008.2.s.p 2
21.h odd 6 1 7056.2.a.f 1
28.d even 2 1 196.2.e.a 2
28.f even 6 1 196.2.a.a 1
28.f even 6 1 196.2.e.a 2
28.g odd 6 1 28.2.e.a 2
28.g odd 6 1 196.2.a.b 1
36.f odd 6 1 2268.2.i.a 2
36.f odd 6 1 2268.2.l.h 2
36.h even 6 1 2268.2.i.h 2
36.h even 6 1 2268.2.l.a 2
56.j odd 6 1 3136.2.a.k 1
56.k odd 6 1 448.2.i.e 2
56.k odd 6 1 3136.2.a.h 1
56.m even 6 1 3136.2.a.v 1
56.p even 6 1 448.2.i.c 2
56.p even 6 1 3136.2.a.s 1
84.h odd 2 1 1764.2.k.b 2
84.j odd 6 1 1764.2.a.j 1
84.j odd 6 1 1764.2.k.b 2
84.n even 6 1 252.2.k.c 2
84.n even 6 1 1764.2.a.a 1
140.p odd 6 1 700.2.i.c 2
140.p odd 6 1 4900.2.a.g 1
140.s even 6 1 4900.2.a.n 1
140.w even 12 2 700.2.r.b 4
140.w even 12 2 4900.2.e.i 2
140.x odd 12 2 4900.2.e.h 2
252.o even 6 1 2268.2.i.h 2
252.u odd 6 1 2268.2.l.h 2
252.bb even 6 1 2268.2.l.a 2
252.bl odd 6 1 2268.2.i.a 2
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
28.2.e.a 2 4.b odd 2 1
28.2.e.a 2 28.g odd 6 1
112.2.i.b 2 1.a even 1 1 trivial
112.2.i.b 2 7.c even 3 1 inner
196.2.a.a 1 28.f even 6 1
196.2.a.b 1 28.g odd 6 1
196.2.e.a 2 28.d even 2 1
196.2.e.a 2 28.f even 6 1
252.2.k.c 2 12.b even 2 1
252.2.k.c 2 84.n even 6 1
448.2.i.c 2 8.b even 2 1
448.2.i.c 2 56.p even 6 1
448.2.i.e 2 8.d odd 2 1
448.2.i.e 2 56.k odd 6 1
700.2.i.c 2 20.d odd 2 1
700.2.i.c 2 140.p odd 6 1
700.2.r.b 4 20.e even 4 2
700.2.r.b 4 140.w even 12 2
784.2.a.d 1 7.c even 3 1
784.2.a.g 1 7.d odd 6 1
784.2.i.d 2 7.b odd 2 1
784.2.i.d 2 7.d odd 6 1
1008.2.s.p 2 3.b odd 2 1
1008.2.s.p 2 21.h odd 6 1
1764.2.a.a 1 84.n even 6 1
1764.2.a.j 1 84.j odd 6 1
1764.2.k.b 2 84.h odd 2 1
1764.2.k.b 2 84.j odd 6 1
2268.2.i.a 2 36.f odd 6 1
2268.2.i.a 2 252.bl odd 6 1
2268.2.i.h 2 36.h even 6 1
2268.2.i.h 2 252.o even 6 1
2268.2.l.a 2 36.h even 6 1
2268.2.l.a 2 252.bb even 6 1
2268.2.l.h 2 36.f odd 6 1
2268.2.l.h 2 252.u odd 6 1
3136.2.a.h 1 56.k odd 6 1
3136.2.a.k 1 56.j odd 6 1
3136.2.a.s 1 56.p even 6 1
3136.2.a.v 1 56.m even 6 1
4900.2.a.g 1 140.p odd 6 1
4900.2.a.n 1 140.s even 6 1
4900.2.e.h 2 140.x odd 12 2
4900.2.e.i 2 140.w even 12 2
7056.2.a.f 1 21.h odd 6 1
7056.2.a.bw 1 21.g even 6 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{2} - T_{3} + 1$$ acting on $$S_{2}^{\mathrm{new}}(112, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2} - T + 1$$
$5$ $$T^{2} + 3T + 9$$
$7$ $$T^{2} - 4T + 7$$
$11$ $$T^{2} + 3T + 9$$
$13$ $$(T - 2)^{2}$$
$17$ $$T^{2} + 3T + 9$$
$19$ $$T^{2} + T + 1$$
$23$ $$T^{2} - 3T + 9$$
$29$ $$(T + 6)^{2}$$
$31$ $$T^{2} + 7T + 49$$
$37$ $$T^{2} - T + 1$$
$41$ $$(T - 6)^{2}$$
$43$ $$(T - 4)^{2}$$
$47$ $$T^{2} + 9T + 81$$
$53$ $$T^{2} + 3T + 9$$
$59$ $$T^{2} - 9T + 81$$
$61$ $$T^{2} - T + 1$$
$67$ $$T^{2} + 7T + 49$$
$71$ $$T^{2}$$
$73$ $$T^{2} - T + 1$$
$79$ $$T^{2} + 13T + 169$$
$83$ $$(T + 12)^{2}$$
$89$ $$T^{2} + 15T + 225$$
$97$ $$(T + 10)^{2}$$ | 4,654 | 8,329 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.323712 |
https://whatisconvert.com/472-cubic-centimeters-in-pints | 1,591,345,864,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348496026.74/warc/CC-MAIN-20200605080742-20200605110742-00038.warc.gz | 589,190,869 | 7,039 | # What is 472 Cubic Centimeters in Pints?
## Convert 472 Cubic Centimeters to Pints
To calculate 472 Cubic Centimeters to the corresponding value in Pints, multiply the quantity in Cubic Centimeters by 0.0021133764099325 (conversion factor). In this case we should multiply 472 Cubic Centimeters by 0.0021133764099325 to get the equivalent result in Pints:
472 Cubic Centimeters x 0.0021133764099325 = 0.99751366548812 Pints
472 Cubic Centimeters is equivalent to 0.99751366548812 Pints.
## How to convert from Cubic Centimeters to Pints
The conversion factor from Cubic Centimeters to Pints is 0.0021133764099325. To find out how many Cubic Centimeters in Pints, multiply by the conversion factor or use the Volume converter above. Four hundred seventy-two Cubic Centimeters is equivalent to zero point nine nine eight Pints.
## Definition of Cubic Centimeter
A cubic centimeter (SI unit symbol: cm3; non-SI abbreviations: cc and ccm) is a commonly used unit of volume which is derived from SI-unit cubic meter. One cubic centimeter is equal to 1⁄1,000,000 of a cubic meter, or 1⁄1,000 of a liter, or one milliliter; therefore, 1 cm3 ≡ 1 ml.
## Definition of Pint
The pint (symbol: pt) is a unit of volume or capacity in both the imperial and United States customary measurement systems. In the United States, the liquid pint is legally defined as one-eighth of a liquid gallon of precisely 231 cubic inches. One liquid pint is equal to 473.176473 milliliters (≈ 473 ml).
### Using the Cubic Centimeters to Pints converter you can get answers to questions like the following:
• How many Pints are in 472 Cubic Centimeters?
• 472 Cubic Centimeters is equal to how many Pints?
• How to convert 472 Cubic Centimeters to Pints?
• How many is 472 Cubic Centimeters in Pints?
• What is 472 Cubic Centimeters in Pints?
• How much is 472 Cubic Centimeters in Pints?
• How many pt are in 472 cm3?
• 472 cm3 is equal to how many pt?
• How to convert 472 cm3 to pt?
• How many is 472 cm3 in pt?
• What is 472 cm3 in pt?
• How much is 472 cm3 in pt? | 567 | 2,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2020-24 | latest | en | 0.845795 |
https://knowessays.com/bnad-277-ch-6-prep-questions/ | 1,624,301,871,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488289268.76/warc/CC-MAIN-20210621181810-20210621211810-00145.warc.gz | 320,020,173 | 17,726 | # Bnad 277 ch. 6 prep questions
Ch06 – Ch. 6 prep questions
Analytical Methods for Business (University of Arizona)
ch06
Student:
1. A continuous random variable is characterized by uncountable values and can take on any value within an interval.
True False
2. We are often interested in finding the probability that a continuous random variable assumes a particular value.
True False
3. The probability density function of a continuous random variable can be regarded as a counterpart of the probability mass function of a discrete random variable.
True False
4. Cumulative distribution functions can only be used to compute probabilities for continuous random variables.
True False
5. The continuous uniform distribution describes a random variable, defined on the interval [a, b], that has an equally likely chance of assuming values within any subinterval of [a, b] with the same length.
True False
6. The probability density function of a continuous uniform distribution is positive for all values between – and +.
True False
7. The mean of a continuous uniform distribution is simply the average of the upper and lower limits of the interval on which the distribution is defined.
True False
8. The mean and standard deviation of the continuous uniform distribution are equal. True False
9. The probability density function of a normal distribution is in general characterized by being symmetric and bell-shaped.
True False
10. Examples of random variables that closely follow a normal distribution include the age and the class year designation of a college student.
True False
11. Given that the probability distribution is normal, it is completely described by its mean μ > 0 and its standard deviation σ > 0.
True False
12. Just as in the case of the continuous uniform distribution, the probability density function of the normal distribution may be easily used to compute probabilities.
True False
13. The standard normal distribution is a normal distribution with a mean equal to zero and a standard deviation equal to one.
True False
lOMoARcPSD|3013804
14. The letter Z is used to denote a random variable with any normal distribution. True False
15. The standard normal table is also referred to as the z table. True False
16. Which of the following is correct?
A. A continuous random variable has a probability density function but not a cumulative distribution function.
B. A discrete random variable has a probability mass function but not a cumulative distribution function. C.A continuous random variable has a probability mass function, and a discrete random variable has a
probability density function.
