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3 years ago

Maths revision 6 can someone help

Mathematics

1 answer:

Inga [223]3 years ago

6 0

20) 13=2f+5

2f=8

f=4

6x<=3

x<=1/2

-f-5(2f-3)=-f-10f+15=-11f+15

answer B

45+0,15*45=45+6,75=51,75$

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Phantasy [73]

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$y=-2.95836 x +234.56159$

Step-by-step explanation:

We assume that th data is this one:

x: 50, 55, 50, 79, 44, 37, 70, 45, 49

y: 152, 48, 22, 35, 43, 171, 13, 185, 25

a) Compute the equation of the least-squares regression line. (Round your numerical values to five decimal places.)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:

$\sum_{i=1}^n x_i =50+ 55+ 50+ 79+ 44+ 37+ 70+ 45+ 49=479$

$\sum_{i=1}^n y_i =152+ 48+ 22+ 35+ 43+ 171+ 13+ 185+ 25=694$

$\sum_{i=1}^n x^2_i =50^2 + 55^2 + 50^2 + 79^2 + 44^2 + 37^2 + 70^2 + 45^2 + 49^2=26897$

$\sum_{i=1}^n y^2_i =152^2 + 48^2 + 22^2 + 35^2 + 43^2 + 171^2 + 13^2 + 185^2 + 25^2=93226$

$\sum_{i=1}^n x_i y_i =50*152+ 55*48+ 50*22+ 79*35+ 44*43+ 37*171+ 70*13+ 45*185+ 49*25=32784$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=26897-\frac{479^2}{9}=1403.556$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=32784-\frac{479*694}{9}=-4152.22$

And the slope would be:

$m=-\frac{-4152.222}{1403.556}=-2.95836$

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

$\bar x= \frac{\sum x_i}{n}=\frac{479}{9}=53.222$

$\bar y= \frac{\sum y_i}{n}=\frac{694}{9}=77.111$

And we can find the intercept using this:

$b=\bar y -m \bar x=77.1111111-(-2.95836*53.22222222)=234.56159$

So the line would be given by:

$y=-2.95836 x +234.56159$

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