The service-time distribution describes the probability p that the service time of the customer will be no more than t hours. if m is the mean number of customers serviced in an hour, then suppose a computer technical support representative can answer calls from 6 customers in an hour. what is the probability that a customer will be on hold less than 30 minutes?

Probability that a customer will be on hold less than 30 minutes is %

Step-by-step explanation:

Complete question is

The service-time distribution describes the probability P that the service time of the customer will be no more than t hours. If m is

the mean number of customers services in an hour, then

. a. Suppose a computer technical support representative can answer calls from 6 customers in an hour. What is the probability

that a customer will be on hold less than 30 minutes?

Solution -

Given the equation of probability -

Where m is the average number of customers served in an hour

and t is the total time

Now, in this case m is equal to 6 as technical support representative ia bale to answer calls from 6 customers in an hour

and time t minutes hours

Substituting the given values in above equation, we get -

Probability that a customer will be on hold less than 30 minutes is %

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)