Answer:
The correct answer to the following question will be Option a (10.5.0.0 255.255.240.0 ).
Explanation:
10.5.0.0 / 20
/20 = 16
⇒ Number of networks = 10.5.0.0.
⇒ The first host is = 10.5.0.1.
⇒ The last host is = 10.5.15.254.
The hosting is from 10.5.0.1 to 10.5.15.254.
And, throughout the 10.5.48.1 255.255.240.0 reference or comparison subnet.
The hosting is from 10.5.48.1 to 10.5.63.254.
That's why 10.5.0.0 255.255.240.0 is accurate.
Answer:
Option A is correct.
Explanation:
This seems to have a Fa0/0 device built from the following Internet Protocol address with that given instructions. 10.5.0.0 255.255.240.0 Subnets are still not known to be incompatible subnets while installed on some other network on the following.
Thus, the following are the reason that also describes the other options are not appropriate according to the scenario.
