http://extraconversion.com/data-storage/kilobytes/kilobytes-to-words.html WebMay 31, 2024 · Using address multiplexing where the address lines are used by the row and column selector of a 3-dimensional memory array with the third dimension being 8 bits, how many address lines are needed for a memory with a bit capacity of 524288 bits? My approach: 524288/8 = 65536. log_2(65536) = 16. Is is 16?
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WebOct 18, 2024 · How many bits are needed to address each location in memory? I know that 1 kb = 1024 bytes, so 64kb = 65536 bytes = 2^16 bytes 2^16 bytes / 8 bytes * 2^3 bits = 2^16 … WebSo, to convert 2 kilobyte (s) to bytes we multiply this quantity by 8000 then divide it by 8. This is the so called 'CONVERSION FACTOR' which, here, is equal to 1000. In this case, to … rawls constructivism
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WebSolution: Converting from kilobyte per day to bit per day is very easy. We know that 1 kB/day = 8000 bit/day. So, to convert 13.2 kB/day to bit/day, multiply 13.2 kB/day by 8000 bit/day. 13.2 kB/day = 13.2 × 8000 bit/day 13.2 kB/day = 105600 bit/day Therefore, 13.2 kilobytes per day converted to bit per day is equal to 105600 bit/day. WebBit Calculator - Convert between bits/bytes/kilobits/kilobytes/megabits/megabytes/gigabits/gigabytes. Enter a number and choose the type of Units WebApr 30, 2016 · The logical address is split up in 3 levels of page tables plus the offset. Due the page table entry size is 8 byte (2^6 = 64 bit), 6 bits of the logical address are used for each stage to address it. The offset will have the size of 30 bits. Each page stage can address 64 bit plus the 30 bits offset. So does this result in the page count? rawls college of business texas tech