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NEW QUESTION: 1
A risk response in a project in your program has actually caused new risks. What term is assigned to the new risks that a risk response has created?
A. Chain risks
B. In flux risks
C. Reactionary risks
D. Secondary risks
Answer: D
NEW QUESTION: 2
A customer needs to build a QlikView application using data in text files. A new data file is created every day and stored in a folder on the customer's network named s. \userfiles.
The file name is in the form dataYYYYMMDD.txt, where YYYYMMDD represents the current date, such as data20111231.txt.
Which data load strategy should a developer use to load all of the customer's data files each day?
A. the 'Wildcard Load' feature by setting the 'File Location' in Document Properties
B. the statement 'Directory s. \userfiles\*;' preceding the Load statement
C. 'Load * From [s. \userfiles\data*.txt] (ansi, txt, delimiter is '\t', embedded labels);'
D. 'Load * From_Field (s. \userfiles\, 'data*.txt') (ansi, txt, delimiter is '\t', embedded labels);'
Answer: C
NEW QUESTION: 3
Which type of 802.11 wireless network provides direct peer-to-peer connection between wireless stations?
A. In-cell relay networking
B. Wireless distribution system
C. Ad hoc networkong
D. Infrastructure network
Answer: C
NEW QUESTION: 4
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 10.0.0.0.0.0.0.1
C. Permit 10.0.0.1.0.0.0.0
D. Permit 10.0.0.1.0.0.0.254
Answer: D
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 ranges from 172.23.16.0 to 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 0000) 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.
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