Asurvey was taken of students in math classes to find out how many hours per day students spend on social media. the survey results for the first-, second-, and third-period classes are as follows: first period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0 second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2 third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3 which is the best measure of center for first period and why?
mean, because there are no outliers that affect the center
median, because there is 1 outlier that affects the center
interquartile range, because there is 1 outlier that affects the center
standard deviation, because there are no outliers that affect the center

Option B)- Median, because there is 1 outlier that affects the center

Step-by-step explanation:

The Outlier is a point that is different from other points. Since 9 is an outlier( its distance from other points in data is large) and when an outlier is present in the data we use the median for finding the best measure of central instead of mean because outlier can affect our analysis.

Option A is incorrect because mean get affected much by outliers.

Option C is not correct because it is used to measure dispersion.

As in First Period outlier is present, So Option D can't be chosen.

5x+2x=49

7x=49

x=7. 35 apples

the graph in the attached figure

step-by-step explanation:

we have

using a graphing tool

the graph in the attached figure

the roots are ( x-intercepts) x=-0.5,x=1,x=3

the y-intercept is the point (0,3)