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3 years ago

What is a irrational number between 12.1 and 12.2

Mathematics

2 answers:

GREYUIT [131]3 years ago

7 0

12.123456234576899122567536492848...

sergij07 [2.7K]3 years ago

7 0

Answer:

The required irrational number between 12.1 and 12.2 is 12.1243536859.....

Step-by-step explanation:

To find : What is a irrational number between 12.1 and 12.2 ?

Solution :

An irrational number is defined as the number written in $\frac{p}{q}$ form or in proper fraction where p and q are integers and q is non-zero.

An irrational numbers are non-terminating and non-repeating.

So, An irrational number between 12.1 and 12.2 is any number which is non-terminating and non-repeating there are so many in between.

Let's take, 12.1243536859......

Therefore, The required irrational number between 12.1 and 12.2 is 12.1243536859......

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Step-by-step explanation:

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$\sigma_{\bar x_m-\bar x_f}=\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

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$E=z_{\alpha/2}\times \sigma_{\bar x_m-\bar x_f}\\\\\\=z_{0.005}\times\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\\\=2.575\times \sqrt{\frac{33^2}{41}+\frac{20^2}{37}}\\\\\\=15.7416$

Hence, the margin of error is 15.7416

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$\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}$

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$CI=\bar x_m-\bar x_f\ \pm z_{\alpha/2}\sqrt{\frac{\sigma_m^2}{n_m}+\frac{\sigma_f^2}{n_f}}\\\\=(135.67-68.64)\pm 15.7416\\\\=67.03\pm 15.7416\\\\=[51.2884, \ 82.7716]$

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