câu 65 khác nhau giữa marketing 1 0 2 0 3 0 philipkotler

Báo cáo toán học: "Generalized Cauchy identities, trees and multidimensional Brownian motions. Part I: bijective proof of generalized Cauchy identities ´ Piotr Sniady" pot

Báo cáo toán học: "Generalized Cauchy identities, trees and multidimensional Brownian motions. Part I: bijective proof of generalized Cauchy identities ´ Piotr Sniady" pot

Ngày tải lên : 07/08/2014, 13:21
... operator Amer J Math., 12 6 (1) : 12 1 18 9, 20 0 4 [DH04b] Ken Dykema and Uffe Haagerup Invariant subspaces of the quasinilpotent DT-operator J Funct Anal., 20 9 (2) :3 32 36 6, 20 0 4 [DY 03] Kenneth Dykema and ... combinatorics 13 ( 20 0 6), #R 62 Figure 3: A graph G corresponding to the sequence = ( +1, 1, +1, +1, 1, 1, +1, 1) The dashed lines represent the pairing σ = {1, 6}, {2, 3} , {4, 5}, {7, 8}} 1. 3 Quotient ... l1 + · · · + lm + For ≤ i ≤ m we set i = +1 , 1 , , ( 1) i 1 , ( 1) i , , li times li 1 times l1 times the electronic journal of combinatorics 13 ( 20 0 6), #R 62 l1 times +1 , 1 , (8) li−1...
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Statistics, Probability and Noise

Statistics, Probability and Noise

Ngày tải lên : 13/09/2012, 09:49
... 0 01 0 0 01 3 0 01 9 00 26 00 35 00 47 00 62 00 82 01 0 7 01 3 9 01 7 9 02 28 02 87 03 59 04 46 05 48 06 68 08 08 09 68 11 51 135 7 15 87 18 41 21 1 9 24 20 27 43 30 8 5 34 46 38 21 4 20 7 4 6 02 500 0 0. 0 0 .1 0. 2 0. 3 0. 4 0. 5 0. 6 0. 7 0. 8 ... -3. 4 -3. 3 -3 .2 -3 .1 -3 .0 -2. 9 -2. 8 -2. 7 -2. 6 -2. 5 -2. 4 -2. 3 -2. 2 -2 .1 -2 .0 -1. 9 -1. 8 -1. 7 -1. 6 -1. 5 -1. 4 -1. 3 -1. 2 -1. 1 -1. 0 -0. 9 -0. 8 -0. 7 -0. 6 -0. 5 -0. 4 -0. 3 -0. 2 -0 .1 0. 0 00 03 00 05 00 07 0 01 0 ... to one 90 10 0 11 0 12 0 13 0 14 0 Value of sample 15 0 16 0 17 0 0 .06 0 c Probability Density Function (pdf) 0. 0 50 Probability density 90 0 .04 0 0. 0 30 0 . 02 0 0. 01 0 0. 000 90 10 0 11 0 12 0 13 0 14 0 15 0 Signal...
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Bài giảng Chapter 17 Free Energy and Thermodynamics

Bài giảng Chapter 17 Free Energy and Thermodynamics

Ngày tải lên : 28/11/2013, 01:12
... 18 8. 83 17 3. 51 198.49 11 6. 73 19 1. 50 2 10 . 62 51. 45 31 . 88 Al2O3(s) Br2(g) C(graphite) CO2(g) CaO(s) CuO(s) Fe2O3(s) H2O2(l) H2O(l) HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) 51. 00 24 5 .3 5.69 21 3 .6 ... + 90. 37 0 Al2O3 Br2(g) C(graphite) CO2(g) CaO(s) CuO(s) Fe2O3(s) H2O2(l) H2O(l) HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) -16 69.8 + 30 . 71 -39 3.5 - 635 .5 -15 6 .1 - 822 .16 -18 7.8 -28 5. 83 - 92. 30 +25 .94 ... Approach 37 Substance S° J/mol-K Substance S° J/mol-K Al(s) Br2(l) C(diamond) CO(g) Ca(s) Cu(s) Fe(s) H2(g) H2O(g) HF(g) HBr(g) I2(s) N2(g) NO(g) Na(s) S(s) 28 .3 15 2. 3 2. 43 19 7.9 41. 4 33 . 30 27 .15 13 0. 58...
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Probability and Mathematical Genetics potx

Probability and Mathematical Genetics potx

Ngày tải lên : 05/03/2014, 22:21
... Subordinated Jacobi diffusion processes Subordinated coalescent process 31 5 31 9 31 9 3 20 32 5 32 9 33 3 34 5 34 5 34 9 35 5 35 8 35 9 3 61 36 8 3 71 37 5 16 Three problems for the clairvoyant demon Geoffrey Grimmett Introduction ... OK Corral Bull Lond Math Soc., 31 ( 5), 6 01 606 [M 1 02 ] Kingman, J F C 20 0 0 Origins of the coalescent: 19 74 19 82 Genetics, 15 6, 14 61 14 63 [M 1 03 ] Kingman, J F C 20 0 2 Stochastic aspects of Lanchester’s ... theory The multitype case Anomalous spreading Discussion of anomalous spreading 11 3 11 3 11 5 11 6 11 9 12 0 12 2 12 4 12 5 13 0 Kingman, category and combinatorics N H Bingham and A J Ostaszewski Introduction...
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PROBABILITY AND MATHEMATICAL STATISTICS pptx

PROBABILITY AND MATHEMATICAL STATISTICS pptx

Ngày tải lên : 17/03/2014, 14:20
... by { (1, 1) (2, 1) (3, 1) S= (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, ... Answer: 23 23 24 + + 11 10 24 24 = + 10 11 25 = 11 25 ! = (14 )! (11 )! = 4, 457, 400 n Example 1. 10 Use the Binomial Theorem to show that ( 1) r r =0 Answer: Using the Binomial Theorem, we get n (1 + ... Mathematical Statistics 31 Answer: The sample space of this random experiment is { (1, 1) (2, 1) S= (3, 1) (4, 1) (1, 2) (2, 2) (3, 2) (4, 2) (1, 3) (1, 4) (2, 3) (2, 4) (3, 3) (3, 4) (4, 3) (4, 4)} Let...
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Probability and Statistics by Example pptx

Probability and Statistics by Example pptx

Ngày tải lên : 28/03/2014, 10:20
... format isbn - 13 isbn - 10 978 -0- 511 - 13 28 3- 4 eBook (NetLibrary) 0- 511 - 13 28 3 -2 eBook (NetLibrary) isbn - 13 isbn - 10 978 -0- 5 21 - 84766-7 hardback 0- 5 21 - 84766-4 hardback isbn - 13 isbn - 10 978 -0- 5 21 - 6 12 33 -3 paperback ... distributions The Central Limit Theorem 10 8 1. 5 1. 6 1. 7 2 .1 2. 2 2. 3 33 54 75 96 10 8 14 2 16 8 Part II Basic statistics 19 1 3 .1 3 .2 3. 3 3. 4 3. 5 19 3 19 3 20 4 20 9 21 3 21 5 Parameter estimation Preliminaries ... probability 10 −5 × 10 1 × 10 1 = 10 −7 A f B w C f, probability 10 −5 × · 10 1 × 10 −5 = · 10 11 A f B f C w, probability 10 −5 × 10 1 × · 10 1 = · 10 −7 A w B f C f, probability − 10 −5 × 10 −5 × 10 1 = − 10 −5...
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Báo cáo khóa học: High level cell-free expression and specific labeling of integral membrane proteins doc

Báo cáo khóa học: High level cell-free expression and specific labeling of integral membrane proteins doc

Ngày tải lên : 30/03/2014, 13:20
... mM 20 mM 20 mM 1. 2 mM 0. 8 mM 0. 8 mM 0. 8 mM mM 0. 2 mM tablet per 10 mL – – – 1 1. 5 mM 20 mM 20 mM 1. 2 mM 0. 8 mM 0. 8 mM 0. 8 mM mM 0. 2 mM tablet per 10 mL 10 0 mM 2. 8 mM 13 mM 2 90 mM 2% 0. 05% 10 0 mM ... A19 pET21a(+) pQB1-T7-gfp pQB1-emrE-gfp pET-gfp pET-emrE pET-sugE pET-tehA pET-yfiK E coli B ompT rne1 31 thr -1 leuB6 thi -1 lacY1 glnV44 rfbD1 recA1 lac[F’Tn 10 (Tetr) lacIq lacZM15] rna19 gdhA2 ... concentrations were 1. 5fold CMC (TX -11 0, TX -11 4, DDM) and twofold CMC (Thesit) Lipid concentrations were mgÆmL )1 DDM, n-dodecyl-b-D-maltoside; TX 100 , Triton X - 10 0; TX -11 4, Triton X -11 4; LPC, L-a-phosphatidylcholine;...
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applied probability and stochastic processes - bryc

