Imagine that a traffic intersection has a stop light that repeatedly cycles through the normal sequence of traffic signals (green light, yellow light, and red light). In each cycle the stop light is green for 30 s, yellow for 3 s, and red for 50 s. Assume that cars arrive at the intersection uniformly, which means that in any one interval of time, approximately the same number of cars arrive at the intersection at any other time interval of equal length. Determine the probability that a car arrives at the intersection while the stop light is yellow. Give your answer as a percentage precise to two decimal places.

The required probability will be "3.6%". A complete solution is provided below.

According to the question,

Throughout the each cycle, the light stop for different lights such as:

→ The total time for three lights will be:

=

=

hence,

→ The probability that a car arrives while the stop light is yellow will be:

=

=

= (%)

Thus the above answer is right.

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3.61%

Step-by-step explanation:

To know the statistics we have to realize that there are 3 options.

green 30s

yellow 3s

red 50s

As each color does not have the same time, we will take the seconds to solve it

first we calculate all the seconds of a cycle

30s + 3s +50s = 83s

the car that arrives will always arrive in some second of those 83 seconds

now of those 83 seconds only 3 seconds the traffic light will be yellow

we make the events that meet the condition on the possible

3s/83s = 3/83

to express it as a percentage, we divide it by 83 and multiply it by 100

100*(3/83)/83 = 3.614457831%

we have to round with two decimal

3.61%

you have 6 cupcakes

x=17

step-by-step explanation:

solve 3x-24=x+10

then that makes it 2x=34 because you subtract an x from both sides to get rid of the one on the right, and then you add 24 to both sides to get rid of the 24 on the left

divide by both sides by 2 to get your answer so that you can get your answer