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1 year ago

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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?

1 answer:

wolverine [178]1 year ago

3 0

Answer:

a) 0.15

b) 0.2

c) 0.6

Step-by-step explanation:

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Answer:

$y = \frac{3}{2} x + 13$

Step-by-step explanation:

1) Use point-slope formula $y-y_1 = m (x-x_1)$ to find the slope-intercept form (or y = mx + b form) of the equation. m represents the slope, and $x_1$ and $y_1$ represent the x and y values of a point the line intersects. So, substitute $\frac{3}{2}$ for m, -4 for $x_1$ , and 7 for $y_1$ . Then, simplify and isolate y on the left side like so:

$y-(7) = \frac{3}{2} (x - (-4)) \\y -7 = \frac{3}{2} (x + 4) \\y - 7 = \frac{3}{2} x + 6\\y = \frac{3}{2} x + 13$

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Step-by-step explanation:

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3 years ago

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2 years ago

The Slow Ball Challenge or The Fast Ball Challenge.

cupoosta [38]

Answer:

Fast ball challenge

Step-by-step explanation:

Given

Slow Ball Challenge

$Pitches = 7$

$P(Hit) = 80\%$

$Win = \$60$

$Lost = \$10$

Fast Ball Challenge

$Pitches = 3$

$P(Hit) = 70\%$

$Win = \$60$

$Lost = \$10$

Required

Which should he choose?

To do this, we simply calculate the expected earnings of both.

Considering the slow ball challenge

First, we calculate the binomial probability that he hits all 7 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 7$ --- pitches

$x = 7$ --- all hits

$p = 80\% = 0.80$ --- probability of hit

So, we have:

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

$P(7) =^7C_7 * 0.80^7 * (1 - 0.80)^{7 - 7}$

$P(7) =1 * 0.80^7 * (1 - 0.80)^0$

$P(7) =1 * 0.80^7 * 0.20^0$

Using a calculator:

$P(7) =0.2097152$ --- This is the probability that he wins

i.e.

$P(Win) =0.2097152$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 -0.2097152$

$P(Lose) = 0.7902848$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.2097152 * \$60 + 0.7902848 * \$10$

Using a calculator, we have:

$Expected = \$20.48576$

Considering the fast ball challenge

First, we calculate the binomial probability that he hits all 3 pitches

$P(x) =^nC_x * p^x * (1 - p)^{n - x}$

Where

$n = 3$ --- pitches

$x = 3$ --- all hits

$p = 70\% = 0.70$ --- probability of hit

So, we have:

$P(3) =^3C_3 * 0.70^3 * (1 - 0.70)^{3 - 3}$

$P(3) =1 * 0.70^3 * (1 - 0.70)^0$

$P(3) =1 * 0.70^3 * 0.30^0$

Using a calculator:

$P(3) =0.343$ --- This is the probability that he wins

i.e.

$P(Win) =0.343$

The probability that he lose is:

$P(Lose) = 1 - P(Win)$ ---- Complement rule

$P(Lose) = 1 - 0.343$

$P(Lose) = 0.657$

The expected value is then calculated as:

$Expected = P(Win) * Win + P(Lose) * Lose$

$Expected = 0.343 * \$60 + 0.657 * \$10$

Using a calculator, we have:

$Expected = \$27.15$

So, we have:

$Expected = \$20.48576$ -- Slow ball

$Expected = \$27.15$ --- Fast ball

<em>The expected earnings of the fast ball challenge is greater than that of the slow ball. Hence, he should choose the fast ball challenge.</em>

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2 years ago