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 = 10 , -12
Step-by-step explanation:
Solution:-
- The given quadratic equation is to be solved using the quadratic formula. The general form of a quadratic equation is:
Where, [ a , b and c are constants ]
- The quadratic formula is given as:
- The given equation is:
Where, a = 1 , b = 2 , c = -120
- Solve using quadratic formula:
we have
Solve for c--------> that means that clear variable c
so
Divide by 
both sides
Adds 
both sides
Multiply by 
both sides
![a[(R/5)+0.3]=c](https://tex.z-dn.net/?f=a%5B%28R%2F5%29%2B0.3%5D%3Dc)
so
![c=a[(R/5)+0.3]](https://tex.z-dn.net/?f=c%3Da%5B%28R%2F5%29%2B0.3%5D)
therefore
<u>the answer is</u>
Answer:
D is correct
Step-by-step explanation:
The opposite of a negative number is the positive version of it which is 3.5 and the absolute value of a number is always the positive version of a number which would make it also 3.5. For example absolute value of -4 is 4 and absolute value of 5 is 5.
