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ohaa [14]
3 years ago
15

Gomez Corp. uses the allowance method to account for uncollectibles. On January 31, it wrote off an $2,200 account of a customer

, C. Green. On March 9, it receives a $1,700 payment from Green. 1. Prepare the journal entry for January 31 2. Prepare the journal entries for March 9; assume no additional money is expected from Green.
Business
2 answers:
MariettaO [177]3 years ago
5 0

Answer::

A journal entry is the act of keeping or making records of any transactions. These transactions are listed in an accounting journal that shows a company's debit and credit balances.

Using the data in the question (above), the journal entries for January 31 and March 09 go as follows:

No ------------------Date----------------General Journal------------- Debit-----------Credit

1. Jan 31 --------Alowance for doubtful accounts ------- $2,200 (Debit)

-- ------------------ Accounts receivable—C. Green -------- $2,200 (Credit)

2. Mar 09 -------Accounts receivable—C. Green --------$1,700 (Debit)

----_----------------Allowance for doubtful accounts ------ $1,700 (Credit)

3. Mar 09 ------- Cash ---------- $1,700 (Debit)

-----------------------Accounts receivable—C. Green -------- $1,700 (Debit)

almond37 [142]3 years ago
3 0

Answer:

31 Jan

Debit allowance for doutful debts $2200 and credit Accounts Receivables $2200

Mar 9 Debit accounts Receivable $1700 and credit Allowance for doubtful debt $ 1700

Mar 9 Debit cash $1700 and Credit Account receivable $1700

Explanation:

When a customer account becomes certain to be uncollectible, it must be written off by debiting the allowance for doubful debt and then when the customer pays we credit the allowance for doubtful debt and debit accounts receivables to reverse the account written of and then finally debit cash and credit account receivables to account for cash received.

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What is the difference between ancient trade and modern trade ​
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The main difference between traditional trade and modern trade is that, distribution in modern trade is more organized. Retailers often deal directly with manufacturers. Many large retail chains have integrated their services to offer their own brands in groceries and other goods.

Explanation:

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3 years ago
Knottworth Gedding Consulting leased machinery from Red Inc. on July 1, 2021. The lease was recorded as a finance lease. The pre
Kobotan [32]

Answer:

The answer is: Knottworth Gedding should report $1,725,000 as interest expense in its 12/31/2021 income statement

Explanation:

The formula for calculating the amount of interest expense is:

interest expense = discount rate x (present value - yearly payment) x time

  • Discount rate = 10%
  • Present value = $40,500,000
  • Yearly  payment = $6,000,000
  • Time = 6 months / 12 months = 0.5

interest expense = 10% x ($40,500,000 - $6,000,000) x 0.5 = $1,725,000

6 0
3 years ago
Mitchell bought 600 shares of centerco two years ago for 34.50 per share. He sold them yesterday for 38.64 per share.
soldi70 [24.7K]

Sure, here is my possible correct answer:

1. 38.64 - 34.50 = 4.14

2. 4.14 x 600 = 2484

So, Mitchell would earn $2484 in (gross) profit.

I hope it helped you!

4 0
4 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
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