4.6 Write a program to prompt the user for hours and rate per hour using input to compute gross pay. Pay should be the normal rate for hours up to 40 and time-and-a-half for the hourly rate for all hours worked above 40 hours. Put the logic to do the computation of pay in a function called computepay() and use the function to do the computation…

4.6.py

def computepay(h,r):
    if h > 40:
        p = 1.5 * r * (h - 40) + (40 *r)
    else:
        p = h * r
    return p
    
hrs = input("Enter Hours:")
hr = float(hrs)
rphrs = input("Enter rate per hour:")
rphr = float(rphrs)

p = computepay(hr,rphr)
print(p)

rate = float(input("Enter Rate of Pay:"))
hrs = int(input("Enter Hours:"))

def computepay(hrs, rate):
hasta_cuarenta = 40*rate
if hrs > 40:
excedente = 1.5 * rate * (hrs - 40)
return hasta_cuarenta + excedente

p = computepay(hrs, rate)
print("Pay", p)

def computepay(h,r):
if h>40:
reg = r * h
otp = (h-40.0)(r0.5)
p = reg + otp

else:
    p = r * h

return p

hrs = input("enter hours:")
hr = float (hrs)
rphrs = input("enter rate:")
rphr = float(rphrs)

p=computepay(hr,rphr)
print('Pay',p)

# Computepay Function
def computepay(h, r) :
    if h <= 40 :
        p = h * r
    else :
        p = (40 * r) + ((h - 40) * r * 1.5)
    return p

# Check input hours
sh = input('Enter Hours: ')
try :
    fh = float(sh)
except :
    print('Error, please enter numeric input')
    quit()

# Check input rate
sr = input('Enter Rate: ')
try :
    fr = float(sr)
except :
    print('Error, please enter numeric input')
    quit()

p = computepay(fh, fr)
print('Pay', p)

