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NEW QUESTION: 1
New LT06 tape drives manufactured by IBM will be attached and defined to a Tivoli Storage Manager (TSM) server.
What should be updated prior to getting the LT06 tape drive working with TSM?
A. device drivers
B. Operations Center Driver database
C. TSM Backup-Archive client
D. TSM storage agent
Answer: A
NEW QUESTION: 2
Your network contains two DHCP servers. The DHCP servers are named DHCP1 and DHCP2. The internal network contains 1,000 DHCP client computers that are located on a single subnet. A router separates the internal network from the Internet. The router has a single IP address on the internal interface.
DHCP1 has the following scope information:
Starting IP address: 172.16.0.1
Ending IP address: 172.16.7.254
Subnet mask: 255.255.240.0 You need to provide a fault tolerant DHCP infrastructure that supports the client computers on the internal network. In the event that a DHCP server fails, all client computers must be able to obtain a valid IP address.
How should you configure DHCP2?
A. Create a scope for the subnet 172.16.0.0/21. Configure the scope to use a starting IP address of
172.16.0.1 and an ending IP address of 172.16.15.254.
B. Create a scope for the subnet 172.17.0.0/16. Configure the scope to use a starting IP address of
172.17.0.1 and an ending IP address of 172.17.255.254.
C. Create a scope for the subnet 172.16.8.0/21. Configure the scope to use a starting IP address of
172.16.8.1 and an ending IP address of 172.16.10.254.
D. Create a scope for the subnet 172.16.0.0/20. Configure the scope to use a starting IP address of
172.16.8.1 and an ending IP address of 172.16.15.254.
Answer: D
Explanation:
Explanation
Explanation/Reference:
Network 255.255.240.0 = /20 -> 4096-2 addresses
255.255.255.0 = /24 -> 256-2 addresses
255.255.254.0 = /23 -> 512-2 addresses
255.255.252.0 = /22 -> 1024-2 addresses
255.255.248.0 = /21 -> 2048-2 addresses
IP address 172.16.0.1 to 172.16.7.254 -> 8*256 -2 = 2048-2 addresses
There are 2048-2 addresses left belind.
Option B (172.16.0.0/21) is ruled out because 172.16.0.1 to 172.16.15.254 are belonged to /20 (not /21) network
Option C (172.16.8.0/21) is ruled out because 172.16.8.1 to 172.16.10.254 contain 3*256-2 = 766
addresses only (insufficient for 1,000 clients)
Option D (172.17.0.0) is ruled out because of different network 172.17.0.0/16 (vs 172.16.0.0/20)
NEW QUESTION: 3
The following SAS program is submitted:
How will the Exam variable value be displayed in the FREQ procedure output?
A. Pass
B. 50.5
C. . (missing numeric value)
D. Fail
Answer: B
NEW QUESTION: 4
Refer to the exhibit.
After redistribution is enabled between the routing protocols; PC2, PC3, and PC4 cannot reach PC1. Which action can the engineer take to solve the issue so that all the PCs are reachable?
A. Filter the prefix 10.1.1.0/24 when redistributed from OSPF to EIGRP.
B. Set the administrative distance 100 under the RIP process on R2.
C. Filter the prefix 10.1.1.0/24 when redistributed from RIP to EIGRP.
D. Redistribute the directly connected interfaces on R2.
Answer: A
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