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NEW QUESTION: 1
Click on the exhibit.

What value would be expected in the state field?
A. Down
B. Exchange
C. 2-Way
D. Up
Answer: A

NEW QUESTION: 2
A university wants to deploy ClearPass with the Guest module. The university has two types that need to use web login authentication. The first type of users are students whose accounts are in an Active Directory server. The second type of users are friends of students who need to self-register to access the network.
How should the service be set up in the Policy Manager for this network?
A. Guest User Repository and Active Directory server both as authentication sources
B. Guest User Repository as the authentication source and the Active Directory server as the authorization source
C. Either the Guest User Repository or Active Directory server should be the single authentication source
D. Active Directory server as the authentication source, and Guest User Repository as the authorization source
E. Guest User Repository as the authentication source, and Guest User Repository and Active Directory server as authorization sources
Answer: A

NEW QUESTION: 3
Corporate headquarters provided your office a portion of their class B subnet to use at a new office location.
Range Given: 172.30.232.0/24
* Sales 57 devices
* HR 23 devices
* IT 12 devices
* Finance 32 devices
* Marketing 9 devices
INSTRUCTIONS
Allocate the minimum number of addresses (using CIDR notation) needed to accommodate each department.
After accommodating each department, identify the unused portion of the subnet by responding to the question on the graphic.
All drop downs must be filled.
If at any time you would like to bring back the initial state of the simulation, please click the Reset All button.


Answer:
Explanation:

Explanation

Sales network - /26 - This will provide up to 62 usable IP addresses (64-2 for subnet and broadcast IP) HR network - /27 - This will provide for up to 30usable IP's (32-2) IT - /28 - This will provide for up to 14 usable IP's (16-2) Finance - /26 - Note that a /27 is 32 IP addresses but 2 of those are reserved for the network and broadcast IP's and can't be used for hosts.
Marketing - /28
If we add up how many IP blocks are used that is 64+32+16+64+16=192.
A /24 contains 256 IP addresses, so 256-192=64.
So the last unused box should be a /26, which equates to 64 addresses