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    20 minus one if you have subnet zero enabled, or minus 2 if not.
    You asked about the number of hosts. That will be 32 minus the number of network bits, minus two. So calculate it as (2

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    NEW QUESTION: 1
    Refer to the exhibit. Hosts in network 192.168.2.0 are unable to reach hosts in network 192.168.3.0. Based on the output from RouterA, what are two possible reasons for the failure? (Choose two.)

    A. Interface S0/0 on RouterB is administratively down.
    B. Interface S0/0 on RouterA is not receiving a clock signal from the CSU/DSU.
    C. Interface S0/0 on RouterA is configured with an incorrect subnet mask.
    D. The IP address that is configured on S0/0 of RouterB is not in the correct subnet.
    E. The cable that is connected to S0/0 on RouterA is faulty.
    F. The encapsulation that is configured on S0/0 of RouterB does not match the encapsulation that is configured on S0/0 of RouterA
    Answer: B,F
    Explanation:
    Explanation/Reference:
    Explanation:
    From the output we can see that there is a problem with the Serial 0/0 interface. It is enabled, but the line protocol is down.
    The could be a result of mismatched encapsulation or the interface not receiving a clock signal from the CSU/DSU.

    NEW QUESTION: 2
    If an Ethernet port on a router was assigned an IP address of 172.16.112.1/20, what is the maximum number of hosts allowed on this subnet?
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    C. 2
    D. 3
    E. 4
    Answer: E
    Explanation:
    Explanation/Reference:
    Explanation:
    Each octet represents eight bits. The bits, in turn, represent (from left to right): 128, 64, 32, 16 , 8, 4, 2, 1 Add them up and you get 255. Add one for the all zeros option, and the total is 256. Now take away one of these for the network address (all zeros) and another for the broadcast address (all ones). Each octet represents 254 possible hosts. Or 254 possible networks. Unless you have subnet zero set on your network gear, in which case you could conceivably have 255. The CIDR addressing format (/20) tells us that 20 bits are used for the network portion, so the maximum number of networks are 2

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