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Chapter 13: Nonlinear and Multiple Regression43.a. x 1 = 2.6, x 2 = 250, and x 1 x 2 = (2.6)(250) = 650, so( 2.6) − 0.3015( 250) + 0.0888( 650) 48. 313y ˆ = 185.49 − 45.97=b. No, it is not legitimate to interpret β1in this way. It is not possible to increase by 1 unitthe cobalt content, x 1 , while keeping the interaction predictor, x 3 , fixed. When x 1changes, so does x 3 , since x 3 = x 1 x 2 .c. Yes, there appears to be a useful linear relationship between y and the predictors. Wedetermine this by observing that the p-value corresponding to the model utility test is

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