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dimulka [17.4K]
4 years ago
9

An agriculturalist working with Australian pine trees wanted to investigate the relationship between the age and the height of t

he Australian pine. A random sample of Australian pine trees was selected, and the age, in years, and the height, in meters, was recorded for each tree in the sample. Based on the recorded data, the agriculturalist created the following regression equation to predict the height, in meters, of the Australian pine based on the age, in years, of the tree. predicted height 0.29+048 (age) Which of the following is the best interpretation of the slope of the regression line?
A. The height increases, on average, by 1 meter each 0.48 year.
B. The height increases, on average, by 0.48 meter each year.
C. The height increases, on average, by 0.29 meter each year.
D. The height increases, on average, by 0.29 meter each 0.48 year.
E. The difference between the actual height and the predicted height is, on average, 0.48 meter for each year.
Physics
1 answer:
SVETLANKA909090 [29]4 years ago
6 0

Answer:

The correct answer is B.

Explanation:

Step 1:

The available regression equation is: Predict height= 0.29 + 0.48 (age).

Here, the predict height is dependent variable and the  age is in-dependent variable.

Intercept = 0.29

Slope      = 0.48

The given regression equation indicates the y on x model and the intercept coefficients of the regression equation is 0.29 and the slope is 0.48.

Step 2:

The height increases, an average, by 0.48 m per year.

Because co-efficient of slope variable indicate the positive sign and we increase 1 year in age then automatically height increased is 0.48 m.

<h3></h3><h3>The height increases, on average, by 0.48 meter each year.</h3>
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Define time period. Give its formula. ​
kondaur [170]

Answer:

Part A

Time period, designated by the symbol, 'T', is defines as the time it takes for a complete cycle of an oscillation or vibration (of a wave) to transit through a given point.

The longer the time period of a wave, the lower the frequency of the wave

The unit of the time period is seconds, 's'

Part B

Mathematically, the formula for the time period is presented as follows;

f = 1/T

∴ T = 1/f

f = v/λ

∴ T = λ/v

Where;

v = The velocity of the wave;

λ = The wavelength of the wave

Explanation:

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3 years ago
An electromagnetic wave has a frequency of 793 Hz. What is<br> its wavelength?
frozen [14]

Answer:

245

Explanation:

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3 years ago
Select the THREE ways that the parallel-plate capacitor differs from a car battery.
Hoochie [10]

Answer:

1. battery is capable of continuous current, capacitor is not

2. battery maintains a potential, capacitor does not

3. battery stores chemical energy, capacitor stores electric energy

Explanation:

1. battery is capable of continuous current, capacitor is not

This is because the battery maintains a constant current throughout while the capacitor maintains an exponential decaying current.

2. battery maintains a potential, capacitor does not

This is because, the battery has a potential at its terminal due to its emf whereas, the capacitor needs a potential to be applied to its terminals.

3. battery stores chemical energy, capacitor stores electric energy

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5 0
3 years ago
. A circular wire loop 40 cm in diameter has 100 Ohm resistance and lies in a horizontal plane. A uniform magnetic field points
sergeinik [125]

Answer:

(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A

Explanation:

Given that,

The diameter of the loop, d = 40 cm

Radius, r = 20 cm

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Final magnetic field, B' = 55 mT

Initial magnetic flux,

\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb

Final magnetic flux,

\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb

Due to change in magnetic field an emf will be generated in the loop. It is given by :

\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V

Let I be the current in the loop. We can find it using Ohm's law such that,

\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A

Hence, this is the required solution.

7 0
3 years ago
A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt
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The total power emitted by an object via radiation is:
P=A\epsilon \sigma T^4
where:
A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW
So, the correct answer is D.
6 0
4 years ago
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