neurologic examination and lesion localization pages 1 7 pdf

... 22 31 23 63 12 23 14 66 0.84 1. 18 0.88 2. 41 0.46 0.88 0.54 2.54 0.55 0 .74 0. 61 1.05 0.25 1. 15 0. 41 1. 32 0.29 1. 23 0 .73 1. 51 1 37 23 11 4 11 1‡ 20 91 5.28 2.24 7. 27 4.29 1. 98 5 .76 0. 81 0.88 0 .79 ... Person-years of follow-up† 670 7 10 8 6599 47. 7 54.6 74 .0 66 .7 37 .1 26 2 47 66 81 104 6 576 47. 7 54.5 73 .7 67. 0 36.4 26 15 4 13 388 212 13 17 5 47. 7 54.6 73 .9 67. 0 36.8 52 4 01 *See text for details †Based ... 21 19 16 0.96 0.39 1. 32 0 .73 0.30 1. 01 0 .76 0 .76 0 .76 0.40 1. 44 0 .11 –4.49 0. 37 1. 53 Pulmonary embolism† ഛ49 y old ജ50 y old 18 16 0.23 0 .10 0. 31 0.69 0.20 1. 00 3. 01 2.03 3 .19 1. 15–9. 27 0 .11 11 9.62...

Ngày tải lên: 15/02/2014, 04:20

... (22. 67)   56 (24.89)   33 (14 . 67)   42 (18 . 67)   18 (8.00) 16 9 (75 .11 )    (0.00)   99 (44.00)   46 (20.44)    (0.00) < 0 01 < 0 01 < 0 01 = 63 < 0 01 H   33 (14 . 67) 15 1 ( 67 .11 ) < 0 01 H L H   17 (7. 56) ... 19 99 See CD for color image and key © 2009 by Taylor & Francis Group, LLC 18 5 An Assessment of Health and Sustainability 18 6 10 9 15 17 6 95 13 92 25 13 11 79 12 Dim 20 51 57 11 8 11 9 12 8 43 12 7 ... Level 19 99 (%) 2000 (%) pa H L H L H L H E L H L E   32 (14 .22)   33 (14 . 67)   30 (13 .33)   31 (13 .78 )   47 (20.89)   50 (22.22)   43 (19 .11 ) 12 4 (55 .11 )   51 (22. 67)   39 ( 17 .33)   40 ( 17 .78 ) 12 5...

Ngày tải lên: 18/06/2014, 19:20

... as for ν =1. 2; 1. 3; 1. 4; 1. 5; 1. 6; 1. 8 and 1= 0 .1; а2 =1. 0 are shown in fig .7- a and 7- b The comparison of these characteristics to the known ones from (Al Haddad et al., 19 86; Cheron, 19 89) shows ... frequency Such characteristics are shown in fig 7- а for ν = 3.0; 3. 316 5; 3.6 and 1= 0 .1; а2=0.2 and in fig 7- b for ν =1. 5; 1. 6; 1. 8 and 1= 0 .1; а2 =1. 0 At the boundary operation mode, the diodes ... + a1 y3 = (U ′ ⋅ (U ′ Cm ) (16 ) ′ ′ ′ − U0UCm + a2U02 − a1 ⋅ Cm ′ ′ ′ ′ + U 0UCm − a2U02 + a1 + 2U0 + (15 ) ) ( 17 ) ′ x = UCm (18 ) y4 = (19 ) For converters with only two reactive elements (L and...

Ngày tải lên: 19/06/2014, 08:20

... Content), mg in mL 325 (65) 300 (60) 19 5 (39) 90 (18 ) Extended release 525 (10 5) Ferrous fumarate 325 (10 7) 19 5 (64) Ferrous 10 0 (33) 325 (39) 300 (35) 15 0 (15 0) 10 0 (10 0) gluconate Polysaccharide iron ... again reveal normal stores and more than an adequate supply to the marrow, despite the microcytosis and hypochromia Iron-Deficiency Anemia: Treatment The severity and cause of iron-deficiency ... blood loss or malabsorption, specific diagnostic tests and appropriate therapy take priority Once the diagnosis of iron-deficiency anemia and its cause is made, there are three major therapeutic...

