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disa [49]
3 years ago
8

Anton needs to identify the eight different scoring statistics for all the players on his school’s basketball team. Which graphi

c organizer would be best for this assignment? A. Chart C. Box B. Venn diagram D. Any of these
Business
2 answers:
loris [4]3 years ago
8 0

Anton needs to identify the eight different scoring statistics for all the players on his school’s basketball team. Which graphic organizer would be best for this assignment?

Answer: A. Chart

Out of the options, A. a chart is the best option because it allows you to break down all of the scoring statisics for each player on the school's basketball team. Organizating the statistics by player and their ranges in a chart makes it easier to understand and utilize the information collected.

Oksanka [162]3 years ago
6 0
A. Chart because there are 8 items that need to be compared. Not sure about a "box" but I know the Venn diagrams are only good for 2 so it crosses out C and D.
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RealTurf is considering purchasing an automatic sprinkler system for its sod farm by borrowing the entire $65,000 purchase price
lesantik [10]

Answer:

The project return is lower than the minimum accepted of 15% thus not profitable for the company

Net Present Value -1.279,86‬

Explanation:

<u>Loan Present value</u>

PMT of the loan:

PV \div \frac{1-(1+r)^{-time} }{rate} = C\\

PV 65,000

time   4

rate 0.12

65000 \div \frac{1-(1+0.12)^{-4} }{0.12} = C\\

C  $ 21,400.238

Present value at MARR:

C \times \frac{1-(1+r)^{-time} }{rate} = PV\\

C $21,400.24

time 4 years

rate 0.15

21400.2383598698 \times \frac{1-(1+0.15)^{-4} }{0.15} = PV\\

PV $61,097.2175

<u>Salvage value:</u>

\frac{Salvage }{(1 + rate)^{time} } = PV  

Salvage $9,000

time  9 years

rate  0.15000

\frac{9000}{(1 + 0.15)^{9} } = PV  

PV   2,558.36

<u>Cost savings present value:</u>

Cost savings per year:           25,000

less maintenance expenses (13,000)

net cash flow                          12,000

C \times \frac{1-(1+r)^{-time} }{rate} = PV\\

C $ 12,000

time 9 years

rate 0.15

12000 \times \frac{1-(1+0.15)^{-9} }{0.15} = PV\\

PV $57,259.0070

Net Present Value

PV cost savings + PV salvage - PV loan payment

57,259 + 2,558.36 - 61,097.22 = -1.279,86‬

3 0
3 years ago
The prepaid insurance account had a balance of $9,400 at the beginning of the year. The account was debited for $10,400 for prem
soldier1979 [14.2K]

Answer:

Following are the response to the given points:

Explanation:

\text{Insurance Expense} =\text{ Beginning Prepaid Insurance} + \text{Cash Premium Paid} -\text{Ending Prepaid Insurance}

                             = \$9,400 + \$10,400 - \$3,730\\\\ = \$16,070

Date  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ General \ Journal  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  Dr.  \  \ \ \ \ \ \ \ \ Cr. \\\\ March \ 31\ \ \ \ \ \ \ \ Insurance \ Expense  \ \ \ \ \ 16,070 \ \ \ \ \ \ \ \ \ \ \ \ \  \\\\

                       Prepaid\ Insurance \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 16,070\\\\

For point b:

Date  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ General \ Journal  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  Dr.  \  \ \ \ \ \ \ \ \ Cr. \\\\ March \ 31\ \ \ \ \ \ \ \ Insurance \ Expense  \ \ \ \ \ 18,100 \ \ \ \ \ \ \ \ \ \ \ \ \  \\\\

                       Prepaid\ Insurance \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 18,100\\\\

5 0
3 years ago
A consumer charges a $2,530. 16 purchase on their credit card. The card has a daily interest rate of 0. 42%. If the consumer pay
aleksandr82 [10.1K]

Answer:

C. $31.88 is the correct answer.

Explanation:

8 0
2 years ago
The demand for subassembly S is 8080 units in week 7. Each unit of S requires 11 unitnothing of T and 11 unitnothing of U. Each
slavikrds [6]

Answer:

Please find the detailed answer as follows:

Explanation:

Consider the table attached in the excel.

The above are the requirements of the units of material if demand of 8080 units of S has to be completed.

The requirement of units of U for maintenance is given in the second table.

As the lead time for S is two weeks the components should be ready by the fifth week.

As the lead time for components T and U both is 2 weeks their components should be ready by the third week.

Hence in the first week the required number of units of V, W, Y and Z are to be ordered as all of them are required in the third week.

The required number of units of X are to be ordered in the second week to be ready in third week.

The required units of U, Y and Z are to be ordered considering the requirement to make S plus the maintenance requirements.

Download xlsx
6 0
3 years ago
Quiz
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Your insurance company may reward you with a lower rate or deductible
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