def computepay(h, r):
    f_h = float(h)
    f_r = float(r)
    if f_h>40:
        return 1.5*f_r*(f_h-40)+40*f_r
    else:
        return f_h*f_r
#
h = input('HOURS:')
r = input('RATE')
p = computepay(h, r)
print("Pay", p)
```

def computepay (h,r):
if h > 40:
return 40 * r + (h - 40) * r * 1.5
else:
return h * r

h = input ("Enter hours:")
h= float (h)
r = input ("Enter rate:")
r= float (r)

p = computepay(h,r)

print("Pay",p)

`//No need to get confused among these many codes, just put this one and get the desired output as same as Coursera is expecting : 👇

def computepay(h,r):
if h > 40:
p = 1.5 * r * (h - 40) + (40 *r)
else:
p = h * r
return p

hrs = input("Enter Hours:")
hr = float(hrs)
rphrs = input("Enter rate per hour:")
rphr = float(rphrs)

p = computepay(hr,rphr)
print("Pay", p)
`

def computepay(h, r):
return 1.5 * r * (h - 40) + (40 * r) if h > 40 else h * r

hrs = input("Enter Hours:")
hr = float(hrs)
rphrs = input("Enter Rate per hour:")
rphr = float(rphrs)
p = computepay(hr, rphr)
print("Pay", p)

def computepay(h, r):
if h<40:
pay=hr
else:
pay=(40r)+(h-40) r 1.5

return pay

hrs = input("Enter Hours:")
rate=input("Enter Rate")
h=float(hrs)
r=float(rate)
p = computepay(h, r)
print("Pay", p)

def computepay(h,r):
if h>40.0:
p = 40.0r
p = p + (h-40.0)1.5r
else:
p = hr
return p
hrs = input("Enter Hours:")
rate = input("Enter rate:")
h = float(hrs)
r = float(rate)
p = computepay(h,r)
print("Pay", p)

why are we using 1.5 value ?

and if we use p= float()*float() instead compute pay will it affect my output?

def computepay(h,r):
if h > 40:
p = 1.5 * r * (h - 40) + (40 *r)
else:
p = h * r
return p
hrs = input("Enter Hours:")
hr = float(hrs)
rphrs = input("Enter rate per hour:")
rphr = float(rphrs)
p = computepay(hr,rphr)
print("Pay",p)

this is right answer

Why did you use 1.5 to multiply? I used 0.5 and i got the same answer as yours. Can you please explain?
Thanks

def computepay(h, r):
if h > 40 :
p = 1.5 * r * (h - 40) + (40 * r)
else :
p = h * r
return p

hrs = input("Enter Hours:")
hr=float(hrs)
r = input("Enter Rate :")
rt = float(r)

p = computepay(hr , rt)
print("Pay",p)

def computepay(h,r):
if h > 40:
p = (1.5 * r) * (h - 40) + (40 *r)
else:
p = h * r
return p

hrs = input("Enter Hours:")
h = float(hrs)
rate = input("Enter rate per hour:")
r = float(rate)

p = computepay(h,r)
print("Pay",p)

def compute_pay(r,h):
if h <= 40:
return r * h
elif h >=40:
return 40*r+(h-40)r1.5

fhours = input('Enter hours: ')
try:
hrs = float(fhours)
frate = input('Enter rate: ')
rate = float(frate)
x = compute_pay(rate,hrs)
print(x)

why are we using 1.5 value ?

It is a good question; I forgot what the question was. :(

def computepay(h,r):
if h <= 40:
return h*r
elif h > 40:
return 40 * r +(h-40) * 1.5 * r

hrs = input("Enter Hours:")
h = float(hrs)
rate = input("Rate")
r = float(rate)

p = computepay(h,r)

print("Pay",computepay(h,r))

hrs = input("Saati gir pezo: ") h = float(hrs) hrt = input("Rate kac:") r = float(hrt)

def computepay(a, b): if (a <= 40): islem = a * b elif (a > 40): islem = 40 * b + (a-40)_b_1.5 return islem

t = computepay (h,r)

print("Pay",t)

#worked thnku

#The specific mnemonics are used intentionally
#Not sure multiplication symbol not updating
def computepay(parameter1,parameter2):
if parameter1<=40:
gpay=(argument1argument2)
elif parameter1>40:
gpay=((40argument2)+(argument1-40)(1.5argument2))
return gpay

argument=input("Enter the hours:")
rargument=input("Enter the rate:")
try :
argument1=float(argument)
argument2=float(rargument)
except :
error=("Please enter the numeric value")
print(error)
quit()
print("pay",computepay(argument1,argument2))

def computepay(h, r):
if h > 40.0:
p = (1.5 * r) * (h - 40.0) + (r * 40.0)
else:
p = r * h
return p

shrs = input('Enter Hours: ')
srate = input('Enter Rate: ')

try:
hours = float(shrs)
rate = float(srate)
except:
print('ERROR, please enter numeric input')
quit()

p = computepay(hours, rate)
print("Pay", p)

def computepay(hrs, rph):
if hrs > 40:
return(1.5rph(hrs-40)+(40rph))
elif hrs <= 40:
return(hrsrph)

hrs = input("Enter Hours:")
rph = input("rate per hour")

hr=float(hrs)
rp=float(rph)

p=computepay(hr,rp)
print("Pay",p)

For sure this will work out

def computepay(h,r):
if h>40:
reg=40r
otp=(h-40.0)1.5r
pay=reg+otp
else:
pay=hr

return pay

hrs =input ("Enter Hours:")
rate=input ("Enter Rate per Hour:")
fh= float(hrs)
fr= float(rate)
p = computepay(fh,fr)
print("Pay",p)

try:
def computepay(h, r):
if h>40:
reg=rh
otp=(h-40.0)(r*0.5)
xp=reg+otp
else:
xp=h+r
return xp

hrs = input("Enter Hours:")
h=float(hrs)
rt = input("Enter rate:")
r=float(rt)
p = computepay(h, r)
print("Pay", p)

except:
print ("bad value")
quit()

can check this:

def computepay(h,r):
if h > 40:
p = 1.5 * r * (h - 40) + (40 *r)
else:
p = h * r
return p

hrs = input("Enter Hours:")
hr = float(hrs)
rphrs = input("Enter rate per hour:")
rphr = float(rphrs)

p = computepay(hr,rphr)
print("Pay", p)

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…

On Thu, May 30, 2024 at 11:58 PM FardadF4 ***@***.***> wrote: ***@***.**** commented on this gist. ------------------------------ can check this: def computepay(h,r): if h > 40: p = 1.5 * r * (h - 40) + (40 *r) else: p = h * r return p hrs = input("Enter Hours:") hr = float(hrs) rphrs = input("Enter rate per hour:") rphr = float(rphrs) p = computepay(hr,rphr) print("Pay", p) — Reply to this email directly, view it on GitHub <https://gist.github.com/ZeroRaven/6c82f2c8e2a2dc327f42c9fe858da3c3#gistcomment-5073892> or unsubscribe <https://github.com/notifications/unsubscribe-auth/AK5MVHJYPN6Y54YRN3EMZLTZE5VTNBFKMF2HI4TJMJ2XIZLTSKBKK5TBNR2WLJDUOJ2WLJDOMFWWLO3UNBZGKYLEL5YGC4TUNFRWS4DBNZ2F6YLDORUXM2LUPGBKK5TBNR2WLJDHNFZXJJDOMFWWLK3UNBZGKYLEL52HS4DFVRZXKYTKMVRXIX3UPFYGLK2HNFZXIQ3PNVWWK3TUUZ2G64DJMNZZDAVEOR4XAZNEM5UXG5FFOZQWY5LFVA4TKNBUGA4DINNHORZGSZ3HMVZKMY3SMVQXIZI> . You are receiving this email because you are subscribed to this thread. Triage notifications on the go with GitHub Mobile for iOS <https://apps.apple.com/app/apple-store/id1477376905?ct=notification-email&mt=8&pt=524675> or Android <https://play.google.com/store/apps/details?id=com.github.android&referrer=utm_campaign%3Dnotification-email%26utm_medium%3Demail%26utm_source%3Dgithub> .

def computepay(h, r):
if h > 40:
p = 0.5 * r * (h - 40 ) + (h * r)
else:
h = h * r
return p

hrs = input("Enter Hours:")
Rate = input("Enter Rate:")
h = float(hrs)
r = float(Rate)

p = computepay(h, r)

print("Pay:", p)

good

ni ce

good