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NEW QUESTION: 1
Note: This question is part of a series of questions that present the same scenario. Each question in the series contains a unique solution that might meet the stated goals. Some question sets might have more than one correct solution, while others might not have a correct solution.
After you answer a question in this section. You will NOT be able to return to it. As a result, these questions will not appear in the review screen.
You have a database that tracks orders and deliveries for customers in North America. The database contains the following tables:
Sales.Customers

Application.Cities

Sales.CustomerCategories

Your company is developing a new social application that connects customers to each other based on the distance between their delivery locations.
You need to write a query that returns the nearest customer.
Solution: You run the following Transact-SQL statement:

The variable @custID is set to a valid customer.
Does the solution meet the goal?
A. Yes
B. No
Answer: B

NEW QUESTION: 2
You are setting up a disaster recovery system using system replication. You want the transaction processing of the primary system to be suspended if the network to the secondary system fails.
What log replication mode must you use?
Please choose the correct answer.
Response:
A. syncmem
B. async
C. sync
D. full sync
Answer: D

NEW QUESTION: 3
Which of the following is a most complete measure of the liquidity gap facing a firm?
A. Liquidity at Risk
B. Marginal liquidity gap
C. Residual liquidity gap
D. Cumulative liquidity gap
Answer: C
Explanation:
Explanation
Marginal liquidity gap measures the expected net change in liquidity over, say, a day. It is just equal to the liquidity inflow minus liquidity outflow. The cumulative liquidity gap measures the aggregate change in liquidity from a point in time, in other words it is just the summation of the marginal liquidity gap for each of the days included in the period under consideration. The residual liquidity gap goes one step further and adds available 'opening balance' of liquidity to the cumulative liquidity gap to reveal the days or times when the net liquidity is most at risk.
Liquidity at Risk measures the expected time to survival at a certain confidence level applied to the firm's cash flows - and is not a measure of the liquidity gap.
Therefore Choice 'a' is the correct answer.

NEW QUESTION: 4
You use SQL Server 2014 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

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 top produce the zipped text file.
- Example: plitting 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);