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2 years ago

PLZ HELP ASAP WILL GIVE BRAINLIEST!!!!!!!!

Mathematics

1 answer:

docker41 [41]2 years ago

8 0

I think it's different by
Group 2 has a 1 yet group 2 didn't sorry if wrong

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Which value of y will make the inequality y<-1 false

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0 or 1 or 2............................................

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3 years ago

Read 2 more answers

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

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3 years ago

HELP PLEASE!!!! solve for x

LenKa [72]

Answer: how do you solve for this

Step-by-step explanation:

7 0

3 years ago

Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably norma

Nataliya [291]

Answer:

The degrees of freedom is 11.

The proportion in a t-distribution less than -1.4 is 0.095.

Step-by-step explanation:

The complete question is:

Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4 if the samples have sizes 1 = 12 and n 2 = 12 . Enter the exact answer for the degrees of freedom and round your answer for the area to three decimal places. degrees of freedom = Enter your answer; degrees of freedom proportion = Enter your answer; proportion

Solution:

The information provided is:

$n_{1}=n_{2}=12\\t-stat=-1.4$