The following table shows the number of hours some high school students in two towns spend riding the bus each week:
1 answer:
Answer: Part A. The IQR for town A is 4.5 and the IQR for town B is 4.5
Part B. No, they are assymetric.
Explanation: To find the interquatile range (IQR) of each set of data, you can do as follows:
1) Put the number in each set of data in order:
A: 3 4 5 5 6 7 8 9 9 20
B: 3 3 5 5 6 8 8 9 9
2) Find the median of each town data:
For A, the median is the number 6. Coincidently, the same can be said for town B: its median is 6.
3) Find Q1, which is the median (or the middle term) of the left side of the median:
A: 3 4 5 5 (set of numbers to calculate)
B: 3 3 5 5
Now, as both set of number are even, to find the median, we have to calculate the mean of the two central numbers:
For A: Q1 = = 4.5
For B: Q1 = = 4
4) Find Q3, which is the median of the right side of the first median:
A: 7 9 9 20 (set of numbers to calculate)
B: 8 8 9 9
As before, the median will be the mean:
For A: Q3 = = 9
For B: Q3 = = 8.5
5) The IQR is the result of a subtraction:
IQR = Q3 - Q1
For A: IQR = 4.5
For B: IQR = 4.5
For the two sets of data the IQR is the same and equals 4.5.
Part B: Box plot is a type of graphic, similar to histogram. To plot it, you need 5 points but for this question, it will be analised just the symmetric aspect. As the graphic is a retangule, to be symmetric, the median or mean has to be in the center of the retangule. In the case of the towns, the box plots are assymetric because the median of each town is closer to the Q1 of each town. In other words, the retangule is not "divided" exactly in half.
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Answer:
Yes, at a time t such that (√2)/2 ≤ t ≤ 2.
Explanation:
To answer the question
Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have
For Chloe's velocity
Finding the boundaries of the function gives;
and
At t = 1, we have
We find the maximum point as follows;
From which we have;
or
∴ 1 = 2·t² and from which t = (√2)/2
Hence the function C(x) is decreasing from t = (√2)/2 to t = 2
For Brandon
For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5
1 ≤ f(x) ≤ 1.5
Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;
For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2
and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5
Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.