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irakobra [83]
3 years ago
14

In how many ways can a list of students names from your class be organized?​

Computers and Technology
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

It can be expressed in 2 ways horizontal and vertical ways.

make sure to follow me and mark me as the brainliest

You might be interested in
Develop a C program that calculates the final score and the average score for a student from his/her (1)class participation, (2)
Ghella [55]

Answer:

#include <iomanip>

#include<iostream>

using namespace std;

int main(){

char name[100];

float classp, test, assgn, exam, prctscore,ave;

cout<<"Student Name: ";

cin.getline(name,100);

cout<<"Class Participation: "; cin>>classp;

while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; }

cout<<"Test: "; cin>>test;

while(test <0 || test > 100){  cout<<"Test: "; cin>>test; }

cout<<"Assignment: "; cin>>assgn;

while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; }

cout<<"Examination: "; cin>>exam;

while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; }

cout<<"Practice Score: "; cin>>prctscore;

while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; }

ave = (int)(classp + test + assgn + exam + prctscore)/5;

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

return 0;}

Explanation:

The required parameters such as cin, cout, etc. implies that the program is to be written in C++ (not C).

So, I answered the program using C++.

Line by line explanation is as follows;

This declares name as character of maximum size of 100 characters

char name[100];

This declares the grading items as float

float classp, test, assgn, exam, prctscore,ave;

This prompts the user for student name

cout<<"Student Name: ";

This gets the student name using getline

cin.getline(name,100);

This prompts the user for class participation. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Class Participation: "; cin>>classp; </em>

<em> while(classp <0 || classp > 100){  cout<<"Class Participation: "; cin>>classp; } </em>

This prompts the user for test. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Test: "; cin>>test; </em>

<em> while(test <0 || test > 100){  cout<<"Test: "; cin>>test; } </em>

This prompts the user for assignment. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Assignment: "; cin>>assgn; </em>

<em> while(assgn <0 || assgn > 100){  cout<<"Assignment: "; cin>>assgn; } </em>

This prompts the user for examination. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Examination: "; cin>>exam; </em>

<em> while(exam <0 || exam > 100){  cout<<"Examination: "; cin>>exam; } </em>

This prompts the user for practice score. The corresponding while loop ensures that the score s between 0 and 100 (inclusive)

<em> cout<<"Practice Score: "; cin>>prctscore; </em>

<em> while(prctscore <0 || prctscore > 100){  cout<<"Practice Score: "; cin>>prctscore; } </em>

This calculates the average of the grading items

ave = (int)(classp + test + assgn + exam + prctscore)/5;

This prints the calculated average

cout <<setprecision(1)<<fixed<<"The average score is "<<ave;  

8 0
3 years ago
g Write a program to sort an array of 100,000 random elements using quicksort as follows: Sort the arrays using pivot as the mid
shtirl [24]

Answer:

header.h->function bodies and header files.

#include<iostream>

#include<cstdlib>

#include<ctime>

using namespace std;

/* Partitioning the array on the basis of piv value. */

int Partition(int a[], int bot, int top,string opt)

{

int piv, ind=bot, i, swp;

/*Finding the piv value according to string opt*/

if(opt=="Type1 sort" || opt=="Type3 sort")

{

piv=(top+bot)/2;

}

else if(opt=="Type2 sort" || opt=="Type4 sort")

{

piv=(top+bot)/2;

if((a[top]>=a[piv] && a[top]<=a[bot]) || (a[top]>=a[bot] && a[top]<=a[piv]))

piv=top;

else if((a[bot]>=a[piv] && a[bot]<=a[top]) || (a[bot]>=a[top] && a[bot]<=a[piv]))

piv=bot;

}

swp=a[piv];

a[piv]=a[top];

a[top]=swp;

piv=top;

/*Getting ind of the piv.*/

for(i=bot; i < top; i++)

{

if(a[i] < a[piv])

{

swp=a[i];

a[i]=a[ind];

a[ind]=swp;

ind++;

}

}

swp=a[piv];

a[piv]=a[ind];

a[ind]=swp;

return ind;

}

void QuickSort(int a[], int bot, int top, string opt)

{

int pindex;

if((opt=="Type3 sort" || opt=="Type4 sort") && top-bot<19)

{

/*then insertion sort*/

int swp,ind;

for(int i=bot+1;i<=top;i++){

swp=a[i];

ind=i;

for(int j=i-1;j>=bot;j--){

if(swp<a[j]){

a[j+1]=a[j];

ind=j;

}

else

break;

}

a[ind]=swp;

}

}

else if(bot < top)

{

/* Partitioning the array*/

pindex =Partition(a, bot, top,opt);

/* Recursively implementing QuickSort.*/

QuickSort(a, bot, pindex-1,opt);

QuickSort(a, pindex+1, top,opt);

}

return ;

}

main.cpp->main driver file

#include "header.h"

int main()

{

int n=100000, i;

/*creating randomized array of 100000 numbers between 0 and 100001*/

int arr[n];

int b[n];

for(i = 0; i < n; i++)

arr[i]=(rand() % 100000) + 1;

clock_t t1,t2;

t1=clock();

QuickSort(arr, 0, n-1,"Type1 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type2 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type3 sort");

t2=clock();

for( i=0;i<n;i++)

arr[i]=b[i];

cout<<"Quick sort time and insertion sort time, with pivot middle element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

cout<<"\n";

t1=clock();

QuickSort(arr, 0, n-1,"Type4 sort");

t2=clock();

cout<<"Quick sort time and insertion sort time, with pivot median element:"<<(t2 - t1)*1000/ ( CLOCKS_PER_SEC );

return 0;

}

Explanation:

Change the value of n in the main file for different array size. Output is in the same format as mentioned, time is shown in milliseconds.

7 0
3 years ago
Hexadecimal to denary gcse method
Anuta_ua [19.1K]

There are two ways to convert from hexadecimal to denary gcse method. They are:

  • Conversion from hex to denary via binary.
  • The use of base 16 place-value columns.

<h3>How is the conversion done?</h3>

In Conversion from hex to denary via binary:

One has to Separate the hex digits to be able to know or find its equivalent in binary, and then the person will then put them back together.

Example - Find out the denary value of hex value 2D.

It will be:

2 = 0010

D = 1101

Put them them together and then you will have:

00101101

Which is known to be:

0 *128 + 0 * 64 + 1 *32 + 0 * 16 + 1 *8 + 1 *4 + 0 *2 + 1 *1

= 45 in denary form.

Learn more about hexadecimal from

brainly.com/question/11109762

#SPJ1

3 0
2 years ago
Which process could you use to add a table to a document?
Lady_Fox [76]
These are the steps you would take..

click on insert at the bar on the top of the document  processor (Word,Google Docs ect.) >>Click table drag as many columns and rows as you need!
I hope this helps!
6 0
3 years ago
How do I change my username/display name
exis [7]

Answer:

.

Explanation:

you do to me on the bottom bar and once you hit that you click the pencil and change it

6 0
3 years ago
Read 2 more answers
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