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Answer :
To solve the problem of finding a linear function that models the data for high school graduates, we'll use the data for the years 1950 and 2000.
### Part (a): Find the Linear Function
1. Identify the Data Points for 1950 and 2000:
- In 1950, the percentage of high school graduates was 34.7%. This corresponds to [tex]\( x = 0 \)[/tex].
- In 2000, the percentage was 87.3%. This corresponds to [tex]\( x = 50 \)[/tex].
2. Calculate the Slope ([tex]\( m \)[/tex]) of the Line:
The formula for the slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Substituting the given data:
[tex]\[
m = \frac{87.3 - 34.7}{50 - 0} = 1.052
\][/tex]
3. Find the Y-intercept [tex]\( (b) \)[/tex]:
Use the point (1950, 34.7), where [tex]\( x = 0 \)[/tex], so:
[tex]\[
y = mx + b \Rightarrow 34.7 = 1.052 \times 0 + b \Rightarrow b = 34.7
\][/tex]
4. Write the Linear Equation [tex]\( f(x) \)[/tex]:
The linear function is:
[tex]\[
f(x) = 1.052x + 34.7
\][/tex]
### Part (b): Approximate the Percentage for 1995
1. Determine [tex]\( x \)[/tex] for the Year 1995:
Since 1950 is represented by [tex]\( x = 0 \)[/tex], 1995 corresponds to [tex]\( x = 45 \)[/tex] (because 1995 - 1950 = 45).
2. Use the Function to Find the Percentage:
Substitute [tex]\( x = 45 \)[/tex] into the linear equation:
[tex]\[
f(45) = 1.052 \times 45 + 34.7
\][/tex]
3. Calculate:
[tex]\[
f(45) = 1.052 \times 45 + 34.7 = 82.04
\][/tex]
Hence, using the linear model, approximately 82.04% of the country's population aged 25 and older were high school graduates in 1995.
### Part (a): Find the Linear Function
1. Identify the Data Points for 1950 and 2000:
- In 1950, the percentage of high school graduates was 34.7%. This corresponds to [tex]\( x = 0 \)[/tex].
- In 2000, the percentage was 87.3%. This corresponds to [tex]\( x = 50 \)[/tex].
2. Calculate the Slope ([tex]\( m \)[/tex]) of the Line:
The formula for the slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\][/tex]
Substituting the given data:
[tex]\[
m = \frac{87.3 - 34.7}{50 - 0} = 1.052
\][/tex]
3. Find the Y-intercept [tex]\( (b) \)[/tex]:
Use the point (1950, 34.7), where [tex]\( x = 0 \)[/tex], so:
[tex]\[
y = mx + b \Rightarrow 34.7 = 1.052 \times 0 + b \Rightarrow b = 34.7
\][/tex]
4. Write the Linear Equation [tex]\( f(x) \)[/tex]:
The linear function is:
[tex]\[
f(x) = 1.052x + 34.7
\][/tex]
### Part (b): Approximate the Percentage for 1995
1. Determine [tex]\( x \)[/tex] for the Year 1995:
Since 1950 is represented by [tex]\( x = 0 \)[/tex], 1995 corresponds to [tex]\( x = 45 \)[/tex] (because 1995 - 1950 = 45).
2. Use the Function to Find the Percentage:
Substitute [tex]\( x = 45 \)[/tex] into the linear equation:
[tex]\[
f(45) = 1.052 \times 45 + 34.7
\][/tex]
3. Calculate:
[tex]\[
f(45) = 1.052 \times 45 + 34.7 = 82.04
\][/tex]
Hence, using the linear model, approximately 82.04% of the country's population aged 25 and older were high school graduates in 1995.
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