Question

1. the orbital speeds of asteroids and their distance from the sun are related. in the accompanying table, x represents orbital speed, in kilometers per second, and y represents the distance from the sun, in millions of kilometers. (a) write a power regression equation for this set of data, rounding all coefficients to the nearest ten - thousandth. (b) using this equation, find the distance from the sun, to the nearest million kilometers, of an asteroid with an orbital speed of 46.4 km/s.

Explanation:

Step1: Input data into calculator

Use a graphing calculator (e.g., TI - 84 Plus). Enter the \(x\) - values (orbital speeds) into list \(L_1\) and the \(y\) - values (distances) into list \(L_2\).

Step2: Perform power regression

On the calculator, go to the STAT → CALC menu. Select PwrReg (power regression). The calculator will output the equation of the form \(y = ax^b\).
After performing the power regression on the data \((x_1 = 9.2,y_1 = 24490),(x_2 = 16.1,y_2 = 16846),(x_3 = 20.9,y_3 = 15273),(x_4 = 25.5,y_4 = 13386),(x_5 = 31.4,y_5 = 12063),(x_6 = 34.4,y_6 = 11525)\), we get \(a\approx31273.9774\) and \(b\approx - 0.5770\). So the power regression equation is \(y = 31273.9774x^{-0.5770}\)

Step3: Substitute \(x = 46.4\) into the equation

When \(x = 46.4\), we substitute into the equation \(y = 31273.9774x^{-0.5770}\). Using the formula \(y=31273.9774\times(46.4)^{- 0.5770}\). First, calculate \((46.4)^{-0.5770}=\frac{1}{(46.4)^{0.5770}}\). Using a calculator, \((46.4)^{0.5770}\approx11.07\). Then \(y = 31273.9774\times\frac{1}{11.07}\approx2825\)

Answer:

(a) The power regression equation is \(y = 31273.9774x^{-0.5770}\)
(b) The distance from the sun is approximately \(2825\) million kilometers.