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2 years ago

Calculate the approximate probability that at least 2 children have been diagnosed with ASD among the 200 selected children.

Mathematics

1 answer:

mixas84 [53]2 years ago

4 0

Answer:

The approximate probability that at least 2 children have been diagnosed with ASD among the 200 selected children is

0.01 or 1%.

Step-by-step explanation:

Number of selected children = 200

The probability of no child been diagnosed with ASD = P(None) = 198/200 = 0.99

Therefore, the probability of at least two children been diagnosed with ASD = 1 - 0.99 = 0.01.

This is the same as:

If 2 children have been diagnosed with ASD,

therefore, the approximate probability that at least 2 children have been diagnosed is:

2/200 = 0.01. This value is equal to 1%.

The above are summed up in:

The probability of at least one = 1/200 = 0.05

Therefore, the probability of at least two = 0.05 * 2 = 0.01

b) Generally, to find the probability of at least one event happening, we calculate the probability of none and then subtract that result from 1. That is, P(at least one) = 1 – P(none). For two events happening, the sum of the probability of at least one in two places is deducted from 1 to get the probability of at least two.

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An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object

Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

$1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2 + \frac{1}{2} at^2\\3) v = v_0 + at$

What do we already know?

$v_0 = 29.4 ms^-1$
The launch is straight up
$a = -9.81 ms^-2$ this is the gravitational acceleration $g$

$s_0 = 0 m$ , since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, $v = 0 ms^-2$

$v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m$

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

$s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2$

this is the equation of the whole trajectory that object makes.

and if you solve this by making $s = 0$ , you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

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