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QUESTION NO: 58

An Agent is configured to run on many Agent Servers with high threading levels expecting a high

volume of traffic. What happens if the volume of traffic became low?

A. The JMS queues would fill up with too many unprocessed transactions.

B. The system would slow down as there are too many processes running without work to do.

C. The Agent Server would automatically reduce the number of threads based on the number of

transactions.

D. The IBM Sterling Selling and Fulfillment Suite would automatically shutdown redundant agent

server instances.

Answer: B

Explanation:

QUESTION NO: 38

What is the function of the DSMUPGRD upgrade utility for IBM Tivoli Storage Manager V6.2

(TSM)?

A. It is run after a new version of TSM is installed.

B. It checks all prerequisites (disk space, memory, software level).

C. It starts the TSM upgrade after a successful check of prerequisites.

D. It prepares and extracts data from the database which will be migrated.

Answer: D

QUESTION NO:16

CSS and XSL can be BEST used together to:

A. Develop advanced transformations between XML DTDs.

B. Extend the capabilities of the HTML formatting model with CSS constructs.

C. Transform XML data on the server into HTML documents that link with CSS style sheets.

D. Allow XSL to directly process HTML documents.

Answer: C

QUESTION NO:98

A customer with a single IBM System Storage DS8100 reports severe performance problems for some of

their applications when a particular data warehousing application creates a heavy load. What is the most

cost effective way to resolve this issue?

A. add additional disks to the DS8100

B. add more cache to the DS8100

C. move affected volumes to a different disk enclosure

D. implement TPC for Disk to monitor performance

Correct Answer: C

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QUESTION NO:131

Given the following code:

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What will the alert(WL.Client.getUserPref(‘key1’)); show?

A. The alert will not show since WL.Client.setUserPref cannot be called in the for loop since it is an

asynchronous function.

B. updatedValue1 since WL.Client.setUserPref({‘key1’ : ‘updatedValue1’}); will replace the value of key1

set in the for loop.

C. value1, updatedValue1 or null since WL.Client.setUserPref is an asynchronous function and there is no

guarantee which call, if any,will return first.

D.

value1 since WL.Client.setUserPref(‘key1’, ‘updatedValue1’); will fail because WL.Client.deleteUserPref

(‘key1’) was notcalled before setting the new value.

Correct Answer: C