Simplify: –3(x – 5) – 8(y + 2) + (–3)(6 – 4b) A. –3x –8y – 12b – 15 B. –3x –8y + 12b + 15 C. –3x –8y + 12b + 20 D. –3x –8y + 12b
2 answers:
You might be interested in
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
