Answer: True
Explanation:
Yes, the given statement is true that the SaaS (Software as a service) provide the different types of services to the organization which basically require the infrastructure like CRM (Customer relationship management) and it is the standard business processing in the organization.
The software as a service is the fundamental technology of the business which basically include the CRM, e-mails and the various types of sale and financial management.
Answer:
Not sure what language, but in python a super basic version would be:
val1 = 98
val2 = 87
val3 = 84
val4 = 100
val5 = 94
sum = val1 + val2 + val3 + val4 + val5
avg = sum / 5
print(avg)
Explanation:
Answer:
import java.util.*;
public class Main {
public static void main(String[] args)
{
Scanner scan = new Scanner();
double budget=0, num=0, total=0;
System.out.println("Your budget for the month? ");
budget=scan.nextDouble();
System.out.println("enter all expense, and after that type -9999 to quit: ");
while(num != -9999)
{
total+=num;
num=scan.nextDouble();
}
if(total<=budget)
{
System.out.println("under budget by ");
System.out.println(budget-total);
}
else
{
System.out.println("over budget by ");
System.out.println(total-budget);
}
}
}
Explanation:
- Take the budget as an input from user and store it to the budget variable.
- Loop until user has entered all his expenses and keep on adding them to the total variable.
- Check If the total is less than or equal to budget or otherwise, and then print the relevant message accordingly.
Local
----------------------------------------------------
Answer:
PROGRAM QuadraticEquation
Solver
IMPLICIT NONE
REAL :: a, b, c
;
REA :: d
;
REAL :: root1, root2
;
//read in the coefficients a, b and c
READ(*,*) a, b, c
WRITE(*,*) 'a = ', a
WRITE(*,*) 'b = ', b
WRITE(*,*) 'c = ', c
WRITE(*,*)
// computing the square root of discriminant d
d = b*b - 4.0*a*c
IF (d >= 0.0) THEN //checking if it is solvable?
d = SQRT(d)
root1 = (-b + d)/(2.0*a) // first root
root2 = (-b - d)/(2.0*a) // second root
WRITE(*,*) 'Roots are ', root1, ' and ', root2
ELSE //complex roots
WRITE(*,*) 'There is no real roots!'
WRITE(*,*) 'Discriminant = ', d
END IF
END PROGRAM QuadraticEquationSolver
