introduction to pdf 1 4 features

Tài liệu Introduction to Java: 1-Abstract Window Toolkit Overview doc

Ngày tải lên : 14/12/2013, 12:15

... containers 10 July 2002 22 :16 14 CHAPTER 1: ABSTRACT WINDOW TOOLKIT OVERVIEW Label North Text Area Button GridLayout Panel Button BorderLayout Panel East Button FlowLayout Panel Button Button Button ... FileDialog Figure 1 19 : Examples of Dialog and FileDialog boxes 10 July 2002 22 :16 1. 5 AND THE REST 17 1. 4. 5 ScrollPane Java 1. 1 introduces the ScrollPane container In version 1. 0, if you want to have ... processing that you want to * http://splash.javasoft.com/beans/spec.html 10 July 2002 22 :16 CHAPTER 1: ABSTRACT WINDOW TOOLKIT OVERVIEW Figure 1 1: Multiple Label instances 1. 1.2 .1 The TextField and...

19
468
1

Introduction to Probability - Chapter 4 doc

Ngày tải lên : 04/07/2014, 10:20

... opened by Monty 1/ 2 Path probabilities 1/ 18 1/ 2 1/ 18 1/ 9 1/ 9 1/ 9 1/ 2 1/ 18 1/ 18 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 1/ 3 1/ 2 1/ 3 1 1/9 1 1/9 1 1/9 1/ 2 1/ 18 1/ 18 1/ 3 1/ 3 1/ 3 1/ 3 1/ 2 Figure 4. 3: The Monty Hall ... 1/ 8 g 1/ 4 Walking with Mr Smith Unconditional probability Conditional probability b 1/ 4 1/ 2 b 1/ 8 1/ 4 b 1/ 8 1/ 4 bb 1/ 2 1/ 4 bg 1/ 4 1/ 2 1/ 4 1/ 2 gb 1/ 4 gg Mr.Smith's children bb 1/ 2 1/ 4 bg 1/ 4 1/ 2 ... second is drawn 15 6 CHAPTER CONDITIONAL PROBABILITY X -1 Y -1 1 /18 1/ 12 1/ 36 1/ 36 1/ 6 1/ 18 1/ 6 1/ 12 1/ 12 1/ 12 1/ 6 Table 4. 6: Joint distribution 36 A die is thrown twice Let X1 and X2 denote the...

50
347
0

An Introduction to Thermodynamics phần 4 ppsx

Ngày tải lên : 08/08/2014, 15:21

... g 91. 3 33.2 11 .1 - 41 1 .2 -40 7.3 -42 5.8 -47 0 .1 142 .7 -296.8 -395.7 - 74. 6 227 .4 52 .4 - 84. 0 -2 01. 0 -239.2 -2 34. 8 -277.6 49 .1 82.9 87.6 51. 3 99.8 -3 84 .1 -393 .1 -379.5 - 41 9 .1 163.2 -300 .1 -3 71. 1 -50.5 ... 26.5 -13 7.2 -3 94. 4 -11 29 .1 -11 28.2 - 748 .8 - 816 -11 75.6 -10 31. 2 -603.3 -897.5 -858 - 244 .3 - 742 .2 -10 15 .4 -53 .4 -10 4. 0 -275 .4 1. 7 -933.6(639.9) H2O H2O2 H2S I2 NH3 NH4Cl N2O NO NO2 N2O4 NaCl NaOH ... 30.9 1. 9 -13 9.3 -96.9 -10 9.8 -66.2 -11 .2 -15 82.3 3 .1 2.9 66.2 -68.6 -11 0.5 -393.5 -12 07.6 -12 07.8 -795 .4 -877.2 -12 28.0 -12 08 -6 34. 9 -985.2 -10 02.8 -272.0 -8 24 -11 18 .4 -36.3 -12 1.6 -273.3 26.5 -13 7.2...

13
461
0

Introduction to Probability phần 4 potx

Ngày tải lên : 09/08/2014, 23:20

... g 1/ 8 g 1/ 4 Walking with Mr Smith Unconditional probability Conditional probability b 1/ 4 1/ 2 b 1/ 8 1/ 4 b 1/ 8 1/ 4 1/ 2 1/ 4 bg 1/ 4 1/ 2 1/ 4 1/ 2 gb 1/ 4 gg Mr.Smith's children bb 1/ 2 1/ 4 bg 1/ 4 1/ 2 ... 3679 18 39 0 613 015 3 00 31 0005 00 01 0000 Binomial n = 10 0 p = 01 3660 3697 18 49 0 610 0 14 9 0029 0005 00 01 0000 Poisson λ = 10 0000 0005 0023 0076 018 9 0378 06 31 09 01 112 6 12 51 12 51 113 7 0 948 0729 ... 05 21 0 347 0 217 012 8 00 71 0037 0 019 0009 00 04 0002 00 01 0000 Binomial n = 10 00 p = 01 0000 00 04 0022 00 74 018 6 03 74 0627 0900 11 28 12 56 12 57 11 43 0952 07 31 0520 0 345 0 215 012 6 0069 0036 0 018 0009...

51
492
0

Introduction to Elasticity Part 4 ppt

Ngày tải lên : 11/08/2014, 09:20

... delivering 10 0 hp (horsepower) at 18 00 rpm (revolutions per minute) to the drive shaft, and we wish to compute the shearing stress From Eqn 8, the torque on the shaft is T = 10 0 hp 1. 3 41 10 −3 W = ... notch cut into the Ludwig Prandtl (18 75 19 53) is best known for his pioneering work in aerodynamics J.P Den Hartog, Advanced Strength of Materials, McGraw-Hill, New York, 19 52 12 Figure 14 : Membrane ... hoop strain as z θ = =− 1 (σθ − νσz ) = E E pr pr −ν b 2b After the French mathematician Simeon Denis Poisson, (17 81 18 40 ) = pr bE 1 ν The radial expansion is then pr ν 1 (9) θ bE Note that the...

25
410
0

Introduction to Electronics - Part 4 pot

Ngày tải lên : 14/08/2014, 02:21

... = VP ⇒ v GS − VP = v DS (11 1) But from eq (11 0) v GS − VP = iD K (11 2) Combining eqs (11 1) and (11 2) gives the boundary: v DS = iD K ⇒ i D = Kv DS (11 3) Fig 13 5 2N3 819 n-channel JFET output characteristics ... B ⇒ i B = (1 − α )i E iC αi E α = = =β iB (1 − α )iE − α (10 0) (10 1) Solving the right-hand half of eq (10 1) for α: α= β β +1 (10 2) For α = 0.99, we have β = 10 0 Rearranging eq (10 1) gives: iC ... IDSS and vGS = into eq (11 4) gives a relationship between K and IDSS : K= IDSS VP (11 5) Metal-Oxide-Semiconductor FETs (MOSFETs) Introduction to Electronics 92 Metal-Oxide-Semiconductor FETs (MOSFETs)...

23
169
0

luận án tiến sĩ lifting from sl(2) to gspin(1,4)

Ngày tải lên : 13/11/2014, 09:14

... a0 + a 1 ···νp i 1 iν2 · · · iνp , 1 1

114
212
0

An Introduction to e2studio (1)

Ngày tải lên : 22/06/2015, 14:04

... 0.3µA standby 44 DMIPS, True Low Power 10 DMIPS, Capacitive Touch  Industrial & Automotive, 13 0nm  14 4 A/MHz, 0.2µA standby  Format Automotive, 13 0nm WideIndustrial &LCDs  350µA/MHz, 1 A standby ... reserved e2studio Release Schedule now Q1 /12 35 Q2 /12 © 2 012 Renesas Electronics America Inc All rights reserved Q3 /12 Q4 /12 Q1 /13 e2studio Live Demo 36 © 2 012 Renesas Electronics America Inc All ... Software Tool * * 13 © 2 012 Renesas Electronics America Inc All rights reserved e2studio Features 14 © 2 012 Renesas Electronics America Inc All rights reserved Supported Families, Toolchains,...

42
384
0

Tài liệu Module 1: Introduction to Exchange 2000 and the Web Storage System pdf

Ngày tải lên : 11/12/2013, 14:15

... Module 1: Introduction to Exchange 2000 and the Web Storage System 11 # Introduction to the Web Storage System Topic Objective To list the topics related to Introduction to the Web Storage System ... easier maintenance ! Storage Groups contain Multiple Stores Storage Group A Store Store Store Store Store Transaction Log Transaction Log Storage Group B Store Store Store Store Store Transaction ... Computers 2 Module 1: Introduction to Exchange 2000 and the Web Storage System # Introduction to Exchange 2000 Topic Objective To list the topics related to introduction to Exchange 2000 ! Versions...

50
671
0

Tài liệu Module 1: Introduction to Windows 2000 Administration pdf

Ngày tải lên : 21/12/2013, 05:17

... permissions to groups rather than to each individual user account Module 1: Introduction to Windows 2000 Administration 11 Assigning Permissions to Gain Access to Resources Slide Objective To identify ... permissions to the accounting group to gain access to Excel and the accounting files Module 1: Introduction to Windows 2000 Administration 13 # Introduction to Windows 2000 Domains Slide Objective To ... gains access to specific resources, and to specify the kind of access that each user has As an administrator, you assign permissions to control access to resources Module 1: Introduction to Windows...

26
403
0

Tài liệu Module 1: Introduction to Microsoft Windows 2000 File, Print, and Web Servers pdf

Ngày tải lên : 17/01/2014, 08:20

... Windows NT 4. 0 network Describe the benefits of Active Directory 2 Module 1: Introduction to Microsoft Windows 2000 File, Print, and Web Servers Introduction to Windows 2000 Slide Objective To highlight ... industry-standard Lightweight Directory Access Protocol (LDAP) rather than proprietary protocols to access and change information in Active Directory 8 Module 1: Introduction to Microsoft Windows 2000 ... Active Directory directory service that provides a way for users, and other clients in a distributed environment, to name, store, and retrieve information 10 Module 1: Introduction to Microsoft...

16
490
0

Tài liệu Module 1: Introduction to Designing a Highly Available Web Infrastructure pdf

Ngày tải lên : 24/01/2014, 10:20

... 19 2 .16 8 .12 .15 AD0 DNS 19 2 .16 8 .12 . 41 Vlan 11 19 2 .16 8 .12 .40 DNS02 SMTP FTP 19 2 .16 8 .12 .23 19 2 .16 8 .11 .17 19 2 .16 8 .11 .16 19 2 .16 8 .12 .12 19 2 .16 8 .11 .15 SQL Cluster -1/ HB 19 2 .16 8 .12 .10 DNS 01 SMTP FTP 19 2 .16 8 .12 . 14 ... 19 2 .16 8 .13 . 41 STG 01 192 .16 8 .13 .46 19 2 .16 8 .13 . 31 Vlan 13 19 2 .16 8 .13 . 21 192 .16 8 .15 .2 54 19 2 .16 8 .13 .30 Vlan 15 19 2 .16 8 .13 .20 IIS05 19 2 .16 8. 21. 51 IIS 04 19 2 .16 8. 21. 50 19 2 .16 8.23 . 14 19 2 .16 8.23 .15 19 2 .16 8.22 .12 ... 19 2 .16 8 .12 . 14 19 2 .16 8 .12 .11 19 2 .16 8 .11 .13 IIS03 19 2 .16 8 .12 .26 19 2 .16 8 .11 .11 19 2 .16 8 .11 .12 19 2 .16 8 .12 .25 19 2 .16 8 .12 . 24 IIS02 19 2 .16 8 .11 . 14 Firewall Firewall IIS 01 VPN 01 ISA ISP1 ISP2 Ba Ne t o rk s y w Storage...

50
383
0

Tài liệu Báo cáo khoa học: Three-dimensional structures of thermophilic b-1,4-xylanases from Chaetomium thermophilum and Nonomuraea ﬂexuosa Comparison of twelve xylanases in relation to their thermal stability pdf

