Potential customers arrive at a full-service, one-pump gas station at a Poisson rate of20 cars per hour. However, customers will only enter the station for gas if there are no more than two cars (including the one currently being attended to) at the pump. Suppose that the amount of time required to service a car is exponentially distributed with a mean of five minutes. Required:
a. What fraction of the attendant’s time will be spent servicing cars?
b. What fraction of potential customers are lost?

a) 19/20

b) 1/20

Step-by-step explanation:

a) calculate the fraction of the attendants time will be spent servicing cars

l / L > 0 =  Ц / [ Ц -  λ ] = 20 --- ( 1 )

Ц (service )  = 15 hour

∴ λ ( arrival rate )  = 14.25

P = λ / Ц  = 14.25 / 15 = 0.95 = 19/20

b) Fraction of potential customers lost

Ls = P / ( 1 - P )

where ; P = 19/20

Ls ( fraction of potential customers lost ) = 19/20 / ( 1 - 19/20 )  = 19

hence proportion of customers lost = 1 - 19/20 = 0.05 = 1/20

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.

according to this law, when unemployment goes up by 1%, gdp drops by 2%. in the question, unemployment rate goes from 2% to 5%, meaning unemployment goes up by 3%, so the gdp drops by 6%. thus the effect is a drop in gdp by 6%.

step-by-step explanation: