I'm really not sure.
Sorry.
If I understand what you are asking, you can use the median as an average since it is not affected by the mean ( median is a measure of center). Since a mean is affected by outliers, it might lower the score if there are very low outliers. But note that if there are outliers way <em>greater </em>it can increase the average. You can use the Inter Quartile Range if shown on a box plot ( measure of variability).
A. The
y-intercept (b) of a linear equation is obtained when x = 0. Therefore from the
given table,
y - intercept
= 8
Since at
time zero the displacement is 8 ft, this means that the horse was already
outside the barn initially.
B. The
average rate of change of the function represents the slope of the linear
equation (m). This can be calculated using the formula:
average rate
of change = m = (y2 – y1) / (x2 – x1)
m = (158 –
58) / (3 – 1)
m = 50
<span>C. Since we have determine the y-intercept and
the slope, we can formulate the linear equation:</span>
y = m x + b
y = 50 x + 8
The domain
is the value of x. When y = 508, x is equivalent to
508 = 50 x +
8
<span>x = 10 hrs</span>
Answer:
1.38
Step-by-step explanation:
The point estimate for the population standard deviation of the length of the walking canes is given as the squareroot of variance. Since variance is provided as 1.9 then the standard deviation will be
Therefore, rounded off to two decimal places, the standard deviation is approximately 1.38
