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.

Beans A= x
Beans B=(110-x)
0.08x+0.32(110-x) = 0.27(110)
Solve for x
0.08x+35.20-0.32x = 29.70
0.08x-0.32x=29.7-35.2
−0.24x=−5.5
X=5.5÷0.24
X=22.9 of Beans A

110−22.9=87.1 of Beans B

This is reasonable if one can weigh beans to the nearest tenth of a pound.

Hope it helps:-)

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 : )

Theoretical probability: Green

Step-by-step explanation:

25 total

5 red

7 blue

10 green

3 yellow

50 times

It would make sense that the 10 green marbles has the best chance of getting picked.  Since you didn't provide any information on the experimental probability, then I can only answer the theoretical one.