Design a combinational circuit that counts the coins placed into an automatic toll coin collector (only nickels, dimes, and quarters). the toll is $0.10 and the machine gives no change. when at least $0.10 are received the go-signal is turned on and the money collected (no change is given). otherwise the stop-light remains on.

10*d+25*q=1480, where d is the number of dimes and q is the number of quarters.If she has 76 coins total, then: 76=q+d Solving for d yields d=76-q, which we substitute into the first equation: 10*(76-q)+25*q=1480 Simplifying and solving for q: 760-10*q+25*q=1480 15*q=720 q=48 Substituting back into the second equation: d=76-48=28 Check solution: 10*28+25*48=1480 Maribel had 28 dimes and 48 quarters.

answer:

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