I'll assume that's an exponent at the end there:
The first derivative gives the velocity. The second derivative gives the acceleration; increasing velocity is the same as positive acceleration. So we want to find when the second derivative is positive.
Let's see if we can use to avoid multiplying this out.
That worked; let's do it again:
That's a nice parabola. It's zero or negative for so positive everywhere else:
Answer: Increasing velocity when t < 1 or t > 3
Answer:
we reject H₀
Step-by-step explanation: Se annex
The test is one tail-test (greater than)
Using α = 0,05 (critical value ) from z- table we get
z(c) = 1,64
And Test hypothesis is:
H₀ Null hypothesis μ = μ₀
Hₐ Alternate hypothesis μ > μ₀
Which we need to compare with z(s) = 2,19 (from problem statement)
The annex shows z(c), z(s), rejection and acceptance regions, and as we can see z(s) > z(c) and it is in the rejection region
So base on our drawing we will reject H₀
