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3 years ago

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Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds a

fter she jumps for the bone can be represented by the function h(t)= -16t^2+20t

1 answer:

kirill115 [55]3 years ago

3 0

Answer:

12 feet per second.

Step-by-step explanation:

Please consider the complete question.

Ian tosses a bone up in the air for his dog, Spot. The height, h, in feet, that Spot is above the ground at the time t seconds after she jumps for the bone can be represented $h(t)=-16t^2+20t$ .

What is Spot's average rate of ascent, in feet per second, from the time she jumps into the air to the time she catches the bone at t=1/2?

We will use average rate of change formula to solve our given problem.

$\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}$

$\text{Average rate of change}=\frac{h(\frac{1}{2})-h(0)}{\frac{1}{2}-0}$

$\text{Average rate of change}=\frac{-16\cdot(\frac{1}{2})^2+20\cdot \frac{1}{2}-(-16\cdot(0)^2+20\cdot (0))}{\frac{1}{2}-0}$

$\text{Average rate of change}=\frac{-16\cdot\frac{1}{4}+10-(0)}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{-4+10}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{6}{\frac{1}{2}}$

$\text{Average rate of change}=\frac{2\cdot 6}{1}$

$\text{Average rate of change}=12$

Therefore, Spot's average rate of ascent is 12 feet per second.

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Answer:

(a): Marginal pmf of x

$P(0) = 0.72$

$P(1) = 0.28$

(b): Marginal pmf of y

$P(0) = 0.81$

$P(1) = 0.19$

(c): Mean and Variance of x

$E(x) = 0.28$

$Var(x) = 0.2016$

(d): Mean and Variance of y

$E(y) = 0.19$

$Var(y) = 0.1539$

(e): The covariance and the coefficient of correlation

$Cov(x,y) = 0.0468$

$r \approx 0.2657$

Step-by-step explanation:

Given

<em>x = bottles</em>

<em>y = carton</em>

<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:

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

$P(0) = 0.63 + 0.09$

$P(0) = 0.72$

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

$P(1) = 0.18 + 0.10$

$P(1) = 0.28$

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:

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

$E(y) = 0 * 0.81 + 1 * 0.19$

$E(y) = 0+0.19$

$E(y) = 0.19$

Variance is calculated as:

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

Calculate $E(y^2)$

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

$E(y^2) = 0^2 * 0.81 + 1^2 * 0.19$

$E(y^2) = 0 + 0.19$

$E(y^2) = 0.19$

So:

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

$Var(y) = 0.19 - 0.19^2$

$Var(y) = 0.19 - 0.0361$

$Var(y) = 0.1539$

Solving (e): The covariance and the coefficient of correlation

Covariance is calculated as:

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

Calculate E(xy)

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

This gives:

$E(xy) = x_0y_0 * P(0,0) + x_1y_0 * P(1,0) +x_0y_1 * P(0,1) + x_1y_1 * P(1,1)$

$E(xy) = 0*0 * 0.63 + 1*0 * 0.18 +0*1 * 0.09 + 1*1 * 0.1$

$E(xy) = 0+0+0 + 0.1$

$E(xy) = 0.1$

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:

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

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

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

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

$r = 0.26569371378$

$r \approx 0.2657$ --- approximated

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