Question

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?

Answer 1

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`.