To estimate the turtle population for 2009:
Let x be the estimated population of turtles in 2009.
The proportion can be set up as:
number of tagged turtles in 2009 / total number of turtles captured in 2009 = number of tagged turtles in 2008 / total number of turtles in 2008
Plugging in the given values, we get:
16 / 50 = 40 / x
Simplifying this proportion, we get:
16x = 2000
x = 125
Therefore, the estimated turtle population for 2009 is 125.
To estimate the turtle population for 2010:
Let y be the estimated population of turtles in 2010.
The proportion can be set up as:
number of tagged turtles in 2010 / total number of turtles captured in 2010 = number of tagged turtles in 2008 / total number of turtles in 2008
Plugging in the given values, we get:
8 / 30 = 40 / y
Simplifying this proportion, we get:
8y = 1200
y = 150
Therefore, the estimated turtle population for 2010 is 150.
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
A researcher tagged 40 giant turtles in 2008. In 2009, she returned and captured 50 turtles, 16 of which were tagged. In 2010, she returned again and captured 30 turtles, 8 of which were tagged.
Use a proportion to estimate the turtle population for 2009.
Use a proportion to estimate the turtle population for 2010.
)))
3 answers
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
Maria has the following scores on exams in her social studies class: 86, 75, 97, 58, 94, and 58.
Find the mean, median, and mode of the scores.
Should Maria’s social studies teacher use the mean, median, or mode of the exam scores to convince Maria that she needs to study more regularly for her exams?
To find the mean:
Add up all the scores and divide by the total number of scores:
(86 + 75 + 97 + 58 + 94 + 58) / 6 = 77.7
Therefore, the mean score is 77.7.
To find the median:
Put the scores in order from least to greatest:
58, 58, 75, 86, 94, 97
Since there are six scores, the median is the average of the two middle scores:
(75 + 86) / 2 = 80.5
Therefore, the median score is 80.5.
To find the mode:
The mode is the score that appears most frequently in the list. In this case, there are two scores that appear twice (58 and 94), so there are two modes: 58 and 94.
Therefore, the modes are 58 and 94.
Based on the scores, Maria’s social studies teacher should use the mean to convince her to study more regularly. The mean score is below 80, which is below average. The median score is above average, but it is still close to 80, which is below average. The mode is not a good measure of central tendency in this case because there are two modes and they do not provide much insight into the overall performance of the class. Therefore, the mean is the best measure to use in this situation.
Add up all the scores and divide by the total number of scores:
(86 + 75 + 97 + 58 + 94 + 58) / 6 = 77.7
Therefore, the mean score is 77.7.
To find the median:
Put the scores in order from least to greatest:
58, 58, 75, 86, 94, 97
Since there are six scores, the median is the average of the two middle scores:
(75 + 86) / 2 = 80.5
Therefore, the median score is 80.5.
To find the mode:
The mode is the score that appears most frequently in the list. In this case, there are two scores that appear twice (58 and 94), so there are two modes: 58 and 94.
Therefore, the modes are 58 and 94.
Based on the scores, Maria’s social studies teacher should use the mean to convince her to study more regularly. The mean score is below 80, which is below average. The median score is above average, but it is still close to 80, which is below average. The mode is not a good measure of central tendency in this case because there are two modes and they do not provide much insight into the overall performance of the class. Therefore, the mean is the best measure to use in this situation.
52 answers