Cats, because the range of their weights is less than that of small dogs.
This can be determined by visually observing the distribution of dots on the dot plot for cats and small dogs. The dot plot for cats shows a more concentrated distribution of weights, with most of the dots clustered around a narrower range of values. In contrast, the dot plot for small dogs shows a wider spread of weights, with dots appearing across a larger range of values. This indicates that there is less variability in the weights of cats compared to small dogs.
Use the image to answer the question.
An illustration shows two sets of dot plots. One is titled Weight in Pounds, Cats and the other is Weight in Pounds, Small Dogs. The plots are shown as an abacus-like representation with dots in a vertical row over each number on a number line. For Cats, a number line with arrows on both ends ranges from 9 to 13 in increments of 1. There is 1 dot above 9, 2 dots above 10, 4 dots above 11, 2 dots above 12, and one dot above 13. For Small Dogs, a number line with arrows on both ends ranges from 8 to 15 in increments of 1. There is 1 dot above 8, 2 dots above 9, 4 dots above 10, 2 dots above 11, 2 dots above 12, 2 dots above 13, 1 dot above 14, and 1 dot above 15.
The dot plots show the weights of randomly selected cats and small dogs in different households. Based on an informal assessment of the visual overlap of these distributions, which animal has the least variability in their weights? Why?
Small dogs, because the range of their weights is greater than that of cats.
Small dogs, because the range of their weights is less than that of cats.
Cats, because the range of their weights is less than that of small dogs.
Cats, because the range of their weights is greater than that of small dogs.
11 answers
Use the image to answer the question.
An illustration shows two sets of dot plots. One is titled Weight in Pounds, Cats and the other is Weight in Pounds, Small Dogs. The plots are shown as an abacus-like representation with dots in a vertical row over each number on a number line. For Cats, a number line with arrows on both ends ranges from 9 to 13 in increments of 1. There is 1 dot above 9, 2 dots above 10, 4 dots above 11, 2 dots above 12, and one dot above 13. For Small Dogs, a number line with arrows on both ends ranges from 8 to 15 in increments of 1. There is 1 dot above 8, 2 dots above 9, 4 dots above 10, 2 dots above 11, 2 dots above 12, 2 dots above 13, 1 dot above 14, and 1 dot above 15.
The dot plots show the weights of randomly selected cats and small dogs in different households. What is the difference between the modal weights of cats and small dogs?
The modal weights are the same for cats and small dogs.
1 pound
0.13 pounds
3 pounds
An illustration shows two sets of dot plots. One is titled Weight in Pounds, Cats and the other is Weight in Pounds, Small Dogs. The plots are shown as an abacus-like representation with dots in a vertical row over each number on a number line. For Cats, a number line with arrows on both ends ranges from 9 to 13 in increments of 1. There is 1 dot above 9, 2 dots above 10, 4 dots above 11, 2 dots above 12, and one dot above 13. For Small Dogs, a number line with arrows on both ends ranges from 8 to 15 in increments of 1. There is 1 dot above 8, 2 dots above 9, 4 dots above 10, 2 dots above 11, 2 dots above 12, 2 dots above 13, 1 dot above 14, and 1 dot above 15.
The dot plots show the weights of randomly selected cats and small dogs in different households. What is the difference between the modal weights of cats and small dogs?
The modal weights are the same for cats and small dogs.
1 pound
0.13 pounds
3 pounds
The difference between the modal weights of cats and small dogs is 1 pound.
This can be determined by looking at the dot plots for both cats and small dogs and identifying the weight at which the highest number of dots (mode) is located. For cats, the mode is 11 pounds, as there are 4 dots above that number. For small dogs, the mode is 10 pounds, with 4 dots above that number. Therefore, the difference between the modal weights of cats and small dogs is 1 pound.
This can be determined by looking at the dot plots for both cats and small dogs and identifying the weight at which the highest number of dots (mode) is located. For cats, the mode is 11 pounds, as there are 4 dots above that number. For small dogs, the mode is 10 pounds, with 4 dots above that number. Therefore, the difference between the modal weights of cats and small dogs is 1 pound.
The data shows the number of miles run per week by randomly selected students from two different classes. Find the difference between the medians. Which class has a higher median? By how much?
Class 1: 6, 8, 10, 11, 14, 4, 5, 8, 2, 7, 7, 5, 10, 12, 11
Class 2: 6, 4, 5, 6, 7, 8, 12, 9, 10, 11, 5, 8, 7, 4, 10
Class 1 has a higher median than Class 2 by 1 mile.
Class 1 has a higher median than Class 2 by 0.5 mile.
Class 2 has a higher median than Class 1 by 0.5 mile.
Class 2 has a higher median than Class 1 by 1 mile.
