Answer: look below :)
Step-by-step explanation:
5 < 11 < 42 < 2015
5 is less than 11 is less than 42 is less than 2015
5/
1
< 11/
1
< 42/
1
< 2015/
1
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.
-12=4(x-7)-8x
One solution was found :
x = -4
Rearrange:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
-12-(4*(x-7)-8*x)=0
Step by step solution :
Step 1 :
Equation at the end of step 1 :
-12 - (4 • (x - 7) - 8x) = 0
Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
4x + 16 = 4 • (x + 4)
Equation at the end of step 3 :
4 • (x + 4) = 0
Step 4 :
Equations which are never true :
4.1 Solve : 4 = 0
Hope this helps you, Have a nice day.
On the set of axes below, graph the line whose equation is To graph your line, click to add your first point and then click again to add a second point. You can either undo or reset to redraw your line. LNE This linear equation contains the point State the value of .
