The Hilbert Drug Store owner plans to survey a random sample of his customers with the objective of estimating the mean dollars spent on pharmaceutical products during the past three months. He has assumed that the population standard deviation is known to be $15.50. Given this information, what would be the required sample size to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00? Question 33 options: 231 15 16 163

a) 231

The sample to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00

n = 231

Step-by-step explanation:

Explanation:-

Given data the population standard deviation is known σ =$15.50

Given the margin of error ±$2.00

we know that  95 percent confidence interval of margin of error is determined by  

cross multiplication √n we get ,

squaring on both sides, we get

the tabulated z- value = 1.96 at 95% of level of significance.

n = 230.7≅231

Conclusion:-

The sample to estimate the population mean with 95 percent confidence and a margin of error of ±$2.00

n = 231

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )

7

step-by-step explanation:

yuo ned the activity or vertical angle of the form of 1,7 but 2,2 has a c4,2 as it verticie so i not have a car