Ask question

Ask question

All categories

jarptica [38.1K]

3 years ago

9

Data was collected on the number of pets and the height of a random group of students. Identify the correlation you would expect

to see between the number of pets and the height.
A) No Correlation
B) Positive
C) Constant
D) Negative

1 answer:

Tcecarenko [31]3 years ago

8 0

Answer:

A) No Correlation

Step-by-step explanation:

i guess this is right

You might be interested in

PLEASE ANSWERRRR!! (4 x 3) + (4 x 7) = 4 x (? + ?)

Vlada [557]

how is the volume of a solution used in calculating concentration?

3 0

2 years ago

Read 2 more answers

On 8(x - 2) + 3 = -61?

DochEvi [55]

Answer:

Hey there!

8(x-2)+3=-61

8(x-2)=-64

x-2=-8

x=-6

Let me know if this helps :)

3 0

2 years ago

Read 2 more answers

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Geometric proofs. I just don’t understand it and my teacher isn’t helping.

pshichka [43]

The midpoint of a line can be represented by the point that is in the very center of the line. A line segment such as AT also represents half of the line. The symbol of the tilde with the equal sign underneath represents congruence meaning the two segments are the same. Therefore each equation shows the same true statement in a different form

5 0

2 years ago

Six more than nine times a Number is 456

Nuetrik [128]

Let the number = x
It would be: 9x + 6 = 456
9x = 456 - 6
9x = 450
x = 450 / 9
x = 50

In short, Your Answer would be 50

Hope this helps!

8 0

3 years ago