Consider a computer with two processes, H, with high priority, and L, with low priority. The scheduling rules are such that H is run whenever it is in ready state. At a certain moment, with L in its critical region, H becomes ready to run (e. g., an I/O operation completes). H now begins busy waiting, but since L is never scheduled with H is running, L never gets the chance to leave its critical region, so H loops forever. This situation is sometimes referred to as the priority | inversion problem. If instead of priority scheduling, we use round-robin scheduling. Will we encounter the same problem? Please explain your answer in detail.

Write a function that draws a pool ball. this function should take as parameters, the color, the number that should go on the pool ball, and the location of the center of the pool ball. the radius of the pool balls should be pool_ball_radius, and the font of the number should be pool_ball_font. the text of the pool ball font should be white. drawpoolball(color.orange, 5, 100, 100); drawpoolball(color.green, 6, 50, 200); drawpoolball(color.red, 3, 150, 350); drawpoolball(color.blue, 2, 250, 140); to center the numbers on the pool ball, you should use the getwidth() and getheight() methods. you are allowed to call these methods on your text object, such as txt.

What does a bonus object do? a. subtracts lives b. keeps track of a player's health c. gives a player an advantage d. makes text appear

Read each statement below. if the statement describes a peer-to-peer network, put a p next to it. if the statement describes a server-based network, put an s next to it. p - peer-to-peer s - server-based