The utility function for x units of bread and y units of butter is f(x,y) = xy. Each unit of bread costs $1 and each unit of butter costs $7. Maximize the utility function f, if a total of $192 is available.
To maximize the utility function f, we need to follow the given steps: We need to find out the budget equation first, which is given by 1x + 7y = 192.
Let's rearrange the above equation in terms of x, we get x = 192 - 7y .....(1).
Now we need to substitute the value of x from equation (1) in the utility function equation (f(x,y) = xy), we get f(y) = (192 - 7y)y = 192y - 7y² .....(2)
Now differentiate equation (2) w.r.t y to find the maximum value of y. df/dy = 192 - 14y.
Setting df/dy to zero, we get 192 - 14y = 0 or 14y = 192 or y = 13.7 (rounded off to one decimal place).
Now we need to find out the value of x corresponding to the value of y from equation (1), x = 192 - 7y = 192 - 7(13.7) = 3.1 (rounded off to one decimal place).
Therefore, the maximum utility function value f(x,y) is given by, f(3.1, 13.7) = 3.1 × 13.7 = 42.47 (rounded off to two decimal places).
Hence, the maximum utility function value f is 42.47.
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Find the sixth term of the expansion of (x+3)8 The sixth term is (Simplify your answer)
To find the sixth term in the expansion of (x + 3)^8, we need to use the binomial theorem. The binomial theorem states that the expansion of (a + b)^n can be found by summing the terms of the form C(n, k) * a^(n-k) * b^k, where C(n, k) represents the binomial coefficient. In this case, a = x, b = 3, and n = 8.
Using the binomial coefficient formula, C(n, k) = n! / (k! * (n-k)!), we can calculate the binomial coefficients for each term in the expansion. The term with k = 6 will give us the sixth term.
In the case of (x + 3)^8, the sixth term is found by plugging in k = 6 into the binomial coefficient formula and multiplying it with the corresponding powers of x and 3. Simplifying the expression, we get:
C(8, 6) * x^(8-6) * 3^6 = 28 * x^2 * 729 = 20,412x^2.
Therefore, the sixth term in the expansion of (x + 3)^8 is 20,412x^2.
The sixth term in the expansion of (x + 3)^8 is 20,412x^2. The binomial theorem and binomial coefficient formula are used to calculate the terms in the expansion. By plugging in k = 6 into the formula and simplifying the expression, we obtain the desired result.
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The following data represent the flight time (in minutes) of a random sample of seven flights from one city to another.
287 270 260 266 257 264 258
Compute the range and sample standard deviation of flight time.
The range of the flight time data is 30 minutes, and the sample standard deviation is approximately 10.03 minutes.
To compute the range and sample standard deviation of the flight time data, we will follow these steps:
Calculate the range:
The range is the difference between the largest and the smallest values in the dataset.
In this case, the largest value is 287, and the smallest value is 257.
Range = 287 - 257 = 30.
Calculate the sample mean (average):
To compute the sample mean, we sum up all the values and divide by the number of observations.
Sum of the values = 287 + 270 + 260 + 266 + 257 + 264 + 258 = 1862.
Number of observations = 7.
Sample mean = 1862 / 7 ≈ 265.86 (rounded to two decimal places).
Calculate the deviations:
The deviation of each data point is the difference between that data point and the sample mean.
Deviation for each data point: (287 - 265.86), (270 - 265.86), (260 - 265.86), (266 - 265.86), (257 - 265.86), (264 - 265.86), (258 - 265.86).
Calculate the sum of squared deviations:
Square each deviation and sum up the squared deviations.
Sum of squared deviations = (287 - 265.86)^2 + (270 - 265.86)^2 + (260 - 265.86)^2 + (266 - 265.86)^2 + (257 - 265.86)^2 + (264 - 265.86)^2 + (258 - 265.86)^2.
Calculate the sample variance:
The sample variance is the sum of squared deviations divided by (n-1), where n is the number of observations.
Sample variance = Sum of squared deviations / (n-1).
Calculate the sample standard deviation:
The sample standard deviation is the square root of the sample variance.
Sample standard deviation = sqrt(sample variance).
Performing these calculations, we find:
Range = 30
Sample standard deviation ≈ 10.03 (rounded to two decimal places).
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check that the following differential forms are exact and find the solutions to the corresponding initial value problems.
(1) y/t+1 dt + (ln(t+1) + 3y^2 )dy = 0, y(0) = 1
(2) (3t^2y - 2t) dt + (t^3 +6y - y^2) dy = 0, y(0) = 3
The solution to the initial value problem is [tex]t^3y - t^2 = 0.[/tex]
What is Potential function?
