Answer:
Explanation:
The following code is written in Python it doesn't use any loops, instead it uses a recursive function in order to continue asking the user for the inputs and count the number of positive values. If anything other than a number is passed it automatically ends the program.
def countPos(number=input("Enter number: "), counter=0):
try:
number = int(number)
if number > 0:
counter += 1
newNumber = input("Enter number: ")
return countPos(newNumber, counter)
else:
newNumber = input("Enter number: ")
return countPos(newNumber, counter)
except:
print(counter)
print("Program Finished")
countPos()
I would say functional and straightforward
,I don't know you all about computer
Answer:
A = 120
B = 40
C = 70
Solution:
As per the question:
Manufacturer forced to make 10 more type C clamps than the total of A and b:
10 + A + B = C (1)
Also, 3 times as many type B as type A clamps are:
A = 3B (2)
The total no. of clamps produced per day:
A + B + C = 330 (3)
The no. of each type manufactured per day:
Now, from eqn (1), and (3):
A + B + 10 + A + B = 330
2A + 2B = 320
A + B = 160 (4)
Now, from eqn (2) and (4):
3B + B = 160
B = 40
Since, A = 3B
A = 
A = 120
Put the values of A and C in eqn (3):
120 + 40 + C = 330
C = 70
Solution:
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This is the required answer.