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Llana [10]
3 years ago
6

What is the range of addresses for conditional branches in MIPS (K = 1024)?

Computers and Technology
1 answer:
sergejj [24]3 years ago
3 0
Addresses up to about 128K before the branch to about 128K after.
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a network administrator for a large oil company has discovered that a host on the company network has been compromised by an att
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My recommendation as an immediate response to prevent further spoofing of the host is to Revoke the host's certificate.

<h3>What does revoke certificate mean?</h3>

Certificate revocation is the process by which one can  invalidate a TLS/SSL and this is one that is often done before its scheduled expiration date.

Note that A certificate need to be revoked immediately  and as such, based on the case above, My recommendation as an immediate response to prevent further spoofing of the host is to Revoke the host's certificate.

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) Which of these will not select all the cells in a document?
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Pressing enter two times I’m pretty sure that’s right cuz like it does nothing when I do it. It’s either that or idk up to u to take an answer from someone who has bad grammar
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A merge is _____.
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The process of combining information from a variety of sources
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onsider a file system on a disk that has both logical and physical block sizes of 512 bytes. assume that the information about e
anygoal [31]

For three disk allocation strategies (contiguous, linked, and indexed),

in this case, the  physical blocks must be read from the disk are 1,4 and 2 respectively.

In the disk, every block has a sequential number. The majority of disks today do logical I/O in hardware. Physical I/O is seldom ever required of operating systems today.

File access uses virtual I/O. Similar to disk logical I/O is disk virtual I/O. The distinction is that with virtual I/O, as opposed to physical I/O, the blocks that make up a file are

You could want to transfer from virtual block 10 to virtual block 4 in a contiguous allocation. Since the file is contiguous and you are at logical block 10 + Z, block four can be accessed by reading logical block 4 + Z. Without any intermediaries, that can be done directly.

Each block in the linked structure has 511 bytes of data and one byte that serves as an offset to the following block.

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If blocks are numbered from 1 to 10, then then is the "4" correct. The remainder of the solutions, however, make the assumption that block numbering begins at 0, using division to obtain offsets.

Assume there is a contiguous index for the indexed way of allocating files. Once more, it is assumed that each offset in the index is one byte in size.

Let's say that Z serves as the index's origin. Then, the byte at block Z + 3 DIV 512 and the offset at 3 MOD 512 (3 = 4 - 1) make up the entry for block 4 (numbered 1 to 4).

You must read the index and the data block in order to locate the block (2 reads).

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