Aprofessor would like to estimate the average number of hours his students spend studying for his class. he wants to estimate μwith 99% confidence and with a margin of error of at most 2 hours. from past experience he knows the standard deviation of number of hours spent studying to be 8 hours. how many students need to be surveyed to meet these requirements? a. 106b. 105c. 107d. 108

n=106.1≅106

Option A which is 106 is correct

Step-by-step explanation:

In order to find the number of students need to be surveyed to meet the requirements we will use the following formula:

where:

n is the number of students need o be surveyed

Z is the distribution

S is the standard Deviation

E is the error Margin

Now:

S=8 ,E=2, CI= 99% or 0.99

For Z we need to find Alpha/2

Alpha=1-0.99

Alpha=0.01

Alpha/2=0.005 or 0.5%

From the cumulative Standard Distribution Table:

Z at alpha/2 =2.576

n= (2.576^2* 8^2)/2^2

n=106.1≅106

Option A which is 106 is correct

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x