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Alex787 [66]
2 years ago
10

En la ruta que un automovilista recorre para trasladarse a su trabajo existen dos intersecciones con señalamientos de tránsito.

La probabilidad de que el automovilista tenga que detenerse en la primera señal es 0.35, la probabilidad análoga para la segunda señal es 0.55, y la probabilidad de que tenga que detenerse en al menos una de las dos señales es 0.75.
(a) ¿Cuál es la probabilidad de que tenga que detenerse en ambas señales?
Mathematics
1 answer:
wolverine [178]2 years ago
3 0

Answer:

a) 0.15

b) 0.2

c) 0.6

Step-by-step explanation:

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Answer:

\left\begin{array}{ccc}A\left( \boxed{3}  , \boxed{8} \right)&B\left( \boxed{9}  , \boxed{6} \right )\\\\C\left( \boxed{1}  ,  \boxed{4} \right)&D\left( \boxed{7}  ,  \boxed{2} \right )\end{array}\right

Step-by-step explanation:

The coordinates are:

\left\begin{array}{ccc}A\left( \boxed{3}  , \boxed{8} \right)&B\left( \boxed{9}  , \boxed{6} \right )\\\\C\left( \boxed{1}  ,  \boxed{4} \right)&D\left( \boxed{7}  ,  \boxed{2} \right )\end{array}\right

The parallelogram is attached below.

To verify if these coordinates form a parallelogram, we show that:

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Using Distance Formula

AB = \sqrt{(6-8)^2+(9-3)^2} = \sqrt{(-2)^2+(6)^2} = \sqrt{40} $ units

CD = \sqrt{(2-4)^2+(7-1)^2} = \sqrt{(-2)^2+(6)^2} = \sqrt{40} $ units

AC = \sqrt{(8-4)^2+(3-1)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{20} $ units

BD = \sqrt{(6-2)^2+(9-7)^2} = \sqrt{(4)^2+(2)^2} = \sqrt{20} $ units

Since AB=CD; and AC=BD, the coordinates A, B, C, and D form the vertex of a parallelogram.

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