Answer:
A trick I did when it comes to multiplying by 10, is just add a 0 at the end of the number your multiplying. Try for yourself!
Answer:
Number of Possible Sequence = 5 x 4 x 3 x 2 x 1 = 120 ways
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"8 in., 15 in., 17 in." is a right triangle and "<span><span>4 in., 15 in., 17 in.</span>" is not a right triangle. Hope this helps!</span>
Here's our equation.

We want to find out when it returns to ground level (h = 0)
To find this out, we can plug in 0 and solve for t.


So the ball will return to the ground at the positive value of

seconds.
What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

I re-orders as 4,5,5,7,8,8,8,10,10.
Mean 7.2222222222222
Median 8
Mode 8
Range 6
Minimum 4
Maximum 10
Count n 9
Sum 65
Quartiles Quartiles:
Q1 --> 5
Q2 --> 8
Q3 --> 9
Interquartile
Range IQR 4
Outliers none