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3 years ago

What is the value of x? show all the work

Mathematics

2 answers:

Usimov [2.4K]3 years ago

7 0

Answer:

9 cm

Step-by-step explanation:

You can use the pythagorean theorem to find x here. Since $AB^2 + BC^2=AC^2, AC^2-BC^2=AB^2$ . $\sqrt{\sqrt{117}^2-6^2}=\sqrt{117-36}=\sqrt{81}=9$ . Hope this helps!

11Alexandr11 [23.1K]3 years ago

5 0

Answer:

9 cm

Step-by-step explanation:

Using pythagoras theorem

(Sqrt(117))² = 6² + x²

117 = 36 + x²

x² = 81

x = 9

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Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

$\hat p=0.52$ estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

$\alpha=0.05$ represent the significance level (no given, but is assumed)

Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error for this case is given by:

$ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

And replacing we got:

$ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259$

And replacing into the confidence interval formula we got:

$0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941$

$0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459$

And the 95% confidence interval would be given (0.4941;0.5459).

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