Which game isn't fair for two players to play? a number cube is rolled. player 1 gets a point for the numbers 3 or less, and player 2 gets a point for the numbers 4 or higher. a coin is flipped. player 1 gets a point for heads side up, while player 2 gets a point for tails side up. a number cube is rolled. player 1 gets a point for an even number, and player 2 gets a point for an odd number. a number cube is rolled. player 1 gets a point for any number that is 3 or higher, and player 2 gets a point for any number less than 3.

wholesale price is the initial price before markup. retail price comes after the markup.

a price of 8 bucks that's marked up by 100%   is

formula: initial price + (percent markup in fraction/decimal)(initial price)

8+(1)(8)

8+8

16

retail price is $16.

so then what are you asking?

step-by-step explanation:

take the logarithm of both sides (the base of the log doesn't matter):

apply the exponent property - this says that :

distribute the log terms on both sides:

group like terms together:

divide through both sides by the coefficient on :

you can do some additional rewriting here using the properties of the logarithm to simplify things if you like:

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)