a spinner is divivded into 4 sections labeled as K I N D someone reproduced the wheel and uses a computer to simulate the outcome of 500 spins what is the aprosimate probability that the spinner will stop on a costonant on the next spin
frequency:
K = 120
I = 140
N = 105
D = 135
To find the approximate probability that the spinner will stop on a consonant on the next spin, we need to first calculate the total number of spins for all consonants (K, N, D).
Total number of spins for consonants = K + N + D = 120 + 105 + 135 = 360
Then, the probability of stopping on a consonant on the next spin is:
Probability = Total number of spins for consonants / Total number of spins
Probability = 360 / 500
Probability = 0.72
Therefore, the approximate probability that the spinner will stop on a consonant on the next spin is 0.72 or 72%.
the letter tiles CARE and s are placed ina box without looking jorelle picks a letter tile from the box and records the result before placing it back. he repeats this prodecure 100 times and observed the following results
C = 20
A = 25
R = 30
E = 10
S = 15
make a probability model for this expiriment express the probbility in decimal form rounded to the nearest hundreth
To create a probability model, we first need to calculate the probability of picking each letter tile.
Total number of trials = 100
Probability of picking C:
P(C) = Number of times C was picked / Total number of trials
P(C) = 20 / 100
P(C) = 0.20
Probability of picking A:
P(A) = Number of times A was picked / Total number of trials
P(A) = 25 / 100
P(A) = 0.25
Probability of picking R:
P(R) = Number of times R was picked / Total number of trials
P(R) = 30 / 100
P(R) = 0.30
Probability of picking E:
P(E) = Number of times E was picked / Total number of trials
P(E) = 10 / 100
P(E) = 0.10
Probability of picking S:
P(S) = Number of times S was picked / Total number of trials
P(S) = 15 / 100
P(S) = 0.15
Therefore, the probability model for this experiment is:
P(C) = 0.20
P(A) = 0.25
P(R) = 0.30
P(E) = 0.10
P(S) = 0.15
Each probability is expressed in decimal form rounded to the nearest hundredth.
the sample space of an expiriment is S = 1 2 3 4 5 if each outcome is equally likely to occur whihc graph models the texpected pobalbility f each outcome
Since each outcome is equally likely to occur in the sample space S = {1, 2, 3, 4, 5}, the expected probability of each outcome is 1/5 or 0.2.
To graphically represent the expected probability of each outcome, you can create a bar graph where the height of each bar represents the probability of that outcome. Since each outcome has an equal probability of 0.2, the height of each bar should be the same.
Below is a description of how to visually represent the expected probabilities of each outcome using a bar graph:
1. Draw a vertical axis (Y-axis) representing the probability scale from 0 to 1.
2. Draw a horizontal axis (X-axis) with labels 1, 2, 3, 4, 5 to represent the outcomes.
3. Draw bars above each label from 1 to 5 with a height of 0.2 (or 1/5) on the vertical axis, representing the expected probability of each outcome.
Each bar should be of equal height (0.2) to visually show that each outcome has an equal probability of occurring. This bar graph effectively models the expected probability of each outcome in the sample space S = {1, 2, 3, 4, 5}.
2 people repeadetly select a ball from both botws and reocrded the ruseults in a table whos result will give you a better indication about the proportion of white and red balls in each bowl explai each answer
Bowl a = 45 and 55
bowl b = 60 and 10
To determine which result gives a better indication of the proportion of white and red balls in each bowl, we need to analyze the breakdown of the selections from each bowl:
Bowl A:
- 45 white balls
- 55 red balls
Bowl B:
- 60 white balls
- 10 red balls
Comparing the proportions in each bowl, we can see that:
- In Bowl A, the proportion of red balls is higher (55 red balls out of 100 total balls) compared to white balls.
- In Bowl B, the proportion of white balls is higher (60 white balls out of 70 total balls) compared to red balls.
Considering the distribution of the white and red balls in each bowl, the result from Bowl B gives a better indication of the proportion of white and red balls in that bowl. This is because the proportion of white balls relative to red balls in Bowl B is significantly different compared to Bowl A.
In Bowl B, the ratio of white balls is much higher than red balls, implying that there are more white balls present in Bowl B. As a result, if you are looking to understand the proportion of white and red balls in each bowl, the selection results from Bowl B give a clearer indication due to the higher proportion of white balls in that bowl.
frist persons results: bowl a = 5 and 15
bowl b = 16 amd 4
With the updated results from the first person's selections:
Bowl A:
- 5 white balls
- 15 red balls
Bowl B:
- 16 white balls
- 4 red balls
Comparing the proportions in each bowl:
- In Bowl A, the proportion of red balls is higher (15 red balls out of 20 total balls) compared to white balls.
- In Bowl B, the proportion of white balls is higher (16 white balls out of 20 total balls) compared to red balls.
Given the updated results, the selection from Bowl B still provides a clearer indication of the proportion of white and red balls in that bowl. The higher proportion of white balls relative to red balls in Bowl B suggests that there are more white balls in Bowl B compared to red balls.
Therefore, the results from the first person's selections continue to support the conclusion that the selection results from Bowl B give a better indication of the proportion of white and red balls in each bowl due to the higher proportion of white balls in Bowl B.
bowls and and b contain a number of white and red balls simon repeadely selected a ball from both bowls and recorded the results in a table using his resuls what is the exiriments probability of pivking up red from bowl a the answer needs to be in precent form
bowl a = 5 and 15
bowl b = 16 and 4