Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A 90% confidence interval for a mean μ if the sample has n-80 with X-22.1 and s = 5.6, and the standard error is SE = 0.63. The 90% confidence interval is to :

(21.064 ; 23.136)

Step-by-step explanation:

Given :

Sample, n = 80

Mean, xbar = 22.1

Standard deviation, s = 5.6

Standard Error, S. E = 0.63

Confidence interval :

Xbar ± Zcritical * S.E

22.1 ± (1.645 * 0.63)

22.1 ± 1.036

Lower boundary = 22.1 - 1.036 = 21.064

Upper boundary = 22.1 + 1.048 = 23.136

(21.064 ; 23.136)

-7

step-by-step explanation:

x^2 + 18x + 74 = (x + 9)^2 - 81 + 74

= (x + 9)^2 - 7.

so q = -7.

16

step-by-step explanation: