executing select statements and tabledirect commands phần 1

... Turbulent flow, 1134.6 Stratified flow, 1174.7 Particle settling, 1194.8 Particle transport by flows, 1214.9 Waves and liquids, 125 4.10 Transport by waves, 1314.11 Granular gravity flow, 1334.12 Turbidity ... Main Text Fig 1.1 NASA/JPL images; 1.2 Nature 350, 55; 1.6 USDA at Kansas State University; 1.7, 1.9 USGS; 1.10 K West/Montserrat Volcanic Observatory; 1.11 NOAA/M Perfit; 1.15, 1.16 www.waterhistory.org; ... Turbidity flows, 1384.13 Flow through porous and granular solids, 1424.14 Fractures, 144 4.15 Faults, 1564.16 Solid bending, buckling, and folds, 1724.17 Seismic waves, 179 4.18 Molecules in motion:

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... ISBN-10: 0-7377-3818-9 Printed in the United States of America 1 2 3 4 5 6 7 12 11 10 09 08 Trang 6Introduction 7The Captive Animals’ Protection Society 2 Animals Do Not Suffer in Captivity 18 ... Dealt With 51 Jesse Donahue and Erik Trump 9 Zoos Play a Key Role in Education 58 The Wildlife Conservation Society 10 Zoos Do Not Play a Key Role in Education 63 Dale Jamieson 11 Captive Breeding ... over 6.5 acres (2.6ha) and grows thick with trees, bamboo, and other lush landscape This exhib-it is home to more than fifty-five wildlife species and twenty west-ern lowland gorillas Some zoos

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... dollars for British pounds in 180 days might be SL2910_frame_FM.fm Page 7 Monday, May 21, 2001 9:06 AM Trang 11SL2910_frame_FM.fm Page 8 Monday, May 21, 2001 9:06 AMTrang 12A ABSOLUTE PURCHASING POWER ... Page 5 Monday, May 21, 2001 9:06 AM Trang 10N OTES AND A BBREVIATIONSKEY NOTES 1 This book has the following features: • Plenty of examples and illustrations • Useful strategies and checklists • ... (FrF 50,000 × $0.21 = $10,500) and foreign exchange gain.] ACCOUNTING FOR MULTINATIONAL OPERATIONS SL2910_frame_CA.fm Page 3 Wednesday, May 16, 2001 4:38 PM Trang 15B.1 The Functional CurrencySFAS

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... -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 1 6 11 16 21 26 31 36 -30 -20 -10 0 10 20 30 1 6 11 16 21 26 31 36 Figure 2.3: For 0 ≤ ω 1 < ··· < ω N ≤ π, q t = P N j=1 q jt ,whereq jt = cos(ω j t)[a ... ρ8 ρ16 Min Max Returns 5.92 - 13. 34 18 .31 0.24 -0.10 5.50 - 13. 83 16.47 0.12 0. 03 6 .35 - 13. 91 15.74 0.09 0. 23 6.08 -15.00 16.97 0. 13 0.18 Deviation from fundamentals 0 .30 0.49 1.82 ... 20 15 exchange rate with bubble 10 5 0 fundamentals -5 -10 1 26 51 76 101 126 151 176 Figure 3. 3:... element and zeros elsewhere and let e2 be a (1 × 2p) row vector with a 1 as the p + 1−th

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... : e t u 1 (c ∗ xt ,c ∗ yt )=βE t [u 1 (c ∗ xt+1 ,c ∗ yt+1 )(x t+1 + e t+1 )], (4.12) ω ∗ yt : e ∗ t u 1 (c ∗ xt ,c ∗ yt )=βE t [u 1 (c ∗ xt+1 ,c ∗ yt+1 )(q t+1 y t+1 + e ∗ t+1 )].(4.13) A set ... (4.6)—(4.13) become q t = 1 − θ θ x t y t , (4.23) e t x t = βE t " µ C t+1 C t ¶ (1−γ ) Ã 1+ e t+1 x t+1 !# , (4.24) e ∗ t q t y t = βE t " µ C t+1 C t ¶ (1−γ ) Ã 1+ e ∗ t+1 q t+1 y t+1 !# ... xt−1 (x t + e t )+ω yt−1 (q t y t + e ∗ t ) − e t ω xt − e ∗ t ω y t − q t c y t ,c y t ) +E t [βu(ω xt (x t+1 + e t+1 )+ω yt (q t+1 y t+1 + e ∗ t+1 ) −e t+1 ω xt+1 − e ∗ t+1 ω y t+1 − q t+1 c

