Answer:
2/3 1/2 3/4 5/6
Step-by-step explanation:
anything that isnt 1/3 or equal to it is an answer
The average speed of Joshua during that time is 2500 m/h.
Explanation:
It is given that Joshua started cycling at 5:15 pm. By 8:09 pm he has covered a distance of 7250 m.
The total time taken by Joshua from 5:15 pm to 8:09 pm is
Dividing we get,
Adding, we have,
Thus, the total time taken by Joshua is 
To determine the average speed we use the formula,
where 
and 
Hence, substituting the values we have,
Dividing, we get,
Thus, the average speed of Joshua during that time is 2500 m/h.
F(x) is most likely the f(f(x)x) where f(x) is g(x))f)) and composition can be 69 + 46 which is the total of 137
<u>Given</u>:
Given that the data are represented by the box plot.
We need to determine the range and interquartile range.
<u>Range:</u>
The range of the data is the difference between the highest and the lowest value in the given set of data.
From the box plot, the highest value is 30 and the lowest value is 15.
Thus, the range of the data is given by
Range = Highest value - Lowest value
Range = 30 - 15 = 15
Thus, the range of the data is 15.
<u>Interquartile range:</u>
The interquartile range is the difference between the ends of the box in the box plot.
Thus, the interquartile range is given by
Interquartile range = 27 - 18 = 9
Thus, the interquartile range is 9.
