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Dropdown 1: 1960-1990
Dropdown 2: 136 million
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
18% (03:23) correct
82%
(03:26)
wrong
based on 438
sessions
History
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Not Attempted Yet
The graph shows the total municipal solid waste (MSW) generated in the United States from 1960 to 2013 (blue line) and the per capita MSW per day (orange line). Total MSW is measured in million tons, while per capita MSW per day is measured in pounds (1 ton = 2,000 pounds).
Select the most accurate option from each drop-down menu to complete the statements based on the information provided.
The greatest percentage change in total solid waste generation for any 30-year period during the study was between .
The municipal population of the United States increased by approximately people from 1960 to 2013.
Attachment:
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Official Answer
Dropdown 1: 1960-1990
Dropdown 2: 136 million
29 Nov 2024, 03:45
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Bunuel
Official Solution:
Drop-down 1:
By examining the graph, we can estimate that the greatest percentage change, represented by the steepest increase, occurred between 1960 and 1990. During this period, total solid waste generation increased from approximately 90 to 210, an increase of about 2.3 times. For other listed 30-year periods, the increase was less than twofold.
Drop-down 2:
The total MSW generated in 1960 was approximately 90 million tons, or 180,000 million pounds (since 1 ton = 2,000 pounds). Per capita waste per year was 2.68 * 365 ≈ 1,000 pounds. Thus, the municipal population in 1960 was (180,000 million pounds) / (1,000 pounds) ≈ 180 million people.
The total MSW generated in 2013 was approximately 250 million tons, or 500,000 million pounds (since 1 ton = 2,000 pounds). Per capita waste per year was 4.40 * 365 ≈ 1,600 pounds. Thus, the municipal population in 2013 was (500,000 million pounds) / (1,600 pounds) ≈ 310 million people.
Therefore, the municipal population of the United States increased by approximately 310 - 180 = 130 million people.
Correct answer:
Dropdown 1: "1960-1990"
Dropdown 2: "135 million"
11 Dec 2024, 16:11
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1. The graph for total MSW generated per year starts out linearly increasing but later stops increasing. So, intuitively the greatest percent change is from 1960 to 1990. We can also check for every answer choice:
- 1960-1990: \(\frac{208.3 - 88.1}{88.1} * 100\% = 136.44\%\).
- 1965-1995: \(\frac{217.3 - 109.4}{109.4} * 100\% = 98.63\%\).
- 1970-2000: \(\frac{243.5 - 127.2}{127.2} * 100\% = 91.43\%\).
- 1975-2005: \(\frac{253.7 - 127.8}{127.8} * 100\% = 98.51\%\).
- 1980-2010: \(\frac{250.6 - 151.6}{151.6} * 100\% = 65.30\%\).
2. Let's write out a formula to figure out the municipal population of a given year using the total and per capita MSW values. \(Per \ capita \ MSW = \frac{\frac{Total \ MSW * 2000 * 1000000}{Municipal \ population}}{Days \ in \ a \ year} = \frac{Total \ MSW * 2000 * 1000000}{Municipal \ population * 365.25}\). Then, \(Municipal \ population = \frac{Total \ MSW * 2000 * 1000000}{Per \ capita \ MSW * 365.25}\). For 2013 and 1960 the municipal population was about \(\frac{88.1 * 2000 * 1000000}{2.68 * 365.25} = 180003473.4 \approx 180\) million and \(\frac{254.1 * 2000 * 1000000}{4.40 * 365.25} = 316221765.9 \approx 316.22\) million, respectively. The difference will be \(316.22 - 180 = 136.22\) million, which is closest to the answer choice 136 million.
3. Our answer will be: 1960-1990 and 136 million.
- 1960-1990: \(\frac{208.3 - 88.1}{88.1} * 100\% = 136.44\%\).
- 1965-1995: \(\frac{217.3 - 109.4}{109.4} * 100\% = 98.63\%\).
- 1970-2000: \(\frac{243.5 - 127.2}{127.2} * 100\% = 91.43\%\).
- 1975-2005: \(\frac{253.7 - 127.8}{127.8} * 100\% = 98.51\%\).
- 1980-2010: \(\frac{250.6 - 151.6}{151.6} * 100\% = 65.30\%\).
2. Let's write out a formula to figure out the municipal population of a given year using the total and per capita MSW values. \(Per \ capita \ MSW = \frac{\frac{Total \ MSW * 2000 * 1000000}{Municipal \ population}}{Days \ in \ a \ year} = \frac{Total \ MSW * 2000 * 1000000}{Municipal \ population * 365.25}\). Then, \(Municipal \ population = \frac{Total \ MSW * 2000 * 1000000}{Per \ capita \ MSW * 365.25}\). For 2013 and 1960 the municipal population was about \(\frac{88.1 * 2000 * 1000000}{2.68 * 365.25} = 180003473.4 \approx 180\) million and \(\frac{254.1 * 2000 * 1000000}{4.40 * 365.25} = 316221765.9 \approx 316.22\) million, respectively. The difference will be \(316.22 - 180 = 136.22\) million, which is closest to the answer choice 136 million.
3. Our answer will be: 1960-1990 and 136 million.
