# Modeling displaystyle{C}{O}_{{2}} emissions a) Use the data from 1960 and 1990 to find a linear function that models the displaystyle{C}{O}_{{2}} b) U

Modeling $C{O}_{2}$ emissions
a) Use the data from 1960 and 1990 to find a linear function that models the $C{O}_{2}$
b) Use your linear function to predict the $C{O}_{2}$ emissions in 2005. Compare your prediction with the actual $C{O}_{2}$ emission in 2005
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Cristiano Sears
a)
Let t represent years after 1960. , $C=C{O}_{2}$ emissions.
Suppose the linear function that gives relation between the year and $C{O}_{2}$ emissions be:
$C=At+B---\left(1\right),$ Where A and B are constants.
Now we need to use data from year 1960 and 1990.
For Year 1960 :
$t=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}C=316.9$
substituting this in (1):
$⇒316.9=A\left(0\right)+B$
$⇒B=316.9$
For Year 1990 :
$t=1990-1960=30\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}C=354.2$
Substituting this in (1):
$⇒354.2=A\left(30\right)+316.9$
$⇒37.3=A\left(30\right)$
$⇒A=1.243$
Hence on Substituting the value of A and B in (1),
$⇒C=1.243t+316.9$
This is our linear function that models $C{O}_{2}$ emissions with time.
b)
Now,
$C=1.243t+316.9$
Now we are given to predict the $C{O}_{2}$ emissions in the year 2005
$⇒t=2005-1960=45$
$⇒C=1.243\left(45\right)+316.9=372.85$
According to our prediction, the $C{O}_{2}$ emissions in 2005 is 372.85
But actually it is 379.7
Difference in Actual and Predicted value $=379.7-372.85=6.85$