Suppose that you are given a decision situation with three possible states of nature: s1, s2, and s3. the prior probabilities are p(s1) = 0.16, p(s2) = 0.57, and p(s3) = 0.27. with sample information i, p(i | s1) = 0.13, p(i | s2) = 0.07, and p(i | s3) = 0.19. compute the revised or posterior probabilities: p(s1 | i), p(s2 | i), and p(s3 | i). if required, round your answers to four decimal places.

b

2

−

b

b

3

+

7

b

2

+

14

b

+

6

%

−

3

b

3

+

7

b

2

+

14

b

+

6

%

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

hi how are you doing today? if you ever need just tell me

step-by-step explanation: