Answer:
D. Add
Explanation:
PowerPoint application can be defined as a software application or program designed and developed by Microsoft, to avail users the ability to create various slides containing textual and multimedia informations that can be used during a presentation. Some of the features available on Microsoft PowerPoint are narrations, transition effects, custom slideshows, animation effects, formatting options etc.
In this scenario, Dione has created a PowerPoint presentation that has several common nouns, names of products, etc. He is running Spell Checker and does not want to be notified in regard to these words in this presentation or in any other presentation created on this computer. Hence, the option he should choose is Add. This would be used to automatically add a list of all the common nouns.
The connector for ethernet cables is called RJ45. for phone jack is RJ11. Although they look the same, the RJ11 has only 4 leads, as opposed to 9 leads in RJ45. Also, RJ11 is not as wide as RJ45, so the connector does not fit.
Even if it would, the signals would be totally incompatible, so no success can be expected.
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
Answer:
ouch what happened to him
congrats you ruined my childhood
(thanks for the free points!!)
Er..... i do not know the answer to this