Ngày tải lên : 20/02/2014, 23:20

... 56.3 13 .1 6.7 6.7 4 .1 3.6 3 .1 0.0 2.6 4 .1 14. 9 6.7 9.3 1. 0 8.8 6.2 4 .1 6.2 4 .1 1.5 4 .1 2 .1 1 94 30.9 51 18 .1 4. 6 6.2 3.6 3 .1 3 .1 0.0 2.6 4 .1 16.0 11 .3 10 .8 1. 0 8.8 7.2 3.6 5.2 2.6 1. 0 3 .1 2 .1 1 94 ... BAX Total number of atoms Atoms/residue ˚ ASA (A2) 15 06 14 95 15 86 15 12 14 93 15 18 14 85 14 48 14 71 1 348 13 94 13 88 14 22 16 64 7.88 817 0 7.83 8250 8 14 0 7.97 8260 8 340 7.79 815 0 7.70 8 14 0 7.83 819 0 ... 58 .4 11 .6 5 .1 10 .1 3 .4 3 .4 3 .4 1. 1 3 .4 3 .4 12 .4 12 .9 10 .1 0.0 5.6 10 .1 6.2 2.8 2.8 0.6 1. 7 1. 7 17 8 33 .1 57.3 9.6 7.6 8.2 2.2 2.7 1. 6 1. 1 4. 9 2.7 10 .3 15 .8 10 .9 1. 1 9.2 6.5 3.3 4. 9 4. 3 0.0 1. 6 1. 1...

14
520
0

CS682-Network Security: Module-1 Introduction to Network Security pdf

Ngày tải lên : 14/03/2014, 22:20

... Matt Bishop, 02 0 14 40 997 CS 918 or EL537 Textbook: “TCP/IP Illustrated, Vol 1, ” Richard W Stevens Programming Reference: “Unix Network Programming, Vol 1, ” Richard W Stevens, 013 49 0 012 X Basic Understanding ... http://citeseer.nj.nec.com/cache/papers/cs/ 2 14 91/ http:zSzzSz www.cs.um.edu.mtzSz~kvelzSzCSA401zSzibm-tcpip .pdf/ tcpip-tutorial-and .pdf (Look for “tcp ip security” at http://www.researchindex.com) From Books 24x7 (http://dibner.poly.edu/) ... MULTICAST MTU :15 00 Metric :1 RX packets:595 017 errors:0 dropped:0 overruns:0 frame:0 TX packets :11 34 01 errors:0 dropped:0 overruns:0 carrier:0 collisions:0 txqueuelen :10 0 Interrupt :10 Base address:0xb800...

19
461
0

INTRODUCTION TO URBAN WATER DISTRIBUTION - CHAPTER 1 pdf

Ngày tải lên : 18/06/2014, 19:20

... London, UK 74 76 78 81 82 84 87 87 88 91 92 98 10 0 10 3 10 3 10 5 10 9 11 7 12 2 12 2 12 3 12 5 13 0 13 0 13 2 13 6 14 3 14 8 15 0 15 2 15 3 15 5 15 8 15 9 16 0 16 0 16 2 16 3 16 5 17 8 18 1 18 2 18 7 19 4 Contents 4. 5.7 Service ... London, UK 19 75 2000 2030 17 2/ 64 547 /52 13 /62 16 7/ 41 140 2 /17 2 21/ 15 2522/28 243 / 74 676/67 21/ 72 322/ 61 240 6/25 40 6/25 40 74/ 38 310 / 71 729/75 30/70 519 /75 3683/37 7 84/ 38 6055 /47 372/ 84 6 91/ 83 41 / 74 726/83 ... Urban, to million inhabitants Urban, below million inhabitants Rural 19 5/5 327/8 10 22/25 2530/62 41 8 /7 7 04 /12 17 23/28 3 210 /53 623/9 10 06 / 14 218 9/ 31 3337 /46 Total 40 74 /10 0 6055 /10 0 715 4 /10 0 is...

32
538
0

Introduction to Modern Liquid Chromatography, Third Edition part 1 pdf

Ngày tải lên : 04/07/2014, 01:20

... Information, 12 1. 4 .1 Books, 12 1. 4. 2 Journals, 13 1. 4. 3 Reviews, 13 1. 4. 4 Short Courses, 13 1. 4. 5 The Internet, 13 References, 15 BASIC CONCEPTS AND THE CONTROL OF SEPARATION 2 .1 Introduction, ... INTRODUCTION 1. 1 Background Information, 1. 1 .1 What Is HPLC?, 1. 1.2 What Can HPLC Do?, 1. 2 A Short History of HPLC, 1. 3 Some Alternatives to HPLC, 1. 3 .1 Gas Chromatography (GC), 1. 3.2 Thin-Layer Chromatography ... 3 .4. 2 Fittings, 99 3 .4. 2 .1 Low-Pressure Fittings, 99 3 .4. 2.2 High-Pressure Fittings, 10 1 3 .4. 2.3 Specialty Fittings, 10 3 3.5 Pumping Systems, 10 4 3.5 .1 Reciprocating-Piston Pumps, 10 4 3.5 .1. 1...

