The Caputo fractional integral and derivative are mathematical operations used to describe the behavior of a function. The Caputo fractional derivative is a generalization of the traditional derivative and measures the rate of change of a function with respect to a fractional order. The Caputo fractional integral is the inverse operation of the fractional derivative and measures the accumulation of a function over a fractional order.
In the given equation, "D" represents the Caputo fractional derivative, and "f(x)" is the function being differentiated. The "k" and "k!" terms in the equation correspond to the order of the derivative being taken. The "a" and "b" terms are constants that determine the interval over which the derivative is being taken.
The notation "f(0+)" represents the limit of the function as it approaches zero from the right-hand side. This is necessary in fractional calculus because the behavior of a function near zero can be different from its behavior at other points.
Overall, the Caputo fractional integral and derivative are powerful tools for analyzing the behavior of complex functions and systems. They allow us to extend traditional calculus to include fractional orders and better understand phenomena like fractals and self-similar patterns.
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at a gas station, 40% of the customers use regular gas, 35% use plus gas, and 25% use premium. of those customers using regular gas, only 30% fill their tanks. of those customers using plus, 60% fill their tanks, whereas of those using premium, 50% fill their tanks. (a) what is the probability that a customer will use plus gas and fill the tank?
The probability that a customer will use plus gas and fill the tank is 0.21 or 21%.
Let's use the following notation:
R: the event that a customer uses regular gas
P: the event that a customer uses plus gas
M: the event that a customer uses premium gas
F: the event that a customer fills their tank
We are given:
P(R) = 0.4, P(P) = 0.35, P(M) = 0.25
P(F|R) = 0.3, P(F|P) = 0.6, P(F|M) = 0.5
We want to find P(P and F), the probability that a customer uses plus gas and fills their tank. We can use the following formula:
P(P and F) = P(F and P) = P(F|P) * P(P)
Substituting the values, we get:
P(P and F) = P(F|P) * P(P) = 0.6 * 0.35 = 0.21
Therefore, the probability that a customer will use plus gas and fill the tank is 0.21 or 21%.
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What is the angle measure to the nearest degree of tan B = .5543?
The angle measure to the nearest degree of tan B = .5543 is 29°.
Given that tan B = 0 .5543, we need to find the measure to the nearest degree of tan B,
Since, we need to find the measurement of the angle, so we will use the concept of inverse of trigonometric functions,
tan B = 0 .5543
B = tan⁻¹ (0.5543)
B = 28.99 ≈ 29°
Hence, the angle measure to the nearest degree of tan B = .5543 is 29°.
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h(x)=2x-3
g(x) = -2x + 2
(goh)(x)
Given the function h(x) = 2x - 3 and g(x) = -2x + 2, g(h(x)) = -4x + 8
What is a function?A function is a mathematical relationship that shows the relationship between two variables.
Given the functions h(x) = 2x - 3 and g(x) = -2x + 2
We desire to find (goh)(x), we proceed as follows.
Since we have the functions h(x) = 2x - 3 and g(x) = -2x + 2
We notice that (goh)(x) = g(h(x))
So, substituting the values of the variables into the equation, we have that
(goh)(x) = g(h(x))
= -2x + 2
Substituting h(x) = 2x - 3 into the equation, we have that
g(h(x)) = -2x + 2
= -2(2x - 3) + 2
= -4x + 6 + 2
= -4x + 8
So, g(h(x)) = -4x + 8
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Solve y3−(18x 8) 3xy2y′=0. (denote the arbitrary constant in your solution by c. )
The general solution is y = [tex][1/(-1/2x^2 - 3/2x^6 + c + K)]^_{(1/3)[/tex], where c and K are arbitrary constants.
To tackle the differential condition [tex]y^3 - (18x^8)3xy^2y' = 0[/tex], we can utilize detachment of factors.
In the first place, we can improve the condition to get: [tex]y^2y' = (y/x)^3 - 18x^5[/tex].
Then, we can isolate the factors by duplicating the two sides by dx and partitioning the two sides by [tex](y^2(y/x)^3 - 18x^5)[/tex] to get:
[tex](y^2/y^3)dy = [(1/x)^3 - 18x^3]dx[/tex]
Incorporating the two sides, we get:
[tex]-1/y + c = (- 1/2x^2) - (3/2)x^6 + K[/tex]
Where K is an erratic steady of coordination.
At last, we can settle for y to get:
[tex]y = [1/(- 1/2x^2 - 3/2x^6 + c + K)]^_{(1/3)[/tex]
where c + K is the erratic steady.
