An automatic car wash takes exactly 5 minutes to wash a car. On average, 10 cars per hour arrive at the car wash. Suppose that, 30 minutes before closing time, five cars are in line. If the number of cars in any time interval follows a Poisson distribution, and if the car was is in continuous operation until closing time, what is the probability that anyone will be in line at closing time

0.9595724

Step-by-step explanation:

Time taken to wash a car (μ) = 5 minutes

Hence, Number of cars washed per hour = 60/5 = 12 cars

If 5 cars arrive, 30 minutes to closing ; it takes 25 minutes to finish up ;

If 1 more car arrives, then it takes exactly 30 minutes

Hence, there will be no one in line at closing ;

IF 0 or 1 car arrives, if more than 1 car arrives, then there will be people in line at closing.

Hence, probability of waiting in line at closing :

1 - [p(0) + p(1)]

Using poisson :

P(x) = [(e^-μ) * (μ^x)] / x!

If x = 0

P(0) = [(e^-5) * (5^0)] / 0!

P(0) = (0.0067379 * 1) / 1

P(0) = 0.0067379

X = 1

P(1) = [(e^-5) * (5^1)] / 1!

P(0) = (0.0067379 * 5) / 1

P(0) = 0.0336897

Hence,

1 - (0.0067379 + 0.0336897)

1 - 0.0404276

= 0.9595724

Hence, the probability that anyone would be in the car washing line after closing is 0.9595724

the answer is a

step-by-step explanation:

39.00 x 24 = total installment price

carrying installment price - cash price

needed to save the .00/

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl