A random sample of 12 recent college graduates reported an average starting salary of $54,000 with a standard deviation of $6,000. a. Construct a 95% confidence interval for the mean starting salary of college graduates.

a. 50188 to 57812$
b. 45,0000 to 50,000$
c. 48,000 to 60,000$
d. 50,000 to 54000$

b. The National Association of Colleges and Employers reports that the mean starting salary for 2017 college graduates is $50,516. Does your confidence interval (part a) indicate that starting salaries have significantly changed since 2017? Why or why not?

a.) $50188 to $57812

Step-by-step explanation: Confidence Interval (CI) is an interval of values in which we are confident the true mean is in.

The interval is calculated as

x ±

a. For a 95% CI, z-value is 1.96.

Solving:

54,000 ±

54,000 ±

54,000 ± 1.96*1732.102

54,000 ± 3395

This means the interval is

50605 < μ < 57395

With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or from 50188 to 57812$.

b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?