Answer:
12.)-5
13.)-16
14)-20
15)6
16)-13
17)1
18)-13
19)15
20)-10
Step-by-step explanation:
12.)-x=2+3
x=5/-1
so,x=-5
13.)-4=(-2×8+x)/8
-4×8=-16+x
-32+16=x
so,x=-16
14)(7×4+m)/4=2
28+m=8
so,m=-20
15)4n=34-10
n=24/4
so,n=6
16)-2a=22+4
a=26/-2
so,a=-13
17)a+3=2×2
a=4-3
so,a=1
18)-133+3=10x
-130=10x
so,x=-13
19)4+1=v/3
5*3=v
so,v=15
20)1=3+r/5
now,
Take LCM,
We get,
1=(3*5+r*1)/5
now,
by moving 5 into LHS
We get,
1*5=15+r
5=15+r
by moving 15 into left hand side,
We get ,
5-15=r
so,r=-10 because when we subtract smaller from bigger we have to take the sign of bigger.
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.
The distance formula is used to find the distance between two points in the coordinate plane. The point that is at the same distance from two points A and B on a line is called the midpoint. You calculate the midpoint using the midpoint formula.
Answer:
65, 123.79
Step-by-step explanation:
Since point Z is where the perpendicular bisector of the three sides of triangle GHJ intersect, point Z is the circumcenter. Circumcenters are equidistant from all three vertices, so GZ, JZ, and HZ all have equal length. Therefore ZJ is 65.
Next, look at triangle ZLJ and notice that it is a right triangle with legs 16 and x long and the hypotenuse 65 long. Using the Pythagorean Theorem, 16^2 + x^2 = 65^2, we learn that x, the length of LJ, is approximately 60.79. Add this with 63 to get 123.79.