A jar contains 11 red marbles, 12 blue marbles, and 6 white marbles. Four marbles from this jar are selected, with each marble being replaced after each selection. What is the expected number of draws until the first red marble? Since about half of the marbles are red, 2 draws is expected until a red marble is drawn.
Since there are 3 colors of marbles, 3 draws is expected until a red marble is drawn.
Since StartFraction 11 Over 29 EndFraction of the marbles are red, StartFraction 11 Over 29 EndFraction draws are expected until a red marble is drawn.
Since StartFraction 11 Over 29 EndFraction of the marbles are red, StartFraction 29 Over 11 EndFraction draws are expected until a red marble is drawn.

final total: 63

step-by-step explanation:

60 times sales tax

60 times 0.05

3

60 + 3

63

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

the answer is:

the correct option is the option b.

to calculate the surface area of the right cylinder, we need to use the following formula:

from the picture we know that:

so, substituting and calculting we have:

hence, we have that the correct option is the option b.

have a nice day!