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ZanzabumX [31]
4 years ago
8

When creating a story in Einstein Discovery, do all potential collinear fields need to be removed before executing the build sto

ry'5 A. No. Einstein Discovery is impervious to collinearity, so the story and subsequent model will be fine. B. No. Although it is ideal to eliminate collinearity as soon as possible, Einstein will give a warning post-build and the ridge regression will prevent collinearity from over-fitting. C. yes. If all collinear variables are not excluded, the model will over-fit and not make any sense. D. Yes. If the collinear variables are not removed, the Einstein Discovery model build will fail.
Physics
1 answer:
Blizzard [7]4 years ago
3 0

Answer:

The correct option is B i.e. <em>No. Although it is ideal to eliminate collinearity as soon as possible, Einstein will give a warning post-build and the ridge regression will prevent collinearity from over-fitting.</em>

Explanation:

With the help of creating a story in Einstein Discovery, the collinears are not needed to be removed because the ride regression, methodology used will automatically prevent overfitting.

Option A is incorrect as the story in Einstein Discovery is not impervious to the collinearity.

Option C  is incorrect as on one hand it does explain the effect of collinearity but as this is catered automatically in the ridge regression thus the need to do this is not satisfied.

Option D is incorrect because it does not depict the correct outcome of the collinear variables. Collinear variables do not fail the model. It just overfit it and make it difficult to be generalized.

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3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
Un móvil, que sale desde un punto situado 3 metros a la izquierda del origen y lleva un movimiento uniforme, se sitúa a 12 metro
Verdich [7]

Answer:

x₁ = 15 m, x₂ = 12 m , x_total = 27m, v₁ = 5 m / s ,  v₂ = - 3 m / s

Explanation:

In this exercise we will use the kinematics of uniform motion

        v = d / t

let's apply this equation for the first move

        v₁ = Δx / t = (x₂-x₀) / t

        v₁ = (12- (-3)) / 3

        v₁ = 5 m / s

the distance traveled is x₁ = 15 m

Now let's analyze the return movement

        v₂ = Δx / dt

        v₂ = (0 - 12) / 4

        v₂ = - 3 m / s

The negative sign indicates that the vehicle is moving to the left

the distance traveled is x₂ = 12 m

The total dystonia is

     x_total = x₁ + x₂

     x_total = 15 +12

     x_total = 27m

In the attached we have the graphics of the movement

8 0
4 years ago
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