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mixer [17]
4 years ago
15

Write the given linear system in matrix form. Assume X = x y z . dx dt = −8x + 7y − 9z dy dt = 5x − y dz dt = 10x + 7y + 8z

Mathematics
1 answer:
Akimi4 [234]4 years ago
6 0

Answer:

The matrix form of given linear system is

\begin{bmatrix}x'\\ y'\\ z'\end{bmatrix}=\begin{bmatrix}-8 & 7 &-9\\ 5 & -1 & 0\\ 10 & 7 & 8\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}

Step-by-step explanation:

Let as assume that

X=\begin{bmatrix}x\\ y\\ z\end{bmatrix}

Given differential equations are

\frac{dx}{dt}=-8x+7y-9z

\frac{dy}{dt}=5x-y

\frac{dz}{dt}=10x+7y+8z

We need to find the matrix form of given linear system.

Write the elements of left side in a column matrix.

Write all the coefficients in one matrix first which is called a coefficient matrix.  Multiply coefficient matrix with the variables matrix and equate left and right side.

\begin{bmatrix}\frac{dx}{dt}\\ \frac{dy}{dt}\\ \frac{dz}{dt}\end{bmatrix}=\begin{bmatrix}-8 & 7 &-9\\ 5 & -1 & 0\\ 10 & 7 & 8\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}

It can be written as

\begin{bmatrix}x'\\ y'\\ z'\end{bmatrix}=\begin{bmatrix}-8 & 7 &-9\\ 5 & -1 & 0\\ 10 & 7 & 8\end{bmatrix}\begin{bmatrix}x\\ y\\ z\end{bmatrix}

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Learning Thoery In a learning theory project, the proportion P of correct responses after n trials can be modeled by p = 0.83/(1
elena-s [515]

Answer:

a)P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536

b) P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666

c) 0.75 =\frac{0.83}{1+e^{-0.2n}}

1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}

e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}

ln e^{-0.2n} = ln (\frac{8}{75})

-0.2 n = ln(\frac{8}{75})

And then if we solve for t we got:

n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials

d) If we find the limit when n tend to infinity for the function we have this:

lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

Step-by-step explanation:

For this case we have the following expression for the proportion of correct responses after n trials:

P(n) = \frac{0.83}{1+e^{-0.2t}}

Part a

For this case we just need to replace the value of n=3 in order to see what we got:

P(n=3) = \frac{0.83}{1+e^{-0.2(3)}}= \frac{0.83}{1+ e^{-0.6}} = 0.536

So the number of correct reponses  after 3 trials is approximately 0.536.

Part b

For this case we just need to replace the value of n=7 in order to see what we got:

P(n=7) = \frac{0.83}{1+e^{-0.2(7)}}= \frac{0.83}{1+ e^{-1.4}} = 0.666

So the number of correct responses after 7 weeks is approximately 0.666.

Part c

For this case we want to solve the following equation:

0.75 =\frac{0.83}{1+e^{-0.2n}}

And we can rewrite this expression like this:

1+ e^{-0.2n} = \frac{0.83}{0.75}= \frac{83}{75}

e^{-0.2n} = \frac{83}{75}-1= \frac{8}{75}

Now we can apply natural log on both sides and we got:

ln e^{-0.2n} = ln (\frac{8}{75})

-0.2 n = ln(\frac{8}{75})

And then if we solve for t we got:

n = \frac{ln(\frac{8}{75})}{-0.2} = 11.19 trials

And we can see this on the plot attached.

Part d

If we find the limit when n tend to infinity for the function we have this:

lim_{n \to \infty} \frac{0.83}{1+e^{-0.2t}} = 0.83

So then the number of correct responses have a limit and is 0.83 as n increases without bound.

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