Write a pseudocode to find the average of any amount of positive numbers.
input numbers
add numbers
divide de numbers by their count
print the average
To find the average of any amount of positive numbers, you can use the following pseudocode:
1. Initialize a variable named sum as 0.
2. Initialize a variable named count as 0.
3. Read the first positive number.
4. While the entered number is positive, do steps 5-6. Otherwise, skip to step 7.
5. Add the entered number to the sum.
6. Increment the count by 1.
7. If count is greater than 0, do steps 8-9. Otherwise, go to step 10.
8. Calculate the average by dividing the sum by the count and store it in a variable named average.
9. Print the average.
10. End the program.
Here is the pseudocode:
```
sum = 0
count = 0
repeat
read number
if number > 0 then
sum = sum + number
count = count + 1
end if
until number <= 0
if count > 0 then
average = sum / count
print average
end if
```
To find the average of any amount of positive numbers, you can follow this pseudocode:
1. Initialize a variable `sum` to 0.
2. Initialize a variable `count` to 0.
3. Read the first number into a variable `num`.
4. Check if `num` is positive. If it is not, skip to step 6.
5. Add `num` to the `sum` variable.
6. Increment the `count` variable by 1.
7. Read the next number into `num`.
8. Repeat steps 4-7 until all the numbers have been inputted.
9. Calculate the average by dividing the `sum` variable by the `count` variable.
10. Display the average.
Here's a Python example implementing this pseudocode:
```
sum = 0
count = 0
while True:
num = int(input("Enter a positive number (or any other number to finish): "))
if num <= 0:
break
sum += num
count += 1
if count == 0:
print("No positive numbers were entered.")
else:
average = sum / count
print("The average is:", average)
```
This pseudocode reads positive numbers from the user until a non-positive number is entered, then it calculates and displays the average of the positive numbers. If no positive numbers were entered, it displays a message indicating that.