There were a total of 102 + 3 + 8 = 113 customers in the last week.
The probability of the next customer paying with a credit card is 8/113.
To express this as a decimal to the nearest hundredth, we divide 8 by 113 and round to two decimal places:
8/113 ≈ 0.07
Therefore, the probability that the next customer will pay with a credit card is approximately 0.07.
Owen owns a small business selling ice-cream. He knows that in the last week 102 customers paid cash, 3 customers used a debit card, and 8 customers used a credit card.
Based on these results, express the probability that the next customer will pay with a credit card as a decimal to the nearest hundredth.
25 answers
In a popular online role playing game, players can create detailed designs for their character's "costumes," or appearance. Leah sets up a website where players can buy and sell these costumes online. Information about the number of people who visited the website and the number of costumes purchased in a single day is listed below.
31 visitors purchased no costume.
371 visitors purchased exactly one costume.
21 visitors purchased more than one costume.
Based on these results, express the probability that the next person will purchase no costume as a percent to the nearest whole number.
31 visitors purchased no costume.
371 visitors purchased exactly one costume.
21 visitors purchased more than one costume.
Based on these results, express the probability that the next person will purchase no costume as a percent to the nearest whole number.
There were a total of 31 + 371 + 21 = 423 visitors to the website in a single day.
The probability that the next person will purchase no costume is the number of visitors who purchased no costume divided by the total number of visitors:
31/423 = 0.073 or 7.3% (rounded to one decimal place).
Therefore, the probability that the next person will purchase no costume as a percent to the nearest whole number is 7%.
The probability that the next person will purchase no costume is the number of visitors who purchased no costume divided by the total number of visitors:
31/423 = 0.073 or 7.3% (rounded to one decimal place).
Therefore, the probability that the next person will purchase no costume as a percent to the nearest whole number is 7%.
In a certain Algebra 2 class of 27 students, 18 of them play basketball and 16 of them play baseball. There are 3 students who play neither sport. What is the probability that a student chosen randomly from the class plays basketball or baseball?
We can use the formula:
P(B or B) = P(B) + P(B) - P(B and B)
where B represents playing basketball and B represents playing baseball.
First, we find P(B) by dividing the number of basketball players (18) by the total number of students (27):
P(B) = 18/27 = 2/3
Similarly, we find P(B) by dividing the number of baseball players (16) by the total number of students (27):
P(B) = 16/27
Next, we need to find P(B and B), the probability that a student plays both basketball and baseball. We can do this by subtracting the number of students who play neither sport (3) from the total number of students (27), and then subtracting the number of students who play only one sport from the result:
27 - 3 - (18 + 16 - n) = n - 2
Simplifying, we get:
6 - n = n - 2
Solving for n, we get:
n = 4
Therefore, there are 4 students who play both basketball and baseball.
Now we can find P(B and B):
P(B and B) = 4/27
Finally, we can plug all of these values into the formula to find P(B or B):
P(B or B) = P(B) + P(B) - P(B and B)
P(B or B) = 2/3 + 16/27 - 4/27
P(B or B) = 22/27
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is 22/27.
P(B or B) = P(B) + P(B) - P(B and B)
where B represents playing basketball and B represents playing baseball.
First, we find P(B) by dividing the number of basketball players (18) by the total number of students (27):
P(B) = 18/27 = 2/3
Similarly, we find P(B) by dividing the number of baseball players (16) by the total number of students (27):
P(B) = 16/27
Next, we need to find P(B and B), the probability that a student plays both basketball and baseball. We can do this by subtracting the number of students who play neither sport (3) from the total number of students (27), and then subtracting the number of students who play only one sport from the result:
27 - 3 - (18 + 16 - n) = n - 2
Simplifying, we get:
6 - n = n - 2
Solving for n, we get:
n = 4
Therefore, there are 4 students who play both basketball and baseball.
Now we can find P(B and B):
P(B and B) = 4/27
Finally, we can plug all of these values into the formula to find P(B or B):
P(B or B) = P(B) + P(B) - P(B and B)
P(B or B) = 2/3 + 16/27 - 4/27
P(B or B) = 22/27
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is 22/27.
There is a spinner with 15 equal areas, numbered 1 through 15. If the spinner is spun one time, what is the probability that the result is a multiple of 3 and a multiple of 2?
