Which data set has an outlier? 25, 36, 44, 51, 62, 77 3, 3, 3, 7, 9, 9, 10, 14 8, 17, 18, 20, 20, 21, 23, 26, 31, 39 63, 65, 66,
umka21 [38]
It's hard to tell where one set ends and the next starts. I think it's
A. 25, 36, 44, 51, 62, 77
B. 3, 3, 3, 7, 9, 9, 10, 14
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Let's go through them.
A. 25, 36, 44, 51, 62, 77
That looks OK, standard deviation around 20, mean around 50, points with 2 standard deviations of the mean.
B. 3, 3, 3, 7, 9, 9, 10, 14
Average around 7, sigma around 4, within 2 sigma, seems ok.
C. 8, 17, 18, 20, 20, 21, 23, 26, 31, 39
Average around 20, sigma around 8, that 39 is hanging out there past two sigma. Let's reserve judgement and compare to the next one.
D. 63, 65, 66, 69, 71, 78, 80, 81, 82, 82
Average around 74, sigma 8, seems very tight.
I guess we conclude C has the outlier 39. That one doesn't seem like much of an outlier to me; I was looking for a lone point hanging out at five or six sigma.
Answer:
x=5/13
Step-by-step explanation:
2/3+7/2=9x+5/6+4x
2/3+7/2-5/6=9x+4x
lcm of 2,3,6=6
2/3=4/6 7/2=21/6 5/6=5/6
4/6+21/6-5/6=9x+4x
30/6=13x
5=13x
x=5/13
Start by writing the system down, I will use 
to represent 
Substitute the fact that 
into the first equation to get,
Simplify into a quadratic form (
),
Now you can use Vieta's rule which states that any quadratic equation can be written in the following form,
which then must factor into
And the solutions will be 
.
Clearly for small coefficients like ours 
, this is very easy to figure out. To get 5 and 6 we simply say that 
.
This fits the definition as 
and 
.
So as mentioned, solutions will equal to but these are just x-values in the solution pairs of a form 
.
To get y-values we must substitute 3 for x in the original equation and then also 2 for x in the original equation. Luckily we already know that substituting either of the two numbers yields a zero.
So the solution pairs are 
and 
.
Hope this helps :)
