Answer:
a) P( 
| X ) = 0.180
b) P(Y | 
) = 0.998
Step-by-step explanation:
Let
P(X) - Probability that he acts hungry
P(Y) - Probability that he had ate dinner,
Given,
P(X | Y) = 0.5
P(X | ) = 0.99
P(Y) = 0.9
a.)
P( | X ) = 
= 
= 0.180
⇒P( | X ) = 0.180
b.)
P(Y | ) = 
= ![\frac{(1 - 0.5)(0.9)}{1 - [(0.99)(0.1) + (0.5)(0.9) ]} = \frac{(0.5)(0.9)}{1 - [(0.099) + (0.45) ]} = \frac{0.45}{1 - [0.540]} = \frac{0.45}{0.451}](https://tex.z-dn.net/?f=%5Cfrac%7B%281%20-%200.5%29%280.9%29%7D%7B1%20-%20%5B%280.99%29%280.1%29%20%2B%20%280.5%29%280.9%29%20%5D%7D%20%3D%20%5Cfrac%7B%280.5%29%280.9%29%7D%7B1%20-%20%5B%280.099%29%20%2B%20%280.45%29%20%5D%7D%20%3D%20%5Cfrac%7B0.45%7D%7B1%20-%20%5B0.540%5D%7D%20%3D%20%5Cfrac%7B0.45%7D%7B0.451%7D)
= 0.998
⇒P(Y | ) = 0.998
Answer:
300 meters
Step-by-step explanation:
250+150+50+x=750
450+x=750
x=750-450=300
Answer:
Step-by-step explanation:
c because 40-180 is140
Answer:
z (min) = 360 x₁ = x₃ = 0 x₂ = 3
Step-by-step explanation:
Protein Carbohydrates Iron calories
Food 1 (x₁) 10 1 4 80
Food 2 (x₂) 15 2 8 120
Food 3 (x₃) 20 1 11 100
Requirements 40 6 12
From the table we get
Objective Function z :
z = 80*x₁ + 120*x₂ + 100*x₃ to minimize
Subjet to:
Constraint 1. at least 40 U of protein
10*x₁ + 15*x₂ + 20*x₃ ≥ 40
Constraint 2. at least 6 U of carbohydrates
1*x₁ + 2*x₂ + 1*x₃ ≥ 6
Constraint 3. at least 12 U of Iron
4*x₁ + 8*x₂ + 11*x₃ ≥ 12
General constraints:
x₁ ≥ 0 x₂ ≥ 0 x₃ ≥ 0 all integers
With the help of an on-line solver after 6 iterations the optimal solution is:
z (min) = 360 x₁ = x₃ = 0 x₂ = 3
