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NEW QUESTION: 1

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

NEW QUESTION: 2
A Wireless LAN Cisco engineer is troubleshooting 802. 11ac WLAN at customer site Doe to interferences with neighboring 802 11n networks the engineer must determine the current primary secondary channel allocation to enhance the current configuration. Channel 60 has been chosen by the customer as primary 20 MHz channel. Which combination determines the resulting primary
40 MHz and primary 80 MHZ channels?
A. primary 40 MHz channel 62, primary 80 MM: channel 58
B. primary 40 MHz channel 54, primary 80 MHz channel 58
C. primary 40 MHz channel 64, primary 80 MHz channel 60
D. primary 40 MHz channel 60, primary 80 MHz channel 56
Answer: A

NEW QUESTION: 3
クライアントデバイスは、アトリウムの異なるフロアにあるアクセスポイント間をローミングします。ローカルモードで同じコントローラと設定に接続されたアクセスポイント。アクセスポイントは異なるIPアドレスにありますが、グループ内のクライアントVLANは同じです。どのタイプのローミングが発生しますか？
A. コントローラー内
B. VLAN内
C. サブネット間
D. コントローラー間
Answer: C
Explanation:
Explanation
Mobility, or roaming, is a wireless LAN client's ability to maintain its association seamlessly from one access point to another securely and with as little latency as possible. Three popular types of client roaming are:
Intra-Controller Roaming: Each controller supports same-controller client roaming across access points managed by the same controller. This roaming is transparent to the client as the session is sustained, and the client continues using the same DHCP-assigned or client-assigned IP address.
Inter-Controller Roaming: Multiple-controller deployments support client roaming across access points managed by controllers in the same mobility group and on the same subnet. This roaming is also transparent to the client because the session is sustained and a tunnel between controllers allows the client to continue using the same DHCP- or client-assigned IP address as long as the session remains active.
Inter-Subnet Roaming: Multiple-controller deployments support client roaming across access points managed by controllers in the same mobility group on different subnets. This roaming is transparent to the client because the session is sustained and a tunnel between the controllers allows the client to continue using the same DHCP-assigned or client-assigned IP address as long as the session remains active.

NEW QUESTION: 4
DRAG DROP
You use SQL Server 2016 Enterprise Edition. Your database contains a partitioned table named AuditData. AuditData is partitioned by year. Partition 1 contains data from the year
2010 and prior.
Management has decided to archive all AUDITDATA records from 2010 and prior.
Management wants the records to be removed from the database entirely and provided to the backup team as a zipped text file. The data must no longer reside in the database.
There is very little tolerance for performance degradation in your environment.
You need to remove all 2010 and prior data from the AuditData table by using the least amount of system resources possible.
Develop the solution by selecting and arranging the required SQL actions in the correct order.
You may not need all of the actions.

Answer:
Explanation:

Explanation:
Box 1: CREATE TABLE
Box 2: SPLIT RANGE
Box 3: SELECT INTO
Box 4: BCP
Box 5: DROP TABLE
Box 6: DROP PARTITION
Note:
* Create a new partitioned table with the partition function you want, and then insert the data from the old table into the new table by using an INSERT INTO...SELECT FROM statement.
* SPLIT RANGE ( boundary_value )
Adds one partition to the partition function. boundary_value determines the range of the new partition, and must differ from the existing boundary ranges of the partition function.
Based on boundary_value, the Database Engine splits one of the existing ranges into two.
Of these two, the one where the new boundary_value resides is considered the new partition.
* BCP can be used to produce the zipped text file.
* Example:
Splitting a partition of a partitioned table or index into two partitions
The following example creates a partition function to partition a table or index into four partitions. ALTER PARTITION FUNCTION splits one of the partitions into two to create a total of five partitions.
CREATE PARTITION FUNCTION myRangePF1 (int)
AS RANGE LEFT FOR VALUES ( 1, 100, 1000 );
GO
--Split the partition between boundary_values 100 and 1000
--to create two partitions between boundary_values 100 and 500
--and between boundary_values 500 and 1000.
ALTER PARTITION FUNCTION myRangePF1 ()
SPLIT RANGE (500);
References:
http://technet.microsoft.com/en-us/library/ms186307(v=sql.110).aspx
http://technet.microsoft.com/en-us/library/ms162802(v=sql.120).aspx