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Sergeeva-Olga [200]
3 years ago
15

In the xy-coordinate system, points (2, 9) and (-1, 0) lie on line k. If the point (n, 21) lies on line k, what is the value of

n?
A. 6
B. 7
C. 8
D. 9
E. 10
Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

A

Step-by-step explanation:

Since they lie on the same line, it means that they are collinear.

Since they are collinear, the value of the slope between each 2 point is the same. Hence, we can have points PQR with the slopes PQ, PR and QR being the same.

The general formula for finding the slope of a line with 2 end points is :

m = (y2 - y1)/(x2 - x1)

using the first two points,

m = (0 - 9)/-1 -2 = -9/-3 = 3

Now to get the value of n. We can use the second point and the point containing n.

This means:

3 = (21 - 0)/(n + 1)

3 = 21/n + 1

3n + 3 = 21

3n = 18

n = 6

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a) r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) m=\frac{64.5}{2000}=0.03225  

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

\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50  

\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425  

b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27  

So the line would be given by:  

y=0.3225 x -0.27  

Step-by-step explanation:

Part a

The correlation coeffcient is given by this formula:

r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}  

For our case we have this:

n=4 \sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501  

r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

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}  

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}=12000-\frac{200^2}{4}=2000  

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5  

And the slope would be:  

m=\frac{64.5}{2000}=0.03225  

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

\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50  

\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425  

And we can find the intercept using this:  

b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27  

So the line would be given by:  

y=0.3225 x -0.27  

4 0
3 years ago
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