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Assoli18 [71]
3 years ago
12

Find the midpoint of the segment between the points (15,−9) and (−2,−18) A. (172,92) B. (13,−27) C. (132,−272) D. (−13,27)

Mathematics
1 answer:
kotegsom [21]3 years ago
5 0

Answer:

from my calculation, the answer is B

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Answer:it would be 4:1


Step-by-step explanation:


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If a + b + c = -6 and x + y = 5<br> what is 9y + 9x - 10a - 10c - 10b?
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A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic
goldfiish [28.3K]

Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event <em>E</em>_{i}, of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}

The law of total probability states that, if events <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = the test is positive

The information provided is:

P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85

Using the above the probabilities compute the remaining values as follows:

P(X^{c})=1-P(X)=1-0.04=0.96

P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10

P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})

          =(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}

             =\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}

             =0.20

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

6 0
3 years ago
Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. The probabi
CaHeK987 [17]

Answer: 0.6375

Step-by-step explanation:

Let's assume that the event that she'll make the first shot is given as P(A) while making the second shot is P(B). Therefore, P(A/B) = 0.85

Therefore, the probability that she makes both free throws will be denoted as:

= 0.75 × 0.85

= 0.6375

3 0
3 years ago
Which best describes the angles of some parallelograms?
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7 0
2 years ago
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