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timama [110]
3 years ago
10

A computer’s memory is composed of 8K words of 32 bits each. How many bits are required for memory addressing if the smallest ad

dressable memory unit is a word?
13

8

10

6

32
Computers and Technology
1 answer:
victus00 [196]3 years ago
7 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^{n}  units. Now you have to find the value of n that approximates to the number of words.  

2^2=4\\2^4=16\\2^7=128\\2^{10}=1024\\2^{12}=4096\\2^{13}=8192

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

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2 years ago
Write a program that reads in a temperature value in °F. Create a function ConvertFahrenheit that takes that value as a paramete
larisa86 [58]

Answer:

#include<stdio.h>

void ConvertFahrenheit(float);

void main()

{

float fahrenheit_temp;

printf("Input the temperature in Fahrenheit: ");

 

scanf("%f", &fahrenheit_temp);

 

ConvertFahrenheit(fahrenheit_temp);

}

void ConvertFahrenheit(float fahren) {

float c, k;

 

c = (fahren - 32)/1.8;

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printf("Celsius = %f\n", c);

printf("Kelvin = %f", k);

}

Explanation:

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Answer:

Explanation:

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;););););););););););););););)

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Answer:

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Explanation:

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Explanation:

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