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NEW QUESTION: 1
What advantage does the HPE SimpliVity deduplication feature provide?
A. SimpliVity saves the data to disk before deduping it, ensuring no data loss.
B. The always-on deduplication feature ensures that data is deduped on the first I/O.
C. SimpliVity uses a proprietary hypervisor, which dedupes the data before it is sent to the Data Virtualization Layer.
D. The deduplication feature runs periodically on the storage devices, continually compressing data and freeing up storage space.
Answer: B

NEW QUESTION: 2
Review the following classes and properties:
MyCo-Data-Shape.Area .Color
MyCo-Data-Shape-Rectangle .Length .Width
MyCo-Data-Shape-Circle .Radius
A page list property is defined at the class group MyCo-FW-ShapeApp-Work. The property name is .Shapes and it is defined as being a Page List of class MyCo-Data-Shape-
Which of the following statements are correct? (Choose Two)
A. At runtime .Shapes(1) could be of class MyCo-Data-Shape-Rectangle while .Shapes(2) is of class MyCo-Data-Shape-Circle
B. At design time PRPC will provide a warning as it is not a good practice to define a page list with an abstract class
C. At design time an architect can create a rule in MyCo-Data-Shape-Circle only if the rule exists in MyCo-Data-Shape-
D. At design time an architect can create a rule in MyCo-Data-Shape-Circle regardless of if the rule exists in MyCo-Data-Shape-
E. At runtime pages of .Shapes can be of either MyCo-Data-Shape-Rectangle or MyCo-Data-Shape-Circle but all pages must be the same
Answer: A,D

NEW QUESTION: 3
Which services allow ECM to connect to content located in competitive content repositories?
A. Business Process Management Services
B. Content Federation and Integration Services
C. Compliance Services
D. Records Management Services
Answer: B
Explanation:
Reference:http://www-01.ibm.com/support/docview.wss?uid=swg27015111&aid=1(page 5)

NEW QUESTION: 4
Which SQL statement should be used to select the minimum and maximum salaries (SALARY), by job code (JOB), from a table EMPLOYEE?
A. SELECT JOB, MIN(salary), MAX(salary)FROM employeeORDER BY job, MIN(salary), MAX(salary);
B. SELECT job, MIN(salary), MAX(salary)FROM employeeGROUP BY job, MIN(salary), MAX(salary);
C. SELECT job, MIN(salary), MAX(salary)FROM employeeGROUP BY job;
D. SELECT job, MIN(salary), MAX(salary)FROM employeeORDER BY job;
Answer: C
Explanation:
Explanation/Reference:
Explanation: