How has the used of computers and advances in technology changed the healthcare industry
1 answer:
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Answer:
Following are code of factorial in python language
def factorial(n): # function
result=1 #variable
for x in range(2,n+1): #iterating the loop
result=result*x #storing result
return result #return result
for n in range(10): #iterating loop
print(n,factorial(n)) #print factorial
Output:
Following are the attachment of output
Explanation:
Missing information :
In the question the information is missing the code is missing which we have to correct it following are the code that are mention below.
def factorial(n): # function
result=1 #variable
for x in range( ): #iterating the loop
result= #storing result
return #return result
for n in range( ): #iterating loop
print(n,factorial(n)) #print factorial
Following are the description of code
- We have create a function factorial in this we have pass the one parameter i.e "n".
- In the for loop we have pass 2,n+1 which has been used to iterating the loop calculating factorial and string the result in the result variable
- In the main function we have pass the range on which we have to calculated the factorial
- Finally the print function will print the factorial .
Following are the attachment snip of code in the python language
Answer:
Explanation:
The following code is written in Python. It creates a method for each one of the questions asked and then tests all three with the same test case which can be seen in the picture attached below.
def alternating_list(lst1, lst2):
lst3 = []
for x in range(len(lst1)):
lst3.append(lst1[x])
try:
lst3.append(lst2[x])
except:
pass
if len(lst2) > len(lst1):
lst3.extend(lst2[len(lst1):])
return lst3
def reverse_alternating(lst1, lst2):
lst3 = []
if len(lst1) == len(lst2):
for x in range(len(lst1) - 1, -1, -1):
lst3.append(lst1[x])
lst3.append(lst2[x])
return lst3
def alternating_list_no_extra(lst1, lst2):
lst3 = []
max = 0
if len(lst1) > len(lst2):
max = len(lst2)
else:
max = len(lst1)
for x in range(max):
lst3.append(lst1[x])
try:
lst3.append(lst2[x])
except:
pass
return lst3
