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In Reference to the Equation y^=0.80+0.12x1+0.08x2\hat { y } = - 0.80 + 0.12 x _ { 1 } + 0.08 x _ { 2 }

question 20

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In reference to the equation y^=0.80+0.12x1+0.08x2\hat { y } = - 0.80 + 0.12 x _ { 1 } + 0.08 x _ { 2 } , the value -0.80 is the y-intercept.


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