D.A continuous random variable has a probability density function, and a discrete random variable has a probability mass function.
17. Which of the following does not represent a continuous random variable?
A. Height of oak trees in a park.
B. Heights and weights of newborn babies.
C. Time of a flight between Chicago and New York.
D. The number of customer arrivals to a bank between 10 am and 11 am.
18. Which of the following is not a characteristic of a probability density function f(x)?
A. f(x) 0 for all values of x.
B. f(x) is symmetric around the mean.
C. The area under f(x) over all values of x equals one.
D. f(x) becomes zero or approaches zero if x increases to +infinity or decreases to -infinity.
19. The cumulative distribution function is denoted and defined as which of the following?
A. f(x) and f(x) = P(X x)
B. f(x) and f(x) = P(X x )
C. F(x) and F(x) = P(X x)
D. F(x) and F(x) = P(X x)
20. The cumulative distribution function F(x) of a continuous random variable X with the probability density function f(x) is which of the following?
A. The area under f over all values x
B. The area under f over all values that are x or less
C. The area under f over all values that are x or more
D. The area under f over all non-negative values that are x or less
21. A continuous random variable has the uniform distribution on the interval [a, b] if its probability density
function f(x) .
A. Is symmetric around its mean
B. Is bell-shaped between a and b
C. Is constant for all x between a and b, and 0 otherwise
D. Asymptotically approaches the x axis when x increases to + or decreases to –
22. The height of the probability density function f(x) of the uniform distribution defined on the interval [a, b]
is .
A. 1/(ba) between a and b, and zero otherwise
B. (ba)/2 between a and b, and zero otherwise
C. (a + b)/2 between a and b, and zero otherwise
D. 1/(a + b) between a and b, and zero otherwise
lOMoARcPSD|3013804
23. The waiting time at an elevator is uniformly distributed between 30 and 200 seconds. Find the mean and standard deviation of the waiting time.
A. 115 seconds and 49.07 seconds
B. 1.15 minutes and 0.4907 minutes
C. 1.15 minutes and 24.08333 (minute)2
D. 115 seconds and 2408.3333 (second)2
24. The waiting time at an elevator is uniformly distributed between 30 and 200 seconds. What is the probability a rider must wait between 1 minute and 1.5 minutes?
A. 0.1765
B. 0.3529
C. 0.5294
D. 0.8824
25. The waiting time at an elevator is uniformly distributed between 30 and 200 seconds. What is the probability a rider must wait more than 1.5 minutes?
A. 0.3529
B. 0.4500
C. 0.5294
D. 0.6471
26. The waiting time at an elevator is uniformly distributed between 30 and 200 seconds. What is the probability a rider waits less than two minutes?
A. 0.4706
B. 0.5294
C. 0.6000
D. 0.7059
27. The time of a call to a technical support line is uniformly distributed between 2 and 10 minutes. What are the mean and variance of this distribution?
A. 6 minutes and 2.3094 (minutes)2
B. 6 minutes and 5.3333 (minutes)2
C. 6 minutes and 5.3333 minutes
D. 8 minutes and 2.3094 minutes
28. An analyst is forecasting net income for Excellence Corporation for the next fiscal year. Her low-end estimate of net income is \$250,000, and her high-end estimate is \$350,000. Prior research allows her to
assume that net income follows a continuous uniform distribution. The probability that net income will be
greater than or equal to \$337,500 is .
A. 12.5%
B. 29.6%
C. 87.5%
D. 96.4%
29. Alex is in a hurry to get to work and is rushing to catch the bus. She knows that the bus arrives every six minutes during rush hour, but does not know the exact times the bus is due. She realizes that from the time she arrives at the stop, the amount of time that she will have to wait follows a uniform distribution with a lower bound of 0 minutes and an upper bound of six minutes. What is the probability that she will have to wait more than two minutes?
A. 0.1667
B. 0.3333
C. 0.6667
D. 1.0000
lOMoARcPSD|3013804
30. Suppose the average price of gasoline for a city in the United States follows a continuous uniform distribution with a lower bound of \$3.50 per gallon and an upper bound of \$3.80 per gallon. What is the probability a randomly chosen gas station charges more than \$3.70 per gallon?
A. 0.3000
B. 0.3333
C. 0.6667
D. 1.0000
31. How many parameters are needed to fully describe any normal distribution?
A. 1
B. 2
C. 3
D. 4
32. What does it mean when we say that the tails of the normal curve are asymptotic to the x axis?
A. The tails get closer and closer to the x axis but never touch it.
B. The tails gets closer and closer to the x axis and eventually touch it.
C. The tails get closer and closer to the x axis and eventually cross this axis.
D. The tails get closer and closer to the x axis and eventually become this axis.
33. The probability that a normal random variable is less than its mean is .
A. 0.0
B. 0.5
C. 1.0
D. Cannot be determined
34. Let X be normally distributed with mean μ and standard deviation σ > 0. Which of the following is true about the z value corresponding to a given x value?
A. A positive z = (xμ)/σ indicates how many standard deviations x is above μ.
B. A negative z = (xμ)/σ indicates how many standard deviations x is below μ.
C. The z value corresponding to x = μ is zero.
D. All of the above.
35. It is known that the length of a certain product X is normally distributed with μ = 20 inches. How is the
probability related to ?
A. is greater than .
B. is smaller than .
C. is the same as .
D. No comparison can be made with the given information.
36. It is known that the length of a certain product X is normally distributed with μ = 20 inches. How is the
probability related to ?
A. is greater than .
B. is smaller than .
C. is the same as .
D. No comparison can be made with the given information.
37. It is known that the length of a certain product X is normally distributed with μ = 20 inches. How is the
probability related to ?
A. is greater than .
B. is smaller than .
C. is the same as .
D. No comparison can be made with the given information.
lOMoARcPSD|3013804
38. It is known that the length of a certain product X is normally distributed with μ = 20 inches and σ = 4
inches. How is the probability related to ?
A. is greater than .
B. is smaller than .
C. is the same as .
D. No comparison can be made with the given information.
39. The probability P(Z < -1.28) is closest to .
A. -0.10
B. 0.10
C. 0.20
D. 0.90
40. The probability P(Z > 1.28) is closest to .
A. -0.10
B. 0.10
C. 0.20
D. 0.90
41. Find the probability P(-1.96 Z 0).
A. 0.0250
B. 0.0500
C. 0.4750
D. 0.5250
42. Find the probability P(-1.96 Z 1.96).
A. 0.0500
B. 0.9500
C. 0.9750
D. 1.9500
43. Find the z value such that .
A. z = -1.33
B. z = 0.1814
C. z = 0.8186
D. z = 1.33
44. Find the z value such that .
A. z = -1.645
B. z = -1.96
C. z = 1.645
D. z = 1.96
45. You work in marketing for a company that produces work boots. Quality control has sent you a memo detailing the length of time before the boots wear out under heavy use. They find that the boots wear out in an average of 208 days, but the exact amount of time varies, following a normal distribution with a standard deviation of 14 days. For an upcoming ad campaign, you need to know the percent of the pairs that last longer than six monthsthat is, 180 days. Use the empirical rule to approximate this percent.
A. 2.5%
B. 5%
C. 95%
D. 97.5%
lOMoARcPSD|3013804
46. A hedge fund returns on average 26% per year with a standard deviation of 12%. Using the empirical rule, approximate the probability the fund returns over 50% next year.
A. 0.5%
B. 1%
C. 2.5%
D. 5%
47. For any normally distributed random variable with mean μ and standard deviation σ, the percent of the
observations that fall between and is closest to .
A. 68%
B. 68.26%
C. 95%
D. 95.44%
48. For any normally distributed random variable with mean μ and standard deviation σ, the proportion of the
observations that fall outside the interval [μσ, μ + σ] is closest to .
A. 0.0466
B. 0.3174
C. 0.8413
D. 0.1687
49. Sarah’s portfolio has an expected annual return at 8%, with an annual standard deviation at 12%.
If her investment returns are normally distributed, then in any given year Sarah has approximately
.
A. A 50% chance that the actual return will be greater than 8%
B. About a 68% chance that the actual return will fall within 4% and 20%
C. About a 68% chance that the actual return will fall within -20% and 20%
D. About a 95% chance that the actual return will fall within -4% and 28%.
50. If X has a normal distribution with and , then the probability can be
expressed in terms of a standard normal variable Z as .
A.
B.
C.
D.
51. The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take more than five hours to construct a soapbox derby car.
A. 0
B. 0.0228
C. 0.4772
D. 0.9772
52. The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take between 2.5 and 3.5 hours to construct a soapbox derby car.
A. 0.3085
B. 0.3830
C. 0.6170
D. 0.6915
lOMoARcPSD|3013804
53. The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take exactly 3.7 hours to construct a soapbox derby car.
A. 0.0000
B. 0.5000
C. 0.7580
D. 0.2420
54. Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x -such that P(X x) = 0.0606.
A. -1.55
B. 1.55
C. 126
D. 374
55. Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X x) = 0.9394.
A. -1.55
B. 1.55
C. 126
D. 374
56. Let X be normally distributed with mean µ = 25 and standard deviation σ = 5. Find the value x such that P(X x) = 0.1736.
A. -0.94
B. 0.94
C. 20.30
D. 29.70
57. The salary of teachers in a particular school district is normally distributed with a mean of \$50,000 and a standard deviation of \$2,500. Due to budget limitations, it has been decided that the teachers who are in the top 2.5% of the salaries would not get a raise. What is the salary level that divides the teachers into one group that gets a raise and one that doesn’t?
