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3 years ago

6

A __________ provides a visual summary of the improvements that a data mining project provides on a binary classification proble

m compared to a random guess
a. Pareto chart
b. lift chart
c. Gantt chart histogram

1 answer:

babymother [125]3 years ago

5 0

Answer:

Option (B)

Explanation:

A lift chart usually refers to a graphical representation that is mainly used in order to improve the drawbacks of a mining model by making a comparison with any random guess, and also helps in determining the changes that occur in terms of lift scores.

It describes the binary classification of the problems associated with the mining activity. This type of chart is commonly used to differentiate the lift scores for a variety of models, and picking the best one out of all.

Thus, the correct answer is option (B).

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The sun delivers an average power of 1.499 w/m2 to the top of neptune's atmosphere. find the magnitudes of vector e max and vect

sesenic [268]

See attachment for answer:

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3 years ago

Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot

Gelneren [198K]

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = $\frac{kQq}{r^{2} }$

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

$N_{x}$ = N cosine 60° + (-N cosine 60°) = 0

for the y-component

$N_{y}$ = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( $N_{x}$ = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

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3 years ago

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

sladkih [1.3K]

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

7 0

3 years ago

How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

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3 years ago

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4

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1 year ago