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Vendor: WatchGuard

Exam Code: Network-Security-Essentials Dumps

Questions and Answers: 104

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NEW QUESTION: 1
What is the maximum number of DIMMs that can be installed in an Integrity rx2660 server?
A. 0
B. 1
C. 2
D. 3
Answer: C

NEW QUESTION: 2
A customer has deployed an OnGuard Solution to all the corporate devices using a group policy rule to push the OnGuard Agents. The network administrator is complaining that some of the agents are communicating to the ClearPass server that is located in a DMZ, outside the firewall The network administrator wants all of the agents System Health Validation traffic to stay inside the Management subnets.
What can the ClearPass administrator do to move the traffic only to the ClearPass Management Ports?
A. Configure a Policy Manager Zone mapping so the OnGuard agent will use the Management Port IP.
B. Edit the agent.conf file being deployed to the clients to use the ClearPass Management Port for SHV updates.
C. Filter TCP port 6658 on the firewall, forcing the OnGuard agent to use the ClearPass Management port.
D. Select the correct OnGuard Agent installer, and use the one configured for Management Port for the clients.
Answer: A

NEW QUESTION: 3

A. Get-StaleMailboxReport -EndDate
B. Get-StaleMailboxReport -Expression
C. Get-MailboxActivityReport -Organization
D. Get-StaleMailboxReport -Organization
Answer: A

NEW QUESTION: 4
A box contains five blue pens, three black pens, and two red pens. If every time a pen is selected, it is
removed from the box, what is the probability of selecting a black pen followed by a blue pen?

A. Option E
B. Option D
C. Option A
D. Option B
E. Option C
Answer: C
Explanation:
Explanation/Reference:
At the start, there are 5 + 3 + 2 = 10 pens in the box, 3 of which are black. Therefore, the probability of
selecting a black pen is3/10After the black pen is removed, there are nine pens remaining in the box, five
of which are blue. The probability of selecting a blue pen second is5/9To find the probability that both
events will happen, multiply the probability of the first event by the probability of the second event: .



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