Recent content by lunds002

right, i forget that. thank you

Okay so I know z1/z2 = r1/r2 cis (theta-\psi) I don't really understand how that applies here though And you're right, I do know that arg(i) = -1

Homework Statement A box contains 22 red apples and 3 green apples. 3 apples are selected at random, one after the other, without replacement. (a) The first two apples are green. What is the probability that the 3rd apple is red? Homework Equations The Attempt at a Solution...

Homework Statement In a bilingual school there is a class of 21 pupils. In this class, 15 of the pupils speak spanish as their first language and 12 of these 15 pupils are Argentine. The other 6 pupils in the class speak english as their first language and 3 of these 6 pupils are argentine...

Okay so then I get arg(z1^k) + arg(z2) = pi 2i (pi/6)^k + 2pi = pi 2i (pi/6)^k + 2pi = 2i(pi) Not sure what to do with the imaginary i

Homework Statement Let z1 = a (cos (pi/4) + i sin (pi/4) ) and z2 = b (cos (pi/3) + i sin (pi/3)) Express (z1/z2)^3 in the form z = x + yi. ]2. Homework Equations [/b] The Attempt at a Solution a(cos (pi/4) + i sin (pi/4)) b (cos (pi/3) + i sin (pi/3)) I then multiplied...

Oops, should be arg ( z1 x z2^k) / (2i)

Thanks, it just seemed too simple!

Homework Statement From January to September, the mean number of car accidents per month was 630. from Oct - Dec the mean was 810 accidents per month. What was the mean number of car accidents per month for the whole year? Homework Equations The Attempt at a Solution...

Homework Statement Given that arg((z1z2)/2i) = \pi, find a value of k. Homework Equations arg z2=2\pi=0 arg z1=\pi/6 The Attempt at a Solution ((\pi/6)^k x (2\pi))/ (2i) = \pi I'm not sure what to do with the imaginary number i..

Great! Thanks so much for helping me through that, it was a bit of a struggle for me.

Wait so my answer was correct?

Okay I'll give it another shot. A=event that the person has the disease B=event that the test is positive (i.e. says that the person has the disease) so P(A|B) = P(A) x P(B|A) P(A) x P(B|A) + P(A') x P(B|A') P(A) = .0001 P(B|A) = .99 P(A') = 1 - P(A) = .9999...

(.0001 x .99) / ((.0001 x .99) + (.01 x .999)) = 0.009 Okay I don't think this is right either, but I got the (.01 x .999) by taking 1 - P(correctly identifies with disease) to get P(incorrectly identifies with disease)=.01. Then I took 1 - P(has disease) to get P(don't have disease)=.999

Yes that was the correct interpretation. Okay, so maybe this makes more sense: (.0001 x .99) / ((.0001 x .99) + (.95 x .01)) = 0.0103