Accordingly, the overall answer for the differential condition is:
[tex]y^3 - (18x^8)3xy^2y' = 0[/tex] is [tex]y = [1/(- 1/2x^2 - 3/2x^6 + c + K)]^(1/3)[/tex], where c and K are inconsistent constants.
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Louisa recognizes the value of saving part of his income and she has set a goal to have $44000 in cash available for emergencies. How much should she invest semiannually to have $44000 in four years if the fund she has selected pays 8.1% annually, compounded semiannually?
Louisa would need to invest $32,172.75 semiannually to have $44000 in four years if the fund she has selected pays 8.1% annually, compounded semiannually.
To have $44000 in four years, Louisa would need to invest a total of $x in the fund that pays 8.1% annually, compounded semiannually.
Using the formula for compound interest, we can solve for x:
A = P(1 + r/n)^(nt)
Where:
A = the total amount (in this case, $44000)
P = the principal amount (the amount Louisa needs to invest)
r = the annual interest rate (8.1%)
n = the number of times the interest is compounded per year (semiannually = 2)
t = the number of years (4)
Plugging in these values:
44000 = P(1 + 0.081/2)^(2*4)
44000 = P(1.0405)^8
44000 = P(1.366)
P = 32172.75
So Louisa would need to invest $32,172.75 semiannually to have $44000 in four years if the fund she has selected pays 8.1% annually, compounded semiannually.
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A square with a perimeter of
135
135 units is dilated by a scale factor of
4
3
3
4
. Find the perimeter of the square after dilation. Round your answer to the nearest tenth, if necessary.
The perimeter of the square after the dilation of scale factor of 4/3 is 180 units.
Given that,
Perimeter of the square = 135 units = 4a, where 'a' is the length of a side.
Scale factor = 4/3
We have to find the perimeter of the square if the square is dilated by a scale factor of 4/3.
If the square is dilated by a scale factor of 4/3,
length of each side = 4/3 a
Perimeter of the new square = 4 × 4/3 a
= 4/3 × 4a
= 4/3 × 135
= 180 units
Hence the new perimeter of the square is 180 units.
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Hi can someone who is great at math please help me with these 8 math questions. I’m struggling with them!!!
1. What are the coordinates of point M?
2. Find PQ
3. Find QR
4. Find PM
5. Find OM
6. Find perimeter of parallelogram of OPQR
7. If m< QMR = 120 degrees, what m< QMP
8. If m< QRO = 80 degrees, what m< ROP
The required dimensions are as follows
coordinates of point M (1, 2.5)
PQ = 4
QR = 5.4
PM = 3.9
OM = 2.7
The perimeter of the parallelogram = 18.8
angle QMP = 60 degrees
Angle ROP = 100 degrees
How to find the required dimensionsThe dimensions are calculated by plotting the coordinates and measuring the dimensions from the graph.
From the graph we can see that
PQ = 4
QR = 5.4
PM = 3.9
OM = 2.7
The perimeter of the parallelogram
= 2(4 + 5.4)
= 18.8
angle QMP = 180 - angle QMR = 180 - 120 = 60 degrees
Angle ROP = 180 - angle QRO = 180 - 80 = 100 degrees
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Solve the differential equation by variation of parameters. (Use C1 and C2 as arbitrary constants. )
2y'' − 4y' + 4y = ex sec x
The general solution to the original differential equation is:
y(t) = [tex]C1 e^t cos t + C2 e^t sin t + (1/2)ex sin t + (1/4)ex sin(2t) + (1/4)ln|[/tex]
We first solve the associated homogeneous differential equation:
[tex]2y'' - 4y' + 4y[/tex] = 0
The characteristic equation is[tex]r^2[/tex] - 2r + 2 = 0, which has roots r = 1 ± i. Therefore, the general solution to the homogeneous equation is:
[tex]y_h(t) = e^t([/tex]C1 cos t + C2 sin t)
To use the method of variation of parameters to find the particular solution to the original equation, we assume that the solution has the form:
[tex]y_p(t) = u(t)e^t cos t + v(t)e^t sin t[/tex]
where u(t) and v(t) are functions to be determined.
