Recent content by SammC

One last thing x1(t) = 5cos(400pi*t +0.5pi) x2(t) = 5cos(400pi*t -0.25pi) x3(t) = 5cos(400pi*t +0.4pi) x4(t) = 5cos(400pi*t - 0.9pi) I need to express each of those as complex exponentials.. then express the sum.. the only way i know how to do that is to use the fact that cos(x) =...

I just tried working 5b through for (1+I)^4: r = sqrt(1^2 + 1^2) = sqrt(2) theta = tan^-1(1/1) = pi/4 ... so (1 + i)^4 = [sqrt(2) * e^(i*pi/4)] ^ 4 apply euler's... = (sqrt(2) * [cos(pi/4) + i * sin(pi/4)])^4 ..simplify = (sqrt(2) * [sqrt(2)/2 + i * sqrt(2)/2])^4...

Okay.. so I'm a little stuck on 5b... [1 + i * sqrt(3)/2]^2 would i do: r = sqrt[1^2 + (sqrt(3)/2)^2] theta = tan^-1[(sqrt(3)/2)/1] = [r * e ^ (i * theta)] ^ 2

Theres a couple problems I am working on that involve complex numbers or euler's formula. e^+-(ix) = cos(x) +- isin(x) ----------------------- 1. A complex number can be written in rectangular coordinates as z = x+ jy. Write the relations to calculate the polar form, z = (r,theta) or z =...

Thanks Much

Homework Statement When the test for steroids is given to soccer players, 98% of players taking steroids test positive and 0.5% of the players not taking steroids test positive. Suppose that 5% of soccer players take steroids, what is the probability that a player who tests positive takes...

ah, this helps a bunch, thanks!

Homework Statement prove or disprove that E[X^2] = E(X)^2 Homework Equations E[X] = \sumxi*pr(xi) The Attempt at a Solution I really don't know where to start, I believe that it is not true, so I should try to disprove it, and the easiest way to do that would be by...

what about pr(1 ace red AND at least 2 aces) one is not necessarily in the other in this case, same with 1 ace diamond

Thanks, that makes sense. How do you go about getting the probability of the union though?

Homework Statement In the following you are given a 5-card hand from a 52 card deck. a) given that you have at least one ace, what is the probability you have at least 2 aces? b) given that you have the ace of diamonds, what is the probability that you have another ace? c) given that you...

But how do you determine what Q is mathematically?

Ah, I see. So you have 26 choices for A, 25 choices for B, 24 choices for C, and 23 choices for D. 26*25*24*23 = 358800? EDIT: Actually, the above is incorrect, A, B, C , and D can all be the same. is 26^4 correct? That seems like it would be over counting, or counting some possibilities...

If you purely want to choose all the possible combinations, it would be n!. The reason n=3 results in 1 is because you assume that the first and last element are connected (like in a necklace) and things that can be made from rotating that necklace or flipping it can't be counted again. So...

The problem statement Suppose you have n beads, each with a different color. You need to string these beads into a necklace. How many distinct necklaces can you make? (A necklace flipped over remains the same and does not count as a distinct necklace.) The attempt at a solution I...