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Alisiya [41]
2 years ago
9

An investor is interested in purchasing an apartment building containing six apartments. The current owner provides the followin

g probability distribution indicating the probability that the given number of apartments will be rented during a given month.
Number of Rented Apartments 0 1 2 3 4 5 6
Probability 5/38 9/38 7/38 2/19 3/38 1/19 4/19

Find the number of apartments the investor could expect to be rented during a given month. (Hint: Find the expected value.)

a) 2.7232 b) 2.7632 c) 2.7432 d) 2.7932 e) 2.7732
Mathematics
2 answers:
Vsevolod [243]2 years ago
4 0

Answer:

b) 2.7632

Step-by-step explanation:

To find the mean, we multiply each value by it's probability. So

E = 0\frac{5}{38} + 1\frac{9}{38} + 2\frac{7}{38} + 3\frac{4}{38} + 4\frac{3}{38} + 5\frac{2}{38} + 6\frac{8}{38} = \frac{1*9 + 2*7 + 3*4 + 4*3 + 5*2 + 6*8}{38} = 2.7632

So the correct answer is:

b) 2.7632

Jlenok [28]2 years ago
4 0

Answer:

<u>The correct answer is B. 2.7632</u>

Step-by-step explanation:

Let's find the number of apartments the investor could expect to be rented during a given month, this way:

0 * 5/38 = 0

1 * 9/38 = 9/38

2 * 7/38 = 14/38

3 * 2/19 = 6/19

4 * 3/38 = 12/38

5 * 1/19 = 5/19

6 * 4/19 = 24/19

In consequence,

9/38 + 14/38 + 6/19 + 12/38 + 5/19 + 24/19 =

38 is the Lowest Common Denominator

(9 + 14 + 12 + 12 + 10 + 48)/38 =

105/38 = 2.7632

<u>The correct answer is B. 2.7632</u>

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Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

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Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

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The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

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With the same end point and route, we have;

l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

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Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

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Answer:

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