A table represents the number of students who passed or failed an aptitude test at two different campuses. South Campus North Campus
Passed 46 38
Failed 54 62

In order to determine if there is a significant difference between campuses and pass rate, the chi square test for association and independence should be performed.

What is the expected frequency of North Campus and failed?

44

step-by-step explanation:

the

it's 100

step-by-step explanation:

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

  n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the  14th  row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100

read more on -

school a:

minimum value is 2

maximum value is 22

the lower quartile is 2.5

the median is 10

the upper quartile is 15.5

school b:

minimum value is 9

maximum value is 20

the lower quartile is 12

the median is 16

the upper quartile is 18

box plot for school a isn't symmetrical. the data tails on the right

box plot for school b isn't symmetrical. the data tails on the left

hope i

-lavira: )

the distance between the school and her home is 3.6km.

12 mins = 0.2 hours = 1/5 hour

54 ÷ 3 = 18km per hour

1/5 of 18 = 3.6km per 12 mins