A random sample of 20 purchases showed the sample mean of $48.77 and the sample standard deviation s is $17.58. How large is the margin of error for a 80 % confidence interval for the mean purchases of all customers

Step-by-step explanation:

Given data          

Sample Mean (M):  $48.77  

Sample Size (n): 20

Standard Deviation (σ) :  $17.58

Confidence Level: 80%

 we know that z*-Values for 80% Confidence  Levels is 1.28

the expression for margin of error is given bellow\

MOE= z*σ/√n

We can now substitute into the expression and solve for the MOE as

MOE= 1.28*17.58/√20

MOE= 22.502/4.47

MOE= 22.502/4.47

MOE= 5.03

The margin of error for a 80 % confidence interval is $5.03

a. 32 degrees

step-by-step explanation:

here we have to use the rule.

m∠e = 1/2 (major arc - minor arc)

m∠e = 1/2 (212 - 148)

m∠e = 1/2 (64)

m∠e = 32 degrees

answer: a. 32 degrees

you

answer: c. is your answer

step-by-step explanation: get me a black leather coat