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satela [25.4K]
3 years ago
11

a train arrives at the station with 150 passenger's on board . 2/5 of the passengers get off the train in Seattle, and then 35 p

assengers board the train. how many passengers are on the train when it leaves the station
Mathematics
1 answer:
hichkok12 [17]3 years ago
5 0
First, We Have Find What Two Fifths Of 150. So, Divide 150 By Five To Find 1/5 Of 150. 
150 ÷ 5 = 30.
 
Now, To Find 2/5 Of 150, Multiply 30 By Two. So:
30 * 2 = 60
So, 60 = 2/5 of 150.

So, subtract 60 from 150. 
150 - 60 = 90.

So, now we need to add 35, because 35 people get back on. 

90 + 35 = 125

There are 125 passengers when the train leaves.
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Each trial is independent.

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P(X=x)=C(100,x)0.5^x(0.5)^{100-x},x=0,1,2,3,...,100 \\ P(X=x)=C(100,x)(0.5)^{100},x=0,1,2,3,...,100

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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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