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 Science, Math, and Philosophy Discussions regarding science, math, and/or philosophy.

09-21-2009, 05:52 PM   #76
smcdonn2
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Wyman This makes no sense, and you got lucky because 1/2 = 1-1/2.
OK here is my reasoning:

P(A)= probability of getting 1 girl and 3 boys

(4C1)(1/2)^1(1/2)^3

This should be the same as getting three girls and 1 boy so multiply by 2

lol, Yeah I have no I dea why I was subtracting off

 09-21-2009, 05:55 PM #77 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread An Urn contains five balls # 1-5 two balls are drawn with replacement. Let X be the larger of the two numbers drawn find px(k) My question here is what to do in the case they are the same? Use the number or redraw? PS Indy or Fins?
09-21-2009, 06:42 PM   #78
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 An Urn contains five balls # 1-5 two balls are drawn with replacement. Let X be the larger of the two numbers drawn find px(k) My question here is what to do in the case they are the same? Use the number or redraw? PS Indy or Fins?
Use the number. Not using it would be "without" replacement; i.e. it's the same as not putting the ball back once you've drawn it.

I like Indy -3.5, but a lot of people seem to be all over MIA +3.5 and under 41.

 09-21-2009, 06:45 PM #79 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread I got teaser Indy +5.5 under 50
 09-22-2009, 12:57 AM #80 mtlchris centurion   Join Date: Jul 2007 Posts: 124 Re: The Official Math/Physics/Whatever Homework questions thread Probability: Total sustained load on the concrete floor is the sum of dead load and occupancy load witch both follow gamma distributions (X1 and X2) with a1=50 b1=2 a2=20 b2=2 assume x1 and x2 to be independent find a value for sustained load that will be exceeded with probability less than 1/16 Attempt at solution: i worked out E(x1+x2)= 140 Var(x1+x2)= 240 now i think i want to use tchebysheff's theorem but how do i use this to find a value rather than a probability? And what is this problem asking for? what is the load exceeding?
 09-22-2009, 05:39 PM #81 PaulieWlnuts veteran     Join Date: Oct 2007 Location: YOYO Posts: 2,046 Re: The Official Math/Physics/Whatever Homework questions thread Simple Chemistry problem that I'm struggling with... We started with a sample of BaCl2 *2H20 mixed with NaCl. Determined mass of mixture before heating and mass of mixture after to find mass of water lost by original. Mass of water lost by the sample = .5151 g Molar Mass of Water = 18.0153 g/mol Moles of water lost by sample = .02860 mol from .5151/18.0153 How many moles of BaCl2 *2H20 were present in the original formula? So here I just take moles of water .02860 and mult by ratio (1molBaCl2/1molH20) to obtain mol of BaCl2 and then add the two together? From this I got .04290 mol of BaCl2 2H20. Then I'm asked to find molar mass of of BaCl2 2H20 which is 244.26 g/mol. Here's where I know I messed up. I'm asked to determine amount of grams of BaCl2 2H2O present in the original sample. And here I just took .04290 (244.26/1) to get 10.48 g of BaCl2 2H20. But this is impossible because the orginal mass of the mixture was determined to be 4.8438g...... I think I'm making an obvious mistake but can't see it...
 09-22-2009, 08:40 PM #82 Myrmidon7328 old hand     Join Date: Dec 2007 Posts: 1,960 Re: The Official Math/Physics/Whatever Homework questions thread Hey guys, I have done all of my Graph Theory problem set minus this question. Since I can't type up in LaTex at a moment, here goes my best attempt in text shorthand: Given a complete graph of order n, Kn, and an edge in it, called e, use Cayley's formula to prove that the number of spanning trees in Kn - e is (n-2)*n^(n-3). If any part of that is unclear (obv the notation I've used isn't very good), please let me know! This is what I have so far: Spoiler: If I break apart the base, I get n^(n-2) - 2n^(n-3). The left part is Cayley's formula. I can't really figure out a justification for the right hand part though. My guess is to say that we have [2n^(n-2)]/n, which turns out to be (n-1)*cayley's number/(total number of edges in Kn), but it is not intuitive why this would be correct. Would there be an easier solution, if I used Prufer's algorithm instead? Thanks!
09-22-2009, 08:57 PM   #83
thylacine
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Myrmidon7328 Hey guys, I have done all of my Graph Theory problem set minus this question. Since I can't type up in LaTex at a moment, here goes my best attempt in text shorthand: Given a complete graph of order n, Kn, and an edge in it, called e, use Cayley's formula to prove that the number of spanning trees in Kn - e is (n-2)*n^(n-3). If any part of that is unclear (obv the notation I've used isn't very good), please let me know! This is what I have so far: Spoiler: If I break apart the base, I get n^(n-2) - 2n^(n-3). The left part is Cayley's formula. I can't really figure out a justification for the right hand part though. My guess is to say that we have [2n^(n-2)]/n, which turns out to be (n-1)*cayley's number/(total number of edges in Kn), but it is not intuitive why this would be correct. Would there be an easier solution, if I used Prufer's algorithm instead? Thanks!
Every spanning tree either contains e or it doesn't. Now average over all edges e. Think about it `cos it's the only hint I'll give, and you can do it in a few lines.

