Abox contains four black pieces of cloth, two striped pieces, and six dotted pieces. a piece is selected randomly and then placed back in the box. a second piece is selected randomly. what is the probability that: (a) both pieces are dotted? (b) the first piece is black and the second piece is dotted? (c) one piece is black and one piece is striped?

a) ⇒

b) ⇒  

c) ⇒  

Step-by-step explanation:

Data provided in the question:

Total Number of pieces = 4 + 2 + 6 = 12

P( Black piece )  =

P( Striped piece ) =

P( Dotted piece ) =

Now,

a) P(Both the pieces are dotted) = P( Dotted piece ) × P( Dotted piece )

⇒  ×  

⇒  

⇒

b) P(the first piece is black and the second piece is dotted)

= P( Black piece ) × P( Dotted piece )

⇒    ×  

⇒  

⇒  

c) P(one piece is black and one piece is striped)

= P( Black piece ) × P( Striped piece )

⇒  ×  

⇒  

⇒  

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !