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3 years ago

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2. In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviati

on of $12,000. Female earnings have a mean of $45,000 per year and a standard deviation of $18,000. The correlation between male and female earnings for a couple is 0.80. Let C denote the combined earnings for a randomly selected couple. What is the mean of C?

1 answer:

Kobotan [32]3 years ago

8 0

Answer: $85,000

Step-by-step explanation:

Given : In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviation of $12,000.

$\mu_M=40,000\ \ ;\sigma_M=12,000$

Female earnings have a mean of $45,000 per year and a standard deviation of $18,000.

$\mu_F=45,000\ \ ;\sigma_F=18,000$

If C denote the combined earnings for a randomly selected couple.

Then, the mean of C will be :-

$\mu_c=\mu_M+\mu_F\\\\=40,000+45,000=85,000$

Hence, the mean of C = $85,000

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Given: △FKL, FK=a

LuckyWell [14K]

Answer:

$FL= \frac{a (\sqrt{3} + 1)}{\sqrt{2} }$

Step-by-step explanation:

As given in figure 1 below:

FK = a, m∠F = 45° and m∠L = 30°

Construction: Draw an altitude KE from point K on FL.

Now, In ΔFEK,

FE = EK (Sides opposite to equal angles of a triangle)

Let FE = EK = x

Now, using pythagoras In ΔFEK,

$(FK)^{2} = (FE)^{2} + (KE)^{2}$

$(a)^{2} = (x)^{2} + (x)^{2}$

$(a)^{2} = 2x^{2}$

$x = \sqrt \frac{a^{2} }{2} = \frac{a}{\sqrt{2} }$

∴ FE = EK = $\frac{a}{\sqrt{2} }$

Now in ΔEKL, EK = $\frac{a}{\sqrt{2} }$

Using trigonometry ratio,

TanФ = Altitude\ Base

$Tan \theta = \frac{KE}{EL}$

Tan 30° = $\frac{a/\sqrt{2} }{EL}$

$\frac{1}{\sqrt{3} } = \frac{a}{\sqrt{2} EL}$

$EL = \frac{\sqrt{3}a}{\sqrt{2} }$

Now FL = FE + EL

FE = $\frac{a}{\sqrt{2} }$ and $EL = \frac{\sqrt{3}a}{\sqrt{2} }$

∴ $FL = \frac{a}{\sqrt{2} } + \frac{\sqrt{3}a }{\sqrt{2} } = \frac{a (\sqrt{3} + 1)}{\sqrt{2} }$

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4 years ago

Find the values of the sine, cosine, and tangent for ZA C A 36ft B <br> 24ft

Reptile [31]

<h2>Question:</h2>

Find the values of the sine, cosine, and tangent for ∠A

a. sin A = $\frac{\sqrt{13} }{2}$ , cos A = $\frac{\sqrt{13} }{3}$ , tan A = $\frac{2 }{3}$

b. sin A = $3\frac{\sqrt{13} }{13}$ , cos A = $2\frac{\sqrt{13} }{13}$ , tan A = $\frac{3}{2}$

c. sin A = $\frac{\sqrt{13} }{3}$ , cos A = $\frac{\sqrt{13} }{2}$ , tan A = $\frac{3}{2}$

d. sin A = $2\frac{\sqrt{13} }{13}$ , cos A = $3\frac{\sqrt{13} }{13}$ , tan A = $\frac{2 }{3}$

<h2>Answer:</h2>

d. sin A = $2\frac{\sqrt{13} }{13}$ , cos A = $3\frac{\sqrt{13} }{13}$ , tan A = $\frac{2 }{3}$

<h2>Step-by-step explanation:</h2>

The triangle for the question has been attached to this response.

As shown in the triangle;

AC = 36ft

BC = 24ft

ACB = 90°

To calculate the values of the sine, cosine, and tangent of ∠A;

<em>i. First calculate the value of the missing side AB.</em>

<em>Using Pythagoras' theorem;</em>

⇒ (AB)² = (AC)² + (BC)²

<em>Substitute the values of AC and BC</em>

⇒ (AB)² = (36)² + (24)²

<em>Solve for AB</em>

⇒ (AB)² = 1296 + 576

⇒ (AB)² = 1872

⇒ AB = $\sqrt{1872}$

⇒ AB = $12\sqrt{13}$ ft

From the values of the sides, it can be noted that the side AB is the hypotenuse of the triangle since that is the longest side with a value of $12\sqrt{13}$ ft (43.27ft).

<em>ii. Calculate the sine of ∠A (i.e sin A)</em>

The sine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the hypotenuse side of the triangle. i.e

sin Ф = $\frac{opposite}{hypotenuse}$ -------------(i)

<em>In this case,</em>

Ф = A

opposite = 24ft (This is the opposite side to angle A)

hypotenuse = $12\sqrt{13}$ ft (This is the longest side of the triangle)

<em>Substitute these values into equation (i) as follows;</em>

sin A = $\frac{24}{12\sqrt{13} }$

sin A = $\frac{2}{\sqrt{13}}$

<em>Rationalize the result by multiplying both the numerator and denominator by </em> $\sqrt{13}$ <em />

sin A = $\frac{2}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }$

sin A = $\frac{2\sqrt{13} }{13}$

<em>iii. Calculate the cosine of ∠A (i.e cos A)</em>

The cosine of an angle (Ф) in a triangle is given by the ratio of the adjacent side to that angle to the hypotenuse side of the triangle. i.e

cos Ф = $\frac{adjacent}{hypotenuse}$ -------------(ii)

<em>In this case,</em>

Ф = A

adjacent = 36ft (This is the adjecent side to angle A)

hypotenuse = $12\sqrt{13}$ ft (This is the longest side of the triangle)

<em>Substitute these values into equation (ii) as follows;</em>

cos A = $\frac{36}{12\sqrt{13} }$

cos A = $\frac{3}{\sqrt{13}}$

<em>Rationalize the result by multiplying both the numerator and denominator by </em> $\sqrt{13}$ <em />

cos A = $\frac{3}{\sqrt{13}} * \frac{\sqrt{13} }{\sqrt{13} }$

cos A = $\frac{3\sqrt{13} }{13}$

<em>iii. Calculate the tangent of ∠A (i.e tan A)</em>

The cosine of an angle (Ф) in a triangle is given by the ratio of the opposite side to that angle to the adjacent side of the triangle. i.e

tan Ф = $\frac{opposite}{adjacent}$ -------------(iii)

<em>In this case,</em>

Ф = A

opposite = 24 ft (This is the opposite side to angle A)

adjacent = 36 ft (This is the adjacent side to angle A)

<em>Substitute these values into equation (iii) as follows;</em>

tan A = $\frac{24}{36}$

tan A = $\frac{2}{3}$

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expeople1 [14]

Simplifying
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Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '0.2x' to each side of the equation.
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