This question was probably confusing because of the language.
It was simply a matter of substituting the value $x=2010$ into the equation
so, $f(2010) = 0.0167*2010^2-65.37x+64440$
$ = 515.97 ~ 516$ ppm
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Friday, 10 January 2003
Applied question - Quadratic model of Concentration of Atmospheric $CO_2$
This question was probably confusing because of the language.
It was simply a matter of substituting the value $x=2010$ into the equation
so, $f(2010) = 0.0167*2010^2-65.37x+64440$
$ = 515.97 ~ 516$ ppm
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