câu 27 phân tích khái niệm nhu cầu thị trường đối với 1 sản phẩm
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
... 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 (2006), #R62 l1 times +1 , 1 , (8) li 1 ... sets B1 , , Bm such that B1 ∪ · · · ∪ Bm = {1, , L} and |B1 | + · · · + |Bn | ≤ l1 + · · · + ln holds true for each ≤ n ≤ m − 1 10 11 12 13 14 label all vertices of T with numbers 1, ... combinatorics 13 (2006), #R62 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...
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Statistics, Probability and Noise
Ngày tải lên :
13/09/2012, 09:49
... 0082 010 7 013 9 017 9 0228 0287 0359 0446 0548 0668 0808 0968 11 51 1357 15 87 18 41 211 9 2420 274 3 3085 3446 38 21 4207 4602 5000 0.0 0 .1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1. 0 1. 1 1. 2 1. 3 1. 4 1. 5 1. 6 1. 7 ... Probability and Noise 31 12 11 pdf a X = RND mean = 0.5, F = 1/ %12 10 Amplitude 0 16 32 48 64 80 Sample number 96 11 2 12 7 12 8 12 11 pdf b X = RND+RND mean = 1. 0, F = 1/ %6 10 Amplitude 0 16 32 48 64 80 ... 96 11 2 12 8 12 7 90 10 0 11 0 12 0 13 0 14 0 Value of sample 15 0 16 0 17 0 15 0 16 0 17 0 10 000 c 256,000 point histogram 8000 Number of occurences FIGURE 2-4 Examples of histograms Figure (a) shows 12 8...
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Bài giảng Chapter 17 Free Energy and Thermodynamics
Ngày tải lên :
28/11/2013, 01:12
... 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 18 8.83 17 3. 51 198.49 11 6.73 19 1.50 210 .62 51. 45 31. 88 Al2O3(s) Br2(g) C(graphite) CO2(g) CaO(s) CuO(s) ... H2O(l) HCl(g) HI(g) I2(g) NH3(g) NO2(g) O2(g) SO2(g) -16 69.8 +30. 71 -393.5 -635.5 -15 6 .1 -822 .16 -18 7.8 -285.83 -92.30 +25.94 +62.25 -46 .19 +33.84 -296.9 Entropy • entropy is a thermodynamic ... Approach 13 Substance ∆H° kJ/mol Substance ∆H° kJ/mol 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) 0 +1. 88 -11 0.5 0 0 -2 41. 82 -268. 61 -36.23...
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Probability and Mathematical Genetics potx
Ngày tải lên :
05/03/2014, 22:21
... 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 ... Delaigle and Peter Hall Introduction 16 9 17 0 17 0 17 2 17 8 18 0 18 1 18 5 18 6 Contents Methodology and theory Relationship to minimum contrast methods ix 19 1 19 5 The coalescent and its descendants ... Martingales in the OK Corral Bull Lond Math Soc., 31( 5), 6 01 606 [M102] Kingman, J F C 2000 Origins of the coalescent: 19 74 19 82 Genetics, 15 6, 14 61 14 63 [M103] Kingman, J F C 2002 Stochastic aspects...
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PROBABILITY AND MATHEMATICAL STATISTICS pptx
Ngày tải lên :
17/03/2014, 14:20
... and r = 1, 2, 3, , n, we have n r Proof: = n 1 n 1 + r r 1 (1 + y)n = (1 + y) (1 + y)n 1 = (1 + y)n 1 + y (1 + y)n 1 n r=0 n r y = r n 1 r=0 n 1 = r=0 n 1 r y +y r n 1 r=0 n 1 r y r n 1 n 1 r n ... 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 + x)n = ... Distribution 11 .3 Bivariate Geometric Distribution 11 .4 Bivariate Negative Binomial Distribution 11 .5 Bivariate Hypergeometric Distribution 11 .6 Bivariate Poisson Distribution 11 .7 Review Exercises 12 .1...
