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NEW QUESTION: 1
What does protocol normalization do?
A. removes application layer data, whether or not it carries protocol-induced anomalies, so that packet headers can be inspected more accurately for signs of abuse
B. compares evaluated packets to normal, daily network-traffic patterns
C. compares a packet to related traffic from the same session, to determine whether the packet is out of sequence
D. removes any protocol-induced or protocol-allowable ambiguities
Answer: D

NEW QUESTION: 2
An organisation's administrator is tasked with setting up and configuring Reputation within a new community. Per the provided requirements, there should be 5 (five) total reputation levels and the point range of the highest level must start at 5001 points. Which three tasks must occur in order to successfully deploy Reputation, per the requirements?
Choose 3 answers
Select one or more of the following:
A. Remove and add reputation levels as required
B. Upload a badge image for each reputation level
C. Provide a custom name for each reputation level
D. Configuration point ranges for the levels
E. Enable setup and display of reputation levels
Answer: A,D,E

NEW QUESTION: 3
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 AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
E. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM 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 (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
H. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
Answer: G
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