The owner of a moving company typically has his most experienced manager predict the total number of labor hours that will be required to complete an upcoming move. In an effort to provide a more accurate method, the owner has decided touse the number of cubic feet moved as the independent variable and has collected data for 20 moves in which the travel time was an insignificant portion of the hours worked. Use the data provided to complete parts (a) through (d) below.E Click the icon to view the data table.....a. Construct a scatter plot. Choose the correct graph below.Data tableO A.OB.С.Cubic Feet MovedLabor Hours60-900-60-61631.5071933.7526313.7545520.500-900Cubic Feet60900Cubic Feet55129.25Hours71234.75b. Assuming a linear relationship, use the least-squares method to determine the regression coefficients b, and b,.41218.7529614.50bo =38421.00b, =45122.5060326.75(Round to three decimal places as needed.)44221.7565131.0033616.2582643.5039617.0032813.2579440.2533015.5047221.00Cubic Feet

Calculate Fitting a straight line - Curve fitting using Least square method X Y 616 31.50 719 33.75 263 13.75 455 20.50 551 29.25 712 34.75 412 18.75 296 14.50 384 21 451 22.50 603 26.75 442 21.75 651 31 336 16.25 826 43.50 396 17 328 13.25 794 40.25 330 15.50 472 21 Solution:Straight line equation is y=a+bx.The normal equations are∑y=a*n+b∑x∑xy=a∑x+b∑x^2The values are calculated using the following table X y x=X-501 x^2 x⋅y 616 31.5 115 13225 3622.5 719 33.75 218 47524 7357.5 263 13.75 -238 56644 -3272.5 455 20.5 -46 2116 -943 551 29.25 50 2500 1462.5 712 34.75 211 44521 7332.25 412 18.75 -89 7921 -1668.75 296 14.5 -205 42025 -2972.5 384 21 -117 13689 -2457 451 22.5 -50 2500 -1125 603 26.75 102 10404 2728.5 442 21.75 -59 3481 -1283.25 651 31 150 22500 4650 336 16.25 -165 27225 -2681.25 826 43.5 325 105625 14137.5 396 17 -105 11025 -1785 328 13.25 -173 29929 -2292.25 794 40.25 293 85849 11793.25 330 15.5 -171 29241 -2650.5 472 21 -29 841 -609 --- --- --- --- --- ∑X=10037 ∑y=486.5 ∑x=17 ∑x^2=558785 ∑x⋅y=29344 Substituting these values in the normal equations20a+17b=486.517a+558785b=29344Solving these two equations using Elimination method,20a+17b=486.5and 17a+558785b=2934420a+17b=486.5→(1)17a+558785b=29344→(2)equation(1)×17⇒340a+289b=8270.5equation(2)×20⇒340a+11175700b=586880Substracting ⇒-11175411b=-578609.5⇒11175411b=578609.5⇒b=578609.5/11175411 ⇒b=0.052Putting b=0.052 in equation (1), we have20a+17(0.052)=486.5⇒20a=486.5-0.88⇒20a=485.62⇒a=485.62/20⇒a=24.281∴a=24.281 and b=0.052Now substituting this values in the equation is y=a+bx, we gety=24.281+0.052xy=24.281+0.052(X-501) y=-1.771+0.052X Hence, b0=-1.771 and b1=0.052