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3 years ago

I need help on this Pls

Mathematics

2 answers:

jolli1 [7]3 years ago

7 0

Y=6x+9 I learned this a while ago

lina2011 [118]3 years ago

6 0

Answer:

y = 6x + 9 :))))))))))((

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What is the Circumference of a circle if its radius in 15 inches? Please help always a procrastinator in math! Thanks!

Tems11 [23]

Circumference Formula:
$C = 2\pi \times R$
R = Radius = 15 in

$C = 2\pi \times 15$
$C = 30\pi \: inches$

Or C = 94,2 inches

7 0

3 years ago

PLEASE HELP WILL GIVE BRAINLY

Bogdan [553]

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:

$\rm MAD = \dfrac{\sum |x-X|}{n}$

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

#SPJ1

6 0

2 years ago

5y-2x=4(solve for y) solve step by step

Hoochie [10]

Your answer will be y= 4/5

4 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

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

4 0

3 years ago

What is the standard deviation of the data set?

Temka [501]

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>

5 0

3 years ago