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Andre45 [30]
2 years ago
12

A snail climbs to the top of a coconut tree 18 meters high. In an hour it will climb 2 meters. But if it go 2 meters, it will sl

ips down one meter. Then how long does it take for the snaik to reach the top of the coconut?
Mathematics
1 answer:
kumpel [21]2 years ago
5 0

Answer:

The snail will reach the top of the coconut in 18 hours.

Step-by-step explanation:

If the snail will climb 2 meters in 1 hour but it will slip down one meter for every 2 meters climbed, then the distance traveled will be:

d = 2 m - 1 m = 1 m                                        

Since in 1 hour, it climbs 2 meters minus 1 meter, the time in which the snail will reach the top is:

t = \frac{1 h}{1 m}*18 m = 18 h      

Therefore, the snail will reach the top of the coconut in 18 hours.

I hope it helps you!                                          

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Given a population with a mean of muμequals=100100 and a variance of sigma squaredσ2equals=3636​, the central limit theorem appl
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Answer:

a) \bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".  

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Let X the random variable that represent the variable of interest on this case, and for this case we know the distribution for X is given by:  

X \sim N(\mu=100,\sigma=6)  

And let \bar X represent the sample mean, by the central limit theorem, the distribution for the sample mean is given by:  

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})  

a. What are the mean and variance of the sampling distribution for the sample​ means?

\bar X \sim N(100,\frac{6}{\sqrt{25}}=1.2)

\mu_{\bar X}=100 \sigma^2_{\bar X}=1.2^2=1.44

b. What is the probability that x overbarxgreater than>101

First we can to find the z score for the value of 101. And in order to do this we need to apply the formula for the z score given by:  

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

If we apply this formula to our probability we got this:  

z=\frac{101-100}{\frac{6}{\sqrt{25}}}=0.833  

And we want to find this probability:

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On this last step we use the complement rule.  

c. What is the probability that x bar 98less than

First we can to find the z score for the value of 98.

z=\frac{98-100}{\frac{6}{\sqrt{25}}}=-1.67  

And we want to find this probability:

P(\bar X

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