The customer service department of a phone company is experimenting with two different systems. On Monday they try the first system which is based on an automated menu system. On Tuesday they try the second system in which each caller is immediately connected with a live agent. A quality control manager selects a sample of seven calls each day. He records the time for each customer to have his or her question answered. The times (in minutes) are listed below. Find the coefficient of variation for each of the two sets of data, then compare the variation. Round results to one decimal place. Automated Menu: 11.7 7.4 3.9 2.9 9.2 6.3 5.5 Live agent: 6.2 2.9 4.4 4.1 3.4 5.2 3.7

Automated menu

Live agent menu

So then we can conclude that we have lower variation for the Live agent menu since the coefficient of variation is lower compared to the Automated menu

Step-by-step explanation:

We can solve this by case

Automated menu

Data: 11.7 7.4 3.9 2.9 9.2 6.3 5.5

We can begin calculating the mean with the following formula:

And replacing we got:

Now we can calculate the standard deviation with the following formula:

And replacing we got:

Now we can calculate the coeffcient of variation with this formula:

And replacing we got:

Live agent menu

Data: 6.2 2.9 4.4 4.1 3.4 5.2 3.7

We can begin calculating the mean with the following formula:

And replacing we got:

Now we can calculate the standard deviation with the following formula:

And replacing we got:

Now we can calculate the coeffcient of variation with this formula:

And replacing we got:

So then we can conclude that we have lower variation for the Live agent menu since the coefficient of variation is lower compared to the Automated menu

ye

step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: