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Alex Ar [27]
3 years ago
5

What is the variance of a portfolio invested 21 percent each in a and b and 58 percent in c?

Mathematics
1 answer:
victus00 [196]3 years ago
4 0
Given the following information:

\begin{tabular}
{|p{1.5cm}|p{1.5cm}|p{1.2cm}|p{1.2cm}|p{1.2cm}|}
\multicolumn{1}{|p{1.5cm}|}{State of economy}\multicolumn{1}{|p{2.6cm}|}{Probability of State of economy}\multicolumn{3}{|p{4.8cm}|}{Rate of Return if State Occurs}\\[1ex] 
\multicolumn{1}{|p{1.5cm}|}{}\multicolumn{1}{|p{2.6cm}|}{}\multicolumn{1}{|c|}{Stock A}&StockB&Stock C\\[2ex]
\multicolumn{1}{|p{1.5cm}|}{Boom}\multicolumn{1}{|p{2.6cm}|}{0.66}\multicolumn{1}{|p{1.27cm}|}{0.09}&0.03&0.34\\
\end{tabular}
\begin{tabular}
{|p{1.5cm}|p{1.5cm}|p{1.2cm}|p{1.2cm}|p{1.2cm}|}
\multicolumn{1}{|p{1.5cm}|}{Bust}\multicolumn{1}{|p{2.6cm}|}{0.34}\multicolumn{1}{|p{1.27cm}|}{0.23}&0.29&-0.14\\
\end{tabular}

Part A:

The expected return on an equally weighted portfolio of these three stocks is given by:

0.66[0.33 (0.09) + 0.33 (0.03) + 0.33(0.34)] \\ +0.34[0.33 (0.23) + 0.33(0.29) +0.33(-0.14)] \\  \\ =0.66(0.0297 + 0.0099 + 0.1122)+0.34(0.0759+0.0957-0.0462) \\  \\ =0.66(0.1518)+0.34(0.1254)=0.1002+0.0426=0.1428=\bold{14.28\%}



Part B:

Value of a portfolio invested 21 percent each in A and B and 58 percent in C is given by

For boom: 0.21(0.09) + 0.21(0.03) + 0.58(0.34) = 0.0189 + 0.0063 + 0.1972 = 0.2224 or 22.24%.

For bust: = 0.21(0.23) + 0.21(0.29) + 0.58(-0.14) = 0.0483 + 0.0609 - 0.0812 = 0.028 or 2.8%

Expected return = 0.66(0.2224) + 0.34(0.028) = 0.1468 + 0.00952 = 0.1563 or 15.63%

The variance is given by

0.66(0.2224-0.1563)^2+0.34(0.028-0.1563)^2 \\  \\ =0.66(0.0661)^2+0.34(-0.1283)^2=0.66(0.00437)+0.34(0.01646) \\  \\ =0.00288+0.0056=\bold{0.00848}
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Answer:

310\text{ feet and }210\text{ feet}

Step-by-step explanation:

GIVEN: A farmer has 520 \text{ feet} of fencing to construct a rectangular pen up against the straight side of a barn, using the barn for one side of the pen. The length of the barn is 310 \text{ feet}.

TO FIND: Determine the dimensions of the rectangle of maximum area that can be enclosed under these conditions.

SOLUTION:

Let the length of rectangle be x and y

perimeter of rectangular pen =2(x+y)=520\text{ feet}

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                                               y=260-x

area of rectangular pen =\text{length}\times\text{width}

                                       =xy

putting value of y

=x(260-x)

=260x-x^2

to maximize \frac{d \text{(area)}}{dx}=0

260-2x=0

x=130\text{ feet}

y=390\text{ feet}

but the dimensions must be lesser or equal to than that of barn.

therefore maximum length rectangular pen =310\text{ feet}

                              width of rectangular pen =210\text{ feet}

Maximum area of rectangular pen =310\times210=65100\text{ feet}^2

Hence maximum area of rectangular pen is 65100\text{ feet}^2 and dimensions are 310\text{ feet and }210\text{ feet}

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Step-by-step explanation:

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Step-by-step explanation:

Formula = K=\frac{y}{x}

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We substitute these known values in the equation,

y = kx

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Dividing both sides of the equation by 5 to find the value for k, we have

1.5/5= k0.3/5

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Having found the value of k,

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y = 0.3x.

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