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lyudmila [28]
3 years ago
15

Which of the following guidelines about the subject line of e-mail messages is most appropriate?

Computers and Technology
2 answers:
alekssr [168]3 years ago
5 0

Answer: 2: Use the subject line wisely to let your reader know the main purpose of the e-mail.

============================================================

Explanation:

The subject line is like the title of a book. You want to give a quick sense of what the email is about, but you also don't want to put the entire contents of the book in the title. The subject line only has so much room. Even less room is available if the reader is checking their email on their mobile device.

Choice 1 is false because subject lines are important. I can't think of a case where you can omit the subject line entirely (ie leave it blank), unless it's a situation where the recipient knows in advance what the topic is about. Though that may be rare.

Choice 3 is false. Yes you want to keep the subject line short, but you also don't want to be cryptic or confusing. If you can get away with using 1 or 2 words, then go ahead and do so. However, you may need more words to convey the topic of the email. Go with what feels right and try to put yourself in the shoes of the person reading your email. Ask yourself "Does this convey the message I want to send?"

Choice 4 is false. As mentioned, you do not want to put the entire email in the subject line. The exception of this rule is that if the email is really really short. For every other case, you should have a fairly short title and then go over the details in the email main body itself.

Vikki [24]3 years ago
4 0

Answer:

Use the subject line wisely to let your reader know the main purpose of the e-mail.

Explanation:

i got it right on edge 2021

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We have said that the average number of comparisons need to find a target value in an n-element list using sequential search is
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Answer:

Part a: If the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b: If the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c: The average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

Explanation:

Suppose the list is such that the starting index is 0.

Part a

If list has 15 elements, the middle item would be given at 7th index i.e.

there are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(8,9,10,11,12,13,14) above it. It will have to run 8 comparisons  to find the middle term.

If list has 17 elements, the middle item would be given at 8th index i.e.

there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons  to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons  to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

If list has 16 elements, there are two middle terms as  one at would be at 7th index and the one at 8th index .There are 7 indices(0,1,2,3,4,5,6) below it and 7 indices(9,10,11,12,13,14,15) above it. It will have to run 9 comparisons  to find the middle terms.

If list has 18 elements, there are two middle terms as  one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons  to find the middle terms.

If list has 20 elements, there are two middle terms as  one at would be at 9th index and the one at 10th index .There are 9 indices(0,1,2,3,4,5,6,7,8) below it and 9 indices(11,12,13,14,15,16,17,18,19) above it. It will have to run 11 comparisons  to find the middle terms.

Now this indicates that if the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

Part c

So the average number of comparisons is given as

((n+1)/2+(n+2)/2)/2=(2n+3)/4

So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

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