255 minutes
<em><u>we </u></em><em><u>know </u></em><em><u>that </u></em>
<em><u>1</u></em><em><u> </u></em><em><u>hour </u></em><em><u>=</u></em><em><u> </u></em><em><u>6</u></em><em><u>0</u></em><em><u> </u></em><em><u>minutes</u></em>
<em><u>then </u></em><em><u>according</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>unitary </u></em><em><u>method</u></em>
<em><u>4</u></em><em><u>*</u></em><em><u>1</u></em><em><u>/</u></em><em><u>4</u></em><em><u> </u></em><em><u>hours </u></em><em><u>=</u></em><em><u> </u></em><em><u>1</u></em><em><u>7</u></em><em><u>/</u></em><em><u>4</u></em><em><u>*</u></em><em><u>6</u></em><em><u>0</u></em>
<em><u>=</u></em><em><u> </u></em><em><u>2</u></em><em><u>5</u></em><em><u>5</u></em><em><u> </u></em><em><u>minutes</u></em>
<em><u>hope</u></em><em><u> it</u></em><em><u> helps</u></em>
Okie dokie gracias por los puntos
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Answer:
5 seconds
Step-by-step explanation:
Suppose the front parts of the trains meet at point A. Since both are the same length and traveling the same speed, each will pass point A in time ...
time = distance/speed
time = (1/18 mi)/(40 mi/h) = (1/720 h) × (3600 s)/(1 h) = 5 s
That is, the rear part of each train will be at point A 5 seconds after the front part.
The rear parts will pass each other 5 seconds after the front parts meet.
Answer:
There are 208 male students in the eight grade.
Step-by-step explanation:
320 students all together - 35% female students= 208 male students
