Two different age groups were surveyed about the amount of time they spent on social media.
The results are shown in the two frequency
polygons below. The same time intervals were
used for both graphs.
11-year old students
15-year old students
a) How many 15-year old
students took part in
the survey?
801
70
60
801
70
60
50
240
240
b) How many more 15-year
old students than 11-year
old students spent
between 35 and 45 mins
on social media?
L 30
30
20
10
20
10
(1)
Total marks: 2
0
0 10 20 30 40 50 60 70 80
Time (mins)
0
0 10 20 30 40 50 60 70 80
Time (mins)
Next Question

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

answer:

24pi or letter c