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bazaltina [42]
3 years ago
9

Question 1 (5 points)

Mathematics
1 answer:
Ivan3 years ago
6 0

Answer:

  f(6) ≈ 31

Step-by-step explanation:

If f(x) varies directly with x, we can write f(x) this way:

  f(x) = kx

Using the given values, we can find k:

  f(25) = k×25 = 130

  k = 130/25 = 5.2

Then for x = 6, we have ...

  f(x) = 5.2x

  f(6) = 5.2×6 = 31.2

Rounded to the nearest whole number, ...

  f(6) ≈ 31

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Bad gums may mean a bad heart. Researchers discovered that 79% of people who have suffered a heart attack had periodontal diseas
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Answer:

(A) 0.297

(B) 0.595

Step-by-step explanation:

Let,

H = a person who suffered from a heart attack

G = a person has the periodontal disease.

Given:

P (G|H) = 0.79, P(G|H') = 0.33 and P (H) = 0.15

Compute the probability that a person has the periodontal disease as follows:

P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.15)+(0.33\times(1-0.15))\\=0.399

(A)

The probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is:

P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.15}{0.399} \\=0.29699\\\approx0.297

Thus, the probability that a person had periodontal disease, what is the probability that he or she will have a heart attack is 0.297.

(B)

Now if the probability of a person having a heart attack is, P (H) = 0.38.

Compute the probability that a person has the periodontal disease as follows:

P(G)=P(G|H)P(H)+P(G|H')P(H')\\=P(G|H)P(H)+P(G|H')(1-P(H))\\=(0.79\times0.38)+(0.33\times(1-0.38))\\=0.5048

Compute the probability of a person having a heart attack given that he or she has the disease:

P(H|G)=\frac{P(G|H)P(H)}{P(G)}\\=\frac{0.79\times0.38}{0.5048}\\ =0.59469\\\approx0.595

The probability of a person having a heart attack given that he or she has the disease is 0.595.

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Step-by-step explanation:

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