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Exam Code: JN0-649
Exam Name: Enterprise Routing and Switching, Professional (JNCIP-ENT)
Vendor: Juniper

60 Questions & Answers
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NEW QUESTION: 1
An administrator is attempting to power on a virtual machine with 32GB of memory. The operation fails with the following error:
Could not power on VM: No space left on device
Checking the space on the virtual machine's datastore, there is 30GB free.
Which action would allow the VM to power on?
A. Enable vSphere HD admission control on the cluster in which the VM resides.
B. Set a 2GB memory limit on the VM.
C. Set a 2GB memory reservation on the VM.
D. Mount the virtual disk from the affected VM on to another virtual machine and free up space from within the OS.
Answer: C
Explanation:
Explanation/Reference:
Explanation:
You must reserve swap space for any unreserved virtual machine memory (the difference between the reservation and the configured memory size) on per-virtual machine swap files.
Reference: https://docs.vmware.com/en/VMware-vSphere/6.5/com.vmware.vsphere.resmgmt.doc/GUID- AC40823A-695B-438A-A8F2-4B7C51A0D014.html

NEW QUESTION: 2
Your network contains an Active Directory domain named contoso.com. The domain contains a Microsoft System Center 2012 infrastructure.
You deploy a service named Service1 by using a service template. Service1 contains two virtual machines.
The virtual machines are configured as shown in the following table.

You need to recommend a monitoring solution to ensure that an administrator can review the availability information of Service1.
Solution: From Operations Manager, you create a Distributed Application and a Monitor Override.
Does this meet the goal?
A. No
B. Yes
Answer: B

NEW QUESTION: 3
Fee schedules are updated by third party payers
A. annually
B. semiannually
C. weekly
D. monthly
Answer: A

NEW QUESTION: 4



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


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