The sales of a grocery store had an average of $7,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 100 days of sales was selected. It was found that the average was $7,280 per day. From past information, it is known that the standard deviation of the population is $1,000.At a 5% significance level one can conclude that a. The advertising campaigns increased the average daily sales. b. The advertising campaigns decreased the average daily sales. c. The average daily sales remained unchanged d. Cannot tell from the information provided

a.The advertising campaigns increased the average daily sales.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the advertising campaigns increased the average daily sales.

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=100.

The sample mean is M=7280.

The standard deviation of the population is known and has a value of σ=1000.

We can calculate the standard error as:

Then, we can calculate the z-statistic as:

This test is a right-tailed test, so the P-value for this test is calculated as:

As the P-value (0.0026) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the advertising campaigns increased the average daily sales.

5.5

step-by-step explanation:

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i