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NEW QUESTION: 1
STP機能を左から右の正しい説明にドラッグアンドドロップします

Answer:
Explanation:

NEW QUESTION: 2
Ein Vertriebsmitarbeiter ändert regelmäßig vertrauliche Dateien auf einem Laptop und benötigt sie, um jederzeit verfügbar zu sein. Der Vertriebsmitarbeiter wird in zwei Wochen eine sechsmonatige Geschäftsreise unternehmen. Ein Techniker wurde gebeten, den Vertriebsmitarbeiter beim Speichern von Dateien zu unterstützen.
Welche der folgenden Methoden ist die BESTE, um Datenverluste für den Vertriebsmitarbeiter zu vermeiden?
A. Erstellen Sie ein redundantes Cloud-Speicherlaufwerk.
B. Verschlüsseln Sie das Image des Laptops auf dem Server.
C. Geben Sie dem Vertriebsmitarbeiter eine externe Festplatte.
D. Geben Sie dem Vertriebsmitarbeiter ein USB-Sicherheitsschloss.
Answer: A

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 deposit and loan accounts.
Which Transact-SQL statement should you run?
A. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
B. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo L.CustNo
C. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
D. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
E. 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
F. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo=L.CustNoWHERE D.CustNo IS NULL
G. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
H. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
Answer: C
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.