applied probability and stochastic processes - bryc

Ngày tải lên : 31/03/2014, 16:23
... : : : : : : : : : : 87 89 89 90 90 90 91 92 93 93 93 94 94 94 97 97 98 99 10 0 10 0 10 0 10 1 10 1 10 2 10 3 10 3 10 5 10 5 10 6 10 7 10 8 10 8 10 9 10 9 11 0 11 0 11 0 11 1 11 3 13 .1 A simple probabilistic modeling ... = n 10 0 0 .07 959 0. 08 20 0 0. 5 627 8 30 0 0. 04 6 03 0. 0 6 10 0 0.5 63 72 500 0. 035 66 0. 03 700 0. 5 63 91 700 0. 03 01 5 0. 02 20 0 0. 5 639 9 Table 1. 2: Probabilities PrX = n in 2n Binomial trials 12 A possible heuristic ... : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 13 1 13 1 13 2 13 3 13 4 13 4 13 5 13 5 13 5 13 6 13 6 13 6 13 7 13 7 13 8 13 9 14 0 14 4 viii Course description Description This course is aimed at...
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crc - standard probability and statistics tables and formulae - daniel zwillinger

crc - standard probability and statistics tables and formulae - daniel zwillinger

Ngày tải lên : 31/03/2014, 16:23
... Figure 2. 2 Class Frequency Relative frequency Cumulative relative frequency 17 0. 0 50 0. 20 0 0. 425 0. 22 5 0. 075 0. 025 0. 0 50 0 .2 50 0. 625 0. 900 0. 975 1. 00 0 [ 40, 50) [ 50, 60) [ 60, 70) [ 70, 80) [ 80, 90) ... follows: Chapter 10 12 14 16 18 Number of pages 18 30 56 36 40 40 26 23 Number of words 4 514 5 426 12 23 4 23 92 9948 18 418 817 9 11 739 518 6 Occurrences of “the” 15 9 14 7 15 9 47 15 3 11 8 26 4 22 3 82 An interactive ... (S1) x = 66.5 , ¯ s2 = 11 5.64 (for grouped data), c = 10 (S2) corrected variance = 11 5.64 − ( 10 2 / 12 ) = 10 7. 31 (S3) r mr mrc mr mrc 66.5 4 535 .0 31 6 9 62. 5 22 69 21 2 5 .0 66.5 4 526 .7 31 5 30 0 .0 22 688 8 02 .9...
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fundamentals of probability and statistics for engineers - t t soong

fundamentals of probability and statistics for engineers - t t soong

Ngày tải lên : 31/03/2014, 16:23
... DISTRIBUTIONS 6 .1 Bernoulli Trials 6 .1. 1 Binomial D istribution Contents 44 46 49 49 51 55 61 66 67 75 76 76 79 83 86 87 88 92 92 93 98 99 10 1 10 8 11 2 11 2 11 9 11 9 12 0 13 4 13 7 14 5 14 7 15 3 15 4 16 1 16 1 16 2 TLFeBOOK ... Problems 25 9 2 61 26 2 26 3 26 4 26 4 26 5 26 6 27 4 27 5 27 7 27 7 29 4 30 6 30 6 30 7 10 MODEL VERIFICATION 10 .1 Preliminaries 10 .1. 1 Type-I and Type-II Errors 10 .2 Chi-Squared Goodness-of-F it Test 10 .2 .1 The ... 20 9 21 1 21 2 21 5 21 9 2 21 22 3 22 5 22 6 22 8 23 3 23 4 23 8 23 8 23 8 23 9 PART B: STATISTICAL INFERENCE, PARAMETER ESTIMATION, AND MODEL VERIFICATION 24 5 OBSERVED DATA AND GRAPHICAL REPRESENTATION 24 7 8.1...
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probability and finance it's only a game

