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3 years ago

The graph shows the distance Julian drives on a trip. What is Julian’s speed

Mathematics

2 answers:

torisob [31]3 years ago

5 0

Where is the graph??
is it me or is it not there?hmm

Mars2501 [29]3 years ago

4 0

Answer:80 apex ;)

Step-by-step explanation:

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miss Akunina [59]

The answer is c . 5.96

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3 years ago

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Find the height of the triangle in which A=192 cm^2

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2 years ago

Statistics show that a certain soccer player has a 63% chance of missing the goal each time he shoots. If this player shoots twi

Kipish [7]

Answer:

<em>The probability of scoring two goals in both times is</em><em> 0.137 or 13.7%</em>

Step-by-step explanation:

Statistics show that a certain soccer player has a 63% chance of missing the goal each time he shoots.

So $P(A)=0.63$

Hence, the probability of getting success in his shoots will be,

$P(A^c)=1-P(A)=1-0.63=0.37$

The probability of scoring two goals in both times is,

$=P(A^c)\cdot P(A^c)$

$=0.37\times 0.37$

$=0.137$

$=13.7\%$

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2 years ago

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16% of 130 is what number

earnstyle [38]

130 • .16 = 20.8 or 20 (4/5) or 104/5
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.16 is 16%

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3 years ago

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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

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

4 0

3 years ago