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Olin [163]
2 years ago
10

TO Cloud

Computers and Technology
1 answer:
scoray [572]2 years ago
7 0

The type of situation would it make sense to use edge computing is   where critical decisions must be made on a split-second basis.

<h3>What is edge computing?</h3>

Edge computing is known to be a kind of computing which often occur on site or near a specific data source, it reduces the need for data to be processed in a distance data center.

Note that The type of scenario  that would it make sense to use edge computing is  when critical decisions must be made on a split-second basis as it requires urgency.

Learn more about edge computing from

brainly.com/question/23858023

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Ammonia Wipes is the answer glad i could help :)
7 0
3 years ago
write a c++program that allows the user to enter the last names of fivecandidates in a local election and the number of votes re
madreJ [45]

Answer:

#include<iostream>

#include<stdlib.h>

using namespace std;

int main(){

   //initialization

   string str1[5];

   int arr_Vote[5];

   int Total_Vote=0,max1_Index;

   int max1=INT_MIN;

   //loop for storing input enter by user

   for(int i=0;i<5;i++){

       cout<<"Enter the last name of candidate: "<<endl;

       cin>>str1[i];

       cout<<"Enter the number of vote receive by the candidate: "<<endl;

       cin>>arr_Vote[i];

       //calculate the sum

       Total_Vote = Total_Vote + arr_Vote[i];

       //find the index of maximum vote

       if(arr_Vote[i]>max1){

           max1=arr_Vote[i];

           max1_Index=i;

       }

   }

   //display

   cout<<"\nThe output is......"<<endl;

   //loop for display data of each candidate

   for(int i=0;i<5;i++){

       cout<<str1[i]<<"\t"<<arr_Vote[i]<<"\t"     <<float(arr_Vote[i]*100)/Total_Vote<<"%"<<endl;

   }

   //display winner

   cout<<"The winner candidate is"<<endl;

   cout<<str1[max1_Index]<<"\t"<<arr_Vote[max1_Index]<<"\t"<<float(arr_Vote[max1_Index]*100)/Total_Vote<<" %"<<endl;

}

Explanation:

Create the main function and declare the two arrays and variables.

the first array for storing the names, so it must be a string type. second array stores the votes, so it must be int type.

after that, take a for loop and it runs for 5 times and storing the values entered by the user in the arrays.

we also calculate the sum by adding the array data one by one.

In the same loop, we also find the index of the maximum vote. take an if statement and it checks the element in the array is greater than the max1 variable. Max1 variable contains the very small value INT_MIN.

If condition true, update the max1 value and update the max1_Index value.

The above process continues for each candidate for 5 times.

After that, take a for loop and display the data along with percentage by using the formula:

Vote\,percent=\frac{Vote\,receive*100}{Total\,vote}

the, finally display the winner by using the max1_Index value.  

5 0
3 years ago
How do I do this??? (Im in 9th)
vlabodo [156]

Answer:

No clue!

can I get brainliest tho-

8 0
3 years ago
Finish and test the following two functions append and merge in the skeleton file:
avanturin [10]

Answer:

Explanation:

#include <iostream>

using namespace std;

int* append(int*,int,int*,int);

int* merge(int*,int,int*,int);

void print(int*,int);

int main()

{ int a[] = {11,33,55,77,99};

int b[] = {22,44,66,88};

print(a,5);

print(b,4);

int* c = append(a,5,b,4); // c points to the appended array=

print(c,9);

int* d = merge(a,5,b,4);

print(d,9);

}

void print(int* a, int n)

{ cout << "{" << a[0];

for (int i=1; i<n; i++)

cout << "," << a[i];

cout << "}\n";

}

int* append(int* a, int m, int* b, int n)

{

int * p= (int *)malloc(sizeof(int)*(m+n));

int i,index=0;

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

p[index++]=a[i];

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

p[index++]=b[i];

return p;

}

int* merge(int* a, int m, int* b, int n)

{

int i, j, k;

j = k = 0;

int *mergeRes = (int *)malloc(sizeof(int)*(m+n));

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

if (j < m && k < n) {

if (a[j] < b[k]) {

mergeRes[i] = a[j];

j++;

}

else {

mergeRes[i] = b[k];

k++;

}

i++;

}

// copying remaining elements from the b

else if (j == m) {

for (; i < m + n;) {

mergeRes[i] = b[k];

k++;

i++;

}

}

// copying remaining elements from the a

else {

for (; i < m + n;) {

mergeRes[i] = a[j];

j++;

i++;

}

}

}

return mergeRes;

}

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Breach..
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hack
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