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Answer:
Horizontal distance = 0 m and 6 m
Step-by-step explanation:
Height of a rider in a roller coaster has been defined by the equation,
y = 
Here x = rider's horizontal distance from the start of the ride
i). 

![=\frac{1}{3}[x^{2}-2(3x)+9-9+24]](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B3%7D%5Bx%5E%7B2%7D-2%283x%29%2B9-9%2B24%5D)
![=\frac{1}{3}[(x^{2}-2(3x)+9)+15]](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B3%7D%5B%28x%5E%7B2%7D-2%283x%29%2B9%29%2B15%5D)
![=\frac{1}{3}[(x-3)^2+15]](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B3%7D%5B%28x-3%29%5E2%2B15%5D)

ii). Since, the parabolic graph for the given equation opens upwards,
Vertex of the parabola will be the lowest point of the rider on the roller coaster.
From the equation,
Vertex → (3, 5)
Therefore, minimum height of the rider will be the y-coordinate of the vertex.
Minimum height of the rider = 5 m
iii). If h = 8 m,


(x - 3)² = 9
x = 3 ± 3
x = 0, 6 m
Therefore, at 8 m height of the roller coaster, horizontal distance of the rider will be x = 0 and 6 m
Let us calculate the median; the 6th observation is 20, so it is 20. We need the 6th observation so that out of the 11 observations we have 5 above the median and 5 below (or equal). We also have that then Q1 is the median of the lowest 5 observations, hence 19 (14,16,19,19,20, the 3rd observation is 19). Similarly, we get that the median for the upper half of the observations, Q3 namely, is 22 (21,21,22,22,23, the 3rd observation is 22). Thus, the interquartile range is 3=Q3-Q1. According to our calculations, all observations are wrong.
Answer: -38,686
Step-by-step explanation:
3 x 4= 12
-19 x 2037= -38,703
12 + 15= 17
17 -38,703= -38,686