BD = 2x-4
But it's also BE - CE + CD
BE is 3x-1
CE is 2x-3
CD is 2
so
BD = 3x-1 -(2x-3) +2
simplified
BD = 1x +4
since BD = BD (obviously), we can also say that the righthand sides of the two equation must be equal
2x -4 = 1x +4
let's solve for x and that put it into either of the 2 equations.
2x -4 = 1x +4
subtract 1x on both sides
x -4 = 4
add 4 on both sides
x = 8
substitute x for its value in 2x -4 = 1x +4
(this way we double check BD in two expressions at once. the equation should come out true, meaning same value for BD in each expression. both sides should be equal).
2*8 -4 = 1*8 +4
12 = 12
seems true
BD is safely 12
It's 57 and 59
Call (n) is the first odd interger => the second odd interger is (n+2)
=> n + (n + 2) = 116
=> n + n + 2 = 116
=> 2n = 114
=> n = 57
First odd interger is 57 and the second odd interger is 59 .
Sorry my English so bad =)))
Answer:
=−6
Step-by-step explanation:
10 pieces because five plus five is ten
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%)
