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Aleonysh [2.5K]
2 years ago
13

How to write a program that prompts the user to input two POSITIVE numbers — a dividend (numerator) and a divisor (denominator).

Your program should then divide the numerator by the denominator, and display the quotient followed by the remainder in python.
Computers and Technology
1 answer:
fomenos2 years ago
3 0

Answer:

num1 = int(input("Numerator: "))

num2 = int(input("Denominator: "))

if num1 < 1 or num2<1:

     print("Input must be greater than 1")

else:

     print("Quotient: "+str(num1//num2))

     print("Remainder: "+str(num1%num2))

Explanation

The next two lines prompts the user for two numbers

<em>num1 = int(input("Numerator: "))</em>

<em>num2 = int(input("Denominator: "))</em>

The following if statement checks if one or both of the inputs is not positive

<em>if num1 < 1 or num2<1:</em>

<em>      print("Input must be greater than 1")-> If yes, the print statement is executed</em>

If otherwise, the quotient and remainder is printed

<em>else:</em>

<em>      print("Quotient: "+str(num1//num2))</em>

<em>      print("Remainder: "+str(num1%num2))</em>

<em />

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Answer:

The code is given below

Explanation:

The correct syntax would be to place appropriate parenthesis.

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Similarly, you can also use the following code:

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Given sphereRadius and piVal, compute the volume of a sphere and assign to sphereVolume. Look up the equation online (e.g., http
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Answer:

  1. #include <stdio.h>
  2. int main()
  3. {
  4.    const double piVal = 3.14159;
  5.    double sphereVolume = 0.0;
  6.    double sphereRadius = 0.0;
  7.    
  8.    sphereRadius = 1.0;
  9.    sphereVolume = 4.0/ 3.0 * piVal * sphereRadius * sphereRadius * sphereRadius;
  10.    
  11.    
  12.    printf("Sphere volume: %lf\n", sphereVolume);
  13.    return 0;
  14. }

Explanation:

Firstly we can identify the formula to calculate volume of sphere which is

Volume = 4/3 \pi r^{3}

With this formula in mind, we can apply this formula to calculate the volume of sphere in Line 10. This is important to perform floating-point division 4.0/3.0 to ensure the resulting value is a floating value as well. Since we have been given piVal and sphereRadius, we can just multiply the result of floating-point division with piVal and sphereRadius and get the sphereVolume value.

At last, display the sphere volume using printf method (Line 13).

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The question is incomplete. It can be found in search engines. However, kindly find the complete question below:

Question

Cites as a pitfall the utilization of a subset of the performance equation as a performance metric. To illustrate this, consider the following two processors. P1 has a clock rate of 4 GHz, average CPI of 0.9, and requires the execution of 5.0E9 instructions. P2 has a clock rate of 3 GHz, an average CPI of 0.75, and requires the execution of 1.0E9 instructions. 1. One usual fallacy is to consider the computer with the largest clock rate as having the largest performance. Check if this is true for P1 and P2. 2. Another fallacy is to consider that the processor executing the largest number of instructions will need a larger CPU time. Considering that processor P1 is executing a sequence of 1.0E9 instructions and that the CPI of processors P1 and P2 do not change, determine the number of instructions that P2 can execute in the same time that P1 needs to execute 1.0E9 instructions. 3. A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, and consider that the processor with the largest MIPS has the largest performance. Check if this is true for P1 and P2. 4. Another common performance figure is MFLOPS (millions of floating-point operations per second), defined as MFLOPS = No. FP operations / (execution time x 1E6) but this figure has the same problems as MIPS. Assume that 40% of the instructions executed on both P1 and P2 are floating-point instructions. Find the MFLOPS figures for the programs.

Answer:

(1) We will use the formula:

                                       CPU time = number of instructions x CPI / Clock rate

So, using the 1 Ghz = 10⁹ Hz, we get that

CPU time₁ = 5 x 10⁹ x 0.9 / 4 Gh

                    = 4.5 x 10⁹ / 4 x 10⁹Hz = 1.125 s

and,

CPU time₂ = 1 x  10⁹ x 0.75 / 3 Ghz

                  = 0.75 x 10⁹ / 3 x 10⁹ Hz = 0.25 s

So, P2 is actually a lot faster than P1 since CPU₂ is less than CPU₁

(2)

     Find the CPU time of P1 using (*)

CPU time₁ = 10⁹ x 0.9 / 4 Ghz

                = 0.9 x 10⁹ / 4 x 10⁹ Hz = 0.225 s

So, we need to find the number of instructions₂ such that  CPU time₂ = 0.225 s. This means that using (*) along with clock rate₂ = 3 Ghz and CPI₂ = 0.75

Therefore,   numbers of instruction₂ x 0.75 / 3 Ghz = 0.225 s

Hence, numbers of instructions₂ = 0.225 x 3 x  10⁹ / 0.75  = 9 x 10⁸

So, P1 can process more instructions than P2 in the same period of time.

(3)

We recall  that:

MIPS = Clock rate / CPI X 10⁶

  So, MIPS₁ = 4GHZ / 0.9 X 10⁶ = 4 X 10⁹HZ / 0.9 X 10⁶ = 4444

        MIPS₂ = 3GHZ / 0.75 X 10⁶ = 3 x 10⁹ / 0.75 X 10⁶ = 4000

So, P1 has the bigger MIPS

(4)

  We now recall that:

MFLOPS = FLOPS Instructions / time x 10⁶

              = 0.4 x instructions / time x 10⁶ = 0.4 MIPS

Therefore,

                  MFLOPS₁ = 1777.6

                  MFLOPS₂ = 1600

Again, P1 has the bigger MFLOPS

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