Pete worked 3 hours and charged Millie $155. Jay worked 6 hours and charged 230. If​ Pete's charge is a linear function of the number of hours​ worked, find the formula for​ Jay?and how much he would charge for working 77 hours for Fred?

1 Answer
Jan 21, 2018

Part A:
C(t) = 25t+ 80

Part B:
$2005

Explanation:

Assuming Pete and Jay both use the same linear function, we need to find their hourly rate.

3 hours of work cost $155, and double that time, 6 hours, cost $230, which is not double the price of 3 hours of work. That implies there was some kind of "up-front charge" added to the hourly rate.

We know that 3 hours of work and the up-front charge costs $155, and 6 hours of work and the up-front charge costs $230.

If we subtract $155 from $230, we would cancel out 3 hours of work and the up-front charge, leaving us with $75 for the other 3 hours of work.

Knowing Pete worked for 3 hours and charged $155, and the fact that 3 hours of work would normally cost $75, we can subtract $75 from $155 to find the up-front charge of $80.

We can now create a function with this information. Let C be the end cost, in dollars, and t be the time worked, in hours.

color(red)(C(t)) = color(green)(25t) color(blue)(+ 80)

color(red)(C(t)) => The cost after t hours of work.
color(green)(25t) => $25 for every hour worked.
color(blue)(+ 80) => $80 up-front charge, regardless of time worked.

Using this function, we can then find out how much 77 hours of work would cost.

C(t) = 25t+ 80

C(77) = 25(77)+ 80

C(77) = 1925 + 80

C(77) = 2005

The cost of 77 hours of work would be $2005.