We start by put the first period in ascending order as shown in diagram below
The measurement of center can use either Mean, Median (Q2), Mode, Range or Interquartile range
The Median, Q2, is the value between 2 and 3 which is 2.5
The mean of the data is the sum of data divided by the number of data = 2.8
The mode of the data is the most frequent value which is the 1, 2, 3, and 4
The range of the data is the difference between the first and the last value which is 8
The interquartile of data is the difference between Q3 and Q1 which is 3
Since the data is widely distributed with a range of 8, hence the best measure of center is the mean since the calculation takes all values within the range
Answer:
the answer is 8
Step-by-step explanation:
i got this right on my question
300 is the LCM of 25 and 30 =)
Answer:
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Step-by-step explanation:
I will use the letter x instead of theta.
Then the problem is, given sec(x) + tan(x) = P, show that
sin(x) = [P^2 - 1] / [P^2 + 1]
I am going to take a non regular path.
First, develop a little the left side of the first equation:
sec(x) + tan(x) = 1 / cos(x) + sin(x) / cos(x) = [1 + sin(x)] / cos(x)
and that is equal to P.
Second, develop the rigth side of the second equation:
[p^2 - 1] / [p^2 + 1] =
= [ { [1 + sin(x)] / cos(x) }^2 - 1] / [ { [1 + sin(x)] / cos(x)}^2 +1 ] =
= { [1 + sin(x)]^2 - [cos(x)]^2 } / { [1 + sin(x)]^2 + [cos(x)]^2 } =
= {1 + 2sin(x) + [sin(x)^2] - [cos(x)^2] } / {1 + 2sin(x) + [sin(x)^2] + [cos(x)^2] }
= {2sin(x) + [sin(x)]^2 + [sin(x)]^2 } / { 1 + 2 sin(x) + 1} =
= {2sin(x) + 2 [sin(x)]^2 } / {2 + 2sin(x)} = {2sin(x) ( 1 + sin(x)} / {2(1+sin(x)} =
= sin(x)
Then, working with the first equation, we have proved that [p^2 - 1] / [p^2 + 1] = sin(x), the second equation.
