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NEW QUESTION: 1

A. Install-AdcsWebEnrollment
B. Install-AdcsCertificationAuthority
C. Add-WindowsFeature
D. Add-CAAuthoritylnformationAccess
E. Install-AdcsOnlineResponder
Answer: A,B
Explanation:
* The Install-AdcsCertificationAuthority cmdlet performs installation and
configuration of the AD CS CA role service.
*The Install-AdcsWebEnrollment cmdlet performs initial installation and configuration of the
Certification Authority Web Enrollment role service.
Topic 3, Volume C

NEW QUESTION: 2
Which of the following should a project manager do for a team member who does not speak the same language?
A. Send the team member to training on the desired language.
B. Ask for a replacement team member that speaks the same language.
C. Hire a translator to assist the team member in communicating.
D. Communicate with the team member using translation software.
Answer: C

NEW QUESTION: 3
Given the code fragment:

What is the result when the result.txt file already exists in c:\student?
A. The program replaces the file contents as well as the file attributes and prints Not equal.
B. An UnsupportedOperationException is thrown at runtime.
C. The program replaces the file contents and the file's attributes and prints Equal.
D. The program replaces only the file attributes and prints Not equal.
Answer: A
Explanation:
Assuming there is a file D:\\faculty\\report.txt then this file will be copied and will be replacing C:\\student\\report.txt.

NEW QUESTION: 4
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。

預金口座とローン口座の合計数を決定する必要があります。
どのTransact-SQL文を実行する必要がありますか？
A. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
C. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
E. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo L.CustNo
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 L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo=L.CustNoWHERE D.CustNo IS NULL
Answer: C
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.