To express the number 3 in binary, we need to determine the smallest power of 2 that can represent it. The binary representation of 3 is "11," which requires 2 bits (as 2^1 = 2 and 2^0 = 1, combining them gives 2 + 1 = 3). Therefore, the least number of bits needed to express 3 is 2.
The number of bits needed to represent one symbol depends on the total number of unique symbols. The formula to calculate the number of bits required is ( n = \lceil \log_2(S) \rceil ), where ( S ) is the number of unique symbols. For example, to represent 256 unique symbols, 8 bits are needed, since ( \log_2(256) = 8 ).
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
To determine the least number of bits required to distinguish among 12 different choices, you can use the formula (2^n \geq 12), where (n) is the number of bits. The smallest (n) that satisfies this is (n = 4), since (2^4 = 16), which is greater than 12. Therefore, at least 4 bits are required to uniquely identify 12 different options.
To represent an eight-digit decimal number in Binary-Coded Decimal (BCD), each decimal digit is encoded using 4 bits. Since there are 8 digits in the number, the total number of bits required is 8 digits × 4 bits/digit = 32 bits. Therefore, 32 bits are needed to represent an eight-digit decimal number in BCD.
Four bits are required to write '12' as a binary number.(12)10 = ( 1 1 0 0 )2
The number of bits needed to represent one symbol depends on the total number of unique symbols. The formula to calculate the number of bits required is ( n = \lceil \log_2(S) \rceil ), where ( S ) is the number of unique symbols. For example, to represent 256 unique symbols, 8 bits are needed, since ( \log_2(256) = 8 ).
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.
To determine the number of check bits needed for a Hamming code to correct single-bit errors in a 64-bit data word, we use the formula (2^r \geq m + r + 1), where (m) is the number of data bits and (r) is the number of check bits. In this case, (m = 64). Solving the inequality, we find that (r) must be at least 7, since (2^7 = 128) is the smallest power of 2 that satisfies (64 + 7 + 1 = 72). Thus, 7 check bits are needed.
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.
5
The number of bits required for data transfer depends on the size of the data being transmitted and the protocol used for communication. For example, transferring a single character typically requires 8 bits (1 byte). However, when considering overhead from headers and error-checking mechanisms in protocols, the actual number of bits needed can be significantly higher. Therefore, the minimum number of bits required for data transfer varies based on the specific scenario and requirements.
There are a number of different depths for a number of different bits. The depth needed depends on the project.
To determine the least number of bits required to distinguish among 12 different choices, you can use the formula (2^n \geq 12), where (n) is the number of bits. The smallest (n) that satisfies this is (n = 4), since (2^4 = 16), which is greater than 12. Therefore, at least 4 bits are required to uniquely identify 12 different options.
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