Use the table to answer the question.
Outcomes K I N D
Frequency 120 140 105 135
A spinner is divided into 4 sections labeled as K , I , N , D . Xavier reproduced the wheel and uses a computer to simulate the outcomes of 500 spins. What is the approximate probability that the spinner will stop on a consonant on the next spin?
(1 point)
Responses
0.28
0.28
0.45
0.45
0.24
0.24
0.72
0.72
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The total frequency of consonants (K, N, D) is 120 + 105 + 135 = 360.
The probability of landing on a consonant on the next spin is the frequency of consonants divided by the total number of spins: 360/500 = 0.72
Therefore, the approximate probability that the spinner will stop on a consonant on the next spin is 0.72.
Answer: 0.72
The letter tiles C, A, R, E , and S are placed in a box. Without looking, Jorelle picks a letter tile from the box and records the result before placing it back. He repeats the procedure 100 times and observes the following results:
Outcomes C A R E S
Frequency 20 25 30 10 15
Based on the observed frequency, develop a probability model for this experiment. Express the probability in decimal form, rounded to the nearest hundredth.
(1 point)
To develop a probability model based on the observed frequency, we first need to find the total number of outcomes, which is the sum of the frequencies of each letter: 20 + 25 + 30 + 10 + 15 = 100.
Then we can find the probability of each outcome by dividing the frequency of each letter by the total number of outcomes:
Probability of picking C = 20/100 = 0.20
Probability of picking A = 25/100 = 0.25
Probability of picking R = 30/100 = 0.30
Probability of picking E = 10/100 = 0.10
Probability of picking S = 15/100 = 0.15
Therefore, the probability model for this experiment, expressed in decimal form rounded to the nearest hundredth, is:
C: 0.20
A: 0.25
R: 0.30
E: 0.10
S: 0.15
The sample space of an experiment is S={1, 2, 3, 4, 5} . If each outcome is equally likely to occur, which graph models the expected probability of each outcome?(1 point)
Since each outcome is equally likely to occur, the expected probability of each outcome is the same. Therefore, the graph that models this situation is a horizontal line at a constant value.
The correct graph would be a horizontal line where each outcome (1, 2, 3, 4, 5) has the same height on the graph, representing the equal probability of each outcome occurring.
Use the tables to answer the question.
Simon’s Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A 5 15
Bowl B 16 4
Clark’s Results
Number of White Balls Selected Number of Red Balls Selected
Bowl A 45 55
Bowl B 60 40
Bowls A and B contain a number of white and red balls. Simon and Clark repeatedly selected a ball from both bowls and recorded the results in a table. Whose results will give you a better indication about the proportion of white and red balls in each bowl? Explain your answer.
(1 point)
Responses
Simon’s experiment. He conducted a greater number of trials than Clark.
Simon’s experiment. He conducted a greater number of trials than Clark.
Clark’s experiment. He conducted a greater number of trials than Simon.
Clark’s experiment. He conducted a greater number of trials than Simon.
Neither; they conducted the same number of trials.
Neither; they conducted the same number of trials.
Both results can be used, as they performed the experiment at random.
Both results can be used, as they performed the experiment at random.
Simon's results will give a better indication about the proportion of white and red balls in each bowl. This is because Simon conducted a greater number of trials for each bowl (total of 20 trials for Bowl A and 20 trials for Bowl B) compared to Clark's results (total of 100 trials for Bowl A and 100 trials for Bowl B).
With a greater number of trials, Simon's results are more likely to be closer to the actual proportion of white and red balls in each bowl. This allows for a more reliable assessment of the proportion of white and red balls in each bowl based on Simon's results.
Therefore, the correct response is: Simon’s experiment. He conducted a greater number of trials than Clark.
Use the table to answer the question.
Number of White Balls Selected Number of Red Balls Selected
Bowl A 5 15
Bowl B 16 4
Bowls A and B contain a number of white and red balls. Simon repeatedly selected a ball from both bowls and recorded the results in a table. Using Simon’s results, what is the experimental probability of picking a red ball from Bowl A? The answer needs to be in percent form.
(1 point)
Responses
78.95%
78.95%
37.5%
37.5%
25%
25%
75%
75%
To find the experimental probability of picking a red ball from Bowl A using Simon's results, we need to use the formula:
Experimental Probability = Number of times event occurs / Total number of trials
From Simon's results:
Number of times a red ball was selected from Bowl A = 15
Total number of trials from Bowl A = 5 + 15 = 20
Experimental Probability of picking a red ball from Bowl A = 15 / 20 = 0.75
Converting this to percent form by multiplying by 100, we get:
0.75 * 100 = 75%
Therefore, the experimental probability of picking a red ball from Bowl A using Simon's results is 75%.
The answer is: 75%