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andrey2020 [161]
3 years ago
15

Suppose two fractions are both less than 1. Can their sum be greater than 1? Greater than 2?

Mathematics
2 answers:
Rashid [163]3 years ago
8 0
Hi there!
Yes, if the two fractions are less than one, their sum can be greater than one. Take the example of 3/4 and 3/4. It is less than 1, but has a sum if 1 1/2.
However, their sum cannot be greater than 2. For that to happen, there would have to be a sum of at least 1 and 1, which are not fractions.

Hope this helps :)
Juliette [100K]3 years ago
4 0
Their sum can be greater than two. There is nothing that states that they couldn't be. All it states it that the two factions sums are less than 1. I hoped this helps 
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Step-by-step explanation:

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6 0
2 years ago
3. Using the sample size of 800 and the proportion 0.47, calculate the margin of error associated with the estimate of the propo
salantis [7]

Answer:

ME=1.96\sqrt{\frac{0.47 (1-0.47)}{800}}=0.0346  

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.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

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

Solution to the problem

We assume for this case a confidence level of 95%. 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 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\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)  

And if we replace the values obtained we got this:

ME=1.96\sqrt{\frac{0.47 (1-0.47)}{800}}=0.0346  

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<h2><em><u>Ans</u></em><em><u>wer</u></em><em><u>:</u></em></h2>

Let m<6 be X

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