A. self-efficacy.
B. overreward inequity.
C. expectancy.
D. cognitive distortion.
Answer:B. overreward inequity.
Explanation: Overreward inequity is a term mostly associated with the feeling of guilt by an employee who believes that his contributions to work or his performance rating is less than what he or she is been paid for.
An employee with this kind of guilt feelings will mostly like do his or her best to engage in in performance improvement Activities which includes enrollment for training and classes to enhance his or her performance.
The best answer to the question that is being presented above would be letter d. One of the biggest risks involved in using electronic mail or e-mail is the amount of junk mail you get from unsolicited, spamming, and phishing e-mail addresses.
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:
int calculate_cost(int quantity) {
double cost = 0;
if (quantity <= 20)
cost = quantity * 23.45;
else if (quantity >= 21 && quantity <= 100)
cost = quantity * 21.11;
else if (quantity > 100)
cost = quantity * 18.75;
return cost;
}
Explanation:
Create a function called calculate_cost that takes one parameter, quantity
Initialize the cost as 0
Check the quantity using if else structure. Depending on the quantity passed, calculate the cost. For example, if the quantity is 10, the cost will be $234.5
Return the cost