10
629
0

Introduction to Modern Liquid Chromatography, Third Edition part 4 pdf

Ngày tải lên : 04/07/2014, 01:20

... (Sections 9.3 .10 , 13 .10 ), Mike Swartz (Chapter 12 ), Tim Wehr (Sections 13 .1 13 .8), Carl Scandella (Section 13 .9), ¨ Wolfgang Lindner, Michael Lammerhofer, and Norbert Maier (Chapter 14 ), Geoff Cox ... ion-exchange chromatography ion-pair chromatography retention factor (same as capacity factor k ); equal to (tR /t0 ) − gradient retention factor; Equation (9.5) column length (mm) liquid chromatography–mass ... referred to as ‘‘conditional’’ or ‘‘sample’’ peak capacity normal-phase chromatography pressure drop across the column (psi); bar or atmospheres = 14 .7 psi; megaPascal (MPa) = 10 bar = 14 7 psi;...

10
488
1

The Oxford Introduction to Proto-Indo-European Part 1 pdf

Ngày tải lên : 05/08/2014, 13:20

... 3.6 3.7 3.8 3.9 3 .10 3 .11 3 .12 4 .1 4. 2 4. 3 4. 4 4. 5 4. 6 4. 7 4. 8 4. 9 4 .10 4 .11 4 .12 4 .13 4 . 14 4 .15 5 .1 5.2 xv Selected cognates in Sanskrit and Avestan 34 Selected cognates in Tocharian, Old English ... 7.9 8 .1 8.2 8.3 8 .4 9 .1 9.2 9.3 9 .4 9.5 10 .1 10.2 10 .3 10 .4 11 .1 11. 2 11 .3 11 .4 11 .5 11 .6 11 .7 Indo-European words for ‘Wre’ 91 Dates of separation from Proto-Indo-European based on the 10 0 and ... 20.6B 20.7 20.8 20.9 21. 1 21. 2 21. 3 21. 4 22 .1 22.2 22.3 22 .4 22.5 22.6 22.7 22.8 22.9 22 .10 22 .11 22 .12 22 .13 22 . 14 22 .15 22 .16 22 .17 22 .18 23 .1 23.2 24 .1 24. 2 24. 3 24. 4 24. 5 25 .1 25.2 Negative qualities...

760
551
0

An Introduction to Database Systems 8Ed - C J Date - Solutions Manual Episode 1 Part 3 pdf

Ngày tải lên : 06/08/2014, 01:21

... REFERENCES J ) ; 4. 2 No answer provided 4. 3 No answer provided 4. 4 a INSERT INTO S ( S#, SNAME, CITY ) VALUES ( S# ('S10'), NAME ('Smith'), 'New York' ) ; STATUS here is set to the applicable ... Date 4 .10 page WITH extensions, which are also intended to address this kind of problem (See the short paper "The Importance of Closure" in my book Relational Database Writings 19 91- 19 94, AddisonWesley, ... only with principles and not pragma But few instructors are likely to enjoy such a luxury One point instructors need to be aware of: Exercise 4 .1 introduces the extended version of the running...

20
923
3

An Introduction to Database Systems 8Ed - C J Date - Solutions Manual Episode 1 Part 4 pptx

Ngày tải lên : 06/08/2014, 01:21

... 5 .17 See Answer 4 .1 and Answer 5 .16 5 .18 No answer provided 5 .19 See Section 5.7, subsection "Structured Types." 5.20 No answer provided 5. 21 Such a type──we call it type omega──turns out to ... OPERATOR DIAMETER ( C CIRCLE ) RETURNS LENGTH ; RETURN ( * THE_R ( C ) ) ; END OPERATOR ; OPERATOR CIRCUMFERENCE ( C CIRCLE ) RETURNS LENGTH ; RETURN ( 3 . 14 159 * DIAMETER ( C ) ) ; END OPERATOR ... ADDWT ( ( ( ( WT, 14 .7 ) 14 .7, WT ) WT, WT ) 14 .7, 3.0 ) (End quote) Note, however, that──even if the current value of WT is WEIGHT(3.0)──the four invocations aren't constrained to return the same...

20
511
3