nd the regression equation, letting the diameter be the predictor (x) variable. Find the best predicted circumference of a marble with a diameter of 1.5 cm. How does the result compare to the actual circumference of 4.7 cm? Use a significance Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball ameter 7.4 24.3 4.2 21.9 7.1 ircumference 23.2 76.3 13.2 68.8 22.3 4.1 21.2 12.9 66.6 Click the icon to view the critical values of the Pearson correlation coefficient r he regression equation is y=+K Cound to five decimal places as needed.) he best predicted circumference for a diameter of 1.5 cm is c cm. Cound to one decimal place as needed.) ow does the result compare to the actual circumference of 4.7 cm? A. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. B. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. OC. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference. D. Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference.