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Aleks [24]
2 years ago
14

Hariette would like to review the income and expenses that were actually paid last month so she can determine how much to set as

ide for her estimated taxes. What accounting method should Hariette choose when she runs the Profit & Loss report? Cash (basis) Accrual (basis) Modified (basis) Hybrid (basis)
Mathematics
1 answer:
Brrunno [24]2 years ago
5 0

Answer:

Hybrid basis

Step-by-step explanation:

There are different methods of accounting used by businesses depending on their peculiar needs. Below are the type of accounting methods:

- Cash basis is when revenues and expenses are recognised when cash is recieved or paid out.

-Accrual basis is when revenue and expenses are recognised when they are earned. For example if services are rendered to a client that will pay in a week's time, since service has already been given it is considered that the future payment has been earned.

- Modified basis combines elements of cash and accrual basis. For example considering short term assets like accounts receivable and accounts payable as cash items. Long term assets are recorded on accrual basis.

- Hybrid basis is used when cash and accrual methods are used for various expenses and tax. Mostly it is used for internal accounting purposes.

In this scenario Hariette would like to review the income and expenses that were actually paid last month. This requires a cash basis that shows actual amount recieved and paid last month. Account receivable and payable are not considered.

In setting aside money for tax she will employ accrual basis accounting. It is an expense that is estimated for future use.

So the hybrid basis is the method that will be most suitable.

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Solve the system of linear equations by graphing. Round the solution to the nearest tenth.
jeyben [28]

Answer:

The approximate solution to the system is (1.2, 4.4)

x = 1.2 and y = 4.4

Step-by-step explanation:

The solution of the system of linear equations equation y = –0.25x + 4.7 and y = 4.9x – 1.64 is shown in the attached graph. The red line represents the equation y = –0.25x + 4.7 and the blue line represents the equation  = 4.9x – 1.64.

The solution of the system of equations is their point of intersection shown on the graph.

The point of intersection is (1.231, 4.392). To the nearest tenth, it is (1.2, 4,4). So x = 1.2 and y = 4.4.

So the approximate solution to the system is (1.2, 4.4)

6 0
1 year ago
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You work on a concrete crew starting a construction project. Once the pouring begins, it must continue non-stop until it is comp
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Answer: B. 6

Step-by-step explanation:

8 0
2 years ago
A pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm, and 96 cm but does not resonate at any wave
erma4kov [3.2K]

Answer:

A Pipe that is 120 cm long resonates to produce sound of wavelengths 480 cm, 160 cm and 96 cm but does not resonate at any wavelengths longer than these. This pipe is:

A. closed at both ends

B. open at one end and closed at one end

C. open at both ends.

D. we cannot tell because we do not know the frequency of the sound.

The right choice is:

B. open at one end and closed at one end .

Step-by-step explanation:

Given:

Length of the pipe, L = 120 cm

Its wavelength \lambda_1 = 480 cm

                         \lambda_2 = 160 cm and \lambda_3 = 96 cm

We have to find whether the pipe is open,closed or open-closed or none.

Note:

  • The fundamental wavelength of a pipe which is open at both ends is 2L.
  • The fundamental wavelength of a pipe which is closed at one end and open at another end is 4L.

So,

The fundamental wavelength:

⇒ 4L=4(120)=480\ cm

It seems that the pipe is open at one end and closed at one end.

Now lets check with the subsequent wavelengths.

For one side open and one side closed pipe:

An odd-integer number of quarter wavelength have to fit into the tube of length L.

⇒  \lambda_2=\frac{4L}{3}                                   ⇒  \lambda_3=\frac{4L}{5}

⇒ \lambda_2=\frac{4(120)}{3}                              ⇒  \lambda_3=\frac{4(120)}{5}

⇒ \lambda_2=\frac{480}{3}                                  ⇒  \lambda_3=\frac{480}{5}

⇒ \lambda_2=160\ cm                           ⇒   \lambda_3=96\ cm  

So the pipe is open at one end and closed at one end .

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Denise is constructing A square in which two of its vertices are points M and N. She has already used her straightedge and compa
yan [13]

Denise is constructing A square.

Note: A square has all sides equal.

We already given two vertices M and N of the square.

And another edge of the square is made by from N.

Because a square has all sides of equal length, the side NO should also be equal to MN side of the square.

Therefore, <em>Denise need to place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass. That would make the NO equals MN.</em>

Therefore, correct option is :

D) place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass.

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