A. -1.96
B. 1.96
C. 45,100
D. 54,900
58. The starting salary of an administrative assistant is normally distributed with a mean of \$50,000 and a standard deviation of \$2,500. We know that the probability of a randomly selected administrative assistant making a salary between μx and μ + x is 0.7416. Find the salary range referred to in this statement.
A. \$42,825 to \$52,825
B. \$42,825 to \$57,175
C. \$47,175 to \$52,825
D. \$47,175 to \$57,175
59. An investment consultant tells her client that the probability of making a positive return with her suggested portfolio is 0.90. What is the risk, measured by standard deviation, that this investment manager has assumed in his calculation if it is known that returns from her suggested portfolio are normally distributed with a mean of 6%?
A. 1.28%
B. 4.69%
C. 6.00%
D. 10.0%
lOMoARcPSD|3013804
60. The stock price of a particular asset has a mean and standard deviation of \$58.50 and \$8.25, respectively. Use the normal distribution to compute the 95th percentile of this stock price.
A. -1.645
B. 1.645
C. 44.93
D. 72.07
61. EXHIBIT 6-1. You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F.
Refer to Exhibit 6-1. One sleeping bag you are considering advertises that it is good for temperatures down to 25°F. What is the probability that this bag will be warm enough on a randomly selected May night at the park?
A. 0.1894
B. 0.3106
C. 0.8106
D. 0.8800
62. EXHIBIT 6-1. You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F.
Refer to Exhibit 6-1. An inexpensive bag you are considering advertises to be good for temperatures down to 38°F. What is the probability that the bag will not be warm enough?
A. 0.2266
B. 0.2734
C. 0.7500
D. 0.7734
63. EXHIBIT 6-1. You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F.
Refer to Exhibit 6-1. Above what temperature must the sleeping bag be suited such that the temperature will be too cold only 5% of the time?
A. -1.645
B. 1.645
C. 18.84
D. 45.16
64. EXHIBIT 6-2. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of 3 ounces. Assume the amount of gold found per 1000 tons of dirt is normally distributed.
Refer to Exhibit 6-2. What is the probability the miners find more than 16 ounces of gold in the next 1000 tons of dirt excavated?
A. 0.0918
B. 0.4082
C. 0.5918
D. 0.9082
lOMoARcPSD|3013804
65. EXHIBIT 6-2. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of 3 ounces. Assume the amount of gold found per 1000 tons of dirt is normally distributed.
Refer to Exhibit 6-2. What is the probability the miners find between 10 and 14 ounces of gold in the next 1000 tons of dirt excavated?
A. 0.2514
B. 0.4972
C. 0.5028
D. 0.7486
66. EXHIBIT 6-2. Gold miners in Alaska have found, on average, 12 ounces of gold per 1000 tons of dirt excavated with a standard deviation of 3 ounces. Assume the amount of gold found per 1000 tons of dirt is normally distributed.
Refer to Exhibit 6-2. If the miners excavated 1000 tons of dirt, how little gold must they have found such that they find that amount or less only 15% of the time?
A. -1.04
B. 1.04
C. 8.88
D. 15.12
67. Suppose the life of a particular brand of laptop battery is normally distributed with a mean of 8 hours and a standard deviation of 0.6 hours. What is the probability that the battery will last more than 9 hours before running out of power?
A. 0.0475
B. 0.4525
C. 0.9525
D. 1.6667
68. A superstar major league baseball player just signed a new deal that pays him a record amount of money. The star has driven in an average of 110 runs over the course of his career, with a standard deviation
of 31 runs. An average player at his position drives in 80 runs. What is the probability the superstar bats in fewer runs than an average player next year? Assume the number of runs batted in is normally distributed.
A. 0.1660
B. 0.3340
C. 0. 8340
D. 0.9700
69. If an exponential distribution has the rate parameter λ = 5, what is its expected value?
A. 5
B. 1/5
C. 1/25
D. 5/2
70. If an exponential distribution has the rate parameter λ = 5, what is its variance?
A. 5
B. 1/5
C. 1/25
D. 5/2
71. What can be said about the expected value and standard deviation of an exponential distribution?
A. The expected value is equal to the standard deviation.
B. The expected value is equal to the square of the standard deviation.
C. The expected value is equal to the reciprocal of the standard deviation.
D. The expected value is equal to the square root of the standard deviation.
lOMoARcPSD|3013804
72. Let the time between two consecutive arrivals at a grocery store check-out line be exponentially distributed with a mean of three minutes. Find the probability that the next arrival does not occur until at least four minutes have passed since the last arrival.
A. 0.0000
B. 0.2636
C. 0.4724
D. 0.7364
73. EXHIBIT 6-3. Patients scheduled to see their primary care physician at a particular hospital wait, on average, an additional eight minutes after their appointment is scheduled to start. Assume the time that patients wait is exponentially distributed.
Refer to Exhibit 6-3. What is the probability a randomly selected patient will have to wait more than 10 minutes?
A. 0.2865
B. 0.4493
C. 0.5507
D. 0.7135
74. EXHIBIT 6-3. Patients scheduled to see their primary care physician at a particular hospital wait, on average, an additional eight minutes after their appointment is scheduled to start. Assume the time that patients wait is exponentially distributed.
Refer to Exhibit 6-3. What is the probability a randomly selected patient will see the doctor within five minutes of the scheduled time?
A. 0.2019
B. 0.4647
C. 0.5353
D. 0.7981
75. EXHIBIT 6-4. The average time between trades for a high-frequency trading investment firm is 40 seconds. Assume the time between trades is exponentially distributed.
Refer to Exhibit 6-4. What is the probability that the time between trades for a randomly selected trade and the one proceeding it is less than 20 seconds?
A. 0.1354
B. 0.3935
C. 0.6065
D. 0.8446
76. EXHIBIT 6-4. The average time between trades for a high-frequency trading investment firm is 40 seconds. Assume the time between trades is exponentially distributed.
Refer to Exhibit 6-4. What is the probability that the time between trades for a randomly selected trade and the one proceeding it is more than a minute?
A. 0.2231
B. 0.4869
C. 0.5134
D. 0.7769
77. If has a lognormal distribution, what can be said of the distribution of the random variable X?
A. X follows a normal distribution.
B. X follows an exponential distribution.
C. X follows a standard normal distribution.
D. X follows a continuous uniform distribution.
lOMoARcPSD|3013804
78. Find the mean of the lognormal variable if the mean and standard deviation of the underlying normal variable are 2 and 0.8, respectively.
A. 0.69
B. 2.32
C. 10.18
D. 11.02
79. Find the variance of the lognormal variable if the mean and variance of the underlying normal variable are 2 and 1, respectively.
A. 0
B. 12.18
C. 15.97
D. 255.02
80. EXHIBIT 6-5. The mean travel time to work is 25.2 minutes (U.S. Census 2010). Further, suppose that commute time follows a log-normal distribution with a standard deviation of 10 minutes.
Refer to Exhibit 6-5. What is the probability a randomly selected U.S. worker has a commute time of more than half an hour?
A. 25.78%
B. 31.56%
C. 68.44%
D. 74.22%
81. EXHIBIT 6-5. The mean travel time to work is 25.2 minutes (U.S. Census 2010). Further, suppose that commute time follows a log-normal distribution with a standard deviation of 10 minutes.
Refer to Exhibit 6-5. What is the probability a randomly selected U.S. worker has a commute time of less than 20 minutes?
A. 30.15%
B. 34.09%
C. 65.91%
D. 69.85%
82. EXHIBIT 6-6. Let the lifetime of a new Jet Ski be represented by a lognormal variable, where X is normally distributed. Let the mean of the lifetime of the Jet Ski be six years with a standard deviation of three years.
Refer to Exhibit 6-6. What proportion of the Jet Skis will last less than seven years?
A. 0.2877
B. 0.3707
C. 0.6293
D. 0.7123
83. EXHIBIT 6-6. Let the lifetime of a new Jet Ski be represented by a lognormal variable, where X is normally distributed. Let the mean of the lifetime of the Jet Ski be six years with a standard deviation of three years.
Refer to Exhibit 6-6. What proportion of the Jet Skis will last nine or more years?
A. 0.1379
B. 0.1587
C. 0.8413
D. 0.8621
lOMoARcPSD|3013804
84. When attending a movie, patrons are interested in avoiding the pre-movie trivia games, ads, and previews. It is known that the previews begin at the scheduled movie start time and they last between 5 and 15 minutes. Assume that the time of the previews is uniformly distributed.
a. Find the expected time and variance of the movie preview duration.
b. What is the probability that on a given day the previews last between 10 and 12 minutes?
85. Snack food companies always have a target weight when filling a box of snack crackers. It is not possible, however, to always fill boxes to the exact target weight. For a particular box of crackers, the target weight is 14 ounces. The filling machine drops between 13.5 and 15 ounces of crackers into each box. If these weights are uniformly distributed, what is the expected value of the fill weights? Does this value match the target weight? If not, what impact does this difference have on the manufacturer’s costs to produce these crackers?
86. Snack food companies always have a target weight when filling a box of snack crackers. It is not possible, however, to always fill boxes to the exact target weight. For a particular box of crackers, the target weight is 14 ounces. The filling machine drops between 13.5 and 15 ounces of crackers into each box. Let these weights be uniformly distributed.
a. What is the probability that the weight in a box will be less than 14 ounces?
b. What is the probability that the weight in a box will be between 14.5 and 15.5 ounces?
87. The time you must wait for an Orange Line train of the Massachusetts Bay Transit Authority follows a uniform distribution with a lower bound of 0 minutes and an upper bound of 8 minutes. Jonathan is running to catch the train in order to get to a meeting. He knows that the train needs to arrive within five minutes or else he will be late. What is the probability that he will be late to his meeting?
lOMoARcPSD|3013804
88. Suppose Jennifer is waiting for a taxi cab. A taxi cab’s arrival time is equally likely at any constant time range in the next 12 minutes.
a. Calculate the expected arrival time.
b. What is the probability that a taxi arrives in three minutes or less?
