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-Dominant- [34]
3 years ago
9

Which tab would you click to access the autosum feature?

Computers and Technology
1 answer:
Bogdan [553]3 years ago
6 0
On Excel, you click the Formulas tab. 

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_____ separation strategies (e.g., attacking and sabotaging others) are used by those for whom co-cultural segregation is an imp
soldier1979 [14.2K]

Answer: Aggressive

Explanation:

Aggressive separation strategies are the technique that involves confrontation and intense feeling for separating something from others.

  • This technique includes action like attacking, assaultive, confronting etc.for segregation.
  • Co-culture segregation is separation of a subset of culture from large and major culture.In this culture, there can be more than two types of culture that are split forcefully through barrier.
  • Thus, co-culture segregation uses the methodology of aggressive separation.

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How many generations of computer languages have there been since the middle of the 20th century?
evablogger [386]
4 generations hahahaha
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Multiple choice
Bezzdna [24]
The answer is D. The colors that we see are all being reflected.
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Mention two hardware groups​
mars1129 [50]

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input Devices, Processing Devices, Output Devices, Memory/Storage Devices

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3 years ago
Write a program to implement problem statement below; provide the menu for input N and number of experiment M to calculate avera
zalisa [80]

Answer:

Explanation:

#include<iostream>

#include<ctime>

#include<bits/stdc++.h>

using namespace std;

double calculate(double arr[], int l)

{

double avg=0.0;

int x;

for(x=0;x<l;x++)

{

avg+=arr[x];

}

avg/=l;

return avg;

}

int biggest(int arr[], int n)

{

int x,idx,big=-1;

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

{

if(arr[x]>big)

{

big=arr[x];

idx=x;

}

}

return idx;

}

int main()

{

vector<pair<int,double> >result;

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

int choice;

cin>>choice;

while(choice!=2)

{

int n,m;

cout<<"Enter N"<<endl;

cin>>n;

cout<<"Enter M"<<endl;

cin>>m;

int c=m;

double running_time[c];

while(c>0)

{

int arr[n];

int x;

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

{

arr[x] = rand();

}

clock_t start = clock();

int pos = biggest(arr,n);

clock_t t_end = clock();

c--;

running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;

}

double avg_running_time = calculate(running_time,m);

result.push_back(make_pair(n,avg_running_time));

cout<<"Enter 1 for iteration\nEnter 2 for exit\n";

cin>>choice;

}

for(int x=0;x<result.size();x++)

{

cout<<result[x].first<<" "<<result[x].second<<endl;

}

}

8 0
3 years ago
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