Asample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.41 mpa and a sample standard deviation of 0.77 mpa.

For this problem, the confidence interval is the one we are looking for. Since the confidence level is not given, we assume that it is 95%.

The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n

Where:

α= 5%

α/2 = 2.5%

t 0.025, 19 = 2.093 (check t table)

n = 20

df = n – 1 = 20 – 1 = 19

So plugging in our values:

8.41 ± 2.093 * 0.77 √ 1 + 1/20

= 8.41 ± 2.093 * 0.77 (1.0247)

= 8.41 ± 2.093 * 0.789019

= 8.41 ± 1.65141676

= 6.7586 < x < 10.0614

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

1. 34

2. 268

3. x=14.4

step-by-step explanation: