Answer:
She payed $269 for the chainsaw :)
Let's assume
people were in the original group of winners as x
A group of winning ticket holders share equally 90,000,000 lottery
Before the money is divided, three more winning ticket holders are declared
so, total number of winning tickes =x+3
each persons share is reduced to 5,000,000
so, total money =5,000,000*(x+3)
and this must be equal to total money
![5000000(x+3)=90,000,000](https://tex.z-dn.net/?f=5000000%28x%2B3%29%3D90%2C000%2C000)
now, we can solve for x
and we get
![x+3=18](https://tex.z-dn.net/?f=x%2B3%3D18)
Subtrac both sides by 3
![x+3-3=18-3](https://tex.z-dn.net/?f=x%2B3-3%3D18-3)
![x=15](https://tex.z-dn.net/?f=x%3D15)
so,
number of people were in the original group of winners is 15 ...........Answer
Answer:
your answer will be<u> A.85°</u>
Step-by-step explanation:
hope it helps you
have a great day!!!
Answer:
0.632
Step-by-step explanation:
Given that a homeowner is three times as likely to purchase additional jewelry coverage as additional electronics coverage
If probability of purchasing additional electronics coverage = p, then prob of purchasing jewelry coverage = 3p
The two events are independent hence joint probability is product of these two.
i.e. P(both) = ![p(3p) =3p^2](https://tex.z-dn.net/?f=p%283p%29%20%3D3p%5E2)
This is given as 0.2
![3p^2 =0.2\\p = 0.258](https://tex.z-dn.net/?f=3p%5E2%20%3D0.2%5C%5Cp%20%3D%200.258)
the probability that a homeowner purchases exactly one coverage.
![=P(AUB)-P(A \bigcap B)](https://tex.z-dn.net/?f=%3DP%28AUB%29-P%28A%20%5Cbigcap%20B%29)
= Prob he purchases I + prob he purchases II-2(prob he purchases both)
![= 0.258+3(0.258)-2(0.2)\\=0.632](https://tex.z-dn.net/?f=%3D%200.258%2B3%280.258%29-2%280.2%29%5C%5C%3D0.632)