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.
K = ln(ending amt/bgng amount)/6 years
k = ln(99,000,000/94,000,000) / 6 years
k = ln(
<span>
<span>
<span>
1.0531914894
</span>
</span>
</span>
) / 6 years
k =
<span>
<span>
<span>
0.0518250679
</span>
</span>
</span>
/ 6
k =
<span>
<span>
<span>
0.0086375113</span></span></span>
Final population = bgng*e^k*years
Where "e" is the mathematical constant 2.718281828
Final population = 94,000,000*e^<span>0.0086375113*12</span>
Final population = 94,000,000*e^<span><span><span>0.1036501356
</span>
</span>
</span>
Final population = 94,000,000*
<span>
<span>
<span>
1.1092123132
</span>
</span>
</span>
<span><span>
</span>
</span>
Final population =
<span>
<span>
<span>
104,265,957
</span>
</span>
</span>
480 + 192 = 672 students total
16 + 12 = 28 classrooms total
672/28 = 24 students per room
24 * 12 = 288 students that need to be in school B
288 - 192 = 96 students who have to transfer
