<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>
We are given the function:
g(x) = 6 (4)^x
Part A.
To get the average rate of change, we use the formula:
average rate of change = [g(x2) – g(x1)] / (x2 – x1)
Section A:
average rate of change = [6 (4)^1 – 6 (4)^0] / (1 – 0) =
18
Section B:
average rate of change = [6 (4)^3 – 6 (4)^2] / (3 – 2) =
288
Part B.
288 / 18 = 16
Therefore the average rate of change of Section B is 16 times
greater than in Section A.
<span>The average rate of change is greater between x = 2 to x = 3 than between
x = 1 and x = 0 because an exponential function's rate of change increases
with increasing x (not constant).</span>
If we plot the data on the graph, we can see that the
data is skewed to the right (positive skew) and there is an outlier. In skewed
data and presence of outlier, the median is most commonly used measure of
central tendency. This is because a positive skew would result in a positive
bias to the mean. Meaning that it would be a lot larger than the median and not
really representing the actual central tendency. The median however is less
affected by the skew and outliers.
Answer: Median, because the data are skewed and there is
an outlier
<span> </span>
How many time that 6688 could into 16 which is 418
The class width for this Frequency Distribution Table is 5.
<h3>What is the class width?</h3>
The class width is the difference between the upper boundary and the lower boundary of the class.
The class width = upper boundary - lower boundary
5 - 0 = 5
To learn more about Frequency tables, please check: brainly.com/question/27344444
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