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8byte addressability: 8bytes can be stored in each memory location. 3bits to access: so 8 memory locations ( 2^3) total size = 8 * 8bytes( each memory location contains) = 64 bytes any queries or discussions, feel free to contact me at: chintala.ganesh@wipro.com

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15y ago

Addressable memory = 2^n bytes where n = # of bits per address 2^4 = 16bytes of addressable memory

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Q: If a particular computer has 8 byte addressability and a 4 bit address space how many bytes of memory does that computer have?
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