The answer for this question is 9 I
we have a maximum at t = 0, where the maximum is y = 30.
We have a minimum at t = -1 and t = 1, where the minimum is y = 20.
<h3>
How to find the maximums and minimums?</h3>
These are given by the zeros of the first derivation.
In this case, the function is:
w(t) = 10t^4 - 20t^2 + 30.
The first derivation is:
w'(t) = 4*10t^3 - 2*20t
w'(t) = 40t^3 - 40t
The zeros are:
0 = 40t^3 - 40t
We can rewrite this as:
0 = t*(40t^2 - 40)
So one zero is at t = 0, the other two are given by:
0 = 40t^2 - 40
40/40 = t^2
±√1 = ±1 = t
So we have 3 roots:
t = -1, 0, 1
We can just evaluate the function in these 3 values to see which ones are maximums and minimums.
w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20
w(0) = 10*0^4 - 20*0^2 + 30 = 30
w(1) = 10*(1)^4 - 20*(1)^2 + 30 = 20
So we have a maximum at x = 0, where the maximum is y = 30.
We have a minimum at x = -1 and x = 1, where the minimum is y = 20.
If you want to learn more about maximization, you can read:
brainly.com/question/19819849
The mean is the average you're used to, where you add up all the numbers and then divide by the number of numbers. The median is the middle value in the list of numbers. Mode is the number which appears most often in a set of numbers.
Hope this helps you :)
Answer: Choice B 
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Explanation:
The endpoint is at 4.5 and this endpoint is a filled in circle. So we'll have "or equal to" as part of the inequality sign. This is because we are including the endpoint as part of the shaded solution region.
The other part of the inequality sign is "less than" because the shading is to the left of the endpoint. Any point in the shaded region is less than 4.5, or it could be equal to 4.5
Put another way: x is either 4.5 or smaller
We write that as 
which is read out as "x is less than or equal to 4.5"
Surrounding this in curly braces tells the reader we're dealing with a set of values. The first part "x |" means "x such that"
All together we end up with the answer 
which translates to "x such that x is less than or equal to 4.5"
