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hodyreva [135]
3 years ago
10

Kelly decided to accept the risk and purchased a high growth stock. Her returns for the past five years are 32 percent, 24 perce

nt, -48 percent, 12 percent, and -9 percent, respectively. What is the standard deviation of these returns
Business
1 answer:
Mazyrski [523]3 years ago
5 0

Answer:

32.03%

Explanation:

The computation of the standard deviation is as follows;

As we know that

Average return = Total return ÷Total time period

= (32 + 24 - 48 + 12 - 9) ÷ 5

= 2.2%

Now

Return         (Return - Average Return)^2

32                  (32 - 2.2)^2 = 888.04

24                 (24 - 2.2)^2 = 475.24

-48                (- 48 - 2.2)^2  = 2520.04

12                  (12 - 2.2)^2 = 96.04

-9                   (-9 - 2.2)^2 = 125.44

Total =                 4104.8%

Now

Standard deviation is

= [Total (Return - Average Return)^2 ÷ (Time period- 1)]^(1 ÷ 2)

= [4104.8 ÷ (5 - 1)]^(1 ÷ 2)

= [4104.8 ÷ 4]^(1 ÷ 2)

= 32.03%

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Answer:

<u>B. shows planned purchase rates of goods and services at various price levels.</u>

Explanation:

  • The aggregate demand is the total demand for final goods and services in the economy over a given period of time. And is often distinguished as the effective demand curve. That is the demand for the GDP of the nation.
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3 years ago
The following transactions and adjusting entries were completed by Robinson Furniture Co. during a three-year period. All are re
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Answer:

<u>Year 1 </u>

Jan. 8. Purchased a used delivery truck for $24,000, paying cash.

  • Dr Truck 24,000
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Mar. 7. Paid garage $900 for changing the oil, replacing the oil filter, and tuning the engine on the delivery truck.

  • Dr Maintenance expenses - Truck 900
  •     Cr Cash 900

Dec. 31. Recorded depreciation on the truck for the fiscal year. The estimated useful life of the truck is four years, with a residual value of $4,000 for the truck.

Depreciation expense = 2 x 0.25 x $24,000 = $12,000

  • Dr Depreciation expense 12,000
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<u>Year 2 </u>

Jan. 9. Purchased a new truck for $50,000, paying cash.

  • Dr Truck new 50,000
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Feb. 28. Paid garage $250 to tune the engine and make other minor repairs on the used truck.

  • Dr Maintenance expenses - Truck 250
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Apr. 30. Sold the used truck for $9,500. (Record depreciation to date in Year 2 for the truck.)

depreciation expense = 2 x 0.25 x 4/12 x $12,000 = $2,000

  • Dr Depreciation expense 2,000
  •     Cr Accumulated depreciation - truck 2,000

truck sold at $9,500 - $10,000 (carrying value) = -$500 loss on sale

  • Dr Cash 9,500
  • Dr Accumulated depreciation 14,000
  • Dr Loss on sale - truck 500
  •     Cr Truck 24,000

Dec. 31. Record depreciation for the new truck. It has an estimated residual value of $12,000 and an estimated life of eight years.

Depreciation expense = 2 x 0.125 x $50,000 = $12,500

  • Dr Depreciation expense 12,500
  •     Cr Accumulated depreciation - truck new 12,500

<u>Year 3 </u>

Sept. 1. Purchased a new truck for $58,500, paying cash.

  • Dr Truck three 58,500
  •     Cr Cash 58,500

Sept. 4. Sold the truck purchased January 9, Year 2, for $36,000. (Record depreciation to date for Year 3 for the truck.)

Depreciation expense = 2 x 0.125 x 8/12 x $37,500 = $6,250

  • Dr Depreciation expense 6,250
  •     Cr Accumulated depreciation - truck new 6,250

truck sold at $36,000 - $31,250 (carrying value) = $4,750 gain on sale

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Dec. 31. Recorded depreciation on the remaining truck. It has an estimated residual value of $16,000 and an estimated useful life of 10 years.

Depreciation expense = 2 x 0.1 x 4/12 x $58,500 = $3,900

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Joseph is reviewing secondary data his company collected about seasonal variations in consumer spending because he is thinking a
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Answer:

The advantages of using secondary data are several, but its main advantage is that it is the cheapest way to gather large sets of information. A lot of secondary data is available on the internet, so it is time saving. Using secondary data saves work, efforts and money.

We can also use secondary data to determine more specifically which primary data we need to gather, again saving resources.

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3 years ago
A stock had returns of 17.88 percent, −5.16 percent, and 20.39 percent for the past three years. What is the variance of the ret
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Answer:

Variance of the return = 0.01983

Explanation:

S^{2}= Σ(X-X)^{2}/ N - 1

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Variance = [0.004683 + 0.026233 + 0.008748]/2

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