1. **State the problem:** We are given data relating orbital speed $x$ (km/s) and distance from the sun $y$ (millions of km). We need to find a power regression equation of the form $$y = a x^b$$ rounded to the nearest ten-thousandth. 2. **Power regression formula:** The power regression model is $$y = a x^b$$ where $a$ and $b$ are constants found by fitting the data. 3. **Data points:** - $(6.565, 86958695)$ - $(12.7127, 42834283)$ - $(17.7177, 37773777)$ - $(24.3243, 24682468)$ - $(27.7277, 24772477)$ - $(30.8308, 23582358)$ 4. **Convert to linear form:** Taking natural logs, $$\ln y = \ln a + b \ln x$$ We perform linear regression on $(\ln x, \ln y)$ to find $\ln a$ and $b$. 5. **Calculate $\ln x$ and $\ln y$ for each point:** - $\ln 6.565 \approx 1.882$, $\ln 86958695 \approx 18.280$ - $\ln 12.7127 \approx 2.544$, $\ln 42834283 \approx 17.570$ - $\ln 17.7177 \approx 2.875$, $\ln 37773777 \approx 17.447$ - $\ln 24.3243 \approx 3.192$, $\ln 24682468 \approx 17.023$ - $\ln 27.7277 \approx 3.321$, $\ln 24772477 \approx 17.027$ - $\ln 30.8308 \approx 3.429$, $\ln 23582358 \approx 16.976$ 6. **Perform linear regression:** Using least squares, - Calculate means: $\overline{\ln x} \approx 2.873$, $\overline{\ln y} \approx 17.055$ - Calculate slope $b = \frac{\sum (\ln x_i - \overline{\ln x})(\ln y_i - \overline{\ln y})}{\sum (\ln x_i - \overline{\ln x})^2} \approx -1.4963$ - Calculate intercept $\ln a = \overline{\ln y} - b \overline{\ln x} \approx 21.3623$ 7. **Rewrite regression equation:** $$y = e^{21.3623} x^{-1.4963}$$ Calculate $a = e^{21.3623} \approx 1.9 \times 10^{9}$ (rounded to nearest ten-thousandth: 1900000000) 8. **Final regression equation:** $$y = 1900000000 \cdot x^{-1.4963}$$ 9. **Find orbital speed for $y=1256$ million km:** Solve for $x$: $$1256 = 1900000000 \cdot x^{-1.4963}$$ $$x^{-1.4963} = \frac{1256}{1900000000} \approx 6.61 \times 10^{-7}$$ $$x = \left(6.61 \times 10^{-7}\right)^{-\frac{1}{1.4963}}$$ Calculate exponent: $$-\frac{1}{1.4963} \approx -0.668$$ So, $$x = (6.61 \times 10^{-7})^{-0.668} \approx 123.7$$ 10. **Answer:** The orbital speed of an asteroid 1256 million km from the sun is approximately **123.7 km/s**.