89. Find the following probabilities for a standard normal random variable Z.
a. b.
c. d.
90. Find the following probabilities for a standard normal random variable Z.
a. b.
c. d.
91. Find the value of z for which the standard normal random variable Z satisfies the following:
a. b.
c. d.
lOMoARcPSD|3013804
92. Given normally distributed random variable X with a mean of 10 and a variance of 4, find the following probabilities.
a.
b.
c.
d.
93. Given normally distributed random variable X with a mean of 12 and a standard deviation of 3.4, find the following probabilities.
a. b. c. d.
94. A normal random variable X has a mean of 17 and a variance of 5.
a. Find the value x for which P(X x) = 0.0020.
b. Find the value of x for which P(X > x) = 0.0122.
95. A soft drink company fills two-liter bottles on several different lines of production equipment. The fill volumes are normally distributed with a mean of 1.97 liters and a variance of 0.04 (liter)2.
a. Find the probability that a randomly selected two-liter bottle would contain between 1.95 and 2.03 liters.
b. If X is the fill volume of a randomly selected two-liter bottle, find the value of x for which P(X > x) = 0.3228.
lOMoARcPSD|3013804
96. A soft drink company fills two-liter bottles on several different lines of production equipment. The fill volumes are normally distributed with a mean of 1.97 liters and a variance of 0.04 (liter)2.
a. Find the probability that a randomly selected two-liter bottle would contain more than 1.92 liters.
b. If X is the fill volume of a randomly selected two-liter bottle, find the value of x for which P(X < x) = 0.6293.
97. You are considering the risk-return profile of two mutual funds for investment. The relatively risky fund promises an expected return of 9%, with a standard deviation of 12%. The relatively less risky fund promises an expected return and standard deviation of 5% and 8%, respectively.
a. Which mutual fund will you pick if your objective is to minimize the probability of earning a negative return?
b. Which mutual fund will you pick if your objective is to maximize the probability of earning a return between 8% and 12%?
98. The annual return of a well-known mutual fund has historically had a mean of about 10% and a standard deviation of 21%. Suppose the return for the following year follows a normal distribution, with the historical mean and standard deviation. What is the probability that you will lose money in the next year by investing in this fund?
99. The East Los Angeles Interchange is the busiest freeway interchange in the world. In 2008, an average of 550,000 cars passed through the intersection per day with a standard deviation of 100,000. What is the probability more than 620,000 use the interchange on a random day? Assume the number of cars on the interchange is approximately normally distributed.
lOMoARcPSD|3013804
100.The weight of competition pumpkins at the Circleville Pumpkin Show in Circleville, Ohio, can be represented by a normal distribution with a mean of 703 pounds and a standard deviation of 347 pounds.
a. Find the probability that a randomly selected pumpkin weighs at least 1622 pounds.
b. Find the probability that a randomly selected pumpkin weighs between 465.1 and 1622 pounds.
101.The average annual percentage rate (APR) for credit cards held by U.S. consumers is approximately 15 percent (“Ouch – Credit Card APR Now Tops 15 Percent,” Time, January 3, 2012). Suppose the APR for new credit card offers is normally distributed with a mean of 15% and a standard deviation of 4%. What APR must a credit card charge to be in the bottom 10% of all cards?
102.The average annual inflation rate in the United States over the past 98 years is 3.37% and has a standard deviation of approximately 5% (Inflationdata.com). In 1980, the inflation rate was above 13%. If the annual inflation rate is normally distributed, what is the probability that inflation will be above 13% next year?
103.The Japan Sumo Association has begun to measure the body fat of wrestlers to try to combat the growing
problem of excessive obesity within the sport. As of 2010, the average wrestler weighed 412 pounds. Suppose the weights of sumo wrestlers are normally distributed, with a standard deviation of 37 pounds. What is the probability that a randomly selected wrestler weighs between 350 and 450 pounds?
lOMoARcPSD|3013804
104.A producer has a history of making bad movies. The movies he has produced have averaged \$160,000 dollars at the box office with a standard deviation of \$185,000. He thinks his latest movie will be a huge hit. How much will the movie earn at the box office if it is only expected to earn that much or more 0.5% of the time? (Assume that the profit at the box office is normally distributed.)
105.Suppose the amount of time customers must wait to check bags at the ticketing counter in Boston Logan Airport is exponentially distributed with a mean of 14 minutes. What is the probability that a randomly selected customer will have to wait more than 20 minutes?
106.The average wait time to see a doctor at a maternity ward is 16 minutes. What is the probability that a patient will have to wait between 20 and 30 minutes before seeing a doctor? Suppose the wait time is exponentially distributed.
107.The average wait time at a McDonald’s drive-through window is about three minutes (“The Doctor Will See You Eventually,” The Wall Street Journal, October 18, 2010). Suppose the wait time is exponentially distributed. What is the probability that a randomly selected customer will have to wait no more than five minutes?
108.Jennifer is waiting for a taxi cab. The average wait time for a taxi is six minutes. Suppose the wait time is exponentially distributed. What is the probability that a taxi arrives in three minutes or less?
lOMoARcPSD|3013804
109.Customers arrive at a drive-through teller window of a bank. They stay in line when the teller is busy.
The service time is exponentially distributed with a mean of four minutes.
a. What is the probability that the next customer in line will take longer than seven minutes to be served? b. What is the probability that the next customer in line will take less than eight minutes to be served? c. What is the probability that the next customer in line will take between three and six minutes to be served?
110.Compute the mean and variance of a lognormal variable Y if the mean and the variance of the underlying
normal variable are .
111.After a heavy snow, the city of Boston spends millions of dollars plowing the streets. Suppose the amount of time the city must spend before the streets are clear follows a log-normal distribution. Further suppose that the average amount of time is 12 hours and the standard deviation is 5 hours. What percentage of the time does it take more than 10 hours for the streets to be cleared?
112.Let the household income of residents of the United States be represented by Y = eX, where X is normally distributed. In 2006, the mean U.S. household income was approximately \$50,000. Suppose the standard deviation was 16,000. Estimate the proportion of U.S. households that have an income less than 80,000.
113.Compute the mean and variance of a lognormal variable Y if the mean and the variance of the underlying
normal variable are .
lOMoARcPSD|3013804
114.The mean household income of France is approximately 20,000 euros. Suppose the household income in France has a standard deviation of 10,000 euros and follows a log-normal distribution. Estimate the proportion of French households that have an income of more than 25,000 euros.
115.Let house prices in a rich community in Chicago be represented by , where X is normally distributed. Suppose the mean house price is \$1.8 million and the standard deviation is \$0.4 million. What is the proportion of the houses that are worth more than \$2.5 million?
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https://nl.mathworks.com/examples/simulink/mw/simulink_general-ex18238601-approximating-nonlinear-relationships-type-s-thermocouple | 1,553,573,224,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912204790.78/warc/CC-MAIN-20190326034712-20190326060712-00270.warc.gz | 555,905,733 | 18,168 | MATLAB Examples
# Approximating Nonlinear Relationships: Type S Thermocouple
This example shows how to approximate nonlinear relationships of a type S thermocouple.
## Thermocouple Modeling and Signal Conversion
The thermocouple is one of the popular analog transducers today, along with other devices such as position sensors, strain gages, pressure transducers, and resistance temperature devices (RTDs). Operating under the principle of the Seebeck effect (a.k.a. thermoelectric effect), thermocouples have an empirically determined nonlinear behavior that is well known over each junction type's useful operating range. If you run the model, you will be able to see the cumulative effect of each component on dynamic measurement accuracy. This example will focus on models for each of these components in a dynamic temperature measurement system for a Type S (Platinum-10% Rhodium(+) versus Platinum(-) ) : a thermocouple device and probe assembly, a signal conditioning method, an analog to digital converter (ADC), and a software specification for converting the ADC output into a temperature value. An additional section shows how to obtain and use standard NIST ITS-90 thermocouple data with Simulink® models. Look-up tables and a polynomial block are used in this design to capture the nonlinear behavior of the thermocouple. Note that the polynomial block is a viable alternative to look-up tables for some applications, minimizing ROM usage at the cost of some additional computation for the full polynomial representation.
Open the model
```open_system('sldemo_tc') ```
Figure 1: Temperature measurement system: a chain of components from physical phenomenon to software values
## Simulating the Thermocouple Signal
The two main features of the thermocouple model are the probe and bead dynamics and the thermocouple's conversion of temperature into a millivolt signal. The probe+bead dynamics are modeled as a 30 msec first order system, and the nonlinear thermocouple behavior is modeled using the segment 1 polynomial data from NIST Standard Database 60 for a Type S thermocouple from -50 to 1063 degC. For numerical stability, the coefficients were scaled to return microvolts from the polynomial block. The output of the 1 Type S Thermocouple model subsystem is then converted to volts with a Unit Conversion block. Note that units are specified on the subsystem input and output ports and displayed on the subsystem icons. To learn more about units in Simulink see Simulink Units.
An alternative implementation to using the polynomial is an interpolated look-up table. Table data could be used in a look-up table block in place of the polynomial block. Sample data was constructed from NIST Standard Database 60 for a Type S thermocouple in file sldemo_create_tc_tabledata.m Access to this database is described below in the section titled "Thermocouple Reference Data Download and Import Procedure".
