Complete Question:
A campus bookstore sells both types and in the last semester sold 56% laptops and 44% desktops. Reliability rates for the two types of machines are quite different, however. In the first year, 5% of desktops require service, while 15% of laptops have problems requiring service.
Given that a computer required service, what is the probability that it was a laptop?
Answer:
Probability = 0.084
Explanation:
Given
Laptops = 56%
Desktop = 44%
Service Required (Laptop) = 15%
Service Required (Desktop) = 5%
Required
Determine the probability that a selected computer is a laptop and it requires service.
The question tests our knowledge of probabilities using "and" condition.
What the question requires is that, we calculate the probability of selecting a LAPTOP that REQUIRES SERVICE
Note the capitalised words.
This will be calculated as follows:
Probability = P(Laptop) and P(Service Required (Laptop))
[Substitute values for P(Laptop) and P(Service Required (Laptop))]
Probability = 56% * 15%
[Convert to decimal]
Probability = 0.56 * 0.15
Probability = 0.084
Answer:
Explanation:
C++ Code
#include <iostream>
#include <cstdlib>
using namespace std;
int main(){
double hour,minute;
cout<<"Enter Hours :";
cin>>hour;
cout<<"Enter Minutes :";
cin>>minute;
minute = minute+15;
if(minute >=60){
hour++;
minute = minute-60;
}
if(hour>23){
hour = 0;
}
cout<<"Hour: "<< hour<< " Minutes: "<<minute;
return 0;
}
Code Explanation
First take hours and minutes as input. Then add 15 into minutes.
If minutes exceeds from 60 then increment into hours and also remove 60 from minutes as hours already incremented.
Then check if hours are greater then 23 then it means new day is start and it should be 0.
Output
Enter Hours :9
Enter Minutes :46
Hour: 10 Minutes: 1
Answer:
HTML Tags: <input/>, <canvas></canvas>, <div></div>
CSS Attributes: padding, font-family, margin
Explanation:
