SOLUTION: In 2006, the median yearly family income was about $48,200. Suppose the average annual rate of change since then is $1240 per year. a.) Write and graph an equality for the annual

Algebra ->  Linear-equations -> SOLUTION: In 2006, the median yearly family income was about $48,200. Suppose the average annual rate of change since then is $1240 per year. a.) Write and graph an equality for the annual      Log On


   



Question 1198863: In 2006, the median yearly family income was about $48,200. Suppose the average annual rate of change since then is $1240 per year.
a.) Write and graph an equality for the annual family incomes, y, are less than the median for x years after 2006.
b.) Determine whether each of the following points is part of the solution set.
(2, 51,000)
(8, 69,200)
(5, 50,000)
(10, 61,000)

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The equation for median family income in terms of years after 2006 is f(x)=48200+1240x

Evaluate that expression for the given x values and see whether the function value is less than or greater than the given y value. Though the wording in your post is terrible, apparently the question is to find the given points that show family incomes that are less than the median.

(2,51000): median is f(2)=50680; given is 51000. NO, the given value is NOT less than the median.
(8,69200): median is f(8)=58120; given is 69200. NO, the given value is NOT less than the median.

etc....