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2 years ago

In a regression analysis, if sse = 200 and ssr = 300, then the coefficient of determination is:_________

Physics

1 answer:

Nataliya [291]2 years ago

8 0

0.600

In a regression analysis, if sse = 200 and ssr = 300, then the coefficient of determination is: <u>0.600</u>

<u></u>

1. You have that the data set having SSR=300 and SSE=200

2. Therefore you have the coefficient of determination is:

r²=SSR/SSTO

SSTO=SSR+SSE

3. Then, when you substitute the values, you obtain:

SSTO=200+300

SSTO=500

r²=300/500

4. So, you have that the result is:

r²=0.6

As a result, as you can see, the solution to the previous exercise is that the coefficient of determination is 0.6

<h3>What exactly is the connection between SST, SSR, and SSE?</h3>

The difference between SST and SSR is the amount of Y's variability that is still unaccounted for after using the regression model, also known as the sum of squared errors (SSE). Sum of squares of residual can be used to directly calculate SSE.

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A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

3 years ago

If earth's mass were half its actual value but its radius stayed the same, the escape velocity of earth would be:________

siniylev [52]

If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

<h3>What is an escape velocity?</h3>

The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.

The formula to calculate the escape velocity of earth is given below:-

$V_e=\sqrt{\dfrac{2GM}{r}}$

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

$V_e=\sqrt{\dfrac{2GM}{r\times 2}}$

$V_e = \sqrt{\dfrac{GM}{r}}$ .

Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be $V_e = \sqrt{\dfrac{GM}{r}}$ .

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8 0

2 years ago

The scientist to first introduce the concept of inertia was

skelet666 [1.2K]

The first scientist to introduce the concept of inertia was Galileo

4 0

3 years ago

200 Coulombs of charge passes through a point in a circuit for 0.6 minutes. what is the magnitude of the current flowing

Tasya [4]

Answer:

5.56 A

Explanation:

From the question,

Q = it.............. Equation 1

Where Q = charges, i = current, t = time.

Make i the subject of the equation

i = Q/t.............. Equation 2

Given: Q = 200 coulombs, t = 0.6 minutes = (0.6×60) seconds

Substitite these values into equation 2

i = 200/(0.6×60)

i = 5.56 A

Hence the magnitude of the current flowing through the circuit is 5.56 A

5 0

3 years ago

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3 Pa s and

icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

g = acceleration due to gravity = 9.8 m/s²,
r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
ρ = density of sphere = 19 g/cm³,
σ = density of water = 1.0 g/cm³ and
η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

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4 0

2 years ago