Recent content by masterchiefo

Not really sure how to answer the load attached question, this is all I know about the problem. I believe there is load. I forgot to mention that it is also possible that some state are impossible. For state 1, that would mean Vo = R*I For state 2, I would say current would go to Vo, as D1...

They are perfectly ideal, with vf = 0.7 question is to find Vo in every state as stated in the first post. Each state is an independent problem.

Each state is independent from each other, its like 4 different problems. In state 1, both are off so its like there is no diode at all and the second resistor is out as well. which means Vo = Vi for state 2, I honestly have no idea how to do it. It confuse me a lot. Can you give me some tips...

Homework Statement I have to find Vo in every diode state possible. We know diode Vf = 0.7 Homework Equations Va-Vc < 0.7 = diode OFF Va-VC > 0.7 = diode ON The Attempt at a Solution State 1: if D1 OFF and D2 OFF Vo will be equal to Vi State 2: if D1 OFF and D2 ON Vo will be equal to...

anyone can help me :(

Homework Statement [/B] A company produces batches of 1800 axle shafts . These are tested to determine the proportion of those with too rough surface according to the standards set in the industry. A) The quality control department of a sample of 150 axle shafts in a lot and concluded that...

I did that: p=(15%+25%)/2 =0.2 s= √p(1−p) = √0.2(1−0.2) = 0.4 t=(ME)/((s/sqrt(n))* (sqrt(1-(n/N))) t = 0.05/((0.4/sqrt(150))* (sqrt(1-(150/1800))) (1-sigma)=tcdf(-t,+t,n-1) (1-sigma) = 1 It does not work :( anyone please help me.

do I still use Student -t formula ?

Hello, I don't understand, what would be p ?

Homework Statement [/B] A company produces batches of 1800 axle shafts . These are tested to determine the proportion of those with too rough surface according to the standards set in the industry. The quality control department of a sample of 150 axle shafts in a lot and concluded that there...

Because the probability is always 0.5 and in Binomial the probability does not change every trials, stays the same. P is constant in Binomial while in Hypergeomtric its not constant.

I do not really appreciate the fact that I proposed I would use binomial with n p q and being told this is not how we do this problem. and also I've said multiple times n=8 and you have told me this is incorrect and made me feel like a retard because at the end now you are saying its a binomial...

This is the answer: Variance = p*n*q = 8*0.5*0.5 =2 I Cant believe it was that sample and I took 3 days... I complicated my life for nothing trying to do some complex prob with multiple formulas.

:OOOO I think I understand now wow