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... t+1 =(−ω 2 L)(−ω −1 2 L −1 )(1 − ω 2 L)(1 − ω 1 L) ˆ k t+1 =(−ω 2 L)(1 − ω −1 2 L −1 )(1 − ω 1 L) ˆ k t+1 , and rearrange to get (1 − ω 1 L) ˆ k t+1 = −ω −1 2 L −1 1 − ω −1 2 L −1 W t = − µ 1 ... kt+1 + d5 kt + d6 kt + d9 At + d10 A∗ t +d12 c∗ + d13 , ˜t (5. 47) where the coefficients are given by j aj 1 βQ31 2 βQ32 3 βQ33 + Q11 4 βQ34 + Q12 5 Q13 6 Q14 7 βQ 35 8 βQ36 9 Q 15 10 Q16 11 ... -3.481 t(β2 = 0) (-2.413) t(β2 = 1) (-3.107) ˆ 4.481 β1 US-JY US-DM DM-BP DM-JY -4.246 -0.796 -1.6 45 -2.731 (-3.6 35) (-0 .54 2) (-1.326) (-1.797) (-4.491) (-1.222) (-2.132) (-2. 455 ) 5. 246 1.796

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... (6.40) and take date-t expectations to get Etst+1 = π1(ft+ p0tδ0+ p1tδ1) + π2p0tδ0+ π3p1tδ1 = π1ft+ (π1 + π2)p0tδ0+ (π1+ π3)p1tδ1 (6.41)Substitute (6.40) and (6.41) into (6.33) to get π1ft+π2p0tδ0+π3p1tδ1 ... proba-bilities ispos-p0,1 = p0P(∆f1|δ0) p0P(∆f1|δ0) + p1P(∆f1|δ1), p0,2 = p0,1P(∆f2|δ0) p0,1P(∆f2|δ0) + p1,1P(∆f2|δ1), p0t = p0,t−1P(∆ft|δ0) p0,t−1P(∆ft|δ0) + p1,t−1P(∆ft|δ1). How long is the learning ... get ∆st+1= [1− k(1 − µ)]xt+ ξt+1, (6.70)where ξt+1 ≡ (1/ρ)[1 − k(1 − µ)]vt+1 − (1 − µ)/ρut+1 − (1 − µ)ut is an error term which is orthogonal to xt If [1− k(1 − µ)] < 0, the Trang 196.6 NOISE-TRADERS

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... using PPIs.VRk 1.00 1.07 0.951 0.906 0.841 0.457 0.323 0.232s.e – 0.152 0.156 0.166 0.169 0.124 0.106 0.0872 can be attributed to a permanent (random walk) component Theasymptotic standard errors ... [0.307]Ger -2.641 (0.000) [0.000] -3.119 (0.000) [0.001] aire τ¯c (p-val) [p-val] τ¯ct (p-val) [p-val] US 66.902 (0.083) [0.088] 40.162 (0.351) [0.346]Ger 101.243 (0.000) [0.000] 102.017 (0.000) [0.000] ... Trang 17.4 LONG-RUN ANALYSES OF REAL EXCHANGE RATES 221Figure 7.1: Real and nominal dollar-pound rate 1871-1997Variance Ratios of Real Exchange Rates We

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... price levels, (9.4) and (9.9) simplifytoPt = [npt(z)1−θ+ (1− n)(Stp∗t(z∗))1−θ] −θ1 , (9.10) Pt∗ = [n(pt(z)/St)1−θ + (1− n)p∗t(z∗)1−θ] −θ1 , (9.11)and that PPP holds for the correct CES price index ... (9.15) to eliminate B∗ to get Ct = (1 + rt−1)Bt−1− Bt− Mt− MP t−1 t − Tt+yt(z)θ−1θ [Ctw+ Gwt]1 , (9.24) ⇐(157)Euler Equations Ct, Mt, and Btare the choice variables for the domes- tic agent and ... agent The budget constraint for home and foreign agents are PtBt+ Mt = (1 + rt−1)PtBt−1+ Mt−1+ pt(z)yt(z)− PtCt− PtTt, (9.22) Pt∗Bt∗+ Mt∗ = (1 + rt−1)Pt∗Bt−1∗ + Mt−1∗ + p∗t(z∗)y∗t(z∗)− Pt∗Ct∗− Pt∗Tt∗

