The data components of a class that belong to every instantiated object are the class's ___

Assume the availability of class named IMath that provides a static method, toThePowerOf which accepts two int arguments and ret

geniusboy [140]

Answer:

cubeVolume = IMath.toThePowerOf(cubeSide, 3);

Explanation:

Following is the explanation for above statement:

Left side:

cubeVolume is a variable with data-type int, it will store the integer value that is the output from right side.

Right side:

IMath is the class name.
toThePowerOf is the built-in function that takes two arguments of data type int. First is the base and second is the power(exponent) separated by comma. In place of first argument that is the base variable we will pass the variable cubeSide that has been declared and initialize.
Now the output will be stored in the variable cubeVolume.

i hope it will help you!

4 0

2 years ago

Write a second constructor as indicated. Sample output: User1: Minutes: 0, Messages: 0 User2: Minutes: 1000, Messages: 5000

Mashcka [7]

Complete Question:

Write a second constructor as indicated. Sample output:User1: Minutes: 0, Messages: 0User2: Minutes: 1000, Messages: 5000// ===== Code from file PhonePlan.java =====public class PhonePlan { private int freeMinutes; private int freeMessages; public PhonePlan() { freeMinutes = 0; freeMessages = 0; } // FIXME: Create a second constructor with numMinutes and numMessages parameters. /* Your solution goes here */ public void print() { System.out.println("Minutes: " + freeMinutes + ", Messages: " + freeMessages); return; }}

Answer:

The second constructor is given as:

//This defines the constructor, the name has to be the same as the class //name

PhonePlan(int numOfMinutes, int numberOfMessages) {

this.freeMinutes = numOfMinutes;

this.freeMessages = numberOfMessages

}

Explanation:

The second constructor is defined using java programming language.

The given class has two constructors This is called "Constructor Overloading) which implements polymophism
In the second constructor that we created, we pass in two arguments of type integer numOfMinutes and numberOfMessages.
In the constructor's body we assign these values to the initially declared variables freeMinutes and freeMessages

6 0

2 years ago

Java languageThe cost to ship a package is a flat fee of 75 cents plus 25 cents per pound.1. Declare a constant named CENTS_PER_

Sophie [7]

Answer:

// Here is code in Java.

// import package

import java.util.*;

// class definition

class Main

{

// main method of the class

public static void main (String[] args) throws java.lang.Exception

{

try{

// variables

final int CENTS_PER_POUND = 25;

final int FLAT_FEE_CENTS = 75;

// Scanner object to read input

Scanner sr=new Scanner(System.in);

System.out.print("Enter the shipping weight:");

// read the input weight

int shipWeightPounds=sr.nextInt();

// calculate total cost

int shipCostCents = (shipWeightPounds * CENTS_PER_POUND) + FLAT_FEE_CENTS;

// print Weight

System.out.println("shipping Weight: " + shipWeightPounds);

// print flat fee

System.out.println("flat fee of shipping in cents : " + FLAT_FEE_CENTS);

// print cents per round

System.out.println("Cents/pound for shipping: " + CENTS_PER_POUND);

// print cost of Shipping

System.out.println("total cost of shipping in cents: " + shipCostCents);

}catch(Exception ex){

return;}

}

Explanation:

Declare and initialize two constant variables "CENTS_PER_POUND=25" & "FLAT_FEE_CENTS = 75". Read the input weight from user with the help of Scanner class object. Then calculate the cost of shipping by multiply weight with cent per pound and add flat fee. Then print each of them.

Output:

Enter the shipping weight:12

shipping Weight : 12

flat fee of shipping in cents: 75

Cents/pound for shipping: 25

total cost of shipping in cents: 375

8 0

2 years ago

Read 2 more answers

A contact list is a place where you can store a specific contact with other associated information such as a phone number, email

NikAS [45]

Answer:

C++.

Explanation:

<em>Code snippet.</em>

#include <map>

#include <iterator>

cin<<N;

cout<<endl;

/////////////////////////////////////////////////

map<string, string> contacts;

string name, number;

for (int i = 0; i < N; i++) {

cin<<name;

cin<<number;

cout<<endl;

contacts.insert(pair<string, string> (name, number));

}

/////////////////////////////////////////////////////////////////////

map<string, string>::iterator it = contacts.begin();

while (it != contacts.end()) {

name= it->first;

number = it->second;

cout<<word<<" : "<< count<<endl;

it++;

}

/////////////////////////////////////////////////////////////////////////////////////////////////////////

I have used a C++ data structure or collection called Maps for the solution to the question.

Maps is part of STL in C++. It stores key value pairs as an element. And is perfect for the task at hand.

8 0

2 years ago

A common fallacy is to use MIPS (millions of instructions per second) to compare the performance of two different processors, an

yulyashka [42]

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

3 0

2 years ago

The data components of a class that belong to every instantiated object are the class's ____ variables.