2007-12-01 17:22:21 · 5 answers · asked by Jack R 1 in Computers & Internet ➔ Programming & Design
4 bits gives 0 to 15.
2007-12-01 17:28:09 · answer #1 · answered by AnalProgrammer 7 · 1⤊ 0⤋
The previous answer is correct.. I would like to give you a generalized formula for it. We can represent 2^n (2 power n) different values with n bits.
2007-12-01 18:01:01 · answer #2 · answered by Ram S 2 · 2⤊ 0⤋
4 bits shops sixteen numbers, 0 via 15: 0000? ( = 0??) 0001? ( = one million??) 0010? ( = 2??) 0011? ( = 3??) 0100? ( = 4??) 0101? ( = 5??) 0110? ( = 6??) 0111? ( = 7??) one thousand? ( = 8??) 1001? ( = 9??) 1010? ( = 10??) 1011? ( = eleven??) 1100? ( = 12??) 1101? ( = 13??) 1110? ( = 14??) 1111? ( = 15??)
2016-12-30 09:01:27 · answer #3 · answered by ? 3 · 0⤊ 0⤋
the first answer is correct.
0 to 15 is 16 values the 0 counts as one of them.
2007-12-01 19:45:18 · answer #4 · answered by ian 3 · 0⤊ 0⤋
Mathematically both the above answers are correct.
If the values have to be represented electronically you will require at least two more bits depending on the system used.
2007-12-01 19:39:40 · answer #5 · answered by The Rugby Player 7 · 0⤊ 1⤋