The line of best t is 1 t 26 we are trying solve c d

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Unformatted text preview: E/∂ C = 2(C − 2D − 4) + 2(C − D − 2) + 2(C + 1) + 2(C + D) + 2(C + 2D) = 10C − 10 and ∂ E/∂ D = 2(C − 2D − 4)(−2)+2(C − D − 2)(−1)+0+2(C + D)(1)+2(C +2D)(2) = 20D +20 so we need to solve 5C = 5, 10D = −10 so C = 1 and D = −1. The line of best fit is 1 − t. 26) We are trying solve C + D = 0, C + E = 1, C − D = 3 and C − E = 4, that is to 1 1 1 0 0 1 0 11 1 1 400 1 1...
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This note was uploaded on 01/30/2014 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell University (Engineering School).

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