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3 years ago

How just how..........

Mathematics

2 answers:

Basile [38]3 years ago

7 0

Answer:

-49/22, decimal form:-2.227

Step-by-step explanation:

vredina [299]3 years ago

5 0

Step One

Change to improper fractions.

1 3/11 = 14/11

1 3/4 = 7/4

Step Two

Find the common denominator

11 = 11 * 1

4 = 2 * 2

The common denominator = 11 * 2 * 2 = 44

Step 3

change the improper fractions to the common denominator.

$\dfrac{14}{11} = \dfrac{14 * 4}{11 * 4} = \dfrac{56}{44}$

$\dfrac{7}{4} = \dfrac{7 * 11}{11 * 4} = \dfrac{77}{44}$

$\dfrac{56}{44}-\dfrac{77}{44} = \dfrac{-21}{44}$

-21/44 That has to be your final answer.

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3 years ago

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

Maksim231197 [3]

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

$P(x_1 + x_2 \ge 2 ) = 1 - P(x_1 + x_2 \le 1 )$

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

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

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

=> $P(x_1 + x_2 \ge 2 ) = 1 - \{ [ \frac{1^0}{ 0! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]] )+ ( [ \frac{1^1}{1! } * e^{-1}] * [[ \frac{1^1}{ 1! } * e^{-1}]] ) + ( [ \frac{1^1}{ 1! } * e^{-1}] * [[ \frac{1^0}{ 0! } * e^{-1}]]) \}$

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

$P(x_1 + x_2 \ge 2 )= 0.5940$

3 0

3 years ago