Answer:
A= 27500 seats, B= 14200 seats, C= 13300 seats
Step-by-step explanation:
a=b+c
a+b+c=55000 seats
30a+24b+18c=1405200 dollars
substitue a for b+c
b+c+b+c=55000
30(b+c)+24b+18c=1405200
seats 2b+2c=55000 --> divide by 2 on both sides b+c=27500 --> c=27500-b
dollars 30b+30c+24b+18c=1405200 --> 54b+48c=1405200 divide by 6 --> 9b+8c=234200
substitute c for 13750-b
9b+8(27500-b)=234200
9b+220000-8b=234,200
b=14,200 seats
c=27500-b=27500-14200=13300 seats
a=55000-b-c=55000-27500=27500 seats
Answer:
p>-1 or p<-19
Step-by-step explanation:
4p-10<6p-8 (add 10 to each side)
4p< 6p+2 (subtract 6p from each side)
-2p<2 (divide by -2 and switch the sign)
p>-1
10p+16<9p-3 (subtract 16 from each side)
10p<9p+-19 (subtract 9p from each side)
p<-19
Answer:
153 times
Step-by-step explanation:
We have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14
Width = 0.14
ME = 
ME = 
ME = 

use p = 0.5
z at 95.8% is 1.727(using calculator)





So, Option B is true
Hence we have to flip 153 times the coin in order to obtain a 95.8% confidence interval of width of at most .14 for the probability of flipping a head