0 = 44-44
1 = 44/44 or (4+4)/(4+4) or (4/4) / (4/4) or [(4! - 4)/ 4] - 4
2 = 4/4+4/4
3 = (4+4+4)/4
4 = 4*(4-4)+4
5 = (4*4+4)/4
6 = 4*.4+4.4
7 = 44/4-4
8 = 4+4.4-.4
9 = 4/4+4+4
10 = 44/4.4
11 = 4/.4+4/4
12 = (44+4)/4
13 = 4!-44/4
14 = 4*(4-.4)-.4
15 = 44/4+4
16 = .4*(44-4)
17 = 4/4+4*4
18 = 44*.4+.4
19 = 4!-4-4/4
20 = 4*(4/4+4)
The total combinations of 5 numbers that can be made from
59 numbers is:
59C5 = 5,006,386
So the probability of winning is:
P = 1 / 5,006,386
P = 1.997 x 10^-7
So the expected payoff is:
P = $1,000,000 (1.997 x 10^-7) - $1 (1 - 1.997 x 10^-7)
<span>P = - $0.80 (loss)</span>
<span>
</span>
<span>D</span>
Solution:
q ∧ r : q and r
q r q∧r
T T T
T F F
F T F
F F F
p ∨ (q∧r)
p OR (q AND r)
p q∧r p∨(q∧r)
F T∧ T=T T
F F∧T = F F
F T ∧ F=F F
F F ∧ F =F F
If P is false , The row which represents when p ∨ (q ∧ r) is true is →→F T T T
Which is Option E.
Answer:
I think the answer is D please forgive me if I'm wrong
The domain is {15,21,18,16} but the range would be {62,65,70,73} because nothing repeats.
