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snow_tiger [21]
3 years ago
9

An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it iswhen she is a nonsmoker. If 32 pe

rcent of women of childbearing age are smokers, what percentageof women having ectopic pregnancies are smokers?
Mathematics
1 answer:
Dmitriy789 [7]3 years ago
8 0

Answer:

48.48%

Step-by-step explanation:

Let's assume that there is a number N of women.

32% of these are smokers, then there are 0.32*N smokers

then 68% of these are non-smokers, then there are 0.68*N non-smokers.

Let's assume that the probability of having a ectopic pregnancy for a non-smoker is p (and the probability for a smoker will be 2*p)

Then the number of women with an ectopic pregnancy that are non-smokers is:

p*0.68*N

The number of women with an ectopic pregnancy that are smokers is:

2*p*0.32*N

Then the total number of women with an ectopic pregnancy will be:

p*0.68*N + 2*p*0.32*N

The percentage of women having an ectopic pregnancy that are smokers is equal to the quotient between the number of women with an ectopic pregnancy that are smokers and the total number of women with an ectopic pregnancy, all that times 100%.

The percentage is:

P = \frac{2*p*0.32*N}{p*0.68*N + 2*p*0.32*N} *100\%

Taking p and N as common factors, we get:

P = \frac{2*p*0.32*N}{p*0.68*N + 2*p*0.32*N} *100\% = \frac{N*p}{N*p} \frac{2*0.32}{0.68 + 2*0.32} *100\%

Then we get:

\frac{2*0.32}{0.68 + 2*0.32} *100\% = 48.48\%

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marishachu [46]

Answer:

n=601

Step-by-step explanation:

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{\hat p(1-\hat 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 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}}

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 0.04 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)  

Since we don't have a prior estimation for the proportion we can use 0.5 as estimation. And replacing into equation (b) the values from part a we got:

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3 0
3 years ago
I’m horrible at geometry, help please?
Korvikt [17]

Answer:

Longest side = 13 inches

Step-by-step explanation:

Perimeter of triangle =  30 inches

x + 4x - 8 + 2x +3    =  30

x + 4x + 2x - 8 + 3 = 30

                   7x - 5 = 30

                          7x = 30 + 5

                          7x = 35

                            x = 35/7

         x = 5

Longest side is the hypotenuse.

Longest side = 2x + 3 = 2*5 + 3 = 10 + 3 = 13 inches

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4 years ago
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Step-by-step explanation:

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3+10+25=38

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joja [24]

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c= 20d

here, c=$20

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or, d= 1 hr

8 0
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