Finding an irrational number between which given pair of numbers supports the idea that irrational numbers are dense
1 answer:
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Answer:
(A)Cost of Rental A, C= 15h
Cost of Rental B, C=5h+50
Cost of Rental C, C=9h+20
(B)
i. Rental C
ii. Rental A
iii. Rental B
Step-by-step explanation:
Let h be the number of hours for which the barbeque will be rented.
Rental A: $15/h
- Cost of Rental A, C= 15h
Rental B: $5/h + 50
- Cost of Rental B, C=5h+50
Rental C: $9/h + 20
- Cost of Rental C, C=9h+20
The graph of the three models is attached below
(b)11.05-4.30
When you keep the barbecue from 11.05 to 4.30 when the football match ends.
Number of Hours = 4.30 -11.05 =4 hours 25 Minutes = 4.42 Hours
- Cost of Rental A, C= 15h=15(4.42)=$66.30
- Cost of Rental B, C=5h+50 =5(4.42)+50=$72.10
- Cost of Rental C, C=9h+20=9(4.42)+20=$59.78
Rental C should be chosen as it offers the lowest cost.
(c)11.05-12.30
Number of Hours = 12.30 -11.05 =1 hour 25 Minutes = 1.42 Hours
- Cost of Rental A, C= 15h=15(1.42)=$21.30
- Cost of Rental B, C=5h+50 =5(4.42)+50=$57.10
- Cost of Rental C, C=9h+20=9(4.42)+20=$32.78
Rental A should be chosen as it offers the lowest cost.
(d)If the barbecue is returned the next day, say after 24 hours
- Cost of Rental A, C= 15h=15(24)=$360
- Cost of Rental B, C=5h+50 =5(24)+50=$170
- Cost of Rental C, C=9h+20=9(24)+20=$236
Rental B should be chosen as it offers the lowest cost.
