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We want to fit v = k1 * exp(-k2 * u) to the data: (0, 1.95), (1, 1.28), (2, 0.70) by least-squares fitting of a straight line to: Ln(v) = c1 + c2 (-u). The matrix b = FT *y is 0.558 0.467 0.558 0.467 3.93 2.68 O[ 0.558 2.680

We want to fit v = k1 * exp(-k2 * u) to the data: (0, 1.95), (1, 1.28), (2, 0.70) by least-squares fitting of a straight line to: Ln(v) = c1 + c2 (-u). The matrix b = FT *y is 0.558 0.467 0.558 0.467 3.93 2.68 O[ 0.558 2.680

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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We want to fit v = k1 * exp(-k2 * u) to the data: (0, 1.95), (1, 1.28), (2, 0.70) by least-squares fitting of a straight line to: Ln(v) = c1 + c2 * (-u). The matrixb = FT * y is %3! 0.558 0.467 0.558 -0. 467. 3. 93 -2.68 0.558 -2.680
Transcribed Image Text:We want to fit v = k1 * exp(-k2 * u) to the data: (0, 1.95), (1, 1.28), (2, 0.70) by least-squares fitting of a straight line to: Ln(v) = c1 + c2 * (-u). The matrixb = FT * y is %3! 0.558 0.467 0.558 -0. 467. 3. 93 -2.68 0.558 -2.680
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