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Dare

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A truck of mass 5000kg is travelling down the highway when it breaks suddenly. The unbalanced force F acting on the truck when
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I think my calculation for the second part was wrong. If the third arrival comes at t=1, then the first two arrivals are uniformly distributed on [0, 1], so we should have E[T] = 1/3. Recalculating based on the order statistics gives 2 - 2t, which matches
Is the first part supposed to be a constant? Is that what I should be looking for?
I think the second part is 3t^2 - 6t + 3.
I think the answer others have gotten for part 3 is slightly off: P(1100 ≤ P ≤ 1200) = P((1100 - 1080)/sqrt(1080) ≤ Z ≤ (1200-1080)/sqrt(1080)) = P(20/sqrt(1080) ≤ Z ≤ 120/sqrt(1080)) = P(Z ≤ 3.65148...) - P(Z ≤ 0.060858...) = 0.99987 -
p01 = P(1 after day i+1 | 0 after day i). This can only happen if Bob receives one more message than he has time to answer, so p01 = q1*p0 + q2*p1 = 0.5*0.2 + 0.4*0.3 = 0.1 + 0.12 = 0.22. p11 = P(1 after day i+1 | 1 after day i). Bob goes off to work on

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