Computer-Networks
August 30, 2024Computer-Networks
August 30, 2024Computer-Networks
| Question 831 |
How many bits are required to encode all twenty six letters, ten symbols, and ten numerals?
| 5 | |
| 6 | |
| 7 | |
| 46 |
Question 831 Explanation:
● It is 26 letters ×2 (Capital and small letters) plus twenty characters in total, so the total is 52+20=72.
● To find how many bits you would need, you would need to find the power of 2 that is equal to or the one immediately larger than your number. The number of bits would be the exponent of the power.
● In our case, since 128=2 7 >72>64=2 6 , the power immediately larger than 72 would be 128=27, so you would need 7 bits, which can codify 128 symbols.
● If you don’t want to codify both capital letters and small letters but only one family of 26 letters and be done with it, then you have 26+20=46 symbols. This time 2 6 =64>46>32=2 5 , so the power immediately larger than 46 would be 64=2 6 ; so 6 bits needed .
● To find how many bits you would need, you would need to find the power of 2 that is equal to or the one immediately larger than your number. The number of bits would be the exponent of the power.
● In our case, since 128=2 7 >72>64=2 6 , the power immediately larger than 72 would be 128=27, so you would need 7 bits, which can codify 128 symbols.
● If you don’t want to codify both capital letters and small letters but only one family of 26 letters and be done with it, then you have 26+20=46 symbols. This time 2 6 =64>46>32=2 5 , so the power immediately larger than 46 would be 64=2 6 ; so 6 bits needed .
Correct Answer: B
Question 831 Explanation:
● It is 26 letters ×2 (Capital and small letters) plus twenty characters in total, so the total is 52+20=72.
● To find how many bits you would need, you would need to find the power of 2 that is equal to or the one immediately larger than your number. The number of bits would be the exponent of the power.
● In our case, since 128=2 7 >72>64=2 6 , the power immediately larger than 72 would be 128=27, so you would need 7 bits, which can codify 128 symbols.
● If you don’t want to codify both capital letters and small letters but only one family of 26 letters and be done with it, then you have 26+20=46 symbols. This time 2 6 =64>46>32=2 5 , so the power immediately larger than 46 would be 64=2 6 ; so 6 bits needed .
● To find how many bits you would need, you would need to find the power of 2 that is equal to or the one immediately larger than your number. The number of bits would be the exponent of the power.
● In our case, since 128=2 7 >72>64=2 6 , the power immediately larger than 72 would be 128=27, so you would need 7 bits, which can codify 128 symbols.
● If you don’t want to codify both capital letters and small letters but only one family of 26 letters and be done with it, then you have 26+20=46 symbols. This time 2 6 =64>46>32=2 5 , so the power immediately larger than 46 would be 64=2 6 ; so 6 bits needed .
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