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.
C = 25h + 100 ......renting for 2 hrs...sub in 2 for h
c = 25(2) + 100
c = 50 + 100
c = 150 <===
if u spend 325...so sub in 325 for c
325 = 25h + 100
325 - 100 = 25h
225 = 25h
225/25 = h
9 = h <=== it was rented for 9 hrs
Step-by-step explanation:

Solving this equation using the quadratic formula, we get two real solutions :
1.1926 or -4.1926
Now we know the values of v , we can calculate x since x is ∛ v

![x = \sqrt[3]{1.1926} = 1.0605 \\ x = \sqrt[3]{ - 4.1926} = - 1.6125](https://tex.z-dn.net/?f=x%20%3D%20%20%5Csqrt%5B3%5D%7B1.1926%7D%20%20%3D%201.0605%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B%20-%204.1926%7D%20%20%3D%20%20-%201.6125)
Answer: D
Step-by-step explanation: A is not possible because the shape and the way it is drawn is nearly the same as the figure given. The shape of B is different from the given shape. C is a little too flat to be considered symmetrical to the figure. Which leaves only option D.
Hope this helped :D