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77julia77 [94]
3 years ago
8

Emma frequently updates a spreadsheet on her computer. How can she ensure that she doesn't lose her data?

Computers and Technology
2 answers:
bixtya [17]3 years ago
8 0

Answer:

OD. do nothing as her computer won't fail

Alla [95]3 years ago
4 0

Answer:

Take regular backups ( A )

Explanation:

Spreadsheets are a form of database management system, this means that important data is been stored, updated and changed there. one of the best ways to do this efficiently without complete or partial loss of stored,updated or changed data is to ensure regular backups of the spreadsheets so that in a case of loss of data on the computer. the backup source can be queried for the information been backed up/stored in it,such backup devices can be an external hard drive.

telling her colleagues about the file location can not help her recover her data when they are lost because the spreadsheet is stored electronically in her computer and not in physical location, making duplicate copies on the same computer won't help as well because if the computer fails the duplicate copy will be lost as well

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A penetration tester is experimenting with Network Mapper (Nmap) on a test network as a privileged user. The tester would like t
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<h3>What is a penetration test?</h3>

A penetration test can be defined as a cybersecurity technique that simulates a cyber attack against an end user's computer system, so as to scan, identify, test and check for exploitable vulnerabilities in the following:

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This ultimately implies that, a penetration test avails an tester the ability to exploit a weakness and potential security threats on a test network as a privileged user, during vulnerability assessments of the system.

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6 0
2 years ago
g Write a program that prompts the user for an integer n between 1 and 100. If the number is outside the range, it prints an err
grin007 [14]

Answer:

The cpp program is given below.

#include<iostream>

#include<iomanip>

using namespace std;

int main() {

   

   // variables declared

   int n;

   int sum=0;

   float avg;

   

   do

   {

       // user input taken for number    

       cout<< "Enter a number between 1 and 100 (inclusive): ";

       cin>>n;

       

       if(n<1 || n>100)

           cout<<" Number is out of range. Enter valid number."<<endl;

       

   }while(n<1 || n>100);

   

   cout<<" "<<endl;

   

   // printing even numbers between num and 50  

   for(int num=1; num<=n; num++)

   {

       sum = sum + num;

   }

   

   avg = sum/n;

   

   // displaying sum and average

   cout<<"Sum of numbers between 1 and "<<n<<" is "<<sum<<endl;

   cout<<"Average of numbers between 1 and "<<n<<" is ";

   printf("%.2f", avg);

   

       return 0;

}

OUTPUT

Enter a number between 1 and 100 (inclusive): 123

Number is out of range. Enter valid number.

Enter a number between 1 and 100 (inclusive): 56

 

Sum of numbers between 1 and 56 is 1596

Average of numbers between 1 and 56 is 28.00

Explanation:

The program is explained below.

1. Two integer variables are declared to hold the number, n, and to hold the sum of numbers from 1 to n, sum. The variable sum is initialized to 0.

2. One float variable, avg, is declared to hold average of numbers from 1 to n.

3. User input is taken for n inside do-while loop. The loop executes till user enters value between 1 and 100. Otherwise, error message is printed.

4. The for loop executes over variable num, which runs from 1 to user-entered value of n.

5. Inside for loop, all the values of num are added to sum.

sum = sum + num;

6. Outside for loop, average is computed and stored in avg.

avg = sum/n;

7. The average is printed with two numbers after decimal using the following code.

printf("%.2f", avg);

8. The program ends with return statement.

9. All the code is written inside main() and no classes are involved.

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Answer:

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Explanation:

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