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QUESTION 1 8 Given the CS: 1 P of 2 5 0 0 : 9 5 F 3 , the physical address will be: a

QUESTION 18
Given the CS:1P of 2500:95F3, the physical address will be:
a.250095F3.0000
b.2E5F4
c.2 E5F3
d. None of the above
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