Certainly! Let’s go through the problem step-by-step.
### Part (a)
To find the percentage of women in the labor force in 2012, we need to determine [tex]\(x\)[/tex] for the year 2012. Since [tex]\(x\)[/tex] represents the number of years since 1950:
[tex]\[ x = 2012 - 1950 = 62 \][/tex]
Now, substitute [tex]\( x = 62 \)[/tex] into the function [tex]\( f(x) = \frac{67.21}{1 + 1.085 e^{-x / 24.71}} \)[/tex]:
[tex]\[ f(62) \approx \frac{67.21}{1 + 1.085 e^{-62 / 24.71}} \][/tex]
Evaluate the expression:
[tex]\[ f(62) = \frac{67.21}{1 + 1.085 e^{-62 / 24.71}} \][/tex]
This simplifies to approximately:
[tex]\[ f(62) \approx 62 \][/tex]
Therefore, in 2012, approximately 62% of the labor force was comprised of women.
### Part (b)
To find the year when women made up 52% of the labor force, we need to solve for [tex]\(x\)[/tex] when [tex]\( f(x) = 52 \)[/tex].
Setting the function equal to 52:
[tex]\[ 52 = \frac{67.21}{1 + 1.085 e^{-x / 24.71}} \][/tex]
Start by isolating [tex]\( e^{-x / 24.71} \)[/tex]:
[tex]\[ 52 + 52 \cdot 1.085 e^{-x / 24.71} = 67.21 \][/tex]
[tex]\[ 52 \cdot 1.085 e^{-x / 24.71} = 67.21 - 52 \][/tex]
[tex]\[ 52 \cdot 1.085 e^{-x / 24.71} = 15.21 \][/tex]
[tex]\[ 1.085 e^{-x / 24.71} = \frac{15.21}{52} \][/tex]
[tex]\[ e^{-x / 24.71} = \frac{15.21}{52 \cdot 1.085} \][/tex]
Take the natural logarithm of both sides:
[tex]\[ -\frac{x}{24.71} = \ln\left(\frac{15.21}{52 \cdot 1.085}\right) \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = -24.71 \cdot \ln\left(\frac{15.21}{52 \cdot 1.085}\right) \][/tex]
[tex]\[ x \approx 32.5 \][/tex]
Since [tex]\(x\)[/tex] is the number of years since 1950:
[tex]\[ \text{Year} \approx 1950 + 32.5 \approx 1982 \][/tex]
Therefore, women made up 52% of the labor force in approximately the year 1982.
So, the answers are:
(a) In 2012, approximately 62% of the labor force was comprised of women.
(b) Women made up 52% of the labor force in approximately the year 1982.
