Also, we just need to set up an equation (without solving) that models the volume obtained by rotation the region bounded by y=x, y=tanx and x=pi/3 around the y axis.

Am i correct here: the integral from 0 to pi/3 of 2pi(tan(x)-x)x

Thanks again.

You are given that (G,*) is an abelian group, with a,b in G. Assume that a has order m, and b has order n, with m and n relatively prime. Prove that Order(a*b) = mn.

I can show that (a*b)^mn = id, but I'm not really sure how to prove that mn is the smallest possible exponent that yields the identity. I was thinking of a proof by construction. Assume d is the order of a*b. Then, 0<d<=mn. So, (a*b)^(mn-d) = id, so mn must be a multiple of d. Without loss of generality, say m<n. Now, I just have to show that n<d. Then, d=mn, and I'm done.

Is there any easy way to show this?

I got 21.(3)

same result with rotated y-x, for easier integrals.

Let d be the order of ab. Then 1=(ab)^d=a^d * b^d => a^(-d)=b^d => a^s = b^t for some 0<=s<m, 0<=t<n, where we have used the fact that a^m=b^n=1. It's easy to show that we cannot have s=0 or t=0. Taking both sides to the power m gives
a^(sm) = b^(tm).
The LHS is (a^m)^s = 1, so 1 = b^(tm) and we must have n | tm. But this is impossible since 1<=t<n and m and n are coprime.

Does this make sense?

Yes, that makes sense. Thanks!

I'll go ahead and be a nit. I know what you mean, but a^(-d) makes no sense (you mean (a^(-1))^d, or a^(|a|-d). Just be careful when you write this up. It only makes sense to take positive integer powers of group elements.

There are nCr = 15 choose 5 = 231 combos of five knickknacks out of fifteen. There is only one combination in which that they are all blue. So, 1 / 231? That's number of acceptable combos (number of combos of {blue, blue, blue, blue, blue}) divided by the number of possible combos (15 C 5)? I think that's right but I'm still new to this.

The second one, I think I just have to find the probabilities that there will only be one color remaining and the probabilities that we will have three colors remaining, and subtract those from 1.

The probability of the last five being a single color is 3/231, because there are still 231 ways to make a combo of five and only three combos will consist of all one color. But how do I tell how many combos of 15C5 will have three different colors?

So far this class seems simple enough, but counting problems are killing me. They seem so easy when I follow my professor as he does them but then I try them on my own and have no idea what to do.

I am solving the following problem:

"A New Car Dealer calculated that the company must average more than 4.8% profit on the sales of its alotted new cars. A random sampling of n=80 cars gave a mean and standard deviation of the percentage profit per car of Xbar = 4.87% and s=3.9%

a) Does the data provide sufficient evidence to indicate that the sales manager's policy in approving sale prices is achieving a mean profit exceeding 4.8% per car? Use alpha=0.01

b) Find the p-value for this test and explain your results.

Therefore:

N=80
Xbar=4.87%
standard deviation (s) = 3.9%
alpha = 0.01

because n is greater than or = to 30 it is a "Z" test.

Ha: M<= to 4.8
Ho: M> 4.8 (claim)

I do the test statistic for Z and get the answer TS= 0.16 which is = to Z score of .0636

Then when you are graphing the bell curve, you do half of the curve (.5-alpha) which is .5-.01=.49

.49 = Z score = 2.33 Therefore Zalpha = 2.33 and .16 is in the AR of .49, so Fail to Reject Ho.

I am okay up until this point. However when my teacher found the P value I am totally unaware of how she got it, she has:

p-value:

p(Z>0.16) = 0.5-0.0636 = 0.4364
P-value = 0.4364 > alpha = 0.01
Do Not Reject Ho

-------------------------------------
I notice that the .0636 is the test statistic Z value, and that .5 is half of the bell curve, but why is this computation happening? and is it because it is larger than alpha that the original hypothesis is not rejected? I don't get it. Please help.

Thanks.

Danny

You're doing the same thing from different angles. You are calculating how high the z-score would have to be to be significant at the .01 level (Z>=2.33). Then you see the Z-score is only .16, and you know the difference isn't significant.

The teacher calculated the z-score of .16. Looking up in a normal distribution table, the probability of being between the mean, and .16 above the mean, is .0636. So the odds of being even higher than that is 1 - .5 (all numbers below the mean) - .0636 (between mean and z=.16 above) = .4364. Since you're looking for a .01 probability of a random sample being that high before calling it significant, it fails.

Since you're actually being asked for the p-value in addition to significant/insignificant, there's not much point in doing what you did, since finding the actual p-value duplicates the significant/insignificant part that you did.

What? I read your post twice and I am more confused. Basically to get the P value all I have to do is take the .5 (the half of the bell curve) and subtract the test statistic value from it (.16 which translates to .0636) in order to get the P value (0.5-0.0636)=0.4364 ??

