A jawbreaker candy machine has 25 red jawbreakers, 30 green jawbreakers, 15 yellow
jawbreakers, and 20 blue jawbreakers. If Loretta puts 1 coin in the machine and gets 1 jawbreaker,
what is the probability the jawbreaker is NOT green?

2/5
1/3
2/3
1/60

Work Shown:

A = # of candies that are not green

A = (# of red)+(# of yellow)+(#  of blue)

A = 25+15+20

A = 60

B = # of candies total

B = (# of red)+(# of green)+(# of yellow)+(# of blue)

B = 25+30+15+20

B = 90

P(not green) = probability of selecting candy that is not green

P(not green) = (# of candies not green)/(# of candies total)

P(not green) = A/B

P(not green) = 60/90

P(not green) = (2*30)/(3*30)

P(not green) = 2/3

answer:   (s+1)(s+4)

step-by-step explanation:

factor s2+5s+4

s2+5s+4

=(s+1)(s+4)

thus, the first answer choice, y = sec x, is the correct one