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2 years ago

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

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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… = 1 + i + i ² (1 + i + i ² (1 + i ))

… = 1 + i + (-1) (1 + i + (-1) (1 + i ))

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Alternatively, notice that

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1 + i + i ² + i ³ + i ⁴ + i ⁵ = (1 - i ⁶) / (1 - i )

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… = 2 / (1 - i )

Multiply the numerator and denominator by the conjugate of 1 - i :

… = 2 (1 + i ) / ((1 - i ) (1 + i ))

… = 2 (1 + i ) / (1² - i ²)

… = 2 (1 + i ) / (1 + 1)

… = 1 + i

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