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NEW QUESTION: 1
Although the SAN appears to be running without any problems, the technician is unable to reach the SAN's web enabled management interface. Which of the following tools can the technician use to QUICKLY find where the communication failure occurred?
A. arp
B. ipconfig
C. traceroute
D. ping
Answer: C

NEW QUESTION: 2
The indenture includes all these covenants that protect the beholder EXCEPT:
A. none of these
B. Debt service reserve covenant
C. Flow of funds covenant
D. Rate covenant
Answer: A

NEW QUESTION: 3
Which of the following is TRUE of the endptEnabler command?
A. It generates IBM-specific deployment classes for a Web service application
B. It adds a HTTP or JMS router module to an EJB-based Web service application
C. It creates a bottom-up Web service from a stateless session EJB
D. It requires an EJB Java archive (JAR) file as input
Answer: B

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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