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