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.
To estimate the turtle population for 2009, we can set up a proportion:
(number of tagged turtles in 2009) / (total number of turtles captured in 2009) = (number of tagged turtles in 2008) / (total number of turtles in the population)
Plugging in the numbers we have:
16/50 = 40/x
Where x is the estimated total number of turtles in 2009. Solving for x:
x = (50*40)/16 = 125
So we can estimate that there were 125 giant turtles in the population in 2009.
To estimate the turtle population for 2010, we can use the same proportion:
8/30 = 40/x
Solving for x:
x = (30*40)/8 = 150
So we can estimate that there were 150 giant turtles in the population in 2010.
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, we add up all the scores and divide by the total number of scores:
(86 + 75 + 97 + 58 + 94 + 58) / 6 = 78
So the mean score is 78.
To find the median, we need to put the scores in order from least to greatest:
58, 58, 75, 86, 94, 97
The median is the middle score, which in this case is 86.
To find the mode, we look for the score(s) that occur most frequently. In this case, there are two modes: 58 and 94.
Based on these measures of central tendency, the social studies teacher should use the mean score to convince Maria that she needs to study more regularly for her exams. The mean shows that her overall performance is below average, and studying more regularly could help improve her scores. However, the median and mode might not be as useful in this situation, as they don't give as clear of a picture of her overall performance.
The stem-and-leaf plot shows the heights in centimeters of Teddy Bear sunflowers grown in two different types of soil.
Soil A Soil B
5 9
5 2 1 1 6 3 9
5 1 0 7 0 2 3 6 7 8
2 1 8 3
0 9
Key: 9|6 means 69 Key: 5|8 means 58
Calculate the mean of each data set.
Calculate the mean absolute deviation (MAD) of each data set.
Which set is more variable? How do you know?
To calculate the mean of each data set, we can use the stem-and-leaf plot to find the individual data points and then add them up and divide by the total number of points.
For Soil A:
Mean = (50 + 51 + 52 + 57 + 58 + 95 + 96) / 12 = 56.167
For Soil B:
Mean = (30 + 32 + 33 + 36 + 37 + 38 + 60 + 63 + 69) / 10 = 44.8
To calculate the mean absolute deviation (MAD) of each data set, we need to find the absolute difference between each data point and the mean, add up those differences, and then divide by the total number of points.
For Soil A:
MAD = (|50 - 56.167| + |51 - 56.167| + ... + |96 - 56.167|) / 12 = 14.333
For Soil B:
MAD = (|30 - 44.8| + |32 - 44.8| + ... + |69 - 44.8|) / 10 = 12.12
Soil B is more variable because it has a larger MAD. The MAD measures the average distance between each data point and the mean, so a larger MAD implies that the data points are more spread out from the mean, indicating more variability in the data.
What is the probability of getting an even number when rolling a six-sided number cube?
A. 0.5
B. 5%
C. StartFraction 2 over 6 EndFraction
D. 30%
There are three even numbers on a six-sided number cube: 2, 4, and 6. There are a total of six possible outcomes, so the probability of rolling an even number is 3/6 or 1/2, which is the same as 0.5. Therefore, the answer is A. 0.5.
You place the letters for the word smart in a bag. What is the probability of choosing a letter that is not a vowel? (Remember, vowels are A, E, I, O, and U.)
A. 0.2
B. StartFraction 1 over 5 EndFraction
C. 80%
D. 0.75
The letters in the word "smart" are S, M, A, R, and T. There is 1 vowel (A) and 4 consonants (S, M, R, and T). Therefore, the probability of choosing a letter that is not a vowel is 4/5 or the same as StartFraction 4 over 5 EndFraction. Therefore, the answer is not listed, but it is equivalent to option B.
You have a bag with 20 marbles: 2 yellow, 5 green, 12 red, and 1 blue. Which of the following statements are true? Select all that apply.
A. The probability of choosing a yellow, green, red, or blue marble is 1.
B. The probability of choosing a yellow marble is less than the probability of choosing a blue marble.
C. The probability of choosing a green marble is 0.25.
D. The probability of choosing a black marble is zero.