log2 200 = ln 200 ÷ ln 2 = 7.6... → need 8 bits. If a signed number is being stored, then 9 bits would be needed as one would be needed to indicate the sign of the number.
5
110 = 00012 110 - This is the number one writen in the decimal system 00012 - This is the number 1 using the binary system. Here, 4 bits are being represented. 00012 = 012
28-bits
As quoted from Google Books, "Word size refers to the number of bits that a microprocessor can manipulate at one time."
Four bits are required to write '12' as a binary number.(12)10 = ( 1 1 0 0 )2
11 bits. 211 = 2048
log2 200 = ln 200 ÷ ln 2 = 7.6... → need 8 bits. If a signed number is being stored, then 9 bits would be needed as one would be needed to indicate the sign of the number.
5
You would need at least 9 bits to borrow. Since 8 bits gives only 255 the additional bit will get you 256. Adding 256 + 128 gives you at least 384 subnets or hosts.
There are a number of different depths for a number of different bits. The depth needed depends on the project.
24 bits are needed for the program counter. Assuming the instructions are 32 bits, then 32 bits are needed for the instruction register.
18 in binary is 10010 Since 18 can't be written in term of 2 to the power x, the number of bits needed is 5. The answer is 5
45 in binary is 101101, so you need at least 6 bits to represent 45 characters.
BY USING FORMULA (M+R+1)<=2r 011110110011001110101 ---- The formula d + p + 1 <= 2^p (where d is the number of data bits and p is the number of check bits) indicates that we need at least 5 check bits in order to correct single-bit errors in blocks of 16 data bits -- a (21,16) code. SECDED requires 6 check bits for blocks of 16 data bits.
Packing a lot of meaning into a small space
8 bits if unsigned, 9 bits if signed