Algebraic Solution of Equations 1: You can use the sum and product properties to find algebraically the exact solutions of certain equations. For 2124, solve the equation by first transforming it to a product equal to zero and then setting each factor equal to zero. Use the domain [0, 360] or x [0, 2 ]. cos 5x cos x = 0
Surname 1 Name: Tutor: Course: Date: a) Fitting a regression model y=348.8458−1.30386x1−0.15122x2 Where y is the wear, x1 the oil viscosity and x2 the load. b) Under the assumption that the error term has a constant variance, then the estimate of the variance is given by 2 SSE σ = n−2 2 1919.244 σ = 3 =639.748 c) ANOVA D f SS MS F Significance F Regression 2 12192.76 6096.378 9.529345 0.050155 Residual 3 1919.244 639.7479 Total 5 14112 d) Surname 2 95% CI for the intercept coefficient 95%CI for bo=bo±z∗SEbo 95%CI for bo=348.8458±1.96∗74.7855 ¿(202.266,495.425) 95% CI for the oil viscosity coefficient 95%CI for bo=bo±z∗SEbo 95%CI for bo=−1.3039±1.96∗1.1654 ¿(−3.5881,0.9803) 95% CI for the load coefficient 95%CI for bo=bo±z∗SEbo 95%CI for bo=−0.1512±1.96∗0.08962 ¿(−0.3269,0.0245) e) 90% CI for the mean wear 90%CI for meanwear=bo±z∗MSE 90%CI forbo=348.8458±2.58∗639.7479 ¿(−1301.7038,1999.395) Surname 3 90% prediction interval for the observed wear at x1=20 and x2=1000 y=348.8458−1.30386x1−0.15122x2 y=348.8458−1.30386(20)−0.15122(1000) y=171.5486 90%CI forobserved wear=observed ±z∗MSE 90%CI forbo=171.5486±2.58∗639.7479 ¿
