Program:-
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2 answers:
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Answer:
Explanation:
This code was written in Java. It creates the Cars class with the requested variables and methods. It also creates the TestCars class which asks the user for the necessary inputs and then creates three Cars objects and passes the input values to the constructors. Finally, it uses the show() method on each object to call the information. A test has been created and the output can be seen in the attached image below.
import java.util.Scanner;
class TestCars{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Enter Car Make: ");
String make1 = in.next();
System.out.print("Enter Car Model: ");
String model1 = in.next();
System.out.print("Enter Car Year: ");
int year1 = in.nextInt();
System.out.print("Enter Car Make: ");
String make2 = in.next();
System.out.print("Enter Car Model: ");
String model2 = in.next();
System.out.print("Enter Car Year: ");
int year2 = in.nextInt();
System.out.print("Enter Car Make: ");
String make3 = in.next();
System.out.print("Enter Car Model: ");
String model3 = in.next();
System.out.print("Enter Car Year: ");
int year3 = in.nextInt();
Cars car1 = new Cars(make1, model1, year1);
Cars car2 = new Cars(make2, model2, year2);
Cars car3 = new Cars(make3, model3, year3);
car1.show();
car2.show();
car3.show();
}
}
class Cars {
String makes, models;
int years;
public Cars(String makes, String models, int years) {
this.makes = makes;
this.models = models;
this.years = years;
}
public void show() {
System.out.println("Car's make: " + this.makes);
System.out.println("Car's model: " + this.models);
System.out.println("Car's year: " + this.years);
}
}
Answer:
z = a.c' + a.b.d' + b.c'.d'
Explanation:
The truth table for this question is provided in the attachment to this question.
N.B - a' = not a!
The rows with output of 1 come from the following relations: 01 > 00, 10 > 00, 10 > 01, 11 > 00, 11 > 01, 11 > 10
This means that the Boolean expression is a sum of all the rows with output of 1.
z = a'bc'd' + ab'c'd' + ab'c'd + abc'd' + abc'd + abcd'
On simplification,
z = bc'd' + ab'c' + ac'd' + ac'd + abc' + abd'
z = ac' + abd' + bc'd'
Hope this helps!
