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jonny [76]
2 years ago
14

Fill in the blank; "As well as their traditional role of computing data, computers are also extensively used for..."

Computers and Technology
1 answer:
Vikki [24]2 years ago
6 0

Answer:

Information output and information storage

Explanation:

Output in the form of images, audio, printed on paper, video, 3D models, etc

Storage - saving information internally on hard drive and externally (HDD, SSD, RAID system, etc

Guest
1 year ago
communications
You might be interested in
This program will store roster and rating information for a soccer team. Coaches rate players during tryouts to ensure a balance
aleksandr82 [10.1K]

Answer:

Question Options :

(1) Prompt the user to input five pairs of numbers: A player’s jersey number (0 – 99) and the player’s rating (1 – 9). Store the jersey numbers in one int array and the ratings in another int array. Output these array(i.e., output the roster).

(2) Implement a menu of options for a user to modify the roster. Each option is represented by a single character. The program initially outputs the menu, and outputs the menu after a user chooses an option. The program ends when the user chooses the option to Quit. For this step, the other options do nothing. (2 pt)

(3) Implement the “Output roster” menu option. (1 pt)

(4) Implement the “Add player” menu option. Prompt the user for a new player’s jersey number and rating. Append the values to the two arrays. (1 pt)

(5) Implement the “Delete player” menu option. Prompt the user for a player’s jersey number. Remove the player from the roster (delete the jersey number and rating). (2 pts)

(6) Implement the “Update player rating” menu option. Prompt the user for a player’s jersey number. Prompt again for a new rating for the player, and then change that player’s rating. (1 pt)

(7) Implement the “Output players above a rating” menu option. Prompt the user for a rating. Print the jersey number and rating for all players with ratings above the entered value. (2 pts)

/******************************************************************************

This program will store roster and rating information for a soccer team.

                             

*******************************************************************************/

#include <iostream>

#include <vector>

using namespace std;

int main() {

   vector< int> jerseyNumber;

   vector< int> rating;

   int temp;

   for (int i = 1; i <= 5; i++) {

       cout << "\nEnter player " << i << "'s jersey number: ";

       cin >> temp;

       if ((temp >= 0) && (temp <= 99)){

           jerseyNumber.push_back(temp);}

       cout << "Enter player " << i << "'s rating: ";

       cin >> temp;

       if ((temp >= 0) && (temp <= 9)){

       rating.push_back(temp);}

       cout << endl;

   } cout << endl;

   cout << "ROSTER" << endl;

   for (int i = 0; i < 5; i++)

cout << "Player " << i + 1 << " -- " << "Jersey number: " << jerseyNumber.at(i) << ", Rating: " << rating.at(i) << endl;

   char option;

   while (true) {

       cout << endl;

       cout << "MENU" << endl;

       cout << "a - Add player" << endl;

       cout << "d - Remove player" << endl;

       cout << "u - Update player rating" << endl;

       cout << "r - Output players above a rating" << endl;

       cout << "o - Output roster" << endl;

       cout << "q - Quit" << endl << endl;

       cout << "Choose an option: " << endl;

       cin >> option;

       switch (option) {

           case 'a':

           case 'A':

               cout << "Enter another player's jersey number: " << endl;

               cin >> temp;

               if ((temp >= 0) && (temp <= 99)){

                   jerseyNumber.push_back(temp);}

               cout << "Enter another player's rating: " << endl;

               cin >> temp;

               if ((temp >= 0) && (temp <= 9)){

                   rating.push_back(temp);}

               break;

           case 'd':

           case 'D':

               cout << "Enter a jersey number: ";

               cin >> temp;

               for (int i = 0; i < jerseyNumber.size(); i++) {

                   if (jerseyNumber.at(i) == temp) {

                       jerseyNumber.erase(jerseyNumber.begin() + i);

                       rating.erase(rating.begin() + i);

                       break; }

               } break;

           case 'u':

           case 'U':

               cout << "Enter a jersey number: ";

               cin >> temp;

               for (int i = 0; i < jerseyNumber.size(); i++) {

                   if (jerseyNumber.at(i) == temp) {

                       cout << "Enter a new rating " << "for player: ";

                       cin >> temp;

                       rating.at(i) = temp;

                       break; }

               } break;

           case 'r':

           case 'R':

               cout << "Enter a rating: ";

               cin >> temp;

               cout << "\nABOVE " << temp << endl;

               for (int i = 0; i < jerseyNumber.size(); i++)

                   if (rating.at(i) > temp)

               cout << "Player " << i + 1 << " -- " << "Jersey number: " << jerseyNumber.at(i)

               << ", Rating: " << rating.at(i) << endl;

               break;

           case 'o':

           case 'O':

               cout << "ROSTER" << endl;

               for (int i = 0; i < jerseyNumber.size(); i++)

               cout << "Player " << i + 1 << " -- " << "Jersey number: " << jerseyNumber.at(i)

               << ", Rating: " << rating.at(i) << endl;

               break;

           case 'q':

               return 0;

               default: cout << "Invalid menu option." << " Try again." << endl;

       }

     }

   }

Explanation:

6 0
2 years ago
JAVA
avanturin [10]

