4. The 92 million Americans of age 50 and over control 50% of all discretionary income. AARP estimates that the average annual expenditure on restaurants and carryout food was $1,873 for individuals in the age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is $550. a. At 95% confidence, what is the margin of error

$120.52

Margin of error M.E = $120.52

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

x+/-M.E

Where M.E = margin of error

M.E = zr/√n

Given that;

Mean x = $1,873

Standard deviation r = $550

Number of samples n = 80

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

M.E = (1.96 × $550/√80) = 120.5240639872

M.E = $120.52

Margin of error M.E = $120.52

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10