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NEW QUESTION: 1
You are in the process of controlling changes to the project schedule. All of the following
are outputs of this process EXCEPT:
A. Schedule baseline updates
B. Activity list updates
C. Performance measurements
D. Schedule model data
E. Schedule model data updates
Answer: D

NEW QUESTION: 2
DRAG DROP


Answer:
Explanation:

NEW QUESTION: 3
Capturing business rules involves which step?
A. Packaging business logic in application code as decision service.
B. Using ruleset parameters to transfer information.
C. Orchestrating business rules in a ruleflow.
D. Representing the business policy as if-then statements.
Answer: B

NEW QUESTION: 4
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 172.17.0.0/16. Configure the scope to use a starting IP address of
172.17.0.1 and an ending IP address of 172.17.255.254.
B. 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.
C. 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.
D. 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.
Answer: B
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)