The number is 2 and if u are not sure how to do this go to this website mathisfun.com
we want correct answer because it will lead or make your standard deviation goes directly
Continuing from the setup in the question linked above (and using the same symbols/variables), we have




The next part of the question asks to maximize this result - our target function which we'll call

- subject to

.
We can see that

is quadratic in

, so let's complete the square.

Since

are non-negative, it stands to reason that the total product will be maximized if

vanishes because

is a parabola with its vertex (a maximum) at (5, 25). Setting

, it's clear that the maximum of

will then be attained when

are largest, so the largest flux will be attained at

, which gives a flux of 10,800.
The answer is 1/6
hope this helped :)
Answer:
2
Step-by-step explanation:
-3 log₂ (-n + 10) = -16+7
-3 log₂ (-n + 10) = -9
log₂ (-n + 10) = -9/-3
log₂ (-n + 10) = 3
-n + 10 = 2^3
-n + 10 = 8
n=2