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MissTica
3 years ago
14

Misleading cost numbers are larger when unit-level assignments and the alternative activity-cost-driver assignments are proporti

onatley similar to each other.
True / False.
Business
1 answer:
frez [133]3 years ago
3 0

Answer:

The correct answer is False.

Explanation:

Misleading cost numbers are considered to be higher when their unit allocations and the alternative activity-cost-driver allocations are proportionally different from each other. This means that it corresponds to the contrary to what is detailed in the statement.

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Pressure and heat are both necessary to create a weld<br><br> true <br> false
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True because heat breaks down the molecular structure and pressure reinforced the new contiguous configuration
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On December 31, Fighting Okra Cooking Services reports the following revenues and expenses. Service revenue $ 78,500 Postage exp
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Explanation:

The preparation of the year-end income statement for Fighting Okra Cooking Services is presented  below:

                                      Fighting Okra Cooking Services

                                                 Income statement  

                                                As on December 31

Revenue  

Service revenue $78,500

Total revenues $78,500 (A)

Less: Expenses

Postage expense $1,500

Legal fees expense $2,600

Rent expense $21,000

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Supplies expense $20,000

Total expenses $67,100 (B)

Net income $11,400 (A- B)

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3 years ago
On an organization's board of directors, inside directors ____; outside directors _____. are supposed to be elected from outside
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On an organization's board of directors, inside directors <span>may be members of the firm; outside directors </span><span>are supposed to be elected from outside the firm.</span>
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4 years ago
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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3 years ago
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Preserved remains of organisms are known as fossils.

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