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  • Exam Number/Code : JN0-664
  • Exam Name : Service Provider, Professional (JNCIP-SP)
  • Questions and Answers : 260 Q&As
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NEW QUESTION: 1

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

NEW QUESTION: 2
Given an IP address of 172.16.0.0, which subnet mask below would
allow us to divide the network into the maximum number of subnets
with at least 600 host addresses per subnet?
A. 255.255.252.0
B. 255.255.224.0
C. 255.255.248.0
D. 255.255.240.0
Answer: A
Explanation:
The last two octets of this class B address, 252.0, gives us binary:
11111100.00000000. The six subnet bits give us 62 (26 -2) subnets, each with 1022 (210 -2) hosts, which allows us to have the maximum number of subnets with almost double the required host addresses.
*, 224.0, is 11100000.00000000 binary, which gives us six (23 -
2) subnets with 8190 (213 -2) hosts each.
* 240.0, is 11110000.00000000 binary, and gives us 14 (24 -2) subnets each with
4 094 (212 -2) hosts.
* 248.0, is 11111000.00000000 binary, which creates 30 (25 -2) subnets with 2046 (211 -2) hosts. Many books give detailed descriptions of IP subnetting.

NEW QUESTION: 3
Your network contains a System Center 2012 Configuration Manager environment. You need to change the organization name displayed by Configuration Manager.
Which settings should you modify from the Configuration Manager console?
A. Computer Agent
B. Compliance Settings
C. Client Policy
D. User and Device Affinity
Answer: A
Explanation:
Explanation/Reference:
The highlighted option is the one used to change the organization name. Administration -> Client Settings -> Default Client Settings -> Computer Agent


 

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