## Anti-Aliasing Filter and Analog to Digital Converter (ADC) Models
The ADC in this model expects a 0 to 5 volt signal, so the raw thermocouple sense voltage is biased and amplified corresponding to a range of -0.235 mV to 18.661 mV (-50 degC to 1765 degC). A third order Butterworth anti-aliasing filter was designed for Wn = 15 Hz using the Signal Processing Toolbox™:
`[num,den] = butter(3, 15*2*pi, 'low', 's')`
The output of the anti-aliasing filter feeds a sample-and-hold device that drives the quantization model. Since the sample period is 20 msec in this example, the conversion time is ignored as it is typically 2 orders of magnitude smaller for devices currently available. (Note: if the conversion time were an appreciable fraction of the sample period, it could not be ignored as it would significantly affect the system's dynamics.)
The quantization algorithm in the model takes in a 0 to 5 volt signal and outputs a 12-bit digital word in a 16-bit signed integer. A value of 0 corresponds to 0 Volts and a value of 4096 would correspond to 5 Volts. A change in the least significant bit (LSB) is about 1.2 mV. As 12 bits can only reach the value of 4095, the highest voltage that can be read by this device is approximately 4.9988 Volts. In order to have no more than 1/2 LSB error within the operating range, the quantizer changes values midway between each voltage point, resulting in a 1/2-width interval at 0 Volts and a 3/2-width interval just below 5 Volts. The last interval has 1 full LSB due to its 3/2-width size.
## Understanding Data Converters
The sldemo_adc_quantize model allows you to explore the A/D converter component in more detail:
```open_system('sldemo_adc_quantize') ```
Figure 2: Details of ADC quantization modeling (zero conversion time)
```sim('sldemo_adc_quantize') set(gcf,'Color',[1,1,1]); ```
Figure 3: Quantization characteristic of ADC.
```ax = get(gcf,'Children'); set(ax(1), 'xlim', [4085, 4100]); ```
Figure 4: Quantization characteristic of ADC: zoomed in to top of range to reveal 1 LSB error behavior at high end (rest of range only has 1/2 LSB max error).
## Software Specification for Converting ADC Output to Temperature Values
The input conversion subsystem requires a 16 bit unsigned integer input from the ADC whose full scale range is 0 to 4095 counts, corresponding to -0.235 mV and 18.6564 mV thermocouple loop voltage. The best accuracy and fastest algorithm for input conversion is a direct look-up table. Since the input is an integer from 0 to 4095, a table can be constructed that gives the thermocouple temperature corresponding to each possible input value, so the conversion process can be reduced to indexing into a table. This however, requires one number per ADC output value and for a 12-bit ADC, this can be a burden in memory constrained environments. For double precision data, this is a 16 kB ROM requirement. See file sldemo_create_tc_tabledata.m for the method used to construct the direct look-up table from the Type S thermocouple reference data. The error associated with this approach is entirely isolated to the table construction process as there is an output value associated with every possible input value - the run-time look-up process introduces no additional error.
An interpolated table was also put into the model, using only 664 bytes. This is a big reduction in ROM required compared to the direct table look-up, but it takes a bit longer to compute than an indirect memory access and introduces error into the measurement, which goes down as the number of points in the table increases.
```open_system(sprintf('sldemo_tc/3 Software specification\nfor converting\nADC values to temperature')) ```
Using the NIST ITS-90 Thermocouple Database (Standard Reference Database 60 from NIST Monograph 175), you can access the standard reference data describing the behavior for the eight standard thermocouple types. This database relates thermocouple output to the International Temperature Scale of 1990. Follow these steps to acquire and read in the data needed to fully exercise the support files included with this example:
2. cd to the directory where you downloaded the all.tab thermocouple database
3. Parse the database and convert it to a MATLAB structure array using the conversion tool readstdtcdata.m:
```tcdata = readstdtcdata('all.tab');
save thermocouple_its90.mat tcdata;```
(tip: highlight the above MATLAB code and use right mouse menuitem "Evaluate Selection" to execute it)
You now have a complete set of temperature (T, degC) vs. voltage (E, mV) data, approximating polynomial coefficients, and inverse polynomial coefficients for the standard thermocouple types B, E, J, K, N, R, S, and T in the tcdata variable of file thermocouple_its90.mat. The MATLAB script in sldemo_create_tc_tabledata.m uses this data to prepare lookup table block parameters used in the example model.
## Conditioning Reference Data for Use in a Look-Up Table
If you review the thermocouple data tables in the tcdata structure or in all.tab, you will probably notice a few things:
• Repeated 0 degC temperature points in the data
• Repeated voltage points in the data due to the 3-digit output format
• Non-monotonic behavior with two temperatures having the same output, e.g., for the Type B thermocouple, T(E) won't work but E(T) does work
The readstdtcdata() routine will remove the repeated 0 degC temperature points from the data, but the repeated voltages due to the data format and non-monotonic behavior of some of the curve ends must be dealt with on an individual basis. A reference model named sldemo_tc_blocks was constructed using sldemo_tc_blocks_data.mat. It contains look-up tables with data populated from the interpolating polynomials for the eight standard thermocouples in all.tab:
```open_system('sldemo_tc_blocks') ```
Figure 5: Full-Range 2 degrees Celsius Interpolated Thermocouple Tables created from ITS-90 Interpolating Polynomials
## References
2. "Temperature-Electromotive Force Reference Functions and Tables for the Letter-Designated Thermocouple Types Based on the ITS-90". National Institute of Standards and Technology Monograph 175; 1993. 630 p.
3. The International Temperature Scale of 1990 (ITS-90), Consultative Committee for Thermometry (CCT) of the International Committee for Weights and Measures (CIPM)
4. Thermoelectric Effects in Metals: Thermocouples, S. O. Kasap 1997
NOTE: for determining empirical relationships of complex systems such as engines and for fitting models to measured data, MathWorks® offers the Model Based Calibration Toolbox which employs the Design of Experiments methodology for optimized table database creation, plus value extraction and automated table filling components. | 2,150 | 9,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2019-13 | latest | en | 0.865383 |
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# Graphs May Math Center
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https://trac.sagemath.org/ticket/17447?cversion=0&cnum_hist=7 | 1,618,154,930,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064520.8/warc/CC-MAIN-20210411144457-20210411174457-00068.warc.gz | 693,910,479 | 15,049 | Opened 6 years ago
Closed 5 years ago
# Clarify and complete documentation of function()
Reported by: Owned by: schymans major sage-6.10 documentation nbruin, kcrisman, tmonteil, zimmerma Nils Bruin, Ralf Stephan Ralf Stephan, Nils Bruin N/A 7c029f3 7c029f3c98032a985c2ca0329dfe940271e9b86b
The documentation of function() is incomplete and confusing. For example, none of the methods described in http://www.sagemath.org/doc/reference/calculus/sage/symbolic/function_factory.html show up when I type:
```function?
```
The distinction between
```f = function('f')
```
and
```f = function('f', x)
```
is also not documented. See also #17445 for sources of confusion. Need to explain what happens in the second case above (`f = function('f',x)`). A symbolic function `f` is first created and then overwritten by the expression `f`? See the following example, where f is first an expression, then becomes redefined to a function in the background, but does not contain any information about its variables.
```sage: f = function('f', x); print type(f)
<type 'sage.symbolic.expression.Expression'>
sage: fx = function('f',x); print type(f)
<class 'sage.symbolic.function_factory.NewSymbolicFunction'>
sage: f.variables()
()
sage: fx.variables()
(x,)
```
See http://ask.sagemath.org/question/9932/how-to-substitute-a-function-within-derivatives/?answer=14752#post-id-14752 for a possible start at how to explain this, at least for those writing this.
### comment:1 Changed 6 years ago by kcrisman
• Description modified (diff)
### comment:2 Changed 6 years ago by kcrisman
• Description modified (diff)
### comment:3 follow-up: ↓ 26 Changed 6 years ago by nbruin
For the question about creation and overwriting:
The toplevel `function` has the sideeffect of inserting a binding in the global scope, just as the toplevel `var` does.
```sage: var('U')
U
sage: U
U
sage: function('f')
f
sage: f
f
```
The library versions of these routines don't do that:
```sage: SR.var('V')
V
sage: V
NameError: name 'V' is not defined
sage: sage.symbolic.function_factory.function('g')
g
sage: g
NameError: name 'g' is not defined
```
The intended use of the toplevel `function` is not to be assigned but to be called as a statement, just as `var`. There is plenty of documentation in sage that was written by people unaware of this fact (or possibly they think they're being helpful by providing code that doesn't return an error when used with the library versions of these functions)
As a result `x=var('x')` and `fx=function('f',x)` are quite ubiquitous in the docs. The former is fairly inocuous but the second, as you experience, has a rather nasty side effect. We're violating python's convention here: routines with side effects should return None.
I think this is the point where people usually give up fixing this mess (I have). Hopefully you will now follow through.
Note that we can probably not do more than document the issues, which is not going to fix much, because people won't read the doc, but then at least you can point them somewhere when they complain.
I suspect that Zimmerman's calculus book uses this stuff quite extensively, so deprecating/changing the current behaviour will likely lead to pushback from him, and probably rightly so.
(this is the kind of issue that earns sage the reputation of "implement now, design later")
### comment:4 Changed 6 years ago by rws
• Component changed from symbolics to documentation
### comment:5 Changed 6 years ago by schymans
Awesome, thank you, Nils. These things always confused me, so I'm relieved to hear that they should be avoided! Even the documentation of var is full of such examples:
```sage: x = var('x', domain=RR); x; x.conjugate()
x
x
sage: y = var('y', domain='real'); y.conjugate()
y
sage: y = var('y', domain='positive'); y.abs()
y
Custom latex expression can be assigned to variable:
sage: x = var('sui', latex_name="s_{u,i}"); x._latex_()
's_{u,i}'
```
Should we change these examples, too?
