Several students are going to ride their bikes from the city park to Lake Kegonsa, a distance of 25 miles. They will ride at an average rate of 10 miles per hour. What is the latest time they can leave the park in order to arrive at the lake by 11

a. M.?

Question

Mathematics

Rylee Maynard

21 November, 05:16

Several students are going to ride their bikes from the city park to Lake Kegonsa, a distance of 25 miles. They will ride at an average rate of 10 miles per hour. What is the latest time they can leave the park in order to arrive at the lake by 11

a. M.?

+5

Answers (1)

Know the Answer?

Heather Gay · Answer 1

Let's answer this question step by step

Distance = d = 25 miles

Average speed = s = 10 miles/hour

The time it takes to reach Lake Kegonsa = t

Now,

d/t=s

d/s=t

Hence, t=25/10

t=2.5 hours

According to our caluclutions:

08:30 A. M is the latest time they can leave the park in order to arrive at the lake by 11 A. M.

Several students are going to ride their bikes from the city park to Lake Kegonsa, a distance of 25 miles. They will ride at an average rate of 10 miles per hour. What is the latest time they can leave the park in order to arrive at the lake by 11a. M.?

Several students are going to ride their bikes from the city park to Lake Kegonsa, a distance of 25 miles. They will ride at an average rate of 10 miles per hour. What is the latest time they can leave the park in order to arrive at the lake by 11

a. M.?