One consequence of the popularity of the Internet is that it is thought to reduce television watching. Suppose that a random sample of 35 individuals who consider themselves to be avid Internet users results in a mean time of 2.01 hours watching television on a weekday. Determine the likelihood of obtaining a sample mean of 2.01 hours or less from a population whose mean is presumed to be 2.45 hours.

sample size , n =55

\sigma = s/\sqrt{n} = 2/\sqrt{55}

= 0.26968

likelihood of obtaining a sample mean of 2.00 hours or less

P(X<=2) = P ( Z <= (2-2.35)/0.26968) = P(Z <= -1.292835)

= 0.0972

i.e likelihood percentage is 9.72%

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°