<u>Answer:</u>
<em>void main ( ) </em>
<em> { </em>
<em> int counter; </em>
<em> cout<<""Even numbers between 1 to 20 are:""<<endl ; </em>
<em> //Method 1
</em>
<em> for (counter = 1; counter <= 20; counter++) </em>
<em> { </em>
<em> if ( counter%2 == 0) </em>
<em> {
</em>
<em> cout<<counter<<""\t""<<endl ; </em>
<em> } </em>
<em> } </em>
<em>//Method 2 – simplest one
</em>
<em>for (counter = 2; counter <= 20;) </em>
<em> { </em>
<em> cout<<counter<<""\t""<<endl ; </em>
<em>counter = counter + 2;
</em>
<em> </em>
<em> }
</em>
<em>
</em>
<em> return 0; </em>
<em>}
</em>
<u>Explanation:</u>
In this, Method 1 runs a for loop and check whether each number is divided by 2. If yes, then printed otherwise it is skipped.
In the second method, it runs for loop only for even numbers. <em>This is obtained by incrementing the counter by 2.
</em>
Answer:
Explanation:
The following procedure is written in Python. It takes the next argument, checks if it is an odd number and if so it adds it to oddsum. Then it asks the user for a new number from the keyboard and calls the accumulator procedure/function again using that number. If any even number is passed the function terminates and returns the value of oddsum.
def accumulator(next, oddsum = 0):
if (next % 2) != 0:
oddsum += next
newNext = int(input("Enter new number: "))
return accumulator(newNext, oddsum)
else:
return oddsum
Answer: Base Stations
Base Stations are responsible for receiving and transmitting signals in a cellular network. These stations are equipped with an antenna or multiple antennas that allow the users to send and receive the calls and texts they send with their cellular phones.
Depending on the size and the range that a base station can reach, these are often stationed in multiple areas to allow cellular signals to constantly be connected to the cellular networks of the different service providers. Without these stations the customers of the providers will not have a signal to send or receive texts or calls.
Answer:
I guess there isnt any password!
Explanation: