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3 years ago

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:

<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:

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Answer:

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Step-by-step explanation:

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sattari [20]

Answer: 1/3

Step-by-step explanation:

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3 years ago

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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.

In-s [12.5K]

Answer:

$y-4=2\,(x+1)$

which agrees with answer C in your list of possible answers.

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)

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