câu 40 thiết kế một câu hỏi trắc nghiệm về khái niệm marketing với 4 lựa chọn

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

... shows points (3) and (4) in the definition of an intermediate point 4. 4 The backward transform We shall prescribe now the inverse of the transform prescribed in Section 4. 3 (i.e a single interation ... Funct Anal., 11(1):693– 741 , 2001 [DH04a] Ken Dykema and Uffe Haagerup DT-operators and decomposability of Voiculescu’s circular operator Amer J Math., 126(1):121–189, 20 04 [DH04b] Ken Dykema and Uffe ... George N Raney Functional composition patterns and power series reversion Trans Amer Math Soc., 94: 441 45 1, 1960 ´ ´ [Sni03] Piotr Sniady Multinomial identities arising from the free probability theory...

27
241
0

Statistics, Probability and Noise

Ngày tải lên : 13/09/2012, 09:49

... 0013 0019 0026 0035 0 047 0062 0082 0107 0139 0179 0228 0287 0359 044 6 0 548 0668 0808 0968 1151 1357 1587 1 841 2119 242 0 2 743 3085 344 6 3821 42 07 46 02 5000 0.0 0.1 0.2 0.3 0 .4 0.5 0.6 0.7 0.8 0.9 ... mean, constant standard deviation 4 Amplitude Amplitude 19 -2 -2 -4 -4 64 128 192 256 320 Sample number 3 84 448 512 511 64 128 192 256 320 Sample number 3 84 448 511 512 FIGURE 2-3 Examples of ... = 0.2 a Mean = 0.5, F = 6 Amplitude Amplitude 13 -2 -2 -4 -4 64 128 192 256 320 Sample number 3 84 448 511 512 64 128 192 256 320 3 84 448 512 511 Sample number FIGURE 2-1 Examples of two digitized...

24
540
0

Bài giảng Chapter 17 Free Energy and Thermodynamics

Ngày tải lên : 28/11/2013, 01:12

... HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) 51.00 245 .3 5.69 213.6 39.75 42 .59 89.96 109.6 69.91 186.69 206.3 260.57 192.5 240 . 45 205.0 248 .5 Relative Standard Entropies States • the gas state ... H2O(g) HF(g) HBr(g) I2(s) N2(g) NO(g) Na(s) S(s) 28.3 152.3 2 .43 197.9 41 .4 33.30 27.15 130.58 188.83 173.51 198 .49 116.73 191.50 210.62 51 .45 31.88 Al2O3(s) Br2(g) C(graphite) CO2(g) CaO(s) CuO(s) ... C3H8(g) + O2(g) → CO2(g) + H2O(g) has ∆Hrxn = -2 044 kJ at 25°C Calculate the entropy change of the surroundings Given: Find: Concept Plan: ∆Hsystem = -2 044 kJ, T = 298 K ∆Ssurroundings, J/K T, ∆H Relationships:...

53
378
0

Probability and Mathematical Genetics potx

Ngày tải lên : 05/03/2014, 22:21

... Conclusion 44 6 44 8 45 0 45 1 46 1 20 Queueing with neighbours V Shcherbakov and S Volkov Introduction Results Asymmetric interaction Symmetric interaction Appendix 46 4 46 4 46 7 47 0 47 5 48 0 21 Optimal ... The case of diﬀusivity ε2 I Monte Carlo computation of the asymptotic growth rate xi 40 0 40 2 40 3 40 5 40 8 41 0 41 1 18 Heavy traﬃc on a controlled motorway F P Kelly and R J Williams Introduction ... Brownian network model A model of a controlled motorway Route choices Concluding remarks 41 6 41 6 41 8 42 2 42 6 43 0 43 8 44 2 19 Coupling time distribution asymptotics for some couplings of the L´vy stochastic...

