SOLUTION: Create an example of a rea life word problem which can be solved using agebraic inequaities. Write the ptoblem abd then solve the problem. SHow the agebraic inequaities that are in
Question 147211: Create an example of a rea life word problem which can be solved using agebraic inequaities. Write the ptoblem abd then solve the problem. SHow the agebraic inequaities that are involved. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Create an example of a real life word problem which can be solved using algebraic inequalities. Write the problem and then solve the problem. Show the algebraic inequalities that are involved.
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The classic cell phone plan problem
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Plan a: $40 a month includes 100 anytime minutes, and .10 a min after that
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Plan b: $60 a month includes 150 anytime minutes, but .05 a min thereafter
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How many minutes would you have to talk in one month to make plan b a better deal?
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Let m = no. of minutes talked
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Assuming you will exceed 150 min a month, we have
Plan a cost > plan b cost
.10(m - 100) + 40 > .05(m-150) + 60
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.10m - 10 + 40 > .05m - 7.5 + 60
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.10m + 30 > .05m + 52.5
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.10m - .05m > 52.5 - 30
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.05m > 22.5
m >
m > 450 min (if you exceed 450 min, plan b is the better deal)
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Prove that by saying you talked 451 min in a month
Plan a cost: .10(451-100) + 40 = $75.10
Plan b cost: .05(451-150) + 60 = $75.05
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