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elixir [45]
1 year ago
13

2. Starting at a fixed time, each car entering an intersection is observed to see whether it turns left (L), right (R), or goes

straight ahead (A). The experiment terminates as soon as a car is observed to turn left. Let X= the number of cars observed. What are possible X values?
Given the outcomes below, find their associated X values.
Outcome: RRL AARRL AARL RRAL ARL
x= ____ ____ ____ ____ ____
Mathematics
1 answer:
jasenka [17]1 year ago
7 0

Answer:

Natural numbers (integers greater than zero)

X = 3,  5,  4,  4,  3

Step-by-step explanation:

The least number of cars that can be observed in this experiment is 1, if the first car turns left. On the other hand, the experiment could go on forever if no car ever turns left, thus the highest number of cars approaches infinite.

The possible values of X are integers greater than zero, which are known as the Natural numbers.

If X = number of cars observed, simply count the number of letters in each outcome for the value of X:

Outcome = RRL, AARRL, AARL, RRAL, ARL

            X = 3,  5,  4,  4,  3

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Let us assume uniform width = x cm wide.

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Subtracting 3500 from both sides, we get

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FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

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Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula =\frac{-b\pm \sqrt{b^2-4ac}}{2a}, we get

=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}

x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)

x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)

We cant take negative value.

So, x=5\left(\sqrt{71}-6\right)=12.13

We could take approximately 12 cm.



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