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vesna_86 [32]
3 years ago
13

Darren has the option of investing in either Stock A or Stock B. The probability of the return of Stock A being 25% is 0.45, 14%

is 0.25, and 4% is 0.30. The probability of the return of Stock B being 30% is 0.30, 9% is 0.25, and 2% is 0.30. Given the probability distributions for the two investments, what is the expected rate of return for Stock A and Stock B?​
A. ​13.65%; 12.85%
B. ​17.82%; 11.95%
C. ​15.95%; 11.85%
D. ​16.80%; 11.45%
E. ​14.75%; 13.75%
Mathematics
1 answer:
Lunna [17]3 years ago
8 0

Answer:

C)A is 15.95% ,B is 11.85%

Step-by-step explanation:

We know that the expected value in probability distribution is given as

Lets X is the expected value then

X= \sum_{i=1}^{i=n} X_i P_i

For stock A

X=0.25 x 0.45+0.14 x 0.25+0.04 x 0.3

X=0.1595

So the expected return for A is 15.95%

For stock 9

X=0.3 x 0.3+0.09 x 0.25+0.02 x 0.3

X=0.1185

So the expected return for B is 11.85%

So our option C will be the answer of that problem.

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No

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-2+4(-5+2m)=5(4-2m)+5m
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7 0
3 years ago
Read 2 more answers
If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf
jonny [76]
With the curve C parameterized by

C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k

with 0\le t\le\dfrac\pi2, and given the vector field

\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k

the work done by \mathbf f on a particle moving on along C is given by the line integral

\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt

where

\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt

The integral is then

\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt
=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t
=269
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What part of the story does this portion of the graph represent?
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Step-by-step explanation:

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3 0
3 years ago
(2x+4)^2 - (x-5)^2 = 26x<br> Please solve (explained)
Zina [86]

Expand the squared binomials:

(2x+4)^2=4x^2+16x+16

(x-5)^2=x^2-10x+25

Then

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5 0
3 years ago
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