The multiples of 3 and 2 are 6 and 12, which appear once each on the spinner. Therefore, the probability of getting a multiple of 3 and a multiple of 2 is the probability of spinning either 6 or 12. Since there are 2 ways to get a 6 or a 12, out of a total of 15 possible outcomes, the probability is:
P(multiple of 3 and multiple of 2) = 2/15
Therefore, the probability of getting a multiple of 3 and a multiple of 2 is 2/15.
P(multiple of 3 and multiple of 2) = 2/15
Therefore, the probability of getting a multiple of 3 and a multiple of 2 is 2/15.
In a certain Algebra 2 class of 27 students, 7 of them play basketball and 18 of them play baseball. There are 7 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?
We cannot determine the probability that a student plays both basketball and baseball, because we do not have enough information about how the 7 students who play neither sport are distributed among the other categories.
For example, it is possible that all 7 students who play neither sport are also not interested in any other extracurricular activities, in which case the probability of a student playing both basketball and baseball would be 0.
On the other hand, it is also possible that some or all of the 7 students who play neither sport also play another sport or participate in another extracurricular activity, in which case the probability of a student playing both basketball and baseball would be nonzero, but we cannot determine its value without more information.
For example, it is possible that all 7 students who play neither sport are also not interested in any other extracurricular activities, in which case the probability of a student playing both basketball and baseball would be 0.
On the other hand, it is also possible that some or all of the 7 students who play neither sport also play another sport or participate in another extracurricular activity, in which case the probability of a student playing both basketball and baseball would be nonzero, but we cannot determine its value without more information.
There is a spinner with 13 equal areas, numbered 1 through 13. If the spinner is spun one time, what is the probability that the result is a multiple of 4 or a multiple of 3?
The multiples of 4 are 4 and 8, and the multiples of 3 are 3, 6, 9, and 12. Notice that 12 is both a multiple of 3 and a multiple of 4. Therefore, to find the probability that the result is a multiple of 4 or a multiple of 3, we need to add up the number of times that 4, 6, 8, 9, and 12 appear on the spinner.
4 and 8 appear once each, 3 and 9 appear twice each, and 12 appears once. Therefore, the number of times that these numbers appear on the spinner is:
1 + 1 + 2 + 2 + 1 = 7
Since there are 13 possible outcomes on the spinner, the probability of getting a multiple of 4 or a multiple of 3 is:
P(multiple of 4 or multiple of 3) = 7/13
Therefore, the probability that the result is a multiple of 4 or a multiple of 3 is 7/13.
4 and 8 appear once each, 3 and 9 appear twice each, and 12 appears once. Therefore, the number of times that these numbers appear on the spinner is:
1 + 1 + 2 + 2 + 1 = 7
Since there are 13 possible outcomes on the spinner, the probability of getting a multiple of 4 or a multiple of 3 is:
P(multiple of 4 or multiple of 3) = 7/13
Therefore, the probability that the result is a multiple of 4 or a multiple of 3 is 7/13.
There is a spinner with 15 equal areas, numbered 1 through 15. If the spinner is spun one time, what is the probability that the result is a multiple of 2 and a multiple of 5?
The multiples of 2 and 5 are 10 and appear once on the spinner. Therefore, the probability of spinning a multiple of 2 and a multiple of 5 is the probability of spinning 10.
Since there is 1 way to get 10, out of a total of 15 possible outcomes, the probability is:
P(multiple of 2 and multiple of 5) = 1/15
Therefore, the probability of spinning a multiple of 2 and a multiple of 5 is 1/15.
Since there is 1 way to get 10, out of a total of 15 possible outcomes, the probability is:
P(multiple of 2 and multiple of 5) = 1/15
Therefore, the probability of spinning a multiple of 2 and a multiple of 5 is 1/15.
In a certain Algebra 2 class of 24 students, 9 of them play basketball and 13 of them play baseball. There are 2 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?
To solve this problem, we need to use the principle of inclusion-exclusion. The number of students who play either basketball or baseball is:
9 + 13 = 22
However, we have counted the 2 students who play both sports twice, so we need to subtract them once to get the total number of students who play either basketball or baseball:
22 - 2 = 20
Therefore, there are 20 students who play either basketball or baseball out of 24 total students.
The probability of choosing a student who plays either basketball or baseball is the number of students who play either sport divided by the total number of students:
P(basketball or baseball) = 20/24
Simplifying, we get:
P(basketball or baseball) = 5/6
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is 5/6.
