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:
Qu. 8 = A (50) = 25w as 125w x 125l = Area ; this because perimeter 500ft would always be a 3:5 ratio to the larger area. Therefore perimeter is 125+125 +125 +125 and = 500ft. However we can prove if 125 is any length or width then A(50)= 1/5 = 25 Qu 9)= I think last 2 are a function. Qu 10) Answer B is correct.
Step-by-step explanation:
(x-1)^3 / 2 = 27
For equations like this, we want to isolate x in order to get the value of it.
First, let's multiply both sides by 2 to eliminate the fraction.
(x-1)^3 = 27(2)
Now, let's take the cube root of both sides
(x-1) = ∛(27 * 2)
Now, simplify the root
x - 1 = 3∛2
Add both sides by 1
x = 1 + 3∛2
That's your answer.
Have an awesome day! :)
Answer:
(1,8) and (-3,0)
Step-by-step explanation:
We simply graph the two functions in the same graph. The solution set to the system is given by the points where the two functions intersect. For this case, the two functions intersect at;
(1,8) and (-3,0)
The solution set is thus; (1,8) and (-3,0)
Check the attachment below;
