A. It corresponds to a Class A address with 13 bits borrowed.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.
Answer: A, C, F
Explanation:
This subnet mask includes the first 5 bits within the third octet, so for a class A address 13 bits will be used for the mask (8 bits in the second octet plus 5 in the third).
Since the first 5 bits are used in this octet, that means that remaining 3 bits in this octet will be available for hosts, so each network will be a factor of 8, making the last availablesubnet with a .248 in the third octet.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.
Answer: A, C, F
Explanation:
This subnet mask includes the first 5 bits within the third octet, so for a class A address 13 bits will be used for the mask (8 bits in the second octet plus 5 in the third).
Since the first 5 bits are used in this octet, that means that remaining 3 bits in this octet will be available for hosts, so each network will be a factor of 8, making the last availablesubnet with a .248 in the third octet.