We know that class starts 11:25 AM (be sure to note that there are only 60 minutes in an hour).
We know that class lasts 30 minutes.
First of all, let's find the time class ends.
11:25 AM (class starts) + 00:30 (minutes) = 11:55 AM (class ends)
Let's start with Yul's time first.
We know that Yul left when class was done, the time was already calculated when we added the time the class started with the time the class lasted and we got (in the previous line) 11:55 AM was when class ended, therefore, Yul left at 11:55 AM.
Sarah stayed an extra 5 minutes.
Class ended 11:55 AM. Simply add 5 minutes (00:05) to 11:55 AM.
11:55 AM (time class ended) + 00:05 (extra time to talk with the teacher) = 12:00 PM. Therefore, Sarah left at 12:00 PM.
As stated in the beginning of this problem, each hour only lasts 60 minutes, not 100 minutes, so every time the minutes are at the 60 tick mark, a new hour starts (in this case, 12:00 PM).
Yul left at 11:55 AM.
Sarah left at 12:00 PM.
<span>Jonathan’s class has 30 boys. 60% are girls
Then
30 x 60/100 =18
Answer:
There are 18 girls in Jonathan's class</span>
Answer:
yes.
simply subtract the coupons from the total.
56.12-9.85= $46.30
Answer:
10(7-4p)
Step-by-step explanation:
First, let's solve the inequality:
Subtracting 4 from all 3 terms, we get -6 < 3x < -2
Solving -6 < 3x results in x>-2, and solving 3x < -2 results in x < -2/3.
Ask: which integers satisfy x>-2? Answers: -1, 0, 1, 2, ....
Also ask: which integers satisfy x < -2/3? Answers: -5, -4, -3, -2, -1
Notice that the only integer that is found in both sets of possibilities is -1. So, only one integer x satisfies the given inequality.