A professional employee in a large corporation receives an average of μ = 39.8 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 38 employees showed that they were receiving an average of x = 33.1 e-mails per day. The computer server through which the e-mails are routed showed that σ = 16.2. Has the new policy had any effect? Use a 10% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee.

The policy has an effect because the null hypothesis rejected since P-value < significance level.

Step-by-step explanation:

Since (P-value = 0.0108) < 0.1 significance level, we have sufficient evidence to show that the mean is not equal to 39.8. This means that the policy has an effect on the average number of the emails received per day.

The workings are clearly written in the file attached below. Please check.

x> 6

step-by-step explanation:

8x+8> 7x+14

8x-7x> 14-8

x> 6

3

step-by-step explanation:

in standard form:

a= dictionaries

b= maps

you have to buy 3 dictionaries, so put a 3 next to the a (dictionaries) so that you can multiply them by each other.

a3+b= 60 (the amount of money you have)

we already know the cost of a=15 so replace a with $15

($15 x 3)+b=60

multiply 15 and 3 to get $45 and replace it with $45

45+b=60

and then replace b with 15 to get the total:

45+15=60

that means you have $15 left after buying 3 dictionaries.

15 divided by the cost of a map ($5) equals 3.

so you can buy 3 maps.

sorry if it was confusing!