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NEW QUESTION: 1
Which of the following allows activities to be done in parallel that would normally be done in sequence?
A. Lag time
B. Lead time
C. Crashing
D. Fast tracking
Answer: D
Explanation:
Explanation/Reference:
Explanation:
Fast tracking is a technique for compressing project schedule. In fast tracking, phases are overlapped that would normally be done in sequence. It is shortening the project schedule without reducing the project scope.
Incorrect Answers:
A: A lag time is a delay between the predecessor and the successor tasks. Sometimes it may be needed to schedule a delay between the predecessor and the successor tasks. For example, if two coats of paint are required to paint a car, then the final coat should be applied only when the first coat dries. This delay is known as the lag time. The lag time is entered as a positive value. The lag time can be entered as a duration or as a percentage of the predecessor's task duration. It is entered on the Predecessor tab in the Task Information dialog box.
B: A lead time is the time that overlaps between the predecessor and the successor tasks. The successor task can start before the predecessor task finishes. For example, if a task can start when its predecessor is one-fourth finished, a finish-to-start dependency with a lead time of 25 percent for the successor task can be specified. The lead time is entered as a negative value. The lead time can be entered as a duration or as a percentage of the predecessor's task duration. It is entered on the Predecessor tab in the Task Information dialog box.
D: Crashing is a process in that the project manager adds more resources to effort-driven activities in an attempt to shorten their duration.

NEW QUESTION: 2
Which IS-IS packet is used to determine the level of an IS-IS neighbor?
A. LSP
B. CSNP
C. PSNP
D. IIH
Answer: D

NEW QUESTION: 3
Companies will increasingly support corporate applications on personal devices, which users will access from both wired and wireless connections. How will this trend affect mid-sized businesses? (Select two.)
A. They will need to support additional protocols such as Voice over IP and Border Gateway Protocol on their networks to enable this access
B. They will need to Implement Identity-based security, rather than port-based security
C. They will need to ensure users have a seamless user experience no matter how they access the network
D. They will need to implement virtualized machines to host these applications in a private cloud or purchase these services from a public cloud provider.
E. They will need to encrypt communications between the personal devices end the network devices that provide either wired or wireless access.
Answer: C,E
Explanation:
Connectivity at the edge of an enterprise network is more than just a wired or wireless LAN infrastructure. Enterprises must chose infrastructure vendors that support network services, including security and management, and can integrate wired and wireless networking products.
Reference: Magic Quadrant for the Wired and Wireless LAN Access Infrastructure, Gartner
http://www.gartner.com/technology/reprints.do?id=1-1AX5XXB&ct=120614&st=sb

NEW QUESTION: 4
You create a table that has the StudentCode, SubjectCode, and Marks columns to record mid-year marks for students. The table has marks obtained by 50 students for various subjects.
You need to ensure that the following requirements are met:
Students must be ranked based on their average marks.

If one or more students have the same average, the same rank must be given to these students.

Consecutive ranks must be skipped when the same rank is assigned.

Which Transact-SQL query should you use?
A. SELECT StudentCode as Code,
RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
B. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY SubjectCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
C. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY StudentCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
D. SELECT StudentCode as Code,
NTILE(2) OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
E. SELECT Id, Name, Marks,
DENSE_RANK() OVER(ORDER BY Marks DESC) AS Rank
FROM StudentMarks
F. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANXO OVER(PARTITION BY StudentCode ORDER BY Marks DESC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
G. SELECT StudentCode as Code,
DENSE_RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
H. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY SubjectCode ORDER BY Marks DESC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
Answer: A
Explanation:
Explanation/Reference:
Reference: http://msdn.microsoft.com/en-us/library/ms189798.aspx