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NEW QUESTION: 1
In den Leitlinien für 2016 bekräftigte das Information Commissioner's Office (ICO) des Vereinigten Königreichs, wie wichtig es ist, eine "mehrschichtige Bekanntmachung" zu verwenden, um betroffenen Personen was zu bieten?
A. Eine Datenschutzerklärung mit kurzen Informationen und Zugriff auf weitere Details.
B. Eine Erläuterung der Sicherheitsmaßnahmen, die bei der Übermittlung personenbezogener Daten an Dritte angewendet werden.
C. Eine Datenschutzerklärung, in der die Konsequenzen für die Deaktivierung der Verwendung von Cookies auf einer Website erläutert werden.
D. Ein effizientes Mittel zur schriftlichen Zustimmung in den Mitgliedstaaten, in denen sie dazu verpflichtet sind.
Answer: A
NEW QUESTION: 2
You are developing an ASP.NET Core Web API.
API methods must use JSON Web Tokens (JWT) to authenticate the method caller.
You need to implement JWT authentication.
How should you complete the code? To answer, select the appropriate options in the answer area.
NOTE: Each correct selection is worth one point.
Answer:
Explanation:
NEW QUESTION: 3
How many interconnects can be created between two Celerras, each with one active Data Mover, and four configured network interfaces on each active Data Mover?
A. Three
B. Four
C. One
D. Two
Answer: C
NEW QUESTION: 4
A. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo=L.CustNoWHERE D.CustNo IS NULL
B. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
C. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
D. 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
E. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
H. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo L.CustNo
Answer: F
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.
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