Questions and Answers: 166
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NEW QUESTION: 1
When starting up your ASM instance, you receive the following error:
SQL> startup pfile=$ORACLE_HOME/dbs/init+ASM.ora
ASM instance started
Total System Global Area 104611840 bytes
Fixed Size 1298220 bytes
Variable Size 78147796 bytes
ASM Cache 25165824 bytes
ORA-15032: not all alternations performed
ORA-15063: ASM discovered an insufficient number of disks for diskgroup "DGROUP3" ORA-15063: ASM discovered an insufficient number of disks for diskgroup "DGROUP2" ORA-15063: ASM discovered an insufficient number of disks for diskgroup "DGROUP1" In trying to determine the cause of the problem, you issue this query:
SQL> show parameter asm
What is the cause of the error?
A. There is insufficient information to solve this problem.
B. The ASM_DISKSTRING parameter is not set; therefore disk discovery is not possible.
C. The format of the ASM_DISKGROUPS parameter is incorrect. It should reference the disk group numbers, not the names of the disk groups
D. The ASM_DISKGROUPS parameter is configured for three disk groups: DGROUP1, DGROUP2, and DGROUP3.The underlying disks for these disk groups have apparently been lost.
E. The ASM_POWER_LIMIT parameter is incorrectly set to 1. It should be set to the number of disk groups being attached to the ASM instance.
Answer: B
Explanation:
Explanation/Reference:
Explanation:
ASM_DISKSTRING specifies an operating system-dependent value used by Automatic Storage Management to limit the set of disks considered for discovery. When a new disk is added to a disk group, each Automatic Storage Management instance that has the disk group mounted must be able to discover the new disk using the value of ASM_DISKSTRING.
In most cases, the default value will be sufficient. Using a more restrictive value may reduce the time required for Automatic Storage Management to perform discovery, and thus improve disk group mount time or the time for adding a disk to a disk group. A "?" at the beginning of the string gets expanded to the Oracle home directory. Depending on the operating system, wildcard characters can be used. It may be necessary to dynamically change ASM_DISKSTRING before adding a disk so that the new disk will be discovered.
An attempt to dynamically modify ASM_DISKSTRING will be rejected and the old value retained if the new value cannot be used to discover a disk that is in a disk group that is already mounted.
NEW QUESTION: 2
You create a custom workflow by using Microsoft Visual Studio 2010.
You need to specify a custom InfoPath workflow initiation form in the workflow element manifest file.
Which attribute should you configure?
A. Instantiation_FormURN
B. Instantiation_FieldML
C. InstantiationUrl
D. Association_FormURN
Answer: A
Explanation:
MNEMONIC RULE: "initiation form = Instantiation_FormURN"
Specifies the URN of the Microsoft InfoPath 2010 form to use to initiate the workflow.
Instantiation_FormURN Element (Workflow) - ECM http://msdn.microsoft.com/en-us/library/aa672715.aspx
NEW QUESTION: 3
Which standard access control entry permits from odd-numbered hosts in the 10.0.0.0/24 subnet?
A. Permit 10.0.0.1.0.0.0.0
B. Permit 10.0.0.1.0.0.0.254
C. Permit 10.0.0.0.255.255.255.254
D. Permit 10.0.0.0.0.0.0.1
Answer: B
Explanation:
Remember, for the wildcard mask, 1's are I DON'T CARE, and 0's are I CARE. So now let's
analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of
the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care
about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16;
maximum: 0001111 = 31).
The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we
only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd
numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even
numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with
a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered
hosts in the 10.0.0.0/24 subnet.