The correct sample space for the experiment if a coin is selected from a bag three times is:
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN
Determine the sample space for an experiment if a coin is selected from a bag three times. There are three coins in the bag: a nickel (N), a quarter (Q), and a dime (D). Each time a coin is selected, it is replaced before selecting the next coin.(1 point)
Responses
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, NNN, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN, DDD, QQD, QQN, QDQ, QDN, QDD, QNQ, QND, QNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, NNN, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN, DDD, QQD, QQN, QDQ, QDN, QDD, QNQ, QND, QNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, QQQ, QQD, QQN, QDQ, QDN, QDD, QNQ, QND, QNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, QQQ, QQD, QQN, QDQ, QDN, QDD, QNQ, QND, QNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN, QQQ, QQD, QQN, QDQ, QDN, QDD, QNQ, QND, QNN
NNN, NNQ, NND, NQN, NQD, NQQ, NDN, NDQ, NDD, DDD, DDQ, DDN, DQD, DQN, DQQ, DND, DNQ, DNN, QQQ, QQD, QQN, QDQ, QDN, QDD, QNQ, QND, QNN
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9 answers
Use the image to answer the question.
An illustration of a circle with eight equal parts labeled from 1 through 8 clockwise. An arrow starting in part 1 passes through the center to point into part 5.
The spinner is spun twice. What is the probability that the spinner lands on both multiples of 4 in both spins?
(1 point)
Responses
364
Start Fraction 3 over 64 End Fraction
14
Start Fraction 1 over 4 End Fraction
116
Start Fraction 1 over 16 End Fraction
132
Start Fraction 1 over 32 End Fraction
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An illustration of a circle with eight equal parts labeled from 1 through 8 clockwise. An arrow starting in part 1 passes through the center to point into part 5.
The spinner is spun twice. What is the probability that the spinner lands on both multiples of 4 in both spins?
(1 point)
Responses
364
Start Fraction 3 over 64 End Fraction
14
Start Fraction 1 over 4 End Fraction
116
Start Fraction 1 over 16 End Fraction
132
Start Fraction 1 over 32 End Fraction
Skip to navigation
To find the probability of landing on multiples of 4 in both spins, we first need to determine the total number of outcomes and the number of favorable outcomes.
Total outcomes when spinning the spinner twice = 8 * 8 = 64
Favorable outcomes for landing on multiples of 4:
- Multiples of 4 on the spinner are 4 and 8. There are two multiples of 4 out of the total 8 parts.
- The probability of landing on a multiple of 4 in one spin = 2/8 = 1/4
- The probability of landing on a multiple of 4 in both spins = (1/4) * (1/4) = 1/16
Therefore, the probability that the spinner lands on both multiples of 4 in both spins is 1/16 or 1 over 16.
The correct response is: Start Fraction 1 over 16 End Fraction
Total outcomes when spinning the spinner twice = 8 * 8 = 64
Favorable outcomes for landing on multiples of 4:
- Multiples of 4 on the spinner are 4 and 8. There are two multiples of 4 out of the total 8 parts.
- The probability of landing on a multiple of 4 in one spin = 2/8 = 1/4
- The probability of landing on a multiple of 4 in both spins = (1/4) * (1/4) = 1/16
Therefore, the probability that the spinner lands on both multiples of 4 in both spins is 1/16 or 1 over 16.
The correct response is: Start Fraction 1 over 16 End Fraction
Leah is playing a board game and needs to roll a 6 on a number cube two times in a row in order to win. What is the probability this will happen if she’s using a normal six-sided number cube?(1 point)
Responses
136
Start Fraction 1 over 36 End Fraction
13
Start Fraction 1 over 3 End Fraction
12
Start Fraction 1 over 2 End Fraction
16
Responses
136
Start Fraction 1 over 36 End Fraction
13
Start Fraction 1 over 3 End Fraction
12
Start Fraction 1 over 2 End Fraction
16
To find the probability of rolling a 6 on a number cube two times in a row, we first calculate the probability of rolling a 6 on a single die:
Probability of rolling a 6 on a fair, six-sided number cube = 1/6
Since the outcome of rolling the die twice is independent (the first roll does not affect the second roll), we can multiply the probabilities:
Probability of rolling a 6 in two consecutive rolls = (1/6) * (1/6) = 1/36
Therefore, the probability of Leah rolling a 6 on a number cube two times in a row is 1/36 or 1 over 36.
The correct response is: Start Fraction 1 over 36 End Fraction
Probability of rolling a 6 on a fair, six-sided number cube = 1/6
Since the outcome of rolling the die twice is independent (the first roll does not affect the second roll), we can multiply the probabilities:
Probability of rolling a 6 in two consecutive rolls = (1/6) * (1/6) = 1/36
Therefore, the probability of Leah rolling a 6 on a number cube two times in a row is 1/36 or 1 over 36.
The correct response is: Start Fraction 1 over 36 End Fraction
Use the image to answer the question.
A scatterplot shows the number of cups of coffee sold versus temperature.