548
902
0

PROBABILITY AND MATHEMATICAL STATISTICS pptx

Ngày tải lên : 17/03/2014, 14:20

... 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, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) ... proof is now complete Example 1.9 Evaluate 23 10 + 23 + 24 11 Answer: 23 23 24 + + 11 10 24 24 = + 10 11 25 = 11 25! = ( 14) ! (11)! = 4, 45 7, 40 0 n Example 1.10 Use the Binomial Theorem to show that ... 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 A denote the event of getting a sum of and B...

703
505
0

Probability and Statistics by Example pptx

Ngày tải lên : 28/03/2014, 10:20

... inequality Exponential families Confidence intervals Bayesian estimation 218 225 229 233 4. 1 4. 2 4. 3 4. 4 4. 5 4. 6 4. 7 4. 8 4. 9 Hypothesis testing Type I and type II error probabilities Most powerful tests ... 344 365 3 64 × ×···× = 49 27 365 365 365 and for n = 23: 1− 365 3 64 343 × ×···× = 5 243 365 365 365 In (ii), the number of outcomes not in the event is 364n and the probability in question − 3 64/ 365 ... with two balls each (probability p4 ) or one cell with one ball and one cell with three balls (probability p4 ) Hence, p4 = p4 + p4 = n n−1 n−2 n × 4+ × nn = 21 64 Here stands for a number of ways...

373
402
0

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

... resulted in the refolding of the pumps Antimicrob Agents Chemother 41 , 44 0 44 4 proteins, and renaturation procedures with strong denatu 14 Aleshin, V.V., Zakataeva, N.P & Livshits, V.A (1999) A new ... (v/v) DDM after h at 45 °C; lane 2, EmrE in 2% (v/v) DDM after h at 75 °C; lane 3, SugE in 3% (v/v) b-OG after h at 40 °C; lane 4, SugE in 20% (v/v) NDSB-201 after h at 40 °C; lane 5, SugE in ... spectra yielded an estimate of 55 ± 4% a-helical content for EmrE, 72 ± 11% (DPC), 60 ± 11% (SDS) and 84 ± 10% (DDM) for SugE and 78 ± 8% (DDM), 49 ± 3% (DPC) and 40 ± 15% (SDS) for TehA The predicted...

13
318
0

applied probability and stochastic processes - bryc

Ngày tải lên : 31/03/2014, 16:23

... : : : : : : : : : : : 28 30 31 31 32 34 34 34 35 35 35 37 38 38 39 39 39 40 40 41 41 42 42 43 45 45 45 47 47 47 Normal distribution 49 Limit theorems 53 4. 1 Herschel's law of errors : : : : : ... its output: 4 4 4 4 1 5 5 5 5 5 1 6 6 6 6 6 Total | 4 4 | 4 4 | 1 | 5 5 | 4 5 | 5 4 | 5 5 | 5 1 | 6 6 | 4 6 | 6 | 6 6 | 6 of 150 four | 4 4 | 4 4 | 5 5 | 2 4 | 5 5 | 5 5 | 5 | 5 6 | 2 4 | 6 6 | ... 6 of a kind 5 5 6 5 5 6 | | | | | | | | | | | | | 4 5 5 6 6 4 5 6 6 4 5 5 6 6 5 5 6 4 5 5 6 | | | | | | | | | | | | | 4 5 5 6 6 4 5 6 6 5 6 6 4 5 6 6 5 6 6 The program is written explicitly for...

163
357
0

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

Ngày tải lên : 31/03/2014, 16:23

... follows: Chapter 10 12 14 16 18 Number of pages 18 30 56 36 40 40 26 23 Number of words 45 14 542 6 122 34 2392 9 948 1 841 8 8179 11739 5186 Occurrences of “the” 159 147 159 47 153 118 2 64 223 82 An interactive ... not in the ith location Example 3.13 : All the derangements of {1, 2, 3, 4} are: 2 143 3 142 41 23 2 341 341 2 43 12 241 3 342 1 43 21 The number of derangements of n elements, Dn , satisﬁes the recursion ... 3.2 Combinatorial methods 3.3 3 .4 Probability Random variables 3.5 3.6 3.7 Mathematical expectation Multivariate distributions Inequalities 4. 1 4. 2 4. 3 4. 4 4. 5 4. 6 4. 7 Functions of Random Variables...

537
1.4K
0

fundamentals of probability and statistics for engineers - t t soong

Ngày tải lên : 31/03/2014, 16:23

... 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 99 101 108 112 112 119 119 120 1 34 137 145 147 153 1 54 161 161 162 TLFeBOOK Contents 6.1.2 ... Deviation 4. 1.3 Conditional Expectation 4. 2 Chebyshev Inequality 4. 3 Moments of Two or More R andom Variables 4. 3.1 Covariance and Correlation Coefficient 4. 3.2 Schwarz Inequality 4. 3.3 The Case ... of Three or More R andom Variables 4. 4 Moments of Sums of R andom Variables 4. 5 Characteristic F unctions 4. 5.1 Generation of Moments 4. 5.2 Inversion F ormulae 4. 5.3 Joint Characteristic F unctions...

408
655
0

probability and finance it's only a game

Ngày tải lên : 31/03/2014, 16:24

... Large Numbers 4. 1 Two Statements of Kolmogorov ’s Strong Law 4. 2 Skeptic’s Strategy 4. 3 Reality’s Strategy 4. 4 The Unbounded Upper Forecasting Protocol 4. 5 A Martingale Strong Law 4. 6 Appendix: ... 121 1 24 126 133 143 144 Lindeberg ’s Theorem 7.1 Lindeberg Protocols 7.2 Statement and Proof of the Theorem 7.3 Examples of the Theorem 7 .4 Appendix: The Classical Central Limit Theorem 147 148 ... 13 .4 Pricing an American Option 31 318 323 328 329 14 Games for Diffusion Processes 14. 1 Game-Theoretic Dijfusion Processes 14. 2 It6 's Lemma 14. 3 Game-Theoretic Black-Scholes Diffusion 14. 4 Appendix:...

417
426
0

probability and its applications - ollav kallenberg

Ngày tải lên : 31/03/2014, 16:24

... Appendices A1 Hard Results in Measure Theory A2 Some Special Spaces 45 5 Historical and Bibliographical Notes 46 4 Bibliography 48 6 Indices Authors Terms and Topics Symbols 509 Chapter Elements of ... functionals as compensators Riesz decomposition 40 9 xii Foundations of Modern Probability 23 Semimartingales and General Stochastic Integration 43 3 predictable covariation and L -integral semimartingale ... result, it is enough to verify the deﬁning condition for a generating subclass 4 Foundations of Modern Probability Lemma 1 .4 (measurable functions) Consider two measurable spaces (S, S) and (T, T ),...

534
378
0

probability and measurements - tarantola a.

Ngày tải lên : 31/03/2014, 16:24

... 4 7 8 9 9 10 10 11 11 11 14 14 14 16 18 19 19 20 23 23 41 42 44 45 46 47 48 50 xii 1.8.10 1.8.11 1.8.12 1.8.13 1.8. 14 Appendix: Appendix: Appendix: Appendix: ... 79 81 82 84 84 85 88 88 89 100 100 102 103 103 1 04 106 116 116 116 120 120 122 123 125 130 131 137 140 143 144 xiii 2.8.11 Appendix: Axioms for the Sum and the Product 148 2.8.12 ... 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 m-2) 0.8 0.2 0 .4 0.6 P/(dyne m-2)...

