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You can have a review of what you have learned through this version.

NEW QUESTION: 1
DRAG DROP
Given:
Runnable r = new Runnable() {
public void run() {
try {
Thread.sleep(1000);
} catch (interruptedException e) {
System.out.println("interrupted");
}
System.out.println("ran");
}
};
Thread t = new Thread(r);
t.start();
System.out.println("started");
t.sleep(2000);
System.out.println("interrupting");
t.interrupt();
System.out.println("ended");
Assue that sleep(n) executes in exactly n milliseconds, and all other code executes in an insignificant amount of time. Place the fragments in the output area to show the result of running this code.
Select and Place:

Answer:
Explanation:

NEW QUESTION: 2
What security management control is MOST often broken by collusion?
A. Separation of duties
B. Least privilege model
C. Increased monitoring
D. Job rotation
Answer: A

NEW QUESTION: 3
You are working for a company that designs mobile applications. They maintain a server where player records are assigned to their different games. The tracking system is new and in development.
The application uses Entity Framework to connect to an Azure Database. The database holds a Player table and Game table.
When adding a player, the code should insert a new player record, and add a relationship between an existing game record and the new player record.
The application will call CreatePlayerWithGame with the correct gameIdand the playerId to start the process. (Line numbers are included for reference only.)

For each of the following statements, select Yes if the statement is true. Otherwise, select No.
NOTE: Each correct selection is worth one point.

Answer:
Explanation:

Explanation:
Many-to-many relationships without an entity class to represent the join table are not yet supported. However, you can represent a many-to-many relationship by including an entity class for the join table and mapping two separate one-to-many relationships.
protected override void OnModelCreating(ModelBuilder modelBuilder)
{
modelBuilder.Entity<PostTag>()
.HasKey(t => new { t.PostId, t.TagId });
modelBuilder.Entity<PostTag>()
.HasOne(pt => pt.Post)
.WithMany(p => p.PostTags)
.HasForeignKey(pt => pt.PostId);
modelBuilder.Entity<PostTag>()
.HasOne(pt => pt.Tag)
.WithMany(t => t.PostTags)
.HasForeignKey(pt => pt.TagId);
}
}