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kykrilka [37]
2 years ago
13

Which extension is appropriate for Word document templates?

Computers and Technology
2 answers:
expeople1 [14]2 years ago
8 0
The answer to this question is A
Anni [7]2 years ago
8 0

Answer:

A - .docx

Explanation:

Hope this helps!!

:)

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Predictive coding software leverages .............................when experts review a subset of documents to teach the softwar
pantera1 [17]

Answer:

A.Human analysis, documents

Explanation:

Predictive coding in softwares is artificial intelligence that works by automating document review. This involves training software with data from "subset of documents" to be generally applied(apply same logic) to a larger group of documents. This is employed by a large group of technologists to ease the task of manually reviewing a huge set of documents.

7 0
3 years ago
Which of the following is the most compelling
Gemiola [76]
To compare data makes the most sense
5 0
2 years ago
Read 2 more answers
In reference to computer communications, what does the term noise mean?
jeka94
Noise usually means static.
6 0
3 years ago
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Determine the number of character comparisons made by the brute-force algorithm in searching for the pattern GANDHI in the text
leonid [27]

Answer:

Total number of character comparison = 43

Explanation:

Using the Brute force algorithm

The string of n characters is known as text, and the string of m characters is known as the pattern.

From the given information:

The text (n)=THERE_IS_MORE_TO_LIFE_THAN_INCREASING_ITS_SPEED

The pattern (m) = GANDHI

The total no of characters that we have in the text = 47

The total number of characters in pattern = 6

For a brute force algorithm;

Since; the first character of the pattern does not exist in the text, then the number of trials made can be attempted can be expressed as = n – m + 1

= 47 – 6 + 1

= 47 – 5

= 42

Thus; the algorithm will attempt the trial 42 times.

Now, for loop in the algorithm to run 42 times, the G in the pattern will have to align against the for T in the text, and in the last case, it will be aligned against the last space.

On each attempted trial, the algorithm will make one unsuccessful comparison.

However, at the trial at which the G in the pattern Is aligned with the G in the text, there will be two successful comparisons.

Hence, we can calculate the total number of character comparison as follows:

Total number of character comparison = \mathbf{\bigg ( ( 42 -  (no. \ of \  failed \ comparison) ) \times 1 + (1 \times ( Two \ successful \  comparisons) ) \bigg ) }

Total number of character comparison = ( (( 42 – 1) × 1 ) + ( 1 × 2) )

Total number of character comparison = 41 + 2

Total number of character comparison = 43

3 0
3 years ago
⦁ Consider transferring an enormous file of L bytes from Host A to Host B. Assume an MSS of 536 bytes. ⦁ What is the maximum val
timurjin [86]

Answer:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

Explanation:

a)  There are approximately  2^32 = 4,294,967,296 possible number of the sequence. This number of the sequence does not increase by one with every number of sequences but by byte number of data transferred. Therefore, the MSS size is insignificant. Thus, it can be inferred that the maximum L value is representable by 2^32 ≈ 4.19 Gbytes.

b)  ceil(2^32 / 536) = 8,012,999

The segment number is 66 bytes of header joined to every segment to get a cumulative sum of 528,857,934 bytes of header. Therefore, overall number of bytes sent will be 2^32 + 528,857,934 =  4.824 × 10^9 bytes.  Thus, we can conclude that the total time taken to send the file will be 249 seconds over a 155~Mbps link.

8 0
2 years ago
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