Answer:
8
Step-by-step explanation:
1 2 3 4 5 6 7 8 quick math
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
From the graph we can see that in the interval [0,1] the value of y is less than 1.
In the interval [1,2] the value of y value is 2 to 4.
In the interval [2,infinity) the graph is going up, and the value of y is greater than or equal to 4.
Therefore, the graph going up after y=4 above the line.
Therefore, the minimum y-value is 4 after which the exponential function will always be greater than the linear function.
Answer:
×
= 
Step-by-step explanation:
× 
To solve the above, we need to follow the steps below;
4k+2 can be factorize, so that;
4k +2 = 2 (2k + 1)
k² - 4 can also be be expanded, so that;
k² - 4 = (k-2)(k+2)
Lets replace 4k +2 by 2 (2k + 1)
and
k² - 4 by (k-2)(k+2) in the expression given
× 
× 
(2k+1) at the numerator will cancel-out (2k+1) at the denominator, also (k-2) at the numerator will cancel-out (k-2) at the denominator,
So our expression becomes;

Therefore,
×
= 
Okay so input would be at 1 and the output at 2.5, making the answer B.(If you're on plato) :D