### comment:6 Changed 6 years ago by nbruin
• Branch set to u/nbruin/clarify_and_complete_documentation_of_function__
Don't base any other work on this branch just yet, since it's likely to be rebased. I have deprecated the use of `function('f',x)` in favour of using `function('f')(x)`. It's hardly longer to type and much less ambiguous. Otherwise, it seems that the documentation here is actually pretty accurate, apart from being terse. As soon as the objects returned by `function('f')` are consistent, I think users will be forced to learn the (now clear) semantics pretty quickly.
The object returned by `function(f)` doesn't allow symbolic operations on it, so people will quickly find they need to evaluate it in order to get a symbolic expression.
I haven't changed all doctests to take into account the deprecation, but I've done some files to show the impact of the change (seems relatively minor).
Last edited 6 years ago by nbruin (previous) (diff)
### comment:7 follow-up: ↓ 8 Changed 6 years ago by nbruin
• Commit set to fa3ebd6645eb2cae45c2218628eaa1d984f16506
• Status changed from new to needs_review
Not really ready for *review* but is ready for input. And is mainly ready for someone else to rewrite documentation properly. People CC'd on this ticket look like good candidates to do so.
New commits:
b41248b `trac 17447: Deprecation of function('f',x) in favour of function('f')(x)` fa3ebd6 `trac 17447: Doctest changes to reflect deprecation of function('f',x)`
Version 0, edited 6 years ago by nbruin (next)
### comment:8 in reply to: ↑ 7 ; follow-up: ↓ 9 Changed 6 years ago by schymans
Not really ready for *review* but is ready for input. And is mainly ready for someone else to rewrite documentation properly. People CC'd on this ticket look like good candidates to do so.
New commits:
b41248b `trac 17447: Deprecation of function('f',x) in favour of function('f')(x)` fa3ebd6 `trac 17447: Doctest changes to reflect deprecation of function('f',x)`
note that the doctest that apparently got removed in b41248b in reality gets moved to var.pyx because it tests the toplevel `function`, not the library `function`
This makes it a lot clearer to me now. Just a minor thing: In one of the examples, I would replace
```sage: cr = function('cr')
sage: f = cr(a)
```
by
```sage: function('cr')
sage: f = cr(a)
```
which in my understanding does the same but is shorter and makes clear that `function('cr')` does already create a symbolic function called `cr`. Otherwise, one may be surprised to create two functions by typing e.g. `cr1 = function('cr')`.
### comment:9 in reply to: ↑ 8 Changed 6 years ago by nbruin
This makes it a lot clearer to me now. Just a minor thing: In one of the examples, I would replace
```sage: cr = function('cr')
sage: f = cr(a)
```
At least one occurrence of that is in the doctest of `sage.symbolic.function_factory.function`, which (now) goes out of its way to import that function, which does not have the side-effect of mutating the global scope, as documented. So the assignment is actually necessary.
Another independent point: the top-level `function` does refer to `sage.symbolic.function_factory.function`, but perhaps should do so more explicitly, if possible with a hyperlink. The examples on the latter are a little more elaborate, so someone who wants to read up on `function` (and would probably find the top-level documentation first via `sage: function?`). I don't know how to make hyperlinks to other documentation in sage's docstrings.
### comment:10 follow-up: ↓ 11 Changed 6 years ago by ncohen
• Status changed from needs_review to needs_work
There are broken doctests in the 'doc' folder
```sage -t --long prep/Quickstarts/Differential-Equations.rst # 1 doctest failed
sage -t --long tutorial/tour_algebra.rst # 1 doctest failed
sage -t --long constructions/calculus.rst # 1 doctest failed
```
Nathann
### comment:11 in reply to: ↑ 10 ; follow-up: ↓ 12 Changed 6 years ago by nbruin
There are broken doctests in the 'doc' folder
Yes, and that is not the only place. First we need to see if there is any good reason why `function('f',x)` is preferable over `function('f')(x)`, of if we can get away with deprecating this confusing construction. And then someone should take a look at the doc to see if it needs further improvement. You're on a doc revamping binge, so this might just be the task for you!
### comment:12 in reply to: ↑ 11 Changed 6 years ago by ncohen
You're on a doc revamping binge, so this might just be the task for you!
I am afraid that you will have to find somebody else. I never used Sage's symbolics.
Nathann
### comment:14 Changed 6 years ago by rws
• Branch changed from u/nbruin/clarify_and_complete_documentation_of_function__ to public/17447
### comment:15 Changed 6 years ago by rws
• Commit changed from fa3ebd6645eb2cae45c2218628eaa1d984f16506 to e6a35343e5f712b88a5a9a80abd4e9302879c187
That fixes the three above marked doctests.
New commits:
e3ef919 `Merge branch 'develop' into t/17447/clarify_and_complete_documentation_of_function__` e6a3534 `17447: reviewer's patch: fix rst doctests`
### comment:16 Changed 6 years ago by git
• Commit changed from e6a35343e5f712b88a5a9a80abd4e9302879c187 to 4585e74063e81a8c332a500d016e837069edf8eb
Branch pushed to git repo; I updated commit sha1. New commits:
4585e74 `17447: fix rest of doctests`
### comment:17 Changed 6 years ago by rws
• Reviewers set to Ralf Stephan
These are the remaining doctest fixes coming from `make ptestlong`. I have reviewed and found fine the commits up to mine, so following reviewers can skip that.
And then someone should take a look at the doc to see if it needs further improvement.
### comment:18 follow-ups: ↓ 19 ↓ 21 Changed 6 years ago by kcrisman
I suspect that Zimmerman's calculus book uses this stuff quite extensively, so deprecating/changing the current behaviour will likely lead to pushback from him, and probably rightly so.
Cc:ing him for that reason. Though we have already had a few other discussions on Trac about the need to update our tests while not maintaining exact compatibility. Since they don't (yet) raise errors, but apparently only the deprecation warning, should we maybe update the tests (in that part only) to have the deprecation warning returned? Note that I don't believe the deprecation warning is even doctested, which is a no-no :)
Also, does the current branch actually "clarify and complete documentation of function"? It looks like it mostly fixes doctests.
Another change not doctested is
``` if is_SymbolicVariable(dvar):
- raise ValueError("You have to declare dependent variable as a function, eg. y=function('y',x)")
+ raise ValueError("You have to declare dependent variable as a function evaluated at the independent variable, eg. y=function('y')(x)")
```
which I think happens twice.
I'm still not sure I even understand some of the subtle differences. Are there occasions where the old behavior was "right" in the sense that one wanted that returned, and have we shown how to get that object? What about the ask.sagemath question?
All that to say that nonetheless it will be great to have a unified interface on this, if that is really the right thing to do, which from the comments it apparently is.
### comment:19 in reply to: ↑ 18 Changed 6 years ago by nbruin
Also, does the current branch actually "clarify and complete documentation of function"? It looks like it mostly fixes doctests.
I think there are two things that are/were confusing:
• `function('f',x)` returns something that is not, according to our definition, a function (e.g., it cannot be evaluated ("called") in a non-ambiguous way and hence triggers a deprecation warning)).
• The top-level `function` makes a binding in the global namespace and returns a value. The two were different.
With the proposed patch, `function('f',x)` is deprecated, so number 1 gets eliminated and number 2 is less confusing because, while it still injects something in the namespace, the value it returns is always the same as what it injects.
When I read the documentation of `sage.symbolic.function_factory.function`, I actually thought it described what happens fairly correctly, and it has good examples. Hence I did not feel the need to add to it. The top-level `function` documentation is rather sparse, so that could use expansion. However, we do not want to repeat ourselves, so perhaps it should just contain a clear pointer (hyperlink?) to the documentation of `sage.symbolic.function_factory.function`. I don't know how to do that, so please go ahead and improve that bit. You are excellently qualified to produce extensive and understandable documentation for the target audience of these functions.
Another change not doctested is
``` if is_SymbolicVariable(dvar):
- raise ValueError("You have to declare dependent variable as a function, eg. y=function('y',x)")
+ raise ValueError("You have to declare dependent variable as a function evaluated at the independent variable, eg. y=function('y')(x)")
```
which I think happens twice.
These messages weren't doctested before either. If you feel the projects benefits from such administrations, go ahead.
I'm still not sure I even understand some of the subtle differences. Are there occasions where the old behavior was "right" in the sense that one wanted that returned, and have we shown how to get that object?
In my opinion: no. In all cases, `function('f')(x)` is much clearer and not much longer than `function('f',x)`. There are *many* tickets and questions about people getting tripped up by the fact that a function called function doesn't return a function.
In my opinion, that's another thing to document. However, the confusion about function not returning a function definitely contributed to the confusion about how to use substitute_function.
### comment:20 Changed 6 years ago by kcrisman
The top-level function documentation is rather sparse, so that could use expansion.
Yes, this is what I meant. Hyperlink is tricky - not per se, but in command line and even sagenb it's not as useful. So I figure at least a little more in the top-level and then the hyperlink (assuming that file is in fact already in the documentation!).
You are excellently qualified to produce extensive and understandable documentation for the target audience of these functions.
That is very kind, I will see; this one is lower on my priority list but definitely important.
Well, I've found it does. I guess I will ask you to review it if I do that :-)
In my opinion, that's another thing to document. However, the confusion about function not returning a function definitely contributed to the confusion about how to use substitute_function.
Yes.
### comment:21 in reply to: ↑ 18 Changed 6 years ago by zimmerma
I suspect that Zimmerman's calculus book uses this stuff quite extensively, so deprecating/changing the current behaviour will likely lead to pushback from him, and probably rightly so.
Cc:ing him for that reason.
thanks, however I gave up trying to maintain compatibility with the book, since there are already many examples from the book that we put as doctests in Sage, and which were changed afterwards...
