Answer
$y=3.4t+5$
Projected in 2006:$\quad {{\$}} 25.4$ billion
Work Step by Step
The regression line is
$\qquad y=mx+b$,
where $m$ and $b$ are computed as follows.
$m=\displaystyle \frac{n(\sum xy)-(\sum x)(\sum y)}{n(\sum x^{2})-(\sum x)^{2}}\qquad b=\frac{\sum y-m(\sum x)}{n},$
$n=$ number of data points.
$\left[\begin{array}{lllll}
& t & y & ty & t^{2}\\
& & & & \\
\hline & 0 & 6 & 0 & 0\\
& 5 & 20 & 100 & 25\\
& 10 & 40 & 400 & 100\\
\hline & & & & \\
\sum & 15 & 66 & 500 & 125\\
& & & &
\end{array}\right]$
$m=\displaystyle \frac{3(500)-(15)(66)}{3(125)-(15)^{2}}=\frac{510}{150}=3.4$
$b=\displaystyle \frac{66-3.4(15)}{3}=5$
$y=3.4t+5$
In the year 2006, $ t=6$ (years after 2000), and we have:
$y=3.4(6)+5=25.4$