Ngày tải lên: 07/07/2014, 04:20

... Hemoglobin >10 gm% Karyotype: Normal 70 Abnormal 78 a From B Dupriez et al Blood 88 :10 13, 19 96 b From F Cervantes et al Br J Haematol 10 2:684, 19 98 c From JT Reilly et al Br J Haematol 98:96, 19 97 ... Low 93 1 2 High 17 Risk group Median survival B Prognostic factorsb Hemoglobin < 10 gm% Constitutional symptoms Blast cells > 1% Number of prognostic factors (months) 0 1 Low 99 2–3 High 21 C Prognostic ... Hemoglobin 10 gm% Karyotype: Normal 54 Abnormal 22 Age 10 gm% Karyotype: Normal 18 0 Abnormal 72 Age >65 years Hemoglobin 10 gm% Karyotype: Normal 44 Abnormal 16 Age >65...

Ngày tải lên: 07/07/2014, 04:20

... specification and validation of the human-machine interface, including the sequencing of tool usage and definition of metrics It is stressed, however, that definition of 13 .7 International Standards 311 ... Technology, and Standards, April 10 , 2003 [11 ] “IV Highway Safety Act of 2004 Introduced in U.S Congress,” IVsource.net, posted August 3, 2004 [12 ] http://www.toyota.com, accessed October 5, 2004 [13 ] ... introduced by 2006 13 .7 International Standards A full treatment of standards issues and activities is beyond the scope of this book However, it is important to note that the standards arena is...

Ngày tải lên: 05/08/2014, 11:20

... pump (Figures 8 -16 and 8 - 17 )—Monel K-500 shaft, 5-in dia, approximately 27 in length overlaid on coupling end, /16 in deep; approximately 21 in length overlaid on thrust end, 1/ 32– /16 in, deep Successful ... areas are overlaid with Inconel 18 2 (AWS A5 .11 ENiCrFe-3) After rough machining, the cases are stress relieved and then machined to final dimensions (Figures 8-20 and 8- 21) The erosion/corrosion problem ... (0.030-in cut), Stellite overlay (GTAW process), and final machining to size (Figures 8 -18 and 8 -19 ) Using this procedure, matched spare sets of impellers and case wear rings are produced, which are...

Ngày tải lên: 05/08/2014, 12:20

... 5.328 -7. 415 (0 .11 5) (0.6 51) [0 .16 8] [0.658] yes -8. 0 17 3 .76 4 -9 .7 01 (0. 018 ) (0 .10 6) Ger[0.022] [0 .12 7] many no -10 .252 3.449 -11 .18 5 (0.000) (0.0 07) [0.0 01] [0.006] PPP Halflife 1. 79 6 τc∗ -1. 878 ... (1) = Since F (1) = C (1) F0 , impose these three restrictions by setting f13 (1) = = c 11 (1) f13,0 + c12 (1) f23,0 + c13 (1) f33,0 , f12 (1) = = c 11 (1) f12,0 + c12 (1) f22,0 + c13 (1) f32,0 , f33 (1) ... G and fij,0 be the ijth element of F0 Writing (8.48) out gives 2 2 g 11 = f 11, 0 + f12,0 + f13,0 , g 11 g12 = f 11, 0 f 21, 0 + f12,0 f22,0 + f13,0 f23,0 , g 11 g13 = f 11, 0 f 31, 0 + f12,0 f32,0 + f13,0...

Ngày tải lên: 05/08/2014, 13:20

... (c) 1 = + z (1 − z) z 1 z 1 + = (z + 1) − − (z + 1) 1 1 = − , (z + 1) − 1/ (z + 1) (z + 1) − 2/(z + 1) = (z + 1) = (z + 1) ∞ = n =1 −∞ ∞ n=0 ∞ n=0 1 − n (z + 1) (z + 1) for |z + 1| > and |z + 1| ... 2n +1 n +1 n f (z) = z − ( 1) n=−∞ n=0 for < |z| < f (z) = 1 − 1+ z 2+z 1 1 ( 1) n +1 n z , 2n +1 n=−∞ ( 1) n +1 z n − = n=−∞ 1 f (z) = ( 1) n +1 n=−∞ 2n +1 − n z , 2n +1 for < |z| for < |z| Solution 12 . 27 ... (z) = 1 − 1+ z 2+z ∞ ∞ ( 1) n z n − = n=0 ∞ n=0 ( 1) n − = n=0 ∞ ( 1) n f (z) = n=0 ( 1) n n z , 2n +1 2n +1 zn, 2n +1 − n z , 2n +1 614 for |z| < for |z| < for |z| < f (z) = 1 − 1+ z 2+z 1 ∞ ( 1) n n...

Ngày tải lên: 06/08/2014, 01:21

Ngày tải lên: 08/08/2014, 00:22

... 11 • What is the role of the timestamp in message in Fig 9-23, and why does it need to be encrypted? 12 Complete Fig 9-23 by adding the communication for authentication between Alice and Bob 13 ... (Diffie and Hellman, 19 76 ) The protocol works as follows Suppose that Alice and Bob want to establish a shared secret key The first requirement is that they agree on two large numbers, nand g that ... important system that is widely used is Kerheros (Steiner et al., 19 88; and Kohl and Neuman, 19 94) Kerberos was developed at M.LT and is based on the Needham-Schroeder authentication protocol we...

Ngày tải lên: 08/08/2014, 21:22

... projects and grants: Hungarian IHM 46 71 / 1/2003 project, Hungarian OTKA T042459 and T04 377 0 grants, OTKA Instrument Grant M04 211 0, Hungarian IKTA OMFB-00580/2003, and EU-GridLab IST-20 01- 3 213 3 13 8 ... (2002) 17 09 - 17 32 [15 ] Goodale, T., et al.: The Cactus Framework and Toolkit: Design and Applications 5th International Conference on Vector and Parallel Processing, 2002, pp 19 7- 2 27 [16 ] Foster, ... Wochenschrift 11 (18 96) 353-362 [7] Brown, P N., Byrne, G D., Hindmarsh, A C.: Vode: A variable coefficient ode solver SIAM Journal of Scientific and Statistical Computing 10 (19 89) 10 38 -10 51 [8] Cohen,...

Ngày tải lên: 08/08/2014, 23:21

... Chapter 23 Gaming with Linux 3 019 0c20.qxd:Layout 12 /18 / 07 12 :48 AM Page 512 3 019 0c20.qxd:Layout 12 /18 / 07 12 :48 AM Page 513 Playing Music and Video O ne of the most popular and enjoyable activities ... names) 510 3 019 0c20.qxd:Layout 12 /18 / 07 12 :48 AM Page 511 Running Applications IN THIS PART Chapter 20 Playing Music and Video Chapter 21 Working with Words and Images Chapter 22 E-Mailing and Web ... (mini-CD) and low RAM Figure 19 -3 shows an example of the DSL desktop 503 19 3 019 0c19.qxd:Layout Part III 12 /18 / 07 12 :45 AM Page 504 Choosing and Installing a Linux Distribution FIGURE 19 -3 Damn...

Ngày tải lên: 09/08/2014, 07:20

... , 11 J3 = a3 d5 s4 (s2 c3 (a3 + d5 c4 ) + d5 c2 s4 ) , 11 200 R Konietschke, G Hirzinger and Y Yan J 11 J 21 J1 22 J4 = 0, 11 J 11 J 21 J2 22 = −a3 d5 c2 s4 (d5 s6 + a3 s4 c5 c6 + a3 c4 s6 ) , J 11 ... ±arccos − d J3 = : 11 φ4 = πk ∨ φ4 = ±arccos − ∨ φ4 = πk ∨ a3 +2πk; d5 s4 d5 c3 (d5 c4 + a3 ) φ2 = ±arctan + πk, J4 = ∀ φi ; 11 J 11 J 21 J1 22 = : φ2 = s6 =0 J 11 J 21 J2 22 J 11 J 21 J3 22 = : φ2 = ... gait,” Gait and Posture, 5, pp 10 8 -11 5 Wilson, D.R., Feikes, J.D., and O’Connor, J.J., 19 98 “Ligament and articular contact guide passive knee flexion,” Journal of Biomechanics, 31, pp 11 2 71 1 36 MODELING...

Ngày tải lên: 10/08/2014, 01:22