Consider a Job of 715 bytes that arrives for processing using a page size of 126 bytes, how many bytes would be loaded onto the
1 answer:
Answer:
Total 6 pages will be consumed and 85 bytes will be on the 6th page.
Step-by-step explanation:
Let the number of pages used for the job = x
If the page size is 126 bytes.
Then total number of bytes used for the job = 126x
If these bytes are equal to 715 bytes then the equation will be
126x = 715
x =
= 5.674 pages
That means 6 pages will be consumed.
and consumed bytes in 5 pages will be = 5×126 = 630 bytes
So remaining bytes will be on sixth page and the number will be = 715 - 630 = 85 bytes.
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Using the normal distribution, it is found that the score needed to qualify is of 88.32.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, by and
.
The top 15% of the applicants are selected, hence the score needed to qualify is the 85th percentile, which is X when Z = 1.04, hence:
X - 80 = 1.04(8)
X = 88.32.
The score needed to qualify is of 88.32.
More can be learned about the normal distribution at brainly.com/question/24663213