Answer:
153m
Step-by-step explanation:
perimeter of an arc = theta/360 × 2πr
P= 17πm
theta = 20°
17πm= 20/360 × 2 × π × r
153πm = πr
r = 153m
B^-2 / b^-3 = b^(-2-(-3) = b^1 = b
answer is d
<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.
15.75+15.75=31.50 you add the two 15.75's because there are two adults then you add the amount of the admission for the adults and the kid 31.50+8.25 and you get $39.75
