Year (X) = 2000 2001 2002 2003 2004

Cost in Dollars (Y) = 56.25 74.30 122.75 200.00 308.50

The table shows the cost of a game from 2000 to 2004, which has been increasing in a quadratic fashion. Let x = 0 in 2000, and find the best-fit quadratic equation. What will game cost in 2010?

A) $417

B) $746

C) $960

D) $1,586

Question

Guest

5 October, 13:20

+3

Answers (1)

Know the Answer?

Joseph Kelly · Answer 1

Here we are going to use the equation y=15x2+3x+56.

x=0 in 2,000 so x=10 in 2010.

Substitute 10 for x.

After this we have the answer: D. $1,586