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Virty [35]
3 years ago
15

Find the probability a molecue will travel a distance at least equal to the mean free path

Mathematics
1 answer:
AnnZ [28]3 years ago
8 0
Exponential probability distribution f(r) = Ae-r/ λ λ where A = a constant, λ λ = mean free path 3. The attempt at a solution P = Integral (limits λ λ to ∞ ∞ )f(r) dr / Integral (limits 0 to ∞ ∞ ) f(r) dr

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Several years​ ago, 50​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students r
galina1969 [7]

Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that​ parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1095, \pi = \frac{478}{1095} = 0.4365

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that​ parents' attitudes toward the quality of education have changed.

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3 years ago
PLEASE HELPPPPPP ME ASAP
dalvyx [7]

Answer: 78

Step-by-step explanation:

8 0
3 years ago
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