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(b) P V>4 = 1−P V ≤ 4 = 1−F V (4) = 1−81/144 = 63/144 (2) (c) P −3 a=1−F V (a)=1−(a5)2/144 = 2/3(4) The unique solution in the range −5 ≤ a ≤ 7isa =4 √ 3−5=1928SF B x = =0 0 x B B= y SM A = 0 50 100 0 45 150 0 9 0 9 0 ( ) ( )( ) ( )( ) ( )N N m N md d =d B dm 50 45 100 135 150 0 9d d d B d B A B = 180 0 9 300 N m ( m) (1) SM B = 0 50 0 9 0 9 100 ( )( ) ( ) (N m m N)(045m) A d = 0 45 0 9 45 0 = A Ad d A A = (09 m) 90 N m (2) Since B˜180 N, EqA positive sign means the wave is traveling in the x direction Part (e) To determine the maximum transverse speed of the string, remember that all parts of the string are experiencing simple harmonic motion We showed that in SHM the maximum speed is v max = Aw In this case we have A = 09 cm and w = 50 rad/s, so v max = 09 * 50 = 45 cm/s

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8 Let W have the density function given by f W(w) = 2w for 0 < w < 1 and f W(w) = 0 for other values of w Set Y = eW (a) Determine the distribution function and quantiles of W (b) Determine the distribution function, density function, and quantiles of Y (c) Determine the mean and variance of Y directly from its density function(b) {w w contains at least three 1s} Σ∗1Σ∗1Σ∗1Σ∗ (c) {w w contains the substring 0101, ie, w = x0101y for some x and y} Σ∗0101Σ∗ (d) {w w has length at least 3 and its third symbol is a 0} ΣΣ0Σ∗ (e) {w w starts with 0and has odd length, or starts with 1and has even length} ∗ (f) {w wGeometrically the probability that X is between a and b, ie, P(a X b), is then represented by the area shown shaded, in Fig 22 The distribution function F(x) P(X x) is a monotonically increasing function which increases from 0 to 1 and is represented by a curve as in Fig 23 d dx3 x a f (u) du f (x) F(x) μ 0 x 1 x 2 1 x 2 1 x 2 dF(x) dx

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Health and Wellness Health and Wellness word search to download and print or play online Add your own words to customize or start creating from scratch Recommended Check out this Advance Word Search Maker to create commercial use printable puzzles NOTE Puzzle contains randomly generated words so be sure to proofread carefully before usingB F(A,B,C,D) = D (A' C') 6 a Since the universal gates {AND, OR, NOT can be constructed from the NAND gate, it is universalIn type theory, a system has inductive types if it has facilities for creating a new type from constants and functions that create terms of that type The feature serves a role similar to data structures in a programming language and allows a type theory to add concepts like numbers, relations, and treesAs the name suggests, inductive types can be selfreferential, but usually

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W @ @ @ @ @ r f (fh) @ ہf u c (ty) @ ہf { (fh) @ o ہf (ty) @ a ʁ@ l @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ @ l ^ (pg) @ r @1, if x ≤ a A simple computation yields E(X) = ab 2 Var(X) = (b−a)2 12 M(s) = esb −esa s(b−a) When a = 0 and b = 1, this is known as the uniform distribution over the unit interval, f(x) = 1 1U ^ \ r q x z q t w _ r w x p q x w ` w a f g hhg i j k g l m b c x x w y t d \ y u r s t o q x ^ w w e o p p v u x w p q x w p n w r s t o u v v w x o y r x o w o p _ r w x p q x w \ f o * 1 # 8 4;

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W w p w w X w C C 0 1 ( ) (413) 45 Here any signal with frequency components smaller than w C will be untouched, whereas all other frequencies will be forced to zero Hence, a discrete time continuous frequency ideal LPF configuration Through the computation of inverse DTFT we obtain C( ) 2 1 p p p w n x n e dw C Sinc0, if x ≥ b;" 7 j > a a d \ x b a w x w z s ` u s w _ ^ u s \ x t y x w z s s z s y x w v u s t s r !

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 Insert the following letters into what is originally an empty Btree of order 5 C N G A H E K Q M F W L T Z D P R X Y S Order 5 means that a node can have a maximum of 5 children and 4 keys All nodes other than the root must have a minimum of 2 keys¨ # 0 ¢ ;Page 2 c) d) wx x′y BD B′D′ A′B 312) Simplify the following Boolean functions in products of sums a) F(w, x, y, z) = ∑ (0, 2, 5, 6, 7, 8, 10)

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(c) Find Q, W, and E int (a) PV P V ifi f 140 1 500 atm 139 atm 300 i fi f V PP V (b) 10 Pa 1 10 m5 3 3 365 K 0 mol14 JmolK ii i PV T nR 10 Pa 300 10 m5 3 3 253 K 0 mol14 JmolK ff f PV T nR (c) The process is adiabatic Q 0 140P V VV C RC CC , 5 V 2 CRN fe a d a c a b a a a s a ܏\ f a a a a a a a a a b ԍ F 1 `50 A 51 `100 A 101 `150 A 151 `0 A 1 `250 A 251 `300 A 301 `350 A 351 `387 A 3 `429 { X 430 `448 ^ ŁEVer BAnswers to problems marked with ~,appear at the end of the book 31'~ Simplify the following Boolean functions, using threevariable maps (a) F(x, y, z) = L(o, 2,6,7) (b) F(x, y, z) =L(o, 1,2,3,7) 32 Simplify the following Boolean functions, using threevariable maps (aY F(x, y, z) =L(o, 1,5,7) (b) F(x, y, z) =L(o, 1,6,7) (c) F(x, y, z) = L(I, 3, 5, 7) 33* Simplify the following Boolean

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£ 9 2 # ;The Fokker CX was originally designed for the Royal Dutch East Indies Army, in order to replace the Fokker CV Like all Fokker aircraft of that time, it was of mixed construction, with wooden wing structures and a welded steel tube frame covered with aluminium plates at the front of the aircraft and with fabric at the rear It typically contains a GH dipeptide 1124 residues from its Nterminus and the WD dipeptide at its Cterminus and is 40 residues long, hence the name WD40 Between the GH and WD dipeptides lies a conserved core It forms a propellerlike structure with several blades where each blade is composed of a fourstranded antiparallel betasheet

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(b) n(Ac), where A = {x xisanevennumberfrom1to10} (c) n(B), where B = {1,3,9} (d) n(?) Addition Rule for Sets Very Useful Formula If A and B are finite sets then n(AB)=n(A)n(B)n(A\B) 2 If n(B)=13,n(AB)=24,andn(A\B)=6,findn(A) 8 Fall 17, Maya Johnson 3 In a survey of 272 people, a pet food manufacturer found that 69 owned a dog but

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