If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
The answer is 2/52 or if you want a rounded up answer it is 1/26
Step-by-step explanation:
these are the functions
and the others are not a functions because the domain can't match to more than one
Answer: Choice E) bimodal skewed
The majority of the data is towards the left side. The right outliers are skewing the mean to be pulled larger than it should be. This is why median < mean, so it's better to use the median as the measure of center. There are 2 modes here. They are the values 4 and 8 since we have these values occur the most (4 times each). That's where the term "bimodal" comes in.
