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elena55 [62]
3 years ago
13

Drag the tiles to the correct boxes to complete the pairs.

Computers and Technology
1 answer:
ra1l [238]3 years ago
7 0

Answer:

ac be df

Explanation:

You might be interested in
C++
nikitadnepr [17]

Answer:

Check the explanation

Explanation:

<u>The Code</u>

#include <fstream>

#include <iostream>

#include <cmath>

#include <cstring>

#include <cstdlib>

#include <ctime>

using namespace std;

//Function Declarations

void fillArrayWithRandNos(int nos[],int size);

void displayArray(int nos[],int size);

double mean(int nos[],int size);

double variance(int nos[],int size);

double median(int nos[],int size);

int mode(int nos[],int size);

void histogram(int nos[],int size);

int main() {

  //Declaring variables

const int size=100;

srand(time(NULL));

 

// Creating array dynamically

int* nos = new int[size];

 

//Calling the functions

fillArrayWithRandNos(nos,size);

displayArray(nos,size);

cout<<"Mean :"<<mean(nos,size)<<endl;

cout<<"variance :"<<variance(nos,size)<<endl;

cout<<"Median :"<<median(nos,size)<<endl;

cout<<"Mode :"<<mode(nos,size)<<endl;

histogram(nos,size);

 

  return 0;

}

void fillArrayWithRandNos(int nos[],int size)

{

  for(int i=0;i<size;i++)

  {

      nos[i]=rand()%(45) + 55;

  }

}

void displayArray(int nos[],int size)

{

  cout<<"Displaying the array elements :"<<endl;

  for(int i=0;i<size;i++)

  {

      cout<<nos[i]<<" ";

      if((i+1)%10==0)

      cout<<endl;

  }

}

double mean(int nos[],int size)

{

  double sum=0;

  for(int i=0;i<size;i++)

  {

      sum+=nos[i];

  }

  return sum/size;

}

double variance(int nos[],int size)

{

  double avg=mean(nos,size);

 

double variance,sum=0;

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

{

sum+=pow(nos[i]-avg,2);

}

//calculating the standard deviation of nos[] array

variance=(double)sum/(size);

return variance;

}

double median(int nos[],int size)

{

      //This Logic will Sort the Array of elements in Ascending order

  int temp;

  for (int i = 0; i < size; i++)

{

for (int j = i + 1; j < size; j++)

{

if (nos[i] > nos[j])

{

temp = nos[i];

nos[i] = nos[j];

nos[j] = temp;

}

}

}

 

int middle;

float med;

middle = (size / 2.0);

if (size % 2 == 0)

med = ((nos[middle - 1]) + (nos[middle])) / 2.0;

else

med = (nos[middle]);

return med;

}

int mode(int nos[],int size)

{

  int counter1 = 0, counter2, modevalue;

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

counter2 = 0;

for (int j = i; j < size; j++) {

if (nos[i] == nos[j]) {

counter2++;

}

if (counter2 > counter1) {

counter1 = counter2;

modevalue = nos[i];

}

}

}

if (counter1 > 1)

return modevalue;

else

return 0;

}

void histogram(int nos[],int size)

{

  int hist[9]={0};

  for(int i=0;i<size;i++)

  {

      if(nos[i]>=55 && nos[i]<=59)

      {

      hist[0]++;

      }

      else if(nos[i]>=60 && nos[i]<=64)

      {

      hist[1]++;

      }

      else if(nos[i]>=65 && nos[i]<=69)

      {

      hist[2]++;

      }

      else if(nos[i]>=70 && nos[i]<=74)

      {

      hist[3]++;

      }

      else if(nos[i]>=75 && nos[i]<=79)

      {

      hist[4]++;

      }

      else if(nos[i]>=80 && nos[i]<=84)

      {

      hist[5]++;

      }

      else if(nos[i]>=85 && nos[i]<=89)

      {

      hist[6]++;

      }

      else if(nos[i]>=90 && nos[i]<=94)

      {

      hist[7]++;

      }

      else if(nos[i]>=95 && nos[i]<=99)

      {

      hist[8]++;

      }

  }

     

  cout<<"Displaying the count of numbers in each interval:"<<endl;

 

      int cnt=0;

  for(int i=55;i<=99;i+=5)

  {

  cout<<i<<"-"<<i+4<<"|"<<hist[cnt]<<endl;

 

      cnt++;

  }

 

  cout<<"Displaying the histogram :"<<endl;

  cnt=0;

  for(int i=55;i<=99;i+=5)

  {

  cout<<i<<"-"<<i+4<<"|";

  for(int j=0;j<hist[cnt];j++)

  {

      cout<<"*";

  }  

      cout<<endl;

      cnt++;

  }          

     

 

 

}

#########

___________________________

The output can be seen in the attached image below.

8 0
3 years ago
Every time I try to look up anything on my laptop it keeps saying my connection is not priavte what do I do
dangina [55]

You can click "Advanced" and still proceed to the website.


Please mark this answer as the Brainliest! Thank you :)




6 0
3 years ago
Read 2 more answers
A security team has downloaded a public database of the largest collection of password dumps on the Internet. This collection co
Scorpion4ik [409]

Answer:

a. Password length, password encryption, password complexity

Explanation:

Under this scenario, the best combination would be Password length, password encryption, password complexity. This is because the main security problem is with the user's passwords. Increasing the password length and password complexity makes it nearly impossible for individuals to simply guess the password and gain access, while also making it extremely difficult and time consuming for hackers to use software to discover the password as well. Password excryption would be an extra layer of security as it encrypts the password before storing it into the database, therefore preventing eavesdroppers from seeing the password and leaked info from being used without decryption.

4 0
3 years ago
La sentencia de ASIGNAR es una sentencia de entrada de datos, verdadero o falso ?
Akimi4 [234]

Answer:

sorry can´t understand langues

Explanation:

cant

7 0
3 years ago
Which of the following should NOT be done to keep people from slipping on floors? A) Have the proper absorbents in an easy-to-fi
RoseWind [281]

The following options helps keep people from slipping on floors

A) Have the proper absorbents in an easy-to-find place to quickly put on spills : So that accidental little spills can be cleaned right away.

C) Wear slip-resistant footwear : It is a precautionary step to follow, no matter where we go.

D) Mark a slippery area with an easy-to-use tent sign that says " Caution, Slippery floor" : This allows people to be aware of the wet area and cross it cautiously.

Leaving oil and fluids to air dry thoroughly, takes a long time and there are chances that people step over it and slip.

So, the answer is

(B) Anytime oil or fluids are spilled on the floor, leave them to thoroughly air dry.

8 0
3 years ago
Read 2 more answers
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