4x=20-8y
x=5-2y now use this value of x in the second equation...
3(5-2y)+6y=15
15-6y+6y=15
15=15 This is true for any y or x value.
So there are infinitely many solutions as the two equations describe the same line.
Answer:
the conditional probability that X = 1 , X = 2 and X = 3 is 0.7333 (73.33%) , 0.25 (25%) and 0.0167 (1.67%) respectively
Step-by-step explanation:
a player wins money when i>0 then defining event W= gain money , then
P(W) = p(i>0) = p(1)+p(2)+p(3)
then the conditional probability can be calculated through the theorem of Bayes
P(X=1/W)= P(X=1 ∩ W)/P(W)
where
P(X=1 ∩ W)= probability that the payout is 1 and earns money
P(X=1 / W)= probability that the payout is 1 given money was earned
then
P(X=1/W)= P(X=1 ∩ W)/P(W) = P(X=1) / P(W) = p(1) /[p(1)+p(2)+p(3)] = 11/40 /(11/40+3/32+1/160
) = 0.7333 (73.33%)
similarly
P(X=2/W)=p(2) /[p(1)+p(2)+p(3)] = 3/32 /(11/40+3/32+1/160
) = 0.25 (25%)
P(X=3/W)=p(2) /[p(1)+p(2)+p(3)] = 1/160 /(11/40+3/32+1/160
) = 0.0167 (1.67%)
This is the concept of sequences and series. To solve the question we proceed as follows:
2] T=-4n-3
to get the 11th term we substitute n=11 in the expression and simplify. Thus, plugging in n=11 we obtain:
T=-4(11)-3
T=-44-3
T=-47
thus the 11th term is -47
3] General term is T=n²-1. The 7th term will be evaluated as above:
plug in n=7 in the expression and simplify:
T7=7²-1
T7=49-1
T7=48
thus the 11th term is 48
4] General sequence is T=1/3n². The term will be evaluated as follows:
plug in n=3 in the expression and simplify:
T3=1/(3×3²)=1/27
Thus the 3rd term is 1/27
