The Official Math/Physics/Whatever Homework questions thread

[smcdonn2]

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

Thanks for your help.

[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?

[Wyman]

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.

[smcdonn2]

I got teaser Indy +5.5 under 50

[mtlchris]

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?

[PaulieWlnuts]

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...

[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!

[thylacine]

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.

[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?

[Cactusmuffin]

[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?

[furyshade]

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.

[Wyman]

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.

[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}

[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)}

[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)