<h3>
Answer:</h3>
x = 1
<h3>
Step-by-step explanation:</h3>
<em>The only solution is an extraneous solution</em>, which is to say the equation has no solution.
The rational expression reduces to -1 (for x≠1), which makes the equation ...
1 = 1/x
The only solution to this is x=1, which is specifically disallowed.
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If you subtract the right side, the equation becomes ...
((-x+1)(x) +2(x)(x -1) -(x -1))/(x -1) = 0
(-x^2 +x +2x^2 -2x -x +1)/(x -1) = 0 . . . . eliminate parentheses in the numerator
(x^2 -2x +1)/(x -1) = 0 . . . . . . . . . . . . . . . collect terms
(x -1)^2/(x -1) = 0 . . . . . . . . . . . . . . . . . . . factor
This is undefined for the only value of x that could possibly be a solution, x=1.
2/3 can also be written as 2:3 or 4:6 or 4/6…8:12…8/12…
Answer:
67 600 000 different licence plates with repetition.
19 656 000 different license plates without repetition.
Step-by-step explanation:
Answer:
The probability of finding the white ball from second Jar = 20/39
Step-by-step explanation:
Jar 1 contains 1 White ball and 4 black balls
P(w | J₁) = 1/5
Jar 2 contains 2 White balls and 1 black ball
P(w | J₂) = 2/3
Jar 1 contains 2 White balls and 1 black ball
P(w | J₃) = 2/3
Also, Given
P(J₁) = 1/2
P(J₂) = 1/3
P(J₃) = 1/6
P(w) = P(J₁)×P(w | J₁) + P(J₂)×P(w | J₂) + P(J₃)×P(w | J₃)
P(w) = 1/2×1/5 + 1/3×2/3 + 1/6×2/3 =13/30
To find: P(J₂ | w)
According to conditional probability,
P(J₂ | w) = P(J₂)×P(w | J₂)/P(w)
P(J₂ | w) = (1/3×2/3)/(13/30)
P(J₂ | w) = (1/3×2/3)/(13/30) = 20/39
<u>The probability of finding the white ball from second Jar = 20/39</u>
