<span>An associate's degree requires two years of academic study and is the highest degree available at a community college</span>
Answer:
def cal(n):
s=0
while n>0:
r=n%10
if(r==0 or r==4 or r==6 or r==9):
s=s+1
elif r==8:
s=s+2
n=n//10
print(s)
n=int(input("enter number:"))
print(n)
cal(n)
Explanation:
- Create a function to calculate count of closed path
.
- Create a variable to store count of closed path
.
- While number is positive
, extract last digit of n
.
- Reduce number by truncating last digit
.
- Make a function call to compute count of path.
Answer:
The correct option to the following question is b.) value-added networks.
Explanation:
A Van, or the value-added network, involves the use of the common carrier's that is phone line to allows the business to business network communications.
Network is the “value-added” because they have various services and the enhancements which improve the way of the business applications which communicate with each other.
<u>Answer:</u>
<em>1. The loop will run unless the value of answer equals to Capital N.</em> The method equals is “case-sensitive” that is lower case letter and upper case letters are considered different. That “a” and “A” means the same to us, but for the method equals it is different and equals method will return false if the case is not matching.
<em>2. a (int)(Math.random() * (upper − lower) ) + lower </em>
Since we need to consider the lower limit value together to get the desired results we need to add the value of lower limit to the multiplied answer.
<em>3. b while(userGuess != secretNumber || userGuess >= lowerLimit && userGuess <= upperLimit)</em>
Here the program is required to check the while loop and exit when user guess the number or we can say the loop should continue until the user guess the number, so that is why we have taken <em>userGuess!=secretNumber. </em>Next the loop should be exited if it is not within the range or we can say that the loop should run only if the guessed number is within the upper and lower limit.<em> That is why we have opted for the condition userGues>=lowerlimit && userGuess<=upperlimit.</em> Why we have taken && as the operator is that, it’s a range of values so && operator best suit for this kind of logical condition.
Answer:
#include <iostream>
#include <cstdlib>
using namespace std;
char grade(double marks){
if(marks>=90)
{
return 'A';
}
else if (marks >=80 && marks<90)
{
return 'B';
}
else if (marks >=70 && marks<80)
{
return 'C';
}
else if (marks >=60 && marks<70)
{
return 'D';
}
else if ( marks<60)
{
return 'F';
}
}
int main()
{
double marks;
cout <<"Ener marks";
cin >>marks;
char grd=grade(marks);
cout<<"Grae is "<<grd;
return 0;
}
Explanation:
Take input from user for grades in double type variable. Write function grade that takes a parameter of type double as input. Inside grade function write if statements defining ranges for the grades. Which if statement s true for given marks it returns grade value.
In main declare a variable grd and store function returned value in it.
