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25-06-2008, 08:31 AM
| #1221 | |
| Administrator | Quote:
1b) .308, z-scores of -4/15 and 8/15 for 96 and 108 respectively 1c) 2:45am, and I can't think of this off the top of my head. Look up confidence intervals and margin of error. >_< 2) A success (S) will be considered finding contraband in a package. A failure (F) will be considered not finding contraband in a package. Let's use a hypergeometric distribution since we are looking at a finite population with selections taken without replacement. Let X = # of successes in 3 selections taken from a finite population of 16 packages where there are a total of 5 possible successes. You can figure out the answers from there. oi... Didn't spot the program on my calculator I made back in college to do this quickly and without me figuring out all the variables in the hypergeometric again... ^.^ 2a) 33/112 2b) 55/112 2c) 22/112 2d) 2/112 No idea if those are right since I'm just trusting the explanations I put in my calculator long ago for the function parameters. >_<
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25-06-2008, 06:38 PM
| #1222 |
| Akatsuki Join Date: Nov 2004 Location: Urahara Store
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For question, 2 how did you came up with those numbers. They are right, since my professor posted the answers, but gave no explanation. and for number one why did you use the standard equation x-mean/sd (standard deviation) and for question 1b, the professor got .8746.
__________________ ![]() ![]() thanks to Tom, ~Mrge and Simonarturo, EE Last edited by Ice(v)an; 25-06-2008 at 06:42 PM. |
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25-06-2008, 09:22 PM
| #1223 | |
| Administrator | Quote:
For 1a, (x - mean)/std.dev. is the equation for z-score, so I solved it for x with the known z-score, mean, and std. dev. filled in. The equation I used was the hypergeometric distribution: N = finite population size (16 in this case) m = "success" possibilities (5 in this case) n = sample size (3 in this case, picked without replacement) k = # successes (out of n picked) For 1a, k = 0. 1b, k = 1. 1c, k = 2. 1d, k = 3 The formula is (mCk)(N-mCn-k)/(NCn). You can see another, similar example here (finally found a site that used the same variables I like to use): The Hypergeometric Distribution
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04-07-2008, 06:26 AM
| #1224 |
| Academy Student Join Date: Oct 2006
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Hey I should be able to help with most calculus problems including integrals, double and triple integrals, first, second, third, and n-order ODEs, Laplace transformations, Gauss Elimination, Eigenvalues, Eigenvectors. All that fun stuff. Also I might be able to help some with physics too depending on what level it is. |
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07-07-2008, 07:09 PM
| #1225 |
| Akatsuki Join Date: Nov 2004 Location: Urahara Store
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Given, Summation n=1, infinity, (-1)^n-1 3^n+2/2^n+3 and Summation n=1, infinity, (-1)^n-1 3^n+3/2^2n+2 One of them converges, the other one diverges, pick the one that converges, use three different tests such that each can show why it converges.
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14-07-2008, 08:07 PM
| #1226 |
| Akatsuki Join Date: Nov 2004 Location: Urahara Store
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Alright, I just use the ratio test, and I get that the second one, summation to infinity, when n=1, (-1)^n-1 3^n+3/2^2n+2 converges. By using the ratio test combined with the absolute convergence test. I need onemore tests, not sure which one would help me. The integral test? That seems to be hard. I think the root test could be my last test, but I'm not sure how that would work out.
__________________ ![]() ![]() thanks to Tom, ~Mrge and Simonarturo, EE Last edited by Ice(v)an; 14-07-2008 at 08:38 PM. |
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23-07-2008, 09:44 PM
| #1227 |
| Akatsuki Join Date: Nov 2004 Location: Urahara Store
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Find the area of one rose leaf of r = 2 sin 3beta. I got pie/12 - 1/6 can anyone check if I did it right? Professor not responding e-mails :/
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19-08-2008, 09:09 PM
| #1228 |
| Raitei Join Date: Nov 2004 Location: Noway
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A math question If a sum (sigma) runs from 0 to 4 is it exactly equal to a sum that runs from 4 to 0? I can imagine it being the same with simple sums as it shouldnt matter if you start adding from left or right in a line of numbers but not so sure about more complex functions.
__________________ All I am about to do is strike you 8 times okay? Last edited by Keerua; 19-08-2008 at 09:10 PM. |
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21-08-2008, 05:25 AM
| #1229 | |
| Administrator | Quote:
Please be mindful of the fact that this doesn't hold for integrals.
