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marta [7]
3 years ago
5

If 8,000 units are produced, what is the total amount of fixed manufacturing cost incurred to support this level of production?

Mathematics
1 answer:
ikadub [295]3 years ago
6 0
F 8,000 units are produced, what is the total amount of manufacturing overhead cost incurred to support this level of production? 

<span>Variable manufacturing overhead $ 1.50 </span>
<span>Fixed manufacturing overhead $ 4.00 x 10,000 = $40,000 </span>
<span>40,000 + (1.50 x 8,000) = $52,000 total overhead </span>

<span>If 8,000 units are produced, What is this total amount expressed on a per unit basis? (Round your answer to 2 decimal places.) </span>
<span>Variable manufacturing overhead $ 1.50 </span>
<span>Fixed manufacturing overhead $40,000 / 8,000 = $5.00 </span>

<span>5,00 + 1.50 = $6.50 manufacturing overhead per unit</span>
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<h3>What is integration?</h3>

It is defined as the mathematical calculation by which we can sum up all the smaller parts into a unit.

The parametric equations for the line segment from (0, 0, 0) to (2, 3, 4)

x(t) = (1-t)0 + t×2 = 2t  

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Finding its derivative;

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The line integral is given by:

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\rm ds = \sqrt{2^2+3^2+4^2} dt

After solving the integration over the limit 0 to 1, we will get;

\rm \int\limits_C {xe^{yz}} \, ds = \dfrac{\sqrt{29}}{12}  (e^{12}-1)   or

= 73037.99 ≈ 73038

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Learn more about integration here:

brainly.com/question/18125359

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y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

Step-by-step explanation:  We are given to find the solution of the following initial value problem :

y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.

Let y=e^{mx} be an auxiliary solution of the given differential equation.

Then, we have

y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.

Substituting these values in the given differential equation, we have

m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.

So, the general solution of the given equation is

y(x)=Ae^x+Be^{-x}, where A and B are constants.

This gives, after differentiating with respect to x that

y^\prime(x)=Ae^x-Be^{-x}.

The given conditions implies that

y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)

and

y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)

Adding equations (i) and (ii), we get

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From equation (i), we get

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y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.

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