There is only one checkout line and the average service time is 5 minutes per customer. there are 3 people in the queue ahead of you. what is the probability that your wait time will exceed 7 minutes?

The probability is 100% because 3*5 is 15, since 5 is the average time and 3 customers are already there.

Step-by-step explanation:

62.7%

Step-by-step explanation:

This is an example of a Poisson process:

Using a Poisson distribution, the probability that the wait time T will exceed a certain time t is:

P(T > t) = e^(-events/time × t)

The average wait time per customer is 5 minutes, so the expected wait time for 3 customers is 15 minutes.

Given that t = 7 min and event/time = 1 / 15 min:

P(T > 7) = e^(-1/15 × 7)

P(T > 7) = 0.627

There is a 62.7% probability that your wait time will exceed 7 minutes.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116