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Sidana [21]
3 years ago
6

Allowance for Doubtful Accounts has a credit balance of $2,100 at the end of the year (before adjustment), and an analysis of cu

stomers' accounts indicates uncollectible receivables of $19,700. Which of the following entries records the proper adjustment for bad debt expense?
a. debit Bad Debt Expense, $21,800; credit Allowance for Doubtful Accounts, $21,800
b. debit Allowance dfor Doubtful Accounts, $17,600; credit Bad Debt Expense, $17,600
c. debit Allowance for Doubtful Accounts, $21,800; credit Debt Expense, $21,800
d. debit Bad Debt Expense, $17,600; crdit Allowance for Doubful Accounts, $17,600

Other receivables includes all of the followoing EXCEPT:

a. taes receivable
b. interest receivable
c. receivables from employees
d. notes receivabe
Business
1 answer:
OleMash [197]3 years ago
8 0

Answer:

1. Analysis of accounts receivables Allowance Required     $19,700

Less: Credit balance available in Allowance account           <u>$2,100</u>

Additional allowance required                                               <u>$17,600</u>

The journal entry will be as follows

                                                              DEBIT        CREDIT

Bad debt expenses                              $17,600

Allowance for doubtful accounts                            $17,600

Hence, the correct option is D.

2. Other receivables include all except "Notes Receivables"

Hence, the correct option is D

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Answer:

  • The marginal gain from Larry's second hour of work is 60 problems
  • The marginal gain from Larry's fourth hour of work is 20 problems
  • The best combination is 1 hour of working problems + 3 hours of reading

Explanation:

To get the <em><u>marginal gain</u></em> we subtract from the latest hour, in this case the second hour (140), the production from the previous hour (80). 140-80=60. <em>It's always the same, the latest minus the previous one.</em>

So let's do the same for the fourth hour:

Noon................200 problems

minus

11:00 AM..........180 problems

200-180= 20 problems

Now to know how many hours he should spend working on problems and reading, let's compare:

An hour of reading equals to 70 problems made; (because working on 70 problems raises a student’s exam score by about the same amount as reading the textbook for 1 hour).

hours of working problems         problems solved

0............................................................0

1.............................................................80

2............................................................140

3............................................................180

hours reading                    problems equivalent to hours read

4...............................................(4*70)=280

3...............................................(3*70)=210

2...............................................(2*70)=140

1................................................(1*70)=70

finally let's add up the two combinations (0 and 4, 1 and 3, 2 and 2, 3 and 1)

0 and 4_______________0+280= 280

1 and 3________________80+210=290

2 and 2_______________140+140=280

3 and 1________________180+70=250

<em>And the best combination is 1 hour of working problems + 3 hours of reading=</em><em>290</em>

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