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The scatterplot above shows the numbers of incidences of me

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Carcass

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The scatterplot above shows the numbers of incidences of me [#permalink]

12 Nov 2020, 10:41

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Attachment:

GRE The scatterplot above shows the numbers of incidences.jpg [ 50.6 KiB | Viewed 498 times ]

The scatterplot above shows the numbers of incidences of melanoma, per 100,000 people from 1940 to 1970. Based on the line of best fit to the data, as shown in the figure, which of the following values is closest to the average yearly increase in the number of incidences of melanoma?

A. 33,000

B. 23,000

C. 13,000

D. 0.33

E. 0.13

Source: 320 GRE Math Questions

anasp99

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Re: The scatterplot above shows the numbers of incidences of me [#permalink]

13 Nov 2020, 07:53

Explanation?

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GreenlightTestPrep

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Re: The scatterplot above shows the numbers of incidences of me [#permalink]

13 Nov 2020, 10:06

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Carcass wrote:

Attachment:

GRE The scatterplot above shows the numbers of incidences.jpg

The scatterplot above shows the numbers of incidences of melanoma, per 100,000 people from 1940 to 1970. Based on the line of best fit to the data, as shown in the figure, which of the following values is closest to the average yearly increase in the number of incidences of melanoma?

A. 33,000
B. 23,000
C. 13,000
D. 0.33
E. 0.13

We just need two data points to determine the average yearly increase in the number of incidences of melanoma

The line of best fit tells us that:
In 1940, the number of incidences of melanoma was about 120,000
In 1970, the number of incidences of melanoma was about 500,000

500,000 - 120,000 = 380,000
So, over a 30-year period (from 1940 to 1970) the number of incidents increased by 380,000

So, the average annual increase ≈ 380,000/30 ≈ 13,000

Answer: C

Cheers,
Brent

hamzazwairy

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Re: The scatterplot above shows the numbers of incidences of me [#permalink]

14 Nov 2020, 10:03

GreenlightTestPrep wrote:

Carcass wrote:

Attachment:

GRE The scatterplot above shows the numbers of incidences.jpg

The scatterplot above shows the numbers of incidences of melanoma, per 100,000 people from 1940 to 1970. Based on the line of best fit to the data, as shown in the figure, which of the following values is closest to the average yearly increase in the number of incidences of melanoma?

A. 33,000
B. 23,000
C. 13,000
D. 0.33
E. 0.13

We just need two data points to determine the average yearly increase in the number of incidences of melanoma

The line of best fit tells us that:
In 1940, the number of incidences of melanoma was about 120,000
In 1970, the number of incidences of melanoma was about 500,000

500,000 - 120,000 = 380,000
So, over a 30-year period (from 1940 to 1970) the number of incidents increased by 380,000

So, the average annual increase ≈ 380,000/30 ≈ 13,000

Answer: C

Cheers,
Brent

but i thought the "per 100,000 " means that we get for example 5 cases out of 100,000 1970 .. amerite? which makes it 5-1.75 / 30 which is closest to 0.13?

Johnr

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Re: The scatterplot above shows the numbers of incidences of me [#permalink]

15 Nov 2020, 05:51

I thought E is the correct answer.

.13

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GreenlightTestPrep

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Re: The scatterplot above shows the numbers of incidences of me [#permalink]

15 Nov 2020, 06:28

hamzazwairy wrote:

but i thought the "per 100,000 " means that we get for example 5 cases out of 100,000 1970 .. amerite? which makes it 5-1.75 / 30 which is closest to 0.13?

Great point!
In my haste, I misread the vertical scale as telling us the NUMBER of cases (in hundred thousands).

Since we don't know the total population, there's no way to determine the average yearly increase in the NUMBER of incidences of melanoma.

For example, if the total population = 100,000, then that means there were 1.2 incidents in 1940, and there were 5 incidents in 1970. This means the average annual increase = (5 - 1.2)/30 = 0.13 per year.

Conversely, if the total population = 1,000,000, then that means there were 12 incidents in 1940, and there were 50 incidents in 1970. This means the average annual increase = (50 - 12)/30 = 1.3 per year.

For this reason, this is a faulty question.