 09-22-2009, 11:10 PM #84 Myrmidon7328 old hand     Join Date: Dec 2007 Posts: 1,960 Re: The Official Math/Physics/Whatever Homework questions thread Thanks Thylacine! A friend of mine had a brainteaser from an interview today. I doubt it's thread worthy, so would this be the right place to post it? Anyways, here goes: You have 8 pool balls (labeled or colored so that you can tell them apart). You know that at least 7 of them are the same exact weight. One of them may be heavier, lighter, or weigh the same as the rest. You can only use a balance that will tell you the relative weights of the ball(s) on each side. What's an algorithm for the least number of weighings to determine the defective ball, or determine that all the balls are the same weight?
 09-23-2009, 02:50 AM #85 Cactusmuffin newbie   Join Date: May 2009 Posts: 40 Re: The Official Math/Physics/Whatever Homework questions thread [QUOTE=PaulieWlnuts;13305639]Simple Chemistry problem that I'm struggling with... We started with a sample of BaCl2 *2H20 mixed with NaCl. Determined mass of mixture before heating and mass of mixture after to find mass of water lost by original. QUOTE] I haven't taken chemistry in a very long time but. For solving for moles of BaCl2*2h20. There is .02860 mol of Water and it takes 2 moles of water for each mole of BaCl2*2h20 so you divide .02860 mol by 2. Then for solving for the mass of BaCl2*2h20 in the sample you take the number given by the previous division and multiply that by the molar mass. Soundright?
09-23-2009, 03:53 AM   #86
Carpal \'Tunnel

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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Myrmidon7328 Thanks Thylacine! A friend of mine had a brainteaser from an interview today. I doubt it's thread worthy, so would this be the right place to post it? Anyways, here goes: You have 8 pool balls (labeled or colored so that you can tell them apart). You know that at least 7 of them are the same exact weight. One of them may be heavier, lighter, or weigh the same as the rest. You can only use a balance that will tell you the relative weights of the ball(s) on each side. What's an algorithm for the least number of weighings to determine the defective ball, or determine that all the balls are the same weight?
i think this is effectively a binary sorting problem. i think the short answer is since you have 2^3 objects it take log2(2^3)=3 sorts to definitely get the ball.

i think the real world interpretation is you put 4 on each side, one side must weigh more than the other. now remove two balls from each side, if the scale is even you know one of the two balls you took from the lighter side is the light one so you weigh those two and know which is the lightest. if the scale is uneven after removing two balls from each side remove one more ball from each side. if the scale is even now then the ball you removed from the light side is the light ball, if the scale is uneven now then you are only weighing two balls so you know the light side ball is the light ball.

09-23-2009, 08:25 AM   #87
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by furyshade i think this is effectively a binary sorting problem. i think the short answer is since you have 2^3 objects it take log2(2^3)=3 sorts to definitely get the ball. i think the real world interpretation is you put 4 on each side, one side must weigh more than the other. now remove two balls from each side, if the scale is even you know one of the two balls you took from the lighter side is the light one so you weigh those two and know which is the lightest. if the scale is uneven after removing two balls from each side remove one more ball from each side. if the scale is even now then the ball you removed from the light side is the light ball, if the scale is uneven now then you are only weighing two balls so you know the light side ball is the light ball.
Though, if you knew that the "odd" ball was heavier, you could identify it in 2 weighings -- so you have to be careful with these kind of arguments.