What I am having the most trouble understanding is by the wording you are using as to discern whether or not the P value accepts or rejects the null hypothesis.

when you say "the probability of being between the mean and .10 above the mean, what is the value of the mean? I am totally confused..

Ignore this; Solved.

This isn't for homework, just my own pondering. Just want to make sure my math isn't defunct.

What is the probability of being dealt at least two clubs in an Omaha poker hand. Assume no dead cards, and assume no flop, turn, or river.

If first card is a club
13/52 * [1 - (39/51 * 38/50 * 37/49) = 1 - (54834/227449) = .76] = .19

If not first, then second card is a club
13/51 * [1 - (38/50 * 37/49) = 1 - (1406/2450) = .43] = .11

If not first or second, then third card is a club
13/50 * 12/49 = .06

1 - [(1-.19) * (1 - .11) * (1- .06)] = .32

Is this correct?

Also, if I were sitting in a game, looking at a board that had exactly 3 clubs on it, then I should change my math to reflect that there were initially only 10 clubs to be dealt, and if I have 2 clubs myself, then 8. Is this also correct? Or should I disregard that and assume that there is a .32 probability that my opponent has at least 2 clubs? That doesn't sound quite correct.

Thanks!

The number of ways to get dealt no clubs is 39c4 (39 cards in the other suits, 4 cards to pick).

The number of ways to get dealt exactly one club is (39c3)*(13c1) (picking 3 cards from a pile of not clubs, and one from a pile of clubs).

The probability of being dealt 2 or more clubs is 1 minus the probability of either of the above happening. There are 52c4 possible omaha hands, so:

So P (>=2 clubs) = 1 - (82251+118807)/270725 = 0.257

If you have 2 clubs, the board has 3 clubs, and you want to know the probability that your opponent has 2 clubs (let's say you have the lowest possible flush), you have to use something called Bayes' Theorem, which helps you calculate the probability of an event given that some other probabilistic event has occured.

I can't exactly solve this problem now, but I'll try and get back to it tomorrow.

Yup! If it's the river and the board has exactly 3 clubs and your hand has no clubs in it, you can calculate the probability that an opponent has at least two clubs exactly the way Myrmidon did, but assume there are only 52-(4+5)=43 cards in the deck, of which 13-3=10 are clubs. Bayes' theorem need not apply.

It's fairly easy to show that the Poisson process has no double jumps a.s., so the only contribution to the quadratic variation are the (single) jumps. Each jump contributes one to the quadratic variation and the number of jumps by time t _is_ N_t, so ...

After some more thinking, this is only the probability that there are exactly two clubs in a hand. I would have to adjust for the times that the hand has exactly 3 clubs and the times that it has exactly 4 clubs. Grrr....

Questions:

Solve the following equations:

(2/x+1)=(2x/(2x-3))+3

I've messed around with the equation to get to:

(2x^2-2x+3)
---------------- = 0
(2x^2-x-3)

How do I proceed?

Well, if a/b = 0, then a=0...

You've lost me.

I could multiply by the denominator and get

(2x^2-2x+3)*(2x^2-x-3)=0

Then do this maybe:?

2x^2(2x^2-x-3)
+
-2x(2x^2-x-3)
+
3(2x^2-x-3)
=0

There must be an easier way.

Factor the numerator and denominator (or at least find their zeros individually... quadratic formula works.)
Since you have a rational polynomial equation, when the numerator=0, the whole thing=0. But you do have to check to make sure a zero for the numerator is not the same for the denominator, otherwise it is not a solution (since you'd be dividing by 0).

Actually, looking at the original problem suggests it already WAS factored, you just multiplied it all out. So go back 1 step to where it was factored and find the zeros there.

If you multiply both sides by the denominator, the LHS is just the numerator, and the RHS is just zero.

(edit: once you find the x's that make the numerator 0, you'll have to plug them into the denominator to make sure that they don't simultaneously make the denominator zero, but this is irrelevant here since there are no real zeros; there are 2 complex solutions)

That was quite an oversight by me to forget that a/b=0 so a=0 so 2x^2-2x+3=0 and I can use the quadratic formula. Thanks.

so my professor kind of explained permutation cycles but didn't really tell us how to do the types of problems we have in homework.

how many permutations on a,b,c,d,e,f have exactly 3 cycles? of these, how many hve structure (ab)(cd)(ef), how many have (abc)(de)(f), how many have (abcd)(e)(f)

the first part i figured out its just stirling number of the first kind for s(6,3) which i got to be 255, the rest i'm not sure if there is a trick or if i just have to go through and enumerate them

the way i understand it i get that there is one of each but i feel i may not completely understand how to generate these

For the structure (ab)(cd)(ef) there are 15 unique ways to choose (ab), then with each of those, 6 ways to choose (cd), then 1 way to choose (ef), so you get 90 different permutations. But, the order of (ab),(cd) and (ef) doesn't matter, so you have to divide the 90 by 6, giving you 15 different permutations of that structure.

The others can all be done similarly.