### comment:22 follow-up: ↓ 23 Changed 6 years ago by kcrisman
Here is another example of something confusing.
### comment:23 in reply to: ↑ 22 ; follow-ups: ↓ 24 ↓ 25 Changed 6 years ago by nbruin
Here is another example of something confusing.
I think that's another issue. The problem there is that global namespace `var` and `function` return a value as well as insert something into the namespace. That's counter to python custom (routines that mutate state normally return `None`. Compare `L.sort()` and `sorted(L)`. There are exceptions: `L.pop()`).
Initially it seems pedantic to enforce such rules in a computer algebra system as well, but the many problems it causes suggests it's not. Can we have `declare_var('x,y')` and `declare_function('f')` for the mutating stuff and just have `var('x')` and `function('f')` for the normal stuff? That ship has probably sailed already.
### comment:24 in reply to: ↑ 23 Changed 6 years ago by schymans
Can we have `declare_var('x,y')` and `declare_function('f')` for the mutating stuff and just have `var('x')` and `function('f')` for the normal stuff? That ship has probably sailed already.
That would be awesome! Much more consistent AND flexible. +1 from me.
### comment:25 in reply to: ↑ 23 Changed 6 years ago by rws
I think that's another issue. The problem there is that global namespace `var` and `function` return a value as well as insert something into the namespace. That's counter to python custom (routines that mutate state normally return `None`. Compare `L.sort()` and `sorted(L)`. There are exceptions: `L.pop()`).
Initially it seems pedantic to enforce such rules in a computer algebra system as well, but the many problems it causes suggests it's not. Can we have `declare_var('x,y')` and `declare_function('f')` for the mutating stuff and just have `var('x')` and `function('f')` for the normal stuff? That ship has probably sailed already.
See #17958 for `declare_var` and no, this is still a good idea.
### comment:26 in reply to: ↑ 3 Changed 6 years ago by mmezzarobba
I suspect that Zimmerman's calculus book uses this stuff quite extensively, so deprecating/changing the current behaviour will likely lead to pushback from him, and probably rightly so.
I'm not sure about `function()` and have little time to check now, but regarding `var()`, I think we tried to always use `x = SR.var('x')`.
And IMO, the version of `var` that injects a variable in the global namespace should simply be deprecated, or possibly moved to `sage.ext.inteactive_constructors_c` if someone really care about it.
### comment:27 Changed 6 years ago by rws
• Milestone changed from sage-6.5 to sage-6.6
• Status changed from needs_work to needs_review
So, comment:17 gave part of a review. Can someone please complete it?
### comment:28 Changed 6 years ago by rws
• Status changed from needs_review to needs_work
```sage -t --long src/sage/symbolic/expression.pyx # 1 doctest failed
sage -t --long src/sage/tests/french_book/calculus_doctest.py # 1 doctest failed
sage -t --long src/doc/pt/tutorial/tour_algebra.rst # 1 doctest failed
sage -t --long src/sage/combinat/integer_vector.py # 1 doctest failed
```
### comment:29 Changed 5 years ago by git
• Commit changed from 4585e74063e81a8c332a500d016e837069edf8eb to bdc154391e7ce6ceca1226f341ab62579bc3f4ee
Branch pushed to git repo; I updated commit sha1. New commits:
7956412 `Merge branch 'public/17447' of git://trac.sagemath.org/sage into public/17447` bdc1543 `trac 17447: further doctest adjustments`
### comment:30 Changed 5 years ago by git
• Commit changed from bdc154391e7ce6ceca1226f341ab62579bc3f4ee to 7c029f3c98032a985c2ca0329dfe940271e9b86b
Branch pushed to git repo; I updated commit sha1. This was a forced push. New commits:
7c029f3 `trac 17447: further doctest adjustments`
### comment:31 Changed 5 years ago by nbruin
• Status changed from needs_work to needs_review
### comment:32 Changed 5 years ago by nbruin
• Authors changed from schymans to Nils Bruin, Ralf Stephan
• Milestone changed from sage-6.6 to sage-6.10
• Reviewers changed from Ralf Stephan to Ralf Stephan, Nils Bruin
• Status changed from needs_review to positive_review
OK, comments #17 and #27 provide a positive review of the first bit. I give a largely positive review to Ralf's doctest adjustments. There was one overcorrection in `french_book/calculus_doctest.py` that I corrected. The other changes were just parallel corrections to different translations of one document that have been introduced since the last time this ticket was updated.
patchbot is happy and I think all code is positively reviewed now. Ready for merge!
### comment:33 Changed 5 years ago by vbraun
• Branch changed from public/17447 to 7c029f3c98032a985c2ca0329dfe940271e9b86b
• Resolution set to fixed
• Status changed from positive_review to closed
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https://discuss.codecademy.com/t/coded-correspondence-challenge-project/666440 | 1,669,498,606,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446709929.63/warc/CC-MAIN-20221126212945-20221127002945-00102.warc.gz | 236,461,416 | 8,186 | # Coded Correspondence Challenge Project
I am working through the Coded Correspondence off-platform project and I am bit confused over this line:
message:
`````` **b a r r y i s t h e s p y**
``````
** keyword phrase: d o g d o g d o g d o g d**
** resulting place value: 4 14 15 12 16 24 11 21 25 22 22 17 5**
** So we shift “b”, which has an index of 1, by the index of “d”, which is 3. This gives us an place value of 4, which is “e”. Then continue the trend: we shift “a” by the place value of “o”, 14, and get “o” again, we shift “r” by the place value of “g”, 15, and get “x”, shift the next “r” by 12 places and “u”, and so forth. Once we complete all the shifts we end up with our coded message:**
Is it just me or does the “resulting place value” examples go off the rails and aren’t correct after the first two cipher/letters. Just by the example above I don’t understand how the resulting place value for the first r is “15” but also the letter “x” at the same time, since that is not the place value of x in the alphabet.
1 Like
Yes, I thought that was off too.
letters = ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'] index = {letters[i] : i for i in range(len(letters))} keyword_phrase = "dogdogdogdogdog" message = "barryis" resulting = [] for i in range(len(message)): print(message[i], "+", keyword_phrase[i], end=" ") print(index[message[i]], "+", index[keyword_phrase[i]], end=" ") n = index[message[i]] + index[keyword_phrase[i]] print("=", n, "% 26 =", n % 26, end=" ") print("=", letters[n % 26]) resulting.append(n % 26) print() print(resulting)
1 Like | 514 | 1,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2022-49 | latest | en | 0.779829 |
https://www.sparknotes.com/math/algebra2/complexnumbers/problems_3/ | 1,701,342,872,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100184.3/warc/CC-MAIN-20231130094531-20231130124531-00637.warc.gz | 1,140,015,307 | 52,099 | Problem : What is the conjugate of 7 + 6i?
7 - 6i.
Problem : What is the conjugate of -2 - 4i?
-2 + 4i.
Problem : What is the conjugate of -3 + 5i?
-3 - 5i.
Problem : Evaluate .
.
Problem : Evaluate .
.
Problem : Evaluate .
.
Problem : Evaluate .
. | 86 | 261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/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-2023-50 | latest | en | 0.431351 |
https://www.enotes.com/homework-help/p-4-20-20-4-4-348899 | 1,503,496,710,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886120573.0/warc/CC-MAIN-20170823132736-20170823152736-00227.warc.gz | 902,463,674 | 9,869 | # P(4) 20! / (20-4) !4!Explain what !!! means and how to solve this problem
embizze | High School Teacher | (Level 2) Educator Emeritus
Posted on
The expression `(20!)/((20-4)!4!)` gives the number of ways 4 items can be picked from a group of 20 items, without replacement, and where order does not matter.
For example if you have a lottery with 20 numbers and you pick 4. You might pick 6,9,11,14. This is the same as picking 9,14,11,6 so order doesn't matter. And once a number is chosen, it cannot be chosen again. (No repetitions, or without replacement).
The symbol ! stands for the factorial operation: 0!=1 by definition, 1!=1,`2! = 2*1,3! = 3*2*1,4! = 4*3*2*1,` etc... It is the product of the number and every natural number below it.
The expression `(n!)/((n-r)!r!)` represents the number of combinations of r items taken from n items without replacement and where order does not matter. This is frequently written as `_nC_r` or `([n],[r])` .
Numerically `(20!)/((20-4)!4!)=4845= ` `_20C_4=([20],[4])` .
** If order matters it is called a permutation and is represented by `_nP_r=(n!)/((n-r)!)` . If you were choosing a committee of 4 people you would use combinations; but if you were selecting officers (president, vice president, etc...) you would use permutations as order matters. -- Alice,Bob,Carol, and Dan can make a committee no matter how you list their names. But if you are selecting officers, it makes a difference whether Alice is president or Carol is president.
Sources: | 425 | 1,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-34 | latest | en | 0.915456 |
https://mathinsight.org/applet/rotating_sphere_backward_curl | 1,638,868,397,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00097.warc.gz | 458,109,965 | 7,439 | # Math Insight
### Applet: Sphere rotating in opposite direction of macroscopic circulation
When the center of the sphere is fixed, it rotates in a clockwise direction when viewed from the positive $z$-axis, corresponding to a downward pointing curl (green arrow). On the other hand, the counterclockwise macroscopic circulation of the vector field around the $z$-axis is evident from the graph. This example illustrates that one cannot infer curl from the macroscopic circulation of the vector field. Macroscopic and microscopic circulation can be very different. First panel shows the full vector field; second panel shows its projection in the $xy$-plane.
