Simplify the expression. 6w+8−4w−1

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

2 years ago

Ruth is knitting hats to sell at a local fair. The booth costs her $63 and the materials for each hat cost $10. She plans on sel

aev [14]

Answer: Ruth have to make 7 hats to break even.

Step-by-step explanation:

Break even point is a point where Cost = Revenue.

Let x be the number of hats sold.

Given: Cost of booth = $63

Each hat cost = $10

Selling price for each cost = $19

As per given,

Total cost = Cost of booth+ (Cost of each hat)(Number of hats)

= 63+10x

Total selling price = 19x

For break even,

63+10x=19x

9x=63

x=7

Hence, Ruth have to make 7 hats to break even.

7 0

3 years ago

In Panama City in January high tide was at midnight. The water level at high tide

Ivanshal [37]

The water levels at high tide was 9 feet and 1 foot at low tide

7 0

2 years ago

Drew is making a dart board.

Nezavi [6.7K]

Answer:

138.16 cm

Step-by-step explanation:

Recall that the formula for calculating the circumference is pi times diameter.

Circumference = 44 x 3.14 cm = 138.16 cm.

5 0

3 years ago

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that th

11111nata11111 [884]

Answer:

The null hypothesis is $H_0: \mu = 14$

The alternate hypothesis is $H_a: \mu < 14$

The test statistic is t = -1.95.

The p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.

Step-by-step explanation:

Suppose a consumer product researcher wanted to find out whether a highlighter lasted less than the manufacturer's claim that their highlighters could write continuously for 14 hours.

At the null hypothesis, we test if the mean is 14 hours, that is:

$H_0: \mu = 14$

At the alternate hypothesis, we test if the mean is less than 14 hours, that is:

$H_a: \mu < 14$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation of the sample and n is the size of the sample.

14 is tested at the null hypothesis:

This means that $\mu = 14$

X = 13.6 hours, s = 1.3 hours. Sample of 40:

In addition to the values of X and s given, we have that $n = 40$

Test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{13.6 - 14}{\frac{1.3}{\sqrt{40}}}$

$t = -1.95$

The test statistic is t = -1.95.

P-value:

The p-value of the test is the probability of finding a sample mean lower than 13.6, which is a left tailed test, with t = -1.95 and 40 - 1 = 39 degrees of freedom.

Using a calculator, the p-value is of 0.0292. This means that for a level of significance of 0.0292 and higher, there is sufficient evidence to conclude that the highlighters wrote for less than 14 continuous hours.

5 0

2 years ago