For an arithmetic sequence, the nth term is the first term plus (n-1)×the differece
the first term is 28, the difference between each adjacent number is 8
so -36=(28)+8(n-1)
8(n-1)=64
n-1=8
n=9
the 9th term is -36
Answer:
D: {(-5, -4, 2, 2, 5)}
R: {(-6, 3, 4, 1, 5)}
The relation is NOT a function.
Step-by-step explanation:
By definition:
A relation is any set of ordered pairs, which can be thought of as (input, output).
A function is a <em><u>relation</u></em> in which no two ordered pairs have the same first component (domain/input/x value) and different second components (range/output/y value).
Looking at the given points in your graph, and in listing down the domain and range, we can infer that the relation is not a function because there is an x-value (2) that has two corresponding y-values: (2, 4) (2, 1).
Another way to tell if a given set of points in a graph represents a function by doing the "Vertical line test." The graph of an equation represents y as a function of x if and only if no vertical line intersects the graph more than once. Looking at the attached image, I drew a vertical line over points (2, 4) (2, 1). The vertical line intersects the two points, which fails the vertical line test. This is an indication that the given relation is not a function.
Your answer is 21 for this one.
Answer:
2 
Step-by-step explanation:
The absolute value of a number is the value of x (the number) without a negative, 2 and a third is already positive, so that is the answer :D
Hope this helps, good luck! :)
We can plot this data on MS Excel and determine the distribution of these data reflected on the graph. Among these numbers, 50 is the outlier since it is very far from the other numbers ranging from 76 to 83. We can perform interquartile range to determine or verify the outliers in the data set. In this respect, we can see that there is not much distribution seen. The average of all data sets is equal to 96.25. When the outlier (50) is removed, we expect the mean to become higher since a low number was ommitted including high numbers only. Outliers are obtained from special causations such as human errors.