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LuckyWell [14K]
3 years ago
15

Peeker Industries is analyzing an average-risk project, and the following data have been developed. Unit sales will be constant,

but the sales price should increase with inflation. Fixed costs will also be constant, but variable costs should rise with inflation. The project should last for 3 years, it will be depreciated on a straight-line basis, and there will be no salvage value. No change in net operating working capital would be required. This is just one of many projects for the firm, so any losses on this project can be used to offset Page 2 of 3 gains on other firm projects. The marketing manager does not think it is necessary to adjust for inflation since both the sales price and the variable costs will rise at the same rate, but the CFO thinks an inflation adjustment is required. What is the difference in the expected NPV if the inflation adjustment is made versus if it is not made? WACC 10.0% Net investment cost (depreciable basis) $200,000 Units sold 50,000 Average price per unit, Year 1 $25.00 Fixed oper. costs excl. depreciation (constant) $150,000 Variable oper. cost/unit, Year 1 $20.20 Annual depreciation rate 33.333% Expected inflation 4.00% Tax rate 40.0%
Business
1 answer:
Anni [7]3 years ago
6 0

Answer:

Explanation:

can you please help me please

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Show that the set of vectors {(−4,1,3),(5,1,6),(6,0,2)} does not span R 3 , but that it does span the subspace of R 3 consisting
Alika [10]

Answer:

The vectors does not span R3 and only span a subspace of R3 which satisfies x+13y-3z=0

Explanation:

The vectors are given as

v_1=\left[\begin{array}{c}-4&1&3\end{array}\right] \\v_2=\left[\begin{array}{c}-5&1&6\end{array}\right] \\v_3=\left[\begin{array}{c}6&0&2\end{array}\right]

Now if the vectors  would span the R^3, the rank of  the consolidated matrix will be 3 if it is not 3 this indicates that the vectors does not span the R^3.

So the matrix is given as

M=\left[\begin{array}{ccc}v_1&v_2&v_3\end{array}\right] \\M=\left[\begin{array}{ccc}-4&5&6\\1&1&0\\3&6&2\end{array}\right]\\

In order to calculate the rank, the matrix is reduced to the Row Echelon form as

\approx  \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{9}{4}&\frac{3}{2}\\ 3&6&2\end{array}\right] R_2 \rightarrow R_2+\frac{R_1}{4}

\approx  \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{9}{4}&\frac{3}{2}\\ 0&\frac{39}{4}&\frac{13}{2}\end{array}\right] R_3 \rightarrow R_3+\frac{3R_1}{4}\\

\approx  \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{39}{4}&\frac{13}{2\\ 0&\frac{9}{4}&\frac{3}{2}}\end{array}\right] R_2\:\leftrightarrow \:R_3

\approx  \left[\begin{array}{ccc}-4&5&6\\ 0&\frac{39}{4}&\frac{13}{2}\\ 0&0&0\end{array}\right] R_3 \rightarrow R_3-\frac{3R_2}{13}\\

As the Rank is given as number of non-zero rows in the Row echelon form which are 2 so the rank is 2.

Thus this indicates that the vectors does not span R^3

<em>Now for any vector the corresponding equation is formulated by using the combined matrix which is given as  for any arbitrary vector and the coordinate as </em>

<em />v=\left[\begin{array}{c}x&y&z\end{array}\right]<em />

<em />c=\left[\begin{array}{c}c_1&c_2&c_3\end{array}\right]<em />

<em />Mc=v<em />

<em />\left[\begin{array}{ccc}-4&5&6\\1&1&0\\3&6&2\end{array}\right]\left[\begin{array}{c}c_1&c_2&c_3\end{array}\right]=\left[\begin{array}{c}x&y&z\end{array}\right]<em />

<em />M=\left[\begin{array}{ccccc}v_1&v_2&v_3& | &v\end{array}\right] \\M=\left[\begin{array}{ccccc}-4&5&6&|&x\\1&1&0&|&y\\3&6&2&|&z\end{array}\right]\\<em />

Now converting the combined matrix as

\approx  \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\\ 3&6&2&|&z\end{array}\right] R_2 \rightarrow R_2+\frac{R_1}{4}\\

\approx  \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\end{array}\right] R_3 \rightarrow R_3+\frac{3R_1}{4}\\

\approx  \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\\ 0&\frac{9}{4}&\frac{3}{2}&|&\frac{4y+x}{4}\end{array}\right] R_3 \leftrightarrow R_2\\

\approx  \left[\begin{array}{ccccc}-4&5&6&|&x\\ 0&\frac{39}{4}&\frac{13}{2}&|&\frac{4z+3x}{4}\\ 0&0&0&|&\frac{13y+x-3z}{13}\end{array}\right] R_3 \rightarrow R_3-\frac{3R_2}{13}\\

From this it is seen that whatever the values of the coordinates does not effect the value of the plane with equation as

\frac{13y+x-3z}{13}=0\\or\\13y+x-3z=0\\

So it is verified that the subspace of R3 such that it satisfies x+13y-3z=0 consists of all vectors.

<em />

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The 2017 balance sheet of Kerber’s Tennis Shop, Inc., showed long-term debt of $1.87 million, and the 2018 balance sheet showed
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Answer:

The firm’s cash flow to creditors during 2018 is -$85,000

Explanation:

The steps to compute the firm’s cash flow to creditors during 2018 is shown below:

Step 1: First the new debt is need to be calculated

Step 2: The step 1 amount is subtracted from interest expense amount. And Finally, the cash flow to creditors came

where,

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