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Mnenie [13.5K]
2 years ago
8

Geometry help please? Find the variables

Mathematics
1 answer:
serious [3.7K]2 years ago
7 0
This is asking what is z and y, so you must know that a triangle is 180 degrees, so if it is 180 degrees, than all three angles that the triangle has must add up to 180 for every single triangle for it to be a triangle. So you have 45 degrees and 47 degrees, which adds up to 92 degrees, so in order to find y you need to subtract 92 from 180, which is 88. 88 is the missing number, and the missing number in this triangle is y, so y= 88 degrees. This can fool you because it looks like 90 degrees, but it is actually 88. Z is a little trickier, because y isn't 90 degrees, if it was than z would be a direct reflection of y, but it isn't, so we must find the exact angle for this variable. I see it like this, if y is 88, than it is smaller than 90 obviously, so it is acute. If that line keeps going down, than really z would be bigger than y, because that line is going down but it is so hard to tell. So if y is smaller by 2 degrees, than z would be bigger by 2 degrees, meaning that z is 92 degrees, since it is slowly decreasing. Hope this helps
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The confidence interval would be given by this formula

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

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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:

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And replacing into the confidence interval formula we got:

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

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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Step-by-step explanation:

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