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Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex · 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ NU ~ j } C Y v Ƃ́A O ̃v C ̃x b g ⃌ C Y ɏ 悹 ă C Y ꍇ ̍Œ z ̂ Ƃ ܂ B ʓI ɁA m ~ b g e L T X z f J ł́A u O ̃v C ̃x b g z { i O ̃v C ̃x b g z 2 O ̃v C ̃x b g z j v Ōv Z ܂ B Ƃ ΁A r b O u C h iBB j 30 h x b g A ̍Let f(x) = Z x 0 (t−1)(t−2)2 e t 2 dt Find all the critical points of f(x), and determine whether the critical point is a local maximum, local minimum, or neither Show

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See the answer Find the series for f(x) = e ^ x^z and determine the radits of convergence Expert Answer Previous question Next question Get more help from CheggM(n)(0) = E(), n ≥ 1 (8) The mgf uniquely determines a distribution in that no two distributions can have the same mgf So knowing a mgf characterizes the distribution in question If X and Y are independent, then E(es(XY )) = E(esXesY) = E(esX)E(esY), and we conclude that the mgf of an independent sum is the product of the individual mgfLearn with content Watch learning videos, swipe through stories, and browse through concepts

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Let \u0393 x Z t x 1 e t dt a Compute \u and \u 8 b Use the fact that lim t t s e t Let γ x z t x 1 e t dt a compute γ1 and γ2 8 b use School University of California, Los Angeles;California' s Dual Eligible Integration Responses to Requests for Information This Request for Information (RFI) due on June 1, 11, the California Department of Health Care Services (DHCS) has solicited for input concerning the ideal model or models for serving beneficiaries who are eligible for both Medicare and MediCal Dually eligible beneficiaries are not a homogenousOr f X(x) = x 1 2 e x 2 21 2 (1 2) This is the pdf of (1 2;2), and it is called the chisquare distribution with 1 degree of freedom We write, X˘˜2 1 The moment generating function of X˘˜2 1 is M X(t) = (1 2t

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X)(Z Z) = EXZ X Z Since there is only one outcome in which both student 1 and student 2 get their correct homework back (namely when all 3 get them back), it has P(XZ= 1) = 1=6 so the expectation is EXZ = 1 1=6 0 5=6 Plugging into our expression for the covariance we ndArtists DENZ x ZE x JIGGZProduced by DNOTEONDABEATLabels VVS MUSIC & TEAM PLATINASpotify https//openspotifycom/track/7rWa98E9KCkY5I78TjY3m4?si=TF8AZVX Z p x) ) F X(x) = F Z(p x) F Z(p x) Therefore f X(x) = 1 2 x 1 2 1 p 2 ˇ e 1 2 x 1 2 x 1 2 1 2ˇ e 1 2 = 1 1 2 px 1 2 e x 2;

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