A random sample of 40 college students has a mean earnings of $3120 with a sample standard deviation of $677 over the summer months. Determine whether a normal distribution or a t-distribution should be used or whether neither of these can be used to test a claim: μ > $3000.

Normal distribution

Step-by-step explanation:

From the statement we have the following information:

Average sample, m = $ 3120

Sample size, n = 40

Population standard deviation, sd = $ 677

Because we know the standard deviation of the population and that in addition the sample size is also greater than 30 (is 40) we can say that the distribution of university earnings is a normal distribution, that is, we can use the z test to build a confidence interval

b. 6

step-by-step explanation:

we are given the table representing an exponential function.

we can see from the table that, as x is increasing, the value of y is increasing in the power of 6.

i.e. the following relation is obtained,

x = 1, y = 6

x = 2, y = 36 =

x = 3, y = 216 =

x = 4, y = 1296 =

x = 5, y = 7776 =

so, the geometric series obtained is .

thus, the common ratio is given by = = 6.

hence, the common ratio is 6.