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musickatia [10]
3 years ago
11

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
Mathematics
1 answer:
Maksim231197 [3]3 years ago
3 0

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 P(x_1 + x_2 \ge 2 )= 0.5940

Step-by-step explanation:

From the question we are told that

     The rate at which fire breaks out every 10 years is  \lambda  =  1

  Generally the probability distribution function for Poisson distribution is mathematically represented as

               P(x) =  \frac{\lambda^x}{ k! } * e^{-\lambda}

Here x represent the number of state which is  2 i.e x_1 \ \ and \ \ x_2

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 \le 1 )

=>        P(x_1 + x_2 \ge 2 ) =  1 - [P(x_1 + x_2 = 0 ) + P( x_1 + x_2 = 1 )]

=>        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)]

=>  P(x_1 + x_2 \ge 2 ) =  1 - P(x_1 = 0)P(x_2 = 0 ) + P( x_1 = 0 ) P( x_2 = 1 )+ P(x_1 = 1 )P(x_2 = 0)

=>    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}]]) \}

=>   P(x_1 + x_2 \ge 2 )= 1- [[0.3678  * 0.3679] + [0.3678  * 0.3679] + [0.3678  * 0.3679]  ]

P(x_1 + x_2 \ge 2 )= 0.5940

               

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Masteriza [31]

Answer:

(2,-5) and (2,1).

Step-by-step explanation:

We need to find the find all the points having an x coordinate of 2 whose distance from the point (-2,-4) is 5.

A circle with center (-2,-4) and radius 5 represents all the points whose distance from the point (-2,-4) is 5.

Standard form of a circle is

(x-h)^2+(y-k)^2=r^2

where, (h,k) is center and r is the radius.

(x-(-2))^2+(y-(-4))^2=(5)^2

(x+2)^2+(y+2)^2=25

Now, put x=2.

(2+2)^2+(y+2)^2=25

(4)^2+(y+2)^2=25

(y+2)^2=25-16

(y+2)^2=9

Taking square root on both sides.

y+2=\pm\sqrt{9}

y+2=\pm3

y+2=-3\text{ or }y+2=3

y=-5\text{ or }y=1

So, the y-coordinates of the points are -5 and 1.

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Answer:

Looking for four values or answers

(A) 5.05

(B) 13

(C) $254,140.33

(D) $254,145.38

Step-by-step explanation:

(A) The value of the margin of error.

Using a 90% confidence level or 0.10 alpha level,

1 - alpha = 1 - 0.10 = 0.90

The degrees of freedom = n - 1 = 14 - 1 = 13

Using the t table, 0.90 under 13 is 1.350

Sample size divided by √n is equal to

14/√14 = 3.742

1.35 × 3.742 = 5.05

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Adding the 14 values together, you have $3,558,000

Dividing by 14 to get the mean;

Mean = $254,142.8571

Lower Limit: $254,140.33

Upper Limit: $254,145.38

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From the figure, we have:

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