A. It corresponds to a Class A address with 13 bits borrowed.
B. It corresponds to a Class B address with 4 bits borrowed.
C. The network address of the last subnet will have 248 in the 3rd octet.
D. The first 21 bits make the host portion of the address.
E. This subnet mask allows for 16 total subnets to be created.
F. The subnetwork numbers will be in multiples of 8.
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A.14
B.16
C.30
D.32
E.62
F.64
You are the network administrator at TestKing. TestKing has been provided withthe network address 165.100.27.0/24. The TestKing CEO wants to know how many subnetworks this address provides, and how many hosts can be supported on each subnet.
What would your reply be?()
A. One network with 254 hosts.
B. 254 networks with 254 hosts per network.
C. 65,534 networks with 255 hosts per network.
D. 30 networks with 64 hosts per network.
E. 254 networks with 65,534 per network.
A.15.234.118.63
B.83.121.178.93
C.134.178.18.56
D.192.168.19.37
E.201.45.116.159
F.217.63.12.192
A. Decimal=160, hexadecimal=00
B. Decimal=170, hexadecimal=AA
C. Decimal=180, hexadecimal=BB
D. Decimal=190, hexadecimal=CC
A./38
B./30
C./27
D./23
E./18
F./32
A. 255.255.255.0
B. 255.255.255.240
C. 255.255.255.248
D. 255.255.255.252
E. 255.255.255.254
A. Deleting unusable addresses through the creation of many subnets.
B. Combining routes to multiple networks into one supernet.
C. Reclaiming unused space by means of changing the subnet size.
D. Calculating the available host addresses in the AS.
A. Static NAT
B. Port loading
C. NAT Overloading
D. Dynamic NAT
E. None of the above
A.172.16.42.0
B.172.16.107.0
C.172.16.208.0
D.172.16.252.0
E.172.16.254.0
A.255.255.255.0
B.255.255.255.128
C.255.255.252.0
D.255.255.255.224
E.255.255.255.192
F.255.255.248.0
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