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NEW QUESTION: 1

A. Option B
B. Option D
C. Option A
D. Option C
Answer: B

NEW QUESTION: 2
View the Exhibit showing steps to create a database resource manager plan.
SQL>executedbms_resource_manager.create_pendingarea(); PL/SQLproceduresuccessfully completed. 3QL>execdbms_resource_manager,create_consumergroup
(consumer_group=>'OLTP,,comment=>,onlineuser') PL/SQLproceduresuccessfullycompleted. SQL>execdbras_resource_raanager.create_plan(plan=>'PRIU3ER3',comment=>'dssprio'); SQL>exec Dbms_resource_manager.create_plan_directive(plan=>'PRIU3ER3',group_or_subplan=>'
OLTP',comraent=>'onlinegrp'CPU_Pl=>60);
PL/3QLproceduresuccessfullycompleted.
After execting the steps in the exhibit you execute this procedure, which results in an error:
SQL> EXECUTEdbms_resource_manager. validate_pending_area ();
What is the reason for the error?
A. The pending area is automatically submitted when creating plan directives.
B. The sys_group group is not included in the resource plan.
C. The other_groups group is not included in the resource plan.
D. The procedure must be executedbefore creating any plan directive.
E. Pending areas can not be validated until submitted.
Answer: D

NEW QUESTION: 3
Which of these is the equation used to derive a 64 Kbps bit rate?
A. 2 x 4-bit code words x 8 kHz
B. 2 x 8 kHz x 4-bit code words
C. 2 x 4 kHz x 8-bit code words
D. 8 kHz x 8-bit code words
Answer: C
Explanation:
Explanation/Reference:
Explanation:
While the human ear can sense sounds from 20 to 20, 000 Hz, and speech encompasses sounds from about 200 to 9000 Hz, the telephone channel was designed to operate at about 300 to 3400 Hz. This economical range carries enough fidelity to allow callers to identify the party at the far end and sense their mood. Nyquist decided to extend the digitization to 4000 Hz, to capture higher-frequency sounds that the telephone channel may deliver. Therefore, the highest frequency for voice is 4000 Hz. According to Nyquist theory, we must double the highest frequency, so 2x4kHz = 8kHz.
Each sample will be encoded into a 8-bit code. Therefore 8kHz x 8-bit code = 64 Kbps (notice about the unit Kbps: 8kHz = 8000 samples per second so 8000 x 8-bit = 64000 bit per second = 64 Kilobit per second = 64 Kbps) Reference: http://encyclopedia2.thefreedictionary.com/Nyquist+theorem Note:
Nyquist theory:
"When sampling a signal (e.g., converting from an analog signal to digital), the sampling frequency must be greater than twice the bandwidth of the input signal in order to be able to reconstruct the original perfectly from the sampled version."