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Probability and Statistics by Example pptx
Ngày tải lên :
28/03/2014, 10:20
... 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 ... + 11 11 10 I win, but not at the 1st throw the value 1 5 5 1 1 13 4 × + × + × + × + × + × = 12 36 11 36 11 12 495 1. 2 Conditional probabilities 21 Then I win = 13 4 244 + = 495 495 Problem 1. 18 ... Y1 ∩ H3 ∩ P3 Y1 C1 P3 C1 ∩ Y1 H3 C1 ∩ Y1 ∩ P3 C1 1 1 = × × × = 3 2 36 If the host doesn’t know where the car is, then = Y1 ∩ H3 ∩ P3 P3 Y1 H3 Y1 ∩ P3 1 1 = × × = 3 18 = and Y1 C1 Y1 ∩ H3 ∩ P3 =...
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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
... Reference BL 21 (DE3) Star C600 XL1-Blue 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’Tn10 (Tetr) ... 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 290 mM 2% 0.05% 10 0 mM 2.8 mM 13 ... lacZM15] rna19 gdhA2 his95 relA1 spoT1 metB1 T7 promoter Apr super glow gfp, Apr emrE NheI in pQB1 Apr, gfp emrE NdeI-HindIII in pET21a(+) sugE NdeI-HindIII in pET21a(+) tehA NdeI-HindIII in pET21a(+)...
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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 in ... : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 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 ... : : : : : 12 5 12 5 12 5 12 5 12 6 12 6 12 6 12 6 12 7 12 7 C Numerical Methods 12 9 D Programming Help 13 1 C .1 Numerical integration : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 12 9 C.2 Solving...
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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 4 514 5426 12 234 2392 9948 18 418 817 9 11 739 518 6 Occurrences of “the” 15 9 14 7 15 9 47 15 3 11 8 264 223 82 An ... Bernoulli trials 11 Regression Analysis 11 .1 Simple linear regression 11 .2 Multiple linear regression 11 .3 Orthogonal polynomials 12 Analysis of Variance 12 .1 12.2 12 .3 12 .4 12 .5 12 .6 13 One-way anova ... Hall/CRC 17 .13 Self-similar processes 17 .14 Signal processing 17 .15 Stochastic calculus 17 .16 Classic and interesting problems 17 .17 Electronic resources 17 .18 Tables 18 Special Functions 18 .1 18.2 18 .3...
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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 ... eferences F urther R eading and Comments Problems ix 16 7 16 9 17 2 17 3 18 1 18 2 18 3 18 4 18 5 19 1 19 1 19 3 19 6 19 9 2 01 205 207 209 211 212 215 219 2 21 223 225 226 228 233 234 238 238 238 239 PART B: STATISTICAL ... P
S1 S2 jS1 S2 PS1 S2
S1 S2 : P
S1 S2 Now, S1 S2 (S1 S2 ) S1 S2 Hence, P
S1 S2 P
S1 S2 P
S1 S2 P
S1 P
S2 À P
S1 S2 p1 p2 : p1 p2 À p1 p2 P
S1 S2 jS1...
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probability and finance it's only a game
Ngày tải lên :
31/03/2014, 16:24
... ISBN 0-4 71- 40226-5 (acid-free paper) Investments-Mathematics Statistical decision Financial engineering Vovk, Vladimir, 19 60- 11 Title 11 1 Series ~ HG4S 15 SS34 20 01 332'. 01' 1 -dc 21 Printed ... Historical Comments 5.6 Appendix: Kolmogorov ’s Finitary Interpretation 99 10 1 10 4 10 8 11 8 11 8 12 0 The Weak Laws 6 .1 Bernoulli’s Theorem 6.2 De Moivre’s Theorem 6.3 A One-sided Central Limit ... 12 1 12 4 12 6 13 3 14 3 14 4 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 14 7 14 8...
