Answer:
Yes. Pseudocode is an improvement over the IPO chart because it lays out the sequence of steps for a particular program
Explanation:
The input–process–output (IPO) chart is a widely used approach in systems analysis and software engineering for describing the structure of an information processing program or other process. The chart has three components (Input, Process and Output), and you write the description of each component in plain English, not code or mathematical formulas.
Pseudocode is a procedure for solving a problem in terms of the actions to be executed and the order in which those actions are to be executed.
Pseudocode is an improvement over the IPO chart because it shows the step by step sequence to be followed by a particular program unlike the IPO chart which just break the program into Input, Process and Output.
import random
i = 1
while i <= 100:
print("#"+str(i)+": "+str(random.randint(1,100)), end=", ")
i+=1
print()
i = 1
while i <= 100:
print("#"+str(i)+": "+str(random.uniform(1,100)), end=", ")
i += 1
I hope this helps!
This gap between user-designer communications <span>can cause a good project to go bad i</span>f the user is not able to process what is required to be fixed in order for the project to run smoothly. The user may have one way of fixing something while the designer has another. In this case, the designer understands how the project fully works while the user does not and this may end up compromising the whole project.
The law that “designers are not users” and “users are not designers” should always be followed.
Answer:
/*
Find Largest and Smallest Number in an Array Example
This Java Example shows how to find largest and smallest number in an
array.
*/
public class FindLargestSmallestNumber {
public static void main(String[] args) {
//array of 10 numbers
int numbers[] = new int[]{32,43,53,54,32,65,63,98,43,23};
//assign first element of an array to largest and smallest
int smallest = numbers[0];
int largetst = numbers[0];
for(int i=1; i< numbers.length; i++)
{
if(numbers[i] > largetst)
largetst = numbers[i];
else if (numbers[i] < smallest)
smallest = numbers[i];
}
System.out.println("Largest Number is : " + largetst);
System.out.println("Smallest Number is : " + smallest);
}
}
/*
Output of this program would be
Largest Number is : 98
Smallest Number is : 23
*/
Explanation:
