1. **State the problem:** We have data points $(20, 57.5)$, $(33, 34.8)$, and $(42, 27.4)$ and want to extrapolate $y$ at $x=49$ assuming the model $y = Ax^p$. 2. **Formula and approach:** The power model is $y = Ax^p$. Taking logarithms on both sides gives: $$\ln y = \ln A + p \ln x$$ This is a linear relation in terms of $\ln y$ and $\ln x$. 3. **Calculate $\ln x$ and $\ln y$ for given points:** $$\ln 20 \approx 2.9957, \quad \ln 57.5 \approx 4.0510$$ $$\ln 33 \approx 3.4965, \quad \ln 34.8 \approx 3.5492$$ $$\ln 42 \approx 3.7377, \quad \ln 27.4 \approx 3.3112$$ 4. **Fit a line $\ln y = p \ln x + \ln A$ using two points (20,57.5) and (33,34.8):** Calculate slope $p$: $$p = \frac{3.5492 - 4.0510}{3.4965 - 2.9957} = \frac{-0.5018}{0.5008} \approx -1.002$$ 5. **Calculate intercept $\ln A$ using point $(20,57.5)$:** $$4.0510 = \ln A + (-1.002)(2.9957)$$ $$\ln A = 4.0510 + 3.001 = 7.052$$ 6. **Rewrite model:** $$y = e^{7.052} x^{-1.002}$$ Calculate $A$: $$A = e^{7.052} \approx 1157.5$$ 7. **Extrapolate $y$ at $x=49$:** $$y = 1157.5 \times 49^{-1.002}$$ Calculate $49^{-1.002}$: $$49^{-1.002} = \frac{1}{49^{1.002}} \approx \frac{1}{49.3} = 0.02028$$ So, $$y \approx 1157.5 \times 0.02028 = 23.47$$ **Final answer:** $$\boxed{y \approx 23.5}$$