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NEW QUESTION: 1
An organization has requested that a penetration test be performed to determine if it is possible for an attacker to gain a foothold on the organization's server segment During the assessment, the penetration tester identifies tools that appear to have been left behind by a prior attack Which of the following actions should the penetration tester take?
A. Attempt to use the remnant tools to achieve persistence
B. Document the presence of the left-behind tools in the report and proceed with the test
C. Discontinue further testing and report the situation to management
D. Remove the tools from the affected systems before continuing on with the test
Answer: B
NEW QUESTION: 2
Which choice below denotes a packet-switched connectionless wide area network (WAN) technology?
A. Frame Relay
B. X.25
C. ATM
D. SMDS
Answer: D
Explanation:
Switched Multimegabit Data Service (SMDS) is a high-speed, connectionless, packet-switching public network service that extends LAN-like performance to a metropolitan area network (MAN) or a wide area network (WAN). It's generally delivered over a SONET ring with a maximum effective service radius of around 30 miles.
*X.25, defines an interface to the first commercially successful connection-oriented
packet-switching network, in which the packets travel over virtual
circuits.
*Frame Relay, was a successor to X.25, and offers a connection-oriented packet-switching
network.
*Asynchronous Transfer Mode (ATM), was developed from an outgrowth of ISDN
standards, and is fast-packet, connection-oriented, cell-switching technology.
Source: Communications Systems and Networks by Ray Horak
(M&T Books, 2000).
NEW QUESTION: 3
CORRECT TEXT
Answer:
Explanation:
On the MGT Router:
Config t
Router eigrp 12
Network 192.168.77.0
NEW QUESTION: 4
Refer to the exhibit.
The network is converged. After link-state advertisements are received from Router_A, what information will
Router_E contain in its routing table for the subnets 208.149.23.64 and 208.149.23.96?
A. O 208.149.23.64 [110/3] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/3] via 190.172.23.10, 00:00:16, Serial 1/0
B. O 208.149.23.64 [110/1] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/3] via 190.173.23.10, 00:00:16, FastEthernet 0/0
C. O 208.149.23.64 [110/13] via 190.173.23.10, 00:00:07, FastEthernet 0/0
O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
D. O 208.149.23.64 [110/13] via 190.172.23.10, 00:00:07, Serial 1/0
O 208.149.23.96 [110/13] via 190.172.23.10, 00:00:16, Serial 1/0
O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
Answer: C
Explanation:
Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface.
The interface cost is calculated with the formula 108 / Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 /
100,000,000 = 1. Therefore the cost is 12 (learned from Router_A) + 1 = 13 for both subnets - B is not correct.
The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps = 64; T3 cost = 108 / 45 Mbps
= 2) so surely OSPF will choose the path through T3 link -> Router_E will choose the path from Router_A through
FastEthernet0/0, not Serial1/0 - C & D are not correct.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -
they are surely incorrect.