22.63 rounded to the nearest whole number

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

victus00 [196]

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

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

4 0

3 years ago

Geometry help please!

Gekata [30.6K]

given the following sets.

A = {0, 1, 2, 3}
B = {a, b, c, d}
C = {0, a, 2, b}

Find B C.

3 0

2 years ago

2. Find the missing side length.<br> 9<br> 37°<br> 6

svetoff [14.1K]

Answer:

I found it! It's over here!!

Now what do I do with it?

8 0

2 years ago

How many of the numbers from 10 through 92 have the sum of their digits equal to a perfect square?

horrorfan [7]

All the numbers in this range can be written as $10d_1+d_0$ with $d_1\in\{1,2,\ldots,9\}$ and $d_2\in\{0,1,\ldots,9\}$ . Construct a table like so (see attached; apparently the environment for constructing tables isn't supported on this site...)

so that each entry in the table corresponds to the sum of the tens digit (row) and the ones digit (column). Now, you want to find the numbers whose digits add to perfect squares, which occurs when the sum of the digits is either of 1, 4, 9, or 16. You'll notice that this happens along some diagonals.

For each number that occupies an entire diagonal in the table, it's easy to see that that number $n$ shows up $n$ times in the table, so there is one instance of 1, four of 4, and nine of 9. Meanwhile, 16 shows up only twice due to the constraints of the table.

So there are 16 instances of two digit numbers between 10 and 92 whose digits add to perfect squares.

7 0

3 years ago

I got this for my classwork does anyone know how to solve this step by step

Nataly_w [17]

Answer:

http://www.cssdrive.com/

Step-by-step explanation:

7 0

2 years ago