A) - 2G = 2 * 2^30 = 2^31 words therefore you need 31 bits to specify a memory location( 31 address line).
<span>- If your word has 32 bits then you need 32 data lines. </span>
<span>- The # of bytes is (2 * 2^30 * 2^5 )/ 8 = (2^36) / 8 = 2^36 / 2^3 = 2^33 bytes </span>
Answer:
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