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If the random variables X, Y, and Z have the means ux = 3, uy = -2, and uz = 2, the variances o = 3, o = 3, o2 = 2, the covar

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ANSWER :Given that Mx=3 2 My = -2 M2=2,642-34-3 con(x, y) = -2 Con(x,2)=-1 and con (712) = = 2 2 U = 4-2 N=x-Y+22 a Mee My - Mz - 2-2< = 3+3+4(2) - (-) + (-)---(() = 10 (b cor Cu andu) a con(y-2, X-4422) = Cor Cyix) - Cor Cy14) +2 cov(412) cow (2x) + cow (2,

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