What can you tell about the mean of eachdistribution
1 answer:
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Answer:
a. 5 b. c. 148.5 d. 1/7
Step-by-step explanation:
Here is the complete question
Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.
Solution
a. f"(2)
f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x
f"(2) = cos(π × 2) + 2 × 2
f"(2) = cos(2π) + 4
f"(2) = 1 + 4
f"(2) = 5
b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0
We first find f(x) by integrating f'(x)
f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C
f(0) = 2 so,
2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C
2 = -cos(0)/π + 0 + 0 + C
2 = -1/π + C
C = 2 + 1/π
f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π
f(x) = [1-cos(πx)]/π + x³/3 +3x + 2
y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)
The tangent to y is thus dy/dx
dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx
dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)
at x = 0,
dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)
dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)
dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)
dy/dx = -(0/π + 2)⁻²(3)
dy/dx = -(0 + 2)⁻²(3)
dy/dx = -(2)⁻²(3)
dy/dx = -3/4
At x = 0,
y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)
y = 1/([1-cos(0)]/π + 0 + 0 + 2)
y = 1/([1 - 1]/π + 2)
y = 1/(0/π + 2)
y = 1/(0 + 2)
y = 1/2
So, the equation of the tangent at (0, 1/2) is
c. If g(x) = f (√(3x² + 4). Find g'(2)
g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2
g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)
g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)
g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)
g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)
g'(2) = [6sin4π + 576 + 18]/4)
g'(2) = [6 × 0 + 576 + 18]/4)
g'(2) = [0 + 576 + 18]/4)
g'(2) = 594/4
g'(2) = 148.5
d. If h be the inverse function of f. Find h' (2)
If h(x) = f⁻¹(x)
then h'(x) = 1/f'(x)
h'(x) = 1/(sin(πx) + x² +3)
h'(2) = 1/(sin(π2) + 2² +3)
h'(2) = 1/(sin(2π) + 4 +3)
h'(2) = 1/(0 + 4 +3)
h'(2) = 1/7
Answer:
a)
X[bar]= 71.8cm
X[bar]= 72cm
b)
M.A.D.= 8.16cm
M.A.D.= 5.4cm
c) The data set for Soil A is more variable.
Step-by-step explanation:
Hello!
The data in the stem-and-leaf plots show the heights in cm of Teddy Bear sunflowers grown in two different types of soil (A and B)
To read the data shown in the plots, remember that the first digit of the number is shown in the stem and the second digit is placed in the leaves.
The two data sets, in this case, are arranged in a "back to back" stem plot, which allows you to compare both distributions. In this type of graph, there is one single stem in the middle, shared by both samples, and the leaves are placed to its left and right of it corresponds to the observations of each one of them.
Since the stem is shared by both samples, there can be observations made only in one of the samples. For example in the first row, the stem value is 5, for the "Soil A" sample there is no leaf, this means that there was no plant of 50 ≤ X < 60 but for "Soil B" there was one observation of 59 cm.
X represents the variable of interest, as said before, the height of the Teddy Bear sunflowers.
a) To calculate the average or mean of a data set you have to add all observations of the sample and divide it by the number of observations:
X[bar]= ∑X/n
For soil A
Observations:
61, 61, 62, 65, 70, 71, 75, 81, 82, 90
The total of observations is n= 10
∑X= 61 + 61 + 62 + 65 + 70 + 71 + 75 + 81 + 82 + 90= 718
X[bar]= ∑X
/n
= 218/10= 71.8cm
For Soil B
Observations:
59, 63, 69, 70, 72, 73, 76, 77, 78, 83
The total of observations is n= 10
∑X= 59 + 63 + 69 + 70 + 72 + 73 + 76 + 77 + 78 + 83= 720
X[bar]= ∑X
/n
= 720/10= 72cm
b) The mean absolute deviation is the average of the absolute deviations of the sample. It is a summary of the sample's dispersion, meaning the greater its value, the greater the sample dispersion.
To calculate the mean absolute dispersion you have to:
1) Find the mean of the sample (done in the previous item)
2) Calculate the absolute difference of each observation and the sample mean |X-X[bar]|
3) Add all absolute differences
4) Divide the summation by the number of observations (sample size,n)
For Soil A
1) X[bar]= 71.8cm
2) Absolute differences |X-X[bar]
|
|61-71.8|= 10.8
|61-71.8|= 10.8
|62-71.8|= 9.8
|65-71.8|= 6.8
|70-71.8|= 1.8
|71-71.8|= 0.8
|75-71.8|= 3.2
|81-71.8|= 9.2
|82-71.8|= 10.2
|90-71.8|= 18.2
3) Summation of all absolute differences
∑|X-X[bar]
|= 10.8 + 10.8 + 9.8 + 6.8 + 1.8 + 0.8 + 3.2 + 9.2 + 10.2 + 18.2= 81.6
4) M.A.D.=∑|X
-X[bar]
|/n
= 81.6/10= 8.16cm
For Soil B
1) X[bar]= 72cm
2) Absolute differences |X-X[bar]
|
|59-72|= 13
|63-72|= 9
|69-72|= 3
|70-72|= 2
|72-72|= 0
|73-72|= 1
|76-72|= 4
|77-72|= 5
|78-72|= 6
|83-72|= 11
3) Summation of all absolute differences
∑ |X-X[bar]
|= 13 + 9 + 3 + 2 + 0 + 1 + 4 + 5 + 6 + 11= 54
4) M.A.D.=∑ |X
-X[bar]
|/n
= 54/10= 5.4cm
c)
If you compare both calculated mean absolute deviations, you can see M.A.D. > M.A.D.
. As said before, the M.A.D. summary of the sample's dispersion. The greater value obtained for "Soil A" indicates this sample has greater variability.
I hope this helps!