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a. 255.255.255.0

b. 255.0.0.0

c. 255.255.224.0

d. 255.255.240.0

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We first change the mask to binary to find the number of 1’s:

a. 11111111 11111111 11111111 00000000/24

b. 11111111 00000000 00000000 00000000 /8

c. 11111111 11111111 11100000 00000000 /19

d. 11111111 11111111 11110000 00000000 /20

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a. The mask 255.255.255.0 has 24 consecutive 1s, so slash notation results :/24
b. The mask 255.0.0.0 has 8 consecutive 1s, so slash notation results :/8
c. The mask 255.255.224.0 has 19 consecutive 1s, so slash notation results :/19
d. The mask 255.255.240.0 has 20 consecutive 1s, so slash notation results :/20

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