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2 years ago

Can anyone help me find the area of these two pls ? ty

Mathematics

2 answers:

Mrrafil [7]2 years ago

8 0

97.2

Yes yes yes yes yes yes

alina1380 [7]2 years ago

7 0

Answer:

1. 92 in²

2. 97.2 yds²

Step-by-step explanation:

1. area of a rectangle: b x h

b = 9.2

h = 10

9.2 x 10 = 92

2. area of paralellogram: b x h

b = 12

h = 8.1

12 x 8.1 = 97.2

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Graph the solution to the system of inequalities in the coordinate plane 3y>2x+122+y<-5

alexdok [17]

The graph is shown in the attached image.

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1 year ago

Solve using substitution y = -4x + 10 y = -x + 1. will give brainleist if correct

gayaneshka [121]

Answer:

Step-by-step explanation:

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2 years ago

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Allowance method entries

Feliz [49]

Using the Allowance Method, the relevant transactions can be completed in the books of Wild Trout Gallery as follows:

1. Allowance for Doubtful Accounts

Accounts Debit Credit

Jan. 1 Beginning balance $53,800

Jan. 19 Accounts Receivable 2,560

Apr. 3 Accounts Receivable $14,670

July 16 Accounts Receivable 19,725

Nov. 23 Accounts Receivable 4,175

Dec. 31 Accounts Receivable 25,110

Dec. 31 Ending balance $56,500

Dec. 31 Bad Debts Expenses $55,470

Totals $116,005 $116,005

Accounts Receivable

Accounts Debit Credit

Jan. 1 Beginning balance $2,290,000

Jan. 19 Allowance for Doubtful 2,560

Jan. 19 Cash $2,560

Apr. 3 Allowance for Doubtful 14,670

July 16 Allowance for Doubtful 19,725

July 16 Cash 6,575

Nov. 23 Allowance for Doubtful 4,175

Nov. 23 Cash 4,175

Dec. 31 Allowance for Doubtful 25,110

Dec. 31 Sales Revenue 8,020,000

Dec. 31 Cash $8,944,420

Dec. 31 Ending balance $1,299,500

Totals $10,316,735 $10,316,735

3. Expected net realizable value of the accounts receivable as of December 31 = $1,243,000 ($1,299,500 - $56,500)

Allowance for Doubtful Accounts ending balance = $40,100 ($8,020,000 x 0.5%)

Allowance for Doubtful Accounts

Accounts Debit Credit

Jan. 1 Beginning balance $53,800

Jan. 19 Accounts Receivable 2,560

Apr. 3 Accounts Receivable $14,670

July 16 Accounts Receivable 19,725

Nov. 23 Accounts Receivable 4,175

Dec. 31 Accounts Receivable 25,110

Dec. 31 Ending balance $40,100

Dec. 31 Bad Debts Expenses $39,070

Totals $99,605 $99,605

4. a. Bad Debt Expense for the year = $39,070

4.b. Balance for Allowance Accounts = $40,100

4.c. Expected net realizable value of the accounts receivable = $1,259,400 ($1,299,500 - $40,100)

Data Analysis:

Jan. 19 Accounts Receivable $2,560 Allowance for Uncollectible Accounts $2,560

Jan. 19 Cash $2,560 Accounts Receivable $2,560

Apr. 3 Allowance for Uncollectible Accounts $14,670 Accounts Receivable $14,670

July 16 Cash $6,575 Allowance for Uncollectible Accounts $19,725 Accounts Receivable $26,300

Nov. 23 Accounts Receivable $4,175 Allowance for Uncollectible Accounts $4,175

Nov. 23 Cash $4,175 Accounts Receivable $4,175

Dec. 31 Allowance for Uncollectible Accounts $25,110 Accounts Receivable $25,110

Accounts Receivable ending balance = $1,299,500

Allowance for Uncollectible Accounts ending balance = $56,500

Learn more: brainly.com/question/22984282

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2 years ago

Hi! I need to get this answered right away asap! Waht is 4500, divided by 23400?

ANEK [815]

According to a calculator, this is the answer: 0.1923076923. Hope this helped?

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3 years ago

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In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

7 0

3 years ago