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Alex787 [66]
3 years ago
11

Aria paid $75,000 for her house. Its property value increased by 2.2% per year. When Aria sold her house after eleven years, how

much was it worth, to the nearest hundred dollars?
Mathematics
1 answer:
astra-53 [7]3 years ago
3 0

Answer:

When Aria sold her house after eleven years it worth was <u>$95,300</u>.

Step-by-step explanation:

Given:

Aria paid $75,000 for her house. Its property value increased by 2.2% per year.

Now, to find the worth of Aria house when sold after eleven years.

Let the amount of house after eleven years be x.

Amount Aria paid for her house (A) = $75,000.

Rate of property increased per year (r) = 2.2%.

Time (t) = 11 years.

Now, to get the amount of house after eleven years we put formula:

x=A\times (1+\frac{r}{100})^{11}

x=75000\times (1+\frac{2.2}{100})^{11}

x=75000\times (1+0.022)^{11}

x=75000\times (1.022)^{11}

x=75000\times 1.2704

x=95280.

<em>The amount of house after eleven years to the nearest hundred dollars is $95,300.</em>

Therefore, when Aria sold her house after eleven years it worth was $95,300.

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Answer:

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b) n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79  

And rounded up we have that n=622

c) n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

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Step-by-step explanation:

Part a

\hat p=\frac{823}{1000}=0.823

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}}

If we replace the values obtained we got:

0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799

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The 95% confidence interval would be given by (0.799;0.847)

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

And on this case we have that ME =\pm 0.03 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)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79  

And rounded up we have that n=622

Part c

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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Step-by-step explanation:

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