Mathematics Standard • Year 12 • Ratios and rates • Lesson 5
Populations and Power Bills, Application
Apply capture-recapture in context and say what its assumptions cost you, then run a power rating all the way through to a quarterly bill and back again.
1. Estimating a population
Show the formula before you substitute. 2 marks each
Q1.1 A ranger tags 55 wallabies. A later survey catches 66 wallabies, of which 11 are tagged. Estimate the population.
Q1.2 90 fish in a lake are tagged. Later 120 fish are caught and 18 are tagged. Estimate the population.
2. When an assumption fails
This is a reasoning question. A number on its own will not score. 3 marks
Q2.1 State two assumptions the capture-recapture method depends on. Then explain what happens to the estimate if being tagged makes an animal harder to catch the second time, and say in which direction the estimate is wrong.
3. What it costs to run
Convert to kilowatts, then hours, then days, then dollars. 3 marks each
Q3.1 A 3200 W oven is used for 1.5 hours a day for 28 days. Electricity costs $0.34 per kWh. Find the cost.
Q3.2 A 900 W fridge runs 24 hours a day for 90 days. Electricity costs $0.28 per kWh. Find the cost.
4. Working backwards
You are given the energy and asked for something else. 2 marks each
Q4.1 A 1.5 kW appliance used 540 kWh over a quarter. For how many hours did it run?
Q4.2 An appliance used 45 kWh over 25 hours of running. Find its power rating in watts.
5. Is the new one worth it?
Compare over a full year. 4 marks
Q5.1 An old fridge is rated at 450 W and a new one at 320 W. Both run 24 hours a day, every day, for 365 days. Electricity costs $0.30 per kWh. How much would the new fridge save in a year?
How did this worksheet feel?
What I'll revisit before next class:
Q1.1, The wallabies
N = (55 × 66) ÷ 11 = 3630 ÷ 11 = 330 wallabies. Check: 11 out of 66 is one sixth, and 55 out of 330 is also one sixth.
Q1.2, The lake
N = (90 × 120) ÷ 18 = 10 800 ÷ 18 = 600 fish. Check: 18 ÷ 120 = 0.15 and 90 ÷ 600 = 0.15.
Q2.1, When an assumption fails
Any two of: the marked animals mix evenly back through the population; marking does not change how easily an animal is caught; the population does not change between the two samples through births, deaths or migration; no marks are lost or become unreadable [2].
If tagging makes an animal harder to catch, then fewer tagged animals turn up in the second sample, so m2 is smaller than it should be. Since m2 is the denominator of N = (n1 × n2) ÷ m2, a smaller denominator makes N larger. The estimate is therefore too high, and the study would report more animals than really exist [1].
Q3.1, The oven
3200 W = 3.2 kW [1]. Energy = 3.2 × 1.5 = 4.8 kWh a day, so over 28 days that is 4.8 × 28 = 134.4 kWh [1]. Cost = 134.4 × 0.34 = 45.696, which is $45.70 to the nearest cent [1]. Keep the full value through the working and round only on the last line.
Q3.2, The fridge
900 W = 0.9 kW [1]. Energy = 0.9 × 24 = 21.6 kWh a day, so over 90 days that is 1944 kWh [1]. Cost = 1944 × 0.28 = $544.32 [1]. A fridge has a small power rating but never switches off, which is why it dominates a bill.
Q4.1, How many hours?
Energy = power × time, so time = energy ÷ power = 540 ÷ 1.5 = 360 hours. Check: 1.5 × 360 = 540 kWh.
Q4.2, What power rating?
Power = energy ÷ time = 45 ÷ 25 = 1.8 kW, which is 1800 W. Check: 1.8 × 25 = 45 kWh. The question asks for watts, so the final conversion is part of the answer.
Q5.1, Old fridge against new
Old: 450 W = 0.45 kW, so 0.45 × 24 = 10.8 kWh a day, and 10.8 × 365 = 3942 kWh a year, costing 3942 × 0.30 = $1182.60 [2].
New: 320 W = 0.32 kW, so 0.32 × 24 = 7.68 kWh a day, and 7.68 × 365 = 2803.2 kWh a year, costing 2803.2 × 0.30 = $840.96 [1].
Saving = 1182.60 − 840.96 = $341.64 a year [1]. Check: 1182.60 ÷ 0.30 = 3942 kWh.