answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
earnstyle [38]
2 years ago
10

What will be displayed after code corresponding to the following pseudocode is run? Main Set OldPrice = 100 Set SalePrice = 70 C

all BigSale(OldPrice, SalePrice) Write "A jacket that originally costs $ " + OldPrice Write "is on sale today for $ " + SalePrice End Program Subprogram BigSale(Cost, Sale As Ref) Set Sale = Cost * .80 Set Cost = Cost + 20 End Subprogram

Computers and Technology
1 answer:
Tatiana [17]2 years ago
7 0

Answer:

A jacket that originally costs $ 100 is on sale today for $ 80                                    

Explanation:

Main : From here the execution of the program begins

Set OldPrice = 100  -> This line assigns 100 to the OldPrice variable

Set SalePrice = 70   -> This line assigns 70to the SalePrice variable

Call BigSale(OldPrice, SalePrice)  -> This line calls BigSale method by passing OldPrice and SalePrice to that method

Write "A jacket that originally costs $ ", OldPrice  -> This line prints/displays the line: "A jacket that originally costs $ " with the resultant value in OldPrice variable that is 100

Write "is on sale today for $ ", SalePrice  -> This line prints/displays the line: "is on sale today for $ " with the resultant value in SalePrice variable that is 80

End Program -> the main program ends

Subprogram BigSale(Cost, Sale As Ref)  -> this is a definition of BigSale method which has two parameters i.e. Cost and Sale. Note that the Sale is declared as reference type

Set Sale = Cost * .80  -> This line multiplies the value of Cost with 0.80 and assigns the result to Sale variable

Set Cost = Cost + 20  -> This line adds 20 to the value of Cost  and assigns the result to Cost variable

End Subprogram  -> the method ends

This is the example of call by reference. So when the method BigSale is called in Main by reference by passing argument SalePrice to it, then this call copies the reference of SalePrice argument into formal parameter Sale. Inside BigSale method the reference &Sale is used to access actual argument i.e. SalePrice which is used in BigSale(OldPrice, SalePrice) call. So any changes made to value of Sale will affect the value of SalePrice

So when the method BigSale is called two arguments are passed to it OldPrice argument and SalePrice is passed by reference.

The value of OldPrice is 100 and SalePrice is 70

Now when method BigSale is called, the reference &Sale is used to access actual argument SalePrice = 70

In the body of this method there are two statements:

Sale = Cost * .80;

Cost = Cost + 20;

So when these statement execute:

Sale = 100 * 0.80 = 80

Cost = 100 + 20 = 120

Any changes made to value of Sale will affect the value of SalePrice as it is passed by reference. So when the Write "A jacket that originally costs $ " + OldPrice Write "is on sale today for $ " + SalePrice statement executes, the value of OldPrice remains 100 same as it does not affect this passed argument, but SalePrice was passed by reference so the changes made to &Sale by statement in method BigSale i.e.  Sale = Cost * .80; has changed the value of SalePrice from 70 to 80 because Sale = 100 * 0.80 = 80. So the output produced is:

A jacket that originally costs $ 100 is on sale today for $ 80                              

You might be interested in
Consider the following two code segments, which are both intended to determine the longest of the three strings "pea", "pear", a
Nikolay [14]

Answer:

e) Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".

Explanation:

<em>if - elseif - else statements work in sequence in which they are written. </em>

  • <em> </em>In case <em>if() statement is true, </em>else if() and else statements will not get executed.
  • In case <em>else if() statement is true</em>, conditions in if() and else if() will be checked and else statement will not be executed.
  • In case <em>if() and else if() both are false</em>, else statement will be executed<em>.</em>

First, let us consider code segment I.

In this, first of all "pea" is checked in if() statement which will look for "pea" only in the String str. So, even if "pearl" or "pear" or "pea" is present in 'str' the result will be true and "pea" will get printed always.

After that there are else if() and else statements which will not get executed because if() statement was already true. As a result else if() and else statements will be skipped.

Now, let us consider code segment II.

In this, "pearl" is checked in if() condition, so it will result in desired output.

Executable code is attached hereby.

Correct option is (e).

Download java
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> java </span>
3 0
2 years ago
For multicore processors to be used effectively, computers must understand how to divide tasks into parts that can be distribute
AnnyKZ [126]

Answer: parallel processing

Explanation:

Parallel processing is the process by which the processor has the capacity to execute parallel instruction at the same time. This is possible only by assigning the different task to the different cores present in the cpu. So a cpu with multiple cores are able to process multiple task as the same time parallel.

