Solve for m:
1.6 m - 4.8 = -1.6 m
Add 1.6 m to both sides:
1.6 m + 1.6 m - 4.8 = 1.6 m - 1.6 m
1.6 m - 1.6 m = 0:
1.6 m + 1.6 m - 4.8 = 0
1.6 m + 1.6 m = 3.2 m:
3.2 m - 4.8 = 0
Add 4.8 to both sides:
3.2 m + (-4.8 + 4.8) = 4.8
4.8 - 4.8 = 0:
3.2 m = 4.8
Divide both sides of 3.2 m = 4.8 by 3.2:
(3.2 m)/3.2 = 4.8/3.2
3.2/3.2 = 1:
m = 4.8/3.2
4.8/3.2 ≈ 1.5:
Answer: m ≈ 1.5
Given:
The data values are
11, 12, 10, 7, 9, 18
To find:
The median, lowest value, greatest value, lower quartile, upper quartile, interquartile range.
Solution:
We have,
11, 12, 10, 7, 9, 18
Arrange the data values in ascending order.
7, 9, 10, 11, 12, 18
Divide the data in two equal parts.
(7, 9, 10), (11, 12, 18)
Divide each parenthesis in 2 equal parts.
(7), 9, (10), (11), 12, (18)
Now,
Median = 
=
=
Lowest value = 7
Greatest value = 18
Lower quartile = 9
Upper quartile = 12
Interquartile range (IQR) = Upper quartile - Lower quartile
= 12 - 9
= 3
Therefore, median is 10.5, lowest value is 7, greatest value is 18, lower quartile 9, upper quartile 12 and interquartile range is 3.
Tamika ran the fastest time at 21.6 seconds in 120 meters because the faster you run the less time you have on your timer.
1st
0.3/100*15000
=45
2nd
40/200*100
=20
=