A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19 and standard deviation of $15.88 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

(60.382 ; 83.998)

Step-by-step explanation:

The confidence interval is given as :

Mean ± Margin of error

Mean = 72.19

The margin of error is : Tcritical * s/√n

n = 13 ; s = 15.88

Tcritical at 98%, df = 13-1 = 12 = 2.681

Margin of Error = 2.681 * 15.88/√13 = 11.808

Confidence interval = 72.19 ± 11.808

(60.382 ; 83.998)

the answer is (2x+8)²= 16

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,