Based on a survey conducted by greenfield online, 25 - 34-year-olds spend the most each week on fast food. the average weekly amount of $44 was reported in a may 2009 usa today snapshot. assuming that weekly fast food expenditures are normally distributed with a standard deviation of $14.50, what is a probability that a person in this age group will spend more than $60 on fast food in a week?

The correct answer is P = 0.1357.

In solving for the probability (P) that the 34-year olds will spend more than $60 on fastfood, the mean and the standard deviation is to be taken into account. The mean (μ) being the $44 average weekly expenditure; while the standard deviation (σ) is $14.50. The formula goes this way P(X > 60) = P( X−μ > 60−44 ) = P ( X−μ/σ > 60−44/14.50). 

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

step-by-step explanation:

the true ones are a b and d