The best prediction for the number of regular spectators that the program will have in 6 years is obtained with the following formula:
Pf = Po×(1-b) ^ t
Where:
Pf = final population
Po = initial population
b = decrease rate
t = timpo in years
So:
Pf = 223.00×(1-0.047) ^ 6
Pf = 167,056
The answer is the last option.
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:
530.66
Step-by-step explanation:
A = πr²
13 x 13 = 169
169 x 3.14 = 530.66
Let me know if this helps!
Answer: Scaline?
Step-by-step explanation:
The first one is 23670-8200=15470 and 8200
and the second answer is 32785-8200=24585 of taxable income so total tax due is 3314
hope this helps! Thank You!