Answer:
- square: 9 square units
- triangle: 24 square units
Step-by-step explanation:
Using a suitable formula the area of a polygon can be computed from the coordinates of its vertices. You want the areas of the given square and triangle.
<h3>Square</h3>
The spreadsheet in the first attachment uses a formula for the area based on the given vertices. It computes half the absolute value of the sum of products of the x-coordinate and the difference of y-coordinates of the next and previous points going around the figure.
For this figure, going to that trouble isn't needed, as a graph quickly reveals the figure to be a 3×3 square.
The area of the square is 9 square units.
<h3>Triangle</h3>
The same formula can be applied to the coordinates of the vertices of a triangle. The spreadsheet in the second attachment calculates the area of the 8×6 triangle.
The area of the triangle is 24 square units.
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<em>Additional comment</em>
We have called the triangle an "8×6 triangle." The intention here is to note that it has a base of 8 units and a height of 6 units. Its area is half that of a rectangle with the same dimensions. These dimensions are readily observed in the graph of the vertices.
Answer:
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Step-by-step explanation:
Previous concepts
The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.
The standard deviation is a measure of dispersion obatined from the sample variance and is given by:
Solution to the problem
Explain the circumstances for which the interquartile range is the preferred measure of dispersion
Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.
What is an advantage that the standard deviation has over the interquartile range?
The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.
Answer:
2⁵× 2ˣ = 256
x= 3
(1/3)² (1/3)ˣ = 1/729
x= 4
Step-by-step explanation:
2⁵× 2ˣ = 256
x= 3
use Product Rule: xᵃxᵇ=xᵃ⁺ᵇ
convert both sides to the same base.
2⁵⁺ˣ = 2⁸
cancel the base of 2 on both sides.
5+x=8
subtract 5 from both sides.
x=8-5
simplify 8-5 3.
x=3
(1/2)² (1/3)ˣ = 1/729
x = 4
use division distributive property: (x/y)ᵃ=xᵃ/yᵃ
i hope this helped i didn't feel like typing the rest out though lol :(
oops i just did :)
Point-slope form is y - y_1 = m (x - x_1) where x_1 and y_1 are the given coordinates and m is the slope. When you plug the given values into the equation you get y - 3 = 6 (x - 8) .
