You have a jar of jelly beans in front of you

with 12 - lime, 17 - papaya, 5 - mango and 13

- bubble gum. What is the probability, as a

fraction, of selecting either a lime or bubble

gum followed by a papaya?

The probability of selecting either a lime or bubble gum followed by a papaya is 425/2209

Step-by-step explanation:

Since, there are four types of items in the jar, namely- lime, papaya, mango and bubble gum that have quantities of 12, 17, 5 and 13 respectively. Before finding the probabilities, we first add up the quantities or number of different items in the jar.

12 + 17 + 5 + 13 = 47 in total.

Since lime is 12 in number, the probability of selecting a lime is = 12/47.

Bubble gums are 13 in number, so the probability of selecting a bubble gum = 13/47.

Then the probability of selecting either a lime or a bubble gum is the addition of the probabilities of selecting a lime and selecting a bubble gum:

= 12/47 + 13/47

= 25/47.

Again, the probability of selecting a papaya is =

number of papaya/total number of items

= 17/47

If the system is with replacement, then the probability of selecting a lime or bubble gum followed by a papaya is:

= Probability of selecting a lime or bubble gum × Probability of selecting papaya

= 25/47 × 17/47

= 425/2209

The probability of selecting a lime or bubble gum followed by a papaya is 425/2209

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18