accessing the wan chapter 2 exam answers

Accessing the WAN – Chapter 3 potx

Accessing the WAN – Chapter 3 potx

... you build a WAN, there is always 3 components, –DTE –DCE ? ?The component sits in the middle, joining the 2 access points.  In the late 1970s and into the early 1990s, the WAN technology ... and Q 933 -A ITE 1 Chapter 6 © 20 06... encapsulates them as the data portion of a Frame Relay frame, and then passes the frame to the physical layer for delivery on the wire –First, Frame ... component: ? ?The physical component defines the mechanical, electrical, functional between the devices. ? ?The link layer component defines the protocol that establishes the connection between the DTE

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Accessing the WAN – Chapter 4 docx

Accessing the WAN – Chapter 4 docx

... the host of the victim makes the request to the host of the attacker's ? ?2 The attacker's host receives the request and fetches the real page from the legitimate website •3 The ... [...]... the legitimate website •3 The attacker can alter the legitimate webpage and apply any transformations to the data they want to make 4 The attacker forwards the requested to the ... © 20 06 Cisco Systems, Inc. All rights reserved. Cisco Public ITE I Chapter 6 1 Network Security Accessing the WANChapter 4 Cisco Thai Nguyen Networking Academy © 20 06 Cisco

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Accessing the WAN – Chapter 5 pdf

Accessing the WAN – Chapter 5 pdf

... . 25 5 . 25 5 .24 0, – take 25 5 . 25 5 . 25 5 . 25 5 and subtract the subnet mask 25 5 . 25 5 . 25 5 .24 0 – The solution this time produces the. .. 0.0.0. 15 Example 3: assume you wanted ... produces the wildcard mask 0.0.0. 25 5 Example 2: Now assume you wanted to permit network access for the 14 users in the subnet 1 92. 168.3. 32 /28 The subnet mask for the IP subnet is 25 5 . 25 5 ... for the number of ports. • 3 protocols X 2 directions X 2 directions = 12 [...]... 25 5 . 25 5 . 25 5 .0, you could take the 25 5 . 25 5 . 25 5 . 25 5 and subtract from the subnet mask – The

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Accessing the WAN – Chapter ppt

Accessing the WAN – Chapter ppt

... data to the computer over an Ethernet connection In the upstream direction, – the CM decodes the digital information from the. .. in either direction at the same time The following ... subscriber to the cable operator, the incoming frequencies are in the range of 5 to 42 MHz ITE 1 Chapter 6 © 20 06 Cisco... between the CPE modem and the CO DSLAM The voice channel ... time, ? ?The remote user, for example, then downloads his email from the mail server at 10.10.0.5, and downloads a document from the Archive at 10 .2. 3.4. Next, without exiting the tunnel, the

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Accessing the WAN – Chapter 7 pot

Accessing the WAN – Chapter 7 pot

... IP addresses from the 20 9.165 .20 1.0 / 27 range The ouput confirms the assigned... Tasks in the Configure tab From the list of tasks, click on the DHCP folder and then select DHCP Pools ... case, IP address 20 9.165 .20 0 .22 6 is used as the inside global address... to the packet, and verify that routers have the correct routing information to move the packet Use the debug ip ... addresses drawn from the three blocks shown in the figure These addresses... 1918 private address In the figure, the IP address 1 92. 168.10.10 is assigned to the host PC1 on the inside network

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Accessing the WAN – Chapter 8 ppsx

Accessing the WAN – Chapter 8 ppsx

... investigation •For example, if users can't access the web server and you can ping the server, then you know that the problem is above Layer 3 •If you can't ping the server, then you know the problem ... failure. – At one extreme is the theorist, or rocket scientist, approach. • The rocket scientist analyzes and reanalyzes the situation until the exact cause at the root of the problem has been identified. ... the common LAN bandwidth The charges for link provision are the major cost element, therefore the WAN implementation must aim to provide maximum bandwidth at acceptable cost ITE 1 Chapter

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PESTICIDES IN AGRICULTURE AND THE ENVIRONMENT - CHAPTER 2 docx

PESTICIDES IN AGRICULTURE AND THE ENVIRONMENT - CHAPTER 2 docx

... wilts [21 ? ?23 ]. Moreover, the incidence of fusarium wilts appears to be related to the relative proportion of the pathogen population within the total population of Fusarium rather than to the absolute ... a 2- year period compared to the control plots. The number of sclerotia of the plant pathogen was significantly reduced by the myco- parasitic activity. The mycoparasite became established in the ... tein... the EPA (Table 1) Bio-Trek, Rootshield, and T -2 2  Planter Box are three products based on T harzianum KRL-AG2 (strain T -2 2 ) that are sold by Bio Works, Inc of Geneva, NY They

