Based on the information given, it is important to give higher priority to the queue that contains the high priority thread.
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What is the Important Fact?</h3>
- A lot of interrupts can take place at any time, and due to that, they cannot always be ignored as the part of code affected by interrupts need to be guarded from constant use.
- So, the load-balancing requirements for keeping about the same number of threads would need to be taken or retained and the important case of top priority thread would also be kept.
- An priority-based scheduling algorithm can handle this situation if one run queue had all high-priority threads and a second queue had all low-priority threads because if giving greater priority to the two queue that has the national priority comment section as well as, so, first method is the thread in all of the queue.
- Multi-level queue scheduling algorithm is used in scenarios where the processes can be classified into groups based on property like process type, CPU time, IO access, memory size, etc. One general classification of the processes is foreground processes and background processes.
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The components that is responsible for providing instructions and processing for a computer is a. CPU.
<h3>What area of the computer executes commands?</h3>
This command center's central processing unit (CPU) is a sophisticated, large-scale collection of electrical circuitry that carries out pre-stored program instructions. A central processing unit is a must for all computers, big and small.
Note that the CPU, RAM, and ROM chips are all located on the motherboard. The "brain" of the computer is known as the Central Processing Unit (CPU). It carries out instructions (from software) and directs other parts.
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Answer:
The display will be 17.
Explanation:
Tracing through the program:
At first, math is called and sent 1 and 2 - so 1 is stored to ans1 and 2 is stored to ans2.
If the user inputs 3 for a and 4 for b, the program then calls function math2 and passed values of 3 and 4.
So now inside of math2, 3 is stored to res1 and 4 is stored to res3. Inside of this function, res1 and res2 are added together - so then 7 is stored to d and then returned back to the original function.
So now 7 was stored back to the variable c. Then a and ans1 are added together (3 + 1 = 4) and b and ans2 are added together (4 + 2 = 6). Each of these values are stored back to e and f.
Then those values, e and f (4 and 6) are again sent to math2, which simply adds the values together and returns it back to the function. So 10 is sent back to math and stored to the value of g.
Then c (7) and g (10) are added together and displayed.
