Six groups of students sell 162 balloons at the school carnival. There are 3 students in each group. If each student sells the s
2 answers:
Each student sell
Further Explanation:
Given:
Six groups of students sell 162 balloons at the school carnival.
There are 3 students in each group.
Each student sells the same number of balloons.
Explanation:
There are 162 balloons that are sold.
There are six groups and each group contains 3 students.
The total number of students in all six groups can be obtained as follows,
Number of balloon each student sell can be obtained as follows,
Each student sell
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Number System
Keywords: six groups, groups, students, sell, six, 162 balloons, school, carnival, 3 students, each group, school carnival, same, number of balloons, student sell, many balloons, 3 students in a group.
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Answer:
Step-by-step explanation:
Hello!
a)
To compare both datasets I've made a box plot for them.
The yellow one belongs to the points scored by the men and the green one is for the points scored by the women.
Women: The box seems symmetrical, the median and the mean (black square) are almost the same. The whiskers are almost the same length, you could say that in general the points scored by the women have a symmetrical distribution.
Men: The box looks a little skewed to the right, the median is closer to the 1st quantile and the mean is greater than the median. The whiskers are the same length, in general the distribution seems almost symmetrical.
b)
As you can see in the box plot the median and mean of the points scored by the women are almost the same. Using the data I've calculated both values:
Me= 24.50
X[bar]= 23.80
The median divides the sample in halves (50-50), it shows you where the middle of the distribution is.
The average shows you the value that centers the distribution, meaning, it is the value around which you'll find most of the data set. It summarizes best the sample information and therefore is a better measure of center.
c.
As you can see in the box plots, there are no outliers in both distributions.
An outlier is an observation that is significantly distant from the rest of the data set. They usually represent experimental errors (such as a measurement) or atypical observations. Some statistical measurements, such as the sample mean, are severely affected by this type of values and their presence tends to cause misleading results on a statistical analysis.
Considering the 1st quartile (Q₁), the 3rd quartile (Q₃) and the interquartile range IQR, any value X is considered an outlier if:
X < Q₁ - 1.5 IQR
X > Q₃ + 1.5 IQR
Or extreme outliers if:
X < Q₁ - 3 IQR
X > Q₃ + 3 IQR
For the women data set:
Q₁= 17; Q₃= 30; IQR= 30 - 17= 13
Lower outliers: X < Q₁ - 1.5 IQR= 17-1.5*13= -2.5 ⇒ The minimum value recorded is 03, so there are no outliers.
Upper outliers: X > Q₃ + 3 IQR= 30 + 1.5*13= 49.5 ⇒ The maximum value registered is 41, so there are no outliers.
I hope this helps!
9514 1404 393
Answer:
5. 88.0°
6. 13.0°
7. 52.4°
8. 117.8°
Step-by-step explanation:
For angle A between sides b and c, the law of cosines formula can be solved to find the angle as ...
A = arccos((b² +c² -a²)/(2bc))
When calculations are repetitive, I find a spreadsheet useful. It doesn't mind doing the same thing over and over, and it usually makes fewer mistakes.
Here, the side opposite x° is put in column 'a', so angle A is the value of x. The order of the other two sides is irrelevant.
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<em>Additional comment</em>
The spreadsheet ACOS function returns the angle in radians. The DEGREES function must be used to convert it to degrees. The formula for the first problem is shown here:
=degrees(ACOS((C3^2+D3^2-B3^2)/(2*C3*D3)))
As you can probably tell from the formula, side 'a' is listed in column B of the spreadsheet.
The spreadsheet rounds the results. This means the angle total is sometimes 179.9 and sometimes 180.1 when we expect the sum of angles to be 180.0.