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Heavy Metals in the Environment - Chapter 2 pot

Heavy Metals in the Environment - Chapter 2 pot

... 2 , Fe 2 , Fe 3ϩ, Cu 2 , Zn 2 , Se 4ϩ, Se 6ϩ, and Mo 6ϩ; adsorptive stripping potentiometry for Co 2 , Ni 2 , and Zn 2 ; potentiometric stripping analysis for Cd 2 and Copyright © 20 ... characteristic wavelength. The elements in the sample are identified by the wavelength/energy of the emitted X-rays while the concentrations are determined by the intensity of the X-rays. Two basic ... electric sector, which acts as an energy filter. The ions Copyright © 20 02 Marcel Dekker, Inc. are then deflected in a single plane by the magnetic field, with the degree of deflection increasing with increasing

Ngày tải lên: 11/08/2014, 15:20

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Radio Propagation and Remote Sensing of the Environment - Chapter 2 pps

Radio Propagation and Remote Sensing of the Environment - Chapter 2 pps

... the general case, S 0 2 ≥ S1 2 + S 2 2 + S 3 2 (2. 62) Therefore, the coefficient m= S1 2 + S 2 2 + S 3 2 S0 2 (2. 63) defines the degree of wave coherence,... p ) 3 (2. ... averaged values Then, S0 = Ex 2 + Ey 2 ˆ ˆ = E2 + E2 + Nx x y 2 + Ny 2 S1 = E x 2 − Ey 2 ˆ ˆ = E2 − E2 + Nx x y 2 − Ny 2 , (2. 57) , (2. 58) ∗ ˆ ˆ S 2 = 2 Re E x E ∗ = 2 E x E ... y 2 2 ˆ ˆ S1 = E 2 − E 2 = E x − E y x y (2. 48) b 2 = S 0 − S1 2 + S 2 2 (2. 49) The ellipse radii are defined as follows: a 2 = S 0 + S1 2 + S 2 2 , where 2 ˆ ˆ S0 = E2

Ngày tải lên: 11/08/2014, 21:21

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Modeling phosphorus in the environment - Chapter 2 ppsx

Modeling phosphorus in the environment - Chapter 2 ppsx

... Runoff 22 2. 2.1 Runoff Volume 22 2. 2.1.1 Curve Number Method 23 2. 2.1 .2 Curve Number Method Implementation 25 2. 2.1.3 Infiltration-Based Approaches 29 2. 2 .2 Hydrograph Development 32 2 .2. 2.1 Kinematic ... Routing 33 2. 2 .2. 2 SCS Unit Hydrograph 34 2. 2 .2. 3 Hydrograph Development Implementation 35 2. 2.3 Streamflow, or Channel, Routing 36 2. 2.3.1 Hydrologic, or Storage, Routing 37 2. 2.3 .2 Muskingum ... 37 2. 2.3.3 Streamflow, or Channel, Routing Implementation 39 2. 2.4 Peak Rate of Runoff 40 2. 2.4.1 Rational Formula 40 2. 2.4 .2 SCS TR-55 Method 41 2. 2.4.3 Peak Runoff Rate Implementation 41 2. 3

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Radioactivity in the environment chapter 2   radiation risks and the ICRP

Radioactivity in the environment chapter 2 radiation risks and the ICRP

... Publication 82 Annals of the ICRP, 29 (1? ?2) ICRP (20 06) The optimisation of radiological protection - broadening the process ICRP Publication 101b Annals of the ICRP, 36(3) ICRP (20 07a) The 20 07 Recommendations ... Underpinning of the Evolution of ICRP Recommendations   26 2. 6 Some Moot Points   29 2. 6.1 Different Dose Limits for Occupational or Public Exposures   29 2. 6 .2 Protecting Average Individuals or the ... Thus the default assumption of a safe threshold was rejected For stochastic effects, primarily the probability rather than the severity of the effect is proportional to the size of the dose Therefore,

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lin download solutions manual for accessing the WAN CCNA exploration labs and study guide 1st edition by john rullan

lin download solutions manual for accessing the WAN CCNA exploration labs and study guide 1st edition by john rullan

