On her way to work, a commuter encounters four traffic signals. Assume that the distance between each of the four is sufficiently great that her probability of getting a green light at any intersection is independent of what happened at any previous intersection. The first two lights are green for forty seconds of each minute; the last two, for thirty seconds of each minute. What is the probability that the commuter has to stop at least three times?

7 in 36 or 0.1944

Step-by-step explanation:

The probability of having to stop at least three times is the probability of getting 3 or 4 red lights.

For the first two lights, the probability of getting them red is 20 in 60 (1/3).

For the last two lights, the probability of getting them red is 30 in 60 (1/2).

The probability of all of them being red is:

The probability of three of them being red (3 red + 1 green) is:

Therefore, the probability of at least three red lights is:

The probability is 7 in 36 or 0.1944.

answer: -8.2562234e+13

54 kilometers per hour

step-by-step explanation:

according to rules:

1 km = 1000 meter

which means that 1 meter = 1/1000 km

1 hour = 3600 second [ as 1 hour = 60 min = 60 *60 seconds ]

which means that 1 second = 1/3600 hour

therefore, 1 kilometer/hour =

thus to convert 15 meter per second to kilometer per hour, we multiply 15 to 3.6 kilometer/hour.

15 meter/second=   which is the required answer.