Agame of “doubles-doubles” is played with two dice. if a player rolls doubles, the player earns 3 points and gets another roll. if the player rolls doubles again, the player earns 9 more points. how many points should the player lose for not rolling doubles in order to make this a fair game? mc011-1.jpg mc011-2.jpg mc011-3.jpg 1

solution:

one double = 5/ 36

two doubles = 1/36

no doubles = 1/6

in a standard form of equation, it will look like this:

12/36 + 5[(3-x)/36] - (5)(x/6) = 0

next, get the value of x, so this will give you the answer of 27/35 that is the probability of not rolling doubles.

3< x< 5

step-by-step explanation:

|x - 4| < 1

there are two solutions, one positive and one negative.   remember to flip the inequality when taking the negative solution

x-4 < 1       x-4 > -1

add 4 to each side

x-4+4 < 1+4       x-4+4 > -1+4

x< 5                 x> 3

3< x< 5

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction