Salaries of 40 college graduates who took a statistics course in college have a mean of $62, 200. assuming a standard deviation of $11,766 construct a 90% confidence interval for estimating the population mean.

round to the nearest integer as needed.

Step-by-step explanation:

We want to determine a 90% confidence interval for the mean salaries of college graduates

Number of sample, n = 40

Mean, u = $62, 200

Standard deviation, s = $11,766

For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean +/- z ×standard deviation/√n

It becomes

62200 +/- 1.645 × 11766/√40

= 62200 +/- 1.645 × 1860.4

= 62200 +/- 3060.358

The lower end of the confidence interval is 62200 - 3060.358 =59139.642

The upper end of the confidence interval is 62200 + 3060.358 =65260.358

Therefore, with 90% confidence interval, the mean mean salaries of college graduates is between $59139.642 and $65260.358

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70

r and t.

both are 64 degrees.