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NEW QUESTION: 1
Refer to the exhibit.

This multilayer Cisco Nexus switch had been the active virtual gateway for Group 1 before it became temporarily unavailable. What will happen to GLBP Group 1 when this device becomes available again?
A. It depends on the priority value that is configured active on the router.
B. The currently active router remains active.
C. It depends on the weighting values that are configured active on the router.
D. The Cisco Nexus switch becomes the active virtual gateway after 600 seconds.
Answer: B
Explanation:
GLBP prioritizes gateways to elect an active virtual gateway (AVG). If multiple gateways have the same priority, the gateway with the highest real IP address becomes the AVG. The AVG assigns a virtual MAC address to each member of the GLBP group. Each member is the active virtual forwarder (AVF) for its assigned virtual MAC address, forwarding packets sent to its assigned virtual MAC address.
The AVG also answers Address Resolution Protocol (ARP) requests for the virtual IP address. Load sharing is achieved when the AVG replies to the ARP requests with different virtual MAC addresses.
Note: Packets received on a routed port destined for the GLBP virtual IP address terminate on the local router, regardless of whether that router is the active GLBP router or a redundant GLBP router. This termination includes ping and Telnet traffic. Packets received on a Layer 2 (VLAN) interface destined for the GLBP virtual IP address terminate on the active router.

NEW QUESTION: 2
What is the job title of the person who is primarily responsible for the quality of deliverables?
A. The design engineer.
B. The project manager.
C. The quality controller.
D. The quality control inspector.
E. The quality manager.
Answer: B

NEW QUESTION: 3
The Ethernet networks connected to router R1 in the graphic have been summarized for router R2 as 192.1.144.0/20. Which of the following packet destination addresses will R2 forward to R1, according to this summary? (Choose two.)

A. 192.1.160.11
B. 192.1.1.144
C. 192.1.143.145
D. 192.1.138.41
E. 192.1.159.2
F. 192.1.151.254
Answer: E,F
Explanation:
This question is about route summary. Example of route summary calculation The routing table stores the following networks: 172.16.12.0/24 172.16.13.0/24 172.16.14.0/24 172.16.15.0/24
Routing protocols summarize or aggregate routes based on shared network numbers within the
network. To calculate the summarized route of the router, it is necessary to determine how many
bits on the left of those addresses are the same. The steps are shown below:
Step onE. Convert addresses to binary format and align them.
Step two: Find out the identical last bit of all addresses.
Step threE. Calculate how many bits are the same. The summarized route is an IP address with a
slash, which will be helpful.
172.16.12.0/24= 172.16.000011 00.00000000
172.16.13.0/24= 172.16.000011 01.00000000
172.16.14.0/24= 172.16.000011 10.00000000
172.16.15.0/24= 172.16.000011 11.00000000
172.16.15.255/24= 172.16.000011 11.11111111
The first 22 bits of IP addresses 172.16.12.0---172.16.15.255 are identical, so the best
summarized route is 172.16.12.0/22.
Using route summary can reduce the routing items on the summarized router to decrease the
memory occupied and network traffic generated by routing protocol.
Route summary implementation:
A. The bits on the left of multiple addresses must be identical.
B. Routing protocol should select route according to the 32-bit IP address and the prefix length of the maximum 32 bit address.
C. Routing update must contain 32-bit IP address and the prefix length (subnet mask)

NEW QUESTION: 4
You are creating an application that manages information about zoo animals. The application includes a
class named Animal and a method named Save.
The Save() method must be strongly typed. It must allow only types inherited from the Animal class that
uses a constructor that accepts no parameters.
You need to implement the Save() method. Which code segment should you use?

A. Option B
B. Option D
C. Option C
D. Option A
Answer: C

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