A company has offices in two different countries. Suppose that the average age (in years) of the employees at location A is 40.0 with a standard deviation of 5.5, and the average at location B is 36.0 with a standard
deviation of 6.3.
Every month, the company takes separate random samples of 50 employees from location A and 75 from
location B for a survey. Each time, they look at the difference in the mean age sampled from each location
(CA - DB).
What are the mean and standard deviation (in years) of the sampling distribution of T A - ib?

$8

step-by-step explanation:

it is pretty much backwards.

you take the 4x2 multiply that, and that is = $8

13.46

step-by-step explanation:

16.80 x .20 = 13.46

3

step-by-step explanation:

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.