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deff fn [24]
3 years ago
7

Does (-2)^n/( 3^n+1) converge or diverge?

Mathematics
1 answer:
hodyreva [135]3 years ago
3 0
\left|\dfrac{(-2)^n}{3^n+1}\right|=\dfrac{2^n}{3^n+1}

As n\to\infty, the sequence a_n=\left(\dfrac23\right)^n converges to zero.

If you're talking about the infinite series

\displaystyle\sum_{n\ge0}\dfrac{(-2)^n}{3^n+1}

well we've shown by comparison that this series must also converge because we know any geometric series \sum\limits_n r^n will converge as long as |r|.
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An educational psychologist wants to test whether a new teaching method negatively affects reading comprehension scores. She ran
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z=\frac{87.8-90.1}{\frac{17.3}{\sqrt{38}}}=-0.820  

The p value for this case would be given by:

p_v =2*P(z  

Since the p value is very high ant any significance level we will have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the true mean is not significantly different from 90.1 and the method not shows a significantly effect in the scores.

Step-by-step explanation:

Information provided

\bar X=87.8 represent the sample mean for the scores on the standardized test in the general population of 6th graders    

\sigma=17.3 represent the population standard deviation

n=38 sample size  

\mu_o =90.1 represent the value that we want to verify

z would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to verify if the new teaching method negatively affects reading comprehension scores, so then the system of hypothesis for this case are:

Null hypothesis:\mu =90.1  

Alternative hypothesis:\mu \neq 90.1  

Since we know the population deviation the statistic would be:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

Replacing the info given we got:

z=\frac{87.8-90.1}{\frac{17.3}{\sqrt{38}}}=-0.820  

The p value for this case would be given by:

p_v =2*P(z  

Since the p value is very high ant any significance level we will have enough evidence to FAIL to reject the null hypothesis, so then we can conclude that the true mean is not significantly different from 90.1 and the method not shows a significantly effect in the scores.

3 0
3 years ago
Read 2 more answers
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