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21-08-2008, 12:54 PM
| #1230 |
| Raitei Join Date: Nov 2004 Location: Noway
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Ah thank you, now that helps me solve a few tasks ![]() New question (I shall be the new Icevan, 2x math class and 1 physix and 1 python) Prove that there is an infinite number of prime numbers. Any natural number N is a product of prime numbers, n= p1*p2*......pn So if I assume that there is a finite number of prime numbers all I have to do is contradict this assumption by finding a number greater than n which naturally must have prime numbers greater than pn. This makes sense in my head but no so much on the paper, inductive reasoning is very new to me and I dont know how to write my reasoning, help appreciated. A question about computers. Not really homework, just curious. For some reason a computer can only calculate the first 16 decimals of a number correctly and the rest will be just gibberish, why? Also how can this be true if you can have millions and millions of decimals for PI, is PI defined in a special way in computers? Edit2: Problem 3 solved, removed it
__________________ All I am about to do is strike you 8 times okay? Last edited by Keerua; 21-08-2008 at 07:34 PM. |
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22-08-2008, 12:27 AM
| #1231 |
| Administrator |
The first question is very easy to find an answer to on the internet, as it is usually one of the first problems solved in a Number Theory course. Google for "infinite primes" or something of that sort (include contradiction in your search for more targeted results that you'll understand). As for the second question, the number of decimal places that can be accurately represented depends on how much space the number has available for storage. Just as there are plenty of large numbers which cannot be represented using standard data types, there are also many decimal numbers that lose their precision after a certain length of expansion. To get around this, there are a number of possibilities, but one of the more common solutions is to use strings of numbers to represent the numbers themselves. This way, you must do a lot more work when doing basic math operations, but you can store numbers with nearly infinite numbers of digits. For calculating Pi and e, you can find special algorithms that make this even easier!
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22-08-2008, 09:05 AM
| #1232 |
| Raitei Join Date: Nov 2004 Location: Noway
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We need the thanks button back
__________________ All I am about to do is strike you 8 times okay? |
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26-08-2008, 05:16 AM
| #1233 |
| Chuunin |
Ummm, stupid question but what is the answer to 4(x+6)+9? I don't know if i got it right, or even got close...
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26-08-2008, 05:46 AM
| #1234 |
| Like a Polaroid Picture Join Date: Oct 2004 Location: Red Light District
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>_> 4x+33 you dont know how to expand brackets? |
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26-08-2008, 07:32 AM
| #1235 |
| Raitei Join Date: Nov 2004 Location: Noway
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I lol'd In other news, who could've thought that 1,2,3,4,5 etc was so fucking complicated, I have to learn 11 axioms by heart (most of which I know already cause I have used them unawarly) but the last one makes absolutely no sense (written in norwegian, with made up words....), I dont know how I should translate it, the principal of completeness? If anyone is familiar with it, can you explain it in good old english please?
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26-08-2008, 04:01 PM
| #1236 | |
| Administrator | Considering the fact that so few students even understand order of operations completely by the end of high school, I think it probably isn't something to be made fun of. Quote:
For a set to be complete, it means that its supremum (the very first value that either exceeds or is equal to every other value in the set) must be in the set. More info on that here: Supremum - Wikipedia, the free encyclopedia The easiest way to explain this in simple terms is that there is some number that is at least as big as every element you are working with in a set of numbers. More importantly still, that number is in your set of numbers. So, if we look at the integers, the number that is at least as big as every element is infinity. Also, infinity is an integer when you consider the highest element in the countably infinite set of integers, so the set is called complete.
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27-08-2008, 07:54 AM
| #1237 |
| Raitei Join Date: Nov 2004 Location: Noway
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Calculus, 700 pages this semester, the workload is hard, but I feel like I am doing something for once.... So let me see if I got this right, if i have the interval [8,10] then the supremum is equal to 10? And the infimus is 8? My professor said that for now we should be content with knowing the axioms and understanding the supremus and infimus because later we will be using it to pull out numbers with whatever quality we desire, I think that is awkward way of doing it as it is generally much more motivating and interesting to see what you are learning can be used for. Thanks for the links, hard to translate this to english so no google ;(
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27-08-2008, 04:34 PM
| #1238 |
| o({})o Join Date: Nov 2006
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Tell your professor to go jump in a lake. There is no reason to learn such things for calculus. Also, your example is correct, but I don't know what the term infimus means. Also, note that the supremum would be the same for the interval (8, 10).
__________________ This is my account for use on public computers. Please do not PM this account unless you have good reason to believe that I will not be on my main account. Last edited by ]MpC[hebe; 27-08-2008 at 04:37 PM. |
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31-08-2008, 01:38 AM
| #1239 |
| A://Stocks |
need quick calc help. question is indicated in the picture, thast from the solutions manual. im get a little lost after that point i pointed to in the picture. http://animestocks.com/img/calchelp.jpg |
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31-08-2008, 02:20 AM
| #1240 |
| Donator Join Date: Nov 2005 Location: In the same house as krozar's stalkers, yep... yep
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Ok the solution is through integration by parts as I assume you already got. The bit where you get lost is a very simple manipulation of the equation in order to make the integration simple. Essentially by adding and subtracting 1 from the denominator (the net effect of which is zero) and grouping the 2x+1 in both the numerator and denomiator kinda like doing the reverse of say for instance if we had 2/7+1/7=3/7. By splitting the fraction into two parts [2x+1 and -1] the first one then simplifies to 1 and the second one becomes -1/(2x+1). The following step is the actual integration and it is a standard solution for both [integral of 1 is x +c and ln(2x+1) is 1/2 ln(2x+1) + c, where combining the constants of integration gives C]. Finally, all the ln(2x+1) terms are gathered and the common factor is removed, then as an additional step the x+1/2 is rewritten as 1/2(2x+1) and you have your final solution. Hope that made sense, if anything doesn't lemme know =]
__________________ Last edited by waqo; 31-08-2008 at 02:22 AM. |
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