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Sliva [168]
3 years ago
7

A manufacturing company is expected to pay a dividend of br. 1.25 per share at the end of the year (D1=br.1.25). The stock sells

for br. 32.50 per share and its required rate of return is 10.5%. The dividend is expected to grow at some constant rate forever. What is the growth rate
Mathematics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

the equilibrium expected growth rate is 6.65%

Step by step Explanation:

We were given stock sold per share of $32.50

Dividend per share =$1.25

Required Return rate = 10.5%

Then we can calculate Percentage of Dividend for share as;

dividend of br. 1.25 per share at the end of the year (D1=br.1.25)

= 1.25×100= 125

Let the dividend percentage = y

stock sold per share × y= 125

125= 32.50y

y = 125/32.50

y= 3.85

y= 3.85*100%

Then the Dividend percentage = 3.85%

Growth rate=(required rate of return -Dividend percentage)

= 10.5 - 3.85 = 6.65

Therefore, the equilibrium expected growth rate is 6.65%

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3 years ago
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2 years ago
5. A yacht is cruising at 20 knots at bearing of 185° when a 15 knot wind starts to blow at a bearing of 290°. Find the directio
kicyunya [14]

Answer:

<em>Direction of the boat is 43.05° and the speed is 21.66 knot</em>

<em></em>

Step-by-step explanation:

yacht speed = 20 knots

yacht bearing = 185° = 85°  below the negative horizontal x-axis

wind speed = 15 knots

wind bearing = 290° = 20° above the negative horizontal x-axis

we find the x and y components of the boat velocities

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x component = -20 cos 85° = -1.74 knots

y component = -20 sin 85° = -19.92 knots

for the wind,

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y component = 15 sin 20° = 5.13 knots

total x component  Vx = -1.74 + (-14.09) = -15.83 knots

total y component Vy =  -19.92 + 5.13 = -14.79 knots

Resultant speed of the boat = \sqrt{Vx^{2} + Vy^{2}  }

==> \sqrt{15.83^{2} + 14.79^{2}  } = <em>21.66 knot</em>

<em></em>

direction of boat = tan^{-1} \frac{Vy}{Vx}

==> tan^{-1} \frac{14.79}{15.83} = <em>43.05°</em>

3 0
3 years ago
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