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USPshnik [31]
2 years ago
13

A computer’s memory is composed of 8k words of 32 bits each. How many bytes does this memory contain?.

SAT
2 answers:
Mandarinka [93]2 years ago
6 0

Answer:

32 000 bytes

Explanation:

algol132 years ago
5 0

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 2 units. Now you have to find the value of n that approximates to the number of words.

2² = 4

24

16

27 = 128

210 1024

212 = 4096

213

8192

So you can see that 13 bits are needed to address 8K words.

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The system of equation that represents this situation is y = 20x + 15 for day and night kernel and y = 30x + 5 for bark time hotel.

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The system of equation that represents this situation is y = 20x + 15 for day and night kernel and y = 30x + 5 for bark time hotel.

Find out more at: brainly.com/question/21835898

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