The horizontal axis shows temperature, in degrees Fahrenheit, ranging from 0 to 70 in increments of 5, with an initial jump from 0 to 25. The vertical axis shows number of cups of coffee sold, ranging from 0 to 60 in increments of 10. A kink is shown on the horizontal axis close to origin. A tightly packed cluster of data plots is plotted between 30 to 37 on the horizontal axis and 42 to 48 on the vertical axis. A loosely packed cluster of data plots is plotted between 40 to 51 on the horizontal axis 30 and 38 on the vertical axis and. A dot is plotted at 62 on the horizontal axis and 5 on the vertical axis.
Numbers of Cups of Coffee Sold Versus Temperature
Interpret the scatterplot by describing any outliers and patterns of clustering.
A scatterplot shows the number of cups of coffee sold versus temperature.
The horizontal axis shows temperature, in degrees Fahrenheit, ranging from 0 to 70 in increments of 5, with an initial jump from 0 to 25. The vertical axis shows number of cups of coffee sold, ranging from 0 to 60 in increments of 10. A kink is shown on the horizontal axis close to origin. A tightly packed cluster of data plots is plotted between 30 to 37 on the horizontal axis and 42 to 48 on the vertical axis. A loosely packed cluster of data plots is plotted between 40 to 51 on the horizontal axis 30 and 38 on the vertical axis and. A dot is plotted at 62 on the horizontal axis and 5 on the vertical axis.
Numbers of Cups of Coffee Sold Versus Temperature
Interpret the scatterplot by describing any outliers and patterns of clustering.
Based on the description provided in the scatterplot:
1. Outliers:
- There is a single data point represented as a dot at 62 on the horizontal axis (temperature) and 5 on the vertical axis (number of cups of coffee sold). This data point is far from the main cluster of points, indicating it may be an outlier compared to the rest of the data.
2. Patterns of Clustering:
- There is a tightly packed cluster of data points between 30 to 37 on the horizontal axis (temperature) and 42 to 48 on the vertical axis (number of cups of coffee sold). This cluster suggests a pattern where within a specific temperature range, there is a consistent number of cups of coffee sold.
- Additionally, there is a loosely packed cluster of data points between 40 to 51 on the horizontal axis (temperature) and 30 to 38 on the vertical axis (number of cups of coffee sold). This cluster might indicate a broader range of temperatures where the number of cups of coffee sold varies more.
In conclusion, the scatterplot shows a clear clustering pattern in terms of temperature and the number of cups of coffee sold, with a possible outlier at the data point representing 62 degrees Fahrenheit temperature and 5 cups of coffee sold.
1. Outliers:
- There is a single data point represented as a dot at 62 on the horizontal axis (temperature) and 5 on the vertical axis (number of cups of coffee sold). This data point is far from the main cluster of points, indicating it may be an outlier compared to the rest of the data.
2. Patterns of Clustering:
- There is a tightly packed cluster of data points between 30 to 37 on the horizontal axis (temperature) and 42 to 48 on the vertical axis (number of cups of coffee sold). This cluster suggests a pattern where within a specific temperature range, there is a consistent number of cups of coffee sold.
- Additionally, there is a loosely packed cluster of data points between 40 to 51 on the horizontal axis (temperature) and 30 to 38 on the vertical axis (number of cups of coffee sold). This cluster might indicate a broader range of temperatures where the number of cups of coffee sold varies more.
In conclusion, the scatterplot shows a clear clustering pattern in terms of temperature and the number of cups of coffee sold, with a possible outlier at the data point representing 62 degrees Fahrenheit temperature and 5 cups of coffee sold.
Elke has a fair spinner with 4 equal-sized sections labeled 1, 2, 3, and 4. Elke plans to spin the spinner 100 times. Predict approximately how many times the spinner will land on 2. Explain your answer.
Based on the information provided, the spinner has 4 equal-sized sections labeled 1, 2, 3, and 4. Since the spinner is fair, each section has an equal probability of being landed on when spun.
To predict approximately how many times the spinner will land on 2 out of 100 spins, we can use the concept of probability. Since there are 4 sections on the spinner and each section has an equal probability of 1/4 (or 25%), the expected number of times the spinner would land on 2 in 100 spins can be calculated as follows:
Expected number of times landing on 2 = Total number of spins * Probability of landing on 2
Expected number of times landing on 2 = 100 * (1/4) = 25 times
Therefore, based on probability calculations, we can predict that the spinner will land on 2 approximately 25 times out of 100 spins.
To predict approximately how many times the spinner will land on 2 out of 100 spins, we can use the concept of probability. Since there are 4 sections on the spinner and each section has an equal probability of 1/4 (or 25%), the expected number of times the spinner would land on 2 in 100 spins can be calculated as follows:
Expected number of times landing on 2 = Total number of spins * Probability of landing on 2
Expected number of times landing on 2 = 100 * (1/4) = 25 times
Therefore, based on probability calculations, we can predict that the spinner will land on 2 approximately 25 times out of 100 spins.