Year (X) =2000 2001 2002 2003 2004Cost in Dollars(Y) = 56.25 74.30 122.75 200.00 308.50The 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
Answer:
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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