Aradio station runs a promotion at an auto show with a money box with 10 $50 tickets, 10 $25 tickets, and 12 $5 tickets. the box contains an additional 20 "dummy" tickets with no value. three tickets are randomly drawn. find the probability that all three tickets have no value.

P = 57/1105

P = 0.052

Step-by-step explanation:

Given:

Total number of dummy tickets = 20

Total number of tickets with value = 10 + 10 + 12 = 32

Total number of tickets in the box = 20 + 32 = 52

The probability of picking 3 tickets randomly from the box without replacement and the three are dummy (with no value) is;

P = 20/52 × 19/51 × 18/50

(As each one is picked it reduces the total number of tickets and also the number of dummy tickets by one)

P = 57/1105

P = 0.052

  21.3

step-by-step explanation:

multiplying the given equation by c/sin(25°), we get

  c = 9/sin(25°) ≈ 21.2958 ≈ 21.3

we assume "c" is the length of ab, the side opposite angle c in triangle abc.

after simplifying it i got: -36.25 as my answer. hope that !