Suppose that a company claims that its air fresheners last 23 days, on average. you take many large samples, and find, each time, that the mean number of days of the sample is outside the 95% confidence interval. based on this information alone, which of the following is probably not the mean number of days that the company's air fresheners last?

a. 19
b. 21
c. 17
d. 23

Option D is the answer.

Explanation:

A company claims that its air fresheners last 23 days, on average. Many large samples are taken each time it is found that the mean number of days of the sample is outside the 95% confidence interval.

This implies:

So, below 21.85 is the probable mean number of days that the company's air fresheners last.

Therefore, 23 is not the mean number of days that the company's air fresheners last.

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."