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lyudmila [28]
3 years ago
6

SOMEONE PLEASE HELP ME WITH THIS!!!!

Mathematics
1 answer:
denis23 [38]3 years ago
7 0

Answer:

D.

Step-by-step explanation:

115x + 170 = 630

Subtract 170 from each side of the equals,

115x = 460

Divide 460 by 115,

4

Thus, D is your correct answer.

<em>Answered by:</em>

matthewarvin06

Hey! Thanks for reading my answer. If this helped you in any way, please rate or say thanks! It's really appreciated. If you have any further questions, please let me know. Thanks!

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What ratio can you use to determine the probability of a compound event?
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Step-by-step explanation:

Favorable outcome. To possible outcome

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Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people ans
vladimir2022 [97]

Answer:

There is no following,  we have data of all the people

Step-by-step explanation:

There where Sixty-five salespersons asked and the if we add all the persons that where asked  it is the same.  So there is no following

Fourteen people answered that they generally sell two cars; 14

nineteen generally sell three cars; 19

twelve generally sell four cars; 12

nine generally sell five cars; 9

eleven generally sell six cars 11

Adding  14+19+12+9+11=65 persons

8 0
3 years ago
find the spectral radius of A. Is this a convergent matrix? Justify your answer. Find the limit x=lim x^(k) of vector iteration
Natasha_Volkova [10]

Answer:

The solution to this question can be defined as follows:

Step-by-step explanation:

Please find the complete question in the attached file.

A = \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right]

now for given values:

\left[\begin{array}{ccc} \frac{3}{4} - \lambda & \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2} - \lambda & 0\\ -\frac{1}{4}& -\frac{1}{4} & 0 -\lambda \end{array}\right]=0 \\\\

\to  (\frac{3}{4} - \lambda ) [-\lambda (\frac{1}{2} - \lambda ) -0] - 0 - \frac{1}{4}[0- \frac{1}{2} (\frac{1}{2} - \lambda )] =0 \\\\\to  (\frac{3}{4} - \lambda ) [(\frac{\lambda}{2} + \lambda^2 )] - \frac{1}{4}[\frac{\lambda}{2} -  \frac{1}{4}] =0 \\\\\to  (\frac{3}{8}\lambda + \frac{3}{4} \lambda^2 - \frac{\lambda^2}{2} - \lambda^3 - \frac{\lambda}{8} + \frac{1}{16}=0 \\\\\to (\lambda - \frac{1}{2}) (\lambda -\frac{1}{4}) (\lambda - \frac{1}{2}) =0\\\\

\to \lambda_1=\lambda_2 =\frac{1}{2}\\\\\to \lambda_3 = \frac{1}{4} \\\\\to A = max {|\lambda_1| , |\lambda_2|, |\lambda_3|}\\\\

       = max{\frac{1}{2}, \frac{1}{2}, \frac{1}{4}}\\\\= \frac{1}{2}\\\\(A) =\frac{1}{2}

In point b:

Its  

spectral radius is less than 1 hence matrix is convergent.

In point c:

\to c^{(k+1)} = A x^{k}+C \\\\\to x(0) =   \left(\begin{array}{c}3&1&2\end{array}\right)  , c = \left(\begin{array}{c}2&2&4\end{array}\right)\\\\  \to x^{(k+1)} =  \left[\begin{array}{ccc} \frac{3}{4}& \frac{1}{4}& \frac{1}{2}\\ 0 & \frac{1}{2}& 0\\ -\frac{1}{4}& -\frac{1}{4} & 0\end{array}\right] x^k + \left[\begin{array}{c}2&2&4\end{array}\right]  \\\\

after solving the value the answer is

:

\lim_{k \to \infty} x^k=o  = \left[\begin{array}{c}0&0&0\end{array}\right]

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3 years ago
Simplify 2(x+3)-4 showing each step
Alex
To simplify this, we first need to distribute the 2 into the parenthesis.

= 2 * x + 2 * 3 - 4

Now, we can simplify:

= 2x + 6 - 4
= 2x + 2

Hope this helps!
7 0
4 years ago
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