Data are same
1 answer:
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Answer:
Step-by-step explanation:
Let
x ----> the length of the third side of triangle in Lauren's sculpture
we know that
The <u><em>Triangle Inequality Theorem</em></u> states that the sum of any 2 sides of a triangle must be greater than the measure of the third side
Applying the Triangle inequality Theorem
1)
Rewrite
2)
therefore
The possible values of x are
Answer:
S(t) = a.sin (b.t) + d
a = -1.5, b = (2π/365), d = 3.47
S(t) = -1.5 sin (2πt/365) + 3.47
Step-by-step explanation:
Complete Question is presented in the attached image to this solution.
- Dingane has been observing a certain stock for the last few years and he sees that it can be modeled as a function S(t) of time t (in days) using a sinusoidal expression of the form
S(t) = a.sin(b.t) + d.
On day t = 0, the stock is at its average value of $3.47 per share, but 91.25 days later, its value is down to its minimum of $1.97.
Find S(t). t should be in radians.
S(t) =
Solution
S(t) = a.sin(b.t) + d.
At t = 0, S(t) = $3.47
S(0) = a.sin(b×0) + d = a.sin 0 + d = 3.47
Sin 0 = 0,
S(t=0) = d = 3.47.
At t = 91.25 days, S(t) = $1.97
But, it is given that T has to be in radians, for t to be in radians, the constant b has to convert t in days to radians.
Hence, b = (2π/365)
S(91.25) = 1.97 = a.sin(b×91.25) + d
d = 3.47 from the first expression
S(t = 91.25) = a.sin (91.25b) + 3.47 = 1.97
1.97 = a.sin (2π×91.25/365) + 3.47
1.97 = a sin (0.5π) + 3.47
Sin 0.5π = 1
1.97 = a + 3.47
a = -1.5
Hence,
S(t) = a.sin (b.t) + d
a = -1.5, b = (2π/365), d = 3.47
S(t) = -1.5 sin (2πt/365) + 3.47
Hope this Helps!!!
<span>Given the table below where Low represents the fewest
points scored and High represents the highest points scored by a single
team member.
![\begin{center} \begin{tabular} {|c||c|c|c|c|c|c|c|} Team & Low & High & Range & Mean & Median & IQR & \sigma \\ [1ex] Team A & 22 & 58 & 36 & 42.1 & 44 & 18.25 & 10.35 \\ Team B & 38 & 49 & 11 & 43.9 & 44.5 & 3.5 & 2.97 \\ Team C & 27 & 36 & 9 & 31.8 & 32 & 3.75 & 2.55 \end{tabular} \end{center}](https://tex.z-dn.net/?f=%5Cbegin%7Bcenter%7D%0A%5Cbegin%7Btabular%7D%0A%7B%7Cc%7C%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7Cc%7C%7D%0ATeam%20%26%20Low%20%26%20High%20%26%20Range%20%26%20Mean%20%26%20Median%20%26%20IQR%20%26%20%5Csigma%20%5C%5C%20%5B1ex%5D%0ATeam%20A%20%26%2022%20%26%2058%20%26%2036%20%26%2042.1%20%26%2044%20%26%2018.25%20%26%2010.35%20%5C%5C%20%0ATeam%20B%20%26%2038%20%26%2049%20%26%2011%20%26%2043.9%20%26%2044.5%20%26%203.5%20%26%202.97%20%5C%5C%20%0ATeam%20C%20%26%2027%20%26%2036%20%26%209%20%26%2031.8%20%26%2032%20%26%203.75%20%26%20%20%20%20%202.55%0A%5Cend%7Btabular%7D%0A%5Cend%7Bcenter%7D)
PART A:
The range, the IQR (inter-quartile range) and the standard deviations are measures which are used to measure the spread of a data set.
A dataset is more consistent relative to another data set when the standard deviation is less than that of the other data set.
From the table, it can be seen that Team C has the least standard deviation.
Therefore, </span><span>If the club wants to award the team that has the most consistent scoring among its team members, team C should be chosen.
PART B
The mean and the median is a measure of center of a data set.
The mean describes the average of a data set.
From the table, it can be seen that Team B has the highest average score.
Therefore, </span><span>If the club wants to award the team with the highest average score, team B should be chosen.</span>
PART A:
The range, the IQR (inter-quartile range) and the standard deviations are measures which are used to measure the spread of a data set.
A dataset is more consistent relative to another data set when the standard deviation is less than that of the other data set.
From the table, it can be seen that Team C has the least standard deviation.
Therefore, </span><span>If the club wants to award the team that has the most consistent scoring among its team members, team C should be chosen.
PART B
The mean and the median is a measure of center of a data set.
The mean describes the average of a data set.
From the table, it can be seen that Team B has the highest average score.
Therefore, </span><span>If the club wants to award the team with the highest average score, team B should be chosen.</span>