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NEW QUESTION: 1
You have a server named Server1 that runs Windows Server 2012. You create a Data Collector Set (DCS)
named DCS1.
You need to configure DCS1 to log data to D:\logs.
What should you do?
A. Right-click DCS1 and click Properties.
B. Right-click DCS1 and click Save template...
C. Right-click DCS1 and click Export list...
D. Right-click DCS1 and click Data Manager...
Answer: A
NEW QUESTION: 2
Which of the following review types would be recommended at the start of each Sprint, for the reason given?
A. A technical review of the test plan, to ensure that the developers are aware that they must produce good quality code.
B. A walkthrough of the set of user-stories, to ensure that the developers and testers understand what is to be built.
C. A management review, to check that the user-stories will accurately reflect the rules for accessibility.
D. An inspection, to find errors in the user-stories.
Answer: B
NEW QUESTION: 3
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.0.255.255.255.254
B. Permit 10.0.0.1.0.0.0.0
C. Permit 10.0.0.1.0.0.0.254
D. Permit 10.0.0.0.0.0.0.1
Answer: C
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).
The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 000 ) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.
NEW QUESTION: 4
The exhibit represents an FCIP tunnel configuration that merges the two FC SANs at either end of the tunnel.
Which types of ports do the labels "X" and "Y" represent on the FCIP gateway shown in the exhibit?
A. X = E_Port and Y = VE_Port
B. X =VF_Port and Y = E_Port
C. X = F_Port and Y = VN_Port
D. X = VN_Port and Y = VE_Port
Answer: A
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