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aleksklad [387]
2 years ago
14

You are training a classification model with logistic regression.

Physics
2 answers:
lions [1.4K]2 years ago
6 0

The correct answers are A) Adding a new feature to the model always results in equal or better performance on examples, not in the training set and B) Adding many new features to the model makes it more likely to overfit the training set.

You are training a classification model with logistic regression.

The following statements are true: Adding a new feature to the model always results in equal or better performance on examples, not in the training set and Adding many new features to the model makes it more likely to overfit the training set.

When we are talking about statistical terms, logistic regression is the analysis used when the dependent variable is binary. This allows the researcher to explain and describe this variable and the interval. This logistic regression can be applied in medical research, engineering or social investigation.

aalyn [17]2 years ago
5 0

Answer:

B) Adding many new features to the model makes it more likely to over fit the training set.

Explanation:

Increasing number of features without increasing the amount of training data, increase over fitting in the model because features may be irrelevant and increase the redundant information in the model.

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_____ of rough surfaces reduces friction
Radda [10]

Answer:

Hey shaikaadil700 !

<u> </u><u>Lubricating</u><u> </u> of rough surfaces reduces friction.

Explanation:

• Lubricating is the smoothening or polishing of the surfaces

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3 0
2 years ago
A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

6 0
2 years ago
GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.
prohojiy [21]

Answer:

<em>They represent kinetic energy</em>

Explanation:

<u>Kinetic Energy </u>

A body can do work due to some of its attributes or states. For example, its mass can do work if used to provide energy, if the object is at a certain height respect to some reference level, it can do work when going downwards (potential energy), if the object moves at a certain speed, it can do work when transferring part of its speed to other objects. It's called kinetic energy and is given by

\displaystyle K=\frac{mv^2}{2}

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

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2 years ago
What is the approximate velocity of the object at 5 seconds ? .
Gnoma [55]

Answer:

do you have an image?

Explanation:

3 0
3 years ago
Match the measurement with the prober SI unit.
padilas [110]

Answer:

Your question is incomplete

8 0
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