All categories

3 years ago

QUESTION 16

Computers and Technology

1 answer:

balandron [24]3 years ago

4 0

Answer:

Report Wizard

Explanation:

Report Wizard is a tool that helps users quickly and easily create reports with charts, tables and drill-down capabilities

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Borrowing money affects both assests and owners equity. True or false?

shtirl [24]

It should be true, hopefully I’m right

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3 years ago

Explain why using infrared telescopes would not be useful for stargazing in the city?

Alexeev081 [22]

Because infrared telescopes use infrared aka anything above absolute zero (0 degrees Celsius or 32 degrees Fahrenheit) meaning that the telescope would not only pick up the stars but also all of the people and lights so on and so forth

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3 years ago

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

Given an int variable n that has been initialized to a positive value and, in addition, int variables k and total that have alre

Fudgin [204]

Answer:

The c++ program to implement the while loop is given below.

#include <iostream>

using namespace std;

int main() {

// declaration of integer variables

int k, n, total;

// initialization of integer variables

k=1, n=4, total=0;

// loop executed till value of k becomes equal to value of n

while( k <= n ){

// cube of every integer is added to the variable total

total = total + ( k * k * k );

// value of k is incremented to go to the next number

k = k + 1 ;

}

return 0;

}

Explanation:

The program begins with the declaration of integer variables.

int k, n, total;

This is followed by initialization of these variables.

k=1, n=4, total=0;

The while loop runs over the variable k which is initialized to 1. The loop runs till value of k reaches the value of integer n.

First, cube of k is computed and added to the variable total.

After first execution of the loop, total is initialized to the cube of 1.

Next, value of variable k is incremented by 1 so that k is initialized to next integer.

After first execution of the loop, k is incremented from 1 to 2.

while( k <= n )

{

total = total + ( k * k * k );

k = k + 1 ;

}

When the value of k reaches the value of integer n, the cube of n is calculated and added to the variable, total.

When k is incremented, it becomes more than n and hence, loop gets terminated.

As the return type of main is int, the program terminates with the statement shown below.

return 0;

No output is displayed as it is not mentioned in the question.

No user input is taken as it is mentioned that integer variables are already initialized.

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3 years ago

How does the zone theory of optical systems resolve the apparent incompatibility of trichromacy and opponency?

lara [203]

Answer:

Money money money

Explanation:

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4 years ago