To find the values for [tex]\( a \)[/tex] and [tex]\( b \)[/tex] in the linear model [tex]\( f(x) = ax + b \)[/tex] that represent the average family income as given, follow these steps:
1. Set up the points:
- The year 1993 corresponds to [tex]\( x = 0 \)[/tex] with an income of 42000.
- The year 2003 corresponds to [tex]\( x = 10 \)[/tex] with an income of 54706.
These points can be written as (0, 42000) and (10, 54706).
2. Calculate the slope [tex]\( a \)[/tex]:
The slope [tex]\( a \)[/tex] of the line can be calculated using the formula:
[tex]\[
a = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Plugging in the coordinates:
[tex]\[
a = \frac{54706 - 42000}{10 - 0} = \frac{12706}{10} = 1270.6
\][/tex]
So, the value of [tex]\( a \)[/tex] is [tex]\( 1270.6 \)[/tex].
3. Calculate the y-intercept [tex]\( b \)[/tex]:
To find [tex]\( b \)[/tex], use one of the points and the slope [tex]\( a \)[/tex] in the equation of the line [tex]\( y = ax + b \)[/tex]. Choosing the point (0, 42000):
[tex]\[
42000 = 1270.6(0) + b \Rightarrow b = 42000
\][/tex]
So, the value of [tex]\( b \)[/tex] is [tex]\( 42000 \)[/tex].
Therefore, the linear model is:
[tex]\[
f(x) = 1270.6x + 42000
\][/tex]
4. Find the average family income in 1998:
- The year 1998 corresponds to [tex]\( x = 5 \)[/tex].
- Substitute [tex]\( x = 5 \)[/tex] into the linear equation:
[tex]\[
f(5) = 1270.6 \cdot 5 + 42000
\][/tex]
Calculate:
[tex]\[
f(5) = 6353 + 42000 = 48353.0
\][/tex]
Hence, the average family income in 1998 is [tex]\( \$48353.0 \)[/tex].
Summarizing:
[tex]\[
\begin{aligned}
a &= 1270.6 \\
b &= 42000.0 \\
\text{Average family income in 1998} &= 48353.0
\end{aligned}
\][/tex]
