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Leokris [45]
3 years ago
5

What type of correlation does the data set display?

Mathematics
2 answers:
Ahat [919]3 years ago
7 0
The answer is a negative correlation. We can observe that as the x-coordinates (inputs) increase, the y-coordinates (outputs) decrease. We can see that this line would have a negative slope, and thus a negative correlation between the points.
Gemiola [76]3 years ago
7 0
<h2>Answers:</h2>

B.

<h2>Examples:</h2>

  • Data sets are often depicted as scatter plot graphs. This is because the purpose of these scatter plots is to check for a linear correlation between the two variables. The two variables are usually denoted as independent and dependent variables. independent variable: x, horizontal axis.

<h2>What's Now?</h2>

  • When you reflect a point across the x-axis, the x-coordinate remains the same, but the y-coordinate is taken to be the additive inverse. The reflection of point (x, y) across the x-axis is (x, -y).

If you need for answers of What type of correlation does the data set display? Just like this answers for me Explantation!!

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Which would be the best measures of center and variation to use to compare the data? The scores of Group B are skewed right, so
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Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun
WITCHER [35]

Answer:

Required solution (a) f(x,y,z)=xyz+3z^2+C (b) 40.

Step-by-step explanation:

Given,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k

(a) Let,

F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}

Then,

f_x=yz,f_y=xz,f_z=xy+6z

Integrating f_x we get,

f(x,y,z)=xyz+g(y,z)

Differentiate this with respect to y we get,

f_y=xz+g'(y,z)

compairinfg with f_y=xz of the given function we get,

g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)

Then,

f(x,y,z)=xyz+h(z)

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

on compairing we get,

h'(z)=6z\implies h(z)=3z^2+C   (By integrating h'(z))  where C is integration constant. Hence,

f(x,y,z)=xyz+3z^2+C

(b) Next, to find the itegration,

\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40

3 0
3 years ago
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