Find the x- and y-intercepts of 6x+3y=30
1 answer:
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The question is incomplete. The compete question is :
Saul decides to use the IQR to measure the spread of data. Saul calculates the IQR of the data set to be 27. Saul asks Jasmine to check his work. Saul copies the data in numerical order from least to greatest for Jasmine. 25, 30, 50, 50, 50, 50, 56, n, 250. What is the value of n that will make the IQR of the data set equal to 27?
Solution :
The inter quartile range is being measured as : 3rd quartile - 1st quartile
The 1st quartile = 25% mark
The 3rd quartile = 75% mark
The data set as given in the question is :
25, 30, 50, 50, 50, 50, 56, n, 250
Therefore, the total number of the sample = 9
And the median is = 50
The median separates the given data sets into two equal halves-- that is the upper half and the lower half.
Let the middle of a lower half is the 1st quartile be
And the middle of a 2nd half is the 3rd quartile be
∴
Now for the IQR of the data set to be equal to ,
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.
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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.
