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NEW QUESTION: 1
Which Open Systems SnapVault tool/command determines the amount of space required on the primary system for transfers?
A. snapvault estimator
B. snapvault installcheck
C. svestimator
D. svspace_estimator
Answer: C

NEW QUESTION: 2
Which statement regarding security profiles is true?
A. The command mqsichangeconfigurableservice can be used to change a security profile.
B. Multiple security providers cannot be created in one security profile.
C. If WS-Trust v1.3.STS is selected for more than one operation, the WS-Trust URL must be different and unique for each of the operations.
D. A security profile can be created for LDAP or a WS-Trust V1.3 STS.
Answer: D
Explanation:
You can create a security profile for use with Lightweight Directory Access Protocol (LDAP) or a WS-Trust V1.3 compliant security token server (STS), such as Tivoli Federated Identity Manager (TFIM) V6.2. Support is also provided for TFIM V6.1, for compatibility with previous versions of IBM Integration Bus. You can create the security profile by using either the mqsicreateconfigurableservice command or an editor in the IBM Integration Explorer.
References:
https://www.ibm.com/support/knowledgecenter/en/SSMKHH_9.0.0/com.ibm.etools.mft.doc/ ap04170_.htm

NEW QUESTION: 3
Which of the following is NOT the step of managing expectations process?
A. Fulfill
B. Align
C. Agree
D. Ask
Answer: C

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,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
B. SELECT StudentCode as Code,
RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
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,
DENSE_RANK() 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,
NTILE(2) OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
G. 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
H. 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
Answer: B
Explanation:
Reference: http://msdn.microsoft.com/en-us/library/ms189798.aspx

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