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 09-17-2009, 04:08 PM #51 Wyman Carpal \'Tunnel     Join Date: Mar 2007 Location: Redoubling with gusto Posts: 12,005 Re: The Official Math/Physics/Whatever Homework questions thread Contradiction. Suppose m > L. When your sequence gets "close enough" to L, it must also be below m. Choose an epsilon based on m and L that defines "close enough".
 09-17-2009, 11:14 PM #52 3kingme3 old hand     Join Date: Mar 2008 Location: dont google me Posts: 1,483 Re: The Official Math/Physics/Whatever Homework questions thread ok idk wtf Im doing wrong but can certainly use some help. ln(x) + ln(x - 1) = 1 I have to find x, heres what i did ln(x(x-1))=1 e^1=x(x-1) x=e x-1=e so x=e+1 I plugged both back into the original equation and neither worked, but when I submitted no solution I was told it was wrong help anyone?
09-17-2009, 11:27 PM   #53
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by 3kingme3 ok idk wtf Im doing wrong but can certainly use some help. ln(x) + ln(x - 1) = 1 I have to find x, heres what i did ln(x(x-1))=1 e^1=x(x-1) x=e x-1=e so x=e+1 <------------------ I plugged both back into the original equation and neither worked, but when I submitted no solution I was told it was wrong help anyone?
The arrow is where you went wrong.

If you had 0 = XY, then you can say X=0 OR Y=0. But if you have e = XY, it's not the case that necessarily X=e or Y=e.

So, starting with e^1=x(x-1), get

0 = x^2 - x - e.

Use the quadratic formula to solve for x.

 09-18-2009, 12:08 AM #54 3kingme3 old hand     Join Date: Mar 2008 Location: dont google me Posts: 1,483 Re: The Official Math/Physics/Whatever Homework questions thread tyty so much, apparently I have some shifting problems as well for some reason I cant get (1/e)^x to shift 10 units right, or -(1/2)x radical(2x-x^2) to shift 3 units right Ive tried the obvious subtracting 10 for the (1/e) part and adding 3 inside the radical, but neither have worked
09-18-2009, 12:18 AM   #55
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by 3kingme3 tyty so much, apparently I have some shifting problems as well for some reason I cant get (1/e)^x to shift 10 units right, or -(1/2)x radical(2x-x^2) to shift 3 units right Ive tried the obvious subtracting 10 for the (1/e) part and adding 3 inside the radical, but neither have worked
To shift right, you replace every (x) with (x-a), where a is the amount of the shift.

Think about why this is the right thing to do.

09-18-2009, 12:32 AM   #56
3kingme3
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Wyman To shift right, you replace every (x) with (x-a), where a is the amount of the shift. Think about why this is the right thing to do.
Ty again. I would say the reason is because since we are shifting with respect to x's, you need to account for each x in the problem. I guess I got lazy and was thinking about when moving up or down, since you are in respect to y, and there is only 1 y, you just tack it onto the end.

Also I would like to thank you for not just giving me the answer, but rather explaining it to me so I can figure it out.

 09-20-2009, 12:26 PM #57 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,244 Re: The Official Math/Physics/Whatever Homework questions thread Just to be sure I'm not missing something huge...if A and B are subsets of ℝ2, where: A = {(x, y) | y < x2} B = {(x, y) | y ≤ x2} then A and B have the same interior, exterior, and boundary; yes?
09-20-2009, 12:39 PM   #58
lastcardcharlie
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Subfallen ...then A and B have the same interior, exterior, and boundary; yes?
A = int(B) and B = cl(A) so they have the same interior and boundary. Not sure what exterior means, but if it's interior of complement then yes.

 09-20-2009, 12:58 PM #59 Subfallen Carpal \'Tunnel     Join Date: Sep 2004 Location: farther back Posts: 7,244 Re: The Official Math/Physics/Whatever Homework questions thread Yeah, Ext A = Int Ac. Thx!
 09-21-2009, 12:58 AM #60 furyshade Carpal \'Tunnel     Join Date: Apr 2006 Location: Pasadena, CA Posts: 10,138 Re: The Official Math/Physics/Whatever Homework questions thread quick set theory question, is U{{a,b,c,d,e,f},{e,f}} equal to {a,b,c,d,e,f} or {a,b,c,d,e,f,e,f}?
09-21-2009, 01:31 AM   #61
bigpooch
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by furyshade quick set theory question, is U{{a,b,c,d,e,f},{e,f}} equal to {a,b,c,d,e,f} or {a,b,c,d,e,f,e,f}?
.

09-21-2009, 01:39 AM   #62
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by bigpooch .
that's what i thought, thanks. also i have a problem involving showing if a set is transitive, "a set A is transitive if each element of A is also a subset of A". maybe i am missing something but i don't see how a set could not be transitive. could someone help me out and maybe give an example of a non-transitive finite set?

09-21-2009, 02:04 AM   #63
bigpooch
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by furyshade that's what i thought, thanks. also i have a problem involving showing if a set is transitive, "a set A is transitive if each element of A is also a subset of A". maybe i am missing something but i don't see how a set could not be transitive. could someone help me out and maybe give an example of a non-transitive finite set?
Let 0 = {}; then, 0 is "vacously" a transitive set. { {} } = {0} is also a transitive set since 0 = {} is a subset of any set. One can define 1 as {0} and define 2 as {0,1} = { {}, { {} } }, etc.

