What multiplies to 72 and adds to -17

A coach is assessing the correlation between the number of hours spent practicing and the average number of points scored in a g

cricket20 [7]

Answer:

a) $r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1$

We have a perfect linear relationship between the two variables

b) $m=\frac{90}{15}=6$

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

$\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2$

$\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17$

And we can find the intercept using this:

$b=\bar y -m \bar x=17-(6*2)=5$

So the line would be given by:

$y=6 x +5$

c) For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

Step-by-step explanation:

We have the following data:

Number of hours spent practicing (x) 0 0.5 1 1.5 2 2.5 3 3.5 4

Score in the game (y) 5 8 11 14 17 20 23 26 29

Part a

The correlation coefficient is given:

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

For our case we have this:

n=9 $\sum x = 18, \sum y = 153, \sum xy = 396, \sum x^2 =51, \sum y^2 =3141$

$r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1$

We have a perfect linear relationship 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:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=51-\frac{18^2}{9}=15$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=396-\frac{18*153}{9}=90$

And the slope would be:

$m=\frac{90}{15}=6$

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

$\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2$

$\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17$

And we can find the intercept using this:

$b=\bar y -m \bar x=17-(6*2)=5$

So the line would be given by:

$y=6 x +5$

Part c

For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

5 0

3 years ago

If the price of a certain stock is $66.50 per share and its earnings are $4.75 per share, what is the ratio of the stock’s price

Blababa [14]

Answer:

I may be wrong but I think it's 14

5 0

2 years ago

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 4 h, and Car B traveled the dista

Kamila [148]

40mph for Car B I think. 45mph for Car A. 45(mph) times 4(h) is 180(distance) and 40 times 4.5 is also 180. Sorry if I'm wrong.

8 0

3 years ago

The value of x is 4 times the value of y.how to writ that in math

kakasveta [241]

I’m not exactly sure what you mean by “write it in math,” however I can tell you what I think.
4y=x
(It says that x is 4 times the value of y which means that y times 4 is x.)

3 0

3 years ago

Is it true that the planes x + 2y − 2z = 7 and x + 2y − 2z = −5 are two units away from the plane x + 2y − 2z = 1?

zhuklara [117]

Lets Find It Out..

First we'll find the equation of ALL planes parallel to the original one.

As a model consider this lesson:

Equation of a plane parallel to other

The normal vector is:
→n=<1,2−2>

The equation of the plane parallel to the original one passing through P(x0,y0,z0)is:

→n⋅< x−x0,y−y0,z−z0>=0
<1,2,−2>⋅<x−x0,y−y0,z−z0>=0
x−x0+2y−2y0−2z+2z0=0
x+2y−2z−x0−2y0+2z0=0

Or

x+2y−2z+d=0 [1]
where a=1, b=2, c=−2 and d=−x0−2y0+2z0

Now we'll find planes that obey the previous formula and at a distance of 2 units from a point in the original plane. (We should expect 2 results, one for each half-space delimited by the original plane.)
As a model consider this lesson:

Distance between 2 parallel planes

In the original plane let's choose a point.
For instance, when x=0 and y=0:
x+2y−2z=1 => 0+2⋅0−2z=1 => z=−12
→P1(0,0,−12)

In the formula of the distance between a point and a plane (not any plane but a plane parallel to the original one, equation [1] ), keeping D=2, and d as d itself, we get:

D=|ax1+by1+cz1+d|√a2+b2+c2
2=∣∣1⋅0+2⋅0+(−2)⋅(−12)+d∣∣√1+4+4
|d+1|=2⋅3 => |d+1|=6First solution:
d+1=6 => d=5
→x+2y−2z+5=0Second solution:
d+1=−6 => d=−7
→x+2y−2z−7=0

8 0

3 years ago