1.
If no changes are made, the school has a revenue of :
625*400$/student=250,000$
2.
Assume that the school decides to reduce n*20$.
This means that there will be an increase of 50n students.
Thus there are 625 + 50n students, each paying 400-20n dollars.
The revenue is:
(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)
3.
check the options that we have,
a fee of $380 means that n=1, thus
250(n+50)(20-n)=250(1+50)(20-1)=242,250 ($)
a fee of $320 means that n=4, thus
250(n+50)(20-n)=250(4+50)(20-4)=216,000 ($)
the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.
Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.
Answer:
The answer is 52
Step-by-step explanation:
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Answer:
There will be 213 songs in each group.
Step-by-step explanation:
Suppose, the number of songs in each group is 
As he has 4 different groups of songs, so the t<u>otal number of songs in all 4 groups</u> will be: 
Given that, the mp3 player has a <u>total of 852 songs</u>. So, the equation will be........
So, there will be 213 songs in each group.
Answer:
29
Step-by-step explanation:
