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3 years ago

What is the difference between divergence and convergence testing outliers?

1 answer:

8 0

A divergence testing just diverts your mind and imagination into several possibilities and scenarios regarding the same issue.
A divergence test will simply test your imagination and forces your mind to go into as many directions as possible.

On the other hand, a converge test will sort of converge your imagination and your mind into a set of possibilities or options.
A convergence test will let you sort between a set of possibilities and decide on the best one.

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Find the coordinates of the vertices of the figure after a rotation of 180°.

jonny [76]

Hey there! I'm happy to help!

First, let's see what the coordinates of each of these points are by counting the squares.

Vertex A is at (-3,-4).

Vertex B is at (1,-3).

Vertex C is at (1,1).

Whenever you rotate a figure 180° about the origin, you find the negative version of each number in the ordered pair. Basically (x,y) turns into (-x,-y) when you rotate a figure 180°.

Let's do this below!

A: (-3,-4)⇒(3,4)

B: (1,-3)⇒(-1,3)

C: (1,1)⇒(-1,-1)

These new coordinates match with the third option: A'(3,4), B'(-1,3), C'(-1,-1). Now you can find the coordinates of translated points!

I hope that this helps! Have a wonderful day!

5 0

3 years ago

I have ten minutes, please help me!!!

MrMuchimi

Use Math ———papa for faster response

8 0

3 years ago

Given the speeds of each runner below, determine who runs the fastest.

Tju [1.3M]

Answer:

Stephanie

Step-by-step explanation:

To compare each runner's speed, we must convert them all to the same units.

Let's convert then all to feet per second.

1. Stephanie

Speed = 13 ft/s

2. Emily

$\text{Speed} = \dfrac{\text{538 ft}}{\text{45 s}} = \text{12.0 ft/s}$

3. Brooke

$\text{Speed} = \dfrac{\text{1 mi}}{\text{560 s}} \times \dfrac{\text{5280 ft}}{\text{1 mi}} = \text{9.4 ft/s}$

4. Katie

$\text{ speed} = \dfrac{\text{747 ft}}{\text{1 min}} \times \dfrac{\text{1 min}}{\text{60 s}} = \textbf{12.4 ft/s}$

Stephanie is the fastest runner.

7 0

3 years ago

Hello can you please help with the problem written below:

Kruka [31]

Answer:

Answer of this question is.

three

4 0

2 years ago

Read 2 more answers

Exercise 3.9.101: Find a particular solution to x 0 = 5x + 4y+ t, y 0 = x + 8y−t, a) using integrating factor method, b) using e

enot [183]

In matrix form, the ODE is given by

$\underbrace{\begin{bmatrix}x'\\y'\end{bmatrix}}_{\vec x'}=\underbrace{\begin{bmatrix}5&4\\1&8\end{bmatrix}}_A\underbrace{\begin{bmatrix}x\\y\end{bmatrix}}_{\vec x}+t\underbrace{\begin{bmatrix}1\\-1\end{bmatrix}}_{\vec f}$

a. Move $A\vec x$ to the left side and multiply both sides by the integrating factor, the matrix exponential of $-A$ , $e^{-At}$ :

$e^{-At}\vec x'-Ae^{-At}\vec x=te^{-At}\vec f$

Condense the left side as the derivative of a product:

$\left(e^{-At}\vec x\right)=te^{-At}\vec f$

Integrate both sides and multipy by $e^{At}$ to solve for $\vec x$ :

$e^{-At}\vec x=\displaystyle\left(\int te^{-At}\,\mathrm dt\right)\vec f\implies\vec x=\displaystyle e^{At}\left(\int te^{-At}\,\mathrm dt\right)\vec f$

Finding $e^{\pm At}$ requires that we diagonalize $A$ .

