To solve the given problem, let's work through each part step-by-step:
### Part (a)
Problem: In 2015, what percent, to the nearest whole number, of the labor force was comprised of women?
The function modeling the percent of women in the labor force is:
[tex]\[ f(x) = \frac{66.23}{1 + \frac{1.084 e^{-x}}{24.77}} \][/tex]
where [tex]\( x \)[/tex] represents the number of years since 1950.
To find the percentage for the year 2015:
1. Calculate the number of years since 1950: [tex]\( 2015 - 1950 = 65 \)[/tex].
2. Substitute [tex]\( x = 65 \)[/tex] into the function:
[tex]\[ f(65) = \frac{66.23}{1 + \frac{1.084 e^{-65}}{24.77}} \][/tex]
Solving this expression will give the percentage of women in the labor force in 2015, which is approximately [tex]\( 61 \% \)[/tex].
### Part (b)
Problem: In what year did women make up 52% of the labor force?
Here we need to find [tex]\( x \)[/tex] when [tex]\( f(x) = 52 \)[/tex]:
[tex]\[ 52 = \frac{66.23}{1 + \frac{1.084 e^{-x}}{24.77}} \][/tex]
Follow these steps to solve for [tex]\( x \)[/tex]:
1. Set up the equation:
[tex]\[ 52 = \frac{66.23}{1 + \frac{1.084 e^{-x}}{24.77}} \][/tex]
2. Isolate the exponential term:
[tex]\[ 52 \left( 1 + \frac{1.084 e^{-x}}{24.77} \right) = 66.23 \][/tex]
[tex]\[ \frac{1.084 e^{-x}}{24.77} = \frac{66.23}{52} - 1 \][/tex]
3. Simplify:
[tex]\[ \frac{1.084 e^{-x}}{24.77} = \frac{66.23}{52} - 1 = 1.2736 - 1 = 0.2736 \][/tex]
[tex]\[ e^{-x} = 0.2736 \cdot \frac{24.77}{1.084} \][/tex]
[tex]\[ e^{-x} \approx 6.24 \times 10^{-3} \][/tex]
4. Solve for [tex]\( x \)[/tex] by taking the natural logarithm:
[tex]\[ -x = \ln(6.24 \times 10^{-3}) \][/tex]
[tex]\[ x \approx - \ln(0.00624) \approx 5.075 \][/tex]
Since [tex]\( x \)[/tex] represents the number of years since 1950:
[tex]\[ \text{Year} = 1950 + 5.075 \approx 1955 \][/tex]
Thus, women made up 52% of the labor force around the year 1955.
