Answer:
The fixed costs per month are $50,000.
Explanation:
The problem can be presented as a system of 2 equations with 2 variables:
Where:
<em>y</em> are the fixed costs,
<em>x </em>are the variable costs per unit produced.
You can solve the system by the method you like. In this case im using the Gaussian Elimination method.
We start with the following AX = b matrix.
![\left[\begin{array}{ccc}1&16000\\1&8000\\\end{array}\right] * \left[\begin{array}{ccc}y\\x\\\end{array}\right] = \left[\begin{array}{ccc}80000\\65000\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2616000%5C%5C1%268000%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy%5C%5Cx%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D80000%5C%5C65000%5Cend%7Barray%7D%5Cright%5D)
We substract the second row by the first row.
![\left[\begin{array}{ccc}1&16000\\0&-8000\\\end{array}\right] = \left[\begin{array}{ccc}80000\\-15000\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2616000%5C%5C0%26-8000%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D80000%5C%5C-15000%5Cend%7Barray%7D%5Cright%5D)
We divide the second row by (-8000):
![\left[\begin{array}{ccc}1&16000\\0&1\\\end{array}\right] = \left[\begin{array}{ccc}80000\\1.875\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2616000%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D80000%5C%5C1.875%5Cend%7Barray%7D%5Cright%5D)
We substract the first row by 16,000 times the second:
![\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] = \left[\begin{array}{ccc}50000\\1.875\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D50000%5C%5C1.875%5Cend%7Barray%7D%5Cright%5D)
Multiplying this reduced matrix by the X matrix to interpret the results:
![\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] * \left[\begin{array}{ccc}y\\x\\\end{array}\right] = \left[\begin{array}{ccc}50000\\1.875\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dy%5C%5Cx%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D50000%5C%5C1.875%5Cend%7Barray%7D%5Cright%5D)
We can say that Mendoza Company's has<em> y = $50,000</em> fixed costs and each unit costs <em>x = $1.875</em> to produce. Therefore the answer to the problem is $50,000.
Answer:
A strength of Generation X managers is likely to be their ability to provide feedback to employees
Explanation:
mark me brainliest!!
Answer:
The correct answer is $5,160.
Explanation:
According to the scenario, the given data are as follows:
Estimated OH = $88,480
Estimated direct labor hour = 2,800 labor hour
So, Estimated OH per labor hour = $88,480 ÷ 2,800 = $31.6 / labor hour
Actual OH = $80,160
Actual Direct labor hour = 2,700 labor hour
Now, Applied OH = Estimated OH per labor hour × Actual Direct labor hour
= $31.6 × 2700
= $85,320
Since, Applied OH is Greater than Actual OH, it is underapplied OH.
Underapplied OH = Applied OH - Actual OH
= $85,320 - $80,160
= $5,160.
Hence, the underapplied OH for the year was $5,160.
Answer:
True
Explanation:
Return from operating activities are returns made from the regular and recurring operations of a business. Since they are from the normal operations of a company, they are less risky than returns made from the non-operating activities of a company which do not re-occur.
As such, a firm that earns more of its return from operating activities which are recurring is usually considered less risky than a firm than earns more of its return from non-operating activities which are usually one-off.
I think it’s higher the risk and the lower present value
