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NEW QUESTION: 1
A. Option C
B. Option A
C. Option D
D. Option B
Answer: B
Explanation:
This question clearly stated there is a WAN connectivity issue between R1 and Branch1 so we should check both of them with the "show ip interface brief" command.
On R1:
On Branch1:
We can see that although the Multilink1 interfaces are in "up/up" state but they are not in the same subnet. According to the IP address scheme shown on the topology we can deduce the Multilink interface on Branch1 has been misconfigured, it should be
192.168.14.2 instead.
NEW QUESTION: 2
Which two licenses must you have to enable OTV on a Cisco Nexus 7000 Series Switch? (Choose two.)
A. Transport Services Package
B. MPLS Services Package
C. Enhanced Layer 2 Package
D. Advanced Services Package
E. Enterprise Services Package
Answer: D,E
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 (DISTINCT L.CustNo) VON tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo = L.CustNoWHERE D.CustNo IS NULL
B. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT CustNoFROM tblLoanAcct) R
C. SELECT COUNT (DISTINCT COALESCE (D.CustNo, L.CustNo)) FROM tblDepositAcct DFULL JOIN
D. SELECT COUNT (*) FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
E. SELECT COUNT (*) FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
F. SELECT COUNT (*) FROM (SELECT CustNoFROMtblDepositAcctUNION ALL SELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT (DISTINCT D.CustNo) FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
H. SELECT COUNT (*) FROM (SELECT CustNoFROM tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
Answer: F
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.
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