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:
-9.5 is larger by 2 units
Step-by-step explanation:
-9.5 - (-11.5)
= -9.5 + 11.5
= 2.
Answer:
(B) Subtract 3x from both sides of the equation, and then divide both sides by 2.
Can't read the second question fully.
(A) 0.53
Step-by-step explanation:
Number 1:
If we have the equation , our first goal is to get rid of the x term on one side.
To do this we can subtract 3x from both sides. This leaves our equation to . To find x, we want to divide both sides by 2 since 2x divided by 2 is just x. Our goal is to isolate x. This leaves .
<em>I couldn't read Number 2 fully - I'm sorry :c</em>
<em></em>
Number 3:
Given the equation , we want to isolate x on one side.
To do this, we first apply the distributive property to the left side.
Now subtract 0.6 from both sides:
And divide both sides by 3.
This rounds to 0.53.
Hope this helped!
Answer: The equation of the circle as specified is (x+5)² + (y -1)² = 25
Step-by-step explanation: Use the Pythagorean Theorem to calculate the length of the radius from the coordinates given for the triangle location: A(-1,4), B(-1,1) and C(-5,1)
subtract y-values of A & B 4-1=3 AB = 3
subtract x-values of B &C -5-(-1) =4 BC=4 3² + 4² = 9 + 16 = 25
√25 = 5 AC = 5 That is the radiuus of the circle.
The sides of the triangle are AB=3, BC=4, AC=5.
Use the equation for a circle:
( x - h )² + ( y - k )² = r², where ( h, k ) is the center and r is the radius.
Use the coordinates of C, (-5, 1) as h and k in the equation
Substituting those values, we have (x -[-5])² + (y -[1])² = 5²
Simplify: (x+5)² + (y -1)² = 25
A graph of the circle is attached. The radius is AC, between the center C (-5,1) and A( -1,4) on the circumference. (Sorry, I still need to learn how to create line segments!)