... 25 5 .25 5 .25 5.0 N/A Fa0/1.30 1 92. 168.30.1 25 5 .25 5 .25 5.0 N/A S0/0/1 10 .2. 2 .2 255 .25 5 .25 5 .25 2 N/A S1 VLAN10 1 92. 168.10 .2 255 .25 5 .25 5.0 1 92. 168.10.1 S2 VLAN20 1 92. 168 .20 .2 255 .25 5 .25 5.0 1 92. 168 .20 .1 ... Fa0/1 .20 1 92. 168 .20 .1 25 5 .25 5 .25 5.0 N/A S0/0/0 10.1.1 .2 255 .25 5 .25 5 .25 2 N/A S0/0/1 10 .2. 2.1 25 5 .25 5 .25 5 .25 2 N/A Lo0 20 9.165 .20 0.161 25 5 .25 5 .25 5 .22 4 N/A Fa0/1 N/A N/A N/A Fa0/1.13 10.13.13.3 25 5 .25 5 .25 5.0 ... 1 92. 168.10.1 25 5 .25 5 .25 5.0 N/A Fa0/1. 12 10. 12. 12. 1 25 5 .25 5 .25 5.0 N/A Fa0/1.13 10.13.13.1 25 5 .25 5 .25 5.0 N/A S0/0/0 10.1.1.1 25 5 .25 5 .25 5 .25 2 N/A Fa0/1 N/A N/A N/A Fa0/1. 12 10. 12. 12. 2 25 5 .25 5 .25 5.0

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A100 luyện giải toán (new a level maths for edexcel year 1  2 exam) answers section one   pure maths

A100 luyện giải toán (new a level maths for edexcel year 1 2 exam) answers section one pure maths

... + 25 x2 + k = - 25 x + 125 x2 + -1 #- 2 (1 – 2x)–1 = + (–1)(–2x) + # (–2x) + (1 – 2x)? ?2 = + (? ?2) (–2x) + -2 #- # (–2x) + = + 4x + 12x2 + Answers a) – 18x ∫ A(1 – 2x )2 + B(5 + 4x)(1 – 2x) ... dy y2 dx dx dy d 3x2 + dx x2y = 2y dx d d 3x2 + x2 dx y + y dx x2 = 2y dy dx dy dy + 2xy = 2y dx dx dy Rearrange to make dx the subject: d y (2y – x2) dx = 3x2 + 2xy 3x2 + 2xy dy dx = 2y - x2 ... constant: 2 2 D = 62x4 - kx2@ #2 ]8x3 - 2kxg dx = : 84x - 2kx 2 = ^2 ]2g4 - k ]2g2h - _2 ^ h4 - k ^ h2 i = ] 32 - 4kg - ]8 - 2kg = 24 - 2k You know that the value of this integral is 2k2, so set this

Ngày tải lên: 01/11/2022, 11:04

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The Aerodynamics of the Aircraft (Chapter 2) Động lực học hàng không của máy bay

The Aerodynamics of the Aircraft (Chapter 2) Động lực học hàng không của máy bay

... currently the biggest and most powerful turbofan engine in the world These big engines can be found on Boeing 777 models JFK Airport, New York, 20 17 Large Aircraft 10 28 29 30 31 Selecting the Engine ... 0933 321 877 – Email: nguyenthientong@gmail.com Các nội dung môn học (15 tuần theo textbook) Introduction - Turbine Engines (Ch 1) – The Atmosphere The Aerodynamics of the Aircraft (Ch 2) – Specific ... Engines - Week 2: Động lực học hàng không máy bay - Động tua-bin khí Tuần – The Aerodynamics of the Aircraft (Chapter 2) Động lực học hàng không máy bay Large Aircraft Sizing the Wing Lift, Drag,

Ngày tải lên: 14/02/2023, 21:05

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Lecture Communication research: Asking questions, finding answers (4e)  Chapter 2: The research process: Getting started

Lecture Communication research: Asking questions, finding answers (4e) Chapter 2: The research process: Getting started

... fo6e 20 0b zof2 a3da pvx4 ijrư hapi p2lt vnu7 slxq cvmp dk1v af5z hvt2 2ft5 2mka d4k4 8who ekdp s48p i91n cfn3 rs4y jt6h 2hte vurn 6hq5 9q 82 y201 yten d7ns omqs kjei 0flq zw5t 43n6 62j3 t2es 51av ... fo6e 20 0b zof2 a3da pvx4 ijrư hapi p2lt vnu7 slxq cvmp dk1v af5z hvt2 2ft5 2mka d4k4 8who ekdp s48p i91n cfn3 rs4y jt6h 2hte vurn 6hq5 9q 82 y201 yten d7ns omqs kjei 0flq zw5t 43n6 62j3 t2es 51av ... fo6e 20 0b zof2 a3da pvx4 ijrư hapi p2lt vnu7 slxq cvmp dk1v af5z hvt2 2ft5 2mka d4k4 8who ekdp s48p i91n cfn3 rs4y jt6h 2hte vurn 6hq5 9q 82 y201 yten d7ns omqs kjei 0flq zw5t 43n6 62j3 t2es 51av

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Accessing the WAN – Chapter 2 docx

Accessing the WAN – Chapter 2 docx

... devices. The DTE is the RS -23 2C interface that a computer uses to exchange data with a modem or other serial device. The DCE is the RS -23 2C interface that a modem or other serial device uses ... request requires no authentication, then PPP progresses to the next level. –If an incoming PPP request requires authentication, then it can be authenticated using either the local database ... call. © 20 06 Cisco Systems, Inc. All rights reserved. Cisco Public ITE I Chapter 6 1 Point-to-Point Protocol Accessing the WANChapter 2 Cisco Thai Nguyen Networking Academy â 20 06 Cisco...

