The table lists data regarding the average salaries of several professional athletes in the years 1991 and 2001.
a) Use the data points to find a linear function that fits the data.
b) Use the function to predict the average salary in 2005 and 2010.
$\begin{array}{cc}\text { Year } & \text { Average Salary } \\ 1991 & \$ 266,000 \\ 2001 & \$ 1,370,000\end{array}$
$2001 \$ 1,370,000$
A linear function that fits the data is $S(x)=110400 x+155600$
(Let $x=$ the number of years since 1990, and let $S=$ the average salary $x$ years from 1990.)
The predicted average salary for 2005 is $\$ 1811600$
(Round to the nearest whole number.)
The predicted avarage salary for 2010 is $\$$ (Round to the nearest whole number.)

Step 1 ：Given data points are (1991, 266000) and (2001, 1370000).

Step 2 ：Find the slope of the line, which is the change in salary divided by the change in years. The slope is \(\frac{1370000 - 266000}{2001 - 1991} = 110400.0\).

Step 3 ：Use one of the points to find the y-intercept of the line, which is the salary when the year is 0. The y-intercept is \(266000 - 110400 \times (1991 - 0) = -219540400.0\).

Step 4 ：The linear function that fits the data is \(S(x) = 110400x - 219540400\), where \(x\) is the year.

Step 5 ：Predict the average salary in 2005 by substituting 2005 into the function: \(S(2005) = 110400 \times 2005 - 219540400 = 1811600.0\).

Step 6 ：Predict the average salary in 2010 by substituting 2010 into the function: \(S(2010) = 110400 \times 2010 - 219540400 = 2363600.0\).

Step 7 ：Final Answer: The predicted average salary for 2005 is \(\boxed{\$1,811,600}\) and for 2010 is \(\boxed{\$2,363,600}\).