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.
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.
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.
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%.
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.
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.
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.