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3 years ago

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Ella S, Lily, Aziza, and Kevin earned a total of $100 shoveling snow. Each of them earned the same amount. How much did each of

them earn? I will give 13 points

1 answer:

Lesechka [4]3 years ago

3 0

Answer:

$50 :)

Step-by-step explanation:

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Any help? Much appreciated

natulia [17]

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$\frac{6+\sqrt{27} }{4-\sqrt{3} } = \frac{r+s\sqrt{3} }{13} \\We\ may\ multiply\ the\ numerator\ and\ denominator\ with\ (4+\sqrt{3} ).\\Hence,\\\frac{(6+\sqrt{27})(4+\sqrt{3}) }{13} = \frac{r+s\sqrt{3} }{13} \\Hence,\\24+6\sqrt{3} +4\sqrt{27} +9= r+s\sqrt{3}\\24+6\sqrt{3} +4*3\sqrt{3} } +9= r+s\sqrt{3}\\24+6\sqrt{3} +12\sqrt{3}+9 = r+s\sqrt{3}\\33+18\sqrt{3} = r+s\sqrt{3}\\Hence,\\r=33, s=18$

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3 years ago

(a) A newspaper conducted a statewide survey concerning the 1998 race for state senator. The newspaper took a SRS of 1200 regist

Sladkaya [172]

Answer:

n=9604

Step-by-step explanation:

1) Notation and definitions

$X=620$ number of people that would vote for the Republican candidate

$n=1200$ random sample taken

$\hat p=\frac{620}{1200}=0.517$ estimated proportion of people that would vote for the Republican candidate

$p$ true population proportion of people that would vote for the Republican candidate

Me= 0.01 represent th margin of error.

Confidence =0.95 or 95%

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

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

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 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 $ME =\pm 0.01$ 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)

The problem says that we need to use $\hat p =0.05$ , and replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.96})^2}=9604$

And rounded we have that n=9604

6 0

3 years ago