# How Many Bits Are Needed In The Mar?

The MAR would be 20 bits long, since that is the size of a single address, 20 bits. 10 bits are sent to the row decoder, and 10 bits to the column decoder. Each decoder would have 2 10 output lines. A decoder has n input lines and 2 n output lines.

## How many bits does a Mar address have?

The MAR would be 20 bits long, since that is the size of a single address, 20 bits. 10 bits are sent to the row decoder, and 10 bits to the column decoder. Each decoder would have 2 10 output lines.

## How many bits are needed for 100 million bytes?

For 100 million bytes, we need 27 bits. 2 26 = 67,108,864 bytes, not enough for 100 million. 2 27 = 134,217,728 and this is enough. d. 1 billion bytes is equal to 1000 times 1 million or 1000 * 2 20 . In base 2 notation, it is (2 20 * 2 10) = 2 30 = 1,073,741,824. So 30 bits are needed to represent 1 billion bytes.

## How many bits are needed for one instruction?

The number of bits necessary for each operation (i.e. one instruction) is (6+18+18)/8 + 1 if there is a remainder. 42 bits / (8 bits per bytes) = 5.25. This rounds up and 6 bytes are necessary per instruction. For b, express the answer in the number of Kilobytes.

## How many bits does 29 have in a binary?

For example: 1 29 has 5 bits because 16 ≤ 29 ≤ 31, or 2 4 ≤ 29 ≤ 2 5 – 1 2 123 has 7 bits because 64 ≤ 123 ≤ 127, or 2 6 ≤ 123 ≤ 2 7 – 1 3 967 has 10 bits because 512 ≤ 967 ≤ 1023, or 2 9 ≤ 967 ≤ 2 10 – 1

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