Answer:
For this case we want to check if the true mean for the depth of groves cut into aluminium by a machine is equal to 1.7 (null hypothesis) and the alternative hypothesis would be the complement different from 1.7. And the best system of hypothesis are:
Null hypothesis: 
Alternative hypothesis ![\mu \neq 1.7[/tx]And the best system of hypothesis are:3. This two-sided test: H0: μ = 1.7 mm H1: μ ≠ 1.7 mmStep-by-step explanation:For this case we want to check if the true mean for the depth of groves cut into aluminium by a machine is equal to 1.7 (null hypothesis) and the alternative hypothesis would be the complement different from 1.7. And the best system of hypothesis are:Null hypothesis: [tex]\mu =1.7](https://tex.z-dn.net/?f=%5Cmu%20%5Cneq%201.7%5B%2Ftx%5D%3C%2Fp%3E%3Cp%3EAnd%20the%20best%20system%20of%20hypothesis%20are%3A%3C%2Fp%3E%3Cp%3E3.%20This%20two-sided%20test%3A%0A%3C%2Fp%3E%3Cp%3EH0%3A%20%CE%BC%20%3D%201.7%20mm%0A%3C%2Fp%3E%3Cp%3EH1%3A%20%CE%BC%20%E2%89%A0%201.7%20mm%3C%2Fp%3E%3Cp%3E%3Cstrong%3EStep-by-step%20explanation%3A%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EFor%20this%20case%20we%20want%20to%20check%20if%20the%20true%20mean%20for%20the%20depth%20of%20groves%20cut%20into%20aluminium%20by%20a%20machine%20is%20equal%20to%201.7%20%28null%20hypothesis%29%20and%20the%20alternative%20hypothesis%20would%20be%20the%20complement%20different%20from%201.7.%20And%20the%20best%20system%20of%20hypothesis%20are%3A%3C%2Fp%3E%3Cp%3ENull%20hypothesis%3A%20%5Btex%5D%5Cmu%20%3D1.7)
Alternative hypothesis [tex]\mu \neq 1.7[/tx]
And the best system of hypothesis are:
3. This two-sided test:
H0: μ = 1.7 mm
H1: μ ≠ 1.7 mm
A. is the answer.
the 2 triangls have AD = AC
and AB as a common side. therefore the triangls are congruent.
Given:
The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).
To find:
The equation of the quadratic function.
Solution:
A quadratic function is defined as:
...(i)
It is passes through the point (0,11). So, substitute
in (i).


Putting
in (i), we get
...(ii)
The quadratic function passes through the point (5,31). So, substitute
in (ii).

Divide both sides by 5.
...(iii)
The quadratic function passes through the point (3,11). So, substitute
in (ii).
Divide both sides by 3.
...(iv)
Subtracting (iv) from (iii), we get




Putting
in (iv), we get



Putting
in (ii), we get
Therefore, the required quadratic equation is
.
Answer: it might be 8
Step-by-step explanation: Write a distance formula: x^2 - x^1
substituted -3.8, 4.2 into x^2 - x^1 and u will get 4.2 + 3.8 and u will calculate and it gets 8
I hope its right