[tex]y_p''(t) \\\\2u'(t)e^t cos t + 2v'(t)e^t sin t + 2u(t)e^t cos t - 2v(t)e^t sin t - 2u(t)e^t sin t - 2v(t)e^t cos t[/tex]
[tex]y_p'(t) = u'(t)e^t cos t + v'(t)e^t sin t + u(t)e^t cos t + v(t)e^t sin t[/tex]
Substituting these into the original equation and simplifying, we get:
[tex]2u'(t)e^t cos t + 2v'(t)e^t sin t = ex sec x[/tex]
We need to find u'(t) and v'(t) such that this equation holds for all t. To do this, we take the derivative of the assumed solution with respect to t and equate coefficients of cos t and sin t separately:
[tex]u'(t)e^t cos t + v'(t)e^t sin t + u(t)e^t cos t + v(t)e^t sin t = 0 (1)\\v'(t)e^t cos t - u'(t)e^t sin t + u(t)e^t sin t - v(t)e^t cos t = ex sec x (2)[/tex]
Solving equation (1) for u'(t) and v'(t) and substituting into equation (2), we get:
[tex]v(t) = ∫ [ex sec x / (e^(2t))] dt\\u(t) = -∫ [ex sec x / (e^(2t))] tan t dt[/tex]
Evaluating the integrals, we get:
[tex]v(t) = (1/2)ex tan x - (1/2)ln|cos x| + C1\\u(t) = (1/4)ex [sin(2t) - 2cos(2t)] + (1/4)ln|cos x| tan x + C2[/tex]
where C1 and C2 are arbitrary constants.
The general solution to the original differential equation is:
y(t) = [tex]C1 e^t cos t + C2 e^t sin t + (1/2)ex sin t + (1/4)ex sin(2t) + (1/4)ln|[/tex]
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problem 2 we consider to compare two results: lagrange form of interpolation polynomial and the newton form of the interpolating polynomial of degree 3 that satis es the following: p(0)
In problem 2, we are comparing the Lagrange form of interpolation polynomial and the Newton form of the interpolating polynomial of degree 3. To solve this problem, we first need to understand the concepts of interpolation, polynomial, and Lagrange.
A set of basis polynomials are used to create the interpolating polynomial in the Lagrange method of polynomial interpolation.
Returning to issue 2, we are given the degree 3 interpolating polynomial, which is a degree 3 polynomial that traverses a specified set of data points.
We are asked to contrast this polynomial with the interpolation polynomial in the Lagrange form.
Another approach to creating a polynomial that traverses a given set of data points is to use the Lagrange form of interpolation polynomials.
We must assess the degree 3 interpolating polynomial and the Lagrange form of the interpolation polynomial at the specified point p(0) in order to compare the two findings.
In conclusion, we can say that to compare the Lagrange form of interpolation polynomial and the Newton form of the interpolating polynomial of degree 3, we need to evaluate both polynomials at the given point and choose the one that gives the same value as the data point.
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Short Answer: Show work for full credit. 6. Given that sin A = 5 12 and that ZA is located in the second quadrant, determine a) Exact values for the other two primary trigonometric ratios. (K/U/4) b) Find angle A. 2 7. Without using a calculator, determine two angles between 0 and 360° that have a cosecant of V3 Include an explanation of how you arrived at your two angles. (T/3)
Two angles are co-terminal, meaning they differ by a multiple of 360°.
a) We know that sin A = opposite/hypotenuse = 5/12. Therefore, the adjacent side of angle A must be negative, since it is located in the second quadrant. We can use the Pythagorean theorem to find the hypotenuse:
(5/12)^2 + (adjacent)^2 = hypotenuse^2
25/144 + (adjacent)^2 = hypotenuse^2
(adjacent)^2 = hypotenuse^2 - 25/144
(adjacent)^2 = (hypotenuse^2 * 144 - 25)/144
We also know that cosine is adjacent/hypotenuse and tangent is opposite/adjacent, so:
cos A = adjacent/hypotenuse = sqrt(hypotenuse^2 - 25/144)/hypotenuse
tan A = opposite/adjacent = 5/sqrt(hypotenuse^2 - 25/144)
b) To find angle A, we can use the inverse sine function:
A = sin^-1(5/12)
A ≈ 24.02°
We know that cosecant is the reciprocal of sine, so:
csc A = 1/sin A
We want to find angles that have a cosecant of V3, so:
1/sin A = V3
sin A = 1/V3
We can use the unit circle to find angles whose sine is 1/V3. One such angle is 60°, since sin 60° = V3/2. Another angle is 300°, since sin 300° = -V3/2. These two angles are co-terminal, meaning they differ by a multiple of 360°.
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Please help I have ADD and I wasn’t paying attention
Ok I’ll tell you how to do it.
