Answer: $1,500
Explanation:
The future value of value using simple interest is:
Future value = Value * ( 1 + rate * time)
2,400 = Value * (1 + 15% * 4)
2,400 = Value * 1.6
Value = 2,400 / 1.6
Value = $1,500
Answer:
C
Explanation:
Putting all government forms on the city web site is the least activity likely to be effective in the purpose of reducing digital divide.
Holding basic computer classes at the community centers will very much help to reduce the digital divide.
Providing free wireless internet connections at locations in low-income neighborhood will also reduce the gap of digital divide
Requiring that every city school has computers that meet a minimum hardware and software will made computing resources available to users thereby reducing digital divide.
Answer:
The answer to this question is given below in the explanation section.
Explanation:
The symbol asterisk (*) in a select query retrieve all result of column in the table. For example, table named "employee" has three column such as <em>id, name, address</em>.
To apply the symbol asterisk (*) in select querry select all the columns in the result.
for example:
select * employee;
this query statement selects all columns of table "employee" in the result.
Answer:
// program in C++ to check leap year.
// include header
#include<iostream>
using namespace std;
// main function
int main() {
// variable
int inp_year ;
// ask user to enter year
cout<<"Enter year:";
// read year
cin>>inp_year;
// check year is leap or not
if (((inp_year % 4 == 0) && (inp_year % 100 != 0)) || (inp_year % 400 == 0))
// if leap year , print leap year
cout<<inp_year<<" is a leap year.";
else
// print not leap year
cout<<inp_year<<" is not a leap year.";
return 0;
}
Explanation:
Read year from user.Then check if year is divisible by 4 and not divisible by 100 or year is divisible by 400 then year is leap year.otherwise year is not leap year.
Output:
Enter year:1712
1712 is a leap year.
Answer:
user_name = input("Please enter a name or type Nope to terminate the program: ")
while( user_name != "Nope" ):
print("Nice to meet you ", user_name )
user_name = input("Please enter a name or type Nope to terminate the program: ")
Explanation:
Get the name from user as an input and store it in the user_name variable.
Run a while loop until user_name is not equal to the value of Nope.
Inside the while loop, display the name of user and repeat the question until user types Nope.
