Answer:
300
Step-by-step explanation:
Find the area of the parallelogram determined by the vectors i +j+k and –2i + 3j + K.
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
Hello!
(Data and full text attached)
The stem and leaf plot is a way to present quantitative data. Considering two-digit numbers, for example 50, the tens digits are arranged in the stem and the units determine the leafs.
So for the stem and leaf showing the digs of the top players of the tournament, the observed data is:
41, 41, 43, 43, 45, 50, 52, 53, 54, 62, 63, 63, 67, 75, 97
n= 15
Note that in the stem it shows the number 8, but with no leaf in that row, that means that there were no "eighties" observed.
a) 6 Players had more than 60 digs.
b)
To calculate the mean you have to use the following formula:
X[bar]= ∑x/n= (41 + 41 + 43 + 43 + 45 + 50 + 42 + 53 + 54 + 62 + 63 + 63 + 67 + 75 + 97)/15= 849/15= 56.6 digs
To calculate the median you have to calculate its position and then identify its value out of the observed data arranged from least to greatest:
PosMe= (n+1)/2= (15+1)/2= 8 ⇒ The median is in the eight place:
41, 41, 43, 43, 45, 50, 42, 53, 54, 62, 63, 63, 67, 75, 97
The median is Me= 53
53 is the value that separates the data in exact halves.
The mode is the most observed value (with more absolute frequency).
Consider the values that were recorded more than once
41, 41
43, 43
63, 63
41, 43 and 63 are the values with most absolute frequency, which means that this distribution is multimodal and has three modes:
Md₁: 41
Md₂: 43
Md₃: 63
The Range is the difference between the maximum value and the minimum value of the data set:
R= max- min= 97 - 41= 56
c)
The distribution is asymmetrical, right skewed and tri-modal.
Md₁: 41 < Md₂: 43 < Me= 53 < X[bar]= 56.6 < Md₃: 63
Outlier: 97
d)
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
PosQ₁= 16/4= 4
Q₁= 43
PosQ₃= 16*3/4= 12
Q₃= 63
IQR= 63 - 43= 20
Q₁ - 1.5 IQR = 43 - 1.5*20= 13 ⇒ There are no values 13 and below, there are no lower outliers.
Q₃ + 1.5 IQR = 63 + 1.5*20= 93 ⇒ There is one value registered above the calculated limit, the last observation 97 is the only outlier of the sample.
The mean is highly affected by outliers, its value is always modified by the magnitude of the outliers and "moves" its position towards the direction of them.
Calculated mean with the outlier: X[bar]= 849/15= 56.6 digs
Calculated mean without the outlier: X[bar]= 752/14= 53.71 digs
I hope this helps.
Answer:
see the explanation
Step-by-step explanation:
we know that
The surface area of a prism is given by the formula
where
B is the area of the base
P is the perimeter of the base
H is the height of the prism
For both prism the formula is the same
How is it different?
Rectangular prism
The area of the base is given by the formula
-----> is the area of rectangle
where
L is the length and W is the width of rectangle
The perimeter of the base is given by
---> is the perimeter of rectangle
The height H is the same in both cases
Triangular prism
The area of the base is given by the formula
-----> is the area of the triangular base
where
b is the base of triangle and h is the height of triangle
The perimeter of the base is given by
---> is the perimeter of triangle
where
a, b and c are the length sides of triangle
The height H is the same in both cases