law and order svu closure part 2 synopsis

... disclosure requirements, 26 2, 28 2 DOJ/FTC Promoting Innovation and Competition, 26 2, 27 4, 28 3–84, 340, 345–48 fair, reasonable, and non-discriminatory (FRAND) terms, 27 3–74 680 Index Standard-setting organizations ... consumer welfare standard, 22 9 damages, 24 0 disclosure, duty of, 22 7? ?28 free consumer choice, 25 4–55 “free riding,” 24 2 injunctive relief against, 24 5 intent, relevance of, 24 0–41 judicial quality ... Copyright Act), 20 0? ?20 2 DOJ See Justice Department Downstream markets circumvention in, 20 1? ?2 DMCA, 20 0? ?20 2 market power in, 199? ?20 0 DRM (Digital rights management), 20 Drug Price Competition and Patent

Ngày tải lên: 02/02/2020, 11:30

... Contents I Law and Order in Contemporary American Politics Setting the Public Agenda 14 Creating the Crime Issue 28 From Crime to Drugs? ?and Back Again 44 Crime and Drugs in the News 62 Crime and Punishment ... Crime Pay: Law and Order in Contemporary American Politics / Katherine Beckett p cm — (Studies in crime and public policy) Includes bibliographical references and index ISBN 0-19-51 128 9-X; ISBN ... Youth Cheryl L Maxson and Malcolm W Klein Making Crime Pay: Law and Order in Contemporary American Politics Katherine Beckett Community Policing, Chicago Style Wesley G Skogan and Susan M Hartnett

Ngày tải lên: 18/01/2020, 19:43

... legumes and farm manure manure with bedding material hay from legumes animal droppings C:N Ratio up to 400 50-150 50 20 -30 20 -25 15 15 Be careful not to use toxic materials For example, plant parts ... for composting 4 .2 Micro-organisms The composting process happens due to the activity of microorganisms and other larger organisms like worms and insects See figure in Section 2. 1 The first condition ... organic components into humus The heap remains clammy and hot inside and the temperature drops from 50oC to 30oC By regulating the temperature, air and water supply, the process can be accelerated

Ngày tải lên: 02/07/2014, 05:20

... Reduction 525 4.1 Total Reduction 525 4 .2 Iterations 526 4.3 Temporal Order 528 4.4 Recursion 529 4.5 Comparison 531 5. Call Graph Based Bug Localization 5 32 5.1 Structural Approaches 5 32 5 .2 Frequency-based ... Approaches 20 3 6.10 Transitive Closure Matrix 20 4 6.11 The 2- hop Distance Aware Cover (Figure 2 in 10) 20 6 6. 12 The Algorithm Steps (Figure 3 in 10) 20 7 6.13 Data Graph (Figure 1(a) in 12) 20 9 6.14 ... gray. 22 2 7.4 Graph (c) is a maximum common subgraph of graph (a) and (b). 22 4 7.5 Graph (a) is a minimum common supergraph of graph (b) and (c). 22 5 7.6 A possible edit path between graph 𝑔 1 and

Ngày tải lên: 03/07/2014, 22:21

... FMS* 0. 028 0.060 0.086 0.086 0.165 0.110 0.15 0 .22 0 .20 0 .27 0 .27 0 .27 0.34 0.34 0 .27 0.34 0.50 0.50 2. 00 1.50 1.50 1.50 2. 50 2. 50 Fig 1 .27 Costs associated with the examples of Figure 1 .26 , a–g ... discounts and taxes) Tool type and size, dimensions in mm Turning solid HSS, x x 100 Brazed carbide carbide insert, plain 12 x 12 x 25 x 25 x carbide insert, coated 12 x 12 x 25 x 25 x ceramic ... 0.0 92 0 .27 15 29 5000 0.65 150 0 .25 0.0 92 0 .27 15 29 5000 0.65 180 0 .25 0.0 92 0 .27 15 Fig 1 .28 Times and costs to remove metal by milling, for the conditions i and ii of Table 1.7 compared with

Ngày tải lên: 21/07/2014, 17:20

... component of 8 Goals and Visions for the Future 2. 1 Government Safety Goals 9 Table 2. 1 Road Safety Goals—National and Regional Road Safety Goals—National and Regional 20 07? ?20 08 20 10 20 13 20 15 Long-term ... institutes and governments worldwide, beginning with safety-focused visions and then expanding into more holistic visions. 2. 2 Visions for the Future 11 > 10,000,000 crashes 4 ,28 2,000 Property ... management system, ABS ……………… Figure 2. 5 Car area networks, component-based design 20 02 2 010 20 20 Evolution of the IV (Source: TNO.) In Figure 2. 4, the evolution of roadside traffic

