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Suppose that X and Y ave random variables with E(X) = 2 , E'(Y) = 5 and E(X*) = 8, E(Y®) = 30 and cov(X – 2Y, 3X – 4Y) = -12, then cov(2X – 3, –5Y + 4) is eqpual to: Hiat: cov(aX + bY.cX + dY) = acV(X) + baV(Y) + (hc + ad)cov(X,Y) -120 O -64 O -59 -30

Suppose that X and Y ave random variables with E(X) = 2 , E'(Y) = 5 and E(X*) = 8, E(Y®) = 30 and cov(X – 2Y, 3X – 4Y) = -12, then cov(2X – 3, –5Y + 4) is eqpual to: Hiat: cov(aX + bY.cX + dY) = acV(X) + baV(Y) + (hc + ad)cov(X,Y) -120 O -64 O -59 -30

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Suppose that X and Y are rundom variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8, E(Y) = 30 and cov(X – 2Y,3X – 4Y) =-12, then cov(2X – 3, -5Y + 4) is equal to: Hint: cov(aX + bY.cX + dY) = acV(X) + baV(Y) + (bc + ad)cov(X,Y) -120 O ID -64 -59 -30
Transcribed Image Text:Suppose that X and Y are rundom variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8, E(Y) = 30 and cov(X – 2Y,3X – 4Y) =-12, then cov(2X – 3, -5Y + 4) is equal to: Hint: cov(aX + bY.cX + dY) = acV(X) + baV(Y) + (bc + ad)cov(X,Y) -120 O ID -64 -59 -30
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