Juniper JN0-351 dumps - in .pdf

JN0-351 pdf
  • Exam Code: JN0-351
  • Exam Name: Enterprise Routing and Switching, Specialist (JNCIS-ENT)
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • PDF Price: $51.98

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  • Exam Code: JN0-351
  • Exam Name: Enterprise Routing and Switching, Specialist (JNCIS-ENT)
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • PDF Version + PC Test Engine + Online Test Engine
  • Value Pack Total: $103.96  $66.98
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Juniper JN0-351 dumps - Testing Engine

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  • Exam Code: JN0-351
  • Exam Name: Enterprise Routing and Switching, Specialist (JNCIS-ENT)
  • Version: V17.95
  • Q & A: 400 Questions and Answers
  • Software Price: $51.98
  • Testing Engine

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NEW QUESTION: 1
A policy template called Customer Credit Card Numbers is being imported into the system.
What is the default result for this action?
A. the policy template will be enabled by default
B. the policy template will be available after logging off and on to Enforce
C. the policy template will be listed under US Regulatory Enforcement Templates and be available
D. the policy template will be listed under Imported Templates
Answer: D

NEW QUESTION: 2
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(*)FROM tblDepositAcct DFULL JOIN tblLoanAcct L ON D.CustNo = L.CustNo
C. SELECT COUNT (DISTINCT D.CustNo)FROM tblDepositAcct D, tblLoanAcct LWHERE D.CustNo = L.CustNo
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 tblDepositAcctUNIONSELECT CustNoFROM tblLoanAcct) R
F. SELECT COUNT(*)FROM (SELECT CustNoFROMtblDepositAcctUNION ALLSELECT CustNoFROM tblLoanAcct) R
G. SELECT COUNT(DISTINCT L.CustNo)FROM tblDepositAcct DRIGHT JOIN tblLoanAcct L ON D.CustNo =L.CustNoWHERE D.CustNo IS NULL
H. SELECT COUNT(*)FROM (SELECT AcctNoFROM tblDepositAcctINTERSECTSELECT AcctNoFROM tblLoanAcct) R
Answer: F
Explanation:
Explanation
Would list the customers with duplicates, which would equal the number of accounts.

NEW QUESTION: 3
You work as an IT security auditor hired by a law firm in Boston to test whether you can gain access to sensitive information about the company clients. You have rummaged through their trash and found very little information. You do not want to set off any alarms on their network, so you plan on performing passive foot printing against their Web servers. What tool should you use?
A. Netcraft
B. Dig
C. Nmap
D. Ping sweep
Answer: A

NEW QUESTION: 4
A patient with lung cancer and bone metastasis is seen for complex treatment planning by a radiation oncologist.
A. 0
B. 1
C. 2
D. 3
Answer: A

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