An urn contains 10 red balls, 6 green balls, 15 orange balls, and 14 blue balls. if one ball is randomly drawn from the urn, what are the odds against the ball being red? state your answer as a ratio using a colon to separate the two numbers.

7:9

Step-by-step explanation:

Number of red balls = 10

Number of green balls = 6

Number of orange balls = 15

Number of blue balls = 14

Total numbers of balls =  45

Probability (odds against the red ball) = P( not red)

P(not red) = P(green) + P (orange) + P (blue)  (1)

P (green) = 6/45

P ( orange) =15/45

P (blue)  = 14/45

Substituting in (1)

P(not red) = 6/45 + 15/45+14/45

                =  35/45

                =  7/9

                7:9

question is incomplete.

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680