I believe that it would actually be when it would be a "<span>codominace" as when both of the rabbits would be have a part in this baby.</span>
Answer: 0.038
Step-by-step explanation:
Given: Total games : n= 375
Number of games won by the team that was winning the game at the end of the third quarter. = 300
The proportion (p) of the team that was winning the game at the end of the third quarter: 
Critical z-value for 90% confidence: z* = 1.645
Margin of error =
The margin of error in a 90% confidence interval estimate of p 
Hence, the required margin of error = 0.038
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:
For number 1: y=2 and x=3
