CompTIA FC0-U71 Q&A - in .pdf

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  • Exam Code: FC0-U71
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NEW QUESTION: 1
Lian has used the Set Value simple action in the Timesheet XPage to set the value of a computed expression.
What does the generated XML look like on the XPage after this simple action has been added?
A. <xp:this.action> <xp:setValue page="/TimesheetApproved.xsp" binding=[ce]value="Approved">
</xp:setValue>
B. <xp:this.action> <xp:setValue docPage="/Timesheet.xsp"
fieldName=[ce]value="Approved"createConflict=false> </xp:setvalue>
C. <xp:this:action> <xp:setValue fieldName=[ce] value="Approved"></xp:setValue>
D. <xp:this.action> <xp:setValue value="Approved"
binding="#{document.Status}"></xp:setValue></xp:this.action>
Answer: D

NEW QUESTION: 2
After migrating a cisco IP Phone to a new cluster the phone continue to register with its old CUCM cluster. IP Phone status ERROR: TRUST LIST UPDATE FAILED Which two actions must you take to troubleshoot the problem?
A. CallManager services are not running on the destination cluster
B. the new TFTP server is not in the ITL file
C. the phone cannot reach its new TFTP server
D. the phone is not provisioned correctly on the destination cluster
Answer: B,C

NEW QUESTION: 3
Siehe Ausstellung. Welche Adress- und Maskenkombination stellt eine Zusammenfassung der von EIGRP gelernten Routen dar?

A. 192.168.25.28 255.255.255.240
B. 192.168.25.16 255.255.255.240
C. 192.168.25.0 255.255.255.240
D. 192.168.25.0 255.255.255.252
E. 192.168.25.28 255.255.255.252
F. 192.168.25.16 255.255.255.252
Answer: B
Explanation:
The binary version of 20 is 10100.
The binary version of 16 is 10000.
The binary version of 24 is 11000.
The binary version of 28 is 11100.
The subnet mask is /28. The mask is 255.255.255.240.
Note:
From the output above, EIGRP learned 4 routes and we need to find out the summary of them:
+ 192.168.25.16
+ 192.168.25.20
+ 192.168.25.24
+ 192.168.25.28
-> The increment should bE. 28 ?16 = 12 but 12 is not an exponentiation of 2 so we must choose 16 (24). Therefore the subnet mask is /28 (=1111 1111.1111 1111.1111 1111.11110000) = 255.255.255.240 So the best answer should be 192.168.25.16 255.255.255.240

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