Answer:
m a t h
Step-by-step explanation:
i am in middle school so i dont know
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:
d = -1/3, 0
Step-by-step explanation:
Subtract the constant on the left, take the square root, and solve from there.
(6d +1)^2 + 12 = 13 . . . . given
(6d +1)^2 = 1 . . . . . . . . . .subtract 12
6d +1 = ±√1 . . . . . . . . . . take the square root
6d = -1 ±1 . . . . . . . . . . . .subtract 1
d = (-1 ±1)/6 . . . . . . . . . . divide by 6
d = -1/3, 0
_____
Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:
f(x) = (6x+1)^ +12 -13
We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.
3/4 + 2/3 = 17/12
7/8 - 2/3 = 5/24
3/5 x 1/3 = 1/5
4/7 x 2/3 = 8/21
