Suppose that X and Y are random variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8, E(Y²) = 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) + bdV(Y) + (bc + ad)cov(X,Y) -59 -120 -64 -30
Suppose that X and Y are random variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8, E(Y²) = 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) + bdV(Y) + (bc + ad)cov(X,Y) -59 -120 -64 -30
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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
Transcribed Image Text:Suppose that X and Y are random 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) + bdV(Y) + (bc + ad)cov(X,Y)
-59
-120
-64
-30
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