It would be considered to be around 0 37
This system can be a bit tricky unless you write it out:
(2n + 2)(2n - 2) ~ First you have to take one variable and multiply it to each of the other two variables, like:
2n * 2n ~ And:
2n * -2 ~ This gives you:
4n^2 - 4n
Now we do this with the other:
2 * 2n
2 * -2
4n - 4
Now we combine them both while adding like terms:
4n^2 - 4
Therefore your answer is: A. 4n^2 - 4
(This is due to how when having the positive 4n subtracted by the negative it cancels it out. Leaving us with the remaining two terms left.)
I hope this helps, have a great rest of your day! ^ ^
~Ghostgate
This is not possible. Why not? Because the smallest the variance can get is 0.
Recall that 's' represents the standard deviation, so s^2 is the variance. It basically measures how spread out the values are. The higher the variance, the more spread out the data. You can think of it as "average distance from the mean". If the variance is 0, then all of the values are at the same point. So you could have a list like {2,2,2,2,2} which has variance 0. We cannot get any smaller variance than that. If your teacher insists all the values in the list are different, then the variance will be greater than 0.
Answer:
Ari's is less than Davids
Ari=18.75
David=24.00
Step-by-step explanation:
Hope this helps! This is how you do it.
