Ellen wishes to mix candy worth $1.45 per pound with candy worth $3.74 per pound to form 27 pounds of a mixture worth $3.06 per pound. How many pounds of the more expensive candy should she use?

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  18.98 pounds

Step-by-step explanation:

Let 'e' represent the number of pounds of expensive candy Ellen should use. Then 27-e is the number of pounds of less expensive candy. Her total cost is ...

  1.45(27 -e) +3.74e = 27(3.06)

  (3.74 -1.45)e = 27(3.06 -1.45) . . . . subtract 1.45(27)

  e = 27(3.06 -1.45)/(3.74 -1.45) = 27(1.61/2.29) ≈ 18.98

Ellen should use about 18.98 pounds of the expensive candy.

Additional comment

We showed the solution this way because it is the generic solution to any mix problem. The fraction of the mix that is the highest-value contributor is the ratio of the difference between the mix value and the lowest contributor to the difference between contributors. Here, that fraction is ...

(3.06 -1.45)/(3.74 -1.45) = (1.61/2.29)

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation:

do you have an easier question cause this shh is hard like theheck

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