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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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11/6 = n + 7/9

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See  below.

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3 years ago
Determine the number of possible solutions for a triangle with B=37 degrees, a=32, b=27
vladimir1956 [14]

Answer:

Two possible solutions

Step-by-step explanation:

we know that

Applying the law of sines

\frac{a}{sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}

we have

a=32\ units

b=27\ units

B=37\°

step 1

Find the measure of angle A

\frac{a}{sin(A)}=\frac{b}{Sin(B)}

substitute the values

\frac{32}{sin(A)}=\frac{27}{Sin(37\°)}

sin(A)=(32)Sin(37\°)/27=0.71326

A=arcsin(0.71326)=45.5\°

The measure of angle A could have two measures

the first measure-------> A=45.5\°

the second measure -----> A=180\°-45.5\°=134.5\°

step 2

Find the first measure of angle C

Remember that the sum of the internal angles of a triangle must be equal to  180\°

A+B+C=180\°

substitute the values

A=45.5\°

B=37\°

45.5\°+37\°+C=180\°

C=180\°-(45.5\°+37\°)=97.5\°

step 3

Find the first length of side c

\frac{a}{sin(A)}=\frac{c}{Sin(C)}

substitute the values

\frac{32}{sin(37\°)}=\frac{c}{Sin(97.5\°)}

c=Sin(97.5\°)\frac{32}{sin(37\°)}=52.7\ units

therefore

the measures for the first solution of the triangle are

A=45.5\° , a=32\ units

B=37\° , b=27\ units

C=97.5\° , b=52.7\ units

step 4    

Find the second measure of angle C with the second measure of angle A

Remember that the sum of the internal angles of a triangle must be equal to  180\°

A+B+C=180\°

substitute the values

A=134.5\°

B=37\°

134.5\°+37\°+C=180\°

C=180\°-(134.5\°+37\°)=8.5\°

step 5

Find the second length of side c

\frac{a}{sin(A)}=\frac{c}{Sin(C)}

substitute the values

\frac{32}{sin(37\°)}=\frac{c}{Sin(8.5\°)}

c=Sin(8.5\°)\frac{32}{sin(37\°)}=7.9\ units

therefore

the measures for the second solution of the triangle are

A=45.5\° , a=32\ units

B=37\° , b=27\ units

C=8.5\° , b=7.9\ units

6 0
3 years ago
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