Two: '0' or '1'
9
Way to many (:
256 (162)
For signed 32 bit values: 2^31-1 = 0x7FFFFFFF = 2,147,483,647 For unsigned 32 bit values: 2^32-5 = 0xFFFFFFFB = 4,294,967,291
0 o 1
Two: '0' or '1'
24, or 16 (0 through 15) One binary digit (bit) can have 21 values (0 or 1). Two bits can have 22 values. Three bits can have 23 values. A five-bit number can have 25 values... and so on...
A 128-bit register can store 2 128th (over 3.40 × 10 38th) different values. The range of integer values that can be stored in 128 bits depends on the integer representation used.
A 4-bit sound allows for 2^4 = 16 levels of amplitude. This means that the sound can represent 16 different discrete values of amplitude.
1. A single bit can represent two different values, 0 and 1. Then simply take the largest of those two possible values, 1, and that's your answer.
An 8-bit sound allows for 256 different levels of amplitude values. This means that the amplitude range can be divided into 256 discrete steps, providing a level of precision in representing sound intensity.
It can have 0 to 1 It can have 0 to 1
65,536
9
Since there are 8 bits in between, and they can assume any of the two values (0 or 1), that results in a total of 28 different combinations.
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