Let x be the no. of oranges and y be the no. of tangerine.
From question, we get,
4y = x
=> 4y = 8260
=> y = 8260/4
=> y = 2065
Therefore, the farmer shipped 2065 tangerines
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.
F(n) = 75*1/5n
This is because the base value would be 75 and it is divided by 5 each part of the sequence.
Answer:
Avg rate of change =
[ f(14 ) - f(8) ] / [ 14 - 8 ] =
[ ( √ [14 - 4] ) - (√ [8 - 4]) ] / 6 =
[ √10 - √4 ] / 6 =
0.19
The answer is 14.58.
I hope this helped, if not, please leave a comment below.