8 0
2 years ago
A laptop gets replaced if there's a hardware issue. Which stage of the hardware lifecycle does this scenario belong to?
Ilya [14]

Answer: Maintenance. This is the stage where software is updated and hardware issues are fixed if and when they occur. Retirement. In this final stage, hardware becomes unusable or no longer needed and it needs to be properly removed from the fleet

Explanation:

3 0
2 years ago
Write a copy assignment operator for CarCounter that assigns objToCopy.carCount to the new objects's carCount, then returns *thi
ollegr [7]

Answer:

Here is the copy assignment operator for CarCounter:

CarCounter& CarCounter::operator=(const CarCounter& objToCopy) {

carCount = objToCopy.carCount;

return *this;  }

The syntax for copy assignment operator is:

ClassName& ClassName :: operator= ( ClassName& object_name)

In the above chunk of code class name Is CarCounter and object name is objToCopy.

Assignment operator = is called which assigns objToCopy.carCount to the new objects's carCount.

This operator basically is called when an object which is already initialized is assigned a new value from another current object.

Then return *this returns the reference to the calling object

Explanation:

The complete program is:

#include <iostream>  //to use input output functions

using namespace std;   //to access objects like cin cout

class CarCounter {  //class name

public:  // public member functions of class CarCounter

CarCounter();  //constructor of CarCounter

CarCounter& operator=(const CarCounter& objToCopy);  //copy assignment operator for CarCounter

void SetCarCount(const int setVal) {  //mutator method to set the car count value

carCount = setVal;  } //set carCount so setVal

int GetCarCount() const {  //accessor method to get carCount

return carCount;  }  

private:  //private data member of class

int carCount;  };    // private data field of CarCounter

CarCounter::CarCounter() {  //constructor

carCount = 0;  //intializes the value of carCount in constructor

return; }  

// FIXME write copy assignment operator  

CarCounter& CarCounter::operator=(const CarCounter& objToCopy){

/* copy assignment operator for CarCounter that assigns objToCopy.carCount to the new objects's carCount, then returns *this */

carCount = objToCopy.carCount;

return *this;}  

int main() {  //start of main() function

CarCounter frontParkingLot;  // creates CarCounter object

CarCounter backParkingLot;   // creates CarCounter object

frontParkingLot.SetCarCount(12);  // calls SetCarCount method using object frontParkingLot to set the value of carCount to 12

backParkingLot = frontParkingLot;  //assigns value of frontParkingLot to backParkingLot

cout << "Cars counted: " << backParkingLot.GetCarCount();  //calls accessor GetCarCount method to get the value of carCount and display it in output

return 0;  }

The output of this program is:

Cars counted = 12

3 0
2 years ago
Compare the memory organization schemes of contiguous memory allocation and paging with respect to the following issues: a. Exte
Free_Kalibri [48]

Answer:

The comparison is based on memory organization schemes of contiguous memory allocation and paging with respect to External fragmentation, Internal fragmentation and Ability to share code across processes.

Explanation:

Memory organization schemes of contiguous memory allocation:

Contiguous memory allocation schemes suffers from external fragmentation. The reason is that address space is distributed contiguously and the holes and gaps keep growing when the old processes die and new processes are introduced. The variable size partition suffers from external fragmentation however the fixed size partitions do not suffer from external fragmentation. Contiguous memory allocation with variable size partitions does not encounter an internal fragmentation but with fixed size partitions suffers from internal fragmentation. Contiguous memory allocation does not support sharing code across processes. This is because the virtual memory segment of a process is not fragmented into non-contiguous fine grained blocks.

Paging:

Paging does not encounter external fragmentation as pages are of the fixed or equal size. So this reduces external fragmentation. However paging suffers from internal fragmentation. This is because a process can request more space or it can request for a less space. When page is allocated to the such a process that page is no longer utilized.This results in internal fragmentation because of the wastage of space even when the page has internal space but cannot be fully utilized. Paging allows to share code across processes.

5 0
2 years ago
Other questions:
  • Which of these is an on-site metric for social media marketing?
    13·1 answer
  • Suppose Dave drops a watermelon off a high bridge and lets it fall until it hits the water. If we neglect air resistance, then t
    12·1 answer
  • The X.500 standard defines a protocol for a client application to access an X.500 directory known as which of the following opti
    10·1 answer
  • LensForAll is a company that sells affordable contact lenses on its website. The CEO of the company realizes that an app must be
    9·1 answer
  • Programmers often author which type of information to guide test runs?
    10·1 answer
  • Cleo finds herself frequently searching for messages from particular senders or with specific subjects. What should she create
    5·1 answer
  • In which of the security mechanism does the file containing data of the users/user groups have inbuilt security?
    6·1 answer
  • A customer opened a file attachment, and now her PC is infected with ransomware. She's unable to open any of her files. Which ac
    11·2 answers
  • Write a class named Employee that has private data members for an employee's name, ID_number, salary, and email_address. It shou
    7·1 answer
  • A Linux systems admin reported a suspicious .py file that ran on a daily schedule after business hours. The file includes shellc
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!