utomobiles arrive at the drive-through window at a post office at the rate of 7 every 10 minutes. The average service time is 7 minutes. The Poisson distribution is appropriate for the arrival rate and service times are exponentially distributed. (Note: the given equations are NOT necessarily in the correct order.) What is the average time a car is in the system?Formulas for 13.17 What is the average number of cars in the system? What is the average time cars spend waiting to receive service? What is the average number of cars in line behind the customer receiving service? What is the probability that there are no cars at the window? What percentage of the time is the postal clerk busy? What is the probability that there are exactly two cars in the system?

The answers to the questions are well numbered below.

Step-by-step explanation:

We have arrival rate

λ = 7 in every 10 minutes = 0.7

Service time u = 7 minutes

1. Average time a car is in system

W = 1/u - λ

= 1/7-0.7

= 0.1587

= 0.16 to 2 decimal places

2. Average number of cars in system

= λ/u-λ

= 0.7/7-0.7

= 0.1111

= 0.11 to 2 decimal places

3. Average time that cars spend to receive service

= λ/u(u-λ)

= 0.7/7(7-0.7)

= 0.01587

= 0.016 to 2 decimal places

4. Average number of cars in line behind customer receiving service

= λ²/u(u-λ)

= 0.7²/7(7-0.7)

= 0.49/44.1

= 0.011 to 2 decimal places

5. Probability no cats at window

1-λ/u

= 1-0.7/7

= 0.9 = 9%

6. Percentage of time postal clerk is busy

= λ/u = 0.7/7 = 0.1 = 10%

7. Probability of exactly 2 cars

= λ²e^-λ/2!

= 0.7²e^-0.7/2!

= 0.1217

= 12.17%

3; 9 is 8 to -4

step-by-step explanation:

30 miles

step-by-step explanation:

we know that 60 seconds equals 1 minute and 60 minutes 1 hour

44 ft       60 sec     60 min     158400   ft

  *   *   =

1 sec       1 min       1 hour         hour

we know that there are 5280 ft per mile

158400   ft       1 mile         30 miles

  *   =

1   hour               5280 ft     1 hour

so in 1 hour   it can travel 30 miles