#### Answer

$y=33.7t+68.3$

#### Work Step by Step

Graphing the approximately linear function,
the points on the graph are (t, y(t)).
Let y be in billions of dollars,
t is the number of years after 2000.
Given points: (1, 102) and (8, 338)
Slope:
$m=\displaystyle \frac{338-102}{8-1}=\frac{236}{7}\approx 33.7$
Point-slope equation:
$y-102=33.7(t-1)$
$y-102=33.7t-33.7\qquad/+102$
$y=33.7t+68.3$