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3 years ago

Hi guys I need help in these two or one I will post one by one please help me in these pleaseeeee.Im so confuseddddd

Mathematics

2 answers:

muminat3 years ago

8 0

The slope is actually 7/5.

Alchen [17]3 years ago

6 0

The slope is 3/2 (use rise over run)

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circumference = diameter x π

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4 years ago

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ziro4ka [17]

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3 years ago

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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

zysi [14]

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.

4 0

3 years ago