not even sure where to begin to solve part E An article gave a scatter plot along with the least squares line of x = rainfall volume (m³) and y= runoff volume (m³) for a particular location. The accompanying values were read from the plot.6 12 14 20 23 30 40 5155 67 72 80 96 112 127y 4 10 13 14 15 25 27 46 38 46 53 71 8299102A USE SALT(a) Does a scatter plot of the data support the use of the simple linear regression model?Yes, the scatterplot shows a reasonable linear relationship.O Yes, the scatterplot shows a random scattering with no pattern.O No, the scatterplot shows a reasonable linear relationship.O No, the scatterplot shows a random scattering with no pattern.(b) Calculate point estimates of the slope and intercept of the population regression line. (Round your answers to four decimal places.)slope0.8466intercept-2.4342(c) Calculate a point estimate of the true average runoff volume when rainfall volume is 55. (Round your answer to four decimal places.)44.1288m3(d) Calculate a point estimate of the standard deviation o. (Round your answer to two decimal places.)(e) What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall? (Round your answer to four decimal places.)

the data shows the rainfall volume and runoff volume.(e) The sums are, x y x2 y2 xy 6 4 36 16 24 12 10 144 100 120 14 13 196 169 182 20 14 400 196 280 23 15 529 225 345 30 25 900 625 750 40 27 1600 729 1080 51 46 2601 2116 2346 55 38 3025 1444 2090 67 46 4489 2116 3082 72 53 5184 2809 3816 80 71 6400 5041 5680 96 82 9216 6724 7872 112 99 12544 9801 11088 127 102 16129 10404 12954 Sum = 805 Sum = 645 Sum = 63393 Sum = 42515 Sum = 51709 The coefficient of variation is, r2=n∑xy-∑x∑y2n∑x2-∑x2n∑y2-∑y2=15×51709-805×645215×63393-805215×42515-6452=6574608810067146279000=0.9791 Thus, the proportion of the observed variation in runoff column can be attributed to the simple linear regression relationship between runoff and rainfall is 0.9791.