The IP address 172.16.10.1 is a Class B address. The default subnet mask for a Class B address is 255.255.0.0. However, in this case, the subnet mask is given in binary as 10111111.11111111.11000000.00000000, which translates to 255.255.192.0 in decimal. This is a valid subnet mask, so option b is incorrect.

The network ID is the bitwise AND of the IP address and the subnet mask. In this case, the network ID is 172.16.0.0, so option a is correct.

The subnet mask 255.255.192.0 means that the first 18 bits are used for the network and the remaining 14 bits are used for hosts. This allows for 2^14 - 2 = 16382 hosts per subnet, not 1260, so option e is incorrect.

The subnet mask also divides the Class B address space into 2^2 = 4 subnets, not 2, so option d is incorrect.

Therefore, the correct options are a and c.

Question

The IP address 172.16.10.1 is a Class B address. The default subnet mask for a Class B address is 255.255.0.0. However, in this case, the subnet mask is given in binary as 10111111.11111111.11000000.00000000, which translates to 255.255.192.0 in decimal. This is a valid subnet mask, so option b is incorrect.

The network ID is the bitwise AND of the IP address and the subnet mask. In this case, the network ID is 172.16.0.0, so option a is correct.

The subnet mask 255.255.192.0 means that the first 18 bits are used for the network and the remaining 14 bits are used for hosts. This allows for 2^14 - 2 = 16382 hosts per subnet, not 1260, so option e is incorrect.

The subnet mask also divides the Class B address space into 2^2 = 4 subnets, not 2, so option d is incorrect.

Therefore, the correct options are a and c.

Knowee AI · Accepted Answer

The IP address 172.16.10.1 is a Class B address. The default subnet mask for a Class B address is 255.255.0.0. However, in this case, the subnet mask is given in binary as 10111111.11111111.11000000.00000000, which translates to 255.255.192.0 in decimal. This is a valid subnet mask, so option b is incorrect.

The network ID is the bitwise AND of the IP address and the subnet mask. In this case, the network ID is 172.16.0.0, so option a is correct.

The subnet mask 255.255.192.0 means that the first 18 bits are used for the network and the remaining 14 bits are used for hosts. This allows for 2^14 - 2 = 16382 hosts per subnet, not 1260, so option e is incorrect.

The subnet mask also divides the Class B address space into 2^2 = 4 subnets, not 2, so option d is incorrect.

Therefore, the correct options are a and c.

A host with an IP address of 172.16.10.1 has network and host portion as:10111111.11111111.11000000.00000000Select one or more:a. Network id is 172.16.0.0 b. invalid bit maskingc. 255.255.192.0 d. 2 subnetse. 1260 host/subnet

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