In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that a student who does not have a dog has a cat?
Has a cat Does not have a cat
Has a dog 9 12
Does not have a dog 3 5
The probability that a student who does not have a dog has a cat can be found by taking the number of students who have a cat but do not have a dog (which is 12) and dividing by the total number of students who do not have a dog (which is 3+5 = 8). Therefore, the probability is:
12/8 = 3/2 or 1.5
However, probabilities cannot be greater than 1, so the answer is not valid. This might be because of some mistake in the data or in the calculation. It could be that there is some overlap between the two categories (some students could have both a cat and a dog), or that the data is incomplete. In any case, without more information, we cannot give a valid probability for this question.
The following table represents the highest educational attainment of all adult residents in a certain town. If an adult is chosen randomly from the town, what is the probability that they have a high school degree or some college, but have no college degree? Round your answer to the nearest thousandth.
Age 20-29 Age 30-39 Age 40-49 Age 50 & over Total
High school only 931 541 855 1242 3569
Some college 2306 1873 1474 2392 8045
Bachelor's degree 719 437 1189 1116 3461
Master's degree 885 656 308 1324 3173
Total 4841 3507 3826 6074 18248
The number of adults who have a high school degree or some college but no college degree is the sum of the values in the "High school only" and "Some college" rows, but not in the "Bachelor's degree" or "Master's degree" rows. Therefore:
High school degree or some college, no college degree = 931 + 541 + 855 + 2306 + 1873 + 1474
= 7979
The total number of adults in the town is 18,248, so the probability that an adult chosen randomly from the town has a high school degree or some college, but no college degree is:
7979 / 18248 ≈ 0.437
Rounding to the nearest thousandth, the answer is 0.437.
The following table represents the highest educational attainment of all adult residents in a certain town. If a resident who is 30-49 years old is chosen at random, what is the probability that they have completed a bachelor's or master's degree? Round your answer to the nearest thousandth.
Age 20-29 Age 30-39 Age 40-49 Age 50 & over Total
High school only 992 816 1155 1266 4229
Some college 1947 912 1275 2171 6305
Bachelor's degree 1197 1320 1272 614 4403
Master's degree 1055 774 213 872 2914
Total 5191 3822 3915 4923 17851
To find the probability that a resident who is 30-49 years old has completed a bachelor's or master's degree, we need to add up the number of residents in the "Bachelor's degree" and "Master's degree" rows that fall in the "Age 30-39" and "Age 40-49" columns:
Bachelor's or master's degree, 30-49 years old = 1272 + 1320 + 213 + 774 = 2579
The total number of residents who are 30-49 years old is the sum of the values in the "Age 30-39" and "Age 40-49" columns:
Total residents, 30-49 years old = 3822 + 3915 = 7737
Therefore, the probability that a resident who is 30-49 years old has completed a bachelor's or master's degree is:
2579 / 7737 ≈ 0.333
Rounding to the nearest thousandth, the answer is 0.333.
A lock has a code of 3 numbers from 1 to 25. If no numbers in the code are allowed to repeat, how many different codes could be made?
There are 25 options for the first number in the code, since it can be any number from 1 to 25. For the second number, there are only 24 options left, since one number has already been used. Similarly, for the third number, there are only 23 options left. Therefore, the total number of possible codes is:
25 × 24 × 23 = 13,800
So there are 13,800 different codes that could be made.
Ms. Bell's mathematics class consists of 6 sophomores, 11 juniors, and 13 seniors. How many different ways can Ms. Bell create a 5-member committee of seniors if each senior has an equal chance of being selected?
There are 13 seniors in the class, and we need to choose a committee of 5. We can use the combination formula to calculate the number of ways to do this:
C(13, 5) = 13! / (5! * (13-5)!) = 1287
Therefore, there are 1,287 different ways for Ms. Bell to create a 5-member committee of seniors.
In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If 12 boys and 7 girls are competing, how many different ways could the six medals possibly be given out?
A committee must be formed with 2 teachers and 6 students. If there are 8 teachers to choose from, and 16 students, how many different ways could the committee be made?
We need to choose 2 teachers out of 8, and 6 students out of 16. We can use the combination formula to calculate this:
C(8, 2) = 8! / (2! * (8-2)!) = 28
C(16, 6) = 16! / (6! * (16-6)!) = 8008
Once the 2 teachers and 6 students have been chosen, there are no further choices to make. Therefore, the total number of ways to form the committee is:
28 * 8008 = 224,224
So there are 224,224 different ways to form the committee with 2 teachers and 6 students.
When Keilantra goes bowling, her scores are normally distributed with a mean of 170 and a standard deviation of 13. Out of the 120 games that she bowled last year, how many of them would she be expected to score less than 169, to the nearest whole number?
We need to use the normal distribution formula to solve this problem:
z = (x - μ) / σ
where z is the z-score, x is the bowling score, μ is the mean score, and σ is the standard deviation.
First, we need to find the z-score for a score of 169:
z = (169 - 170) / 13 = -0.077
Next, we need to find the area under the normal distribution curve to the left of this z-score. We can use a standard normal distribution table to find this. The table gives us the area to the left of a given z-score, so we need to look for the closest z-score in the table. A z-score of -0.08 has an area of 0.4681.
Therefore, the probability of Keilantra scoring less than 169 in a single game is 0.4681.
To find the expected number of games out of 120 where she would score less than 169, we can multiply the probability of scoring less than 169 in a single game by the total number of games:
Expected number of games = 0.4681 * 120 = 56.17
Rounding to the nearest whole number, Keilantra would be expected to score less than 169 in about 56 games out of 120.
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Out of a randomly selected 1350 people from the population, how many of them would have an IQ less than 130, to the nearest whole number?
We need to use the normal distribution formula to solve this problem:
z = (x - μ) / σ
where z is the z-score, x is the IQ score, μ is the mean score, and σ is the standard deviation.
First, we need to find the z-score for an IQ of 130:
z = (130 - 100) / 15 = 2
Next, we need to find the area under the normal distribution curve to the left of this z-score. We can use a standard normal distribution table to find this. The table gives us the area to the left of a given z-score, so we need to look for the closest z-score in the table. A z-score of 2 has an area of 0.9772.
Therefore, the proportion of people with an IQ less than 130 in the population is 0.9772.
To find the number of people out of 1350 who would have an IQ less than 130, we can multiply the proportion by the total number of people in the sample:
Number of people = 0.9772 * 1350 = 1319.22
Rounding to the nearest whole number, about 1319 people out of 1350 people would have an IQ less than 130.
For the girls, we need to choose 3 out of 7 to receive medals. We can use the combination formula to calculate this:
C(7, 3) = 7! / (3! * (7-3)!) = 35
Similarly, for the boys, we need to choose 3 out of 12 to receive medals:
C(12, 3) = 12! / (3! * (12-3)!) = 220
Once the medalists have been chosen for the girls and for the boys, there are 3 possible medals to distribute to the girls and 3 possible medals to distribute to the boys. Therefore, the total number of ways to award the 6 medals is:
35 * 220 * 3 * 2 * 1 * 3 * 2 * 1 = 9,979,200
So there are 9,979,200 different ways to award the six medals to the top three girls and top three boys.