The answer is A, or Locating bottlenecks in a network.
If 29 bits of the 32 available addressing bits are used for the subnet, then only 3 bits giving 2^3=8 combinations remain for the host addresses.
In reality, the all zeros and all ones addresses are reserved. So, 8 addresses can exist, but 6 of those are available.
The way the question is formulated it seems the answer 8 is what they're after.
Answer:
procedimento
Explanation:
faz se a montagem conforme mostra a figura.
