CPhill

P(1 three)  =  C(10, 1) (1/6) (5/6)^9  ≈  .323  ≈   32.3%

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, CC....that's a great explanation   !!!!

Good job, tertre   !!!

Correct !!!

81^(2x)  = 9^(x + 4)

(9^2)^(2x) = 9^(x + 4)         (3^4)^(2x) =   (3^2)^(x + 4)   

9^(4x) = 9^(x + 4)                 3^(8x) = 3^(2x + 8)

6 = ab^1 ⇒  6 = ab  ⇒   6/a = b     (1)

d = ab^3    (2)

Sub (1) into (2)  for   b    and we have

d = a(6/a)^3

d = 216/a^2

a^2  =  216/d

a = √[216/d]  =  6√[6/d]   

And b =   6 / ( 6 √[6/d] )   =   1 / √[6/d]  =    √[d / 6]  =  √d / √6   =  √ [6d ] / 6

So....the function is

y =  6√[6/d ] * [√ [ 6d ] / 6 ]^x   

100 = ab^(-2)      ⇒ 100 =  a / b^2   (1)

.8 = ab^1    ⇒  .8 = ab   ⇒  .8/b  = a   (2)

Sub (2) into (1)  for a

100 =  .8 / b^3

b^3 =  .8/100

b^3 = (8/10)(1/100)

b^3  = 8/1000

b =  2 /10  =  .2

And

.8/.2 = a  = 4

So....the function is

y = 4(.2)^x