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NEW QUESTION 48
Refer to the exhibit.

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.

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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