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3 years ago

5

L'età media di Aldo, Bruno, Carlo e Davide è 16 anni. Se non si tiene conto di Davide, l'età media dei tre rimanenti sale a 18.

Qual è l'età di Davide?

1 answer:

photoshop1234 [79]3 years ago

4 0

Answer:

Davide's age is 10.

L'età di Davide è 10.

Step-by-step explanation:

The average age of Aldo, Bruno, Carlo and Davide is 16 years.

I am going to say that:

x is Aldo's age.

y is Bruno's age.

z is Carlo's age.

w is Davide's age.

Average of the 4 is 16 years.

This means that:

$16 = \frac{x + y + z + w}{4}$

$x + y + z + w = 64$

If we do not consider Davide, the mean is 18. So

$18 = \frac{x + y + z}{3}$

$x + y + z = 54$

Replacing in the first equation:

$x + y + z + w = 64$

$54 + w = 64$

$w = 10$

Davide's age is 10.

L'età di Davide è 10.

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Answer:

a) $y=-0.317 x +46.02$

b) Figure attached

c) $S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

Step-by-step explanation:

We assume that th data is this one:

x: 30, 30, 30, 50, 50, 50, 70,70, 70,90,90,90

y: 38, 43, 29, 32, 26, 33, 19, 27, 23, 14, 19, 21.

a) Find the least-squares line appropriate for this data.

For this case we need to calculate the slope with the following formula:

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

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

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

So we can find the sums like this:

$\sum_{i=1}^n x_i = 30+30+30+50+50+50+70+70+70+90+90+90=720$

$\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324$

$\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200$

$\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540$

$\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=49200-\frac{720^2}{12}=6000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=17540-\frac{720*324}{12}{12}=-1900$

And the slope would be:

$m=-\frac{1900}{6000}=-0.317$

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

$\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60$

$\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27$

And we can find the intercept using this:

$b=\bar y -m \bar x=27-(-0.317*60)=46.02$

So the line would be given by:

$y=-0.317 x +46.02$

b) Plot the points and graph the line as a check on your calculations.

For this case we can use excel and we got the figure attached as the result.

c) Calculate S^2

In oder to calculate S^2 we need to calculate the MSE, or the mean square error. And is given by this formula:

$MSE=\frac{SSE}{df_{E}}$

The degred of freedom for the error are given by:

$df_{E}=n-2=12-2=10$

We can calculate:

$S_{y}=\sum_{i=1}^n y^2_i -\frac{(\sum_{i=1}^n y_i)^2}{n}=9540-\frac{324^2}{12}=792$

And now we can calculate the sum of squares for the regression given by:

$SSR=\frac{S^2_{xy}}{S_{xx}}=\frac{(-1900)^2}{6000}=601.67$

We have that SST= SSR+SSE, and then SSE=SST-SSR= 792-601.67=190.33[/tex]

So then :

$S^2=\hat \sigma^2=MSE=\frac{190.33}{10}=19.03$

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2 years ago

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Ooh, fun

what I would do is to make it a piecewise function where the absolute value becomse 0

because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up

so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points

we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5

A.
$\int\limits^{-5}_{-4} {x^2+x-12} \, dx + \int\limits^{-4}_3 {-x^2-x+12} \, dx + \int\limits^3_5 {x^2+x-12} \, dx$

B.
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next one
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the last one you can do yourself, it is $\frac{50}{3}$
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so the area under the curve is $\frac{233}{3}$

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