Suppose you roll a pair of honest dice. if you roll a total of 7, you win $20; if you roll a total of 11, you win $50; if you roll any other total, you lose $10 find the expected payoff for this game. the expected payoff for this game is s (round to the nearest cent as needed)

- $ 1.67

Step-by-step explanation:

Since, when two dices are rolled then the total number of possible ways = 6 × 6 = 36,

The possible ways of getting a total of 7,

{ (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

So, the possibility of getting a total of 7 = =

The possible ways of getting a total of 11,

{ (5,6), (6,5) }

So, the probability of getting a total of 11 = =

Now, other possible rolls = 36 - 6 - 2 = 36 - 8 = 28,

So, the probability of getting the sum of numbers other than 7 or 11 = =

Since, for the sum of 7, $ 20 will earn, for the sum of 11, $ 50 will earn while for any other total loss is $ 10,

Hence, the expected value for this game =

≈ -$ 1.67

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.