Answer:
The minimum number of bits necessary to address 8K words is 13.
Explanation:
You have the number of words to address that is 8000 words, a word is the smallest addressable memory unit.
8000 words can be addressed with [tex]2^{n}[/tex] units. Now you have to find the value of n that approximates to the number of words.
[tex]2^2=4\\2^4=16\\2^7=128\\2^{10}=1024\\2^{12}=4096\\2^{13}=8192[/tex]
So you can see that 13 bits are needed to address 8K words.
