IO.7(4.2)+15.9=<br><br> 51.25+43.85+14=

Hope someone can help me

Advocard [28]

Hi!

Let's find out how much he has to pay for each of the deals.

Deal 1

£1.75 per mile.

192 x 1.75 = 336

£336

Deal 2

15% off of £680

$\frac{x}{680} \frac{15}{100}$

680 x 15 = 10200

10200/100 = 102

680 - 102 = 578

£578

Deal 3

£90 per day - 1/10

90 x 7 = 630

1/10 of 630 = 63

630 - 63 = 567

£567

The cheapest deal would be Deal 1

Hope this helps! :)

-Peredhel

5 0

3 years ago

Question 7 (Essay Worth 5 points)

kogti [31]

Answer:

x = -2

y=-3

(-2,-3)

Step-by-step explanation:

Both equations are equal to y

We can set them equal to each other

y = 3x + 3

y = x − 1

3x+3 = x-1

Subtract x from each side

3x+3 -x = x-1-x

2x+3 = -1

Subtract 3 from each side

2x+3-3 = -1-3

2x = -4

Divide each side by 2

2x/2 = -4/2

x = -2

Now we need to find y

y = x-1

y = -2-1

y = -3

y = x-1

3 0

3 years ago

Read 2 more answers

Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x

Anna11 [10]

Answer:

a) Figure attached

b) $y=1.31 x +98.57$

c) The correlation coefficient would be r =0.47719

d) $y=1.31 x +98.57=1.31*21 + 98.57 =126.08$

Step-by-step explanation:

(a) Draw a scatter diagram for the data.

See the figure attached

(b) Find x, y, b, and the equation of the least-squares line. (Round your answers to three decimal places.) x =__ y =__ b =__ y =__ + __x

$m=\frac{S_{xy}}{S_{xx}}$

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:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6576-\frac{300^2}{14}=147.429$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=38186-\frac{300*1773}{14}=193.143$

And the slope would be:

$m=\frac{193.143}{147.429}=1.31$

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

$\bar x= \frac{\sum x_i}{n}=\frac{300}{14}=21.429$

$\bar y= \frac{\sum y_i}{n}=\frac{1773}{14}=126.643$

And we can find the intercept using this:

$b=\bar y -m \bar x=126.643-(1.31*21.429)=98.571$

So the line would be given by:

$y=1.31 x +98.57$

(c) Find the sample correlation coefficient r and the coefficient of determination r?2. (Round your answers to three decimal places.)

n=14 $\sum x = 300, \sum y = 1773, \sum xy=38186, \sum x^2 =6576, \sum y^2 =225649$

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

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

$r=\frac{14(38186)-(300)(1773)}{\sqrt{[14(6576) -(300)^2][14(225649) -(1773)^2]}}=0.9534$

So then the correlation coefficient would be r =0.47719

What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.)

The % of variation is given by the determination coefficient given by $r^2$ and on this case $0.47719^2 =0.2277$ , so then the % of variation explaines is 22.8%.

(d) If a female baby weighs 21 pounds at 1 year, what do you predict she will weigh at 30 years of age? (Round your answer to two decimal places.) ___ lb

So we can replace in the linear model like this:

$y=1.31 x +98.57=1.31*21 + 98.57 =126.08$