Bunuel wrote:

If Ted brought n baseball cards costs $3 each, then brought twice as many cards baseball cards at $9 each, then the average (arithmetic mean) cost, in dollars per baseball card, is equal to

A. 4n

B. 9

C. 7n

D. 6/n

E. 7

Average cost in dollars = \(\frac{TotalCost}{TotalQty}\)

Total Cost = (Qty\(_1\))(Cost\(_1\)) + (Qty\(_2\))(Cost\(_2\))

(Qty\(_1\))(Cost\(_1\)): \(n\) baseball cards that cost $3 each = n($3)

(Qty\(_2\))(Cost\(_2\)): twice as many cards (as \(n\)) that cost $9 each = 2n($9)

Total Cost in dollars: n(3) + 2n(9)

Total Qty: (n + 2n) = 3n

Average cost in dollars =

\(\frac{n(3) + 2n(9)}{3n}=\frac{3n+18n}{3n}=\frac{21n}{3n}=7\)

Answer E

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