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NEW QUESTION: 1
You have several SQL Server queries.
You plan to optimize the queries to improve performance.
You run the queries in SQL Server Management Studio.
You need to compare query runs to expose the indexing issues of the queries.
Which three actions should you perform from SQL Server Management Studio? Each correct answer presents part of the solution.
A. Enable the Include Actual Execution Plan option.
B. Add the STATISTICS TIME execution setting to the query.
C. Add the STATISTICS PROFILE execution setting to the query.
D. Enable the Debug option.
E. Add the STATISTICS IO execution setting to the query.
Answer: A,B,E
Explanation:
E: An execution plan is theresult of the query optimizer's attempt to calculate the most efficient way to implement the request represented by the T-SQL query you submitted. To generate the first execution plan, you can enable the Include Actual Execution Plan option.
B: SET STATISTICS TIME displays the number of milliseconds required to parse, compile, and execute each statement.
C: STATISTICS IO causes SQL Server to display information regarding the amount of disk activity generated by Transact-SQL statements. This is useful information for optimizing queries.
The information include Scan count:
Number of seeks/scans started after reaching the leaf level in any direction to retrieve all the values to construct the final dataset for the output.
Scan count is 0 if the index used is a unique index or clustered index on a primary key and you are seeking for only one value. For example WHERE Primary_Key_Column = <value>.
Scant count is 1 when you are searching for one value using a non-unique clustered index which is defined on anon-primary key column. This is done to check for duplicate values for the keyvalue that you are searching for. For example WHERE Clustered_Index_Key_Column = <value>.
Scan count is N when N is the number of different seek/scan started towards the leftor right side at the leaf level after locating a key value using the index key.

NEW QUESTION: 2
What is a contract modification?
A. an increase or decrease in expected collectability
B. a change to the contract caused by negotiation with the customer
C. a change (modification) to the contract data
D. a revision or correction to the estimate of variable consideration made at inception
Answer: C

NEW QUESTION: 3
A researcher has a sample of 700 observations from a population whose standard deviation is known to be
1,235.6. The mean of the sample is calculated to be 219.2. The null hypothesis is stated as Ho: mean
150 and the alternative is non-directional. The p-value in this case equals ________.
A. 10.16%
B. 13.88%
C. 6.94%
D. 12.82%
Answer: B
Explanation:
Explanation/Reference:
Explanation:
The z-statistic under the null equals (219.2 - 150)/(1235.6/(700

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.5)) = 1.48. Since the alternative is non- directional, you must use a two-tailed test and ask the question, "What is the probability of observing a z- statistic at least as large in magnitude as 1.48?" The answer to this will give you the p- value for this sample under the two-tailed test. The right-tailed probability of observing a z-statistic which is at least as big as 1.48 equals 1.0 - 0.9306 = 0.0694 = 6.94%. The left-tailed probability of observing a z-statistic which is at least as small as - 1.48 also equals 6.94% since the standard normal distribution is symmetrical about zero. Therefore, the p-value of this two-tailed test in this sample equals 6.94%*2 = 13.88%.