A repair team is responsible for a stretch of oil pipeline 2 miles long. The distance (in miles) at which any fracture occurs can be represented by a uniformly distributed random variable, with probability density function f(x) = 0.5. What is the probability that the next fracture will occur in the last half mile of the two mile interval?

If p(x) is the total value of the production when there are x workers in a plant, then the average productivity of the workforce at the plant is a(x) = p(x) x . (a) find a'(x). a'(x) = xp'(x) − p(x) x a'(x) = xp'(x) − p(x) x2 a'(x) = p'(x) − p(x) x a'(x) = xp'(x) − p'(x) x2 a'(x) = p'(x) − xp(x) x2 why does the company want to hire more workers if a'(x) > 0? a'(x) > 0 ⇒ a(x) is ; that is, the average productivity as the size of the workforce increases. (b) if p'(x) is greater than the average productivity, which of the following must be true? p'(x) − xp(x) > 0 p'(x) − xp(x) < 0 xp'(x) − p'(x) > 0 xp'(x) − p(x) < 0 xp'(x) − p(x) > 0