Answer:
b=0
c=0
lol=list()
while True:
salary=input("Enter your salary: ")
if salary=="e" or salary=="E":
print("Thankyou!!")
break
else:
a=int(salary)
if a>=300000 and a<=400000:
c=c+1
if a<0 or a>400000:
print("Your salary is either too small or too big ")
if a>0 and a<=400000:
a=a-(25/100*a)-(5/100*a)-(2/100*a)
lol.append(a)
b=b+1
if a>=300000:
c=c+1
print(lol)
print("The number of salaries entered is: "+ str(b))
print("The number of salaries that exceeded 300000 is: "+str(c))
Answer:
The explaination fo this question is given below in the explanation section. However, the correct option of this question is D.
Explanation:
The correct option of this question is D
keep_going = 'y'
while keep_going == 'y':
hours = float(input('How many hours did this worker work? '))
gross_pay = hours * 10
print('Gross pay:', gross_pay)
keep_going = input('Calculate another gross pay? [Enter y for yes] ')
# this program will print in a sequence and produce the correct result as #shown in attached picture.
<u>Why Other options are not correct</u>
<u>A</u>. This option is not correct becuase it not running in a sequence. It asked first "Calculate another gross pay enter y for yes" . then it get started in unterminated loop and do not finish the loop.
The following output is produce by this code
Calculate another gross pay? [Enter y for yes] y How many hours did this worker work? 12 Gross pay: 120.0 How many hours did this worker work? 10 Gross pay: 100.0 How many hours did this worker work? 13 Gross pay: 130.0 How many hours did this worker work?
<u> B. </u>This option produce the following output
How many hours did this worker work? 10 Gross pay: 100.0 Calculate another gross pay? [Enter y for yes] y
In this option as you enter Y. the program get close and terminated.
<u></u>
<u>C. </u>This option produce the following out put.
while keep_going == 'y': NameError: name 'keep_going' is not defined
This option is not correct because in it keep_going is not defined.