probability and finance it's only a game

Ngày tải lên : 31/03/2014, 16:24
... Stochastic Theory 2 01 20 3 21 5 2 21 22 6 22 9 2 31 10 Games for Pricing Options in Discrete Time 10 .1 Bachelier’s Central Limit Theorem 10 .2 Bachelier Pricing in Discrete Time 10 .3 Black-Scholes Pricing ... index ISBN 0- 4 71- 4 02 26 -5 (acid-free paper) Investments-Mathematics Statistical decision Financial engineering Vovk, Vladimir, 19 60- 11 Title 11 1 Series ~ HG4S 15 SS34 20 01 3 32 ' . 01 ' 1 -dc 21 Printed ... Instruments 13 .3 Weak and Strong Prices 13 .4 Pricing an American Option 31 31 8 32 3 32 8 32 9 14 Games for Diffusion Processes 14 .1 Game-Theoretic Dijfusion Processes 14 .2 It6 's Lemma 14 .3 Game-Theoretic...
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probability and its applications - ollav kallenberg

probability and its applications - ollav kallenberg

Ngày tải lên : 31/03/2014, 16:24
... xn ) → x1 · · · xn The convolution is said to be associative if ( 12 ) ∗ 3 = 1 ∗ ( 23 ) whenever both 12 and 23 are σ-finite and commutative if 12 = 21 1 A measure ... · × Sn 1 × B) = 1 (s, ds1 ) ··· 2 (s1 , ds2 ) · · · µn 1 (sn 2 , dsn 1 )µn (sn 1 , B) Exercises Prove the triangle inequality µ(A∆C) ≤ µ(A∆B) + µ(B∆C) (Hint: Note that 1A∆B = |1A − 1B |.) Show ... If ϑ is U (0, 1) , we get P {f (s, ϑ) ≤ x} = P {ϑ ≤ µ(s, [0, x])} = µ(s, [0, x]), x ∈ [0, 1] , and so f (s, ϑ) has distribution µ(s, ·) by Lemma 2. 3 ✷ Proof of Theorem 2 .19 : By Lemma 2. 22 there exist...
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probability and measurements - tarantola a.

probability and measurements - tarantola a.

Ngày tải lên : 31/03/2014, 16:24
... 56 61 63 65 67 69 70 70 70 71 73 75 78 78 79 79 81 82 84 84 85 88 88 89 10 0 10 0 10 2 10 3 10 3 10 4 10 6 11 6 11 6 11 6 12 0 12 0 12 2 12 3 12 5 13 0 13 1 13 7 14 0 14 3 14 4 ... REDRAWN y x0 x 1. 2 T/K 1. 2 T/K A A 0. 8 0. 8 B 0. 6 B 0. 6 0. 4 0. 4 0. 2 0. 2 0 0 .2 0. 4 0. 6 0. 8 0. 2 P/(N m -2) 0. 4 0. 6 0. 8 0. 8 P/(dyne m -2) 1. 2 T/K 1. 2 T/K 1 0. 8 0. 8 0. 6 0. 6 0. 4 0. 4 0. 2 0. 2 0 0 .2 0. 4 0. 6 P/(N ... 15 3 15 4 15 5 15 7 15 8 15 8 15 9 16 0 16 1 16 2 16 2 16 3 16 4 16 4 16 5 16 8 16 9 16 9 17 1 17 6 17 6 17 8 18 3 18 5 18 6 18 7 18 9 18 9 19 1 19 4 19 5 19 5 19 5 19 7 20 0 ...
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probability and statistical inference - nitis mukhopadhyay

probability and statistical inference - nitis mukhopadhyay

Ngày tải lên : 31/03/2014, 16:25
... Procedure 3 41 3 41 3 42 3 42 34 4 3 51 3 51 35 4 35 8 35 8 36 5 36 6 3 71 37 4 37 5 37 7 37 7 3 80 3 82 39 5 39 5 39 6 39 9 4 01 4 01 4 13 416 417 417 4 20 422 425 425 426 428 429 4 41 4 41 4 43 444 Contents xvii 9 .2. 2 9 .2. 3 9 .2. 4 ... 88 89 99 99 10 0 10 1 10 3 10 7 10 7 11 5 11 9 12 4 12 5 13 1 13 9 14 1 14 1 14 4 14 5 14 5 14 8 14 9 15 2 15 7 15 8 15 9 17 7 17 7 17 9 17 9 18 1 18 2 18 5 18 7 19 0 19 2 19 5 20 6 Contents xv 4.6 4.7 4.8 4.9 4.5 .1 The Student’s ... Complements 2 41 2 41 24 2 25 3 25 6 25 7 26 4 26 4 26 4 26 5 26 5 2 70 Sufficiency, Completeness, and Ancillarity 2 81 2 81 28 2 28 4 28 8 29 4 29 5 30 0 3 01 30 4 30 9 31 4 31 6 31 8 3 20 32 4 32 7 6 .1 6 .2 6 .3 6.4 6.5 6.6...
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probability and combinations

