2 years ago

Can someone solve this

Mathematics

1 answer:

RSB [31]2 years ago

7 0

a) 110°

b) 39°

️️️️️️️️

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Which part of the graph best represents the solution set to the system of inequalities y ≤ x + 1 and y + x ≤ –1?

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<span>The part of the graph that best represents the solution set to the system of inequalities y ≤ x + 1 and y + x ≤ –1 is Part C.</span>

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Cuanto es 100+100+0+0+0+0+0

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Answer:

es 200

Step-by-step explanation:

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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

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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:

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

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

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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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