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Charra [1.4K]
3 years ago
8

Hey lol *simping noises* anyways. im in Rain but without the \ so SpEaK to me haha

SAT
2 answers:
Alecsey [184]3 years ago
6 0

Answer:

Thx for points again meanie :)

Explanation:

Brrunno [24]3 years ago
3 0
Uhmmm ma’am/sir i am not sure what’s going on exactly
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Youmin has accounts receivable of $95,400 at march 31, 2021 (amounts in thousands). Credit terms are 2/10, n/30. At march 31, 20
Elodia [21]

Based on the information given bad debt expense is $8,675 and the journal entry is: Debit Bad debt  $6,575; Credit Allowance for Doubtful Account $6,575.

<h2>a. Bad debt expense:</h2>

Accounts Receivable   Amount       %          Estimated Uncollectible

Current                          $65,000         2%              $1,300

1–30 days past due      $12,900          5%              $645

31–90 days past due    $10,100          30%            $3,030

Over 90 days past due $7,400          50%            $3,700

Total                                                                         $8,675

<h2>b. Journal entry:</h2>

March 31 2021

Debit Bad debt  $6,575

Credit Allowance for Doubtful Account $6,575

($8,675-$2,100)

Inconclusion bad debt expense is $8,675 and the journal entry is: Debit Bad debt  $6,575; Credit Allowance for Doubtful Account $6,575.

Learn more about bad debt expense here:brainly.com/question/24871617

6 0
2 years ago
I need the full thing solved
Nesterboy [21]

Answer:

x-axis: Amount of candy sold

y- axis: Amount of money made

Billy is making $0.20 per candy bar.

Linear function: y=3/5x-6

Billy sold 50 candy bars if he made $24.

The x-intercept represents the point when he will earn a profit because at point (5,-3) he sold 5 candy bars but he still made a loss of $3.

5 0
2 years ago
Assuming Leakey's hypothesis of a sixth mass extinction to be true, how will we expect the model to change?
aivan3 [116]
If you are using USA test prep the answer would be:
<span>
C). <span>a sharp spike in the graph immediately following the "0" location of the X axis</span></span>
6 0
2 years ago
What mass of glucose (C6H12O6) should be dissolved in 12. 0 kg of water to obtain a solution with a freezing point of -5. 8 ∘C?.
bogdanovich [222]

The mass of glucose solute dissolved in the solution is 6.739 Kg.

Recall that;

ΔT = K m i

ΔT = Freezing point depression

K =Freezing point depression constant = 1.86°C/mol

m = molality of the solution

i = Van't Hoff factor = 1 (molecular solution)

We have to find the freezing point depression from;

Freezing point depression = Freezing point of pure solvent - Freezing point of solution

Freezing point of pure water = 0°C

Freezing point of solution = -5. 8 ∘C

Freezing point depression = 0°C - (-5. 8 ∘C) = 5. 8 ∘C

Now;

m = ΔT/K i

m = 5. 8 ∘C/ 1.86°C/mol × 1

m = 3.12 m

But molality = number of moles of solute/mass of solvent in Kg

Molar mass of solute = 180 g/mol

Let the mass of solute be m

3.12 = m/180/12

3.12 = m/180 × 12

m = 3.12 × 180 × 12

m = 6739 g or 6.739 Kg

Learn more: brainly.com/question/6249935

8 0
2 years ago
Read 2 more answers
In order to ensure your message is clear and easily understood you should ____________.
guajiro [1.7K]
I think you should review your message, make sure there aren't a lot or any mistakes and stuff like that. 

7 0
3 years ago
Read 2 more answers
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