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storchak [24]
3 years ago
15

If you aren't

Advanced Placement (AP)
1 answer:
marishachu [46]3 years ago
7 0
<h2>nay nay killua nay nay killua</h2>
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Stock Y has a beta of 1.0 and an expected return of 12.4 percent. Stock Z has a beta of .6 and an expected return of 8.2 percent
Harlamova29_29 [7]

Answer: 1.9%

Explanation:

First derive the Market return as this is needed in the Capital Asset Pricing Model by using the same model:

Required return = Risk free rate + Beta * ( market return - Risk free rate)

Using stock Y:

12.4% = Risk free rate + 1 * (market return - Risk free rate)

12.4% = Rf + market return - Rf

Market return = 12.4%

Use this to calculate the Risk free rate:

Stock Z:

8.2% = Rf + 0.6 * (12.4% - Rf)

8.2% = Rf + 7.44% - 0.6Rf

Rf - 0.6Rf = 8.2% - 7.44%

0.4Rf = 0.76%

Rf = 0.76% / 0.4

= 1.9%

7 0
3 years ago
The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0
3 years ago
What is the net charge of two ions bonded together ?
makvit [3.9K]

the net charge of an ion is non-zero due to its total number of electrons being unequal to its total number of protons.

4 0
3 years ago
The most effective way to help a child learn new behavior is to A. teach it during story time. B. read about the behavior to the
Oliga [24]
The correct answer is D: model the behavior whenever appropriate.
8 0
3 years ago
I have no question at all :)
Firlakuza [10]

Answer:

nice :)

Explanation:

7 0
2 years ago
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