Ngày tải lên: 05/08/2014, 11:20

... the parts replacement purchase order The authorized parts level is periodically and automatically reviewed to prevent reordering of parts with a low turnover A block diagram showing the spare parts ... stock requirements, and improved warehouse response to maintenance requirements for materials and spare parts Identification of obsolete parts and materials is far easier and far more thorough ... a planned basis 26 Machinery Component Maintenance and Repair Figure 2- 1 Maintenance as part of Asset Management (Source: SKE Publication 51605 20 03) Maintenance Organization and Control for

Ngày tải lên: 05/08/2014, 12:20

... Networks, 12( 5), 987–997 TITSIAS, M K and LIKAS, A (20 02) : Mixtures of Experts Classification Using a Hierarchical Mixture Model Neural Computation, 14, 22 21? ?22 44 TUTZ, G and BINDER H (20 05): Localized ... Proc of 26 th DAGM-Symposium Springer, 22 0? ?22 7 HAASDONK, B and BURKHARDT, H (20 07): Invariant kernels for pattern analysis and machine learning Machine Learning, 68, 35–61 SCHÖLKOPF, B and SMOLA, ... (1983): The Comparison and Evaluation of Forecasters The Statistician 32, 12? ? ?22 GEBEL, M and WEIHS, C (20 07): Calibrating classifier scores into probabilities In: R Decker and H Lenz (Eds.): Advances

Ngày tải lên: 05/08/2014, 21:21

... Algorithms on the New-Item and User-Bias Problem 527 information with CF (Burke (20 02) , Melville et al (20 02) , Kim and Li (20 04), Tso and Schmidt-Thieme (20 05)) However, there has ... 411- 426 . MIKA, P. (20 05): Ontologies Are Us: A Unified Model of Social Networks and Semantics. In: Y. Gil, E. Motta, V. R. Benjamins and M. A. Musen (Eds.), ISWC 20 05, vol. 3 729 of LNCS, pp. 522 ... Wide Web : 625 ˚ U6 32. DESHPANDE, M. and KARYPIS, G. (20 04): Item-based top-n recommendation algorithms. ACM Transactions on Information Systems, 22 (1):1-34. GOLBER, S., HUBERMAN, B.A. (20 05): "The

Ngày tải lên: 05/08/2014, 21:21

... Semiconductor Processes and Devices, pp 43–46, 20 02 Chapter 1 Technology Challenges Motivating Adaptive Techniques 23 [20 ] N... Takano, Y Akiyama, and K Imai, “Ultra-Low Standby Power (U-LSTP) ... [11 ] [ 12 ] [13 ] [14 ] [15 ] [16 ] [17 ] [18 ] [19 ] David Scott, Alice Wang Symposium on Low Power Electronics and Design, pp 78–83, August 25 27 , 20 03, Seoul, Korea K Shakeri and J ... HfSiON Dielectric and Body-Biasing Scheme,” Symposium on VLSI Technology, Digest of Tech Papers, pp 21 8 – 21 9 , June 20 05 Chapter 2 Technological Boundaries of Voltage and Frequency Scaling

Ngày tải lên: 06/08/2014, 01:21

... x→∞ ax 2 x 2 + βx + χ = lim x→∞ 2ax 2x + β = lim x→∞ 2a 2 = a f(x) = 2x 2 x 2 + βx + χ Now we use the fact that f(x) is even to conclude that q(x) is even and thus β = 0. f(x) = 2x 2 x 2 + χ Finally, ... 12 2 2 2 2 y (3,7) (2, 3) (4 ,2) (1, 1) x Figure 2 .15 : Quadrilateral Solution 2. 7 The tetrahedron is determined by the three vectors with tail at (1, 1, 0) and heads at (3, 2, ... (−k) + a2 b3 i + a3 b1 j + a3 b2 (−i) = (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k Next we evaluate the determinant i j k a a a a a a a1 a2 a3 = i 2 3 − j 1 3 + k 1 2 b2 b3

