If a silver alloy that costs $6.75 an ounce is going to be mixed with 55 ounces of a silver
alloy that costs $10 an ounce to make a
mixture that costs $8 an ounce, how many
ounces of the $6.75 an ounce alloy must be
used?

9514 1404 393

  88 ounces

Step-by-step explanation:

Let x represent the number of ounces of the less expensive alloy in the mix. Then the cost of the mix will be ...

  6.75x +10(55) = 8(55+x)

  110 = 1.25x . . . . . . subtract 440+6.75x

  88 = x . . . . . . . . divide by 1.25

88 ounces of $6.75/oz silver must be used.

i dont know it

step-by-step explanation: