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NEW QUESTION: 1
U+ Bank offers two credit cards to its customers: the Cold Card and the Silver Card. The bank wants the
annual fee property of the credit cards to be dynamic based on the customer lifetime value property.
That is, customers with customer lifetime value greater than 65 receive a discount of 20% on the annual fee.
As a Strategy Designer, how do you implement this requirement in the decision strategy?
A. By using the If function to set the customer value
B. By using the When rule to set the annual fee
C. By using the If function to set the annual fee
D. By using the When rule to set the customer value
Answer: A

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. CustNo = L.CustNo
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
C. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN
tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
D. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo
E. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT
CustNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
G. CustNo IS NULL
H. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE
I. CustNo
J. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
K. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT
CustNoFROM tblLoanAcct) R
Answer: B
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
A DLP administrator has determined that a Network Discover server is unable to scan a remote file server.
Which action should the administrator take to successfully scan the remote file server?
A. verify that the target file server is a Windows 2008 server
B. verify that the file server has .NET services running
C. restart the discover scan
D. use the fully qualified name (FQDN) of the server
Answer: D
Explanation:
Explanation/Reference:
Explanation: