255.255.248.0
the subnetmask is 255.255.248.0
140.128.0.0 or 10001100 11111000 00000000 00000000
140.128.0.0 (10001100 10000000 00000000 00000000).
11111000 (from the Windows Calculator)
Surprise: it is -10002. (If you wanted to ask 1000(2), then it is 11111000(2))
One jk flip-flop with j=k=1 should be added to the system so that it's modulus becomes 16 instead of 8.
it depends , wether it is igned or unsigned number 2^7 +2^6 +2^5 + 2^4 + 2^3 =248 0r (-2^7) + 2^6 +2^5 + 2^4 + 2^3 = -8
1
Let's break this into binary first:11111111 11111111 11110000 00000000For a class b network, there are 16 bits available (the right 16 numbers). We see that 4 of those have been borrowed already (they have 1's) so we have 12 left.The formula to find the maximum number of hosts is 212 - 2 (the 12 comes from the number above).So, the answer is 4096.----------------------------------------------------------------------------------------------------------------Great binary breakdown explanation!!!However you forgot to subtract 2 decimals to fully apply the formula.(1 Network address, and 1 Broadcast range).So the correct answer is 4094