 09-23-2009, 12:21 PM #88 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread What is this basic shorthand operation: Z_30 where Z are the integers So if we have a generator 25. then is the set {5,20,15,10,} correct? So when I see Z_n and any given generator a, do I divide n by a then put the remainder in the set? then for step 2 add a+a divide it by n and again put the remainder in the set? Is Z always closed under addition unless already specified? actually I think I first need to add a+a / n to get my first term and then continue a+a+a for second giving the set {20,15,10,5,0,25}
 09-23-2009, 12:44 PM #89 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread OK let me know if I got this one correct: find the number of elements in the indicated cyclic group C* under multiplication generated by (1+i)/sqrt(2) I got 8 {i,(i-1)/sqrt(2),-1,(-1-i)/sqrt(2),-i,(1-i/sqrt(2),1,(1+i)/sqrt(2)} Last edited by smcdonn2; 09-23-2009 at 12:46 PM. Reason: forgot an important element
 09-23-2009, 01:04 PM #90 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread Inverse permutations a= (123) (321) b= (123) (213) is a^-1 (321) (123) so If I have to find a^-1*b*a (123) (121) Last edited by smcdonn2; 09-23-2009 at 01:28 PM.
09-23-2009, 01:29 PM   #91
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 What is this basic shorthand operation: Z_30 where Z are the integers So if we have a generator 25. then is the set {5,20,15,10,} correct? So when I see Z_n and any given generator a, do I divide n by a then put the remainder in the set? then for step 2 add a+a divide it by n and again put the remainder in the set? Is Z always closed under addition unless already specified? actually I think I first need to add a+a / n to get my first term and then continue a+a+a for second giving the set {20,15,10,5,0,25}
Yes, but you don't "divide by n"; you take a remainder modulo n.

<25> = {0, 25, 20, 15, 10, 5}. This is a (cyclic) subgroup of Z/(30Z) (or if you write it that way, Z_30) of order 6 and index 5 (=30/6).

09-23-2009, 01:33 PM   #92
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 OK let me know if I got this one correct: find the number of elements in the indicated cyclic group C* under multiplication generated by (1+i)/sqrt(2) I got 8 {i,(i-1)/sqrt(2),-1,(-1-i)/sqrt(2),-i,(1-i/sqrt(2),1,(1+i)/sqrt(2)}
i.e. Find the smallest positive number N such that x^N = 1, where x is your generator. Note that I'm writing the group multiplicatively. In your last post, you were in Z/30Z, which is an additive group, so you'd be solving Nx = 0.

So (1+i)/sqrt2 has order 8. We can see this because, viewing it in polar, it has r=1, \theta = pi/4 (and 2pi /(pi/4) = 8).

09-23-2009, 01:45 PM   #93
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 Inverse permutations a= (123) (321) b= (123) (213) is a^-1 (321) (123) so If I have to find a^-1*b*a (123) (121)
I'm not going to use 2 line notation, like you did. Rather, I'll just write down the image of (123). That is: (312) will denote the permutation where f(1)=3, f(2)=1, and f(3)=2, and (123) will denote the identity permutation.

First note that (121) can't be a^(-1) * b * a, since, for one, it's NOT A PERMUTATION!

a = (321)
b = (213)
a^(-1) = (321)

Since a^-1 = a, we don't have to worry about the right- or left-handedness of things, so let's figure out what the image of 1 is.

Apply a. 1-->3
Apply b. 3-->3
Apply a^(-1). 3-->1.
So 1-->1.

Image of 2:
Apply a. 2-->2
Apply b. 2-->1
Apply a^(-1). 1-->3.
So 2-->3.

Image of 3:
Apply a. 3-->1
Apply b. 1-->2
Apply a^(-1). 2-->2.
So 3-->2.

So a^(-1)*b*a = (132).

 09-23-2009, 01:54 PM #94 smcdonn2 centurion   Join Date: Jul 2009 Posts: 165 Re: The Official Math/Physics/Whatever Homework questions thread ok , thank you for your help, So why does a^-1 = a? if our inage is (123456) and a = (314562) does a = a^-1? basically inverse says it will bring us back to our image right?
09-23-2009, 01:57 PM   #95
Pyromantha
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 ok , thank you for your help, So why does a^-1 = a?
[Speaking of the permutation (321)]

Because it's a transposition - it leaves 2 fixed and swaps 1 and 3. Then obviously if you do it again you swap 1 and 3 back to where they were, and 2 is still fixed. Any transposition - a permutation which swaps two elements and leaves everything else fixed has order two. Similarly, any permutation which is the product of disjoint transpositions has order two, like (12)(34)(5)(67), for example.

