The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 51 inches. Is the result close to the actual weight of 362 pounds? Use a significance level of 0.05. Chest size (inches) 45 50 43 43 52 52 Weight (pounds) 352 374 275 314 440 367 Click the icon to view the critical values of the Pearson correlation coefficient r. . . . . . What is the regression equation? y= [+x(Round to one decimal place as needed.) What is the best predicted weight of a bear with a chest size of 51 inches? The best predicted weight for a bear with a chest size of 51 inches is pounds. (Round to one decimal place as needed.) Is the result close to the actual weight of 362 pounds? O A. This result is close to the actual weight of the bear. O B. This result is exactly the same as the actual weight of the bear. O C. This result is very close to the actual weight of the bear. O D. This result is not very close to the actual weight of the bear.Find 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 level of 0.05. Baseball Basketball Golf Soccer Tennis Ping-Pong Volleyball Diameter 7.3 23.9 4.3 21.9 6.9 3.9 21.3 Circumference 22.9 75.1 13.5 68.8 21.7 12.3 66.9 Click the icon to view the critical values of the Pearson correlation coefficient r. The regression equation is y = + x. (Round to five decimal places as needed.) The best predicted circumference for a diameter of 1.5 cm is cm. (Round to one decimal place as needed.) How does the result compare to the actual circumference of 4.7 cm? O A. Even though 1.5 cm is beyond the scope of the sample diameters, the predicted value yields the actual circumference. O B. Since 1.5 cm is within the scope of the sample diameters, the predicted value yields the actual circumference. O C. Since 1.5 cm is beyond the scope of the sample diameters, the predicted value yields a very different circumference. O D. Even though 1.5 cm is within the scope of the sample diameters, the predicted value yields a very different circumference