An economist wants to estimate the mean per capita income (in thousands of dollars) for a major city in California. He believes that the mean income is $24.8, and the variance is known to be $125.44. How large of a sample would be required in order to estimate the mean per capita income at the 85% level of confidence with an error of at most $0.59

747 samples

Step-by-step explanation:

Given:

Standard deviation = √125.44 = 11.2

Zcritical = 85℅

Margin of error, E = 0.59

The sample size, n required cnanbe obtained using the relation :

n = [(Zα/2 * σ) / E]²

Zcritical at 85% = 1.44

n = [(1.44 * 11.2) / 0.59]²

n = (16.128 / 0.59)²

n = 747.23

n = 747 samples

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !