Suppose thatann selects a ball by first picking one of two boxes at random and then selecting a ball from this box. the first box contains three orange balls and four black balls, and the second box contains five orange balls and six black balls. what is the probability that ann picked a ball from the second box if she has selected an orange ball?

35/68

Step-by-step explanation:

The first box contains 3 orange balls and 4 black balls.

The second box contains 5 orange balls and 6 black balls

Total number of boxes = 2

Let Pr(F) be the probability that the first box is picked.

Let Pr(S) be the probability that the second ball is picked

Let O be the orange ball selected

Pr(F) = 1/2

Pr(S) = 1/2

Pr(O|S) = n(O) / n(S)

= 5/ 6+5

5/11

Pr(O|F) = n(O) / n(F)

= 3/ 4+3

= 3/7

Pr(S|O) = [Pr( O|S).P(S)] / [Pr( O|S).P(S) + Pr( O|F).P(F)]

= (5/11*1/2) / (5/11*1/2) +(3/7*1/2(

= (5/22) / (5/22 + 3/14)

= (5/22) / (136/308)

= 35/68

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

c

step-by-step explanation:

it would be see because your formula is f(x)=6x, so 6 times 1 is 6 and you go on and the best answer is c.