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Svetradugi [14.3K]
2 years ago
15

Why can a failure in a database environment be more serious than an error in a nondatabase environment?

Computers and Technology
2 answers:
Elina [12.6K]2 years ago
5 0
Failure in a database environment is more serious then a non data base because if you lose important information you may not get it back and failure in a nondatabase environment the problem may be more easier to solve
Anestetic [448]2 years ago
3 0

Answer:

Failures in database environments can occur due to hardware failures, power failures and software/programming error.

Of these three errors, the non-database environment can survive the hardware failure due to recovery processes and the Software/programming error.

Explanation:

Databases are quite fragile when they hit a hardware failure if you try to recover them (Most of them have raw files that can only be read by certain types of Database Management Softwares) this does not exist in the Non-Database Scenario as most of these records are kept in files.

Software failures can occur when a wrong query is used to retrieve data from a DBMS. This can be frustrating except the user gets the right query.

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
2 years ago
Write a program in c or c++ to perform different arithmeticoperation using switch statement .the program will take two inputinte
baherus [9]

Answer:

C code :

#include<stdio.h>

int main()

{

int j;

float k,l, x;  //taking float to give the real results.

printf("Enter your two operands: ");  //entering the numbers on which      //the operation to be performed.

scanf("%f %f", &k, &l);

 

printf("\n Now enter the operation you want to do: ");

printf("1 for Addition, 2 for Subtraction, 3 for Multiplication, 4 for Division ");

scanf("%d", &j);  //j takes the input for opearation.

 

switch(j)  

{

 case 1:  

  x=k+l;

  printf("%.2f+%.2f=%.2f",k,l,x);   //we write %.2f to get the result //upto 2 decimal point.

  break;

   

 case 2:

  x=k-l;

  printf("%.2f-%.2f=%.2f",k,l,x);

  break;

 case 3:

  x=k*l;

  printf("%.2f*%.2f=%.2f",k,l,x);

  break;

 case 4:

  if(l!=0) //division is not possible if the denominator is 0.

  {

   x=k/l;

   printf("%.2f/%.2f=%.2f",k,l,x);

  }

  else  

   printf("Division result is undefined");

               default:

   printf("\n invalid operation");

}

}

Output is in image.

Explanation:

At first we take two numbers.

Then we take integers from 1 to 4 for Addition, subtraction, multiplication, division respectively.

Then accordingly the case is followed and the operation is performed.

6 0
2 years ago
What prevents someone who randomly picks up your phone from sending money to themselves using a messenger-based payment?
Gekata [30.6K]

Answer:

Explanation:

There are various safety features in place to prevent such scenarios from happening. For starters phones usually have a pin code, pattern code, or fingerprint scanner which prevents unauthorized individuals from entering into the phone's services. Assuming that these features have been disabled by the phone's owner, payment applications usually require 2FA verification which requires two forms of approval such as email approval and fingerprint scanner in order for any transactions to go through. Therefore, an unauthorized individual would not have access to such features and would not be able to complete any transactions.

6 0
2 years ago
Which of the following statements is true? Using existing exceptions makes the program less robust. Always create your own excep
hjlf

Answer:

The third option is correct.

Explanation:

The following option is true because it's a derived with that Throwable class. More than that exception type, it is also other category called Error originating through that Throwable class. As any other class, the exception class will also include fields as well as functions. So, the following are the reason that describes the following answer is true according to the exception class.

The other options are not appropriate according to the following scenario.

8 0
2 years ago
A new company starts up but does not have a lot of revenue for the first year. Installing anti-virus software for all the compan
Feliz [49]

WAN domain which stands for Wide Area Network and consists of the Internet and semi-private lines. The <span>RISKS are: Service provider can have a major network outage, Server can receive a DOS or DDOS attack</span> and A FTP server can allow anonymously uploaded illegal software.





7 0
2 years ago
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