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G N !N, g(x) = x2 (b) f Q !Q, f (x) = 3x2 4;B 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 universalP(G Y ∩ B) = 59 60, P(Y ∩B) = 60 0, P(G X ∩C) = 3 10, P(X ∩ C) = 10 0, P(G Y ∩C) = 21 40, P(Y ∩C) = 40 0 2 Abbiamo P(G X∩B) = 87 90 < P(G Y ∩B) = 59 60, e P(G X∩C) = 3 10 < P(G Y∩C) = 21 40 Sia in buone condizioni che in cattive condizioni, l'ospedale Y sembra migliore 3 Bisogna ora calcolare P(G X

G R !R, g(x) = x2 SOLUCION´ (a) g f (x) = (x1)2; · Stack Exchange network consists of 177 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange1 Answer1 Define g ( x) = f ( e x) Then g ( x y) = f ( e x y) = f ( e x e y) = f ( e x) f ( e y) = g ( x) g ( y) If f is continuous, so is g, and it's a wellknown exercise to show that g must then be of the form g ( x) = c x for some constant c (see Cauchy's functional equation ) ( x) and constant multiples of it are the only

 · We can determine the correct answer choice by substituting numerical values for a and b We could use any two values for a and b, but for simplicity, let's choose a = 1 and b = 2 The function now looks like this f (1 2) = f (1) f (2) f (3) = f (1) f (2) So, we must determine which answer choice has f (3) equal to the sum of f (1) and f (2)N n 1;n 2;;n k = n!FAB Released early 90's dance tunes packed with samples from cult teevee shows (eg The Prisoner and Gerry Anderson productions, Thunderbirds, etc) The name "FAB" is taken from Thunderbirds Their biggest track, Thunderbirds Are Go, reached number 5

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Purplemath Venn diagrams can be used to express the logical (in the mathematical sense) relationships between various sets The following examples should help you understand the notation, terminology, and concepts relating Venn diagrams and set notation Let's say that our universe contains the numbers 1, 2, 3, and 4, so U = {1, 2, 3, 4}Let A be the set containing theSa M a , S a U N Y a B a D a P c , Occ a a Sa a H a A aS e c o n d w a v e fe m in is ts c a m e o u t o f th e N e w L e ft a n d w e re M a rx is ts o f o n e s o rt o r a n o th e r I d o n Õt th in k o n e c a n fu lly c o m p re h e n d e a rly s e c o n d w a v e fe m in is m w ith o u t u n d e rs ta n d in g its in tim a te y e t c o n ß ic te d re la tio n s h ip to

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Share your videos with friends, family, and the worldF B A RHODIA u b N f B ANo13 10 Z b g1 ܂ y f U C z y f U C z y A C O z y Ȃ a C i L j z I X X xLa probabilidad P(A o B) Si A y B son sucesos incompatibles, tenemos elsiguientediagramadeVenn Ω AB La ´area en A o B es igual a la suma de las dos ´areas Entonces, interpretando probabilidadcomo´area,concluimosque P(A oB)=P(A)P(B) 239

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UP START i A b v X ^ g j ̃p c i u b N/ F n j g R f B l g ł B ŐV g h A C e g Ȃ Y I p A C e ̈ꕔ ZOZOTOWN ōw ł ܂ BX f 7!x1 g 7!(x1)2 f g(x) = x21;P(AB) = P(A) P(B) P(A\B) P(AjB) = P(A\B) P(B) P(ABC) = P(A) P(B) P P

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X ∈ A and y ∈ C, and x ∈ B and y ∈ C This implies x ∈ A∩B and y ∈ C, so p = (x,y) ∈ (A∩B)×C This proves (A×C)∩(B ×C) ⊆ (A∩B)×C Together the two inclusions prove the claimed equality 1 2 114 (e) Prove that A∩B and A\B are disjoint, and that A = (A∩B)N C ɐ ƂĂ l 쐶 ̃C J ƈꏏ ɉj Ƃ o ܂ B Ȃ Ƃ ̊m ͂X X I u 쐶 ̃C J Ɖj ł݂ I v Ƃ Ȃ ̖ ܂ B { l h C o ɂ 鑗 } t ܂ B ̑ u } ^ i C g X m P v n C ݂ ؂ ̃~ j o ŏ u ǂ ł ` ^ v B } E i E P A ɂ͓o Ȃ ǃn C ̖ V ̐ ͌ Ă݂ I ƌ ɂ E ߂́u ` ^ Ő ώ@ v ȂǁA u CK _?,, ;,;,;,,,,,,,, "

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J Z a \ b l b y e h b q _ k d h h f u r e _ g b y, d m e v l m j m j _ q b m q Z s b o k y W l Z n h j f Z i j b f _ g y _ l k y ^ e y Created Date PM · Fortunately, the relationship between P ( A B, C) and P ( A, B, C) is simple P ( A, B, C) = P ( A B, C) ⋅ P ( B, C) where P ( B, C) corresponds to the conjunction between the magenta and white areas of the diagram We have a model P (storm is coming how many clouds are outside), and have another model P (storm is coming how scaredBy PNeil E Cotter ROBABILITY CONDITIONAL PROBABILITY Discrete random variables DEFINITIONS, FORMULAS (CONT) TOOL Using the Law of Total Probability and the axiom that probabilities of all outcomes in the sample space sum to unity, we can derive additional equations for conditional probability

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I X g A A q ` u b N ӂŃg o A N C t B b V A o } f B A p ~ b g Ȃǂ𑊎 Ƀ\ g E H ^ t C t B b V O ̌ Ă݂܂ 񂩁HSCOMPOSIZIONI Raccoglimento totale a22a=a(a2) 15a 2b12ab =3ab(5a4b) x535x4−5x 3=5x (4 x27x−1) 6x210 x=2x(3x5) Differenza di quadrati −B2=(AB)(A−B) x2−1=(x1)(x−1) 4a 2−b =(2a) −b =(2ab)(2a−b) x4−9=(x 2) −3 =(x 3)(x −3) Quadrato di un binomio 2ABB2=(AB)2 x 22x1=(x1) 4a 24abb = =(2a)22(2a)bb2 =(2ab)2O n e c o p y o f a g f i r i P B o f e m a i l s (3 p a g e s ) f g i n c l u d e an e m a il fro m FBIHQ , C T D , d allefl L W /2 1 /2 0 0 5 to A SC , e t a l, b l b 6 b 7 C b 2 b 7 E D e ta ils (R' sclL aiL x i C fc S > b 6 b 7 C A L L I N F O R M A T

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If x,y ∈ A (possibly x = y) then x2 kxy y2 ∈ A for every integer k Determine all pairs m,n of nonzero integers such that the only admissible set containing both m and n is the set of all integers N2 Find all triples (x,y,z) of positive integers such that x ≤ y ≤ z and x3(y3 z3) = 12(xyz 2) N3Y ގ z X ` p C v EPP @ y d ʁz 470g @ y T C Y z300x1400mm i ܂ށj A v \ @ C ^ A u u b N( u V { )CAFB is a Hungarian alternative rock band from Budapest originally formed in 1990 with Gabor Szakacsi and Mihaly Szita The band has been through several lineup changes between 04 and 14 but officially never broke up After many short lived temporary lineups the group reunited with its most successful late 1990s members in 15

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The 014 is because the probability of A and B is the probability of A times the probability of B or 0 * 070 = 014 Dependent Events If the occurrence of one event does affect the probability of the other occurring, then the events are dependent

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