The <em>horizontal</em> asymptote of the <em>exponential</em> function f(x) = 0.2ⁿ is represented by the <em>horizontal</em> line y = 0 , to which the curve tends for n → + ∞.
<h3>How to find the horizontal asymptote of an exponential function</h3>
<em>Exponential</em> functions of the form f(x) =aⁿ have an asymptote, a <em>horizontal</em> one. For 0 < a < 1,The <em>horizontal</em> asymptote exists for n → + ∞ and tends to be 0, and the no asymptote exists for n → - ∞. Now we proceed to present a graph in the figure attached below.
Hence, the <em>horizontal</em> asymptote of the <em>exponential</em> function f(x) = 0.2ⁿ is represented by the <em>horizontal</em> line y = 0 , to which the curve tends for n → + ∞.
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Answer:
-23
Step-by-step explanation:
We note that as x increases by 36 from -72 to 36, y decreases by 24 from 25 to 1.
Increasing x from -36 by another 36 to zero will correspond to a decrease in y by another 24 from 1 to -23.
The y-intercept is -23.
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You could go to the trouble to use these observations to compute the slope as -24/36 = -2/3. Then you could pick one of the points and write the equation in point-slope form as ...
y = -2/3(x +36) +1 . . . point-slope form with m=-2/3, (h, k) = (-36, 1)
y = -2/3x -24 +1 . . . . . eliminate parentheses
y = -2/3x -23 . . . . . . . y-intercept is -23
Answer:

Now we can estimate the population variance with the sample variance given by:

And replacing we got:

And the estimator for the population deviation
is given by :

Step-by-step explanation:
For this case we have the following data given:
1.83,1.85,1.79,1.73,1.69,1.74,1.76,1.70
First we need to calculate the mean with the following formula:

And replacing we got:

Now we can estimate the population variance with the sample variance given by:

And replacing we got:

And the estimator for the population deviation
is given by :

Answer:
The one on the middle left.
Explanation:
x goes infinitely in the positive direction.
y does not show any sign of not going infinitely positively or negatively.