Debra is buying prizes for a game at her school's fundraiser. the game has three levels of prizes, and she has already bought the second and third prizes. she wants the first prize to be nice enough to attract people to the game. the game's manufacturer has supplied her with the probabilities of winning first, second, and third prizes. tickets cost $3 each, and she wants the school to profit an average of $1 per ticket. how much should she spend on each first prize? (round your answer to two decimal places.)
prize cost of prize probability
1st $ .15
2nd $1.15 .30
3rd $0.74 .45

$ 9.22

Step-by-step explanation:

1st cost = $0.15 and p is unknown

2nd cost = $1.15 and Probability of 0.3

3rd cost = $0.74 and Probbility of 0.45

1st prize value variable is = p

to gain $1 on each ticket the gain for  the ticket purchaser must be $-1

we have that the probabilities of winning  add to 0.9  so, the probability of losing $3 is 0.1

formula:

Probability of winning:

W(p) = 0.15p + 0.3*1.15 + 0.45*0.74 - 3*0.1 = -1

0.15p + 0.3825 = -1

0.15p = - 1.3825

p = - 9.2166

p = -9.22

The 1st prize must be $ 9.22

area of rectangle a is 2 times that of rectangle b that means 2x+x =   36cm^2

3x = 36cm ^2

x = 36/3

x= 12cm^2

and 2x = 24cm^2

now side of square = are divided by sides = 36/4 = 6cm

that means length of rectangle a and b is 6cm too

now for rectangle a,

l*b = area

= 6*y=24

y=24/6 = 4

therefore dimensions of rectangle a are- length = 6cm, breadth = 4cm

and for rectangle b, length = 6cm and breadth = 6-4 = 2cm.

hope this .

i believe that the answer is y+6=1/3(x+2)