Suppose that X andY are random variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8, E(Y?) = 30 and cov(2X – Y,4X – 3Y) = -12, then cov(-5X – 3, 2Y + 4) is equal to: Hint: cov(aX + br, сх + dr) %3 асV(х) + bdv (Y) + (be + ad)cov(X,Y) -64 -30 -120 -59

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.8: Probability
Problem 19E
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Suppose that X andY are random variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8,
E(Y?) = 30 and cov(2X – Y,4X – 3Y) = -12, then cov(-5X – 3, 2Y + 4) is equal to:
Hint: cov(aX + br, сх + dr) %3 асV(х) + bdv (Y) + (be + ad)cov(X,Y)
-64
-30
-120
-59
Transcribed Image Text:Suppose that X andY are random variables with E(X) = 2 , E(Y) = 5 and E(X²) = 8, E(Y?) = 30 and cov(2X – Y,4X – 3Y) = -12, then cov(-5X – 3, 2Y + 4) is equal to: Hint: cov(aX + br, сх + dr) %3 асV(х) + bdv (Y) + (be + ad)cov(X,Y) -64 -30 -120 -59
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