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
(-3,2)))))))))))))))))))))))))))))))))))
Answer:
Si me dieras una imagen con esto, podría responderla fácilmente, pero no hay una imagen o modelo para ayudar a disculparme.
Step-by-step explanation:
lo siento, tal vez vuelva a publicarlo con un modelo o una imagen para que pueda ayudar
Sorry this is a tricky one IDK I'll ask some others to help you
Answer: e. Substitution Property
g. Substitution Property
Step-by-step explanation:
