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Ghella [55]
3 years ago
8

Nix’It Company’s ledger on July 31, its fiscal year-end, includes the following selected accounts that have normal balances (Nix

’It uses the perpetual inventory system). Merchandise inventory $ 44,800 Sales returns and allowances $ 5,100 Retained earnings 129,300 Cost of goods sold 109,200 Dividends 7,000 Depreciation expense 11,700 Sales 161,600 Salaries expense 39,500 Sales discounts 4,300 Miscellaneous expenses 5,000 A physical count of its July 31 year-end inventory discloses that the cost of the merchandise inventory still available is $42,950.Prepare journal entries to close the balances in temporary revenue and expense accounts. Remember to consider the entry for shrinkage.
Business
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer and Explanation:

The Journal entries are shown below:-

1. Sales Dr, $161,600

          To Income summary $161,600

(Being To close a temporary account with credit balances is recorded)

2. Income summary Dr, $176,650

      To Sales discount $4,300

       To Sales return and allowance $5,100

        To Cost of good sold $111,050

        To Depreciation expenses $11,700

        To Salaries expenses $39,500

        To Miscellaneous expenses $5,000

(Being to close a temporary account with a debit balance is recorded)

Working note:-

shrinkage based on physical count = $44,800 - $42,950

= $1,850

Cost of good sold = $109,200 + $1,850

= $111,050

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A. Reverse logistics systems are usually less cost- efficient than forward-based systems.

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8 0
2 years ago
​U(X,Y)equals=20Xplus+80Yminus−Upper X squaredX2minus−2Upper Y squaredY2 where X is his consumption of CDs with a price of ​$11
ankoles [38]

Answer:

The number of CDs = 111.36

The number of movie videos = 242.72

N/B: I choose not to round up the answers.

Explanation:

The method used is the Lagrangian method. Basically, the optimization problem we are trying to solve is  the utility function u(x,y) = 20x+80y -x^2 -y^2

subject to the constraint

11x + 22y = 6565.

So the optimization problem(Lagrangian) is

\Delta = 20x + 80y -x^2 -y^2- \lambda(11x+22y-6565),

where \lambda is a constant called the Lagrange multiplier.

To find the optimal consumption, we need to maximize the Lagrangian with respect to the variables x,y,\lambda. This we do by differentiating \Delta with respect to each variable and then equate to 0.

\Delta_x : 11\lambda = 20 - 2x ........................(1) \\\Delta_y: 11\lambda = 40 -y .........................(2) \\\Delta_\lambda = 11x + 22y = 6565............................(3) \\

Equate (1) and (2), to get y = 20+2x and substitute into (3) to get x = 111.36. Substituting x = 111.36 into 20+2x to get the corresponding value of y.

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4 years ago
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