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probability and its applications - ollav kallenberg
Ngày tải lên :
31/03/2014, 16:24
... relation p 1 + q 1 = implies (p − 1) (q − 1) = 1, and so the equations y = xp 1 and x = y q 1 are equivalent for x, y ≥ By calculus, |f g| ≤ and so fg |f | xp 1 dx + ≤ p 1 |g| y q 1 dy = p 1 |f |p ... f 1 B c = (f 1 B)c , f 1 f 1 Bk , Bk = k k f 1 f 1 Bk Bk = k (1) k The next result shows that f 1 also preserves σ-fields, in both directions For convenience we write f 1 C = {f 1 B; ... (s, ds1 ) ··· µ2 (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 that Lemma 1. 9 is...
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probability and measurements - tarantola a.
Ngày tải lên :
31/03/2014, 16:24
... 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 ... 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 ... 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:...
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probability and statistical inference - nitis mukhopadhyay
Ngày tải lên :
31/03/2014, 16:25
... 4.4 .1 Several Variable Situations 4.5 Special Sampling Distributions 84 86 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 ... STATISTICS: Textbooks and Monographs D B Owen, Founding Editor, 19 72 19 91 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 The Generalized Jackknife Statistic, H ... 9.2 .1 Inversion of a Test Procedure 3 41 3 41 342 342 344 3 51 3 51 354 358 358 365 366 3 71 374 375 377 377 380 382 395 395 396 399 4 01 4 01 413 416 417 417 420 422 425 425 426 428 429 4 41 4 41 443...
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probability and combinations
Ngày tải lên :
14/05/2014, 16:55
... Favourable outcomes = 10 C1 as there are 10 pairs and we need ONE from these 10 pairs Total # of outcomes= 20C2 as there are 20 shoes and we are taking from them P =10 C1/20C2 =10 / (19 *10 ) =1/ 19 Answer: C ... and S A 2 419 200 B 254 016 00 C 18 14400 D 19 26300 E 13 215 00 There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice Choosing letters out of 10 (12 -2(P and S) =10 ) to place ... probability that Kate has more than $10 but less than $15 ? A 5 /16 B 15 /32 C 1/ 2 D 21/ 32 E 11 /16 After tries Kate to have more than initial sum of 10 $ and less than 15 $ must win or times (if she wins...
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against coherence truth probability and justification jun 2005
Ngày tải lên :
10/06/2014, 21:52
... Paul 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 ... Connection: A Reply to Bovens and Hartmann 9 .1 9.2 9.3 9.4 77 77 79 81 84 88 90 Cartesian Scepticism Jamesian Wagering 17 3 17 3 17 6 contents xiii 10 .3 10 .4 10 .5 10 .6 10 .7 Peirce’s Reply to Scepticism More ... BonJour’s 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...
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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
Ngày tải lên :
22/06/2014, 19:20
... have ⎛ T UT U T = −T T ⎝T 11 T 11 0⎠ (A .13 ) T T T − = UT1 T11T T 11 T 11 VT1 (A .14 ) T − − T = UT1 T 11 T 111 T11T T 11 T 11 − mη2 Ik VT1 T − − = UT1 T 11 Ik − mη2 T 111 T11T VT1 This is the result given ... 12 4000 i = 15 10 4000 4000 10 i = 13 2000 2000 i = 11 10 0 i=9 10 4000 10 i = 10 0 4000 i=8 10 2000 i=6 10 0 10 10 10 0 4000 i=5 i=4 10 i=3 10 10 10 2000 (a) 4000 (b) 2000 4000 (c) Figure 11 : ... T T T use E = EV T1 V T1 + EV T2 V T2 to obtain H = ZV T with Z = Z1 , Z2 = U T1 T 11 + EV T1 , EV T2 (A .10 ) We have T T T T T Z1 Z1 = T 11 U T1 U T1 T 11 + T 11 U T1 EV T1 T Wtdc = V T T T...
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