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Anuta_ua [19.1K]
3 years ago
14

Round 8.1845062 to the nearest hundred thousandths

Mathematics
1 answer:
balandron [24]3 years ago
6 0
8.1845062 

8.1845   << your answer.

the number 5 was in the hundred thousandths so to round it look at the next number and if it's less than 5, it stays the same but if it's more than 5 you round it up to the next number.

hope this helps, God bless!
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Step-by-step explanation:

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Answer: there were 12 chocolates in each box. (Originally, before any got eaten)

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My brother wants to estimate the proportion of Canadians who own their house.What sample size should be obtained if he wants the
AVprozaik [17]

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a) n=\frac{0.675(1-0.675)}{(\frac{0.02}{1.64})^2}=1475.07

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b) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

The margin of error for the proportion interval is given by this formula:  

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

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n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2} (b)  

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In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.9=0.1 and \alpha/2 =0.05. And the critical value would be given by:  

z_{\alpha/2}=\pm 1.64  

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.02 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

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Part b

For this case since we don't have a prior estimate we can use \hat p =0.5

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.64})^2}=1681

And rounded up we have that n=1681

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