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
<u>ANSWER</u>
Only the equation is a function
<u>EXPLANATION</u>
is graphed above with its inverse
.
We perform the vertical line test for both the equation and its inverse.
The equation passed the vertical line hence it is a function.
However, failed the vertical line test, hence it is not a function.
Answer:
f(r) = (x -1)(x -4+√7)(x -4-√7)
Step-by-step explanation:
The signs of the terms are + - + -. There are 3 changes in sign, so Descartes' rule of signs tells you there are 3 or 1 positive real roots.
The rational roots, if any, will be factors of 9, the constant term. The sum of coefficients is 1 -9 +17 -9 = 0, so you know that r=1 is one solution to f(r) = 0. That means (r -1) is a factor of the function.
Using polynomial long division, synthetic division (2nd attachment), or other means, you can find the remaining quadratic factor to be r^2 -8r +9. The roots of this can be found by various means, including completing the square:
r^2 -8r +9 = (r^2 -8r +16) +9 -16 = (r -4)^2 -7
This is zero when ...
(r -4)^2 = 7
r -4 = ±√7
r = 4±√7
Now, we know the zeros are {1, 4+√7, 4-√7), so we can write the linear factorization as ...
f(r) = (r -1)(r -4 -√7)(r -4 +√7)
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<em>Comment on the graph</em>
I like to find the roots of higher-degree polynomials using a graphing calculator. The red curve is the cubic. Its only rational root is r=1. By dividing the function by the known factor, we have a quadratic. The graphing calculator shows its vertex, so we know immediately what the vertex form of the quadratic factor is. The linear factors are easily found from that, as we show above. (This is the "other means" we used to find the quadratic roots.)
