Direct mail advertisers send solicitations ("junk mail") to thousands of potential customers in the hope that some will buy the company's product. The response rate is usually quite low. Suppose a company wants to test the response to a new flyer and sends it to 1160 people randomly selected from their mailing list of over 200,000 people. They get orders from 144 of the recipients. Create a 90% confidence interval for the percentage of people the company contacts who may buy something.

The 90% confidence interval for the percentage of people the company contacts who may buy something is between 10.82% and 14%

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

90% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for the percentage of people the company contacts who may buy something is between 10.82% and 14%

21

step-by-step explanation:

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47