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posledela
3 years ago
8

If the coefficient of determination for a data set is 0.75 and the SSE for the data set is 9, what is the SST for the data set?

Mathematics
2 answers:
OverLord2011 [107]3 years ago
6 0
0.1873 aaaaaaaaaaaaaaaaaaaaaaaa
Gemiola [76]3 years ago
4 0
<h2>Answer:</h2>

Hence, the SST for the data set is:

                              36

<h2>Step-by-step explanation:</h2>

We know that the formula or the expression that relates coefficient of determination i.e. (r²) of the set and SSE and SST for the data set is:

            r^2=1-\dfrac{SSE}{SST}

i.e.

\dfrac{SSE}{SST}=1-r^2

We are given:

        r^2=0.75

SSE=9

Hence, we get:

\dfrac{9}{SST}=1-0.75\\\\\\i.e.\\\\\\\dfrac{9}{SST}=0.25

i.e.

SST=\dfrac{9}{0.25}\\\\\\SST=36

                   Hence, SST for the data set is:

                                     36

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We know that after dilation with scale factor k the coordinates of pre-image (x,y) map to (kx,ky) of the image.

From the given graph, the coordinate of A =(9,4) and the coordinate of A' (3,2)

Then x = 9 and kx = 3

\Rightarrow k(9)=3\\\\\Rightarrow\ k=\frac{3}{9}\\\\\Rightarrow\ k=\frac{1}{3}

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4 0
3 years ago
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ser-zykov [4K]
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3 years ago
Write an equation of the line that passes through the point (–1, 4) with slope 2. A. y+1=−2(x−4) B. y+1=2(x−4) C. y−4=2(x+1) D.
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y-4=2\,(x+1)

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Step-by-step explanation:

We can use the general point-slope form of a line of slope m and going through the point (x_0, y_0):

y-y_0=m(x-x_0)

which in our case, given the info on the slope (2) and the point (-1, 4) becomes:

y-y_0=m\,(x-x_0)\\y-4=2\,(x-(-1))\\y-4=2\,(x+1)

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3 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

3 0
3 years ago
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