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3 years ago

Lucas is offered either 15% or 21$ off his total shopping bill. How much would have to be spent to make 15% option the best one?

1 answer:

8 0

Assume that the amount of money that Lucas would have to spend is "x".

Now, Lucas is offered:
either 15% of his total shopping which can be written as 0.15x
or 21$

We want the 15% to be a better option, therefore:
0.15x > 21

Solve the inequality for x:
0.15x > 21
x > 21/0.15
x > 140

Therefore, for the 15% back to be a better option, Lucas has to spend more than $140

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3 years ago

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