Eight bungee cord for $10 or 20 bungee cords for $20

At an airport, it cost 7 to park for one hour and 5 per hour for each additional hour.Let x represent the number hours parked.wr

kompoz [17]

The cost of parking is an initial cost plus an hourly cost.
The first hour costs $7.
You need a function for the cost of more than 1 hour,
meaning 2, 3, 4, etc. hours.
Each hour after the first hour costs $5.

1 hour: $7
2 hours: $7 + $5                  = 7 + 5 * 1          = 12
3 hours: $7 + $5 + $5          = 7 + 5 * 2          = 17
4 hours: $7 + $5 + $5 + $5 = 7 + 5 * 3          = 22

Notice the pattern above in the middle column.
The number of $5 charges you add is one less than the number of hours.
For 2 hours, you only add one $5 charge.
For 3 hours, you add two $5 charges.

Since the number of hours is x, according to the problem, 1 hour less than the number of hours is x - 1.

The fixed charge is the $7 for the first hour.
Each additional hour is $5, so you multiply 1 less than the number of hours,
x - 1, by 5 and add to 7.

C(x) = 7 + 5(x - 1)

This can be left as it is, or it can be simplified as

C(x) = 7 + 5x - 5

C(x) = 5x + 2

Answer: C(x) = 5x + 2

Check:
For 2 hours: C(2) = 5(2) + 2 = 10 + 2 = 12
For 3 hours: C(3) = 5(3) + 2 = 15 + 2 = 17
For 4 hours: C(3) = 5(4) + 2 = 20 + 2 = 22
Notice that the totals for 2, 3, 4 hours here
are the same as the right column in the table above.

6 0

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

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

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

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

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

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