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G B hèeight Pem"> P—(al rviceŒ ³o³o Preface Docu„Ê”úA—šibil. Copyright © 09 ̃R ~ j P V @ b z y W b I C X g W. O B N F T J U F !.

J p ̃v X ` b N p b h p ӂ Ă Ȃ Ƃ J ܂ B Ȓ A w قƂ ǂ ׂĂ̈ R K l ɕt v X ` b N p b h x. X Check the box if the filing relates solely to preliminary communications made before the commencement of a tender offer Check the appropriate boxes below to designate any transactions to which the statement relates x thirdparty tender offer subject to Rule 14d1 o issuer tender offer subject to Rule 13e4. Deze dienst is gratis en bedacht en ontwikkeld door DJP Ieder jaar worden er nieuwe afkortingen verzonnen Vooral via MSN , SMS , Twitter en Whatsapp wordt veel gebruik gemaakt van een afkorting Bekende afkortingen zijn cu (see you = tot ziens), w8 (wacht), 2M (tomorrow = morgen), FYI (for your information = voor jouw informatie).

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Editorial contacts David Shane, HP 1 650 857 3859 corpmediarelations@hpcom Christina Schneider, HP 1 650 857 22 corpmediarelations@hpcom. P قɂ z ̂ q l ݂̂łȂ A n ̕ C y ɏW 悤 ɂ Ɓu I I I r t H A t ^ v ŏ Ɋ A. E B t @ O u v ~ A E H b N p 26cm C W j I l I L ̏ڍ Ɠd ʔ̂ EC J g B ݌ɏ󋵂 A J g i A i r Ȃǂ̏ m F ł ܂ B Ȃ 5,500 ~( ō ) ȏ w ő I.

Pm k=1 k 2, and therefore f(m1) = f(m)(m1)2 = mX1 k=1 k2 19 Decide for which n the inequality 2n > n2 holds true, and prove it by mathematical induction The inequality is false n = 2,3,4, and holds true for all other n ∈ N Namely, it is true by inspection for n = 1, and the equality 24 = 42 holds true for n = 4. X~n p ܅ K, yM _ \ = q )S^ 熒Xb 5 jt Eȸ P Ȳ es l w 7 _ *@ k pWkr #6 , K M} {P Q ʇ~ I_H, e\ = 9k Xψ g ރ Yy s G o ( EP O D T 94 u V > UW F yc G C D M _ 7 n 㺺 #% { {p & eƗ u $ G V 񍂭` ~ W̽ & o #_ 9 ki yS &6>` d%ԧ z aS qMt> "߂ G pO O_ m P 6& ~ ^ & b 70 3w l o 1. P) = p=100 La résolution de l'équation donne alors pour p6= 0 , x p= r 2 2 ln 1 p 100 ;.

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P lim n!1 X n = X = 1 There are measure theoretic subtleties to be aware of here In particular, the sample space inside the probability statement here is set of in nite sequences and it requires some machinery to be precise here There are other equivalent (this is somewhat di cult to see) ways to de ne almost sure. P 2 Q R S T U V W X Y E D Z \ E ^ P _ ‘ a b S c d V W e f Y E g h / i j k G & l m n G & / o p & q r s % E & Z t F D / u v w x;. Et pour p= 0 tout x p 0 convient 3 On admettra le résultat Z 1 0 e x2dx= p ˇ 2 L'espérance de Xexiste car la fonction t7!t2= 2e t2=(2 2) est intégrable sur 0;1 De plus, E(X) = Z 1 0 t2 2 exp t2 2 2 dt = 2 p 2 Z 1 0 x2e x2dx en faisant le.

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