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NEW QUESTION: 1
Ethernet data frame is the last field in the frame check sequence (FCS), the field is occupied by () bytes
A. 0
B. 1
C. 2
D. 3
Answer: D

NEW QUESTION: 2
The Cisco InterCloud Fabric Director provides what functionality?
A. It is the single point of management and consumption for private Cloud solutions.
B. It is the single point of management and consumption for hybrid Cloud solutions.
C. It is the single point of management and consumption for public Cloud solutions.
D. It is a plugin of a Virtual Machine Manager to provide management and configuration for hybrid Cloud solutions.
Answer: B

NEW QUESTION: 3
以下の各ステートメントについて、ステートメントが真である場合は「はい」を選択します。それ以外の場合は、「いいえ」を選択します。
注:それぞれの正しい選択には1ポイントの価値があります。

Answer:
Explanation:


NEW QUESTION: 4
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 tblDepositAcctUNIONSELECT
CustNoFROM tblLoanAcct) R
B. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT
CustNoFROM tblLoanAcct) R
C. SELECT COUNT (DISTINCT COALESCE(D.CustNo, L.CustNo))FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL OR
D. CustNo IS NULL
E. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN
tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
F. CustNo
G. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE
H. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
I. SELECT COUNT(*)FROM (SELECT CustNoFROM tblDepositAcctEXCEPTSELECT
CustNoFROM tblLoanAcct) R
J. SELECT COUNT(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo
K. CustNo = L.CustNo
Answer: B
Explanation:
Would list the customers with duplicates, which would equal the number of accounts.


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C-SEN-2305 Exam Info

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C-SEN-2305 Exam Topics

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C-SEN-2305 Offcial Page

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