All Answers 358187 Answers

P( no fives)  =  (5/6)^9

P( 1 five)  = C(9,1)(1/6)(/6)^8

P(2 fives) = C(9,2) ( 1/6)^2 ( 5/6)^7

P(3 fives) =  C(9,3) (1/6)^3(5/6)^6

P(4 fives) = C(9,4)(1/6)^(4)(5/6)^5

So....the probability is just the sum of these

(5/6)^9 + C(9,1)(1/6)(5/6)^8   + C(9,2)(1/6)^2 (5/6)^7 + C(9, 3)(1/6)^3(5/6)^6 + C(9, 4) (1/6)^4 (5/6)^5

≈ .99   ≈  99%

A = { 1, 3, 5 , 7 }

B = { 1, 2, 4, 8}

A∩ B    = { 1 }

 H ⊂ G   means that H is a proper subset of G

H = {1, 2, 4 }

G = {1, 2, 3, 4, 6, 12 }

So....this is true

The first three are  decimal  values  between 0 and 1, inclusive.....these are all valid

D is not

thanks!

Flip the x and the y, so it becomes \(x=3y+12\)

Then just solve for y

\(x-12=3y+12-12\)

\(\frac{x-12}{3}=\frac{3y}{3}\)

\(\frac{x}3-4=y\)

So the inverse is \(\frac{x}3-4=y\)

27^(1/3) x - 81 - 5 =

3x - 86

y=3x+12    Solve for 'x'

(y-12)/3 = x       then switch the 'x' and 'y's

(x-12)/3 = y

or    y = 1/3 x -4

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