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NEW QUESTION: 1
Which software program controls the installation of the Modular Messaging Services using the Data Collection Tool (DCT) input file?
A. asconfig.exe
B. miscm.exe
C. gvconfig.exe
D. setup.exe
Answer: B

NEW QUESTION: 2
A Unity storage system is installed with 10Gb optical SFPs in CNA ports 4 and 5 of both Storage Processors.
The storage administrator receives from a user to connect to the storage system through Fibre Channel.
How can the administrator address the request?
A. Configure CNA port 4 and 5 of SPA as 10 Gb copperConfigure CNA ports 4 and 5 of SPB as 16 Gb
Fibre Channel
B. Configure all CNA ports for 16Gb Fibre Channel
C. Once committed, CNA ports cannot be reconfigured without data loss.
D. Configure CNA port 4 of SPA and SPB as 10Gb opticalConfigure CNA port 5 of SPA and SPB as 16
Gb Fibre Channel
Answer: C
Explanation:
Explanation
References: https://www.emc.com/collateral/TechnicalDocument/docu60144.pdf
(page 5)

NEW QUESTION: 3
What does a router do if it has no EIGRP feasible successor route to a destination network and the successor route to that destination network is in active status?
A. It routes all traffic that is addressed to the destination network to the interface indicated in the routing table.
B. It sends a copy of its neighbor table to all adjacent routers.
C. It sends a multicast query packet to all adjacent neighbors requesting available routing paths to the destination network.
D. It broadcasts Hello packets to all routers in the network to re-establish neighbor adjacencies.
Answer: C
Explanation:
Explanation/Reference:
Explanation:
http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a0080093f07.shtml

NEW QUESTION: 4
Which two statements describe the IP address 10.16.3.65/23? (Choose two.)
A. The network is not subnetted.
B. The subnet address is 10.16.3.0 255.255.254.0.
C. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
D. The last valid host address in the subnet is 10.16.2.254 255.255.254.0
E. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
Answer: C,E
Explanation:
Explanation/Reference: The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254