Answer:
Step-by-step explanation:
The profitability index of an investment with cash flows in years 0 thru 4 of -340, 120, 130, 153, and 166, respectively, and a discount rate of 16 percent is: 15%.
<h3>
Profitability index</h3>
First step is to find the Net present value (NPV) of the given cash flow using discount rate PVF 16% and PV of cash flow which in turn will give us net present value of 49.7.
Second step is to calculate the profitability index
Profitability index = 49.7/340
Profitability index = .15×100
Profitability index=15%
Therefore the profitability index of an investment with cash flows in years 0 thru 4 of -340, 120, 130, 153, and 166, respectively, and a discount rate of 16 percent is: 15%.
Learn more about Profitability index here:brainly.com/question/3805108
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Answer:
Infinite number of solutions.
Step-by-step explanation:
We are given system of equations
Firs we find determinant of system of equations
Let a matrix A=![\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%264%265%5C%5C1%261%262%5C%5C2%261%26-1%5Cend%7Barray%7D%5Cright%5D)
and B=![\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%5C%5C1%5C%5C-3%5Cend%7Barray%7D%5Cright%5D)
Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.
We are finding rank of matrix
Apply 
and 
![\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C1%261%262%5C%5C0%26-1%26-3%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C1%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply
![\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C0%26-1%26-3%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C-5%5Cend%7Barray%7D%5Cright%5D)
Apply 
![\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%261%5C%5C0%260%26-2%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
Apply 
and 
![\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%261%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Apply 
![\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%260%5C%5C0%261%260%5C%5C0%260%261%5Cend%7Barray%7D%5Cright%5D)
:![\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-%5Cfrac%7B9%7D%7B2%7D%5C%5C%5Cfrac%7B13%7D%7B2%7D%5C%5C-%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.
Therefore, rank of matrix is equal to rank of B.
Answer:
Step-by-step explanation:
Graph y = -3x. Start at the origin (0, 0). Place a dark dot there. Now, starting with your pencil point on that dot, move 1 space to the right and from this new point move 3 spaces down. Place a dark dot at this new point.
Now draw a straight line through both dark dots. This is the desired graph.