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Accessing the WAN – Chapter 1 ppt

Accessing the WAN – Chapter 1 ppt

... ITE 1 Chapter 6 12 Cisco Thai Nguyen Networking Academy WAN Physical Layer Standards  The WAN physical layer also describes the interface between the DTE and the DCE. –EIA/TIA -23 2 - This ... all the VCs require maximum demand simultaneously, the capacity of the leased line can be smaller than the sum of the individual VCs.  Examples of packet- or cell-switched include: –X .25 ... on a 25 -pin D-connector over short distances. It was formerly known as RS -23 2. The ITU-T V .24 specification is effectively the same. –EIA/TIA-449/530 - This protocol is a faster (up to 2 Mb/s)...

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Accessing the WAN

Accessing the WAN

... Define the three phases of PPP session establishment © 20 06 Cisco Systems, Inc. All rights reserved. Cisco Public 1 Version 4.0 Access Control Lists Accessing the WANChapter 5 © 20 06 ... Describe the key WAN technology concepts.  Identify the appropriate WAN technologies to use when matching ECNM best practices with typical enterprise requirements for WAN communications. © 20 06 ... terms of the OSI Reference Model © 20 06 Cisco Systems, Inc. All rights reserved. Cisco Public 23 Configuring PPP with Authentication  Outline the PPP encapsulation and authentication...

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AEROSOL CHEMICAL PROCESSES IN THE ENVIRONMENT - CHAPTER 2 docx

AEROSOL CHEMICAL PROCESSES IN THE ENVIRONMENT - CHAPTER 2 docx

... 2 41 4 41 =− () −− + ()                 =− () −− + ∞ ∞ ∞ ∞ ∞ π β πβ π β , , II rkRT T 2 2 4 ()                 ∞ πβ I BbA A BbA A BBBBBB gg 1 22 1 1 1 12 2 2 2 22 11 11 22 12 21 1 1 = − () − () +− () −−+ − I BbA A BbA A BBBBBB gg 2 11 2 2 2 21 1 1 1 22 11 11 22 ... bl 11 22 0àà+= nd nd Ad sl sl 11 22 0àà++= à i i v r += 2 0 * 4 3 3 11 2 2 rnvnv=+ àà 12 21 vv= r v i i * = 2 à Gr**= 4 3 2 nd nd Ad ll 11 22 0àà++= L 829 /frame/ch 02 Page 30 Monday, January 31, 20 00 ... rRm pp =+ () 22 12 2 / Grfmx* * (,)= 2 3 2 12 L 829 /frame/ch 02 Page 34 Monday, January 31, 20 00 2: 03 PM â 20 00 by CRC Press LLC Physical Chemistry of Aerosol Formation 35 where (2. 43) and (2. 44) and (2. 45) Nucleation...

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Radionuclide Concentrations in Foor and the Environment - Chapter 2 docx

Radionuclide Concentrations in Foor and the Environment - Chapter 2 docx

... 10 10 y 22 8 Ra 5.8 y 6.1 h 22 8 Ac 23 2 22 8 1.9 y 56 s 0.15 s 3.7 d 21 2 Pb 21 6 Po 22 0 Rn 22 4 Ra 11 h 21 2 Bi 21 2 Po 20 8 Tl 61 m (64%) 3.1 m 61 m (36%) 3.1 ì 10 7 s 20 8 Pb DK594X_book.fm Page 27 Tuesday, ... h 23 5 U 23 1 Pa 22 7 Ac 22 y (99%) 4 s 11 d 19 d 21 9 Rn 22 3 Ra 21 5 Po 22 7 21 1 Pb 8 ì 10 4 s 21 1 Bi 20 7 Tl 20 7 Pb 36 m 2. 1 m 4.8 m 22 3 Fr 22 y (1%) 22 m 3.3 × 10 4 y DK594X_book.fm Page 28 Tuesday, June 6, 20 06 ... m 23 8 U 23 4 Pa 23 4 U 2. 4 ì 10 5 y 3.8 d 1.6 ì 10 3 y 7.7 ì 10 4 y 22 2 Rn 22 6 Ra 21 8 Po 23 0 21 4 Pb 3 m 21 4 Bi 21 4 Po 21 0 Pb 20 m 6.4 ì 10 5 s 21 0 Bi 22 .3 y 5 d 21 0 Po 138 d 27 m 20 6 Pb DK594X_book.fm...

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