The minimum number of bits the address bus must have in the byte addressable memory system is 12
<h3>How to determine the minimum number of bits?</h3>
The memory size is given as;
Size = 4 KiB
Convert to bytes
Size = 4 * 2^10 bytes
Express 4 as 2^2
Size = 2^2 * 2^10 bytes
Multiply
Size = 2^12 bytes
The minimum number of bits is then calculated as;
2^n = Size
This gives
2^n = 2^12
Cancel out the base (2)
n = 12
Hence, the minimum number of bits is 12
Read more about bits and bytes at:
brainly.com/question/20594719
The least common multiple (LCM) of 78, 90, and 140 is: 16,380
78 × 210 = 16,380
90 × 182 = 16,380
140 × 117 = 16,380
Answer:
101011100
Explanation:
its just putting them next each other with out spacing.
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Answer:
Im sure its called Single User OS