Answer:
0.30
Step-by-step explanation:
Probability of stopping at first signal = 0.36 ;
P(stop 1) = P(x) = 0.36
Probability of stopping at second signal = 0.54;
P(stop 2) = P(y) = 0.54
Probability of stopping at atleast one of the two signals:
P(x U y) = 0.6
Stopping at both signals :
P(xny) = p(x) + p(y) - p(xUy)
P(xny) = 0.36 + 0.54 - 0.6
P(xny) = 0.3
Stopping at x but not y
P(x n y') = P(x) - P(xny) = 0.36 - 0.3 = 0.06
Stopping at y but not x
P(y n x') = P(y) - P(xny) = 0.54 - 0.3 = 0.24
Probability of stopping at exactly 1 signal :
P(x n y') or P(y n x') = 0.06 + 0.24 = 0.30
Answer:
x = 0
Step-by-step explanation:
To 'solve' means to find the x values that make the equation equal zero, so
0 = (3x)/[(x + 5)(x - 4)
Multiply both sides by the denominator to get rid of the fraction
0[(x + 5)(x - 4)] = [(3x)(x + 5)(x - 4)]/[(x + 5)(x - 4)]
0 = 3x
0 = x (divide both sides by 3)
So x = 0 is the solution for this equation
Ok so ummmm idk how to speck in that but well ummmmm idk
Answer:
iIm sorry dont report but i do not understand what im answering, but ill take the points thank you