Test Cybersecurity-Architecture-and-Engineering Valid - Test Cybersecurity-Architecture-and-Engineering Centres, WGU Cybersecurity Architecture and Engineering (D488) Dumps - Cads-Group

  • Exam Number/Code : Cybersecurity-Architecture-and-Engineering
  • Exam Name : WGU Cybersecurity Architecture and Engineering (D488)
  • Questions and Answers : 260 Q&As
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NEW QUESTION: 1
注:この質問は、同じまたは類似の回答の選択肢を使用する一連の質問の一部です。 回答の選択肢は、シリーズの複数の質問に対して正しいかもしれません。 各質問は、このシリーズの他の質問とは独立しています。 質問に記載されている情報や詳細がその質問に適用されます。
あなたには、銀行システム用のデータベースがあります。 データベースには、預金口座とローン口座をそれぞれ格納する2つのテーブルtblDepositAcctとtblLoanAcctがあります。 どちらの表にも次の列があります。

預金口座またはローン口座のいずれかを持つ顧客の総数を決定する必要がありますが、両方の種類の口座を決定する必要はありません。
どのTransact-SQL文を実行する必要がありますか?
A. 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
B. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
D. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
E. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo L.CustNo
H. CustNoWHERE D.CustNo IS NULL
I. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo
Answer: A
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: 2
IP 地址 133.18.2.110 是 ______ 地址。
A. C 类
B. D 类
C. A 类
D. B 类
Answer: D

NEW QUESTION: 3
Refer to the graphic. Users on the Holyoke router are unable to access the intranet server attached to interface E0 of the Chicopee router. Inspection of the routing table of the Holyoke router shows that an entry for the Chicopee E0 network is missing. Which command will configure the Holyoke router with a path to the intranet server network?

A. Holyoke(config)# ip network 202.18.38.0
B. Holyoke(config)# ip network 202.18.18.0 255.255.255.0
C. Holyoke(config)# ip host Chicopee 201.73.127.0 255.255.255.0
D. Holyoke(config)# ip route 202.18.18.0 255.255.255.0 201.73.127.2
E. Holyoke(config)# ip route 201.73.127.2 255.255.255.0 202.18.18.0
F. Holyoke(config)# ip host Chicopee 201.73.127.2
Answer: D

 

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