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Svet_ta [14]
3 years ago
12

xFormula1" title="\lim_{x\to \ 4} \frac{x-4}{\sqrt{x}-\sqrt{4} }" alt="\lim_{x\to \ 4} \frac{x-4}{\sqrt{x}-\sqrt{4} }" align="absmiddle" class="latex-formula"> Please answer this one
Mathematics
1 answer:
Hoochie [10]3 years ago
3 0

Answer:

\large \boxed{\sf \ \  \lim_{x\to \ 4} \dfrac{x-4}{\sqrt{x}-\sqrt{4} }=4 \ \ }

Step-by-step explanation:

Hello,

We need to find the following limit.

\displaystyle \lim_{x\to \ 4} \dfrac{x-4}{\sqrt{x}-\sqrt{4} }

First of all, for any x real number different from 4 and positive, we can write

\dfrac{x-4}{\sqrt{x}-\sqrt{4}} = \dfrac{(x-4)(\sqrt{x}+\sqrt{4})} {(\sqrt{x}-\sqrt{4})(\sqrt{x}+\sqrt{4})}} ==\dfrac{(x-4)(\sqrt{x}+\sqrt{4})}{x-4}=\sqrt{x}+\sqrt{4}

so the limit is

\sqrt{4}+\sqrt{4}=2+2=4

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago
In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home. Give the value of th
astra-53 [7]

Answer:

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Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

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This means that n = 100, p = \frac{81}{100} = 0.81

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Hope this helps!! :)
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