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____ [38]
3 years ago
9

Which of the following statements describes an overfitted model the best. a. Performs worse on training data as the training pro

cess proceeds, while performing better on a held-out test data b. Its performance doesn't improve by additional training neither on the training data nor on the test data c. Performs better on training data as the training process proceeds, while performing worse on a held-out test data d. Performs worse on both the training data and the test data as the training process proceeds.
Physics
1 answer:
Mumz [18]3 years ago
7 0

Answer:

c. Performs better on training data as the training process proceeds, while performing worse on a held-out test data

Explanation:

An over-fitted model is one that will perform best on training but would fail or do worse on a held-out test data.

Such models are optimum for a just a particular set of data but would grossly failed when extrapolated to some other data set not novel to it.

  • Over-fitting a model implies that a model closely corresponds to a set of data but would not perform well with others.
  • It is usually as a result of a model adapting the noise and other details of a particular data set and thereby incorporates it.
  • This makes it difficult for the model to fit into another data set.
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How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at
tatuchka [14]

Answer:

0.66

Explanation:

By using the formula

u = v^2 / r g

Where u is coefficent of friction

u = 23.5 × 23.5 / (85 × 9.8)

u = 0.66

4 0
3 years ago
Cozebilir misiniz !!!​
GREYUIT [131]

Answer:

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5 0
3 years ago
While participating in a blood drive at school, Keona learns that blood has a density of 1.06 g/mL. She donates one pint of bloo
Sergio [31]

Answer:

m≈501.57 g

Explanation:

The density formula is:

d=m/v

Let’s rearrange the formula for m. m is being divided by v. The inverse of division is multiplication, so multiply both aides by v.

d*v= m/v*v

d*v=m

The mass can be found by multiply the density and the volume.

m=d*v

The density is 1.06 grams per milliliter and the volume is 473.176 milliliters.

d= 1.06 g/mL

v= 473.176 mL

Substitute the values into the formula.

m= 1.06 g/mL * 473.176 mL

Multiply. When multiplying, the mL will cancel out.

m= 501.56656 g

Let’s round to the nearest hundredth. The 6 in the thousandth place tells us to round the 6 to a 7 in the hundredth place.

m ≈501.57 g

The mass is about 501.57 grams.

7 0
3 years ago
Read 2 more answers
In an electric field, 0.90 joule of work is required to bring 0.45 coulomb of charge from point a to point
jarptica [38.1K]
The difference in electric potential energy between the two points is
\Delta U = q \Delta V
where q is the magnitude of the charge and \Delta V is the electric potential difference.

But for energy conservation, the difference in electric potential energy \Delta U between the two points is equal to the work done to move the charge between A and B:
W=\Delta U
so we have
W=q \Delta V

and by substituting the numbers of the problem, we find the value of \Delta V:
\Delta V =  \frac{W}{q}= \frac{0.90 J}{0.45 C}=2 V
3 0
2 years ago
A woman throws a javelin 35 mph at an angle 30 degrees from the ground. Neglecting wind resistance or the height the javelin thr
Anna71 [15]

Answer:

35 mph

Explanation:

The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.

When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.

When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.

So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.

5 0
2 years ago
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