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Maurinko [17]
4 years ago
10

__________ is an example of a capability that is based in the functional area of distribution. Group of answer choices Effective

control of inventories through point-of-purchase data collection Effective organizational structure Effective use of logistics management techniques Product and design quality
Business
1 answer:
ICE Princess25 [194]4 years ago
6 0

<u>Effective use of logistics management techniques</u> is an example of a capability that is based in the functional area of distribution.

<u>Option: C</u>

<u>Explanation:</u>

An aspect of supply chain management that is utilized to fulfill consumer expectations by planning, monitoring and enforcing the efficient transportation and storage of relevant information, goods and services from source to destination, thus understood as a logistic management.

This is accompanied by a logistics approach that is a collection of guiding principles, attitudes and driving forces that will help you manage plans, priorities and initiatives through any supply chain among different partners. It allows companies to increase performance in the supply chain while enhancing supply chain management overall.

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Compute the new national income given MPC = 0.9, and an autonomous injection of $100B from federal government stimulus spending.
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4 years ago
On July 1, Year 5, Eagle Corp. issued 600 of its 10%, $1,000 bonds at 99 plus accrued interest. The bonds are dated April 1, Yea
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E) $609,000

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7 0
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 />

6 0
3 years ago
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