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cestrela7 [59]
3 years ago
12

Eloise surveyed her classmates to see who goes to the movies at least twice a week. She recorded the results in the table separa

ting by gender. Is gender independent of going to the movies at least twice a week? Explain. A) Yes, the probability of being male is the same as the probability of being male given you go to the movies twice a week. B) No, the probability of being male is the same as the probability of being male given you go to the movies twice a week. C) No, the probability of being male is different from the probability of being male given you go to the movies twice a week. D) Yes, the probability of being male is different from the probability of being male given you go to the movies twice a week.
Mathematics
1 answer:
Margaret [11]3 years ago
5 0

Answer:

The answer is C, i got it wrong on USAtestprep so i can answer it for you :)

Step-by-step explanation:

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Is this correct ? Help
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Read 2 more answers
According to the Center for Disease Control and Prevention (CDC), up to 20% of Americans contract the influenza virus each year,
12345 [234]

Answer:

(1) a. 0.0009

(2) d. 0.640

(3)

  • a. P(A and B) = 0.06.
  • b. P(A or B) = 0.70.

(4)Not disjoint

(5) a. nearly 0.

(6)b. 0.919

Step-by-Step Explanation:

(1)Probability of a baby being born with a birth defect =3%=0.03

The probability that both babies have birth defects=0.03 X 0.03= 0.0009.

(2)The probability of contracting the influenza virus each year = 20%=0.2

Therefore, the probability of not contracting the influenza virus =1-0.2=0.8

The probability that neither baby catches the flu in a given year:

=0.8 X  0.8

=0.64

(3)

P(A)=0.1

P(B)=0.6

P(A or B)=P(A)+P(B)=0.1 + 0.6 =0.7

P(A and B)=P(A)XP(B)=0.1 X 0.6 =0.06

(4)

P(A)=0.2

P(B)=0.9

Event A and B cannot be disjoint.

(5)

The probability of an American woman aged 20 to 24 having Chlamydia infection  =\dfrac{2791.5}{100000}

The probability that three randomly selected women in this age group have the infection

=\dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \times \dfrac{2791.5}{100000} \\\\=0.00002175\\\approx 0

(6)The probability of an American woman aged 20 to 24 not having Chlamydia infection  =1-\dfrac{2791.5}{100000}

The probability that three randomly selected women in this age group do not have the infection

=\left(1-\dfrac{2791.5}{100000}\right)^3\\\\=0.9186\\\approx 0.919

7 0
4 years ago
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