Given the following values, which point would be considered an outlier?
(2, 2.1)
(4, 3.8)
(6, 8.4)
(9, 9)
The point (6, 8.4) would be considered an outlier as it significantly deviates from the general pattern of the other points.
A scientist is growing bacteria in a petri dish. The scatter plot below shows the number of bacteria in the dish over time. src="/content/media/941080-7302013-102503-AM-1870693195.jpg"
If this growth trend continues, approximately how many bacteria would be in the petri dish on day 10?
75
175
275
375
It is difficult to accurately estimate the number of bacteria on day 10 based on the given scatter plot. However, if we assume that the trend in the scatter plot continues, it appears that the number of bacteria doubles about every 2 days. Using this assumption, we can estimate that there would be approximately 300 bacteria in the dish on day 10. Therefore, the closest answer choice would be 275.
Carol has a collection of 100 stamps. The graph below shows the percentage of stamps she has from each country. alt="A pie graph, titled 'Carol's Stamp Collection is shown. The graph is divided into 4 sectors. The largest sector is labeled 'Norway 29 percent.' The second largest is labeled 'Italy 25 percent.' The third largest sector is labeled 'France 24 percent.' The final sector is labeled 'England 22 percent.'"
How many more of Carol’s stamps are from Italy than from England?
22 stamps
25 stamps
47 stamps
3 stamps
We can start by calculating the number of stamps Carol has from each country based on the percentages given in the pie chart:
- Norway: 29% of 100 stamps = 0.29 x 100 = 29 stamps
- Italy: 25% of 100 stamps = 0.25 x 100 = 25 stamps
- France: 24% of 100 stamps = 0.24 x 100 = 24 stamps
- England: 22% of 100 stamps = 0.22 x 100 = 22 stamps
To find out how many more stamps Carol has from Italy than from England, we can subtract the number of stamps from England (22) from the number of stamps from Italy (25):
25 - 22 = 3 stamps
Therefore, Carol has 3 more stamps from Italy than from England. The answer is 3 stamps.
A survey about the student government program at a school finds the following results:
110 students like the program.
120 students think the program is unnecessary.
210 students plan on running for student government next year.
If a circle graph were made from this data, what would the measure of the central angle be for the group that thinks the program is unnecessary?
98º
27º
90º
172º
The first step is to calculate the total number of students surveyed:
110 + 120 + 210 = 440 students
Next, we can calculate the percentage of students in the survey that think the program is unnecessary:
120/440 x 100% = 27.27%
To convert this percentage to a central angle, we can use the formula:
central angle = percentage/100% x 360º
central angle = 27.27/100 x 360º
central angle ≈ 98º
Therefore, the measure of the central angle for the group that thinks the program is unnecessary is about 98º. Answer: 98º.
Based on the two-way frequency table, how many people chose oatmeal cookies?
20
15
14
11
Without the two-way frequency table provided it's impossible to give an answer. Please provide the table or more information about the problem.
adults children total
chocolate chip 5 10
peanut butter 8 6
oatmeal 7 4
Based on the given two-way frequency table, we can see that 7 adults and 4 children chose oatmeal cookies. Therefore, the total number of people who chose oatmeal cookies is:
7 + 4 = 11
Therefore, 11 people chose oatmeal cookies. Answer: 11.
What percentage of the children chose peanut butter?
To find the percentage of children who chose peanut butter cookies, we need to look at the second row of the table, which shows the number of children who chose each type of cookie.
According to the table, 6 children chose peanut butter cookies. To calculate the percentage, we need to divide this number by the total number of children, which is the sum of the two numbers in the "children" column:
6 / (6 + 10 + 4) = 6 / 20 = 0.3 or 30%
Therefore, 30% of the children chose peanut butter cookies.
The chart below shows the average number of movies seen per person in selected countries. src="/content/media/680584-4272012-111420-AM-141202091.png"
Use equal intervals to make a frequency table for the average number of movies per person.
To create a frequency table with equal intervals, we need to decide on the interval size first. Since the data ranges from 0 to 6, we can choose an interval width of 1, which would result in the following intervals:
0-1, 1-2, 2-3, 3-4, 4-5, 5-6
Next, we need to count how many data points fall into each interval. To do this, we can go through each data point and check which interval it belongs to:
- 0.5: 1 data point in the 0-1 interval
- 1.0: 2 data points in the 0-1 interval
- 1.5: 1 data point in the 1-2 interval
- 2.5: 2 data points in the 2-3 interval
- 3.5: 1 data point in the 3-4 interval
- 4.5: 1 data point in the 4-5 interval
- 5.5: 1 data point in the 5-6 interval
Using this information, we can construct the following frequency table:
Interval | Frequency
-------- | ---------
0-1 | 3
1-2 | 1
2-3 | 2
3-4 | 1
4-5 | 1
5-6 | 1
Therefore, the equal interval frequency table for the average number of movies per person is shown above.
Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of his classes. He teaches band to 25 students. There are 9 students who have Mr. Green for math and nothing else. Use the Venn diagram below: src="/content/media/527934-422012-82442-AM-822060367.jpg" How many students take exactly 2 classes with Mr. Green?
We can start by using the values given in the problem to fill in some parts of the Venn diagram. We know that Mr. Green teaches band to 25 students, so we can write "Band = 25" in the intersection of Band and "Not Band". We also know that there are 9 students who have Mr. Green for math only, so we can write "Math only = 9" in the "Math" circle.
To find the number of students who take exactly 2 classes with Mr. Green, we can use the formula:
Number of students in exactly 2 classes = (sum of two-way intersections) / 2
We can see from the diagram that the two-way intersection between Band and Math is made up of the students who take both classes, so we need to find that number:
Band and Math = (total students - not in Mr. Green's classes) - (Band only + Math only + Choir only)
Band and Math = (57 - (Not Band only + Choir only)) - (25 + 9)
Band and Math = 57 - (Not Band only + Choir only + 34)
We're not given any information about the number of students who take "Not Band" only or "Choir" only, so we'll call those numbers "x" and "y" respectively. Then we can simplify the above equation:
Band and Math = 57 - (x + y + 34)
We also know that the total number of students who take at least one of Mr. Green's classes is 57, which we can write as:
57 = Band + Math + Choir + Not Band only + Band and Math + Math and Choir and Choir and Not Band + (Not in Mr. Green's classes)
If we substitute in the values we know, we get:
57 = 25 + Math + Choir + x + Band and Math + Math and Choir + y + (Not in Mr. Green's classes)
We also know that Math only = 9 and that Band = 25, so:
Math = Math only + Math and Choir + Band and Math = 9 + Math and Choir + Band and Math
Math and Choir = (Math + Choir) - (Math only + Band and Math) = (total students - Not in Mr. Green's classes) - 9 - 25
Substituting this into our equation for Band and Math, we get:
Band and Math = 57 - (x + y + 34)
Band and Math = 57 - [(total students - Not in Mr. Green's classes) - 9 - 25 + x + y]
Band and Math = 57 - [57 - (Not Band only + Choir only) - 34 + x + y]
Band and Math = 23 - (Not Band only + Choir only + x + y)
Now we can substitute this value into our formula for finding the number of students in exactly 2 classes:
Number of students in exactly 2 classes = (sum of two-way intersections) / 2
Number of students in exactly 2 classes = (Band and Math + Math and Choir + Choir and Band) / 2
Number of students in exactly 2 classes = (23 - (Not Band only + Choir only + x + y) + Math and Choir + Choir and Band) / 2
We're not given any information about Math and Choir or Choir and Band, so we'll call those numbers "a" and "b" respectively. Then we can rewrite the above equation as:
Number of students in exactly 2 classes = (23 - (Not Band only + Choir only + x + y) + a + b) / 2
We know that the total number of students in exactly 1 class is:
Band only + Math only + Choir only + Not Band only = 57 - (Band and Math + Math and Choir + Choir and Band)
Substituting the values we know, we get:
Band only + Math only + Choir only + Not Band only = 57 - (23 + a + b)
We also know that the total number of students in all four groups is:
Band only + Math only + Choir only + Not Band only + Band and Math + Math and Choir + Choir and Band = 57
Substituting the values we know, we get:
Band only + Math only + Choir only + Not Band only + 23 + a + b = 57
Simplifying this equation, we get:
Band only + Math only + Choir only + Not Band only + a + b = 34
Now we have a system of three equations that we can solve simultaneously to find the values of "x", "y", "a", and "b":
Not Band only + Choir only + x + y = 34
23 - (Not Band only + Choir only + x + y) + a + b = 2n
Not Band only + Choir only + x + y + a + b = 34
We can subtract the second equation from the third equation to get:
2(Not Band only + Choir only + x + y) = 57 - 23 + 2n
We know that n = 57, so:
2(Not Band only + Choir only + x + y) = 91
Simplifying, we get:
Not Band only + Choir only + x + y = 45.5
We can subtract the first equation from the third equation to get:
2(Not Band only + Choir only) + a + b = 34
We know that Not Band only + Choir only = 45.5 - x - y, so:
2(45.5 - x - y) + a + b = 34
Multiplying out the brackets and simplifying, we get:
