Colin and jezebel are employees at game zone. they recorded the number of computer games they sold each week for the past 9 weeks. colin 15 20 21 9 3 16 9 14 17 jezebel 10 14 20 11 4 26 5 8 20 (a) all of the games sold of which person had the greatest spread? explain how you know. (b) the middle 50% of the games sold of which person had the least spread? explain how you know. (c) what do the answers to parts 2(a) and 2(b) tell you about colin’s and jezebel’s sold games?

a) The sold games of Jezebel had the greatest spread.

b) The middle 50% of the games sold by Colin has the least spread.

c) Jezebel sold more games than colin.

Step-by-step explanation:

Range is sued to calculate the spread of the given data.

Range is given by:

So,

Part a)

For Collin:

Range=21-3=18

For Jezebel:

Range=26-5=21

Part b)

Middle 50% of the values will be the values between 1st quartile and 3rd quartile

So, to find quartiles:

For Colin:

=3,9,9,14,15,16,17,20,21

The median is 15.

The lower half is 3,9,9,14

Q1 = (9+9)/2= 18/2 = 9

The upper half is 16,17,20,21

Q3 = (17+20)/2= 37/2= 18.5

The middle 50% is first quartile subtracted from third quartile

So, the spread is:

18.5-9=9.5

For Jezebel:

4,5,8,10,11,14,20,20,26

The median is 11.

The lower half is 4,5,8,10

Q1 = (5+8)/2=13/2=6.5

The upper half is 14,20,20,26

Q3 = (20+20)/2=40/2=20

The middle 50% values' spread is:

20-6.5=13.5

Part c) The answer to part a tells us that Jezebels sold games have more spread than Colin's sold games. Similarly the answer of part b tells us that the spread of middle 50% values of Jezebel's sold games was more than the spread of middle 50% values of Colin's sold games ..

answer: a clock

step-by-step explanation:

answer: yes, since the line has a steeper rate of change than the exponential the y-values will be greater and they will only intersect one time.