$A$ has eigenvalues 4 and 9, with corresponding eigenvectors $\begin{bmatrix}-4&1\end{bmatrix}^\top$ and $\begin{bmatrix}1&1\end{bmatrix}^\top$ (explanation for this in part (b)), so we have

$A=\begin{bmatrix}-4&1\\1&1\end{bmatrix}\begin{bmatrix}4&0\\0&9\end{bmatrix}\begin{bmatrix}-4&1\\1&1\end{bmatrix}^{-1}$

$\implies A^n=\begin{bmatrix}-4&1\\1&1\end{bmatrix}\begin{bmatrix}4^n&0\\0&9^n\end{bmatrix}\begin{bmatrix}-4&1\\1&1\end{bmatrix}^{-1}$

$\implies A^n=\dfrac15\begin{bmatrix}4^{n+1}+9^n&4\cdot9^n-4^{n+1}\\9^n-4^n&4^n+4\cdot9^n\end{bmatrix}$

$\implies e^{\pm At}=\dfrac15\begin{bmatrix}4e^{\pm4t}+e^{\pm9t}&4e^{\pm9t}-4e^{\pm4t}\\e^{\pm9t}-e^{\pm4t}&e^{\pm4t}+4e^{\pm9t}\end{bmatrix}$

$\implies\vec x=\dfrac15e^{At}\begin{bmatrix}C_1\\C_2\end{bmatrix}-\dfrac1{216}\begin{bmatrix}72t+20\\-36t-7\end{bmatrix}$

b. Find the eigenvalues of $A$ :

$\det(A-\lambda I_2)=\begin{vmatrix}5-\lambda&4\\1&8-\lambda\end{vmatrix}=\lambda^2-13\lambda+36=0$

$\implies(\lambda-4)(\lambda-9)=0\implies\lambda_1=4,\lambda_2=9$

Let $\vec\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top$ and $\vec\theta=\begin{bmatrix}\theta_1&\theta_2\end{bmatrix}^\top$ be the corresponding eigenvectors.

For $\lambda_1=4$ , we have

$\begin{bmatrix}1&4\\1&4\end{bmatrix}\begin{bmatrix}\eta_1\\\eta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

which means we can pick $\eta_1=-4$ and $\eta_2=1$ .

For $\lambda_2=9$ , we have

$\begin{bmatrix}-4&4\\1&-1\end{bmatrix}\begin{bmatrix}\theta_1\\\theta_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

so we pick $\theta_1=\theta_2=1$ .

Then the characteristic solution to the system is

$\vec x_c=C_1e^{\lambda_1t}\vec\eta+C_2e^{\lambda_2t}\vec\theta$

$\vec x_c=C_1e^{4t}\begin{bmatrix}-4\\1\end{bmatrix}+C_2e^{9t}\begin{bmatrix}1\\1\end{bmatrix}$

c. Now we find the particular solution with undetermined coefficients.

The nonhomogeneous part of the ODE is a linear function, so we can start with assuming a particular solution of the form

$\vec x_p=\vec at+\vec b\implies\vec x_p'=\vec a$

Substituting these into the system gives

$\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}5&4\\1&8\end{bmatrix}\left(\begin{bmatrix}a_1\\a_2\end{bmatrix}t+\begin{bmatrix}b_1\\b_2\end{bmatrix}\right)+\begin{bmatrix}1\\-1\end{bmatrix}t$

$\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}5&4\\1&8\end{bmatrix}\begin{bmatrix}a_1t+b_1\\a_2t+b_2\end{bmatrix}+\begin{bmatrix}t\\-t\end{bmatrix}$

$\begin{bmatrix}a_1\\a_2\end{bmatrix}=\begin{bmatrix}(5a_1+4a_2+1)t+(5b_1+4b_2)\\(a_1+8a_2-1)t+(b_1+8b_2)\end{bmatrix}$

$\implies\begin{cases}5a_1+4a_2=-1\\5b_1+4b_2=a_1\\a_1+8a_2=1\\b_1+8b_2=a_2\end{cases}\implies a_1=-\dfrac13,a_2=\dfrac16,b_1=-\dfrac5{54},b_2=\dfrac7{216}$

Put everything together to get a solution

$\vec x=\vec x_c+\vec x_p$

that should match the solution in part (a).

8 0

3 years ago