1. **State the problem:** We have the function $$f(x) = \frac{68.35}{1 + 0.036e^{-0.117x}}$$ modeling the percent of women in the labor force, where $x$ is years since 1950. 2. **Part (a):** Find the percent of women in 2012. - Calculate $x$ for 2012: $$x = 2012 - 1950 = 62$$ - Substitute into the function: $$f(62) = \frac{68.35}{1 + 0.036e^{-0.117 \times 62}}$$ 3. **Evaluate the exponent:** $$-0.117 \times 62 = -7.254$$ 4. **Calculate $e^{-7.254}$:** $$e^{-7.254} \approx 0.00071$$ 5. **Calculate denominator:** $$1 + 0.036 \times 0.00071 = 1 + 0.00002556 = 1.00002556$$ 6. **Calculate $f(62)$:** $$f(62) = \frac{68.35}{1.00002556} \approx 68.348$$ 7. **Round to nearest whole number:** $$\boxed{68\%}$$ --- 8. **Part (b):** Find the year when women made up 30% of the labor force. - Set $f(x) = 30$: $$30 = \frac{68.35}{1 + 0.036e^{-0.117x}}$$ 9. **Solve for the denominator:** $$1 + 0.036e^{-0.117x} = \frac{68.35}{30} = 2.2783$$ 10. **Isolate the exponential term:** $$0.036e^{-0.117x} = 2.2783 - 1 = 1.2783$$ 11. **Divide both sides by 0.036:** $$e^{-0.117x} = \frac{1.2783}{0.036} = 35.5083$$ 12. **Take natural logarithm:** $$-0.117x = \ln(35.5083)$$ 13. **Calculate the logarithm:** $$\ln(35.5083) \approx 3.569$$ 14. **Solve for $x$:** $$x = \frac{-3.569}{0.117} = -30.5$$ 15. **Interpret $x$:** Negative $x$ means before 1950. 16. **Calculate the year:** $$1950 + (-30.5) = 1919.5 \approx 1920$$ **Final answers:** - (a) In 2012, women composed approximately $\boxed{68\%}$ of the labor force. - (b) Women made up 30% of the labor force around the year $\boxed{1920}$.