Answer:
203.20 millimeters
Step-by-step explanation:
StartFraction 8 inches Over question mark millimeters EndFraction = StartFraction 40 inches Over 1,016 millimeters
8 inches / ? Millimeters = 40 inches / 1,016 millimeters
8 × 1,016 = ? × 40
8,128 = 40?
Divide both sides by 40
? = 8,128 / 40
= 203.20
? = 203.20 millimeters
There are 203.20 millimeters in 8 inches
Complete Question
The Brown's Ferry incident of 1975 focused national attention on the ever-present danger of fires breaking out in nuclear power plants. The Nuclear Regulatory Commission has estimated that with present technology there will be on average, one fire for every 10 years for a reactor. Suppose that a certain state has two reactors on line in 2020 and they behave independently of one another. Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that by 2030 at least two fires will have occurred at these reactors?
Answer:
The value is 
Step-by-step explanation:
From the question we are told that
The rate at which fire breaks out every 10 years is
Generally the probability distribution function for Poisson distribution is mathematically represented as

Here x represent the number of state which is 2 i.e 
Generally the probability that by 2030 at least two fires will have occurred at these reactors is mathematically represented as

=> ![P(x_1 + x_2 \ge 2 ) = 1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5BP%28x_1%20%2B%20x_2%20%3D%200%20%29%20%2B%20P%28%20x_1%20%2B%20x_2%20%3D%201%20%29%5D)
=> ![P(x_1 + x_2 \ge 2 ) = 1 - [ P(x_1 = 0 , x_2 = 0 ) + P( x_1 = 0 , x_2 = 1 ) + P(x_1 , x_2 = 0)]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5B%20P%28x_1%20%20%3D%200%20%2C%20%20x_2%20%3D%200%20%29%20%2B%20P%28%20x_1%20%3D%200%20%2C%20x_2%20%3D%201%20%29%20%2B%20P%28x_1%20%2C%20x_2%20%3D%200%29%5D)
=> 
=> ![P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%20%3D%20%201%20-%20%5C%7B%20%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%20%29%2B%20%28%20%5B%20%5Cfrac%7B1%5E1%7D%7B1%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E1%7D%7B%201%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%20%29%20%2B%20%28%20%5B%20%5Cfrac%7B1%5E1%7D%7B%201%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%20%2A%20%5B%5B%20%5Cfrac%7B1%5E0%7D%7B%200%21%20%7D%20%2A%20e%5E%7B-1%7D%5D%5D%29%20%5C%7D)
=> ![P(x_1 + x_2 \ge 2 )= 1- [[0.3678 * 0.3679] + [0.3678 * 0.3679] + [0.3678 * 0.3679] ]](https://tex.z-dn.net/?f=P%28x_1%20%2B%20x_2%20%5Cge%202%20%29%3D%201-%20%5B%5B0.3678%20%20%2A%200.3679%5D%20%2B%20%5B0.3678%20%20%2A%200.3679%5D%20%2B%20%5B0.3678%20%20%2A%200.3679%5D%20%20%5D)
