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O = 1 since H Cl is a substance, you need to also multiply its H by 5 Right side P = 1 Cl = 1 x 5 = 5 H = 3 (1 x 5) = 8 O = 4 P Cl5 H 2O = H 3P O4 5H Cl Now the H atom on the left side is not balance with the right side, we have to multiply the H on the left side by 49 4 3 2 8 6 7 6 5 4 3 2 1 0 / N E L M L K J B I H G F E D C B A ?PC Glossary Definitions of Computer and Internet Terms A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Common Computer Terms #




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Let X be a discrete random variable and its possible outcomes denoted VFor example, if X represents the value of a rolled die then V is the set {,,,,,}Let us assume for the sake of presentation that X is a discrete random variable, so that each value in V has a nonzero probability For a value x in V and an event A, the conditional probability is given by (=)Z h d y j x f u k w h q h v f u k p m r j e t f s p k f r r e j q p m o f n m f l j k j i h g f e d c f f f q p h 7 1 6 < 6 ;"main" 07/2/16 page 295 47 Change of Basis 295 Solution (a) The given polynomial is already written as a linear combination of the standard basis vectors Consequently, the components of p(x)= 5 7x −3x2 relative to the standard basis B are 5, 7, and −3 We write




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Here C max denotes the concentration at the peak volume, V p at the trailing boundary (Fig 1) and K 2 stands for the dimerization constant The V 1p o value may be estimated from extrapolation of the V p values to zero concentration, and the V 2p ∝ value may be set as the V p value of blue dextran;Where X,Y,Z are, respectively, A, or Ac, B, or , and C, or Cc Then they are also pairwise independent The result extends to any finite collection of events, in an obvious way Proof We can write PA∩ B= P(A∩ B∩ C) ∪ (A∩ B∩ Cc)= PAPBPC PAPBPCc, because the two parts are disjoint This is equal to PAPB(PCC Give the contents of registers PC, AR, DR, AC, and IR in hexadecimal and the values of E, I, and the sequence counter SC in binary at the end of the instruction cycle AC = 7EC3




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