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andrezito [222]
3 years ago
7

What is the solution to y=6x+25 y=12x-9

Mathematics
1 answer:
prisoha [69]3 years ago
8 0

Answer:

x=17/3, y=59. (17/3, 59).

Step-by-step explanation:

y=6x+25

y=12x-9

--------------

6x+25=12x-9

12x-6x-9=25

6x-9=25

6x=25+9

6x=34

x=34/6=17/3

y=12(17/3)-9=4*17-9=68-9=59

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Suppose commute times in a large city are normally distributed and that 63.70% of commuters in this city take more than 22 minut
USPshnik [31]

Answer:

The mean is 24.205

Step-by-step explanation:

Firstly, we need to get the z-score

From the question, the probability we have is the probability that commuters take more than 22 minutes to commute one-way

So the probability we had was;

P( x > 22)

So we need the z-score corresponding to 63.70% or simply 0.637

We can use the standard normal distribution table to get this

Mathematically, from the standard normal distribution table this is -0.35

z-score = (x- mean)/SD

-0.35 = (22-mean)/6.3

22-mean = -2.205

mean = 22 + 2.205 = 24.205

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3 years ago
1whole and 1 half plus 3 and 2 thirds
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Add a 3ed to the pie
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A clothing store donated a percent of every sale to hospitals. The total sales were $6,500 so the store
ivanzaharov [21]
The answer is 4%

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8 0
3 years ago
Is the relationship proportional or not? Explain. y=0.25x
givi [52]
Yes, it is Proportional.
Y=0.25x
For example:
If x=1, y=0.25
If x=2, y=0.5
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5 0
3 years ago
Simultaneous ODE
leonid [27]
\mathbf y'=\mathbf A\mathbf y\iff\begin{bmatrix}{y_1}'\\{y_2}'\end{bmatrix}=\begin{bmatrix}\frac52&-\frac32\\-\frac32&\frac52\end{bmatrix}\begin{bmatrix}y_1\\y_2\end{bmatrix}

Find the eigensystem corresponding to the coefficient matrix.

\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}\frac52-\lambda&-\frac32\\-\frac32&\frac52-\lambda\end{vmatrix}=0
\left(\dfrac52-\lambda\right)^2-\left(-\dfrac32\right)^2=0
\lambda^2-5\lambda+4=0
(\lambda-1)(\lambda-4)=0
\implies \lambda_1=1,\lambda_2=4

For \lambda_1=1, the associated eigenvector satisfies

(\mathbf A-\mathbf I)\mathbf v_1=\mathbf 0\iff\begin{bmatrix}\frac32&-\frac32\\-\frac32&\frac32\end{bmatrix}\begin{bmatrix}v_{11}\\v_{12}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\implies v_{11}-v_{12}=0\implies\mathbf v_1=\begin{bmatrix}1\\1\end{bmatrix}

For \lambda_2=4, we have

(\mathbf A-4\mathbf I)\mathbf v_2=\mathbf 0\iff\begin{bmatrix}-\frac32&-\frac32\\-\frac32&-\frac32\end{bmatrix}\begin{bmatrix}v_{21}\\v_{22}\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}
\implies v_{21}+v_{22}=0\implies\mathbf v_2=\begin{bmatrix}1\\-1\end{bmatrix}

The general solution for the ODE system is then

\mathbf y=C_1e^{\lambda_1t}\mathbf v_1+C_2e^{\lambda_2t}\mathbf v_2
\iff\begin{bmatrix}y_1\\y_2\end{bmatrix}=C_1e^t\begin{bmatrix}1\\1\end{bmatrix}+C_2e^{4t}\begin{bmatrix}1\\-1\end{bmatrix}
\implies\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}C_1e^t+C_2e^{4t}\\C_1e^t-C_2e^{4t}\end{bmatrix}
7 0
4 years ago
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