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3 years ago

14

Jack and Jill got tired of falling down that hill with a pail of water and decided to try flying a kite instead. Jack is holding

the kite string, and Jill is standing directly under the kite. The kite string forms a 61° angle with the line of sight between Jack and Jill, who are the same height. The two BFFs are standing 12 yards apart. How long is the kite string?
(this is Geometry, Trigonometric Ratios)

1 answer:

lapo4ka [179]3 years ago

8 0

Because Jill is standing directly under the kite the angle between jack Jill and the kite is 90°
so
a=12 yards
A°=61°
B°=90°
b=?
to find b we have a formula:
a devided by the sinus of A equals b devided by the sinus of B

12÷sin(61°)=b÷sin(90°)
12÷0.8=b÷1
15=b

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(a): Marginal pmf of x

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Step-by-step explanation:

Given

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<em>See attachment for complete question</em>

<em />

Solving (a): Marginal pmf of x

This is calculated as:

$P(x) = \sum\limits^{}_y\ P(x,y)$

So:

$P(0) = P(0,0) + P(0,1)$

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$P(1) = 0.18 + 0.10$

$P(1) = 0.28$

Solving (b): Marginal pmf of y

This is calculated as:

$P(y) = \sum\limits^{}_x\ P(x,y)$

So:

$P(0) = P(0,0) + P(1,0)$

$P(0) = 0.63 + 0.18$

$P(0) = 0.81$

$P(1) = P(0,1) + P(1,1)$

$P(1) = 0.09 + 0.10$

$P(1) = 0.19$

Solving (c): Mean and Variance of x

Mean is calculated as:

$E(x) = \sum( x * P(x))$

So, we have:

$E(x) = 0 * P(0) + 1 * P(1)$

$E(x) = 0 * 0.72 + 1 * 0.28$

$E(x) = 0 + 0.28$

$E(x) = 0.28$

Variance is calculated as:

$Var(x) = E(x^2) - (E(x))^2$

Calculate $E(x^2)$

$E(x^2) = \sum( x^2 * P(x))$

$E(x^2) = 0^2 * 0.72 + 1^2 * 0.28$

$E(x^2) = 0 + 0.28$

$E(x^2) = 0.28$

So:

$Var(x) = E(x^2) - (E(x))^2$

$Var(x) = 0.28 - 0.28^2$

$Var(x) = 0.28 - 0.0784$

$Var(x) = 0.2016$

Solving (d): Mean and Variance of y

Mean is calculated as:

$E(y) = \sum(y * P(y))$

So, we have:

$E(y) = 0 * P(0) + 1 * P(1)$

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$E(y) = 0+0.19$

$E(y) = 0.19$

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$Var(y) = E(y^2) - (E(y))^2$

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$E(y^2) = \sum(y^2 * P(y))$

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$Var(y) = E(y^2) - (E(y))^2$

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Covariance is calculated as:

$COV(x,y) = E(xy) - E(x) * E(y)$

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So:

$COV(x,y) = E(xy) - E(x) * E(y)$

$Cov(x,y) = 0.1 - 0.28 * 0.19$

$Cov(x,y) = 0.1 - 0.0532$

$Cov(x,y) = 0.0468$

The coefficient of correlation is then calculated as:

$r = \frac{Cov(x,y)}{\sqrt{Var(x) * Var(y)}}$

$r = \frac{0.0468}{\sqrt{0.2016 * 0.1539}}$

$r = \frac{0.0468}{\sqrt{0.03102624}}$

$r = \frac{0.0468}{0.17614266944}$

$r = 0.26569371378$

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