All categories

3 years ago

Dakota Company had net sales (at retail) of $260,000.

Business

1 answer:

disa [49]3 years ago

8 0

Answer:

$35,860

Explanation:

The computation of the ending inventory using the retail inventory method is shown below

Particulars Cost Retail

Opening Inventory(A) $63,800 $128,400

Purchases(B) $115,060 $196,800

Goods available

C=(A-B) $178,860 $325,200

Cost ratio

($178,860 ÷ $325,200 × 100) 55%

Sales at retail (D) $260,000

End, Inventory at Retail $65,200

($325,200 - $260,000)

End, Inventory at Cost $35,860

($65,200 × 55%)

You might be interested in

Managers tell employees what to do because employees are not seen as capable of solving problems themselves. employees mostly fe

STALIN [3.7K]

The first two scenarios are Hierarchical control and the third scenario is Decentralized control.

6 0

3 years ago

Suppose during the course of a year an economy produces $7 trillion of consumer goods, $1 trillion of investment goods, $5 trill

In-s [12.5K]

Answer:

D) $12 trillion.

Explanation:

GDP is the sum of all final goods and services produced in an economy within a given period which is usually a year.

GDP = Consumption + Investment + Government Spending + Net Export

Net Export = Export - Import

Net Export = $1 - $2 = -$1

GDP = $7 + $1 + $5 - $1 = $12

All calculations are in trillion

I hope my answer helps you.

4 0

3 years ago

How much will we end up buying?

makvit [3.9K]

I think its 500 cause yes

4 0

2 years ago

QS 11-9 Recording warranty repairs LO P4 On September 11, 2016, Home Store sells a mower for $490 cash with a one-year warranty

prisoha [69]

Answer:

On September 11, 2016, Home Store sells a mower for $490 cash with a one-year warranty that covers parts

Recording of revenue:

Cash $490 (debit)

Revenue $490 (credit)

Recording of Warranty granted :

Assurance Warranty expense $49.00 (debit)

Warranty Provision $49.00 (credit)

$490 × 10% = $49.00

On July 24, 2017, the mower is brought in for repairs covered under the warranty requiring $34 in materials taken from the Repair Parts Inventory

When warranty is subsequently received:

Warranty Provision $ 34 (debit)

Repair Parts Inventory $ 34 (credit)

Explanation:

On September 11, 2016, Home Store sells a mower for $490 cash with a one-year warranty that covers parts

Recording of revenue:

Cash $490 (debit)

Revenue $490 (credit)

We Recognise Revenue to depict transfer of control of mower

Recording of Warranty granted :

Assurance Warranty expense $49.00 (debit)

Warranty Provision $49.00 (credit)

$490 × 10% = $49.00

There is no option for customer to take the warranty or not, so this is a service warranty.The warranty is measured at the best estimate of expenditure required to settle the obligation that is at 10% of sales.

On July 24, 2017, the mower is brought in for repairs covered under the warranty requiring $34 in materials taken from the Repair Parts Inventory

When warranty is subsequently received:

Warranty Provision $ 34 (debit)

Repair Parts Inventory $ 34 (credit)

Utilise the Warranty Provision when the warranty claim is subsequently received

7 0

3 years ago

Suppose 20.0 g pieces of gold and iron, both initially at 100oC, are added to different containers of water, both initially at 2

Sauron [17]

Answer:

The final temperature in the container with Gold is 27.49 ⁰C

The final temperature in the container with Iron is 33.01 ⁰C

Therefore, the highest final temperature is obtained in the container with Iron.

Explanation:

Q = mcΔT

Where;

Q is the quantity of heat gained or lost

m is the mass of the metals or water

c is the specific heat capacity

ΔT is the change in temperature, T₂ - T₁

T₂ is the final temperature and T₁ is the initial temperature

Heat lost by metals at 100°C is equal to heat gained by water at 25°C

$-Q_{metal} = Q_{water}$

$-Q_{Au} = Q_{H_2O} \\-Q_{Fe} = Q_{H_2O}$

Specific heat capacity of water = 4.18 J/g°C

Specific heat capacity of gold = 0.129 J/g°C

Specific heat capacity of iron = 0.45 J/g°C

⇒For Gold

-20*0.129*(T₂ - 100) = 18*4.18 (T₂ - 25)

-2.58T₂ +258 = 75.24T₂ - 1881

77.82T₂ = 2139

T₂ = 2139/77.82

T₂ = 27.49 ⁰C

⇒For Iron

-20*0.450*(T₂ - 100) = 18*4.18 (T₂ - 25)

-9T₂ +900 = 75.24T₂ - 1881

84.24T₂ = 2781

T₂ = 2781/84.24

T₂ = 33.01 ⁰C

Therefore, the highest final temperature is obtained in the container with Iron.

5 0

3 years ago