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... The short-run versions of (9.141) and(9.142) are ˆ Mt− ˆMt∗ = 1 ²( ˆCt− ˆCt∗)− β ²(1− β)( ˆS− ˆSt). (9.174)From (9.153) and (9.154) you get ˆ Ct = ˆδt+ ˆC + ˆP , (9.175)ˆ Trang 2298CHAPTER 9 THE ... steady state isattained at t + 1, so that ˆSt+1= ˆS, ˆPt+1= ˆP , and ˆP∗ t+1= ˆP∗.Date t nominal goods prices are set and Þxed one-period in advance By (9.10) and (9.11), it follows that the general ... from (9.171) and plug into(9.174) to get " ²(1− β) #³ˆ Mt− ˆMt∗´= 1 ² ³ˆ Ct− ˆCt∗´+ β ²2(1− β) ³ˆ C− ˆC∗´+ β ²(1− β)Sˆt.Using (9.177) to eliminate ˆC− ˆC∗, you get ˆ Ct = nˆxt(z) + (1− n)[

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... of (11.31). Noting that δ e = w − s 0 and collecting terms giv es δ = λ α [αδ e +¯y + u] . (11.32) Equating (11.31) and (11.32) you get the real wage w −s = θδ e − α(¯y + u) α 2 + θ . (11.33) ... . (11.33) Substitute (11.33) into (11.23) to get the deviation of output from the target ¯y − y = θ θ + α 2 [αδ e +¯y + u] . (11.34) Substitute (11.32) and (11.34) into (11.29) to get the cost ... 11.2. A SECOND GENERATION MODEL 341 -0.002 0 0.002 0.004 0.006 0.008 0. 01 0.012 0.014 0.016 -0.15 -0.13 -0. 11 -0.09 -0.07 -0.05 -0.03 -0.01 0.01 0.03 u u Figure 11.2: Realignment thresholds for

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... 3.2 10 –6 (10 m a –1 ) Antarctic ice stream 3.2 10 –4 (1,000 m a –1 ) Lithospheric plate 1.6 10 –9 (0.05 m a –1 ) Pyroclastic flow >100 Magma in volcanic vent 8.3 10 –3 (30 m h –1 ) Magma ... u 5 ms 1for positions x1x5. Time, t t1 t2 t3 t4 t5 Steady motion t1 t5 t4 t3 t2 Object 1 5 5 5 5 5 x5 x4 x3 x2 Object 1 5 5 5 5 5 u constant Speed–distance graph 5 (a) (b) Trang 102.4.7 The ... suchsuspensions (see Fig 2.12) is a feature of interest and importance in considering the flow dynamics of suchsystems Fig 2.10 Variation of seawater density with salinity. at 0°C and 1 atm 1,028 kg m –3

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... s3 s1 Uniaxial compression t s1 s1 = s2 = sn 83 sn t s1 = s2= s3 sn Triaxial stresses (f ) s1 > s2 > s3 (e) t t s3 sn (c) Uniaxial tension s2 s1 Pure shear sn s3 s1 0 s2 0 s3 s2 s1 s2 = s1 = ... l1 and the original length of the line l0 before deformation (Fig 3.81) The stretch can be also obtained by adding to the extension e (since e ϭ (l1 Ϫl0/l0) ϭ (l1/l0 Ϫ l0/l0) ϭ (l1/l0 Ϫ 1) and ... Homogeneous and inhomogeneous strain (a) is the object before deformation; (b), (c), and (d) show homogeneous strain; and (e) and (f) inhomogeneous strain LEED-Ch-03.qxd 11/27/05 4:26 Page 87 Forces and

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... object and represents the homogeneous deformation of a circle of radius 1 transformed into an ellipsoid As I0is 1, S1 I1/1 I1 which represents the stretch S of the long axis. Similarly S3 I1 giving ...  10 –3 0.08 10 2 10 8 10 16 10 22 Fluid h (Pa) (h constant) (h variable) Trang 10ideal conditions; they are considered homogeneous (therock has the same composition in all its volume) and ... strain is in thor-G.H Davies and S.J Reynolds’s Structural Geology of Rocks and Regions (Wiley, 1996); R.J Twise and E.M Moores’s Structural Geology (1992) and J.G Ramsay and M Huber’s The Techniques