Class 1: 6, 8, 10, 11, 14, 4, 5, 8, 2, 7, 7, 5, 10, 12, 11
Class 2: 6, 4, 5, 6, 7, 8, 12, 9, 10, 11, 5, 8, 7, 4, 10
Class 1 has a higher median than Class 2 by 1 mile.
Class 1 has a higher median than Class 2 by 0.5 mile.
Class 2 has a higher median than Class 1 by 0.5 mile.
Class 2 has a higher median than Class 1 by 1 mile.
To find the median in each class, we first arrange the data in order:
Class 1: 2, 4, 5, 5, 6, 7, 7, 8, 8, 10, 10, 11, 11, 12, 14
Class 2: 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10, 10, 11, 12
Calculating the medians:
For Class 1, we have 10 numbers, so the median is the average of the middle two numbers: (7 + 8)/2 = 7.5
For Class 2, we have 16 numbers, and the median is the middle number in the ordered list, which is 7.
Thus, the difference between the medians of Class 1 and Class 2 is given by:
7.5 - 7 = 0.5
Therefore, Class 1 has a higher median than Class 2 by 0.5 mile.
The correct answer is: Class 1 has a higher median than Class 2 by 0.5 mile.
Class 1: 2, 4, 5, 5, 6, 7, 7, 8, 8, 10, 10, 11, 11, 12, 14
Class 2: 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 10, 10, 11, 12
Calculating the medians:
For Class 1, we have 10 numbers, so the median is the average of the middle two numbers: (7 + 8)/2 = 7.5
For Class 2, we have 16 numbers, and the median is the middle number in the ordered list, which is 7.
Thus, the difference between the medians of Class 1 and Class 2 is given by:
7.5 - 7 = 0.5
Therefore, Class 1 has a higher median than Class 2 by 0.5 mile.
The correct answer is: Class 1 has a higher median than Class 2 by 0.5 mile.
Two fertilizers are applied to randomly selected potato plants. After several weeks, the potatoes are weighed. The weights in kilograms are as follows. Which fertilizer is generally more effective?
Fertilizer A: 27, 20, 16, 18, 22, 19, 23, 21, 17, 19
Fertilizer B: 28, 19, 18, 21, 24, 20, 25, 27, 29, 21
Fertilizer A is generally more effective, because the mean and median weights of the potatoes treated with it are higher.
Fertilizers A and B are equally effective, because the ranges of the weights of the potatoes are the same.
Fertilizers A and B are both ineffective, because there is no difference in the ranges of the weights of the potatoes.
Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.
Fertilizer A: 27, 20, 16, 18, 22, 19, 23, 21, 17, 19
Fertilizer B: 28, 19, 18, 21, 24, 20, 25, 27, 29, 21
Fertilizer A is generally more effective, because the mean and median weights of the potatoes treated with it are higher.
Fertilizers A and B are equally effective, because the ranges of the weights of the potatoes are the same.
Fertilizers A and B are both ineffective, because there is no difference in the ranges of the weights of the potatoes.
Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.
To determine which fertilizer is generally more effective, we can compare the mean and median weights of the potatoes treated with each fertilizer.
For Fertilizer A:
Mean = (27 + 20 + 16 + 18 + 22 + 19 + 23 + 21 + 17 + 19) / 10 = 20.2
Median = the middle value when the data is arranged in ascending order = 19.5
For Fertilizer B:
Mean = (28 + 19 + 18 + 21 + 24 + 20 + 25 + 27 + 29 + 21) / 10 = 23.2
Median = the middle value when the data is arranged in ascending order = 22.5
Comparing the mean and median weights, we see that for both Fertilizer A and Fertilizer B, the median and mean are both higher for Fertilizer B. This indicates that Fertilizer B is generally more effective in producing heavier potato plants.
Therefore, the correct answer is: Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.
For Fertilizer A:
Mean = (27 + 20 + 16 + 18 + 22 + 19 + 23 + 21 + 17 + 19) / 10 = 20.2
Median = the middle value when the data is arranged in ascending order = 19.5
For Fertilizer B:
Mean = (28 + 19 + 18 + 21 + 24 + 20 + 25 + 27 + 29 + 21) / 10 = 23.2
Median = the middle value when the data is arranged in ascending order = 22.5
Comparing the mean and median weights, we see that for both Fertilizer A and Fertilizer B, the median and mean are both higher for Fertilizer B. This indicates that Fertilizer B is generally more effective in producing heavier potato plants.
Therefore, the correct answer is: Fertilizer B is generally more effective, because the mean and median weights of the potatoes that were treated with it are higher.
