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3 years ago

Find the indicated probability using the venn diagram

Mathematics

1 answer:

AURORKA [14]3 years ago

4 0

Answer:

P(A n B) = 5

Step-by-step explanation:

From the venn diagram given, we can see that;

P(A) = 25

P(B) = 15

P(A n B) = 5

Thus,the correct answer of P(A n B) which is the intersection between A and B is 5

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A dumpling machine automatically inserts mixed vegetables and meat filled wontons into a plastic package. Past experience reveal

Tanya [424]

Answer:

c. 0.810

Step-by-step explanation:

Given, total % of satisfactory packages = 90% = 0.90

Now, since selecting a satisfactory package is independent of any other trials (i.e. all the selection trials are independent to each other)

Hence, Probability of selecting two packages that re satisfactory

P(Selecting 2 satisfactory packages) = (0.90)^2

P(Selecting 2 satisfactory packages) = 0.90*0.90

P(Selecting 2 satisfactory packages) = 0.810

Note: We are selecting product since the chance of selecting in each trials are independent.

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3 years ago

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3 years ago

Read 2 more answers

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.

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3 years ago

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Lesechka [4]

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3 years ago

M is the midpoint of PQ, N is the midpoint of MQ, and L is the midpoint of MN. IfPQ = 64, find LN.

alisha [4.7K]

LN=8 because every time there is a midpoint you have to divide it in half. So 64,32,16,8

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3 years ago