Answer:

   public ArrayList onlyBlue(String[] clothes){

       ArrayList<String> blueCloths = new ArrayList<>();

       for(int i =0; i<clothes.length; i++){

           if(clothes[i].equalsIgnoreCase("blue")){

               blueCloths.add(clothes[i]);

           }

       }

       return blueCloths;

   }

Explanation:

  • Create the method to accept an Array object of type String representing colors with a return type of an ArrayList
  • Within the method body, create and initialize an Arraylist
  • Use a for loop to iterate the Array of cloths.
  • Use an if statement within the for loop to check if item equals blue and add to the Arraylist.
  • Finally return the arrayList to the caller
3 0
2 years ago
Write a method that accepts a String object as an argument and displays its contents backward. For instance, if the string argum
Nata [24]

Answer:

// importing the necessary libraries

import java.util.Scanner;

public class BackwardString {

public static void main(String[] args) {

String input;

// Scanner object to read string from user input

Scanner scanner = new Scanner(System.in);

System.out.print("Enter String here : ");

input=scanner.next();

// calling backward method to reverse passed string

backward(input);

// closing Scanner object

scanner.close();

}

// Method to Reverse the input string

private static void backward(String source) {

int i, len = source.length();

StringBuilder dest = new StringBuilder(len);

for (i = (len - 1); i >= 0; i--){

dest.append(source.charAt(i));

}

System.out.println("Reversed String : "+dest);

}

}

Explanation:

// importing the necessary libraries

import java.util.Scanner;

public class BackwardString {

public static void main(String[] args) {

String input;

// Scanner object to read string from user input

Scanner scanner = new Scanner(System.in);

System.out.print("Enter String here : ");

input=scanner.next();

// calling backward method to reverse passed string

backward(input);

// closing Scanner object

scanner.close();

}

// Method to Reverse the input string

private static void backward(String source) {

int i, len = source.length();

StringBuilder dest = new StringBuilder(len);

for (i = (len - 1); i >= 0; i--){

dest.append(source.charAt(i));

}

System.out.println("Reversed String : "+dest);

}

}

The program above accepts a String object as an argument and displays its contents backward.

It reverses the any string inputed while producing the output.

6 0
2 years ago
PLEASE HELP!!!
ollegr [7]
What are the type groups?
8 0
2 years ago
Read 2 more answers
For the (pseudo) assembly code below, replace X, Y, P, and Q with the smallest set of instructions to save/restore values on the
Dimas [21]

Answer:

Explanation:

Let us first consider the procedure procA; the caller in the example given.

  Some results: $s0,$s1,$s2, $t0,$t1 and $t2 are being stored by procA. Out of these registers, few registers are accessing by procA after a call to procB. But, procB might over-write these registers.

       Thus, procA need to save some registers into stack first before calling procB, .

      only $s1,$t0 and $t1 are being used after return from procB in the given example,

       Caller saves and restores only values in $t0-$t9, according to MIPS guidelines for caller-saved and callee-saved registers, .

       Thus, procA needs to save only $t0 and $t1.

    jal instruction overwrites the register $ra by writing the address, to which the control should jump back, after completing the instructions of procB, when procB is called,.

       Therefore, procA also need to save $ra into stack.

 ProcA is writing new values into $a0,$a2, procA must save $a0 and $a1 first before calling procB, .

     In the given example, after return from procB, only $a0 is being used. It is therefore enough to save $a0.

   Also, procA needs to save frame pointer, which points the start of the stack space for each procedure.

       Generally, as soon as the procedure begins, frame pointer is set to the current value of the stack pointer,.

Let us consider the procedure procB; the callee in the given example.

 The callee is responsible for saving values in $s0-$s7 and restoring them before returning to caller, this is according to MIPS guidelines for caller-saved and callee-saved registers,

   procB is expected to over-write the registers $s2 and $t0. Nonetheless, in the first two lines, procB might over-write the registers $s0 and $s1.

   Thus, procB is responsible for saving and restoring $s0,$s1 and $s2.

X:

We need to create space for 5 values on the stack since procA needs to save $a0,$ra,$t0,$t1 and $fp(frame pointer), . Each value(word) takes 4 bytes.

$sp = $sp – 20 # on the stack, create space for 5 values

sw $a0, 16($sp) # store the result in $a0 into the memory address

               # indicated by $sp+20

sw $ra, 12($sp) # save the second value on stack

sw $t0, 8($sp) # save the third value on stack

sw $t1, 4($sp) # save the fourth value on stack

sw $fp, 0($sp) #  To the stack pointer, save the frame pointer

$fp = $sp      #  To the stack pointer, set the frame pointer

Y:

lw $fp, 0($sp) #  from stack, start restoring values

lw $t1, 4($sp)

lw $t0, 8($sp)

lw $ra, 12($sp)

lw $a0, 16($sp)

$sp = $sp + 20 # decrease the size of the stack

P:

$sp = $sp – 12 #  for three values, create space on the stack

sw $s0, 0($sp) # save the value in $s0

sw $s1, 0($sp) # save the value in $s1

sw $s2, 0($sp) # save the value in $s2

Q:

lw $s0, 0($sp) #  from the stack, restore the value of $s0

lw $s1, 0($sp) #  from the stack, restore the value of $s1

lw $s2, 0($sp) #  from the stack, restore the value of $s2

$sp = $sp + 12 # decrease the stack size

8 0
2 years ago
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