Two number lines extend from 1 to 10 in increments of 1 and are labeled as Number of States Visited, Hiking Club, and Number of States Visited, Swimming Club. The number lines show an abacus-like representation with columns of open circles over the numbers. Above the number line labeled as Number of States Visited, Hiking Club, there are no circles over 1, two circles over 2, three circles over 3, five circles over 4, six circles over 5, five circles over 6, four circles over 7, three circles over 8, two circles over 9, and one circle over 10. Above the number line labeled as Number of States Visited, Swimming Club, there is one circle over 1, two circles over 2, one circle over 3, two circles over 4, two circles over 5, three circles over 6, four circles over 7, five circles over 8, six circles over 9, and six circles over 10.
For a geography project, you have surveyed the people in your hiking club and swimming club to find the number of states they have visited. You make a table and dot plots to show the results. Use the dot plots to compare the shapes and spreads of the dot plots. What conclusions can you draw about the shapes and the spreads?
The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has peaks at 9 and 10. The swimming club has a peak at 5.
The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has a peak at 5. The swimming club has peaks at 9 and 10.
The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed right. The hiking club has a peak at 5. The swimming club has peaks at 9 and 10.
The hiking club has a greater spread than the swimming club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has a peak at 5. The swimming club has peaks at 9 and 10.
For a geography project, you have surveyed the people in your hiking club and swimming club to find the number of states they have visited. You make a table and dot plots to show the results. Use the dot plots to compare the shapes and spreads of the dot plots. What conclusions can you draw about the shapes and the spreads?
The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has peaks at 9 and 10. The swimming club has a peak at 5.
The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has a peak at 5. The swimming club has peaks at 9 and 10.
The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed right. The hiking club has a peak at 5. The swimming club has peaks at 9 and 10.
The hiking club has a greater spread than the swimming club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has a peak at 5. The swimming club has peaks at 9 and 10.
Based on the description provided, we can conclude that the swimming club has a greater spread than the hiking club: There are more dots spread out across a wider range of states visited in the swimming club compared to the hiking club.
The hiking club appears to be symmetrical and has peaks at 9 and 10 states visited. This implies that there are more members in the hiking club who have visited 9 or 10 states compared to other numbers of states visited.
In contrast, the swimming club is skewed left and has peaks at 9 and 10 states visited. This indicates that the majority of members in the swimming club have visited 9 or 10 states, with a few members visiting fewer states on the left side of the plot.
Therefore, the correct conclusion is: The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has peaks at 9 and 10. The swimming club has peaks at 9 and 10.
The hiking club appears to be symmetrical and has peaks at 9 and 10 states visited. This implies that there are more members in the hiking club who have visited 9 or 10 states compared to other numbers of states visited.
In contrast, the swimming club is skewed left and has peaks at 9 and 10 states visited. This indicates that the majority of members in the swimming club have visited 9 or 10 states, with a few members visiting fewer states on the left side of the plot.
Therefore, the correct conclusion is: The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed left. The hiking club has peaks at 9 and 10. The swimming club has peaks at 9 and 10.
no
I apologize for the confusion in my previous responses. Let's revisit the dot plots for the hiking club and swimming club:
- Hiking Club: The dot plot for the hiking club peaks at 5 and is symmetrical, with decreasing counts towards 1 and 10 states visited.
- Swimming Club: The dot plot for the swimming club peaks at 10 and displays a relatively uniform spread over states visited.
Based on this description, the conclusions we can draw about the shapes and spreads of the dot plots are as follows:
The hiking club has a symmetrical distribution with a peak at 5 states visited and a gradual decrease towards 1 and 10 states. The shape is bell-shaped and centered around the peak.
The swimming club has a right-skewed distribution with a peak at 10 states visited and a more uniform spread of members across different numbers of states visited. The individuals in the swimming club have visited a wider range of states overall compared to the hiking club, leading to a greater spread.
Therefore, the correct conclusion is: The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed right. The hiking club has a peak at 5 states visited, while the swimming club has a peak at 10 states visited.
- Hiking Club: The dot plot for the hiking club peaks at 5 and is symmetrical, with decreasing counts towards 1 and 10 states visited.
- Swimming Club: The dot plot for the swimming club peaks at 10 and displays a relatively uniform spread over states visited.
Based on this description, the conclusions we can draw about the shapes and spreads of the dot plots are as follows:
The hiking club has a symmetrical distribution with a peak at 5 states visited and a gradual decrease towards 1 and 10 states. The shape is bell-shaped and centered around the peak.
The swimming club has a right-skewed distribution with a peak at 10 states visited and a more uniform spread of members across different numbers of states visited. The individuals in the swimming club have visited a wider range of states overall compared to the hiking club, leading to a greater spread.
Therefore, the correct conclusion is: The swimming club has a greater spread than the hiking club. The hiking club is symmetrical, and the swimming club is skewed right. The hiking club has a peak at 5 states visited, while the swimming club has a peak at 10 states visited.