A grocer buys food from three suppliers. the suppliers deliver every five days, six days, and seven days. all three came today.
1 answer:
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Part A
<h3>Answer: T(n) = 58n+1800</h3>-------------
Work Shown:
1800 = initial cost or starting cost
58*1 = 58 = cost for one person
58*2 = 116 = cost for two people
58*n = cost for n people
58n+1800 = total cost
58n+1800 = T(n)
T(n) = 58n+1800
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Part B
<h3>Answer: D(T) = 0.85T</h3>-------------
Work Shown:
T = total bill
15% discount means you still pay 85% of the bill (85%+15% = 100%)
D = discounted cost
D = 85% of T
D = 0.85*T
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Part C
<h3>Answer: D(T(n)) = 49.3n + 1530</h3><h3>See attached image below for the table.</h3>-------------
Work Shown:
We combine parts A and B.
T(n) = 58n+1800 from part A
D(T) = 0.85*T from part B
D(T(n)) = 0.85*( T(n) )
D(T(n)) = 0.85*( 58n+1800 ) ... plug in T(n) = 58n+1800
D(T(n)) = 0.85*( 58n+1800 )
D(T(n)) = 0.85*(58n)+0.85*(1800) ... distribute
D(T(n)) = 49.3n + 1530
If we plugged in n = 0, then,
D(T(n)) = 49.3n + 1530
D(T(0)) = 49.3*0 + 1530
D(T(0)) = 1530
Or we could plug n = 0 into T(n) to get T(0) = 1800
Then plug T = 1800 into D(T) = 0.85*T to get D(1800) = 1530
Either way you'll get the same answer.
Repeat this (using either method) for n = 50, 100, 150, 200 and you'll get the table of values you see below in the attached image.
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Part D
The D( T(n) ) function allows us to find the discounted cost D for any number n of people that show up.
The input is n, where n is some positive integer less than or equal to 200 (because this is the reception hall's max capacity).
The output is the final discounted cost.
Example: n = 50 is one input which leads to its corresponding output of D( T(n) ) = 3995 as the table shows. So 50 people would have a discounted total cost of $3,995.
