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Wittaler [7]
3 years ago
13

Suppose the scores on a trivia quiz are normally distributed with a mean of 80 and a standard deviation of 5.

Mathematics
1 answer:
Kitty [74]3 years ago
8 0

I think it would be scores below 75

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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
Which is equal to 12.74 divided by 13
Shkiper50 [21]
12.74 divided by 13 is 0.98. 

Hope that helped:)
7 0
3 years ago
Read 2 more answers
A fisherman can row upstream at 4mph and downstream at 6mph. He started rowing upstream until he got tired and then started down
ikadub [295]
Total time = upstream time + downstream time
he rowed the same distance both ways.
upstream time = D/4
downstream time = D/6
total time = D/4 + D/6 = 3D/12 + 2D/12 = 5D/12 = 3 hours
D = 3*12/5 = 36/5 or 7.2 miles
where D is the one-way distance. The total distance he rowed is 2*D = 72/5 or 14.4 miles
6 0
3 years ago
Read 3 more answers
A rectangle is 11 inches long and 8 inches wide. Find its area.
poizon [28]

Answer:

the answer is A:88 square inches

8 0
3 years ago
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Please help Which of the following inequalities is graphed on the coordinate plane?
Gekata [30.6K]

Answer:

The inequality which represents the graph is y ≤ -2x + 1 ⇒ A

Step-by-step explanation:

To solve the question you must know some facts about inequalities

  • If the sign of inequality is ≥ or ≤, then it represents graphically by a solid line
  • If the sign of inequality is > or <, then it represents graphically by a dashed line
  • If the sign of inequality is > or ≥, then the area of the solutions should be over the line
  • If the sign of inequality is < or ≤, then the area of the solutions should be below the line

Let us study the graph and find the correct answer

∵ The line represented the inequality is solid

∴ The sign of inequality is ≥ or ≤

→ That means the answer is A or B

∵ The shaded area is the area of the solutions of the inequality

∵ The shaded area is below the line

∴ The sign of inequality must be ≤

→ That means the correct answer is A

∴ The inequality which represents the graph is y ≤ -2x + 1

3 0
3 years ago
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