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19 July, 12:03

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

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  1. 19 July, 12:13
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    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
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