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2 years ago

An investigation of a number of automobile accidents revealed the following information:

Mathematics

1 answer:

seropon [69]2 years ago

3 0

Answer:

59 accidents were investigated.

Step-by-step explanation:

The question above is a probability question that involves 2 elements: causes of accidents.

Let

A = Alcohol

E = Excessive speed

In the question, we are given the following information:

18 accidents involved Alcohol and Excessive speed =P(A ∩ E)

26 involved Alcohol = P(A)

12 accidents involved excessive speed but not alcohol = P( E ) Only

21 accidents involved neither alcohol nor excessive speed = neither A U B

We were given P(A) in the question. P(A Only) = P(A) - P(A ∩ E)

P(A Only) = 26 - 18

= 8

So, only 8 accident involved Alcohol but not excessive speed.

The Total number of Accidents investigated = P(A Only) + P( E only) + P(A ∩ E) + P( neither A U B)

= 8 + 12 + 18 + 21

= 59

Therefore, 59 accidents were investigated.

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How many times larger is 4 x 1012 than 8 x 107? A) 2 x 104 B) 2 x 105 C) 5 x 104 D) 5 x 105

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The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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ra1l [238]

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----------------
4x= 8
-- --
4 4
------------
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Answer:

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