On the other hand, {1} = { {0} } = { { {} } } is not transitive since 1 = { {} }, the only element, is not a subset of {1}; the only subsets of {1} are {1} and 0 = {}.

There are a lot of "mundane" examples too. For example, any nonempty set S of objects that are not sets will do since each object (by virtue of not being a set) can not be a subset of S.

09-21-2009, 02:12 AM   #64
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by bigpooch Let 0 = {}; then, 0 is "vacously" a transitive set. { {} } = {0} is also a transitive set since 0 = {} is a subset of any set. One can define 1 as {0} and define 2 as {0,1} = { {}, { {} } }, etc. On the other hand, {1} = { {0} } = { { {} } } is not transitive since 1 = { {} }, the only element, is not a subset of {1}; the only subsets of {1} are {1} and 0 = {}. There are a lot of "mundane" examples too. For example, any nonempty set S of objects that are not sets will do since each object (by virtue of not being a set) can not be a subset of S.
so would the the set of natural numbers be a transitive set?

09-21-2009, 02:33 AM   #65
bigpooch
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by furyshade so would the the set of natural numbers be a transitive set?
If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.

09-21-2009, 02:37 AM   #66
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by bigpooch If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.
alright, that makes sense. the questions says to show that {null,{null}} is transitive and then asks for an example of an infinitely large transitive set so i figured that is what they were getting at. thanks a lot for the help!

09-21-2009, 04:04 AM   #67
thylacine
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by bigpooch If by "natural numbers" you mean nonnegative integers and as defined in the aforementioned way ( 0 = {}, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. ), yes.
Are you sure?

 09-21-2009, 01:36 PM #68 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread If probability of having a boy is 1/2. then what is the probability of having three children of the same sex? I think it is: 1-2[(4 choose 1)(1/2)^1(1/2)^3] I multiplied it twice becaue it can come 3 boys 1 girl or 3 girls 1 boy Is this right?
 09-21-2009, 02:25 PM #69 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread Lost on this one, Just a move in the right direction would be great Suppose a series of n independent trials can end in one of three posibilities. Let k_1 and k_2 denote the number of trials that result in outcomes 1 and 2 respectively. Let p_1 and p_2 denote the probabilities associated with outcomes k_1 and k_2. Use theorem 3.2.1 to deduce a formula for the probability of getting k_1 and k_2 occurences of outcomes 1 and 2 respectively. Theorem 3.2.1 Binomial Distribution (n choose k)(p)^k(1-p)^n-k
09-21-2009, 03:41 PM   #70
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 If probability of having a boy is 1/2. then what is the probability of having three children of the same sex? I think it is: 1-2[(4 choose 1)(1/2)^1(1/2)^3] I multiplied it twice becaue it can come 3 boys 1 girl or 3 girls 1 boy Is this right?
Clarify: You're assuming that there are 4 children, and you want to know the probability that exactly 1 child is of one sex and 3 are of the other?

If so, p = (2C1) * (4C1) * (1/2)^4 = 1/2.

Which is also what you got, but idk why you subtracted from 1.

09-21-2009, 03:45 PM   #71
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 Lost on this one, Just a move in the right direction would be great Suppose a series of n independent trials can end in one of three posibilities. Let k_1 and k_2 denote the number of trials that result in outcomes 1 and 2 respectively. Let p_1 and p_2 denote the probabilities associated with outcomes k_1 and k_2. Use theorem 3.2.1 to deduce a formula for the probability of getting k_1 and k_2 occurences of outcomes 1 and 2 respectively. Theorem 3.2.1 Binomial Distribution (n choose k)(p)^k(1-p)^n-k
You've got N trials, k_1 of which need outcome 1: N choose k_1
Now you've got N-k_1 remaining, k_2 of which need outcome 2: (N-k_1) choose k_2

The probability of getting k_1 outcome 1's, then k_2 outcome 2's, then outcome 3's is: (p_1)^(k_1) * (p_2)^(k_2) * (1-p_1-p_2)^(N-k_1-k_2)

(N choose k_1) * ((N-k_1) choose k_2) * (p_1)^(k_1) * (p_2)^(k_2) * (1-p_1-p_2)^(N-k_1-k_2)

09-21-2009, 03:47 PM   #72
smcdonn2
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Wyman Clarify: You're assuming that there are 4 children, and you want to know the probability that exactly 1 child is of one sex and 3 are of the other? If so, p = (2C1) * (4C1) * (1/2)^4 = 1/2. Which is also what you got, but idk why you subtracted from 1.
Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.

 09-21-2009, 03:50 PM #73 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread a college awards 5 scholarships there are 8 men and 10 woman all equally likely to win, what is the probabilty that there will be both men and women in the award? I did it like this but got a difference answer than the book (8C0)(10C5)/(18C5)=2.94 1-2.94=97.06% the book says 96.4
09-21-2009, 04:15 PM   #74
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 Yes that is correct, I wanted to use binomial distribution thats why I subtracted by one.
This makes no sense, and you got lucky because 1/2 = 1-1/2.

09-21-2009, 04:16 PM   #75
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 a college awards 5 scholarships there are 8 men and 10 woman all equally likely to win, what is the probabilty that there will be both men and women in the award? I did it like this but got a difference answer than the book (8C0)(10C5)/(18C5)=2.94 1-2.94=97.06% the book says 96.4
p(all men) = 8C5 / 18C5

p(all women) = 10C5 / 18C5

p(both men and women) = 1 - p(all men) - p(all women)

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