Acoffee mixture contains beans that sell for $0.08 per pound and $0.32 per pound. if 110 pounds of beans create a mixture that sells for $0.27 per pound, to the nearest tenth of a pound, how much of each bean is used in the mixture? model the scenario with an equation and solve. use complete sentences to explain whether or not your solution is reasonable.

0.08x + 0.32(110 - x) = 0.27(110)

0.08x + 35.20 - 0.32x = 29.7

           -35.2                   -35.2

0.08x - 0.32x = -5.5

-0.24x = -5.5

/-0.24    /-0.24

x = 22.9 pounds of beans that sell for $0.08 per pound

110 - 22.9 = 87.1 pounds of beans that sell for $0.32 per pound

This is reasonable because the beans can be weighed to the nearest tenth.

Please rate and thank : )

19

step-by-step explanation:

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497