The mean absolute deviation of the scuba diver’s depth over the entire 25 min is 2.4 meters
<h3>What is mean absolute deviation?</h3>
It is defined as the measure to show the variation in data set in other words between the mean and every data value, the distance known as the MAD.
We have depths of the scuba diver, in meters
{−9, −18, −16, −13, −14}
We know the formula for the mean absolute deviation:

Here n is total number of observation
x is the elements in the data set
X is the mean of the data.
n = 5
x = -70/5 = -14
∑|x-X| = 12
MAD = 12/5 = 2.4 meters
Thus, the mean absolute deviation of the scuba diver’s depth over the entire 25 min is 2.4 meters.
Learn more about the mean absolute deviation here:
brainly.com/question/10528201
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Your answer will be y= 4/5
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]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D)
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 \\\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%20-%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%3D0%20%5C%5C%5C%5C)
![\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\\\\](https://tex.z-dn.net/?f=%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B-%5Clambda%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%20-0%5D%20-%200%20-%20%5Cfrac%7B1%7D%7B4%7D%5B0-%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7B1%7D%7B2%7D%20-%20%5Clambda%20%29%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B4%7D%20-%20%5Clambda%20%29%20%5B%28%5Cfrac%7B%5Clambda%7D%7B2%7D%20%2B%20%5Clambda%5E2%20%29%5D%20-%20%5Cfrac%7B1%7D%7B4%7D%5B%5Cfrac%7B%5Clambda%7D%7B2%7D%20-%20%20%5Cfrac%7B1%7D%7B4%7D%5D%20%3D0%20%5C%5C%5C%5C%5Cto%20%20%28%5Cfrac%7B3%7D%7B8%7D%5Clambda%20%2B%20%5Cfrac%7B3%7D%7B4%7D%20%5Clambda%5E2%20-%20%5Cfrac%7B%5Clambda%5E2%7D%7B2%7D%20-%20%5Clambda%5E3%20-%20%5Cfrac%7B%5Clambda%7D%7B8%7D%20%2B%20%5Cfrac%7B1%7D%7B16%7D%3D0%20%5C%5C%5C%5C%5Cto%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%28%5Clambda%20-%5Cfrac%7B1%7D%7B4%7D%29%20%28%5Clambda%20-%20%5Cfrac%7B1%7D%7B2%7D%29%20%3D0%5C%5C%5C%5C)


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] \\\\](https://tex.z-dn.net/?f=%5Cto%20c%5E%7B%28k%2B1%29%7D%20%3D%20A%20x%5E%7Bk%7D%2BC%20%5C%5C%5C%5C%5Cto%20x%280%29%20%3D%20%20%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D3%261%262%5Cend%7Barray%7D%5Cright%29%20%20%2C%20c%20%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%29%5C%5C%5C%5C%20%20%5Cto%20x%5E%7B%28k%2B1%29%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B3%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B4%7D%26%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%26%200%5C%5C%20-%5Cfrac%7B1%7D%7B4%7D%26%20-%5Cfrac%7B1%7D%7B4%7D%20%26%200%5Cend%7Barray%7D%5Cright%5D%20x%5Ek%20%2B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%262%264%5Cend%7Barray%7D%5Cright%5D%20%20%5C%5C%5C%5C)
after solving the value the answer is
:
![\lim_{k \to \infty} x^k=o = \left[\begin{array}{c}0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Clim_%7Bk%20%5Cto%20%5Cinfty%7D%20x%5Ek%3Do%20%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%260%260%5Cend%7Barray%7D%5Cright%5D)
The Standard Deviation is a measure of how spread out numbers are. Standard deviation can be calculated by the expression as follows:
s = √∑(x - x(mean))^2/n-1
x(mean) = 24
s = √∑(x - 24)^2 / 7-1
<span>s = 6.11
Hope this answers the question. Have a nice day.</span>