Recent content by mrmt

ooops...sorry The question is determining the curvature of a curve defined by a vector valued function K = curvature = ll T'(x) ll / ll r'(x) ll = ( ll r'(x) * r"(x) ll ) / ll r'(x) ll ^3

Stephen, That's still slowly sinking in - I think I need more coffee To be completely honest my brain is stuck on how we go from here: = P((X≥2) and (X≥1))/P(X≥1) to here: = P(X≥2)/P(X≥1). I know that's 101 somewhere but my brain just won't work - it's been something that's been bugging me...

Proof: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2) for y=f(x) Prove: K(x)= l f"(x) l / (1+(f'(x)^2)^3/2) r(x)= xi + f(x)j = <x , f(x)> r'(x)= 1i + f'(x)j= <1, f'(x)> T(x) = r'(x)/llr'(x)ll = <1, f'(x)> / ((1^2+(f'(x))^2)^1/2) This is where I start to get even more lost: T'(x)...

Hi guys, I can't get my head around this, if anyone could help that would be great. "A robotic assembly line contains 20 stations. Suppose that the probability that each individual station will fail is 0.3 and that the stations fail indepen- dently of each other. Given that at least one...

Thank you chiro, Your response helped me greatly and solidified my (lack of) understanding of a cdf. However, in the interest of discussion/further learning only I do have take you up on one point: The definition of a cdf is in fact a splitting of the probabilities. From wikipedia...

Hi Guys, Long time reader first time poster... This simple question has stumped me all day and I think I've finally cracked it! I'm hoping someone can confirm that, or tell me how wrong I am - either is fine :) One in 1000 cows have a rare genetic disease. The disease is not contagious...