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Softa [21]
3 years ago
8

Is the following relation a function? {(1,2), (3,4),(-2,4),(3,0)}

Mathematics
2 answers:
Nana76 [90]3 years ago
8 0

Answer:

Step-by-step explanation:

KengaRu [80]3 years ago
3 0

Answer:

yes

Step-by-step explanation:

a function has one uniqe output to a input

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Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl
amm1812

Answer:

1. P(s_1|I)=\frac{1}{11}

2. P(s_2|I)=\frac{8}{11}

3. P(s_3|I)=\frac{2}{11}

Step-by-step explanation:

Given information:

P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3

P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1

(1)

We need to find the value of P(s₁|I).

P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}

P(s_1|I)=\frac{0.015}{0.165}

P(s_1|I)=\frac{1}{11}

Therefore the value of P(s₁|I) is \frac{1}{11}.

(2)

We need to find the value of P(s₂|I).

P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}

P(s_2|I)=\frac{0.12}{0.165}

P(s_2|I)=\frac{8}{11}

Therefore the value of P(s₂|I) is \frac{8}{11}.

(3)

We need to find the value of P(s₃|I).

P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}

P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}

P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}

P(s_3|I)=\frac{0.03}{0.165}

P(s_3|I)=\frac{2}{11}

Therefore the value of P(s₃|I) is \frac{2}{11}.

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