The length of time between breakdowns of an essential piece of equipment is important in the decision of the use of auxiliary equipment. An engineer thinks that the best model for time between breakdowns of a generator is the exponential distribution with a mean of 14 days. If the generator has just broken down, what is the probability that it will break down in the next 21 days

The probability that it will break down in the next 21 days is 0.777

Step-by-step explanation:

Let the random variable x represent

time between breakdowns of a generator

Given that mean = 14 days,

Decay parameter, m = 1/14

The probability density function for exponential distribution is expressed as

F(x) = me-mx

It becomes

f(x) = 1/14e-(x × 1/14)

P(X < x) = 1 - e- mx

The probability that it will break down in the next 21 days is expressed as

P(x ≤ 21) = P(x < 21)

Therefore,

P(x < 21) = 1 - e-(1/14 × 21)

= 1 - e-1.5

= 1 - 0.223 = 0.777

P(x ≤ 21) = 0.777

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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b.my answer is b.

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