324
314
0

probability and statistical inference - nitis mukhopadhyay

Ngày tải lên : 31/03/2014, 16:25

... 341 341 342 342 344 351 351 3 54 358 358 365 366 371 3 74 375 377 377 380 382 395 395 396 399 40 1 40 1 41 3 41 6 41 7 41 7 42 0 42 2 42 5 42 5 42 6 42 8 42 9 44 1 44 1 44 3 44 4 Contents xvii 9.2.2 9.2.3 9.2 .4 ... Comparisons 9 .4. 1 Estimating a Multivariate Normal Mean Vector 9 .4. 2 Comparing the Means 9 .4. 3 Comparing the Variances Exercises and Complements 44 6 45 1 45 2 45 5 45 6 45 6 46 0 46 3 46 3 46 5 46 7 46 9 10 Bayesian ... Sampling from a Finite Population, J Hajek 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 Statistical Modeling Techniques, S S Shapiro...

690
285
0

probability and statistics textbook

Ngày tải lên : 23/04/2014, 22:19

508
319
0

probability and combinations

Ngày tải lên : 14/05/2014, 16:55

... one out of left 4/ 4, total=1/5 *4/ 4=1/5; Select any member but Anthony as second member out of - 4/ 5 and Anthony as a third out of left 1 /4, total =4/ 5*1 /4= 1/5; Sum=1/5+1/5=2/5 =40 % Fourth approach: ... is the greatest number of distinct teams to which the project could be assigned? A 4^ 3 B 4^ 4 C 4^ 5 D 6 (4^ 4) E 4( 3^6) # of ways to choose which department will provide employee for the team and ... of the word PERMUTATIONS be arranged if there are always letters between P and S A 241 9200 B 2 540 1 600 C 181 44 00 D 1926300 E 1321500 There are 12 letters in the word "PERMUTATIONS", out of which...

22
195
0

against coherence truth probability and justification jun 2005

Ngày tải lên : 10/06/2014, 21:52

... 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 Problem ... on Lewis’s Definition of Independence Laurence BonJour’s Radical Justification of Belief 4. 1 4. 2 4. 3 4. 4 4. 5 The Problem of Justifying Beliefs BonJour on Justification from Scratch Lying and Individual ... index ISBN 0-19-927999-3 (alk paper) Truth–Coherence theory Skepticism I Title BD171.O47 2005 121–dc22 20 040 2 9060 Typeset by Newgen Imaging Systems (P) Ltd., Chennai, India Printed in Great Britain...

247
306
0

morris h degroot probability and statistics 2nd edition 1986

Ngày tải lên : 12/06/2014, 16:09

730
354
0

michael mitzenmacher, eli upfal probability and computing an introduction to randomized algorithms and probabilistic analysis 2005

Ngày tải lên : 12/06/2014, 16:14

366
458
0

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

... 2000 40 0 0 i =4 2000 40 0 0 i=7 40 0 0 2000 40 0 0 i = 10 2000 40 0 0 i=8 2000 40 0 0 i = 13 2000 40 0 0 i = 11 2000 40 0 0 2000 40 0 0 i=9 2000 40 0 0 2000 40 0 0 2000 40 0 0 i = 12 2000 40 0 0 i = 14 i = 15 2 i=6 40 0 0 ... 2000 40 0 0 i =4 2000 40 0 0 i=7 2000 40 0 0 i = 10 2000 40 0 0 i=5 0 2000 40 0 0 2000 40 0 0 i = 13 0 2000 40 0 0 i = 11 2000 40 0 0 i=6 0 2000 40 0 0 i=9 0 2000 40 0 0 2000 40 0 0 2000 40 0 0 i = 12 2000 40 0 0 i = 14 0 ... 0 2000 40 0 0 2000 2000 40 0 0 2000 i=7 2000 40 0 0 2000 2000 40 0 0 40 0 0 2000 2000 40 0 0 2000 i = 14 40 0 0 i = 12 40 0 0 i = 15 10 40 0 0 40 0 0 10 i = 13 2000 2000 i = 11 10 0 i=9 10 40 0 0 10 i = 10 0 40 0 0 i=8...

24
349
0