Previous studies reported that teenagers spent 8.5 hours per week, on average, on the phone. This mean is supposed to have increased, to test this hypothesis, a random sample of 500 teenagers were interviewed about how many hours per week they spent on their phones. The sample mean was 10.5 hours with a sample standard deviation of 2.35. Conduct a hypothesis test. The null and alternative hypotheses are:

Since the calculated value of Z= 19.029 is much  greater than Z (0.05) = 1.645 and falls in the rejection region we reject the null hypothesis and conclude that mean of teenagers spending time on phone per week  has increased.

Step-by-step explanation:

1) Formulate the null and alternate hypothesis as

H0 : μ≤ 8.5  against the claim Ha:  μ > 8.5

2) Choose the significance level ∝ =0.05

3) The test statistic under H0  is

Z= x`-  μ/ s/√n

4) The rejection region is Z≥ Z(0.05)= 1.645

5) Computing the value of Z from sample information we get

Z= 10.5-8.5/ 2.35/ √500

Z= 2/ 2.35/ 22.361

Z= 19.029

Since the calculated value of Z= 19.029 is much  greater than Z (0.05) = 1.645 and falls in the rejection region we reject the null hypothesis and conclude that mean of teenagers spending time on phone per week  has increased.

it's d.

step-by-step explanation:

because: -1(4x + 1y) +2 (1x-3y) distribute each number into the parenthesis.

then you would get: -4x - 1y + 2x -6y combine like terms

and get: 2x -7y

hope my answer has you if not then i'm sorry.