A slot machine has 3 dials. Each dial has 30 positions, one of which is "Jackpot". To win the jackpot, all three dials must be in the "Jackpot" position. Assuming each play spins the dials and stops each independently and randomly, what are the odds of one play winning the jackpot?

1/30 = 0.03 or 3%

3/(30+30+30) = 3/90 = 0.33 or 33%

3/(30×30×30) = 3/27000 = 0.0001 or 0.01%

1/(30×30×30) = 1/27000 = 0.00003 or 0.003%

3/(30×30×30) = 3/27000 = 0.0001 or 0.01%

Step-by-step explanation:

here we have 3 dials in the slot machines and each dial has 30 position

so probability of getting a number of the each dial   =  

30

1

since all dials are independent so probability of getting same number would be

P ( getting jack pot) =  

30

1

30

1

30

1

   =    

27000

1

   = 0.003% option D is correct.

Step-by-step explanation:

x = 12

y = 15

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.