Last visit was: 05 Nov 2024, 15:14 It is currently 05 Nov 2024, 15:14

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Verbal Expert
Joined: 18 Apr 2015
Posts: 29894
Own Kudos [?]: 36129 [0]
Given Kudos: 25919
Send PM
avatar
Intern
Intern
Joined: 21 Aug 2020
Posts: 39
Own Kudos [?]: 32 [0]
Given Kudos: 0
Send PM
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12176 [0]
Given Kudos: 136
Send PM
avatar
Intern
Intern
Joined: 11 Nov 2020
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
Re: The scatterplot above shows the numbers of incidences of me [#permalink]
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?
User avatar
Intern
Intern
Joined: 15 Aug 2020
Posts: 23
Own Kudos [?]: 22 [0]
Given Kudos: 0
Send PM
Re: The scatterplot above shows the numbers of incidences of me [#permalink]
I thought E is the correct answer.

.13
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Own Kudos [?]: 12176 [0]
Given Kudos: 136
Send PM
Re: The scatterplot above shows the numbers of incidences of me [#permalink]
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.
Prep Club for GRE Bot
Re: The scatterplot above shows the numbers of incidences of me [#permalink]
Moderators:
Retired Moderator
6218 posts
Moderator
1111 posts
Retired Moderator
187 posts
Retired Moderator
348 posts
Retired Moderator
160 posts
GRE Instructor
117 posts
Retired Moderator
63 posts
Retired Moderator
1307 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne