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G d ±The two factors are Q(g(x)) and (g(x) − g(a)) / (x − a) The latter is the difference quotient for g at a, and because g is differentiable at a by assumption, its limit as x tends to a exists and equals g′(a) As for Q(g(x)), notice that Q is defined wherever f is Furthermore, f is differentiable at g(a) by assumption, so Q is continuous at g(a), by definition of the derivative TheF z « H F I Ó
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Title INBPANUApdf Author kkasprzak Created Date 117 PM$$=f(\lim_{h\rightarrow 0}xbh)f(\lim_{h\rightarrow 0}xah)$$ $$=f(xb)f(xa)$$ Share Cite Follow answered Jun 1 ' at 1814 crystal_math crystal_math 1,507 4 4 silver badges 16 16 bronze badges $\endgroup$ Add a comment Your Answer Thanks for contributing an answer to Mathematics Stack Exchange!T o p r o v id e in p u t d a t a t h a t a r e alw a ys needed a n d w o u ld b e b o r in g to sp ecify with the v a r i a b l e n a m e = v a r i a b l e v a l u e syntax used b y NAMELIST IN P U T CARDS require data in sp eciÞc o r der ( which m a y dep e nd on the s ituation and o n t he value o f a c a r d fo rmat sp eciÞer ) F o r in s t a n c e I N P U T _ C A R D c a r d _ f o r
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E(X i) =μand V(X i) = σ2 are assumed Consider arithmetic average X =(1/n) n i=1 X i Then, mean and variance of X are given by E(X) =μ, V(X) = σ2 n 138 Proof The mathematical expectation of X is given by E(X) =E(1 n n i=1 X i) = 1 n E(n i=1 X i) = 1 n n i=1 E(X i) = 1 n n i=1 μ= 1 n nμ=μ E(aX) =aE(X) in the second equality and E(X Y) = E(X)E(Y) in the third equality are utilized' % c ¹C X Y g V z G I r X ¥
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To Z and Y we obtain EE(ZjY)= E(Z) which is the same as EE(g(X)jY)= E(g(X)) 2 This property may seem to be more general statement than (i) in Theorem 1 The proof above shows that in fact these are equivalent statements 3 E(XYjY)=YE(XjY) Proof E(XYjY =y)=E(yXjY =y)=yE(XjY =y)(because y is a constant) Hence, E(XYjY)= YE(XjY) by the definition of the conditionalA c = v 2 /r m/s 2 Acceleration (gravity) g = F/m m/s 2 Force F = ma N or kgm/s 2 Weight F wt = mg N or kgm/s 2 Force (gravity) N or kgm/s 2 Force (Coulomb) F = k q 1 q 2 /(d 2) N or kgm/s 2 Force (magnetic) F m = Bqv N or kgm/s 2 Force (magnetic) F m = BIL N or kgm/s 2 Force (friction) F f = mF N N or kgm/s 2 Torque Τ = Fl Nm Momentum p = mv kgm/s ImpulseL g h ?
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H 9 3 ¿Department of Computer Science and Engineering University of Nevada, Reno Reno, NV 557 Email Qipingataolcom Website wwwcseunredu/~yanq I came to the USS T D A U F E E L V W N X M Y Z A B K C U D B E L I E V E F Word search puzzle words to find S E E H E A R R E A D F E E L T A L K H O P E K N O W T H I N K G U E S S D R E A M B E L I E V E U N D E R S T A N D Reprinted with permission by the Dana Alliance for Brain Initiatives "FORMING WORDS" This puzzle gives you five key words to work with Each key word can be inserted in
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