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Kamila [148]
2 years ago
10

Y+2=2/5(x+6) write in standard form​

Mathematics
1 answer:
icang [17]2 years ago
8 0

y+2=(2/5)(x+6)

5(y+2)=2(x+6)

5y+10=2x+12

-2x+5y+10-12=0

-2x+5y-2=0

2x-5y+2=0

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An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by th
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Answer:

a) n =1623

b) ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208    

Step-by-step explanation:

1) Notation and definitions

n random sample taken

\hat p=0.19 estimated proportion of customers are not satisfied with the service provided by the local dealer

Confidence =0.96 or 96%

Me= 0.02 represent the margin of error

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}})

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 96% of confidence, our significance level would be given by \alpha=1-0.96=0.04 and \alpha/2 =0.02. And the critical value would be given by:

z_{\alpha/2}=-2.0538, z_{1-\alpha/2}=2.0538

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 0.02 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{0.19(1-0.19)}{(\frac{0.02}{2.0538})^2}=1622.912  

And rounded up we have that n=1623

Part b

The margin of error is given by:

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

So then if w replace the value of n obtained from part a we got:

ME=2.0538\sqrt{\frac{0.21 (1-0.21)}{1623}}=0.0208    

3 0
3 years ago
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