Answer:
The code is given in the explanation section
Explanation:
//Class Airconditioner
public class AirConditioner {
private boolean turnOnOff;
//The Constructor
public AirConditioner(boolean turnOnOff) {
this.turnOnOff = turnOnOff;
}
//method turn_on
public void turn_on(){
this.turnOnOff = true;
}
//method turn_off
public void turn_off( ){
this.turnOnOff = false;
}
}
// A new class to test the airconditional class
class AircondionTest{
public static void main(String[] args) {
//Creating an object of the Aircondional class
AirConditioner office_a_c = new AirConditioner(false);
//Using the reference varible to call method turn_on
office_a_c.turn_on();
}
}
Answer:
(a)The CPU B should be selected for the new computer as it has a low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A.
(b) The CPU B is faster because it executes the same number of instruction in a lesser time than the CPU A
.
Explanation:
Solution
(a)With regards to the MIPS performance metric the CPU B should be chosen for the new computer as it low clock cycle time which implies that it will implement the process or quicker when compared to the CPU A and when we look at the amount of process done by the system , the CPU B is faster when compared to other CPU and carries out same number of instruction in time.
The metric of response time for CPU B is lower than the CPU A and it has advantage over the other CPU and it has better amount as compared to CPU A, as CPU B is carrying out more execution is particular amount of time.
(b) The execution can be computed as follows:
Clock cycles taken for a program to finish * increased by the clock cycle time = the Clock cycles for a program * Clock cycle time
Thus
CPU A= 5*10^6 * 60*10^-9 →300*10^-3 →0.3 second (1 nano seconds =10^-9 second)
CPU B= 3 *10^6 * 75*10^-9 → 225*10^-3 → 0.225 second
Therefore,The CPU B is faster as it is executing the same number of instruction in a lesser time than the CPU A
Answer:
myDoubles.add(75.6);
Explanation:
ArrayList<Double> myDoubles = new ArrayList<Double>();
myDoubles.add(10.8);
myDoubles.add(92.46);
myDoubles.add(75.6);
The above code creates a double ArrayList called myDoubles. We add 10.8 and 92.46 initially. After these numbers, 75.6 is added to the myDoubles using <em>add</em> method (You need to type the list name, ".", and the method name to add the number)
I believe that you either press the help icon or the F1 key, so Elisa will be the one to access General Help.
Answer:
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