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Rom4ik [11]
3 years ago
12

What number is one fifth of one fourth of one ninth of 900? 2 3 4 5 6 7 none of these i don't know?

Mathematics
1 answer:
Shalnov [3]3 years ago
3 0
First, we want to write an expression for what is given:

(1/5)(1/4)(1/9)(900)

Now, we can solve this. Remember that to multiply fractions, we can just multiply the numerators and the denominators. Also, whole numbers always have a denominator of 1.

(900/180)
=5
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17*5= 85

15*25= 375

375+85= 460

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17* 5 cents= $.85

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4267 divided by 31 in mixed number form
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The Gotemba walking trail up Mount Fuji is about 9 kilometres (km) long. Walkers need to return from the 18 km walk by 8 pm. Tos
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Step-by-step explanation:

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Suppose that a fast food restaurant decides to survey its customers to gauge interest in a breakfast menu. After surveying multi
Harlamova29_29 [7]

Answer:

ME= \frac{Width}{2}=\frac{0.078}{2}=0.039

\hat p =0.688+0.039=0.727

\hat p =0.766-0.039=0.727

Step-by-step explanation:

Assuming that the confidence interval is (0.688; 0.766)

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}})  

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  

Use the confidence interval to find the point estimate and margin of error for the proportion

The margin of error is given by :

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

And for our case we can find the width of the confidence interval like this:

Width =0.766-0.688=0.078

And the estimation for the margin of error would be given by:

ME= \frac{Width}{2}=\frac{0.078}{2}=0.039

Now we can find th point of estimate adding the margin of error to the lower limit of the interval or subtracting the margin of error to the upper limit, like this:

\hat p =0.688+0.039=0.727

\hat p =0.766-0.039=0.727

4 0
3 years ago
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