probability and combinations

Ngày tải lên : 14/05/2014, 16:55
... http://gmatclub.com/forum/permutation-question-884 92. html 11 In how many ways can the letters of the word PERMUTATIONS be arranged if there are always letters between P and S A 2 419 20 0 B 25 4 01 6 00 C 18 14 400 D 19 26 30 0 E 13 21 5 00 There are 12 letters ... Favourable outcomes = 10 C1 as there are 10 pairs and we need ONE from these 10 pairs Total # of outcomes= 20 C2 as there are 20 shoes and we are taking from them P = 10 C1 /20 C2 = 10 / (19 * 10 ) =1/ 19 Answer: C ... to 10 2 inclusive C From 12 2 to 12 6 inclusive D From 12 8 to 13 2 inclusive E From 19 6 to 20 0 inclusive Tickets can be distributed in the following ways: {8 ,0} {6 ,2} {4,4} {2, 6} {0, 8} - 8C8 =1 8C6*2C2 =28 ...
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against coherence truth probability and justification jun 2005

against coherence truth probability and justification jun 2005

Ngày tải lên : 10/06/2014, 21:52
... Thagard Part IV 11 2 11 2 11 6 11 9 12 3 12 5 13 4 14 3 15 6 15 6 15 7 15 9 16 2 Scepticism and Incoherence 10 Pragmatism, Doubt, and the Role of Incoherence 10 .1 10. 2 95 95 10 1 10 2 10 5 10 8 Other Views How ... Anti-scepticism 9 12 16 21 24 26 27 30 31 32 34 34 35 36 38 39 48 49 52 55 58 61 61 66 69 72 74 xii contents C A J Coady’s Radical Justification of Natural Testimony 5 .1 5 .2 5 .3 5.4 5.5 5.6 The ... Includes bibliographical references and index ISBN 0 -19 - 927 999 -3 (alk paper) Truth–Coherence theory Skepticism I Title BD1 71. O47 20 0 5 12 1–dc 22 20 0 4 02 906 0 Typeset by Newgen Imaging Systems (P) Ltd.,...
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Báo cáo hóa học: " Research Article Subspace-Based Noise Reduction for Speech Signals via Diagonal and Triangular Matrix Decompositions: Survey and Analysis" docx

Báo cáo hóa học: " Research Article Subspace-Based Noise Reduction for Speech Signals via Diagonal and Triangular Matrix Decompositions: Survey and Analysis" docx

Ngày tải lên : 22/06/2014, 19:20
... 0 20 0 0 400 0 10 20 0 0 i=4 20 0 0 400 0 20 0 0 400 0 20 0 0 400 0 i=9 20 0 0 400 0 20 0 0 400 0 i = 11 i = 12 10 20 0 0 10 i = 10 400 0 i=8 10 0 400 0 10 20 0 0 20 0 0 10 i=7 0 i=6 10 10 0 400 0 i=5 10 i =3 10 400 0 10 20 0 0 ... 400 0 i=7 20 0 0 400 0 i = 10 20 0 0 400 0 i=5 0 20 0 0 400 0 20 0 0 400 0 i = 13 0 20 0 0 400 0 i = 11 20 0 0 400 0 i=6 0 20 0 0 400 0 i=9 0 20 0 0 400 0 20 0 0 400 0 20 0 0 400 0 i = 12 20 0 0 400 0 i = 14 0 i = 15 2 400 0 20 0 0 ... Jensen 21 i =1 0 20 0 0 400 0 i=4 20 0 0 400 0 i=7 400 0 20 0 0 400 0 i = 10 20 0 0 400 0 i=8 20 0 0 400 0 i = 13 20 0 0 400 0 i = 11 20 0 0 400 0 20 0 0 400 0 i=9 20 0 0 400 0 20 0 0 400 0 20 0 0 400 0 i = 12 20 0 0 400 0 i = 14 i = 15 ...
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