I'm not entirely sure, but I remember something about needing to calculate usable addresses. Is it 128 or 126? I feel like I've seen similar questions before.
I think the subnet mask 255.255.255.0 gives us a /24, which means we have 256 total addresses, but we need to subtract 2 for the network and broadcast addresses, right?
I've got this one! The subnet mask 255.255.255.0 means we have 8 bits for the host portion. 2^8 = 256 possible host addresses, but we need to subtract the network address (172.16.0.0) and the broadcast address (172.16.0.255), so the answer is 254 available IP addresses.
This looks straightforward. The subnet mask is 255.255.255.0, so the network has 256 possible IP addresses (2^8). Subtracting the network and broadcast addresses, we get 254 available IP addresses.
Okay, let's think this through step-by-step. The IP segment is 172.16.0.0 with a subnet mask of 255.255.255.0, so the network portion is 172.16.0.0 and the host portion is 0.0.0.0. The number of host bits is 8, so the number of possible host addresses is 2^8 - 2 = 256 - 2 = 254.
I'm with Jeannetta on this one. A) 126 has got to be the correct answer. Who would waste two IP addresses for the network and broadcast addresses? Silly question.
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