An agency is studying the income of store managers in the retail industry. A random sample of 25 managers reveals a sample mean of $45,420. The sample standard deviation is $2,050. Use 90% confidence level to determine the confidence interval.

The Confidence Interval = ($44,745.55 , $46,094.45)

Step-by-step explanation:

The formula for Confidence Interval =

Confidence Interval = Mean ± z × Standard deviation/√n

Where n = number of samples = 25 managers

Standard deviation = $2,050

Mean = $45,420

z = z score of the given confidence interval

= z score of 90% confidence interval

= 1.645

Confidence Interval = $45,420 ± 1.645 × $2,050/√25

= $45,420 ± 1.645 × $2,050/5

= $45,420 ± 674.45

Confidence Interval =

$45,420 - 674.45 = $44,745.55

$45,420 + 674.45 = $46,094.45

Therefore, the Confidence Interval = ($44,745.55 , $46,094.45)

javeir earned 56 dollors more than kyle so kyle and javier had 14w but javier made 56 dollors more so 14w plus 56 equals 70w so kyle has 14w

step-by-step explanation: hope i plz give me brainliest with the points so this is not constant cause it says last week so its clearly not every week

step-by-step explanation: s