A survey 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 second period and why? (5 points)
1. Mean, because there are no outliers that affect the center
2. Median, because there is 1 outlier that affects the center
3. Interquartile range, because there is 1 outlier that affects the center
4. Standard deviation, because there are no outliers that affect the center

1. Mean, because there are no outliers that affect the center

Step-by-step explanation:

Second period: 3,2,3,1,3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Sorted values : 0, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4

The mean = ΣX / n

n = sample size, n = 15

Mean = 34 / 15 = 2.666

The median = 1/2(n+1)th term.

1/2(16)th term = 8th term.

The 8th term = 2

The best measure of centre is the mean because the values for the second period has no outliers that might have affected the centre of the distribution.

Both interquartile range and standard deviation are measures of spread and not measures of centre.

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50