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3 years ago

Ethan wants to change the font in his document. He should _____.

Computers and Technology

2 answers:

SSSSS [86.1K]3 years ago

8 0

HEYA!!!!

The answer to your question would be

C) Highlight the text and select a new font.

Hope it helps you.

:)

Elena L [17]3 years ago

7 0

The answer is : C. <span>highlight the text and select a new font</span>

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What is the purpose of an arraignment?

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At an arraignment, the court officially illuminates a litigant regarding charges contained in a prosecution or data, gives the respondent a duplicate of the charging instrument, and takes the litigant's response to those charges as a request.

Explanation:

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3 years ago

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A(n) ________ software installation enables you to decide which features you want to install on the hard drive. Select one: A. f

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Answer:

D.Custom

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4.In order for a driver to graduate from a learner’s license to an operator’s license, how many hours of driving need to be docu

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3 years ago

QUESTION 56 Use the Windows ________ to check on a nonresponsive program. Backup utility Error-checking System Restore Task Mana

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Answer: Task manager

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3 years ago

Computer system with a 32-bit logical address and 4k-byte page size. assume that each entry of a page table consists of 4bytes.

Annette [7]

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:

32-bit logical address
Page size = 4 KB = $\mathbf{2^{12} \ bytes }$
Size of each Page table entry = 4 bytes

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:

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

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

= 26 bits

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

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

= 4 MB

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

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2 years ago