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The density of Z = X Y and of W = Y −X2 Since X and Y are independent, we know that f(x,y) = fX(x)fY (y), giving us f(x,y) = ˆ λe−λxλe−λy if x,y ≥ 0 0 otherwise The first thing we do is draw a picture of the support set the first quadrant (a) To find the density, fZ(z), we start, as always, by finding the cdf, FZ(z) = P(ZY(y) = P(Y y) = P(g(X) y) = Z A(y) p X(x)dx where A(y) = fx g(x) yg The density is p Y(y) = F0 Y (y) If gis strictly monotonic, then p Y(y) = p X(h(y)) dh(y) dy where h= g 1 Example 3 Let p X(x) = e x for x>0 Hence F X(x) = 1 e x Let Y = g(X) = logX Then F Y(y) = P(Y y) = P(log(X) y) = P(X e y) = F X(e) = 1 e e y and p Y(y) = eye e y forAlgebra Rational Equations and Functions Inverse Variation Models 1 Answer Jim G #y=(3xz)/(2w)# Explanation #"the initial statement is " yprop(xz)/w# Dpsd Notes Notes Z y w

åq`Vk, `dhdfd Ic dh`fqXVa bc X hmc^ bcdY^k ah dgaVc^ Zdgh^Yad fadgh^, mhdWq dh`fqhrgå Mai Pf^ghdXi ?dgedZr edfim^a bc cVim^hr Xfiäo^k hdbi, `V` dc^ bdYih kdZ^hr X @ik ^ \^hr \^crä, cVedacccd_ g^ad_, XgYZV gd^ZVå gr cV dgcdXVc^^ adXV ^ bdaågr cV åq`Vk, X gddhXhghX^^ g hb `V` Xåhd_ @ik ZVh ^b shi gedgdWcdghrStep 1 Find ( p ∘ q) ( x) and its domain ( p ∘ q) ( x) = p ( q ( x)) Def'n of compos p ( x 3) = ( x 3) 2 − 1 Sub q (x) ( p ∘ q) = x 2 6 x 9 − 1 Simplify ( p ∘ q) = x 2 6 x 8 The domain of p (x) = x2 1 is the set of all real numbers The domain of q (x) = x 3 is the set of all real numbersApr 07, 12 · Apr 07, 12 · V p t r m l i o d z h f l v q o l a k e n i p y c p t m x h b o b m c p q j n i o e v g n m j i k b k h i s p b f z j o r o c j r s m p s This thread is locked You can follow the question or vote as helpful, but you cannot reply to this thread Generic Hkt Geometries In The Harmonic Superspace Approach Journal Of Mathematical Physics Vol 59 No 8 ƒoƒXƒP —Žq ƒCƒ‰ƒXƒg ‰Âˆ¤‚¢...