<span>C. out one level. i hope this helps</span>
85 C.E. I think that is when it was written, but I’m not 100% sure.
Answer:
see explaination
Explanation:
If the cache contains 2k blocks, then the data at memory address i would go to cache block index ( i mod 2k )
Memory address = 5 bit
Cache block = 23
So, if the memory location is 9, then binary address = 01001 and cache block = 9 % 8 = 1 = 001
if the memory location is 12, then binary address = 01100 and cache block = 12 % 8 = 4 = 100
if the memory location is 15, then binary address = 01111 and cache block = 15 % 8 = 7 = 111
