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Daniel Jones NT1210
Lab 1.1 Review 1. Convert the decimal value 127 into binary. Explain the process of conversion that you used. 127 | 127 | 63 | 31 | 15 | 7 | 3 | 1 | 128 | - 64 | - 32 | - 16 | - 8 | - 4 | - 2 | - 1 | | = 63 | = 31 | = 15 | = 7 | = 3 | = 1 | = 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
The answer is: 01111111
If the decimal number is less than the greatest power of 2 than you must put a 0 for that number than carry that same decimal number over to the right one decimal place. For example. 127 is less than 128 so I cannot be subtracted. So you will put a zero and move that number over. Since 127 is greater than 64, you will subtract 64 from 128 and put a 1 for that value. Carry it over to the right. 32 is less than 63 so you will subtract and put a 1 for that value, and continue down the line. 2. Explain why the values 102 and 00102 are the same.
Because when you plug the binary numbers into the 8 bit conversion table, the two zeros before the 10 equal nothing. So 10 and 0010 have the same decimal number.
128 64 32 16 8 4 2 1 128 64 32 16 8 4 2 1 1 0 = 2 0 0 1 0 = 2 3. Based on the breakdown of the binary and decimal systems in this lab, describe the available digit values and the first four digits of a base 5 numbering system. You can use the binary system as a reference, where the available digit values are 0 and 1 and the first four digits are 1, 2, 4, and 8. The available digit values in a base 5 numbering system are 1s and 0s. But we only use 5 bits of the 8 bits on the binary system. For example, you are using 16 8 4 2 1 instead of 128 64 32 16 8 4 2 1. 4. Using the internet and Help files in Excel, explain why creating a converter from decimal to binary would be more difficult

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