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NEW QUESTION 48
Refer to the exhibit.
100-101-3c6c7dca0106a88037ac46e7ba93fb8f.jpg
The network is converged.After link-state advertisements are received from Router_A, what information will Router_E contain in its routing table for the subnets 208.149.23.64 and 208.149.23.96?

  • A. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, Serial1/0 208.149.23.96[110/13] via 190.173.23.10, 00:00:16, Serial1/0 208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
  • B. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, FastEthemet0/0 208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
  • C. 208.149.23.64[110/1] via 190.172.23.10, 00:00:07, Serial1/0 208.149.23.96[110/3] via 190.173.23.10, 00:00:16, FastEthemet0/0
  • D. 208.149.23.64[110/3] via 190.172.23.10, 00:00:07, Serial1/0 208.149.23.96[110/3] via 190.173.23.10, 00:00:16, Serial1/0

Answer: B

Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface. The interface cost is calculated with the formula 108 /
Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1. Therefore the
cost is 12 (learned from Router_A) + 1 = 13 for both subnets ->
The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps =
64; T3 cost = 108 / 45 Mbps = 2) so surely OSPF will choose the path through T3 link ->
Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0.
In fact, we can quickly eliminate answers B, C and D because they contain at least one
subnet learned from Serial1/0 -> they are surely incorrect.

 

NEW QUESTION 49
Refer to the exhibit.
100-101-7c4f6307bf89b34d7b22e21f5ed7646b.jpg
After configuring two interfaces on the HQ router, the network administrator notices an error message. What must be done to fix this error?

  • A. The address of the FastEthernet interface should be changed to 192.168.1.66
  • B. The serial interface must be configured first.
  • C. The subnet mask of the FastEthernet interface should be changed to 255.255.255.240
  • D. The serial interface must use the address 192.168.1.2
  • E. The subnet mask of the serial interface should be changed to 255.255.255.0

Answer: C

 

NEW QUESTION 50
Which IP addresses are valid for hosts belonging to the 10.1.160.0/20 subnet? (Choose three.)

  • A. 10.1.176.1
  • B. 10.1.160.255
  • C. 10.1.174.255
  • D. 10.1.160.0
  • E. 10.1.175.255
  • F. 10.1.168.0

Answer: B,C,F

Explanation:
Explanation/Reference:
All IP address in IP ranges between: 10.1.160.1 and 10.1.175.254 are valid as shown below Address: 10.1.160.0 00001010.00000001.1010 0000.00000000 Netmask: 255.255.240.0 = 20 11111111.11111111.1111 0000.00000000 Wildcard: 0.0.15.255 00000000.00000000.0000 1111.11111111
Which implies that: Network: 10.1.160.0/20 00001010.00000001.1010 0000.00000000 HostMin: 10.1.160.1 00001010.00000001.1010 0000.00000001 HostMax: 10.1.175.254 00001010.00000001.1010 1111.11111110 Broadcast: 10.1.175.255 00001010.00000001.1010 1111.11111111

 

NEW QUESTION 51
What action would a business want its customers to take if it selects Conversions as its campaign objective?

  • A. Install an app
  • B. View a video
  • C. Provide an email address
  • D. Like a post
  • E. Make a purchase

Answer: E

 

NEW QUESTION 52
......

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