Two: '0' or '1'
Way to many (:
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
Infinitely many. If you take a ball of dough, for example, and gently push a bit in, you will have a different shape. Do it again, another shape. And so on.Infinitely many. If you take a ball of dough, for example, and gently push a bit in, you will have a different shape. Do it again, another shape. And so on.Infinitely many. If you take a ball of dough, for example, and gently push a bit in, you will have a different shape. Do it again, another shape. And so on.Infinitely many. If you take a ball of dough, for example, and gently push a bit in, you will have a different shape. Do it again, another shape. And so on.
The answer depends on division of WHAT. And since you have not bothered to share that crucial bit of information, I cannot provide a more useful answer.
2 to the power 16.
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.
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.
0 o 1
210 = 1024, so there are 1024 different bit configurations in a 10-bit code.
4, which is equal to 2 to the power 2.In general, with "n" bits, you can have "2 to the power n" different states (or represent that many different numbers).
4.1 bit for 2,2 bits for 4,3 bits for 8,4 bits for 16.
Neither of the following are true about 1 bit, it can not represent decimal values 0 and 9 nor can it be used to represent one character in the lowercase English alphabet and one binary digit four binary. A true statement would be that 1 bit is represented by the decimal values 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...
Binary PSKQPSK.1. Two different phases are used torepresent two binary values.1. Four different phases are used to represent two binary values.2. Each signal element represents only one bit.2. Each signal element representstwo bits
We need signed integers in order to represent both negative and positive values. However, some numbers can never be negative. For instance, the size of a file must always be greater than or equal to zero so we use unsigned integers to represent file sizes. Also, natural numbers must be greater than 0 so there's no point in using a signed value to represent a natural number. Signed integers also use one bit to denote the sign, but unsigned integers do not thus unsigned integers can effectively represent twice the range of positive values than an unsigned integer can. For instance, an 8-bit signed value can represent values in the range -128 to +127 using twos complement notation, but an 8-bit signed value can represent values in the range 0 to 255.
A "bit" is the most basic unit of digital information. It is a binary digit, representing either a 0 or a 1. A bit can have only two possible values, and it is used to represent the presence or absence of a certain property or condition. A string of bits is used to represent larger units of data, such as characters, numbers, and images.
No and Yes... No because as such the two numbers are different and convention will define what the bit pattern should represent. However, with 16 bits, the UNSIGNED bit pattern for 32768 is the same as the SIGNED pattern for -32768 BUT the two numbers are being represented in different ways (ie in different number systems). If the binary representation is a SIGNED number then the top bit will be set if it is negative and with 16 bits the possible range of numbers is -32768 to +32767 and so it is IMPOSSIBLE to represent +32768. With an UNSIGNED number of 16 bits the top bit has no such special meaning and the range possible is 0 to +65535, so it is IMPOSSIBLE to represent -32768. ie it is IMPOSSIBLE with 16 bits to represent -32768 and +32768 in the same number system as they have the same bit pattern, BUT in different number systems the bit pattern can be used to represent the two numbers as the different number systems cannot represent BOTH -32678 and +32768.