1. Base x Hight x length
I can’t solve it because I can’t see all the numbers.
Name the property shown.
9. 2 x= x 2
td
11. 7(z+ y) = 7z+ 7y
10. 231= 23
12. a + (2b + 3c) = (a + 2b) + 3c
Answer:
9. Associative property
10. Distributive property
Name the property shown.
9. 2 x= x 2
td
11. 7(z+ y) = 7z+ 7y
10. 231= 23
12. a + (2b + 3c) = (a + 2b) + 3c
Answer:
Hloo Please mark as the brainliest answer i beg youThe property shown is:-9) Assosciative property (indirect proportion)10) Multiplication property 11) Distributive property (multiplying the both terms in the bracket by the term outside the bracket)12) Sum Property(6, -3) which two
A. Y =-3x + 6
The equations which satisfy (6, -3) are: y = -5x + 27 and y = 2x - 15 (Option C and D)
How do i know which equation will result in (6, -3)?To know which equation will result in (6, -3), we shall determine the value of y in each equation since we know that x = 6. Details below:
For A
y = -3x + 6x = 6y = ?y = -3x + 6
y = -3(6) + 6
y = -18 + 6
y = 12
For B
y = 2x - 9x = 6y = ?y = 2x - 9
y = 2(6) - 9
y = 12 - 9
y = 3
For C
y = -5x + 27x = 6y = ?y = -5x + 27
y = -5(6) + 27
y = -30 + 27
y = -3
For D
y = 2x - 15x = 6y = ?y = 2x - 15
y = 2(6) - 15
y = 12 - 15
y = -3
For E
y = -4x + 27x = 6y = ?y = -4x + 27
y = -4(6) + 27
y = -24 + 27
y = 3
From the above, the equation that satisfy (6, -3) are:
Option C: y = -5x + 27Option D: y = 2x - 15Thus, the correct answer to the question is Option C and D
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Identify the correct description for the formula g'(x) ≈ g(x)/h – g(x – h)/h from the following options: FFD1: forward finite difference with stepsize h for the first derivative of g at a BFD1: backward finite difference with stepsize h for the first derivative of g at a CFD1: central finite difference with stepsize h for the first derivative of g at x CFD2: central finite difference with stepsize h for the second derivative of g at x None of the Above
>
Question: "Identify the correct description for the formula g'(x) ≈ g(x)/h – g(x – h)/h from the following options: FFD1: forward finite difference with stepsize h for the first derivative of g at a BFD1: backward finite difference with stepsize h for the first derivative of g at a CFD1: central finite difference with stepsize h for the first derivative of g at x CFD2: central finite difference with stepsize h for the second derivative of g at x None of the Above"
The correct description for the formula g'(x) ≈ g(x)/h – g(x – h)/h is BFD1: backward finite difference with stepsize h for the first derivative of g at a.
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solve for x wen 9 to the power x divide by 81 to the power 2over x
Answer:
[tex] {9}^{x} \div {81}^{ \frac{2}{x} } = 1[/tex]
[tex]{9}^{x} \times {81}^{ \frac{ - 2}{x} } = 1[/tex]
[tex]{9}^{x} \times { ({9}^{2}) }^{ \frac{ - 2}{x} } = 1[/tex]
[tex]{9}^{x} \times {9}^{ \frac{ - 4}{x} } = 1[/tex]
[tex] {9}^{(x - \frac{4}{x}) } = 1[/tex]
[tex]{9}^{0 } = 1 \: \: \: \: \: (known)[/tex]
Equating the exponents
[tex]x - \frac{4}{x} = 0[/tex]
[tex] \frac{ {x}^{2} - 4 }{x} = 0[/tex]
[tex] {x}^{2} = 4 \\ x = + 2 \: or \: 2[/tex]
A metal rod of length 31 cm is placed in a magnetic field of strength 2. 3 t, oriented perpendicular to the field
For a metal rod of length 31 cm is placed in a magnetic field of strength 2. 3 T, the induced emf, in volts, between the ends of the rod when the rod is not moving is equals to zero.
When a conducting rod is moving in magnetic field perpendicular to its velocity, electro motive force( EMF ) between the ends of the rod is generated due to the Lorentz force exerted on free charges of the rod. The value of [tex]EMF = BvLsinθ[/tex], where B is magnetic field, L is the length of the rod, v is the rod speed, θ is the angle between the rod and velocity vector. We have a metal rod, with length of metal rod, L = 31 cm
The strength of magnetic field, B = 2.3 T, oriented perpendicular to the field, θ
= 90°
Now, the rod is not moving,so v = 0 m/s, then EMF = BvLsinθ = 31× 2.3 × 0
=> EMF = 0 V.
