Suppose that a die-hardly-ever battery has an exponential time-to-failure distribution with a mean of 48 months. at 60 months, the battery is still operating. what is the probability that the battery is going to last for the next 12 months?

Since the time to failure distribution is exponential, the distribution is F (n) = 1 – e ^ (-x/48)

The probability of failure between months 60 and 72 is the same as the probability that the battery will fail in 12 months, since the exponential distribution is memoryless F (12) = 0.221

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2

maybe showing the graph would

step-by-step explanation:

-387129 iq