Wouldn’t it just be 56.25 bc it holds that much bc v=l x w x h and there is no height
Is there any sums that you needed to solve this with?
Answer:
0.8 Not Outlier
1.1 Not Outlier
10.2 Not Outlier
10.9 Not Outlier
Solution:
Arranging the numbers in ascending order:
0.8 1.1 4.9 5.2 5.8 5.9 6.1 6.1 7.4 10.2 10.9
we can see that the median is 5.9.
We can find the first quartile Q1 by getting the median in the lower half of the data
0.8 1.1 4.9 5.2 5.8
that is, Q1 = 4.9
We can find the third quartile Q3 by getting the median for the upper half of the data
6.1 6.1 7.4 10.2 10.9
that is, Q3 = 7.4
We subtract Q1 from Q3 to find the interquartile range IQR.
IQR = Q3 - Q1 = 7.4 - 4.9 = 2.5
We can now calculate for the upper and lower limits:
upper limit = Q3 + 1.5*IQR = 7.4 + (1.5*2.5) = 11.15
lower limit = Q1 – 1.5*IQR = 0.8 - (1.5*2.5) = -2.95
There is no data point that lies above the upper limit and below the lower limit, therefore, there are no outliers in the data set.
Answer: Option D. 13
Solution:
The data is:
16, 6, 8, 12, 14, 14
1) Sort the data from the least to the greatest:
6, 8, 12, 14, 14, 16
2) Find the median (middle value)
6, 8, 12, 14, 14, 16
1 2 3 4 5 6
We have to middle values, the third and fourth values (12 and 14), then the median is the average of these two values:
Median=(12+14)/2
Median=(26)/2
Median=13
Answer: No
Step-by-step explanation:
The answer to this question is quite simple in the sense that all you need to do is graph the points on a graph and you can easily see that it is not a right triangle.
