Please solve the whole page please please
1 answer:
1d) ↑ (Do the same but turned if that's easier)
1a) 199
1b) 193
1c) 217
1e) Based on the box plot, we can for example, find the lowest and greatest values, the range (largest - smallest), the mean (all values / # values), and the median middle value or q2 given by the second line in the box.
2b) 7
2c) 9.5
2d) ↑
2e) because the data is not ordered, it is difficult to determine the quartile data. Once it is ordered like 6, 7, 7, 7, 8, 8, 9, 10, 10. The data can be visualized alot easier. With the box plot we can determine q3 of the data by looking at the 3rd line, using the whiskers of the box to find the minimum and maximum values.
3)
You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.
In this case the ordered data would be: 4,6,8,8,9,11,12,14,14,16.
[4, 6, 8, 8, 9] [] [11, 12, 14, 14, 16]
4 is the minimum, 16 is the maximum, q1 is 8, q3 is 14, and median is 10.
4) ↑
5)
You need to find the minimum, maximum, the quartile data obtained by having an even amount of data on both sides, order it so that values can be grouped. this includes q1, q2 (or median), and q3.
In this case the ordered data would be:
45, 45, 50, 55, 60, 70, 70, 75, 80, 85
[45, 45, 50, 55, 60] [] [70, 70, 75, 80, 85]
45 is the minimum, 85 is the maximum, q1 is 50, q3 is 75, and median is 65.
6) ↑
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9514 1404 393
Answer:
1.3363
Step-by-step explanation:
The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.
__
Let P be a point on L1. Then the parametric equation for P is ...
P = (6t, 0, -t) . . . . . . origin + t × direction vector
Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...
Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector
The direction vector for PQ is ...
Q -P = (3s+1-6t, 2s-1, 5s+1+t)
The dot product of this and the two lines' direction vectors will be zero:
(3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1
(3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2
The solution to these equations is ...
s = -157/1237
t = 112/1237
Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...
|PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363
The distance between the two lines is about 1.3363 units.
