For the data in the table, find the line of best fit, rounding values to three places if necessary. A) y = 4.88 + 0.625x B) y = 4.98 + 0.725x C) y = 4.88 + 0.525x D) y = 4.98 + 0.425x

Respuesta :

Answer: y = 4.88 + 0.525x

The equation of the line of best fit for the data shown in the picture is y=0.894x+0.535

What is a straight line?

A straight line is a combination of endless points joined on both sides of the point.

We have a data in the table:

Let's suppose the equation for the line is:

y = mx + c

Where m is the slope of the line and c is the y-intercept.

The value of m is given by:

[tex]\rm m = \frac{S_{xy}}{S_{xx}} }[/tex]    and

[tex]\rm S_{xy} = \sum_{i=1}^{n}x_iy_i-\frac{(\sum_{i=1}^{n}x_i)(\sum_{i=1}^{n}y_i)}{n}[/tex]

[tex]\rm S_{xx} = \sum_{i=1}^{n}x_i^2-\frac{(\sum_{i=1}^{n}x_i)^2}{n}[/tex]

From the table(data is not mentioned we have assumed the data) the values of  [tex]\rm \sum_{i=1}^{n}x_iy_i= 415[/tex], [tex]\rm {\sum_{i=1}^{n}x_i= 44[/tex], [tex]\rm {\sum_{i=1}^{n}y_i= 42[/tex],

[tex]\rm {\sum_{i=1}^{n}x_i^2= 438[/tex], and n = 5.

[tex]\rm S_{xy}= 415-\frac{44\times42}{5} \Rightarrow 45.4\\\\\rm S_{xx}= 438- \frac{44^2}{5} \Rightarrow 50.8[/tex]

[tex]\rm m = \frac{45.4}{50.8} } \Rightarrow 0.8937 \approx0.894[/tex]

For c we have to calculate the means of x and y for this:

[tex]\rm \bar{x} = \frac{\sum x}{n} \Rightarrow \frac{44}{5} \Rightarrow8.8[/tex]

[tex]\rm \bar{y} = \frac{\sum y}{n} \Rightarrow \frac{42}{5} \Rightarrow8.4[/tex]

Now, put the above values in the equation of a line to find the value of c

8.4 = (0.894)(8.8)+c

c = 0.5328 ≈ 0.535

Now the line is y = 0.894x+0.535

Thus, the line of best fit for the data shown in the picture is y=0.894x+0.535

Learn more about the straight line here:

brainly.com/question/3493733

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