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Zigmanuir [339]
2 years ago
12

The data set shows the numbers of hours students study the week before a test. Find and interpret the mean absolute deviation of

the data. Write your answer as a decimal. 1, 1, 1, 3, 4, 5, 6, 7, 8, 10 The data values differ from the mean by an average of hours.
Mathematics
1 answer:
tatiyna2 years ago
6 0

Answer:

4.6

Step-by-step explanation:

to find the mean, you add up all given numbers and divide by how many numbers is given. so if there are five numbers, you divide the sum of all of the numbers by five.

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18%

Step-by-step explanation:

0.12 is 12 percent.

We also have the 0.06 percent which would just be 6 percent

We need to add up those percentages to get 18 percent, for the people surveyed that cannot swim would be 18 percent.  

Our final answer is D.

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2 years ago
HELP PLSS QUICKLY!!!!
tigry1 [53]
3 cans!!!

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5x6=30 (divided by 2 = 15)
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During statistical analysis, the investigator may consider searching for statistical associations among various groups that may
IceJOKER [234]

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Answer:

This is known as Data dredging.

Step-by-step explanation:

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2 years ago
5/10 x 9/12 will give brainliest if correct
mote1985 [20]

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3 years ago
Read 2 more answers
Let y 00 + by0 + 2y = 0 be the equation of a damped vibrating spring with mass m = 1, damping coefficient b > 0, and spring c
stira [4]

Answer:

Step-by-step explanation:

Given that:    

The equation of the damped vibrating spring is y" + by' +2y = 0

(a) To convert this 2nd order equation to a system of two first-order equations;

let y₁ = y

y'₁ = y' = y₂

So;

y'₂ = y"₁ = -2y₁ -by₂

Thus; the system of the two first-order equation is:

y₁' = y₂

y₂' = -2y₁ - by₂

(b)

The eigenvalue of the system in terms of b is:

\left|\begin{array}{cc}- \lambda &1&-2\ & -b- \lambda \end{array}\right|=0

-\lambda(-b - \lambda) + 2 = 0 \ \\ \\\lambda^2 +\lambda b + 2 = 0

\lambda = \dfrac{-b \pm \sqrt{b^2 - 8}}{2}

\lambda_1 = \dfrac{-b + \sqrt{b^2 -8}}{2} ;  \ \lambda _2 = \dfrac{-b - \sqrt{b^2 -8}}{2}

(c)

Suppose b > 2\sqrt{2}, then  λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.

(d)

From λ² + λb + 2 = 0

If b = 3; we get

\lambda^2 + 3\lambda + 2 = 0 \\ \\ (\lambda + 1) ( \lambda + 2 ) = 0\\ \\ \lambda = -1 \ or   \  \lambda = -2 \\ \\

Now, the eigenvector relating to λ = -1 be:

v = \left[\begin{array}{ccc}+1&1\\-2&-2\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

\sim v = \left[\begin{array}{ccc}1&1\\0&0\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

Let v₂ = 1, v₁ = -1

v = \left[\begin{array}{c}-1\\1\\\end{array}\right]

Let Eigenvector relating to  λ = -2 be:

m = \left[\begin{array}{ccc}2&1\\-2&-1\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

\sim v = \left[\begin{array}{ccc}2&1\\0&0\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

Let m₂ = 1, m₁ = -1/2

m = \left[\begin{array}{c}-1/2 \\1\\\end{array}\right]

∴

\left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= C_1 e^{-t}  \left[\begin{array}{c}-1\\1\\\end{array}\right] + C_2e^{-2t}  \left[\begin{array}{c}-1/2\\1\\\end{array}\right]

So as t → ∞

\mathbf{ \left[\begin{array}{c}y_1\\y_2\\\end{array}\right]=  \left[\begin{array}{c}0\\0\\\end{array}\right] \ \  so \ stable \ at \ node \ \infty }

5 0
2 years ago
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