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FinnZ [79.3K]
3 years ago
12

In a regression class, a student suggested the following: If extremely influential outlying cases are detected in a data set, si

mply discard these cases from the data set. Provide your comments as the validity of this statement.
Mathematics
1 answer:
mel-nik [20]3 years ago
6 0

Answer:

Discarding the influential outlying cases when detected  is also known as flagging outliers in a data set, and this is because outliers do not follow the rest of the dataset's pattern. if this outliers are not discarded they would have a negative effect on any model attached to the dataset

Step-by-step explanation:

In a regression class ; If extremely influential outlying cases are detected in a Data set, discarding this influential outlying cases is the right way to go about it

Discarding the influential outlying cases when detected  is also known as flagging outliers in a data set, and this is because outliers do not follow the rest of the dataset's pattern. if this outliers are not discarded they would have a negative effect on any model attached to the dataset

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Answer:

No estoy muy seguro de lo que significan los puntos y comas, pero la respuesta para A entre paréntesis son (18):(51):(7) B es -26:12. C es 12:2. D es 101. Espero que esto ayude, ¡buena suerte! :)

Step-by-step explanation:

4 0
4 years ago
a number has 7 at the tens place .there is zero in the thousand place. the number 5 is at the hundreds place .there is number 1a
crimeas [40]
10570 - ten thousand, five hundred and seventy
4 0
3 years ago
I need help on this asap please!
son4ous [18]

Answer:

-8 5/20 , 4/25 , 9/20

Step-by-step explanation:

If there are negative numbers, those are the least. Let's say the number was -2 and -1. With a positive integer, 2 would be greater, but since it's negative, it's the least. SO -8 and 5/20 would be the least. Then putting the other two fractions into decimal form, or you can find the greatest common denominator, you'll find that 9/20 is greater than 4/25.

3 0
3 years ago
Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1
Yuri [45]
Let's check if the ODE is exact. To do that, we want to show that if

\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0

then M_y=N_x. We have

M_y=2(x+y)
N_x=2y+2x=2(x+y)

so the equation is indeed exact. We're looking for a solution of the form \Psi(x,y)=C. Computing the total differential yields the original ODE,

\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0
\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}

Integrate both sides of the first PDE with respect to x; then

\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)

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\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)
\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)
f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C

So the solution to this ODE is

\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C

i.e.


\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C
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3 years ago
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MatroZZZ [7]
(-3.5, 8)

You figure this out with the midpoint formula
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3 years ago
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