Considering the situation described above, the fraction of the CPU execution that is devoted to handling clock interrupts "<u>12 percent</u>."
<h3>CPU Execution process.</h3>
The process of CPU Execution involves the execution of an instruction which in life fetches an instruction from memory through its ALU to carry out an operation and then saves the result to memory.
To illustrate the fraction of the CPU execution that is devoted to handling clock interrupts, we have:
60 × 2 msec = 120 msec ÷ 1 sec = 12 percent.
Hence, in this case, it is concluded that the correct answer is <u>12 percent CPU</u> devoted to the clock.
Learn more about the CPU execution here: brainly.com/question/14400616
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Answer:
When Δ is set to large value, then a process's resident set is overestimated and this might prevent many processes from being scheduled even though their required pages are resident
Explanation:
When Δ is set to a small value, then the set of resident pages for a process might be underestimated, allowing a process to be scheduled even though all of its required pages are not resident.
This could result in a large number of page faults. When Δ is set to large value, then a process's resident set is overestimated and this might prevent many processes from being scheduled even though their required pages are resident.
Hoewever, once a process is scheduled, it is unlikely to generate page faults since its resident set has been overestimated.
Answer:
the second dlc of the second installment of the Black ops series
Explanation: