Cars arrive at an automatic car wash system every 10 minutes on average. The cars inter-arrival times are exponentially distributed. Washing time for each is 6 minutes per car and is purely deterministic (i. e., the waiting line system is M/D/c). Assuming that the car wash has a single bay to serve the cars, what is the average number of cars waiting in line (L.)?

the average number of cars waiting in line L is 0.45

Step-by-step explanation:

Given the data in the question;

Cars arrive at an automatic car wash system every 10 minutes on average.

Car arrival rate λ = 1 per 10 min = [ 1/10 × 60 ]per hrs = 6 cars per hour

Washing time for each is 6 minutes per car

Car service rate μ = 6min per car = [ 1/6 × 60 ] per hrs = 10 cars per hour

so

P = λ/μ = 6 / 10 = 0.6

Using the length of queue in M/D/1 system since there is only one service bay;

L = [ P² / ( 1 - P ) ]

so we substitute

L = [ (0.6)² / ( 1 - 0.6 ) ]

L = [ 0.36 / 0.4 ]

L = [ 0.9 ]

L = 0.45

Therefore, the average number of cars waiting in line L is 0.45

answer: since the question is talking about growth you have to use the growth formula which is inital amount (1 + growth)^time

step-by-step explanation:

the intial amount is 1200, the growth is 4.8, and the time is 7. plug those numbers into the equation, also i think you have to change 4.8 into a percent. but, i have forgotton how to do that. if this is a online class go on to the lesson.

the correct answer is you spend $55.78 more in a brand store that buy ing in a normal store.

step-by-step explanation:

brand = spend   $ 135.85 total

store = spend $80.07 total