Suppose you have 6 dice in a bag. you draw a single die and roll it. one die is a standard fair die, numbered 1-6 two dice are have an extra pip drawn in the middle of the side with 4 pips, such that there is no 4 on these two dice, but there are two 5's (that is, these dice are numbered (1, 2, 3, 5, 5, 6)) three dice are loaded such that it is twice as likely that an even number is rolled than an odd what is the probability that you roll a 4? what is the probability that you roll a 5? you should input your answers in decimal form, rounded to the nearest thousandth. for example, if you find 0.0148 as an answer, you should enter "0.015".

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )

what statement? is there an image?

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)