9 + 13 = 22
However, we have counted the 2 students who play both sports twice, so we need to subtract them once to get the total number of students who play either basketball or baseball:
22 - 2 = 20
Therefore, there are 20 students who play either basketball or baseball out of 24 total students.
The probability of choosing a student who plays either basketball or baseball is the number of students who play either sport divided by the total number of students:
P(basketball or baseball) = 20/24
Simplifying, we get:
P(basketball or baseball) = 5/6
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is 5/6.
In a certain Algebra 2 class of 26 students, 16 of them play basketball and 9 of them play baseball. There are 5 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?
To find the probability that a student chosen randomly from the class plays basketball or baseball, we need to use the principle of inclusion-exclusion.
The total number of students who play either sport is:
16 + 9 - 5 = 20
This is because we can't just add the number of basketball players (16) and baseball players (9) together, since some of them play both sports. By adding them together, we've counted the 5 students who play both sports twice (once in each category), so we need to subtract them once.
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is the number of basketball players + the number of baseball players (without counting the students who play both) divided by the total number of students:
P(basketball or baseball) = 20/26
Simplifying, we get:
P(basketball or baseball) = 10/13
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is 10/13.
The total number of students who play either sport is:
16 + 9 - 5 = 20
This is because we can't just add the number of basketball players (16) and baseball players (9) together, since some of them play both sports. By adding them together, we've counted the 5 students who play both sports twice (once in each category), so we need to subtract them once.
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is the number of basketball players + the number of baseball players (without counting the students who play both) divided by the total number of students:
P(basketball or baseball) = 20/26
Simplifying, we get:
P(basketball or baseball) = 10/13
Therefore, the probability that a student chosen randomly from the class plays basketball or baseball is 10/13.
A bag contains 3 red marbles, 6 blue marbles and 4 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 10th of a percent, that all three marbles drawn will be blue?
The total number of marbles in the bag is:
3 + 6 + 4 = 13
We want to find the probability of drawing three blue marbles in a row. To do this, we need to multiply the probabilities of drawing a blue marble on each of the three draws, since the draws are independent of each other.
On the first draw, there are 6 blue marbles out of 13 total marbles, so the probability of drawing a blue marble on the first draw is:
6/13
On the second draw, there will be 5 blue marbles left out of 12 total marbles (since we already drew one marble), so the probability of drawing a blue marble on the second draw is:
5/12
On the third draw, there will be 4 blue marbles left out of 11 total marbles (since we already drew two marbles), so the probability of drawing a blue marble on the third draw is:
4/11
To find the probability of drawing three blue marbles in a row, we multiply these probabilities:
(6/13) * (5/12) * (4/11) = 120/5148
This simplifies to:
5/213
To express this as a percent to the nearest 10th of a percent, we divide the numerator by the denominator, then multiply by 100 and round to the nearest tenth:
5/213 ≈ 0.0234
0.0234 * 100 ≈ 2.3%
Therefore, the probability, to the nearest 10th of a percent, that all three marbles drawn will be blue is 2.3%.
3 + 6 + 4 = 13
We want to find the probability of drawing three blue marbles in a row. To do this, we need to multiply the probabilities of drawing a blue marble on each of the three draws, since the draws are independent of each other.
On the first draw, there are 6 blue marbles out of 13 total marbles, so the probability of drawing a blue marble on the first draw is:
6/13
On the second draw, there will be 5 blue marbles left out of 12 total marbles (since we already drew one marble), so the probability of drawing a blue marble on the second draw is:
5/12
On the third draw, there will be 4 blue marbles left out of 11 total marbles (since we already drew two marbles), so the probability of drawing a blue marble on the third draw is:
4/11
To find the probability of drawing three blue marbles in a row, we multiply these probabilities:
(6/13) * (5/12) * (4/11) = 120/5148
This simplifies to:
5/213
To express this as a percent to the nearest 10th of a percent, we divide the numerator by the denominator, then multiply by 100 and round to the nearest tenth:
5/213 ≈ 0.0234
0.0234 * 100 ≈ 2.3%
Therefore, the probability, to the nearest 10th of a percent, that all three marbles drawn will be blue is 2.3%.
A bag contains 3 red marbles, 5 blue marbles and 6 green marbles. If three marbles are drawn out of the bag, what is the probability, to the nearest 1000th, that all three marbles drawn will be blue?