Ngày tải lên: 06/08/2014, 01:21

... point d 2 2 ex /2 y = x2n+1 ex /2 dx y = e−x x 2 /2 ξ 2n+1 eξ 2 /2 dξ + c e−x 2 /2 1 We choose the constant of integration to satisfy the initial condition y = e−x x 2 /2 ξ 2n+1 eξ ... exact if and. .. tA +2 t A = 0, 2  A + A +2 + c, A 1 2 t y = ln t + 2 t + c, A=0  1 2  − t + ln t + c, A = 2  2 2 1 t  A + A +2 + ct−A , A = 2  1 y = ln t + 2 t2 + c, A=0 ... condition y = e−x x 2 /2 ξ 2n+1 eξ 1 826 2 /2 dξ + e(1−x 2 ) /2 If n ≥ 0 then we... a function of integration ux = 3x2 − 2xy + 2 u = x3 − x2 y + 2x + f (y) We substitute this into the

Ngày tải lên: 06/08/2014, 01:21

... an−1 an = − 2 + 2( n − 1 /2) − 1 8(n − 1 /2) an−1 an = − 2n(4n − 3) 12 34 an... = − 2 dt t d d = −t2 dz dt w + d2 d d = −t2 −t2 2 dz dt dt 2 d d = t4 2 + 2t3 dt dt 1 24 0 The equation ... x2 /2 + O(x3 ))(x + a2 x2 + a3 x3 + O(x4 )) = 0 (2a2 x + 6a3 x2 ) + (2x + 4a2 x2 ) + (6x + 6(1 + a2 )x2 ) = O(x3 ) = 0 17 a2 = 4, a3 = 3 17 y1 = x − 4x2 + x3 + O(x4 ) 3 Now we... , n ≥ 0 n +2 ... a2n = − a0 (2n)(2n − 2) · · · 4 · 2 a0 = (−1)n n , n≥0 m=1 2m = (−1)n a2n−1 2n + 1 a2n−3 = (2n + 1)(2n − 1) a2n+1 = − a1 (2n + 1)(2n − 1) · · · 5 · 3 a1 = (−1)n n , n≥0 m=1 (2m + 1) = (−1)n

Ngày tải lên: 06/08/2014, 01:21

... 2 1 /2 π −1 /2 z −3 /2 sin z − 2 1 /2 π −1 /2 cos z 2 π 1 /2 = 2 π 1 /2 = z −3 /2. .. 1 −1 = 2 2 ζ ζ 1 1 2 4 c2 = − 2 3 = + 3 ζ ζ ζ ζ 1 −1 12 3 24 c3 = 2 − 2 − 4 = 2 ... n−1  2n−1 n! 1 + ζ 2 + + · · · + 2 4···(n−1)·(2n 2) ···(2n−(n−1)) ζ n+1 2( 2n 2) 2 4·(2n 2) (2n−4) 16 42 for even n for odd... nz n−1 2 2n 1 z n = 2 2 Jn (z) = 1 2 1 = 2 1 = 2 Jn ... find J3 /2 (z), 1 /2 J1 /2 (z) − J1 /2 (z) z 1 /2 1 /2 2 1 = z −1 /2 sin z − − z π 2 J3 /2 (z) = 2 π 1 /2 z −3 /2 sin z − 2 π 1 /2 z −1 /2 cos z = 2 1 /2 π −1 /2 z −3 /2 sin

Ngày tải lên: 06/08/2014, 01:21

... β) 1 ((a − α )2 + (b − β )2 ) u= ln 4π ((a − α )2 + (b + β )2 ) 1 ((x2 − y 2 − ξ 2 + η 2 )2 + (2xy − 2 η )2 ) ln u= 4π ((x2 − y 2 − ξ 2 + η 2 )2 + (2xy + 2 η )2 ) u= 1 ln ... 1 2? ? 1  k 2 + β 2  e − √ k 2 +β 2 |x−ξ| + e − √ k 2 +β 2 (x+ξ)  G(x, y; ξ, ψ) = − 1 π  ∞ 0 sin(βy) sin(βψ)  k 2 + β 2  e − √ k 2 +β 2 |x−ξ| + e − √ k 2 +β 2 (x+ξ)  dβ 20 17 Solution 45 .20 ... )2 + (y − η )2 ) ((x + ξ )2. .. η )2 + ln (x + ξ )2 + (y + η )2 u= ((x − ξ )2 + (y − η )2 ) ((x + ξ )2 + (y + η )2 ) ((x + ξ )2 + (y − η )2 ) ((x − ξ )2 + (y + η )2 ) 1 ln 4π 2

