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3 years ago

What is 3/5 of 21? plz answer the question

Mathematics

2 answers:

aleksley [76]3 years ago

7 0

The answer is 12.6, the fractional answer is 12 3/5

WITCHER [35]3 years ago

5 0

It's 12.6.
You divide 21 by the denominator, 5, and multipy that (4.2) by the numerator, 3.

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Triangle ABC and Triangle DEF are similar. For triangle ABC, AB = 2x , AC = 24 inches and BC = 20. For triangle DEF, EF = 5 and

Norma-Jean [14]

X=8.

Following the similarity statement, we know that AC is to DF as BC is to EF. The ratio would be:

24/(x-2) = 20/5

Cross multiply:
24*5 = 20(x-2)

Use the distributive property:
120 = 20x - 40

Add 40 to both sides:
120+40 = 20x - 40 + 40
160 = 20x

Divide both sides by 20:
160/20 = 20x/20
8 = x

8 0

3 years ago

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RoseWind [281]

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3 years ago

You estimated the average rental cost of a 2 bedroom apartment in Denton. A random sample of 16 apartments was taken. The sample

Zigmanuir [339]

Answer:

The new sample size required in order to have the same confidence 95% and reduce the margin of erro to $60 is:

n=28

Step-by-step explanation:

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".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=160)$

And the distribution for $\bar X$ is:

$\bar X \sim N(\mu, \frac{160}{\sqrt{n}})$

We know that the margin of error for a confidence interval is given by:

$Me=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The next step would be find the value of $\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$

Using the normal standard table, excel or a calculator we see that:

$z_{\alpha/2}=\pm 1.96$

If we solve for n from formula (1) we got:

$\sqrt{n}=\frac{z_{\alpha/2} \sigma}{Me}$

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And we have everything to replace into the formula:

$n=(\frac{1.96(160)}{78.4})^2 =16$

And this value agrees with the sample size given.

For the case of the problem we ar einterested on Me= $60, and we need to find the new sample size required to mantain the confidence level at 95%. We know that n is given by this formula:

$n=(\frac{z_{\alpha/2} \sigma}{Me})^2$

And now we can replace the new value of Me and see what we got, like this:

$n=(\frac{1.96*160}{60})^2 =27.32$