The best way I can explain the elimination method is, it's a method where you either add or subtract equations in order to get rid of a variable.
Answer:
![z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.872-0.917%7D%7B%5Cfrac%7B0.303%7D%7B%5Csqrt%7B37%7D%7D%7D%3D%20-0.903)
And we can find this probability to find the answer:
![P(z](https://tex.z-dn.net/?f=%20P%28z%3C-0.903%29)
And using the normal standar table or excel we got:
![P(z](https://tex.z-dn.net/?f=%20%20P%28z%3C-0.903%29%3D0.1833%20)
Step-by-step explanation:
Let X the random variable that represent the amounts of nocatine of a population, and for this case we know the distribution for X is given by:
Where
and
We have the following info from a sample of n =37:
the sample mean
And we want to find the following probability:
![P(\bar X \leq 0.872)](https://tex.z-dn.net/?f=%20P%28%5Cbar%20X%20%5Cleq%200.872%29)
And we can use the z score formula given by;
![z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
And if we find the z score for the value of 0.872 we got:
![z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.872-0.917%7D%7B%5Cfrac%7B0.303%7D%7B%5Csqrt%7B37%7D%7D%7D%3D%20-0.903)
And we can find this probability to find the answer:
![P(z](https://tex.z-dn.net/?f=%20P%28z%3C-0.903%29)
And using the normal standar table or excel we got:
![P(z](https://tex.z-dn.net/?f=%20%20P%28z%3C-0.903%29%3D0.1833%20)
I got -28.08695652. It doesn't repeat nor does it terminate
Answer:
8 months
Step-by-step explanation:
194-3m = 217-6m
3m = 23
m=7.666666..... (8 months