One of the wagers in roulette is to bet that the ball will stop on a number that is a multiple of 3. (both 0 and 00 are not included.) if the ball stops on such a number, the player wins double the amount bet. if a player bets $2, compute the player's expectation.

The players expectation is -0.1052

Step-by-step explanation:

Consider the Roulette numbers to from 1-36, also the two green zeroes 0 and 00,

Thus, there are 38 places on which the ball can land.

On first 36 numbers, 12 of them are divisible by 3. Also there are 26 numbers which are not divisible by 3.

The probability of a wager winning is: 12/38 or 6/19

Now the probability of a wager to loss is: 26/38 or 13/19

So for a 6/19 chance of winning $4 and a 13/19 chance of losing $2.

Therefore, the expected outcome is:

4 × 6/19 + -2 × 13/19 = 1.2632 - 1.3684 = -0.1052

The expected value comes up with a negative sign. Hence, he will lose money.

The players expectation is -0.1052

the answer is a

step-by-step explanation:

39.00 x 24 = total installment price

carrying installment price - cash price

needed to save the .00/

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl