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NEW QUESTION: 1
If the primary object on a custom report is a custom object, and the custom object is deleted, then the custom report type and any reports created from it will automatically be deleted.
A. False
B. True
Answer: B
NEW QUESTION: 2
Click on the exhibit button below.
Given the configuration below, which of the following scenarios is FALSE?
A. The primary path is active and the secondary path is in hot standby.
B. Both the primary and secondary paths are explicitly defined.
C. The LSP defined will use the primary path and will signal the secondary path to become active if the primary path fails.
D. The LSP has the same source and destination for both primary and secondary paths.
Answer: A
NEW QUESTION: 3
Hinweis: Diese Frage ist Teil einer Reihe von Fragen, bei denen die gleichen oder ähnliche Antwortmöglichkeiten verwendet werden. Eine Antwortauswahl kann für mehr als eine Frage in der Reihe richtig sein. Jede Frage ist unabhängig von den anderen Fragen in dieser Reihe. In einer Frage angegebene Informationen und Details gelten für diese Frage.
Sie haben eine Datenbank für ein Bankensystem. Die Datenbank enthält zwei Tabellen mit den Namen tblDepositAcct und tblLoanAcct, in denen Einlagen- und Darlehenskonten gespeichert sind. Beide Tabellen enthalten folgende Spalten:
Sie müssen die Gesamtzahl der Einlagen- und Darlehenskonten ermitteln.
Welche Transact-SQL-Anweisung sollten Sie ausführen?
A. SELECT COUNT (*) FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
B. SELECT COUNT (*) FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
C. SELECT COUNT (DISTINCT D.CustNo) FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
D. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
E. SELECT COUNT (*) FROM (SELECT CustNoFROMtblDepositAcctUNION ALL SELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT (DISTINCT L.CustNo) VON tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
G. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
H. SELECT COUNT (DISTINCT COALESCE (D.CustNo, L.CustNo)) FROM tblDepositAcct DFULL JOIN
Answer: E
Explanation:
Erläuterung
Listen Sie die Kunden mit Duplikaten auf, die der Anzahl der Konten entsprechen.