Higher standard deviation shows a greater spread of the data.

2005 - sd=3.5
2006 - sd=2.8
2007 - sd=1.0
2008 - sd=4.0

The correct statement is the last statement which says the game scores of season 2005 have a greater spread than the game scores of season 2006

6 0

3 years ago

Read 2 more answers

Hi can anyone help me with my math? thanks x

Katen [24]

It's multily
f(x) time g(x) is basically what they wnat

so

(x^2+8x+15) times (5/(x^2-9))
hold a sec, first factor each
(x+3)(x+5) times 5/((x-3)(x+3))
we get
((x+3)(x+5)(5))/((x-3)(x+3))
(x+30 cancels out
((x+5)(5))/(x-3)
(5x+25)/(x-3)

3rd option

7 0

4 years ago

Simplify the following <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%20%5Cfrac%7B1%7D%7Bx%7D%20%20%2B%201%20%7D%20%7B%20%5Cfrac%2

Anon25 [30]

Good evening, Riley!

Used properties:

$\star~\boxed{\mathsf{a = \dfrac{ax}{x}, \ \ x\neq 0}}\\ \\ \\ \star~\boxed{\mathsf{\dfrac{\frac{a}{b}}{\frac{c}{d}} = \dfrac{ad}{bc}}}$

Now, note that:

$\mathsf{1 = \dfrac{x}{x},\ x\neq 0}$

So:

$\mathsf{\dfrac{\frac 1 x + \frac x x }{\frac 1 x} = \dfrac{\frac{x+1}{x}}{\frac 1 x}}=} \mathsf{\dfrac{x}{1}\dfrac{x+1}{x} = \boxed{x + 1}}$

8 0

3 years ago