Hey there!
A prepaid tuition plan is a plan that allows you to pay the current rate of tuition now (say, years in advance), even if it's much higher than the time when the payment for tuition would actually be paid. This plan is great for anyone who wants to pay a lower price for high–cost education now (even if their kid isn't college age yet) and not have to worry about economic standing or inflation in the future that could drive the tuition prices up.
The plan only covers tuition and other similar fees. You can not purchase books or room and board with it in advance. So, your answer should be: C. Tuition and Fees.
Hope this helped you out! :-)
<h2>Answer:</h2>
<h2>intelligence - centered</h2>
<h2>
Explanation:</h2>
I hope it helps you
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
Answer:
Following are the program in the C++ Programming Language.
//set header file
#include <iostream>
//set namespace
using namespace std;
//define class
class format
{
//set access modifier
public:
//set string type variable
string res;
//define function
void names(string first_name, string last_name)
{
//set if-else if condition to check following conditions
if(first_name.length()>0 && last_name.length()>0)
{
res="Name: "+last_name+", "+first_name;
}
else if(first_name.length()>0 and last_name.length()==0)
{
res="Name: "+first_name;
}
else if(first_name.length()==0 and last_name.length()==0)
{
res="";
}
}
//define function to print result
void out(){
cout<<res<<endl;
}
};
//define main method
int main() {
//set objects of the class
format ob,ob1,ob2;
//call functions through 1st object
ob.names("John","Morris");
ob.out();
//call functions through 2nd object
ob1.names("Jhon","");
ob1.out();
//call functions through 3rd object
ob2.names("", "");
ob2.out();
}
<u>Output</u>:
Name: Morris, John
Name: Jhon
Explanation:
<u>Following are the description of the program</u>:
- Define class "format" and inside the class we define two void data type function.
- Define void data type function "names()" and pass two string data type arguments in its parameter "first_name" and "last_name" then, set the if-else conditional statement to check that if the variable 'first_name' is greater than 0 and 'last_name' is also greater than 0 then, the string "Name" and the following variables added to the variable "res". Then, set else if to check that if the variable 'first_name' is greater than 0 and 'last_name' is equal to 0 then, the string "Name" and the following variable "first_name" added to the variable "res".
- Define void data type function "out()" to print the results of the variable "res".
- Finally, we define main method to pass values and call that functions.
Answer:
a. 118.20.210.254
Explanation:
Here are the few characteristics of Class A:
First bit of the first octet of class A is 0.
This class has 8 bits for network and 24 bits for hosts.
The default sub-net mask for class A IP address is 255.0.0.0
Lets see if the first bit of first octet of 118.20.210.254 address is 0.
118 in binary (8 bits) can be represented as : 1110110
To complete 8 bits add a 0 to the left.
01110110
First bit of the first octet of class A is 0 So this is class A address.
For option b the first octet is : 183 in the form of bits = 10110111 So it is not class A address
For option c the first octet is 215 in the form of bits = 11010111 So it is not class A address
For option d the first octet is 255 in the form of bits = 11111111. The first bit of first octet is not 0 so it is also not class A address.
