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GenaCL600 [577]
2 years ago
14

A long distance runner started on a course running at an average speed of 6mph. half an hour later, a second runner began the sa

me course at an average speed of 7mph. how long after the second runner started will the second runner overtake the first runner?
Mathematics
1 answer:
Stells [14]2 years ago
6 0

<u> </u><u>9 hours</u><u> long after the second runner started will the second runner overtake</u><u> the first runner.</u>

What does it mean to do speed?

  • Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement.
  • Put another way, speed is a scalar value, while velocity is a vector.

Runner I = 6 mph

Runner II = 7 mph

Difference in time= 90 minutes

Train A will have covered 9 miles before Runner II starts

catch up distance= 9 miles

catch up speed = 7-6 mph

catch up speed = 1 mph

Catchup time = catchup distance/catch up speed

catch up time= 9/1

catch up time= 9 hours

Learn more about speed

brainly.com/question/14964958

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it already is in its simple form.

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Hopefully I helped answer your question.

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3 years ago
I'm giving brainliest to the most helpful answer. For 50 points:
Alenkinab [10]

Answer:

1) \displaystyle\frac{5}{18}\approx27.78\%

2) \displaystyle\frac{4}{7}\approx57.14\%

3) \displaystyle  \frac{13}{16}=81.25\%

Step-by-step explanation:

We are given a two-way frequency table. Using the table, we will determine the probability for each case.

Question 1)

P(A Student With A Part Time Job Without A Car)

Using the first column, the total number of students that have a part time job is 78+30=108.

Likewise, using the first column, we can see that out of those 108 students, 30 do not have a car.

Hence, the probability that a student with a part time job without a car is:

\displaystyle P=\frac{30}{108}=\frac{5}{18}\approx27.78\%

Question 2)

P(No Car | Does Not Have A Part Time Job)

Remember that the vertical line means conditional probability.

So, we want the probability of a student having no car given that they do not have a part time job.

Using the second column, we can see that the total number of students that do not have a part time job is 18+24=42.

Likewise, using the second column, 24 do not have a car.

Hence, the probability that a student with a part time job without a car is:

\displaystyle P=\frac{24}{42}=\frac{4}{7}\approx57.14\%

Question 3)

P(Part Time Job | Car)

So, we want to probability of a student having a part time job given that they have a car.

Using the first row, the total students that have a car is 78+18=96.

And of those 96 students, 78 have a part time job.

Hence, the probability that a student with a car has a part time job is given by:

\displaystyle P=\frac{78}{96}=\frac{13}{16}=81.25\%

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Answer:

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