Multiplying ,you get 18x^2+21x-24x-28. You simplify to get:
18x^2-3x-28.
Given:
Normal price of a tv = $200
Coupon = 25% off
To find:
The money saved by Katherine.
Solution:
Katherine buys a tv with a normal price of $200 and she has a 25% off coupon. It means, the money saved by Katherine is 25% of normal price of tv, i.e., $200.




Therefore, the money saved by Katherine is $50.
Answer:
Adult tickets sold= 9x4 = 36
Senior tickets sold = 9x3= 27
Child tickets sold = 9 x 1 = 9
Step-by-step explanation:
If the parts for adult tickets changed within the 2 ratios given, we should have equated Ratio further. But here that is not 5e case. So, put both ratios together:
Senior:Adult:Child
3:4:1
Now,
Total number of parts 3+4+1=8
Number of tickets sold per part = 72/8 = 9
Therefore number of
Adult tickets sold= 9x4 = 36
Senior tickets sold = 9x3= 27
Child tickets sold = 9 x 1 = 9
Hope this helps.
Good Luck
Answer: 1) 0.6561 2) 0.0037
Step-by-step explanation:
We use Binomial distribution here , where the probability of getting x success in n trials is given by :-

, where p =Probability of getting success in each trial.
As per given , we have
The probability that any satellite dish owners subscribe to at least one premium movie channel. : p=0.10
Sample size : n= 4
Let x denotes the number of dish owners in the sample subscribes to at least one premium movie channel.
1) The probability that none of the dish owners in the sample subscribes to at least one premium movie channel = 

∴ The probability that none of the dish owners in the sample subscribes to at least one premium movie channel is 0.6561.
2) The probability that more than two dish owners in the sample subscribe to at least one premium movie channel.
= ![P(X>2)=1-P(X\leq2)\\\\=1-[P(X=0)+P(X=1)+P(X=2)]\\\\= 1-[0.6561+^4C_1(0.10)^1(0.90)^{3}+^4C_2(0.10)^2(0.90)^{2}]\\\\=1-[0.6561+(4)(0.0729)+\dfrac{4!}{2!2!}(0.0081)]\\\\=1-[0.6561+0.2916+0.0486]\\\\=1-0.9963=0.0037](https://tex.z-dn.net/?f=P%28X%3E2%29%3D1-P%28X%5Cleq2%29%5C%5C%5C%5C%3D1-%5BP%28X%3D0%29%2BP%28X%3D1%29%2BP%28X%3D2%29%5D%5C%5C%5C%5C%3D%201-%5B0.6561%2B%5E4C_1%280.10%29%5E1%280.90%29%5E%7B3%7D%2B%5E4C_2%280.10%29%5E2%280.90%29%5E%7B2%7D%5D%5C%5C%5C%5C%3D1-%5B0.6561%2B%284%29%280.0729%29%2B%5Cdfrac%7B4%21%7D%7B2%212%21%7D%280.0081%29%5D%5C%5C%5C%5C%3D1-%5B0.6561%2B0.2916%2B0.0486%5D%5C%5C%5C%5C%3D1-0.9963%3D0.0037)
∴ The probability that more than two dish owners in the sample subscribe to at least one premium movie channel is 0.0037.