Will give more points if answered !

emily leaves her house at 8: 25 a. m to bike to her school, which is 3.42 miles away. when she passes the post office, which is 3/4 mile away from her home, she looks at her watch and sees that it is 30 seconds past 8: 29 a. m.

if emily's school starts at 8: 50 a. m., can emily make it to school on time without increasing her rate of speed? show/and explain the work necessary to support your answer.

From 8:25 to 30 seconds past 8:29 is 4.5 minutes.

This means she rode her bike 3/4 mile in 4.5 minutes.

Calculate her speed per mile:

1 mile / 0.75 = 1.333

4.5 minutes * 1.333 =  6 minutes per mile.

She needs to ride 3.42 miles, so multiply distance by speed:

3.42 miles x 6 minutes per mile = 20.52 minutes.

0.52 * 60 = 31.2 seconds.

This means it will take a total of 20 minutes and 31 seconds to get from her house to school.

She left at 8:25, add her travel time:

8:25 + 20 minutes and 31 seconds  = 8:45 and 31 seconds.

This is sooner than the time school starts so she makes it on time.

1

step-by-step explanation:

it's the only multiple of both and the least