You are given the following address: 15.16.193.6/21. Determine the subnet mask, address class, subnet address, and broadcast address. * 255.255.248.0, A, 15.16.192.0, 15.16.199.255 * 255.255.248.0, B, 15.16.192.0, 15.16.199.255 * 255.255.248.0, A, 15.16.199.255, 14.15.192.0 * 255.255.242.0, A, 15.16.192.0, 15.16.199.255
Description : You are given the following address: 15.16.193.6/21. Determine the subnet mask, address class, subnet address, and broadcast address. * 255.255.248.0, A, 15.16.192.0, 15.16.199.255 * 255.255.248.0, B, 15.16.192.0, ... 0, A, 15.16.199.255, 14.15.192.0 * 255.255.242.0, A, 15.16.192.0, 15.16.199.255
Last Answer : Correct answer: A,B & D Subnet addresses in this situation are all in multiples of 8. In this example, 201.222.5.16 is the subnet, 201.22.5.31 is the broadcast address. The rest are valid host IDs on subnet 201.222.5.16.