8
0.0000001
4 bits. 24 = 16, so you have 16 different combinations.4 bits. 24 = 16, so you have 16 different combinations.4 bits. 24 = 16, so you have 16 different combinations.4 bits. 24 = 16, so you have 16 different combinations.
65,535
At least 8 bits are needed to represent the number 231. 231 ÷ 2 = 115 r 1 115 ÷ 2 = 57 r 1 57 ÷ 2 = 28 r 1 28 ÷ 2 = 14 r 0 14 ÷ 2 = 7 r 0 7 ÷ 2 = 3 r 1 3 ÷ 2 = 1 r 1 1 ÷ 2 = 0 r 1 → 231 is 1110 0111 in binary and has 8 binary digits. Thus 231 can be represented in 8 bits, but if more are provided, eg 16, it can still be represented (in 16 bits it would be 0000 0000 1110 0111, unless there is a binary point, with say 8 bits after it, then 231 would be 1110 0111 . 0000 0000).
5
8 bits if unsigned, 9 bits if signed
how many bits are needed to represent decimal values ranging from 0 to 12,500?
9 bits
1200
8
8
103
Four bytes represent 32 bits. 32 bits represent 4,294,967,296 possibilities.
17 bits would allow a value up to 131071.
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.
6 bits 111110 = 32 + 16 + 8 + 4 + 2 + 0