09-23-2009, 02:00 PM   #96
Pyromantha
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by smcdonn2 if our inage is (123456) and a = (314562)
The inverse of a here is (261345). Just look where everything gets mapped by a. 1 gets sent to position 3, so the 3rd entry of a^-1 must be 1 to send it back again, etc.

 09-23-2009, 07:44 PM #97 iheartleeks journeyman   Join Date: May 2008 Posts: 309 Re: The Official Math/Physics/Whatever Homework questions thread I have no idea how to attempt this: -Circle (culvert) has diameter of 6. -Bottom two units of circle is clogged with mud -Area? (of what remains open) Any help appreciated. Edit: not a math major Edit2: here is my attempt: pi*(d/2)^2 - (The part idk how to do) Last edited by iheartleeks; 09-23-2009 at 08:10 PM.
09-23-2009, 09:14 PM   #98
Wyman
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by iheartleeks I have no idea how to attempt this: -Circle (culvert) has diameter of 6. -Bottom two units of circle is clogged with mud -Area? (of what remains open) Any help appreciated. Edit: not a math major Edit2: here is my attempt: pi*(d/2)^2 - (The part idk how to do)

An elementary way to compute this:

We have to add the blue sector and the pink triangle. The area of a sector is (1/2) r^2 * angle. In our case, blue = 9B/2. We'll solve for B later.

The pink triangle has area (1/2)(base)(height) = C * 1 = C.

Now, C = sqrt(8) by looking at half of the pink triangle and using pythagorean theorem. Angle A is 2*arccos(1/3), again from the pink triangle. So B = 2pi - 2arccos(1/3), so the area of the pink + blue is

9 * (pi-arccos(1/3)) + sqrt(8)

Of course, we could use calculus instead:

A = integral from 2 to 6 of (2 sqrt(9 - (3-h)^2) dh) should give the same thing, but I have not verified this.

09-24-2009, 12:46 AM   #99
Myrmidon7328
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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Wyman Though, if you knew that the "odd" ball was heavier, you could identify it in 2 weighings -- so you have to be careful with these kind of arguments.
I figured it out I think. I'm getting 3 weighings, which is consistent with log base 2 of 8.

So, if the balls are numbered 1-8, take 1-3 and weigh it against 4-6.

If they are equal, weigh 7v8. If these are equal, we're done. If they are different, weigh one against ball 1. Then you can figure out which one is the odd one out, and whether or not its heavier or lighter.

If the 3-3 weighing is different, without loss of generality, assume 123 is the heavier group of balls.

Now, weigh 1 and 2 against 3 and 7 (since we know 7 is a normal ball). If they are different, (say 1 and 2 are heavier), we can weigh them against each other to find the wrong ball.

If they are the same, weigh 4 and 5 (two of the lighter balls). If they're the same, 6 is the light ball. Otherwise, you will know which ball is lighter.

If we generalize this to n balls, with n-1 identical, would the number of weighings be the smallest integer greater than or equal to log base 2 of n?

09-24-2009, 03:20 AM   #100
Carpal \'Tunnel

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Re: The Official Math/Physics/Whatever Homework questions thread

Quote:
 Originally Posted by Myrmidon7328 I figured it out I think. I'm getting 3 weighings, which is consistent with log base 2 of 8. So, if the balls are numbered 1-8, take 1-3 and weigh it against 4-6. If they are equal, weigh 7v8. If these are equal, we're done. If they are different, weigh one against ball 1. Then you can figure out which one is the odd one out, and whether or not its heavier or lighter. If the 3-3 weighing is different, without loss of generality, assume 123 is the heavier group of balls. Now, weigh 1 and 2 against 3 and 7 (since we know 7 is a normal ball). If they are different, (say 1 and 2 are heavier), we can weigh them against each other to find the wrong ball. If they are the same, weigh 4 and 5 (two of the lighter balls). If they're the same, 6 is the light ball. Otherwise, you will know which ball is lighter. If we generalize this to n balls, with n-1 identical, would the number of weighings be the smallest integer greater than or equal to log base 2 of n?
unless there is something i'm missing i think it is always going to be ceiling(log2(n)) because it looks an awful lot like a binary sort problem and i'm pretty sure that is a theoretical maximum on efficiency

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