Asked by Nguyen
A firm is assigned the network part 128.171. It selects a 8-bit subnet part. a) Draw the bits for the four octets of the IP address of the first host on the first subnet. (Hint: Use Windows Calculator.)
b) Convert this answer into dotted decimal notation.
c) Draw the bits for the second host on the third subnet. (In binary, 2 is 10, while 3 is 11.)
d) Convert this into dotted decimal notation.
e) Draw the bits for the last host on the third subnet.
f) Convert this answer into dotted decimal notation.
b) Convert this answer into dotted decimal notation.
c) Draw the bits for the second host on the third subnet. (In binary, 2 is 10, while 3 is 11.)
d) Convert this into dotted decimal notation.
e) Draw the bits for the last host on the third subnet.
f) Convert this answer into dotted decimal notation.
Answers
Answered by
PEX-TUIKE
Remember for any type of address the first address 00000000(0) and last address 11111111(255) are not used, therefore first host address begins with 1 and last usuable host is 254 same goes for the subnet address.
Solution:
a)10000000.10101011.00000001.00000001
b)128.171.1.1
c)10000000.10101011.00000011.00000010
d)128.171.3.2
e)10000000.10101011.00000011.11111110
f)128.171.3.254
Solution:
a)10000000.10101011.00000001.00000001
b)128.171.1.1
c)10000000.10101011.00000011.00000010
d)128.171.3.2
e)10000000.10101011.00000011.11111110
f)128.171.3.254
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