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mamaluj [8]
3 years ago
11

Rectangle ABCD is transformed to A’B’C’D’ by a dilation. What is the scale factor used for the dilation?

Mathematics
1 answer:
jekas [21]3 years ago
6 0

Answer:

Scale factor = 3

Step-by-step explanation:

Scale factor used for the dilation = the ratio of the corresponding side length of the new figure to the side length of the original figure

Scale factor = B'C'/BC

B'C' = 9 units

BC = 3 units

Scale factor = 9/3 = 3

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National data in the 1960's showed that about 44% of the adult population had never smoked cigarettes. In 1995 a national health
Marta_Voda [28]

Answer:

a) 0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487

0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553

The 95% confidence interval would be given by (0.487;0.553)

b) After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated  proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes.

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".  

Solution to the problem

Part a

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:

0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487

0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553

The 95% confidence interval would be given by (0.487;0.553)

Part b

After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated  proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes at 5% of significance.

4 0
4 years ago
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