Three $1 bills, four $5 bills, and two $10 bills.
⇒total=3+4+2=9
p($5 then $10)=p($5)×p($10 after p($5))
p($5)=49 p($10)=28=14
p($5 then $10)=49×14=19
Answer:
The first 4 guests can arrive in 32,760 ways.
Step-by-step explanation:
The order in which the guests arrive is important. For example, A,B,C,D is a different outcome to B,A,C,D. So we use the permutations formula to solve this question.
Permutations formula:
The number of possible permutations of x elements from a set of n elements is given by the following formula:
![P_{(n,x)} = \frac{n!}{(n-x)!}](https://tex.z-dn.net/?f=P_%7B%28n%2Cx%29%7D%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-x%29%21%7D)
In this question:
4 guests from a set of 15. So
![P_{(15,4)} = \frac{15!}{(15-4)!} = 32760](https://tex.z-dn.net/?f=P_%7B%2815%2C4%29%7D%20%3D%20%5Cfrac%7B15%21%7D%7B%2815-4%29%21%7D%20%3D%2032760)
The first 4 guests can arrive in 32,760 ways.
Divide 4 and 5/8 by 5/8. you will get 7 and 2/5, or 7.4. since this you need whole pieces of wood, round down to 7.
final answer=7 pieces of wood
Solve for x by simplifying both sides of the equation, then isolating the variable.
x = -54
4420 people in 1993
5600 people in 1998
(5600 - 4420) = 1180....so they gained 1180 people in 5 years (from 1993 to 1998)
1180/5 = 236.....thats 236 people per year.
So in 2000.....which is 2 years from 1998, at 236 per yr =
5600 + 2(236) = 5600 + 472 = 6072 (population in 2000)