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sweet-ann [11.9K]
3 years ago
13

Data plays a crucial role in decision-making because it can reveal relationships between different quantities. We often use line

ar equations to model this relationships and make predictions about the data.
Think about a situation when you needed to analyze data. What types of trends did you find in the data? How did noticing the trends help you make a decision related to the situation?
Mathematics
2 answers:
Natasha2012 [34]3 years ago
7 0

Answer:

A situation when I needed to analyze data was when I was researching the different prices for ( Input your own thing ) . The type of trend I found in the data was different prices across all websites. I noticed most of the prices are close but some were crazy. I found that the prices resembled a scatter plop. There was a strong ( weak ) positive ( negative ) correlation of 89.667 ( Anything number ). Noticing the trends helped me make a decision related to the situation by allowing me to make the best choice "price-wise" for the NVIDIA GeForce RTX 3090 Graphics card ( enter the same thing you put above ) .

Step-by-step explanation:

Use A Website called Smodin.io

It will rewrite and check this for plagiarism and you won't get in trouble for plagiarism. Make sure you use your own situation in the ( ) and take out my answers.

mote1985 [20]3 years ago
5 0

Answer:

(a×b)×c = 0 means that vector c parallel to vector (a×b). But because   (a × b) ⊥ a and   (a -b) ⊥ b so

c ⊥ a and c ⊥ b.

Hope this helps, have a nice day! :D

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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}

6 0
3 years ago
Please help fast its quiz. :(
dezoksy [38]

Answer:

What is the question??

Thank you for the points

8 0
2 years ago
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Donna has a $300 loan through the bank she is charged a simple rate The total interest she paid on the loan was $63 As a percent
Black_prince [1.1K]

Answer:

21%

Step-by-step explanation:

($63*100%) / $300 = 21%

3 0
2 years ago
2. What percent of cal Rs 80 is Rs 20 ? ​
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Answer:

Hope this is helpful to you

7 0
2 years ago
What is the quotient of 1/2÷1/6=
Sphinxa [80]
1/2 <span>÷ 1/6 =

Divide fractions:

a x d /  b x c

 Therefore:

1 x 6 / 2 x 1 =

6 / 2 = 3

hope this helps!

</span>
3 0
3 years ago
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