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NEW QUESTION: 1
A. 50; new
B. 2; old
C. 2; new
D. 50; old
Answer: A
Explanation:
https://docs.citrix.com/en-us/netscaler/12/load-balancing/load-balancing-customizing-algorithms.html
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NEW QUESTION: 2
Note: This question is part of a series of questions that use the same or similar answer choices. An answer choice may be correct for more than one question in the series. Each question is independent of the other questions in this series. Information and details provided in a question apply to that question.
You have a database for a banking system. The database has two tables named tblDepositAcct and tblLoanAcct that store deposit and loan accounts, respectively. Both tables contain the following columns:
You need to determine the total number of customers who have either deposit accounts or loan accounts, but not both types of accounts.
Which Transact-SQL statement should you run?
A. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL OR L.CustNo IS NULL
C. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
E. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
F. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
Answer: B
Explanation:
Explanation
SQL Server provides the full outer join operator, FULL OUTER JOIN, which includes all rows from both tables, regardless of whether or not the other table has a matching value.
Consider a join of the Product table and the SalesOrderDetail table on their ProductID columns. The results show only the Products that have sales orders on them. The ISO FULL OUTER JOIN operator indicates that all rows from both tables are to be included in the results, regardless of whether there is matching data in the tables.
You can include a WHERE clause with a full outer join to return only the rows where there is no matching data between the tables. The following query returns only those products that have no matching sales orders, as well as those sales orders that are not matched to a product.
USE AdventureWorks2008R2;
GO
-- The OUTER keyword following the FULL keyword is optional.
SELECT p.Name, sod.SalesOrderID
FROM Production.Product p
FULL OUTER JOIN Sales.SalesOrderDetail sod
ON p.ProductID = sod.ProductID
WHERE p.ProductID IS NULL
OR sod.ProductID IS NULL
ORDER BY p.Name ;
References: https://technet.microsoft.com/en-us/library/ms187518(v=sql.105).aspx
NEW QUESTION: 3
In response to the threat of ransomware, an organization has implemented cybersecurity awareness activities.
The risk practitioner's BEST recommendation to further reduce the impact of ransomware attacks would be to implement:
A. two-factor authentication.
B. encryption for data in motion.
C. continuous data backup controls.
D. encryption for data at rest.
Answer: C
NEW QUESTION: 4
You are a server administrator of a SQL Server 2008 Analysis Services (SSAS) instance. The instance
contains a database that is used by the members of the Sales group.
You configure a new role named Northern Region by using the "{[Customers].[Region].[Region].[Northern]}"
allowed permission set.
You assign the Sales group to the Northern Region role.
You need to verify that users in the Northern Region role can view data only for their region.
What should you do?
A. Temporarily add your account to the Northern Region role.
B. Add the User ID=Northern Region; parameter to the connection string.
C. Add the Roles=Northern Region; parameter to the connection string.
D. Select the Enable Visual Totals check box for the Northern Region role.
Answer: C
