How many bits are needed to count up to maximum of 1024? Explain your answer.
Answers
Answered by
0
Answer:
12 Answers
For instance, in i), 3 decimal digits -> 10^3 = 1000 possible numbers so you have to find the lowest power of 2 that is higher than 1000, which in this case is 2^10 = 1024 (10 bits)
Answered by
0
Answer:
To realize Y = CD + EF + G, two AND gates are required and two OR gates are required.
Similar questions