Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. What is the planning value for the population standard deviation (to the nearest whole number)?

How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number)?

a. $500?
b. $200?
c. $100?

a) 217

b) 1351

c) 5403

Step-by-step explanation:

Given that:

confidence interval (c) = 0.95

The Z score of is from the z table is given as:

Range = $45000 - $30000 = $15000

The standard deviation (σ) is given as:

Sample size (n) is given as:

a) E = $500

≈ 217

b) ≈ 1351

c) ≈ 5403

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.