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... rocks and form-ing a high-angle step inclined backward with respect to the com-10 km duplex structure Foreland Hinterland Thrust and fold belt Thrust and fold belt Decollement Foreland Hinterland ... s3 E F1 F1 Fig 4.101 Strike–slip faults form to accommodate deformation in situations in which an extension and compression occur in the horizontal surface in some section of the crust (a) Anderson’s ... the foreland, where a subsiding basin can form by tectonic loading (Section 5.2) The area located at the back of the thrust belt is the hinterland (Fig 4.105) Structures in Trang 11166 Chapter

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... world seismic belts (14 year record of M 4.5) and the location of selected Holocene-active volcanoes and the major volcanic arcs. Trang 155.1.1 Difficult initial questions and early cluesWe need ... Counter-rotating cells at Ra > c.1,700 Plan view Side view Side view Fig 4.152 Convection in vertical slots and in horizontal slots and reservoirs. Trang 11200 Chapter 4Fig 4.153 View from above of Bénard ... Currents (Cambridge, 1997) Folds and faults are related to stressand strain as in G H Davies’ and S J Reynolds’s Structural Geology of Rocks and Regions (Wiley, 1996), R. J Twiss and E M Moores’ Structural

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... Total area ϫ 106 km2 Land area ϫ 106 km2 Speed mm yearϪ1 Periphery ϫ 102 km Ridge length ϫ 102 km Trench length ϫ 102 km 60 41 108 59 60 79 69 15 5 36 20 15 15 31 51 0 0 0.6 11 13 80 17 61 76 86 24 ... 11 13 80 17 61 76 86 24 64 42 25 388 305 499 356 420 418 421 187 88 88 103 98 28 146 87 152 208 124 230 90 76 40 0 30 12 124 91 10 53 25 41 Plates: NA – North America, SA – South America, PAC ... spreading in 1963 The seismological recognition and significance of the LVZ (Sections 1.5 and 4.17) in defining the mechanical layers of lithosphere and asthenosphere LEED-Ch-05.qxd 11/27/05 2:21 Page

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... sometimes up to 4 m or more above 10 11 12 0 1 2 3 4 5 6 7 10 1 2 3 10 9 8 12 1 2 3 3 1.5 0.5 3 4 5 6 Trang 12rota-mean high-water level, as in Hurricane Carla on PadreIsland, Gulf of Mexico The magnitude ... kind (Fig 6.51), inresponse to settling into the salt layer and subsequent 5 10 sed conc (mg l –1 ) Salinity (‰) Flow velocity (m s–1 ) Salt Wedge River Water Mixing Z one 5 10 15 20 25 Estuary ... shallow-water wave to the tude of the Coriolis parameter, that is, R  gh/f At t = 9 10 11 0 1 2 8 7 6 5 Times in 1/n of 12 h tidal period Pressure force Pressure force Coriolis force Coriolis force

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... southern hemisphere stars John Goldingham (17 96 18 05, 18 12 18 30), T.G Taylor (18 30 18 48), W.S Jacob (18 49 18 58), and Norman R Pogson (18 61 18 91) were successive Government Astronomers ... .536 Conference Photograph xv xvi 1 2 3 4 5 6 7 8 9 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 Conference Photograph Robert J Rutten 35 K B Ramesh 36 ... of 11 ,000 southern stars produced by the Madras Observatory in 18 44 under Taylor’s direction using the new 5-ft transit... .536 Conference Photograph xv xvi 1 2 3 4 5 6 7 8 9 10 11 12

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... Present, and Future xvii 1.1 Three Revolutions / 1.2 Computational Imaging / 1.3 Overview / 1.4 The Fourth Revolution / Problems / Geometric Imaging 11 2.1 Visibility / 11 2.2 Optical Elements / 14 ... Elements / 115 4.6 Wave Propagation Through Thin Lenses / 121 4.7 Fourier Analysis of Wave Imaging / 124 4.8 Holography / 130 Problems / 141 Detection 147 The Optoelectronic Interface / 147 Quantum ... B-Splines / 89 3.10 Wavelets / 96 Problems / 100 Wave Imaging 103 4.1 Waves and Fields / 103 4.2 Wave Model for Optical Fields / 104 4.3 Wave Propagation / 106 4.4 Diffraction / 109 4.5 Wave Analysis

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