To find the probability that all three marbles drawn will be blue, we need to use combinations. There are a total of 14 marbles in the bag, so the total number of ways to draw 3 marbles out of the bag is:
C(14, 3) = 364
To count the number of ways to draw 3 blue marbles out of the bag, we need to use combinations again. There are 5 blue marbles in the bag, so the number of ways to draw 3 blue marbles out of 5 is:
C(5, 3) = 10
Therefore, the probability that all three marbles drawn will be blue is:
10/364
Simplifying, we get:
1/36.4
Expressing this as a decimal to the nearest 1000th, we get:
0.0275
Therefore, the probability, to the nearest 1000th, that all three marbles drawn will be blue is 0.0275.
C(14, 3) = 364
To count the number of ways to draw 3 blue marbles out of the bag, we need to use combinations again. There are 5 blue marbles in the bag, so the number of ways to draw 3 blue marbles out of 5 is:
C(5, 3) = 10
Therefore, the probability that all three marbles drawn will be blue is:
10/364
Simplifying, we get:
1/36.4
Expressing this as a decimal to the nearest 1000th, we get:
0.0275
Therefore, the probability, to the nearest 1000th, that all three marbles drawn will be blue is 0.0275.
A bag contains 7 red marbles, 2 blue marbles and 3 green marbles. If two marbles are drawn out of the bag, what is the exact probability that both marbles drawn will be blue?
The total number of marbles in the bag is:
7 + 2 + 3 = 12
We want to find the probability of drawing two blue marbles in a row. To do this, we need to multiply the probabilities of drawing a blue marble on each of the two draws, since the draws are independent of each other.
On the first draw, there are 2 blue marbles out of 12 total marbles, so the probability of drawing a blue marble on the first draw is:
2/12
On the second draw, assuming that a blue marble was drawn on the first draw, there will only be 1 blue marble left out of 11 total marbles, so the probability of drawing another blue marble on the second draw is:
1/11
Therefore, the probability of drawing two blue marbles in a row is:
(2/12) * (1/11) = 1/66
Therefore, the exact probability that both marbles drawn will be blue is 1/66.
7 + 2 + 3 = 12
We want to find the probability of drawing two blue marbles in a row. To do this, we need to multiply the probabilities of drawing a blue marble on each of the two draws, since the draws are independent of each other.
On the first draw, there are 2 blue marbles out of 12 total marbles, so the probability of drawing a blue marble on the first draw is:
2/12
On the second draw, assuming that a blue marble was drawn on the first draw, there will only be 1 blue marble left out of 11 total marbles, so the probability of drawing another blue marble on the second draw is:
1/11
Therefore, the probability of drawing two blue marbles in a row is:
(2/12) * (1/11) = 1/66
Therefore, the exact probability that both marbles drawn will be blue is 1/66.
A bag contains 2 red marbles, 4 blue marbles and 7 green marbles. If two marbles are drawn out of the bag, what is the exact probability that both marbles drawn will be blue?
The total number of marbles in the bag is:
2 + 4 + 7 = 13
We want to find the probability of drawing two blue marbles in a row. To do this, we need to multiply the probabilities of drawing a blue marble on each of the two draws, since the draws are independent of each other.
On the first draw, there are 4 blue marbles out of 13 total marbles, so the probability of drawing a blue marble on the first draw is:
4/13
On the second draw, assuming that a blue marble was drawn on the first draw, there will only be 3 blue marbles left out of 12 total marbles, so the probability of drawing another blue marble on the second draw is:
3/12
Therefore, the probability of drawing two blue marbles in a row is:
(4/13) * (3/12) = 1/13
Therefore, the exact probability that both marbles drawn will be blue is 1/13.
2 + 4 + 7 = 13
We want to find the probability of drawing two blue marbles in a row. To do this, we need to multiply the probabilities of drawing a blue marble on each of the two draws, since the draws are independent of each other.
On the first draw, there are 4 blue marbles out of 13 total marbles, so the probability of drawing a blue marble on the first draw is:
4/13
On the second draw, assuming that a blue marble was drawn on the first draw, there will only be 3 blue marbles left out of 12 total marbles, so the probability of drawing another blue marble on the second draw is:
3/12
Therefore, the probability of drawing two blue marbles in a row is:
(4/13) * (3/12) = 1/13
Therefore, the exact probability that both marbles drawn will be blue is 1/13.
1 answer