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)

The area of every circle is (pi) (radius²) .
Radius = 1/2 of the diameter, so the area of your circle is
(pi) x (8 cm)² = 201 cm² . (rounded)
(pi) is not 3 . There's nothing wrong with approximating (pi),
but 3 is more than 4.5% wrong, and that's too much. There's
no reason why 3.14 should be too hard to handle.
You can use the slope intercept form for the line equation
Domain is (-infinity, infinity)
Range is [8, infinity)
Answer:
The length is 16 ft, and the width is 3 ft.
Step-by-step explanation:
Let L = length & let W = width.
The perimeter of a rectangle is
P = 2(L + W)
The area of a rectangle is
A = LW
We know the perimeter and the area, so we substitute those values int he equations above and we switch sides in both equations.
Perimeter: 2(L + W) = 38
Divide both sides by 2:
L + W = 19
Area: LW = 48
We have a system of two equations in two unknowns:
L + W = 19
LW = 48
Solve the first equation for L and substitute it into the second equation.
L = 19 - W
(19 - W)W = 48
19W - W^2 - 48 = 0
Multiply both sides by -1, and rearrange the order of the terms.
W^2 - 19W + 48 = 0
(W - 16)(W - 3) = 0
W - 16 = 0 or W - 3 = 0
W = 16 or W = 3
Use W = 3 to find L
L = 19 - W
L = 19 - 3
L = 16
Answer: The length is 16 ft, and the width is 3 ft.