In a survey of 5100 randomly selected T. V. viewers, 40% said they watch network news programs. Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T. V. viewers who watch network news programs.

Margin of error = 0.01344

Step-by-step explanation:

Margin of error = critical value × standard deviation.

critical value for 95% confidence interval = 1.96

Standard deviation = √[(p)(q)/n]

p = 0.4, q = 1 - p = 1 - 0.4 = 0.6, n = 5100

Standard deviation = √[(0.6×0.4)/5100] = 0.00686

Margin of error = 1.96 × 0.00686 = 0.0134

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00