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SVETLANKA909090 [29]
3 years ago
8

Write the definition of a function that evaluates three double numbers and returns true if the floor of the product of the first

two numbers equals the floor of the third number; otherwise it returns false.
Computers and Technology
1 answer:
Mama L [17]3 years ago
4 0

Answer:

The function is written in C++

bool products(double num1, double num2, double num3) {

if(floor(num1 * num2) == floor(num3)){

    return true;

}

else

{

    return false;

}

}

Explanation:

I answered this question using C++ programming language.

This line defines the function with three parameters; num1 to num3

bool products(double num1, double num2, double num3) {

This checks if the floor of num1 * num2 equals floor of num3

if(floor(num1 * num2) == floor(num3)){

If yes, this returns true

    return true;

}

else {

If otherwise, this returns true

    return false;

}

The method ends here

}

<em>I added the full program as an attachment that include the main method</em>

Download cpp
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Answer:

def power(base, expo):

   if expo == 0:

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Explanation:

*The code is in Python.

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2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of
allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

=  n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0
3 years ago
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