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Alex Ar [27]
4 years ago
10

Please help me with these. Thanks

Mathematics
1 answer:
DIA [1.3K]4 years ago
6 0
5.

f(K) = D^3 => f(25) = 125 => 25 * t = 125 ( because K is directly proportional with D^3 )=> t = 125 / 25 => t = 5 => f(25) = 25 * 5 => K * 5 = D^3 ;

6. 

f(L) = F^3 => f(2) =3^3 =>f(2) = 27 => 2 / t =27 => t = 2 / 27 => t = 0.074 => f(2) = 2 / 0.074 => K / 0.074 = F^3 ;


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Answer:

See below

Step-by-step explanation:

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Answer:

a) y=-0.317 x +46.02

b) Figure attached

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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.

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

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

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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}

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\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

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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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