1. 0.4*0.4 / (0.2*0.2 + 0.3*0.3 + 0.4*0.4 + 0.1*0.5) = 0.47

2. All figures should be divided by ten
``` CF(A) = 0.6*0.9 = 0.54
CF(A&B) = min(0.54, 0.8) = 0.54
CF(Hv(A&B)) = max(0.8, 0.54) = 0.8
CF(C) = 0.9*0.8 = 0.72
CF(D) = 0.8*0.7 = 0.56
CF(CvD) = max(0.72, 0.56) = 0.72
CF(E) = 0.75*0.72 = 0.54
```
3. ```                    I     II      III
1  has no car      OUT    IN      OUT
2  < 18 years      OUT    IN      IN
3  millionaire     OUT    OUT     IN
4  win on horses   OUT    OUT     OUT
5  univ. teacher   OUT    OUT     IN

1 is in if 2 is in and 3 is out
3 is in if 4 or 5 is in
```
4. See page 4 in this document.
5. ```      eval[ (foo (quote (a b)) (quote (b c))) ]
eval[ (quote (a b)) ]
(a b)
eval[ (quote (b c)) ]
(b c)
apply[ foo, ((a b) (b c)) ]
apply[ (lambda (x y) (cons (car x) y)),((a b) (b c)) ]
eval[ (cons (car x) y)), ((x (a b))(y (b c))) ]
eval[ (car x), ((x (a b))(y (b c))) ]
eval[ x, ((x (a b))(y (b c))) ]
(a b)
apply[ car, ((a b)) ]
a
eval[ y, ((x (a b))(y (b c))) ]
(b c)
apply[ cons, (a (b c)) ]
(a b c)
```