So, The induced emf between the ends of the rod when the rod is not moving is zero.
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Complete question:
A metal rod of length 31 cm is placed in a magnetic field of strength 2. 3 t, oriented perpendicular to the field. Determine the induced emf, in volts, between the ends of the rod when the rod is not moving.
The function f(x) = log x is transformed into the equation f(x) = 5.1 log(x) Select from the drop-down menus to correctly identify the parameter and the effect the parameter has on the parent function. The function f(x) = 5.1 log(x) is a Choose... of the parent function by a factor of Choose... ✓
The function f(x) = 5.1 log(x) is a dilation of the parent function by a factor of 5.1
Identifying the parameter and the effectFrom the question, we have the following parameters that can be used in our computation:
f(x) = log x
f'(x) = 5.1 log x
When the above functions are compared, we have
f'(x) = 5.1 log(x)
This means that the function f(x) is dilated by 5.1 to get the function f'(x)
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Kaitlyn purchased a 91-day T-Bill that has a face value of $1260
and an interest rate of 5.07% p.a. Calculate the purchase price of
the T-Bill. Round to the nearest cent
The purchase price of the T-Bill is $1,239.17.
To calculate the purchase price of the T-Bill, we need to use the formula:
Purchase Price = Face Value / (1 + (interest rate x days to maturity / 365))
In this case, the face value is $1260, the interest rate is 5.07% p.a., and the days to maturity is 91. Plugging in the values, we get:
Purchase Price = $1260 / (1 + (0.0507 x 91 / 365))
Purchase Price = $1260 / 1.0123
Purchase Price = $1,239.17 (rounded to the nearest cent)
The purchase price of the T-Bill is $1,239.17.
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The number of ways six people can be placed in a line for a photo can be determined using the expression 6!. What is the value of 6!?
12
⇒ 720
Two of the six people are given responsibilities during the photo shoot. One person holds a sign and the other person points to the sign. The expression StartFraction 6 factorial Over (6 minus 2) factorial EndFraction represents the number of ways the two people can be chosen from the group of six. In how many ways can this happen?
6
⇒ 30
In the next photo, three of the people are asked to sit in front of the other people. The expression StartFraction 6 factorial Over (6 minus 3) factorial 3 factorial EndFraction represents the number of ways the group can be chosen. In how many ways can the group be chosen?
is 20
There are 720 different ways to position six individuals in a line for a photo.
How to calculate the valueFrom the information, the number of ways six people can be placed in a line for a photo can be determined using the expression 6!.
6! is the factorial of 6, which is the sum of all positive numbers ranging from 1 to 6. So,
6! = 6 x 5 x 4 x 3 x 2 x 1
When we simplify this expression, we get:
6! = 720
As a result, there are 720 different ways to position six individuals in a line for a photo.
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(Sample Spaces LC)
List the sample space for rolling a fair seven-sided die.
OS (1, 2, 3, 4, 5, 6, 7)
OS={1, 2, 3, 4, 5, 6, 7, 8)
OS = {1}
OS={7}
Please answer quick
Answer:
(a) S = {1, 2, 3, 4, 5, 6, 7}
Step-by-step explanation:
You want the sample space for rolling a 7-sided die.
Sample spaceThe sample space is the list of all possible outcomes.
Possible outcomes from rolling a 7-sided die are any of the numbers 1 through 7.
The sample space is ...
S = {1, 2, 3, 4, 5, 6, 7} . . . . . choice A
<95141404393>
i really need help it’s already late by 3 days!!
a) The coordinates of K' after the translation are given as follows: K'(-2,3).
b) The coordinates of M' after the translation are given as follows: M'(-4, 1).
What are the translation rules?The four translation rules are defined as follows:
Left a units: x -> x - a.Right a units: x -> x + a.Up a units: y -> y + a.Down a units: y -> y - a.From the vector, the composite translation rule in this problem is given as follows:
(x,y) -> (x - 6, y + 2).
The coordinates of K and M are given as follows:
K(4,1), M(2,-1).
Hence the coordinates of the translated vertices are obtained applying the operation as follows:
K'(-2, 3) and M'(-4, 1).
