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astra-53 [7]
3 years ago
14

Which line is parallel to a line that has a slope of 3 and a y-intercept at (0, 0)?

Mathematics
2 answers:
stich3 [128]3 years ago
8 0
Thats gonna be HJ <===
anyanavicka [17]3 years ago
7 0

we know that

If two lines are parallel , then their slopes are the same

we will proceed to calculate the slope of each line to determine the solution.

The formula to calculate the slope m between two points of the line is equal to

m=\frac{(y2-y1)}{(x2-x1)}

<u>case N 1) </u>line AB

Let

A(-4,3)\\B(4,3)

substitute the values

m=\frac{(3-3)}{(4+4)}

m=\frac{(0)}{(8)}

m=0

0\neq 3

therefore

<u>The line AB is not the solution</u>

<u>vase N 2)</u> line FG

Let

F(-3,-1)\\G(3,-3)

substitute the values

m=\frac{(-3+1)}{(3+3)}

m=\frac{(-2)}{(6)}

m=-1/3

-1/3\neq 3

therefore

<u>The line EG is not the solution</u>

case N 3) line CD

Let

C(-3,0)\\D(3,2)

substitute the values

m=\frac{(2-0)}{(3+3)}

m=\frac{(2)}{(6)}

m=1/3

1/3\neq 3

therefore

<u>The line CD is not the solution</u>

case N 4) line HJ

Let

H(-1,-4)\\J(1,2)

substitute the values

m=\frac{(2+4)}{(1+1)}

m=\frac{(6)}{(2)}

m=3

3=3

therefore

<u>The line HJ is  the solution</u>

therefore

<u>the answer is the option</u>

line HJ

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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

               

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Step-by-step explanation:

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