Answer: The range is 13. The mean is 10. The standard deviation is 4.62 .
Step-by-step explanation:
The given data : 3, 6, 9, 11, 15, 16
Total number of data values : n = 6
The mean of data is given by :-

The standard deviation is given by :-

The range of the data : Maximum value -Minimum value

Answer: 0.796 meters
Step-by-step explanation:
Circumference of circle =
, where r=radius
Here, Circumference of loop = 2.5 meters
i.e. 


Diameter = 2r = 2(0.398) = 0.796 meters
Hence, The diameter of Taiga's exercise hoop = 0.796 meters
Answer:
A
Step-by-step explanation:
So we have the two functions:

And we want to find:

This is the same as:

So, to find the answer, find g(1) first:

Now, substitute this value for g(1):

And plug this into f(x):

Therefore:

Our answer is A
The answer is independent