D. The males have a suspected significant high outlier.
The outlier will make the box plot be uneven and this will affect the spread and the center.
I know isn't A, because if anything that would make it more balanced.I know it isn't B because this would not affect the spread or the center.I know it isn't C because the question is not asking you to determine the spread and center, just to see if there is anything that would affect them.
X + y + z = 50
y + 10 = x, solving for y, y = x - 10
x = 2z, solving for z, z = x/2
Substituting for y and z,
x + x - 10 + x/2 = 50
(5/2)x - 10 = 50
(5/2)x = 60
x = (2/5) * 60 = 24
y = x - 10 = 24 - 10 = 14
Z = x/2 = 24/2 = 12
Checking:
x + y + z = 50, 24 + 14 + 12 = 50, 50 = 50. Solution meets the initial constraint.
Answer:
5/9
Step-by-step explanation:
The equation for slope is y2 - y1 / x2 - x1. The first step is that you subtract 9 from 14. Which will give you the sum of -5. So -5 is going to be your numerator. Then you subtract 10 from 1, and will give you the sum of -9. If you have two negatives it will make the fraction a positive. And will give you the answer of 5/9.
S= a(1-r ⁿ) /(1-r) & when r is < 1 & n →∞, the sum becomes S= a/(1-r)
Progression P is 2,1, 1/2,1/4,.. ==> r = 1/2
Sum of P when n--> ∞ = 2/(1-1/2) ==> S =4
Progression Q is 3,1,1/3,1/9,....==> r = 1/3
Sum of Q when n--> ∞ = 3/(1-1/3) ==> S =4.5
GIVEN THAT 4, 9/2 & X (to be calculated later is a geometric Progression, hence 9/2 - 4 = 0.5 =d (common difference , so X = 4.5+0.5 & X = 5
Sum of R =5 Then => 5= 4 /(1-r) & r=1/5
Then the sum of R = 4/(1-1/5) ==> S of R =5
2(3/2c) = 21
3c = 42
c = 14