What is the interquartile range of the data? With the numbers 2,3,3,4,5,7,8,8,8,10,10, and 12
harina [27]
Answer:
The interquartile range = 5.5
Step-by-step explanation:
Given the data
2 3 3 4 5 7 8 8 8 10 10 12
As we know that
The interquartile range is the difference between the third and first quartiles.
- The first quartile is the median of the bottom half of the numbers.
So, to find the first quartile, we need to place the numbers in value order and find the bottom half.
2 3 3 4 5 7 8 8 8 10 10 12
So, the bottom half is
2 3 3 4 5 7
The median of these numbers is 3.5.
- The third quartile is the median of the upper half of the numbers.
So, to find the third quartile, we need to place the numbers in value order and find the upper half.
2 3 3 4 5 7 8 8 8 10 10 12
So, the upper half is
8 8 8 10 10 12
The median of these numbers is 9
so
The third quartile is 9.
The first quartile is 3.5.
The interquartile range is the difference between the third and first quartiles.
The interquartile range = 9 - 3.5 = 5.5
Therefore, the interquartile range = 5.5
97.55500 would round to 97.56
Y^3=x and 3y=x^2 is nonlinear
the others are linear
0.10r + 0.20b = 24
3r = b + 20....b = 3r - 20
0.10r + 0.20(3r - 20) = 24
0.10r + 0.60r - 4 = 24
0.70r = 24 + 4
0.70r = 28
r = 28/0.70
r = 40 <=== there are 40 reds
0.10r + 0.20b = 24
0.10(40) + 0.20b = 24
4 + 0.20b = 24
0.20b = 24 - 4
0.20b = 20
b = 20/0.20
b = 100 <=== there are 100 blues
Answer:
A and D
Step-by-step explanation:
To find the x intercepts, substitute 0 into y.
2x = 6
(3,0)
To find the y intercept, substitute 0 into x;
3y = 6
(0, 2)
So a and d
