Agreed. This was a fairly quick one as no ambiguity in the possible answers.2^9=512, so 9 bits must be left out of the possible 16 for hosts (address is a class B with a /16 mask) which leaves 7 bits for networks. /16 plus 7 = /23 /24 is 255 . 255 . 255 . 0 1 bit less (which also only has a weight of1) makes 255 . 255 . 254 . 0Only 1 answer has 255 . 255 . 254 . 0 in the answer anywhere. Didn't ...
Agreed. 6 of 8 available bits needed to create 64 subnets(2^6 = 64). This leaves 2 bitsfor hosts which would only allow for 4 ip addresses(2^2 =4). Cant use first or last, so 2 hosts and 64 networks (assuming you can use the zero networks)
Agreed. 9 bits needed for 512 subnet (2^9=512). 16 bits available - 9 leaves 7 bits for hosts. 2^7=128Class B has a mask of /16, plus the 9 bits required for networks makes /25 or 255.255.255.128
Agreed.2^5=32. Means we need to LEAVE 5 bits at least out of the possible 8 for hosts. This leaves 3 for network. 2^3=8networks. 32 addresses per network.First network .0 to .31 first useable .1Second network .32 to .63 first useable .33