Question * 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,-3Y+4) is equal to: Hint: cov(ax + by.cx+dY) = acV(X) + bdv(Y) + (bc + ad)cov(X,Y) O-120 -30 -59 -64 O
Question * 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,-3Y+4) is equal to: Hint: cov(ax + by.cx+dY) = acV(X) + bdv(Y) + (bc + ad)cov(X,Y) O-120 -30 -59 -64 O
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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![Question *
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,-3Y+4) is equal to:
Hint: cov(ax + by.cx+dY) = acV(X) + bdv(Y) + (bc + ad)cov(X,Y)
O -120
-30
-59
-64
O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf50eeea-ab68-4e53-9b5b-756342f418dc%2F99dadcd6-ce67-4cfc-afd6-632c38ac9ef6%2Fhsvfe7q_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Question *
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,-3Y+4) is equal to:
Hint: cov(ax + by.cx+dY) = acV(X) + bdv(Y) + (bc + ad)cov(X,Y)
O -120
-30
-59
-64
O
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