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NEW QUESTION: 1
A Citrix Administrator needs to ensure that a revoked certificate is NOT being used for client certificate authentication.
Which two entities can the administrator select on the NetScaler? (Choose two.)
A. Server Name Indication
B. Subject Alternative certificates
C. Certificate Revocation List
D. Online Certification Status Protocol
Answer: C,D
Explanation:
https://support.citrix.com/article/CTX138304

NEW QUESTION: 2
Ein Projektmanager, der neu in einem Projekt ist, bemerkt schnell eine gewisse Motivation bei den Mitgliedern des Projektteams. Was soll der Projektmanager tun?
A. Implementieren Sie die Maslow-Pyramide
B. Entwickeln Sie Teambuilding-Übungen.
C. Weisen Sie die Ressourcen anderen Projekten zu
D. Grundregeln festlegen.
Answer: C

NEW QUESTION: 3
Which of the following cannot be used to transfer data?
A. Local variables
B. Semaphores
C. Queues
D. Notifiers
Answer: B
Explanation:
Explanation/Reference:

NEW QUESTION: 4
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

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