Which wildcard mask would be used to match all the addresses that exist within the 8.47.26.0/19 network?

A) 0.0.15.255
B) 0.0.0.15
C) 0.0.31.255
D) 0.0.0.31


C
Explanation: C) The wildcard mask required is based on the subnet mask given. In this case, the subnet mask was 255.255.224.0; the wildcard equivalent is 0.0.31.255.
The following is the math for the last two octets:
224 = 11100000
31 = 00011111
255 = 11111111
0 = 00000000
The wildcard is just the inverse of the subnet mask.

Computer Science & Information Technology

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