Considering that the addresses of memory locations are specified in hexadecimal.
a) The number of memory locations in a memory address range ( 0000₁₆ to FFFF₁₆ ) = 65536 memory locations
b) The range of hex addresses in a microcomputer with 4096 memory locations is ; 4095
<u>applying the given data </u>:
a) first step : convert FFFF₁₆ to decimal ( note F₁₆ = 15 decimal )
( F * 16^3 ) + ( F * 16^2 ) + ( F * 16^1 ) + ( F * 16^0 )
= ( 15 * 16^3 ) + ( 15 * 16^2 ) + ( 15 * 16^1 ) + ( 15 * 1 )
= 61440 + 3840 + 240 + 15 = 65535
∴ the memory locations from 0000₁₆ to FFFF₁₆ = from 0 to 65535 = 65536 locations
b) The range of hex addresses with a memory location of 4096
= 0000₁₆ to FFFF₁₆ = 0 to 4096
∴ the range = 4095
Hence we can conclude that the memory locations in ( a ) = 65536 while the range of hex addresses with a memory location of 4096 = 4095.
Learn more : brainly.com/question/18993173
Answer:
Zahir.
Step-by-step explanation:
Zahir has 98p more money than Korede.
Answer:
17 units
Step-by-step explanation:
add 8 to get to 0.
and 9 more to get to point 9.
add 8 and 9 to get the answer of 17.
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