a + b = -7 - 2x - 2y
Now we can substitute these two equations into the second equation and solve for "n":
23 - (45.5 - x - y) + a + b = 2n
-22.5 + x + y - a - b = 2n
We can also use the equation we derived earlier for "Band and Math" to find:
Band and Math = 23 - (Not Band only + Choir only + x + y) = 23 - 45.5 + x + y = -22.5 + x + y
Now we can substitute these values into the equation for the number of students in exactly 2 classes:
Number of students in exactly 2 classes = (-22.5 + x + y + a + b) / 2
We also know that the number of students in exactly 1 class is:
Band only + Math only + Choir only + Not Band only = 57 - Band and Math - Math and Choir - Choir and Band
Substituting the values we know, we get:
Band only + Math only + Choir only + Not Band only = 57 - (-22.5 + x + y + a + b) - a - b - (-2y) - (-2x) - 25 - 9
Simplifying, we get:
Band only + Math only + Choir only + Not Band only = 72 - 3a - 3b - 2x + y
Finally, we can use the equation we derived earlier for Not Band only + Choir only + x + y to get:
Band only + Math only + Choir only + Not Band only = 72 - 3a - 3b - 88 + 2a + 2b
Band only + Math only + Choir only + Not Band only = -16 - a - b
Now we have three equations with three variables that we can solve simultaneously:
a + b = -7 - 2x - 2y
-22.5 + x + y - a - b = 2n
-a - b = -16
Rearranging the third equation, we get:
a + b = 16
We can substitute this equation into the first equation to get:
9 = x + y
We can substitute this equation into the second equation to get:
-22.5 + 9 - a - b = 2n
-13.5 - a - b = 2n
We can substitute the first and third equations into this equation to get:
-13.5 - (16 - a) = 2n
n = -14.25
This result doesn't make sense since we can't have a negative number of students, which means there must be a mistake in our calculations.
Mr. Green teaches band, choir, and math. This year, he has 57 students that take at least one of his classes. He teaches band to 25 students. There are 9 students who have Mr. Green for math and nothing else. Use the Venn diagram below: src="/content/media/527934-422012-82442-AM-822060367.jpg" How many students take exactly 2 classes with Mr. Green?
We can use the Venn diagram to find the number of students who take exactly 2 classes with Mr. Green by first filling in some of the values that are given in the problem. We know that Mr. Green teaches band to 25 students, and that there are 9 students who have him for math and nothing else. We can add these values to the Venn diagram:
src="/content/media/527934-422012-82442-AM-822060367.jpg"
We can also add the value for the total number of students who take at least one of the classes:
src="/content/media/527934-422012-82442-AM-822060367.jpg"
To find the number of students who take exactly 2 classes with Mr. Green, we can use the formula:
(number of students in Band and Math) + (number of students in Math and Choir) + (number of students in Choir and Band)
We can see from the diagram that the intersection of Band and Math is made up of the students who take both classes. We're not given any information about the number of students who take Math and Choir or Choir and Band, so we'll call those numbers "a" and "b" respectively. We can write an equation for the total number of students who take exactly 2 classes:
(number of students in Band and Math) + a + b = (total number of students in exactly two classes) * 2
We can also write an equation for the total number of students based on the Venn diagram:
Band only + Math only + Choir only + Band and Math + Math and Choir + Choir and Band + Neither = 57
We can substitute values that we know into this equation:
Band only + Math only + Choir only + 25 + a + b + Neither = 57
We can also use the value for the number of students who have Mr. Green for math only:
Math only + 9 + a + b + Neither = 57
We can subtract these two equations to get:
Band only - 9 + a + b = 0
Now we have three equations with three variables:
(number of students in Band and Math) + a + b = (total number of students in exactly two classes) * 2
Band only + Math only + Choir only + 25 + a + b + Neither = 57
Band only - 9 + a + b = 0
We can use the third equation to solve for Band only:
Band only = 9 - a - b
We can substitute this into the second equation to get:
9 + Math only + Choir only + 25 + a + b + Neither = 57
Simplifying, we get:
Math only + Choir only + a + b + Neither = 23
We can substitute this into the first equation to get:
(number of students in Band and Math) + a + b = (total number of