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Veronika [31]
2 years ago
5

Solve two sevenths times four sixths . (2 points) Group of answer choices six forty seconds eight forty seconds six thirteenths

eight thirteenths
Mathematics
1 answer:
Maksim231197 [3]2 years ago
8 0

Given:

The statement is "two sevenths times four sixths".

To find:

The value of the product.

Solution:

Two sevenths times four sixths can be written as:

\dfrac{2}{7}\times \dfrac{4}{6}

It can be rewritten as:

\dfrac{2}{7}\times \dfrac{4}{6}=\dfrac{2\times 4}{7\times 6}

\dfrac{2}{7}\times \dfrac{4}{6}=\dfrac{8}{42}

The answer of given expression is eight forty seconds.

Therefore, the correct option is B.

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Assuming the incident of fires for individual reactors can be described by a Poisson distribution, what is the probability that
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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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