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July 24, 2014

July 24, 2014

Posted by **Yadira M** on Sunday, August 16, 2009 at 3:10pm.

- statistics -
**MathMate**, Sunday, August 16, 2009 at 6:02pmThis is really a problem of evaluation of functions.

As indicated, for year

2000, x=1, so for

2008, x=2008-1999=9, and for

2010, x=2010-1999=11

Let

y = f(x) = 90+0.9x+3x²

for 2000, f(1) = 90 + 0.9*(1) + 3*1² = 93.9

for 2008, f(9) = 90 + 0.9*(9) + 3*9² = 341.1

for 2010, f(11)= 90 + 0.9*(11) + 3*11² = 462.9

Note that in general, the number of repairs is an integer (whole number), so the estimate should be rounded to the nearest integer.

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