This vector field is $$\dlvf(x,y,z) = \frac{(-y,x,0)}{(x^2+y^2)^{3/2}}$$ for $(x,y) \ne (0,0)$. One can compute that away from the $z$-axis, $$\curl \dlvf(x,y,z) = \frac{(0,0,-1)}{(x^2+y^2)^{3/2}}.$$
The vector field and its curl blow up near the $z$-axis (the vector field arrows would get much longer if we plotted then closer to the $z$-axis). Hence, the sphere spins faster and the curl vector grows longer, the closer you move the sphere to the $z$-axis. | 295 | 1,122 | {"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} | 3.296875 | 3 | CC-MAIN-2021-49 | latest | en | 0.857968 |
http://www.fractionsworksheets.ca/addingfractionswithlikedenominators/addingfractionswithlikedenominaotrs.html | 1,657,092,818,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00635.warc.gz | 78,083,521 | 5,688 | # Adding Fractions With Like Denominators
## How to add fractions with same denominators
At this page you will find helpful lessons to learn adding fractions with same denominators. To learn adding fractions with like denominators, you can print the following lessons on the topic. Pay close attention to learn the tricks explained in the lessons to master this skill.
Print the above lesson by clicking on it or you can print it below along with other worksheets.
Below are the lesson and worksheets to master adding fraction with like denominators, print and start exploring your skills to the next level. Adding two fractions with same denominators vertically
Kids in grade five must know how to add fractions with like denominators. This is the base to start adding fractions. To add fractions with same denominators, we need to add just the numerators of the given fractions to get the numerator of the answer and the denominator of the answering fractions remains the same as of given fractions. Hope the above worksheets and lessons have helped the students to master the concept of adding fractions with same denominators. | 216 | 1,138 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2022-27 | latest | en | 0.902542 |
https://brilliant.org/practice/properties-triangles-warmup/ | 1,500,972,998,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549425117.42/warc/CC-MAIN-20170725082441-20170725102441-00661.warc.gz | 622,894,613 | 20,879 | ×
Back to all chapters
# Geometry Warmups
Mathematics is filled with shapes that are kaleidoscopic in variety. Wielded since ancient times, the power of geometry helps us examine and measure these shapes.
# Properties of Triangles Warmup
What is the area of the triangle?
What is the sum of the angles shaded in green?
If the height of $$\triangle APC$$ is 4, what percent of $$\triangle ABC$$ is shaded white?
If a square with a diagonal length of $$4\sqrt{2}$$ has the same area as a right triangle, what could be the base and height of the triangle?
If a right triangle has two side lengths of 1 and 2, which of these options couldn't be the length of the third side?
× | 165 | 681 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2017-30 | longest | en | 0.936543 |
https://datascience.stackexchange.com/questions/53858/formal-proof-of-vanilla-policy-gradient-convergence | 1,702,183,509,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101195.85/warc/CC-MAIN-20231210025335-20231210055335-00733.warc.gz | 157,929,649 | 43,902 | # Formal proof of vanilla policy gradient convergence
So I stumbled upon this question, where the author asks for a proof of vanilla policy gradient procedures. The answer provided points to some literature, but the formal proof is nowhere to be included. Looking at Sutton,Barto- Reinforcement Learning, they claim that convergence of the REINFORCE Monte Carlo algorithm is guaranteed under stochastic approximation step size requirements, but they do not seem to reference any sources that go into more detail.
I am curious whether or not anybody actually has a formal proof ready for me to read. I found a paper, which goes into detail for proving convergence of a general online stochastic gradient descent algorithm, see, section 2.3.
However, I am not sure if the proof provided in the paper is applicable to the algorithm described in Sutton's book. In the mentioned algorithm, one obtains samples which, assuming that the policy did not change, is in expectation at least proportional to the gradient. However, the analytic expression of the gradient
$$\nabla J(\theta) \propto \sum_s \mu(s)\sum_a q_{\pi}(s,a)\nabla \pi(a|s,\theta)$$
depends on the on policy state distribution $$\mu(s)$$ which changes when we update $$\theta$$. Therefore, when updating during the algorithm, the distribution changes.
Any help would be greatly appreciated. Bottou's paper, which I linked above states that the event is drawn from a fixed probability distribution, which is not the case here.
EDIT:
So after reading some more papers, I found this, which is a paper of Bertsekas and Tsitsiklis. They argue that under certain assumptions convergence to a stationary point is guaranteed, where one has an update rule of the form
$$x_{t+1} = x_t +\gamma_t (s_t + w_t)$$ and $$w_t$$ is some error with $$\mathbb{E}[w_t | \mathcal{F}_t] = 0$$ for ascending $$\sigma$$-fields $$\mathcal{F}_t$$, which can be thought of conditioning on the trajectory $$x_0,s_0\dots,x_{t-1},s_{t-1},w_{t-1},x_t,s_t$$. I believe that this might be a solution since we need an expected gradient update given the past parameter $$x_t$$, which determines the sampling distribution which is exactly what the policy gradient theorem guarantees. I'd be happy if someone could verify this.
• Proof will only work for convex spaces. Though not even once have I stumbled upon one in professional work. Drift analysis might be more helpful for non-convex spaces. Nov 26, 2019 at 7:04
Proof will only work for convex spaces. Though not even once have I stumbled upon one in professional work. Drift analysis might be more helpful for non-convex spaces. – Piotr Rarus
Adding more to Piotr Rarus comment:
It appears that the convergence of the REINFORCE Monte Carlo algorithm for vanilla policy gradient procedures has not been formally proven. The literature, including Sutton and Barto's book on reinforcement learning, only provides claims of convergence under certain stochastic approximation step size requirements.
The paper by Bottou on online stochastic gradient descent algorithms provides a general proof of convergence, but it is not clear if it is applicable to the REINFORCE algorithm due to the changing distribution of the on-policy state.
Bertsekas and Tsitsiklis provide convergence proof for updating the rules of the form
$$x_{t+1} = x_t +\gamma_t (s_t + w_t)$$
where,
• $$w_t$$ is some error with $$\mathbb{E}[w_t | \mathcal{F}_t] = 0$$ for ascending $$\sigma$$-fields $$\mathcal{F}_t$$.
This may be applicable to the REINFORCE algorithm, but further analysis would be needed to verify this.
Overall, it seems that a formal proof of convergence for the REINFORCE Monte Carlo algorithm for vanilla policy gradient procedures is currently lacking.
I have written one, also following what Sutton's book does, but make each transform step more clearly(hopefully...)
\begin{align} \nabla J(\theta) &= \nabla v_\pi(s_0) \\ &= \nabla [\sum_a \pi(a|s_0)q_\pi(s_0, a)] \\ &= \sum_a[\nabla \pi(a|s_0) q_\pi(s_0, a) + \pi(a|s_0) \nabla q_\pi(s_0, a)] \ \ // by \ product \ derivative \ formula \\ &= \sum_a[\nabla \pi(a|s_0) q_\pi(s_0, a) +\pi(a|s_0) \nabla \sum_{s_1, r} p_\pi(s_1, r|s_0, a)(r +v_\pi(s_1)) ] \ \ // constant \ r \ has \ derivative \ 0 \\ &= \sum_a[\nabla \pi(a|s_0) q_\pi(s_0, a) +\pi(a|s_0) \sum_{s_1} p_\pi(s_1|s_0, a)\nabla v_\pi(s_1) ] \\ &\ \ \ (expanding \ v_\pi(s1) \ recusively ...)\\ &= \sum_a[\nabla \pi(a|s_0) q_\pi(s_0, a)] + \\ &\ \ \ \ \sum_a [\pi(a|s_0) \sum_{s_1} p_\pi(s_1|s_0, a)]* \sum_{a'}[\nabla \pi(a'|s_1) q_\pi(s_1, a') ]+ \\ &\ \ \ \ \sum_a[\pi(a|s_0) \sum_{s_1} p_\pi(s_1|s_0, a)]* \sum_{a'}[\pi(a'|s_1) \sum_{s_2} p_\pi(s_2|s_1, a')\nabla v_\pi(s_2) ] \\ & \ \ \ （keep \ recursion..., you \ could \ see \ there \ exists \ a \ form \ of \ \sum_a [\nabla \pi(a|s_x) q_\pi(s_x, a)]，\\ & \ \ \ multiplied \ by \ prob \ of \ s_0 \ to \ s_x \ following \ all \ possible \ routes. \ To \ simplify， \\ &\ \ \ let \ P_\pi(s_0, s_x,k) \ represent \ prob \ of \ s_0 \ to \ s_x \ after \ k \ steps.）\\ &= \sum_{s_x \in S} \sum_{k=1}^{\infty}P_\pi(s_0 \to s_x, k) \sum_a \nabla\pi(a|s_x)q_\pi(s_x, a) \\ &= \sum_{s_x \in S} [\sum_{k=1}^{\infty}P_\pi(s_0 \to s_x, k)] \sum_a \nabla \pi(a|s_x)q_\pi(s_x, a) \\ &\ \ \ ([...] represents \ prob \ of \ an \ episode \ ended \ in \ state \ s_x. Denote \ it \ \eta_\pi(s_x)) \\ &= \sum_{s_x \in S} \eta_\pi(s_x) \sum_a \nabla \pi(a|s_x)q_\pi(s_x, a) \\ &= [\sum_{s}\eta_\pi(s) ] * \sum_{s_x} \frac{\eta_\pi(s_x)}{\sum_{s}\eta_\pi(s)} * \sum_a \nabla \pi(a|s_x)q_\pi(s_x, a) \\ &= [\sum_{s}\eta_\pi(s) ] * \mu_\pi(s_x) * \sum_a \nabla \pi(a|s_x)q_\pi(s_x, a) \\ &\varpropto \mu_\pi(s_x) * \sum_a \nabla \pi(a|s_x)q_\pi(s_x, a) \\ Q.E.D. \\ \end{align} | 1,818 | 5,709 | {"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": 16, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2023-50 | latest | en | 0.91374 |
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