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How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes?
- a)12
- b)14
- c)16
- d)18
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
How many bits would be required to encode decimal numbers 0 to 9999 in...
Total number of decimals to be represented = 10000 = 104 = 2n (where n is the number of bits required) = 213.29. Therefore, the number of bits required for straight binary encoding = 14.
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How many bits would be required to encode decimal numbers 0 to 9999 in...
To determine the number of bits required to encode decimal numbers from 0 to 9999 in straight binary codes, we need to consider the range of values and calculate the number of bits needed to represent the largest value in that range.
Range of Values:
The range of decimal numbers is from 0 to 9999.
Calculation:
To calculate the number of bits required to represent a decimal number in binary, we use the formula log2(n), where n is the decimal number.
For the largest number in the range, which is 9999, we calculate the number of bits required as follows:
Number of bits = log2(9999)
Using the logarithmic identity log2(x) = log10(x) / log10(2), we can rewrite the calculation as:
Number of bits = log10(9999) / log10(2)
Using a scientific calculator or logarithmic tables, we find:
Number of bits ≈ 13.29
Answer:
The number of bits required to encode decimal numbers 0 to 9999 in straight binary codes is approximately 13.29. Since we cannot have fractional bits, we round up to the nearest whole number, which is 14.
Therefore, the correct answer is option 'B' - 14.
Range of Values:
The range of decimal numbers is from 0 to 9999.
Calculation:
To calculate the number of bits required to represent a decimal number in binary, we use the formula log2(n), where n is the decimal number.
For the largest number in the range, which is 9999, we calculate the number of bits required as follows:
Number of bits = log2(9999)
Using the logarithmic identity log2(x) = log10(x) / log10(2), we can rewrite the calculation as:
Number of bits = log10(9999) / log10(2)
Using a scientific calculator or logarithmic tables, we find:
Number of bits ≈ 13.29
Answer:
The number of bits required to encode decimal numbers 0 to 9999 in straight binary codes is approximately 13.29. Since we cannot have fractional bits, we round up to the nearest whole number, which is 14.
Therefore, the correct answer is option 'B' - 14.
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How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes?a)12b)14c)16d)18Correct answer is option 'B'. Can you explain this answer?
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