The fraction of the total number of marbles is left in the bag after removing and giving part to her friend is 3/8
<h3>Fraction</h3>
let
- Number of marbles in her bag = x
- Number of marbles removed from her bag = 1/2x
Remaining marbles = x - 1/2x
= (2x-x) / 2
= x/2
= 1/2x
Number of marbles given to her friend = 1/4 of 1/2x
= 1/4 × 1/2x
= 1/8x
Number of marbles left in the bag = Remaining marbles - Number of marbles given to her friend
= 1/2x - 1/8x
= (4x-x) / 8
= 3x/8
= 3/8x
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Where is the triangles ???
step by step
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:




(c)
Suppose
, then λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.
(d)
From λ² + λb + 2 = 0
If b = 3; we get

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]](https://tex.z-dn.net/?f=v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2B1%261%5C%5C-2%26-2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dv_1%5C%5Cv_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Csim%20v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dv_1%5C%5Cv_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let v₂ = 1, v₁ = -1
![v = \left[\begin{array}{c}-1\\1\\\end{array}\right]](https://tex.z-dn.net/?f=v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=m%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C-2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dm_1%5C%5Cm_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Csim%20v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dm_1%5C%5Cm_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let m₂ = 1, m₁ = -1/2
![m = \left[\begin{array}{c}-1/2 \\1\\\end{array}\right]](https://tex.z-dn.net/?f=m%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%2F2%20%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
∴
![\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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy_1%5C%5Cy_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%3D%20C_1%20e%5E%7B-t%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20C_2e%5E%7B-2t%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%2F2%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
So as t → ∞
Answer:
It would be B
Step-by-step explanation:
You need to add them both together and then once you do that divide everything and you'll get the answer of b.

Recall that the PDF is given by the derivative of the CDF:

The mean is given by
![\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5E2%5Cleft%28x-%5Cdfrac%7Bx%5E2%7D2%5Cright%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac23)
The median is the number

such that

. We have

but both roots can't be medians. As a matter of fact, the median must satisfy

, so we take the solution with the negative root. So

is the median.