Probabilities with possible states of nature: s1, s2, and s3. suppose that you are given a decision situation with three possible state of nature: s1, s2, and s3.the prior probabilities are p(s1) =.1, p(s2) = .6, and p(s3) = .3.with sample information i, p(i|s1) =.15, p(i|s2) = .2, and p(i|s3) = .1.compute p(s1|i)compute p(s2|i)compute p(s3|i)

1.

2.

3.

Step-by-step explanation:

Given information:

(1)

We need to find the value of P(s₁|I).

Therefore the value of P(s₁|I) is .

(2)

We need to find the value of P(s₂|I).

Therefore the value of P(s₂|I) is .

(3)

We need to find the value of P(s₃|I).

Therefore the value of P(s₃|I) is .

step-by-step explanation:

i could only see the first page. the other pages were blurry. ; -;

x hot dogs for 2$, y drinks for 3$. 29 items and total 70

2x+3y=70

x+y=29

if x=17and y=12

2(17)+3(12)=70

34+36=70

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w