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%

segment bisector

step-by-step explanation:

line segment bisector. definition: a line, ray or segment which cuts another line segment into two equal parts. try this drag one of the orange dots at a or b and note the the line ab always divides the line pq into two equal parts.

mark brainliest!