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NEW QUESTION: 1
Your network contains a System Center 2012 R2 Configuration Manager Service Pack 1 (SP1) environment.
The environment contains a primary site server named Server1 and a server named Server2 that runs Microsoft SQL Server 2012. Server2 contains the Configuration Manager database.
Server2 fails.
You install SQL Server 2012 on a new server. You name the server Server3.
You need to restore the Configuration Manager database to the new server.
What should you do?
A. From Server1, run the Configuration Manager 2012Setup Wizard.
B. From Server2, run Microsoft SQL Server Management Studio, and then restore the backed up SQL Server database andlog files.
C. From Server1, run the Site Repair Wizard.
D. From Server2, run Microsoft SQL Server Management Studio, and then attach the backed up SQL Server database and log files.
Answer: A
Explanation:
Explanation/Reference:
Explanation:
Answer is From Server1, run the Configuration Manager 2012 Setup Wizard.
Recover a Configuration Manager Site
A Configuration Manager site recovery is required whenever a Configuration Manager site fails or data loss occurs in the site database. Repairing and resynchronizing data are the core tasks of a site recovery and are required to prevent interruption of operations. Site recovery is started by running the Configuration Manager Setup Wizard from installation media or by configuring the unattended installation script and then using the Setup command /script option. Your recovery options vary depending on whether you have a backup of the Configuration Manager site database.
Site Database Recovery Options
When you run Setup, you have the following recovery options for the site database:
* Recover the site database using a backup set: Use this option when you have a backup of the Configuration Manager site database that was created as part of the Backup Site Server maintenance task run on the site before the site database failure. When you have a hierarchy, the changes that were made to the site database after the last site database backup are retrieved from the central administration site for a primary site, or from a reference primary site for a central administration site. When you recover the site database for a stand-alone primary site, you lose site changes after the last backup.
When you recover the site database for a site in a hierarchy, the recovery behavior is different for a central administration site and primary site, and when the last backup is inside or outside of the SQL Server change tracking retention period.
NEW QUESTION: 2
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.255.255.255.254
B. 許可10.0.0.1.0.0.0.254
C. 許可10.0.0.0.0.0.0.1
D. 許可10.0.0.1.0.0.0.0
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.
NEW QUESTION: 3
Arnold and Danny are two twin brothers that are celebrating their birthday. The product of their ages today is smaller by 9 from the product of their ages a year from today. What is their age today?
A. 5.
B. 7.
C. 2.
D. 9.
E. 4.
Answer: E
Explanation:
Explanation/Reference:
Explanation:
Back solve using the answers. Take the age 4.
4 x 4 = 16. 16 + 9 = 25. And in one year they'll be 5 so 5 x 5 = 25.
NEW QUESTION: 4
Which two have a media and/or density type? (Choose two.)
A. drives
B. media set
C. volume pools
D. robots
E. drives
Answer: A,E