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Elis [28]
3 years ago
6

A married man gets a new job in a company. After three months, he meets a younger woman in the finance department and they begin

having an affair. The affair carries on openly in the workplace over the course of 6 months, then begins to sour when she discovers that he is already married. The relationship quickly changes to bickering and name-calling while at work. The woman eventually breaks off the relationship, quits her job, and sues the company for fostering a hostile work environment. Of the seven IT infrastructure domains, the USER domain was most at risk.
a. True
b. False
Computers and Technology
1 answer:
pav-90 [236]3 years ago
5 0

Answer:

False ( B )

Explanation:

Of  the seven IT  infrastructure domains The USER domain was not at risk because the User Domain is not task with handling the sharing of data or Mutual communication between users in a Typical IT infrastructure.

The domain charged with such responsibility is The LAN domain because the LAN domain is charged with the sharing of information between the USERS and maintaining good mutual communication, and since the relationship has become sour, the sharing of information between the users will suffer the most and it will be at a high risk

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The number of bits in the physical address is 26 bits. The number of entries in a page table is \mathbf{2^{20}} entries. The size of the page table in a one-level paging scheme is 4MB.

<h3>
What is paging in Operating System?</h3>

Paging is a storage method used in operating systems to recover activities as pages from secondary storage and place them in primary memory. The basic purpose of pagination is to separate each procedure into pages.

We are given the following parameters:

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Suppose the system supports physical memory size = 64 MB = \mathbf{2^{26} \ bytes}

Thus, the number of bits in the physical address is computed as:

= \mathbf{log_2 \{Physical-memory-size\}}

=\mathbf{log_2(2^{26})}}

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The number of entries in a page table = logical address space size/page size

The number of entries in a page table \mathbf{= \dfrac{2^{32}}{2^{12}}}

\mathbf{=2^{20}} entries

In a one-level paging scheme, the size of the table is:

= entire no. of page entries × page table size

= \mathbf{2^{20}\times 4 \ bytes}

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b.

suppose that this system supports up to 2^30 bytes of physical memory.

  • The size of the page table will be the same as 4 MB due to the fact that the number of entries, as well as, the page table entry size is the same.

Since the size of the page table surpasses that of a single page. A page cannot include a whole page table. Therefore, the page table must be broken into parts to fit onto numerous pages, and an additional level of the page table is required to access this page table.

  • This is called the Multi-Level Paging system.

Therefore, we can conclude that the number of bits in the physical address is 26 bits, the number of entries in a page table is \mathbf{2^{20}} entries, and the size of the page table in a one-level paging scheme is 4MB.

Learn more about Paging in Operating System here:

brainly.com/question/17004314

#SPJ1

4 0
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Answer:

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In this question except option (d), all options are wrong.

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