To answer this question, we need to understand the concept of importing packages and static members.

In the given code, the class Beta and the enum Color are in different packages. To access the enum Color in the class Beta, we need to import it. Additionally, since the enum Color is defined using the "enum" keyword, we can import its static members individually.

Let's go through each option to understand why it is correct or incorrect:

Option A) import sun.scjp.Color.*; This option is incorrect because it imports all the members (including non-static members) of the Color enum. It does not allow us to access the enum values directly without specifying the enum name.

Option B) import static sun.scjp.Color.*; This option is incorrect because it imports all the static members of the Color enum. It does not import the enum itself. Therefore, we won't be able to access the enum values directly without specifying the enum name.

Option C) import sun.scjp.Color; import static sun.scjp.Color.*; This option is correct because it imports the Color enum and its static members individually. It allows us to access the enum values directly without specifying the enum name.

Option D) import sun.scjp.; import static sun.scjp.Color.; This option is incorrect because it imports all the members of the sun.scjp package, including non-static members of the Color enum. It does not import the enum itself. Therefore, we won't be able to access the enum values directly without specifying the enum name.

Option E) import sun.scjp.Color; import static sun.scjp.Color.GREEN; This option is correct because it imports the Color enum and only the GREEN static member. It allows us to access the GREEN enum value directly without specifying the enum name.

The correct answers are C and E.

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• Wrong explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) collie - This option is incorrect. If we follow the code execution, we see that the switch statement matches the value of myDog with the cases. The value of myDog is Dogs.collie, so it matches the case collie on line 17. However, there is no break statement after the System.out.print("collie ") on line 18, so the execution continues to the next case.

Option B) harrier - This option is incorrect. As explained in Option A, after the execution of the System.out.print("collie ") on line 18, there is no break statement, so the execution continues to the next case. Therefore, the output will include both "collie" and "harrier".

Option C) Compilation fails - This option is incorrect. The code provided does not contain any compilation errors.

Option D) collie harrier - This option is correct. As explained in Option A and Option B, the output will include both "collie" and "harrier". So the correct answer is D.

The correct answer is D.

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• Wrong explanation

To determine the result of the given code, let's go through each line:

1. String test = "a1b2c3"; - This line declares and initializes a String variable named "test" with the value "a1b2c3".

2. String[] tokens = test.split("\d"); - This line splits the string "test" using the regular expression "\d", which matches any digit. The result is an array of strings, where each element contains the substrings separated by digits. In this case, the array will contain ["a", "b", "c"].

3. for(String s : tokens) System.out.print(s + " "); - This line iterates over each element in the "tokens" array and prints it followed by a space. The output will be "a b c".

Therefore, the result of the given code is:

A) a b c

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• Wrong explanation

To answer this question, let's go through each option:

Option A) An encapsulated, public class promotes re-use - This option is true. Encapsulation is a concept in object-oriented programming that promotes re-usability by hiding the internal implementation details of a class and exposing only the necessary public methods or properties. By encapsulating a class, other developers can use the class without needing to know the intricate details of how it works, thus promoting re-use.

Option B) Classes that share the same interface are always tightly - This option is false. Classes that share the same interface may or may not be tightly coupled. The level of coupling between classes depends on how they interact with each other and the dependencies they have. It is possible to have loosely coupled classes that share the same interface.

Option C) An encapsulated class allows subclasses to overload methods, but - This option is false. Encapsulation is a mechanism that hides the internal implementation details of a class. Overloading methods is a concept related to polymorphism, where a class can have multiple methods with the same name but different parameters. Encapsulation and method overloading are separate concepts.

Option D) An encapsulated class allows a programmer to change an - This option is true. Encapsulation provides a level of abstraction and encapsulates the internal implementation details of a class. This allows a programmer to change the internal implementation of a class without affecting the code that uses the class. The external code only interacts with the public methods or properties of the encapsulated class, so changes to the internal implementation can be made without impacting the code that uses the class.

Based on the explanations above, the correct options are A) An encapsulated, public class promotes re-use and D) An encapsulated class allows a programmer to change an.

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• Wrong explanation