Please help asap 25 pts
1 answer:
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Given:
![p= \frac{RT}{vm} + \frac{a+bT}{vm^{2}} \\or\\v= \frac{1}{p}[ \frac{R}{m} + \frac{a+bT}{m^{2}}]](https://tex.z-dn.net/?f=p%3D%20%5Cfrac%7BRT%7D%7Bvm%7D%20%2B%20%5Cfrac%7Ba%2BbT%7D%7Bvm%5E%7B2%7D%7D%20%5C%5Cor%5C%5Cv%3D%20%5Cfrac%7B1%7D%7Bp%7D%5B%20%5Cfrac%7BR%7D%7Bm%7D%20%20%2B%20%5Cfrac%7Ba%2BbT%7D%7Bm%5E%7B2%7D%7D%5D%20)
Answer:
When p is held constant,
![\frac{\partial v}{\partial T} |_{p}= \frac{1}{p}[ \frac{R}{m}+ \frac{b}{m^{2}}]](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5Cpartial%20v%7D%7B%5Cpartial%20T%7D%20%7C_%7Bp%7D%3D%20%5Cfrac%7B1%7D%7Bp%7D%5B%20%5Cfrac%7BR%7D%7Bm%7D%2B%20%5Cfrac%7Bb%7D%7Bm%5E%7B2%7D%7D%5D%20%20%20)
Answer:
When p is held constant,
Answer:
A sample size of at least 1,353,733 is required.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
The margin of error is:
98% confidence level
So , z is the value of Z that has a pvalue of
, so
.
You would like to be 98% confident that you esimate is within 0.1% of the true population proportion. How large of a sample size is required?
We need a sample size of at least n.
n is found when M = 0.001.
Since we don't have an estimate for the proportion, we use the worst case scenario, that is
So
Rounding up
A sample size of at least 1,353,733 is required.