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write the network ip and subnet mask in binary:
192.168.100001(10).00000000
255.255.111111(00).00000000
so what are the propabale bits for the last 2 bits in octet 3? it is 00,01,10,and 11
Those are the 4 networks which will be 192.168.(132,133,134,135).0
answer is 4 for sure.
write the network ip and subnet mask in binary:
192.168.100001(10).00000000
255.255.111111(00).00000000
so what are the propabale bits for the last 2 bits in octet 3? it is 00,01,10,and 11
Those are the 4 networks which will be 192.168.(132,133,134,135).0
answer is 4 for sure.
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