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dimulka [17.4K]
3 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]3 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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alisha [4.7K]

Answer:

a)    x = v₀² sin 2θ / g

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This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

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         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

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we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

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          x = 13.5 2 sin (2 32) / 9.8

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