Ngày tải lên: 06/08/2014, 01:21

... Petroleum, EJ/Year 14 0 12 0 10 0 80 60 27 0 billion gallons of ethanol per year 40 20 0 20 00 20 20 20 40 20 60 20 80 21 0 0 21 2 0 21 4 0 21 6 0 21 8 0 22 00 Fig 2. 4 The estimated decline ... Logistic projection RFA data 2 017 − Bush’s Goal Billion gallons per year 30 25 20 15 10 5 0 19 80 19 85 19 90 19 95 20 00 20 05 2 010 2 015 20 20 Fig 2. 2 By an exponential extrapolation ... 1000 Dollars $ Corn grain 2, 690 kg b 2, 550 b 28 7.36 Corn transport 2, 690 kg b 322 c 21 .40 d Water 15,000 L e 90 f 21 .16 f Stainless steel 3 kg g 165 h 10.60 d Steel 4 kg g 92 h 10.60 d Cement 8 kg

Ngày tải lên: 06/08/2014, 04:20

... and  ’s. By using the first two terms in the Taylor series expansion A i e  it ; i = 1 ,2. Equation (22 ) turns out to be                   11 111 22 2 22 2 22 2 22 2 cos ...   2 2 (23 ) where the Taylor series expansion is given by: 1 t et     Making the following substitutions in equation (23 ), equation (26 ) can be obtained, 11 21 1 32 2 422 2 52 2 622 ... equation (21 ) are considered, then the error e(t) can be written as (21 )       12 1 122 2 cos cos tt et Ae wt Ae wt    (21 ) Using the well-known trigonometric identity   22 2 2 2 2 cos...

Ngày tải lên: 19/06/2014, 08:20

... 1 Heater 2 Re = 100, d = 1 t 24 6810 121 4 NU 5 10 15 20 25 30 Heater 1 Heater 2 Re = 1000, d = 1 t 24 6810 121 4 NU 5 10 15 20 25 30 Heater 1 Heater 2 Re = 10, d = 2 t 5101 520 NU 5 10 15 20 25 30 Heater ... 2 t 5101 520 NU 5 10 15 20 25 30 Heater 1 Heater 2 Re = 100, d = 2 t 5101 520 NU 5 10 15 20 25 30 Heater 1 Heater 2 Re = 1000, d = 2 t 5101 520 NU 5 10 15 20 25 30 Heater 1 Heater 2 Re = 10, d = 3 t 5101 520 NU 5 10 15 20 25 30 Heater ... 1 θ 0,00 0,05 0,10 0,15 0 ,20 0 ,25 d = 1 d = 2 d = 3 Re = 100, Gr = 10 5 Heater 2 θ 0,06 0,08 0,10 0, 12 0,14 0,16 0,18 0 ,20 0 ,22 0 ,24 d = 1 d = 2 d = 3 Re = 1000, Gr = 10 5 Heater 2 θ 0,00 0,05 0,10 0,15 0 ,20 0 ,25 d...

Ngày tải lên: 19/06/2014, 10:20

... radiation and chlorophyll found in trees, plants, and algae. A product of this process is carbohydrates of the form C n H 2n O n . For example, when n=6, glucose is derived: chlorophyll-l 22 6 126 2 6CO ... M. et al 20 02) . Both the inhabited and uninhabited environment is threatened and one such indication is the changes of the climate (Theophanides, T. et al. 20 02) . Furthermore, as of 20 06, the ... al. 20 03; Filleul L, et al 20 03, Basu R. & Samet J. M., 20 02, Le Tertre A. et al. 20 02, Dominici F., 20 02, Sunyer & Basagana 20 01) of occupational diseases on the working population are...

Ngày tải lên: 19/06/2014, 10:20

...  (8) or   42 2 2 210xm x m m    (9) From this equation one can find that: 2 22 22 2 11 22 mm xm m m m          (10) But:  2 2 22 22 2 2 3 33 1 dB dB dB xm ... 23     22 2 42 2 322 21 2 14 2 2 xxmxmxm xxmxmxm     (21 ) One of possible solutions of this equation is x = 0. Other solutions can be obtained from the equation: 22 22 10xm ... ( 12) to zero, one can obtain:     2 222 22 12 1 22 4mm m mm mm m   (13) Squaring both sides of this equation, then:     2 2 122 11mmm mm        (14) And...

Ngày tải lên: 19/06/2014, 19:20