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2 years ago

Any body, Waiting for answer

Mathematics

1 answer:

mart [117]2 years ago

3 0

The maximum height of the rocket is 43.89 feet

<h3>How to write the function</h3>

The general function is given as:

h(t) = -16t^2 + vt + h

The initial velocity is

v = 53

So, we have:

h(t) = -16t^2 + 53t + h

The initial height is

h = 0

So, we have:

h(t) = -16t^2 + 53t

Hence, the function of the height is h(t) = -16t^2 + 53t

<h3>The maximum height of the rocket</h3>

In (a), we have:

h(t) = -16t^2 + 53t

Differentiate the function

h'(t) = -32t + 53

Set to 0

-32t + 53 = 0

This gives

-32t = -53

Divide by -32

t = 1.65625

Substitute t = 1.65625 in h(t) = -16t^2 + 53t

h(1.65625) = -16 * 1.65625^2 + 53 * 1.65625

Evaluate

h(1.65625) = 43.890625

Approximate

h(1.65625) = 43.89

Hence, the maximum height of the rocket is 43.89 feet

<h3>Time to hit the ground</h3>

In (a), we have

h(t) = -16t^2 + 53t

Set to 0

-16t^2 + 53t = 0

Divide through by -t

16t - 53 = 0

Add 53 to both sides

16t = 53

Divide by 16

t = 3.3125

Hence, the time to hit the ground is 3.3125 seconds

<h3>The graph of the function h(t)</h3>

See attachment for the graph of the function h(t)

Read more about height functions at:

brainly.com/question/12446886

#SPJ1

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Answer:

a) $y=-0.317 x +46.02$

b) Figure attached

c) $S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

Step-by-step explanation:

We assume that th data is this one:

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

a) Find the least-squares line appropriate for this data.

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720$

$\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324$

$\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200$

$\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540$

$\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900$

And the slope would be:

$m=-\frac{1900}{6000}=-0.317$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60$

$\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27$

And we can find the intercept using this:

$b=\bar y -m \bar x=27-(-0.317*60)=46.02$

So the line would be given by:

$y=-0.317 x +46.02$

b) Plot the points and graph the line as a check on your calculations.

For this case we can use excel and we got the figure attached as the result.

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In oder to calculate S^2 we need to calculate the MSE, or the mean square error. And is given by this formula:

$MSE=\frac{SSE}{df_{E}}$

The degred of freedom for the error are given by:

$df_{E}=n-2=12-2=10$

We can calculate:

$S_{y}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}=9540-\frac{324^2}{12}=792$

And now we can calculate the sum of squares for the regression given by:

$SSR=\frac{S^2_{xy}}{S_{xx}}=\frac{(-1900)^2}{6000}=601.67$

We have that SST= SSR+SSE, and then SSE=SST-SSR= 792-601.67=190.33[/tex]

So then :

$S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

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