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SOVA2 [1]
3 years ago
5

Midwest Water Works estimates that its WACC is 10.45%. The company is considering the following capital budgeting projects.Assum

e that each of these projects is just as risky as the firm's existing assets and that the firm may accept all the projects or only some of them. Which set of projects should be accepted?Project Size Rate of Return A $1 million 12.0% B 2 million 11.5 C 2 million 11.2 D 2 million 11.0 E 1 million 10.7 F 1 million 10.3 G 1 million 10.2 Accept or Dont Accept A through G.
Mathematics
1 answer:
irinina [24]3 years ago
7 0

Answer:

Don't accept A-G

Accept only A-E

Step-by-step explanation:

The company would those projects with a return of return equal to or higher than its cost of capital of 10.45%

Project A with a 12% return is acceptable.

Project B with a 11.5% rate of return is also acceptable

Project C has a  rate of return of 11.2% , hence acceptable.

Project D has 11% rate of return and it is therefore acceptable.

Project E has 10.7% return rate and it is acceptable.

Project F has a lower rate of return of 10.3%, hence rejected, as well as projects G

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A (0, 2) and B (6,6) are points on the straight line ABCD.
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Answer:

(18, 14)

Step-by-step explanation:

We know that C and D lie on the line AB and BC = CD = AB. Then we need to use the distance formula and equation of the line AB to find the other two coordinates.

The distance formula states that the distance between two points (x_1,y_1) and (x_2,y_2), the distance is denoted by: \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. Let's find the distance between A and B:

d = \sqrt{(6-0)^2+(6-2)^2}=\sqrt{6^2+4^2} =\sqrt{36+16} =\sqrt{52} =2\sqrt{13}

Now say the coordinates of D are (a, b). Then the distance between D and B will be twice of 2√13, which is 4√13:

4√13 = \sqrt{(6-a)^2+(6-b)^2}

Square both sides:

208 = (6 - a)² + (6 - b)²

Let's also find the equation of the line AB. The y-intercept we know is 2, so in y = mx + b, b = 2. The slope is (6 - 2) / (6 - 0) = 4/6 = 2/3. So the equation of the line is: y = (2/3)x + 2. Since (a, b) lines on this line, we can put in a for x and b for y: b = (2/3)a + 2. Substitute this expression in for b in the previous equation:

208 = (6 - a)² + (6 - b)²

208 = (6 - a)² + (6 - (2/3a + 2))² = (6 - a)² + (-2/3a + 4)²

208 = a² - 12a + 36 + 4/9a² - 16/3a + 16 = 13/9a² - 52/3a + 52

0 = 13/9a² - 52/3a - 156

13a² - 156a - 1404 = 0

a² - 12a - 108 = 0

(a + 6)(a - 18) = 0

a = -6 or a = 18

We know a can't be negative so a = 18. Plug this back in to find b:

b = 2/3a + 2 = (2/3) * 18 + 2 = 12 + 2 = 14

So point D has coordinates (18, 14).

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