Answer: Transaction processing system
Explanation:
A transaction processing system basically records all the data in the system. It is the set of data that basically monitor all the transaction process in the system. It basically perform various transaction oriented applications in the system.
It also allow time delay in the system when the item is supposed to being sold to the actually selling of the item. There are basically various types of transaction processing system arr payroll, account payable and the inventory control.
Answer:
//Define class
public class Main {
//define main method
public static void main(String[] args)
{
//declare and initialize the double type variable to 128
double mexico = 128;
//declare and initialize the double type variable to 323
double us = 323;
//declare and initialize the integer type variable to 0
int yr = 0;
//set the while loop to check which is greater
while (mexico < us)
{
//increment in the variable by 1
yr++;
//initialize in the variables acc. to the percentage
mexico *= 1.0101;
us *= 0.9985;
}
//print the following results
System.out.println("Population of the Mexico will be exceed the population U.S. in " + yr + " years");
System.out.println("Population of the Mexico will be " + mexico + " million");
System.out.println("and population of the U.S. will be " + us + " million");
}
}
<u>Output</u>:
Population of the Mexico will be exceed the population U.S. in81 years
Population of the Mexico will be 288.88435953355025 million
and population of the U.S. will be 286.0198193927948 million
Explanation:
<u>Following are the description of the program</u>.
- Firstly, we define the class 'Main' and inside it, we define the main method.
- Then, declare two double data type variables which are 'mexico' and 'us' and initialize in it to 128 and 323.
- Declare integer data type variable 'yr' and initialize in it to 0.
- Set the while loop and pass the condition to check that the variable 'mexico' is less than the variable 'us' then, increment in the variable 'yr' by 1 and multiply the variables 'us' and 'mexico' by the following percentage.
- Finally, print the following results with the message.
A. Changing the company name when the same letter is sent to different companies
The find and replace tool is meant to help replace all instances of a certain piece of text with a different piece of text.
For example, if a letter was sent to Company A, the find and replace tool could change every time the letter says “Company A” and make it say “Company B” instead so the same letter could be sent to Company B.
Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires is True.
True
<u>Explanation:</u>
In network packet loss is considered as connectivity loss. In this scenario host A send two segment to host B and acknowledgement from host B Is awaiting at host A.
Since first acknowledgement is lost it is marked as packet lost. Since in network packet waiting for acknowledgement is keep continues process and waiting or trying to accept acknowledgement for certain period of time, once period limits cross then it is declared as packet loss.
Meanwhile second comes acknowledged is success. For end user assumes second segments comes first before first segment. But any how first segment expires.
Answer:
#include<stdio.h>
//declare a named constant
#define MAX 50
int main()
{
//declare the array
int a[MAX],i;
//for loop to access the elements from user
for(i=0;i<MAX;i++)
{
printf("\n Enter a number to a[%d]",i+1);
scanf("%d",&a[i]);
}
//display the input elements
printf("\n The array elements are :");
for(i=0;i<=MAX;i++)
printf(" %d ",a[i]);
}
Explanation:
PSEUDOCODE INPUTARRAY(A[MAX])
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT “Enter a number to A[I]”
READ A[I]
[END OF LOOP]
REPEAT FOR I EQUALS TO 1 TO MAX
PRINT A[I]
[END OF LOOP]
RETURN
ALGORITHM
ALGORITHM PRINTARRAY(A[MAX])
REPEAT FOR I<=1 TO MAX
PRINT “Enter a number”
INPUT A[I]
[END OF LOOP]
REPEAT FOR I<=1 TO MAX
PRINT A[I]
[END OF LOOP]
