Recent content by geno678

So it ends up becoming s^2 X(s) - s X(0) - \dot{x}(0) Which in turn becomes s^2 X(s) - 2s +3

Oh I see it now that's why I've been messing up. Thanks.

Solving an IVP using Laplace Transforms. HELP! Ok I'm supposed to Solve this problem using Laplace Transforms. \frac{d^2 x}{dt^2}+2\frac{dx}{dt}+x = 5e^{-2t} + t Initial Conditions x (0) = 2 ; \frac{dx}{dt} (0) = -3 so I transformed the the IVP and it looks like this s^2 x(s) - s...

Ok I plugged in all of my variables, but the equation is set up to find the success of the parts. Do I need to switch up some variables?

Ok I get it now. Their are 8 trials because, 1 wheel bearing can be good or bad. That's 2 trials for 1 wheel bearing. For four wheel bearings, you get 8 trials. And the number of successes are 4.

I get it. It's just 90%. Because it's consistent for every trial.

Am I way off??

If they aren't all good it will be. (.1)*(.1)*(.1)*(.1) = .01%

For some reason I was thinking you had 1 out 4 chances of a bearing being right, that's why I said 1/4. My mistake.

Ok wait a minute. So 90% probability for each bearing that it is good. Then I guess for four bearings it would be (9/10)*(9/10)*(9/10)*(9/10) = 65.61%

The probability that all four are good is. (1/4)*(1/4)*(1/4)*(1/4) = 1/256 = .4 chance that all four are good

Ok I think I got it, they said 10 percent were defective out of a large bin. They chose 4. If I multiply 4 * .1 = .4 that is 4 out of 10 that are defective. Then this must mean 5/10 are successful. That means my p isn't .9 it is .5

Help! Binomial Distribution: Statistics for M.E's Homework Statement Four wheel bearings are to be replaced on a company vehicle. The mechanic has selected the four replacement parts from a large supply bin in which 10% of the bearings are defective and will fail within the first 100 miles...

Awesome. Thank you for your help!

Then that means the answer is E(y) = 3(3.5) + 5(4.5) + 4(5.5) + 6(6.5) = 94 Y = 3 X1+5 X2+4 X3+6 X4.