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
(9+8+2)/(4+9+8+2)=19/23=82%
Therefore very likely
44x10^-2
or in other words
44 times 10 to the -2nd power
Answer:
Domain= 2,3,4,6
Range=4,7,9,11
or
Domain [2,6]
Range [4,11]
Step-by-step explanation:
