(3x squared)to the power of -2

Mathematics

2 answers:

Mazyrski [523]3 years ago

7 0

$(3x^2)^{-2}=\frac{1}{9}x^{-4}=\frac{1}{9x^4}$

Nady [450]3 years ago

3 0

$(3x^2)^{-2}\to\boxed{1/(3x^{2})^2\to\frac{1}{9x^4}}$

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Five observations taken for two variables follow.

I am Lyosha [343]

Answer:

a) Figure attached

b) If we see the scatter plot we can conclude that the possible relation between x and y is linear and with a positive correlation since when the values of x increases the values for y increases as well.

c) $Cov (X,Y) = \frac{\sum_{i=1}^n (x_i -\bar X)(y_i -\bar Y)}{n-1}$

We can find the numerator like this:

$\sum_{i=1}^5 (6-16)(6-10)+(11-16)(9-10)+(15-16)(6-10)+(21-16)(17-10)+(27-16)(12-10)=106$

And then:

$Cov(X,Y) = \frac{106}{5-1}=26.5$

d) $r=\frac{5(906)-(80)(50)}{\sqrt{[5(1552) -(80)^2][5(586) -(50)^2]}}=0.693$

Step-by-step explanation:

Part a

For this part we use excel in order to create the scatterplot and we got the result on the figure attached.

Part b

If we see the scatter plot we can conclude that the possible relation between x and y is linear and with a positive correlation since when the values of x increases the values for y increases as well.

Part c

The sample covariance is defined as:

$Cov (X,Y) = \frac{\sum_{i=1}^n (x_i -\bar X)(y_i -\bar Y)}{n-1}$

We can find the numerator like this:

$\sum_{i=1}^5 (6-16)(6-10)+(11-16)(9-10)+(15-16)(6-10)+(21-16)(17-10)+(27-16)(12-10)=106$

And then:

$Cov(X,Y) = \frac{106}{5-1}=26.5$

Part d

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

And in order to calculate the correlation coefficient we can use this formula:

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