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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. TSM storage agent
C. TSM Backup-Archive client
D. Operations Center Driver database
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 172.16.0.0/20. Configure the scope to use a starting IP address of
172.16.8.1 and an ending IP address of 172.16.15.254.
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 172.16.0.0/21. Configure the scope to use a starting IP address of
172.16.0.1 and an ending IP address of 172.16.15.254.
Answer: A
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. Fail
B. 50.5
C. Pass
D. . (missing numeric value)
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. Set the administrative distance 100 under the RIP process on R2.
B. Filter the prefix 10.1.1.0/24 when redistributed from RIP to EIGRP.
C. Filter the prefix 10.1.1.0/24 when redistributed from OSPF to EIGRP.
D. Redistribute the directly connected interfaces on R2.
Answer: C
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