<span>n/2 = average number of items to search.
Or more precisely (n+1)/2
I could just assert that the answer is n/2, but instead I'll prove it. Since each item has the same probability of being searched for, I'll simulate performing n searches on a list of n items and then calculate the average length of the searches. So I'll have 1 search with a length of 1, another search looks at 2, next search is 3, and so forth and so on until I have the nth search looking at n items. The total number of items looked at for those n searches will be:
1 + 2 + 3 + 4 + ... + n
Now if you want to find the sum of numbers from 1 to n, the formula turns out to be n(n+1)/2
And of course, the average will be that sum divided by n. So we have (n(n+1)/2)/n = (n+1)/2 = n/2 + 1/2
Most people will ignore that constant figure of 1/2 and simply say that if you're doing a linear search of an unsorted list, on average, you'll have to look at half of the list.</span>
Answer:
$413,000
Explanation:
Calculation to determine the carrying value of the note as of September 30, 2021
Carrying value=[$420,000 - ($420,000 .010*6/12)]+ [($420,000 .010*6/12)*4/6]
Carrying value=[$420,000-$21,000]+ ($21,000*4/6)
Carrying value=[$420,000-$21,000]+ $14,000
Carrying value=$399,000+ 14,000
Carrying value=$413,000
Therefore the carrying value of the note as of September 30, 2021 is $413,000
The blank will be filled by services.
<h3>What do you mean by services?</h3>
Services are intangible activities or advantages that a business offers to meet customers' demands in exchange for cash or other valuables.
<h3>Which should be fill in blank?</h3>
There may be a great deal of interaction between a service provider and a consumer in which they co-create value together. In such situations, the customer perceives a high degree of Blank services between the service provider and the company he or she represents.
Learn more about services here brainly.com/question/24553900
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<span>25 years: No Payment, but total is 250000
6 months earlier. Payment of "P". It's value 1/2 year later is P(1+0.03)
6 months earlier. Payment of "P". It's value 1 year later is P(1+0.03)^2
6 months earlier. Payment of "P". It's value 1½ years later is P(1+0.03)^3
6 months earlier. Payment of "P". It's value 2 years later is P(1+0.03)^4
</span><span>We need to recognize these patterns. Similarly, we can identify the accumulated value of all 50 payments of "P". Starting from the last payment normally is most clear.
</span>
<span>P(1.03) + P(1.03)^2 + P(1.03)^3 + ... + P(1.03)^50
That needs to make sense. After that, it's an algebra problem.
P[(1.03) + (1.03)^2 + (1.03)^3 + ... + (1.03)^50]
</span>
P(<span><span>1.03−<span>1.03^51)/(</span></span><span>1−1.03) </span></span>= <span>250000</span>
Answer:
Units to be produced will be 540
So option (a) will be the correct answer
Explanation:
We have given number of units sold = 500 units
Beginning inventory is given = 60 units
And ending Inventory= 100 units
We have to find units to be produced
Units to be produced is given by
Units to be Produced= Ending Inventory + Units to be Sold - Beginning Inventory = 500 + 100 - 60 = 540 units
So 540 units are produced
So option (a) will be the correct answer
