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NEW QUESTION: 1
Which of the following characteristics pertaining to databases is NOT true?
A. All relations must have a specific cardinality.
B. Justifications must exist for normalized data.
C. No NULLs should be allowed for primary keys.
D. A data model should exist and all entities should have a significant name.
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Data normalization is the process of reducing data to its canonical form. Database normalization is the process of organizing the fields and tables of a relational database to minimize redundancy and dependency. Justification is not a term that is used for normalized data.
Incorrect Answers:
A: A database model, such as a relational database model, is a type of data model that determines the logical structure of a database and fundamentally determines in which manner data can be stored, organized, and manipulated. Within a database model the entities must be named properly.
C: A primary key cannot have a NULL value.
D: A database relation could be either one-to-one, one-to-many, or many-to-many.
References:
https://en.wikipedia.org/wiki/Data_normalization

NEW QUESTION: 2
Which statement best describes the function of Hot Standby Router Protocol (HSRP)?
A. HSRP specifies a single IP address that all routers in the group must use.
B. HSRP defines a set of routers that represent one virtual, fault-tolerant router.
C. HSRP provides a round-robin gateway selection process to increase fault tolerance.
D. HSRP defines a frame-tagging scheme that allows end stations to use any router as a gateway.
Answer: B

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 tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
C. 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
D. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
E. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
F. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
Answer: C
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: 4
1つの300GB EBSボリュームを持つm1.small EC2インスタンスを使用して、リレーショナルデータベースをホストしています。データベースへの書き込みスループットを増やす必要があると判断しました。次のアプローチのどれがこれを達成するのに役立ちますか？ 2つの答えを選んでください
A. インスタンスをAuto Scaling Groupに配置します
B. データベースをElastic Load Balancerの背後に配置します
C. EC2インスタンスのサイズを増やします
D. EBSボリュームの配列を使用します
E. EBSボリュームを追加し、RAID 5に配置します
F. マルチAZモードを有効にする
Answer: C,D

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