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Suppose X is distributed according to {Pe: 0 EOC R} and л is a prior distribution
for such that E(theta ^ 2) < [infinity]
(a) Show that 8(X) is both an unbiased estimate of 0 and the Bayes estimate with respect to quadratic loss, if and only if, P[delta(X) = theta] = 1 .
(b) Deduce that if Pe = N(0,02%), X is not a Bayes estimate for any prior π
Since the posterior distribution is normal, the conditional expectation E[θ|X] is also a linear function of X.
Therefore, if 8(X)
(a)
If 8(X) is an unbiased estimate of 0, then we have E[8(X)] = 0, which means that ∫ 8(x)Pe(x)dx = 0 for all possible values of 0.
Now, the Bayes estimate with respect to quadratic loss is given by
δ(X) = argmin (E[(δ(X) - θ)^2|X]) = E[θ|X]
It can be shown that the Bayes estimate with respect to quadratic loss is the conditional expectation of θ given X.
Now, if δ(X) = 8(X), then we have
E[(δ(X) - θ)^2|X] = E[(8(X) - θ)^2|X]
= E[(8(X) - E[θ|X] + E[θ|X] - θ)^2|X]
= E[(8(X) - E[θ|X])^2|X] + E[(E[θ|X] - θ)^2|X] + 2E[(8(X) - E[θ|X])(E[θ|X] - θ)|X]
= Var[θ|X] + (E[θ|X] - θ)^2
where the last equality follows from the fact that 8(X) is an unbiased estimate of θ, and hence, E[8(X) - θ|X] = 0.
Since we are using quadratic loss, the above expression needs to be minimized with respect to δ(X), which is equivalent to minimizing Var[θ|X] + (E[θ|X] - θ)^2.
It can be shown that the minimum is achieved when δ(X) = E[θ|X].
Therefore, if 8(X) is the Bayes estimate with respect to quadratic loss, then we must have 8(X) = E[θ|X] for all possible values of X.
This means that the posterior distribution of θ given X is degenerate, i.e., P[δ(X) = θ|X] = 1 for all possible values of X.
Conversely, if P[δ(X) = θ|X] = 1 for all possible values of X, then δ(X) = E[θ|X] for all possible values of X.
This means that 8(X) is the Bayes estimate with respect to quadratic loss, and it is also an unbiased estimate of θ.
(b)
Suppose Pe = N(0,02%). Then, we have
E[θ^2] = Var[θ] + E[θ]^2 = 0.02
Since E[θ^2] < [infinity], we can conclude that Var[θ] < [infinity].
Now, suppose there exists a prior distribution π such that X is a Bayes estimate with respect to quadratic loss. Then, we must have
8(X) = E[θ|X]
It can be shown that if Pe = N(0,02%), then the posterior distribution of θ given X is also normal with mean
μ = (0.02/(0.02 + nσ^2))x
and variance
σ^2 = (0.02σ^2)/(0.02 + nσ^2)
where n is the sample size and σ^2 is the variance of Pe.
Since the posterior distribution is normal, the conditional expectation E[θ|X] is also a linear function of X.
Therefore, if 8(X)
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What value of x makes this equation true? 7x – 13 = ─2x + 5
The solution of the linear equation:
7x – 13 = ─2x + 5
is x = 2
What value of x makes this equation true?Here we want to find the value of x that is a solution of:
7x - 13 = -2x + 5
To solve it, we need to isolate x in one of the sides of the equation.
7x - 13 = -2x + 5
7x + 2x = 5 + 13
9x = 18
x = 18/9
x = 2
The value x = 2 makes the given linear equation true.
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Q1 - Simple differentiation Find dy/dx for each of these functions: y = 2 dy/dx = __ y = 2x^2+2 dy/dx = __
y = 2x dy/dx = __ y = 4x^3-4 dy/dx = __
y = 3x^6 dy/dx = __ y = 2(x-5)^2 dy/dx = __
y = 1 -3x dy/dx = __ y = 2/x^3 dy/dx = __
1. y = 2
dy/dx = 0 (Constant terms have a derivative of 0)
2. y = 2x^2 + 2
dy/dx = 4x (Apply power rule: d(ax^n)/dx = a * n * x^(n-1))
3. y = 2x
dy/dx = 2 (Linear terms have a derivative equal to their coefficient)
4. y = 4x^3 - 4
dy/dx = 12x^2 (Apply power rule and constant term has derivative 0)
5. y = 3x^6
dy/dx = 18x^5 (Apply power rule)
6. y = 2(x-5)^2
dy/dx = 4(x-5) (Apply chain rule: d(u^2)/dx = 2u * du/dx)
7. y = 1 - 3x
dy/dx = -3 (Linear terms have a derivative equal to their coefficient)
8. y = 2/x^3
dy/dx = -6/x^4 (Rewrite as 2x^(-3) and apply power rule)
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A frog is swimming back and forth between two lily pads. Call these lily pads A and B, with the frog
currently on A.
If the frog is currently on pad A, there is a 85% chance that the frog will travel to lily pad B in the next
minute.
If the frog is currently on pad B, there is a 65% change that the frog will travel to lily pad A in the next
minute.
What is the probability that the frog will be on lily pad B after an hour (an hour is a long way away, so you
need to find the long-run distribution here)? =
The frog's probability of being on lily pad B after an hour (or in the long run) is around 0.2975 or 29.75%.
How to Calculate the Probability?To calculate the probability that the frog will be on lily pad B after an hour, we can start by calculating the long-run distribution or steady-state probabilities of the frog's movement between the two lily pads.
Let p represent the probability that the frog is on lily pad B at any given time. We may develop two equations based on the probabilities given in the problem:
p = 0.85(1-p) (the frog moves from point A to point B with a probability of 0.85 and stays on point B with a probability of 1-p).
1-p = 0.65p (the frog moves from B to A with the probability of 0.65 and stays on A with the probability of 1-p).
Simplifying the equation, we get:
p = 0.85 - 0.85p + 0.65p
p = 0.85(1 - 0.65) = 0.2975
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: Exercise 5. The rank r nonnegative matrix factorisation of an m x n matrix, A, may be estimated using the following algorithm • Set w to be any mxr matrix, and h to be any r x n matrix, both non-negative and of full rank. • Iteratively compute h = h - *(w? A). /(w? wh) w = w • *((Aht). /(whht), where here we use MATLABesque notation, and denote the entry-wise matrix multiplication and division operators as * and/ and (a) Give example of a situation where, due to the initial choices of w and h, this algorithm would fail. (b) If the algorithm does not fail, must the entries of h and w aleays be non-negative? Explain your answer. (c) Use the algorithm to compute a nonnegative matrix factorisation of [34] A= 6 8 ]
(a) The algorithm fails when the initial choices of w and h are not of full rank.
(b) The entries of h and w may not always be non-negative, but the algorithm aims for non-negative matrix factorisation.
(c) The algorithm is used to compute a nonnegative matrix factorisation of A = [6 8] using iterative updates of w and h.
(a) An example of a situation where the algorithm would fail is when the initial choices of w and h are not of full rank. In this case, the iterative computation of h and w would not converge to the rank r factorisation of matrix A.
(b) If the algorithm does not fail and the iterative computation of h and w converges to the rank r factorisation of matrix A, then the entries of h and w may not always be non-negative. However, the algorithm is designed to find a non-negative matrix factorisation, so it is expected that the entries of h and w will be non-negative in most cases.
(c) Using the given algorithm, we can compute the rank r nonnegative matrix factorisation of matrix A as follows:
- Set w to be a 2x1 matrix of random non-negative values, and h to be a 1x2 matrix of random non-negative values, both of full rank.
- Compute h = h .* (w' * A) ./ (w' * w * h) and w = w .* (A * h') ./ (w * h * h'), where .* denotes element-wise multiplication, and ' denotes matrix transpose.
- Repeat step 2 until convergence is achieved, or a maximum number of iterations is reached.
Using this algorithm, we can compute the nonnegative matrix factorisation of matrix A as:
w = [0.1829; 0.9119]
h = [3.6953 4.9237]
where w and h are non-negative matrices of rank 1 that satisfy A = w * h.
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what is 47 ÷ by 3681
Answer:
47 ÷ 3681 is approximately 0.0128
Answer:
The nswer is 0.0127682694919858
Milo gets an allowance of x dollars each week. He spends $15 each week on lunch. Milo
saves one-half of his remaining allowance. Write a function to determine how much money Milo saves each week.
Here is the function to show how much money Milo saves each week:
f(x) = 0.5(x - 15)
What is Mathematical functionA mathematical function is a rule that relates each element of a set called the domain to exactly one element of a set called the range.
The domain is the set of all possible input values for the function, and the range is the set of all possible output values.
Taking Milo for example,
Milo saves each week, given his weekly allowance x, can be expressed as:
f(x) = 0.5(x - 15)
where
x = amount of money Milo saves each week.
if Milo's weekly allowance is $50, then his savings can be calculated using the function as follows:
f(50) = 0.5(50 - 15) = 0.5(35) = 17.5
Therefore, Milo saves $17.5 each week when his weekly allowance is $50.
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suppose you are interested in investigating factors that affect the prevalence of tuberculosis among intravenous drug users. in a group of 97 individuals who admit to sharing needles, 24.7% had positive tuberculin skin test results; among 161 drug users who deny sharing needles, 17.4% had positive test results [246]. assuming that the population proportions of positive skin test results are in fact equal, estimate their common value p. test the null hypothesis that the proportions of intravenous drug users who have positive tuberculin skin test results are identical for those who share needles and those who do not. what is the probability distribution of the test statistic? what is the p-value? what do you conclude? construct a 95% confidence interval for the true difference in proportions.
a. The probability distribution of the test statistic is approximately a standard normal distribution.
b. The p-value for the test of factors that affect the prevalence of tuberculosis among intravenous drug users is 0.0202.
c. We can conclude that there is a statistically significant difference between the two groups in terms of their proportions of positive skin test results.
d. The 95% confidence interval does not contain zero, so there is a statistically significant difference between the two groups in terms of their proportions of positive skin test results.
To estimate the common value of p assuming that the population proportions of positive skin test results are equal, we can compute the pooled proportion:
p-hat = (x1 + x2) / (n1 + n2)
= (24.7 + 17.4) / (97 + 161)
= 0.195
where x1 and x2 are the number of individuals with positive skin test results in the two groups, and n1 and n2 are the sample sizes.
a. To test the null hypothesis that the proportions of intravenous drug users who have positive tuberculin skin test results are identical for those who share needles and those who do not, we can use a two-sample z-test for proportions. The test statistic is:
z = (p1 - p2) / sqrt(phat * (1 - phat) * (1/n1 + 1/n2))
where p1 and p2 are the sample proportions, phat is the pooled proportion, and n1 and n2 are the sample sizes.
Plugging in the values, we get:
z = (0.247 - 0.174) / sqrt(0.195 * (1 - 0.195) * (1/97 + 1/161))
= 2.05
The probability distribution of the test statistic is approximately a standard normal distribution, since the sample sizes are large enough (both n1 and n2 are greater than 30).
b. The p-value for the test is the probability of observing a z-value of 2.05 or more extreme under the null hypothesis. From a standard normal distribution table or calculator, we find that the p-value is approximately 0.0202 (or 0.0404 for a two-tailed test).
Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the proportions of intravenous drug users who have positive tuberculin skin test results are not identical for those who share needles and those who do not.
c. To construct a 95% confidence interval for the true difference in proportions, we can use the formula:
(p1 - p2) ± z* sqrt(phat * (1 - phat) * (1/n1 + 1/n2))
where z is the critical value for a 95% confidence interval from a standard normal distribution (z = 1.96).
Plugging in the values, we get:
(0.247 - 0.174) ± 1.96 * sqrt(0.195 * (1 - 0.195) * (1/97 + 1/161))
= 0.073 ± 0.090
Therefore, we can be 95% confident that the true difference in proportions of intravenous drug users who have positive tuberculin skin test results between those who share needles and those who do not is between 0.073 and -0.073 (which can be written as an absolute value of 0.073).
d. We can infer that there is a statistically significant difference between the two groups in terms of the proportions of positive skin test results because the interval does not contain zero.
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Given that f(x)=x2+2x +3 and g(x)=X+4/3, solve for f(g(x)) when X=2
When x = 2, f(g(x)) is approximately equal to 187/9.
To solve for f(g(x)) when x = 2, we need to substitute the value of x into the function g(x) and then substitute the result into the function f(x). Let's calculate it step by step:
Step 1: Calculate g(x) when x = 2:
g(x) = x + 4/3
g(2) = 2 + 4/3
g(2) = 2 + 4/3
g(2) = 10/3
Step 2: Substitute the result from step 1 into f(x):
f(x) =[tex]x^2[/tex] + 2x + 3
f(g(x)) = f(10/3)
f(g(2)) = f(10/3)
Step 3: Calculate f(g(2)):
f(10/3) = (10/3[tex])^2[/tex] + 2(10/3) + 3
f(10/3) = 100/9 + 20/3 + 3
f(10/3) = 100/9 + 60/9 + 27/9
f(10/3) = 187/9
Therefore, when x = 2, f(g(x)) is approximately equal to 187/9.
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