Please help me with this and dont play around pls
2 answers:
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Given:
The data values are:
7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19
To find:
The outliers of the given data set.
Solution:
We have,
7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19
Divide the data set in two equal parts.
(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)
Divide each parenthesis in two equal parts.
(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)
Now,
And
The interquartile range is:
The data values lies outside the interval are known as outliers.
All the data values lie in the interval [6,22]. So, there are no outliers.
Hence, the correct option is 4.
<h3>Answer: x = 7 and y = 3</h3>
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Explanation:
Apply the difference of squares rule
x² - 4y² = 13
x² - (2y)² = 13
(x - 2y)(x + 2y) = 13
Since x and y are positive integers, this means x-2y and x+2y are both integers as well.
The value 13 is prime. Its only factors are 1 and 13
Since the above equation shows 13 factoring into x-2y and x+2y, then we have two cases:
- A) x-2y = 1 and x+2y = 13
- B) x-2y = 13 and x+2y = 1
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Let's consider case A
We have this system of equations
Add the equations straight down
- x+x becomes 2x
- -2y+2y becomes 0y = 0 which goes away
- 1+13 becomes 14
Therefore we have 2x = 14 solve to x = 7
From here, plug this into either equation to solve for y
x-2y = 1
7 - 2y = 1
-2y = 1-7
-2y = -6
y = -6/(-2)
y = 3
You should get the same result if you used x+2y = 13
----------------
Since we've found that x = 7 and y = 3, notice how case B is not possible
Example: x-2y = 13 becomes 7-2(3) = 13 which is false.
Also, x+2y = 1 would turn into 7+2(3) = 1 which is also false.
-----------------
Let's check those x and y values in the original equation
x² - 4y² = 13
7² - 4*(3)² = 13
49 - 4(9) = 13
49 - 36 = 13
13 = 13
The answer is confirmed.