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Flauer [41]
3 years ago
15

I need help with this tysm​

Mathematics
1 answer:
erica [24]3 years ago
8 0

9514 1404 393

Answer:

  1. -51
  2. -30
  3. -60
  4. 2 or -2

Step-by-step explanation:

The function puts a minus sign on 3 times the magnitude of the input.

1. y = -|3×17| = -51

2. y = -|3×10| = -30

3. y = -|3×20| = -60

4. -6 = -|3x|

  2 = |x| . . . . divide by -3

  x = ±2 . . . . two possibilities

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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.

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