Recent content by stevecallaway

Homework Statement 10 face cards are face down in a row on a table. Exactly one of them is an ace. You turn the cards over one at a time, moving from left to right. Let X be the random variable for the number of cards turned before the ace is turned over. What is the probability function...

Homework Statement How many different bridge hands are possible containing five spaids, three diamonds, three clubs, and two hearts? Homework Equations The Attempt at a Solution Total number of hands in which I can get 5 spaids is 13C5 Total number of hands in which I can get...

.002xe^(-.002x) - .2e^(-.002x). Ok. u=.002x du=dx v= -(1/.002)e^(-.002x) dv=e^(-.002x) so then, (.oo2x)(-(1/.002)e^(-.002x)- Integral of -(1/.002)e^(-.002x) dx from 100 to infiniti. (.oo2x)(-(1/.002)e^(-.002x)...

(x-100)(.002e^(-.002*x)) I first multiplied out these two and got .002xe^(-.002x) - .2e^(-.002x). Then integrating the .002xe^(-.002x) I integrated the e part first and then the x part to get -xe^(-.002x) * e^(-.002x) and then I integrated the .2e^(-.002x) and got -100e^(-.002x).

Homework Statement find the integral from 100 to infiniti for (x-100)(.002e^(-.002*x)) Homework Equations The Attempt at a Solution.(xe^-(.002x))(e^(-.002x))-(-100e^(-.002x)) |from 100 to infiniti = -(100e^-.2)(e^-.2)+100e^(-.2)=14.84 the answer should be 409.37. What am I...

Homework Statement the lifetime of a machine part has a continuous distribution on the interval(0,40) with probability density function f, where f(x) is proportional to (10 + x)^(-2) Calculate the probability that the lifetime of the machine part is less than 6. Homework Equations...

Homework Statement Ten cards are face down in a row on a table. Exactly one of them is an ace. You turn the cards over oen at a time, moving from left to right. Let X be the random variable for the number of cards turned before the ace is turned over. What is the probability function for...

Homework Statement Let S be the sample space for rolling a single die. Let A={1,2,3,4}, B={2,3,4}, and C={3,4,5}. Which of the pairs (A,B), (A,C), and (B,C) is independent? Homework Equations P(A|B)=P(A) P(A|B)=P(A&B)/P(B) P(A&B)=P(A)*P(B) The Attempt at a Solution P(A)=2/3...

Homework Statement .002x - .000001x^2 = .50 Homework Equations -b+-sq.rt.((b^2)-(4ac))/2a [b]3. The Attempt at a Solution Plugging a=-.000001, b=.002, and c=-.5 does not get the the correct answer. x is supposed to be 292.89. I can't remember any other way of going about...

Homework Statement Two cards are drawn from a standard deck with replacement. A=first card is an ace. B=second card is an ace. Show that A and B are independent Homework Equations P(A and B)=P(A given B)/P(B) P(A given B)=P(A) The Attempt at a Solution P(A)=4/52 P(B) =4/52 P(A...

u^2=sqrt(z)...got it. Brilliant. Thanks.

Sorry about the garbled-ness. I thought I put more spaces in between the equations. Yes, I did substitute u=z and then I dividev both sides by 16 to get (x^2+y^2)/16=z^2. Then I took the sqrt of both sides which would give z=sqrt((x^2+y^2)/16) but the book gives an answer of z=(x^2+y^2)^2/16

I was in the middle of editing when you two responded, will you please re-look at my attempt at a solution?

Homework Statement r(u,v)=2ucosv i + 2usinv j + u^4 0<=u<=1 0<=v<=2pi Homework Equations don't care about the graph as much as how the book got to the answer equation z=(x^2+y^2)^2/16 [b]3. The Attempt at a Solution [/b cosu=x/2v sinu=y/2v cosu^2+sinu2=1 (x/2u)^2+(y/2u)^2=1...

I think I see a relation between y and z and that is y=2z. So is that all that I'm supposed to do is find a relationship from among the original equation and have that equal to zero? Because y-2z=0 is supposed to be the answer, but what happens to the u i?