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Oksanka [162]
1 year ago
6

What is the core content included in each section of the report?

Business
1 answer:
Inga [223]1 year ago
5 0

Title page, Table of contents, Executive summary, Introduction, Discussion, Conclusion, Recommendations and References are the core content included in each section of the report.

A report is a written document that organizes information for a particular audience and use. Complete reports are nearly always supplied in the form of written papers, despite the fact that summaries of reports may be given verbally.

Report writing is a formal way of writing in-depth about a subject. Reports always have an official tone. Always consider who the intended audience is before writing a section. Writing a report about a school event, a business case, etc. are a few examples.

Learn more about core content of report here

brainly.com/question/13171394

#SPJ4

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The new Fore and Aft Marina is to be located on the Ohio River near Madison, Indiana. Assume that Fore and Aft decides to build
Nataly [62]

Po = 0.5385, Lq = 0.0593 boats, Wq = 0.5930 minutes, W = 6.5930 minutes.

<u>Explanation:</u>

The problem is that of Multiple-server Queuing Model.

Number of servers, M = 2.

Arrival rate, \lambda= 6 boats per hour.

Service rate, \mu= 10 boats per hour.

Probability of zero boats in the system,\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\} = 0.5385

<u>Average number of boats waiting in line for service:</u>

Lq =[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0

= [\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385 = 0.0593 boats.

The average time a boat will spend waiting for service, Wq  =  0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.

The average time a boat will spend at the dock, W =  0.009883 plus (1 divide 10) = 0.109883 hours = 6.5930 minutes.

4 0
3 years ago
How can you value our farmers in your own simple way​
SOVA2 [1]

Answer:

By helping them

Explanation:

We need to cooperate

3 0
3 years ago
A coffee distributor needs to mix a(n) Costa Rican coffee blend that normally sells for $9.50 per pound with a Kenya coffee blen
snow_tiger [21]

Answer:

5,11 pounds of Costa Rican coffee and 64,89 pounds of Kenya coffee

Explanation:

First, we need to know the proportion of every coffee in the mix trying with different percentages until getting the result of $13,49  

 

($9,50*25%)+($13,80*75%)=$12,73  

 

($9,50*10%)+($13,80*90%)=$13,37  

 

($9,50*7,5%)+($13,80*92,5%)=$13,48  

 

($9,50*7,3%)+($13,80*92,7%)=$13,49  

 

So, the percentage of Costa Rican coffee is 7,3% and Kenya coffee is 92,7%  

And we can get the pounds required to get 70 pounds  

 

70*7,3%=5,11

 

70*92,7%=64,89

   

We need 5,11 pounds of Costa Rican coffee and 64,89 pounds of Kenyan coffee to create 70 pounds of mixed coffee that can sell for $13,49 per pound  

8 0
3 years ago
What is likely to happen if a borrower misses a payment on a credit card account?
julsineya [31]
The borrower will get a late fee for not paying on the due date. 
6 0
3 years ago
Read 2 more answers
Suppose the data have a bell-shaped distribution with a mean of 25 and a standard deviation of 5. Use the empirical rule to dete
Zielflug [23.3K]

Answer:

a) 15 to 35 approximately 95%

(b) 10 to 40 approximately almost all

(c) 20 to 30 approximately 68%

Explanation:

The data have a bell-shaped distribution which means the data is equally distributed on both sides of the mean.

We have the mean at 25 and a standard deviation of 5 which means that the interval is for each of the values of 5 .

The mean would be u and

The first value would be u ±σ = 25 ± 5= 20 and 30 (68 % )

The second value will be u ± 2σ= 25± 10 = 15 and 35 (95%)

The third value will be u ± 3σ= 25 ± 15 = 10 and 40 (99.7 % almost all)

In the figure below the light blue region gives u ±σ on both sides of the mean

, dark blue gives u ± 2σ values on both sides of the mean and grey gives

u ± 3 σ values on both sides of the mean.

It is obvious that 68 % of the data is contained in the u ±σ light blue region, 95 % of the data in the  u ± 2σ dark blue including light blue and 99.7 % in the u ± 3σ all colored regions.

3 0
3 years ago
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