What is 10/28 in simplest form

Evaluate the limit of lim/n->infinity 6n^2+5/3n^2

castortr0y [4]

we are given

$\lim_{n \to \infty} \frac{6n^2+5}{3n^2}$

we can see that

degree of numerator = degree of denominator =2

so, we can divide both numerator and denominator by n^2

and we get

$\lim_{n \to \infty} \frac{(6n^2+5)/n^2}{(3n^2/n^2)}$

$\lim_{n \to \infty} \frac{(6n^2/n^2)+(5/n^2)}{(3n^2/n^2)}$

now, we can simplify it

$\lim_{n \to \infty} \frac{6+(5/n^2)}{3}$

now, we can plug n=inf

$= \frac{6+(5/(\infty))}{3}$

$= \frac{6+0}{3}$

$= \frac{6}{3}$

$= 2$ ...............Answer

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3 years ago

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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:

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