A potential function, also known as a scalar potential or simply a potential, is a concept used in vector calculus to describe a vector field in terms of a scalar field. In the context of differential forms, a potential function is a scalar function that, when differentiated with respect to the variables involved, yields the coefficients of the differential form.
To check whether the given differential forms are exact, we can use the necessary and sufficient condition for exactness: if the partial derivative of the coefficient of dt with respect to y is equal to the partial derivative of the coefficient of dy with respect to t, then the form is exact.
Let's start with the first differential form:
[tex](1) y/t+1 dt + (ln(t+1) + 3y^2) dy = 0[/tex]
The coefficient of dt is y/(t+1), and the coefficient of dy is ln[tex](t+1) + 3y^2.[/tex]
Taking the partial derivative of the coefficient of dt with respect to y:
[tex]∂/∂y (y/(t+1)) = 1/(t+1)[/tex]
Taking the partial derivative of the coefficient of dy with respect to t:
[tex]∂/∂t (ln(t+1) + 3y^2) = 1/(t+1)[/tex]
Since the partial derivatives are equal, the form is exact.
To find the solution to the corresponding initial value problem, we need to find a potential function F(t, y) such that the partial derivatives of F with respect to t and y match the coefficients of dt and dy, respectively.
For (1), integrating the coefficient of dt with respect to t gives us the potential function:
[tex]F(t, y) = ∫(y/(t+1)) dt = y ln(t+1)[/tex]
To find the solution to the initial value problem y(0) = 1, we substitute y = 1 and t = 0 into the potential function:
F(0, 1) = 1 ln(0+1) = 0
Therefore, the solution to the initial value problem is y ln(t+1) = 0.
Moving on to the second differential form:
[tex](2) (3t^2y - 2t) dt + (t^3 + 6y - y^2) dy = 0[/tex]
The coefficient of dt is [tex]3t^2y - 2t[/tex], and the coefficient of dy is [tex]t^3 + 6y - y^2.[/tex]
Taking the partial derivative of the coefficient of dt with respect to y:
[tex]∂/∂y (3t^2y - 2t) = 3t^2[/tex]
Taking the partial derivative of the coefficient of dy with respect to t:
[tex]∂/∂t (t^3 + 6y - y^2) = 3t^2[/tex]
Since the partial derivatives are equal, the form is exact.
To find the potential function F(t, y), we integrate the coefficient of dt with respect to t:
[tex]F(t, y) = ∫(3t^2y - 2t) dt = t^3y - t^2[/tex]
The solution to the initial value problem y(0) = 3 is obtained by substituting y = 3 and t = 0 into the potential function:
[tex]F(0, 3) = 0^3(3) - 0^2 = 0[/tex]
Therefore, the solution to the initial value problem is[tex]t^3y - t^2 = 0.[/tex]
In summary:
(1) The given differential form is exact, and the solution to the corresponding initial value problem is y ln(t+1) = 0.
(2) The given differential form is exact, and the solution to the corresponding initial value problem is [tex]t^3y - t^2 = 0.[/tex]
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Luke and Bertha Johnson file jointly. Their dependent son, aged 5, live with all. The Johsons only income was $19,442 from dividends and interest. Which of the following statements is true regarding the earned income credit?
A. The Johnsons cannot claim the credit because their earned income is too high
B. The Johnsons cannot claim the credit because their AGI is too high
C. The Johnsons cannot claim the credit because their investment income is too high
D. The Johnsons cannot claim the credit because their son is too old
The Johnsons cannot claim the earned income credit because their investment income is too high.
The earned income credit is a tax credit designed to provide relief to low-income working individuals and families. To be eligible for the credit, taxpayers must have earned income, such as wages or self-employment income. Investment income, such as dividends and interest, does not count as earned income for the purposes of the earned income credit. In this case, the Johnsons' only income is from dividends and interest, which means they do not have any earned income and therefore cannot claim the credit. It is important to note that the Johnsons' AGI and the age of their son are not relevant factors for determining eligibility for the earned income credit.
Option C is the correct answer of this question.
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Solve the following system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent 2x - 3y - 5z = 2 6x + 10y +422 = 0 - 2x + 2y + 2z=1
To solve the system of equations 2x - 3y - 5z = 2, 6x + 10y + 422 = 0, and -2x + 2y + 2z = 1 using matrices and row operations, we represent the system augmented matrix form and perform row operations to simplify.
Let's represent the system of equations in augmented matrix form:
| 2 -3 -5 | 2 |
| 6 10 422 | 0 |
| -2 2 2 | 1 |
Using row operations, we can simplify the matrix to bring it to row-echelon form. By performing operations such as multiplying rows by constants, adding or subtracting rows, and swapping rows, we aim to isolate the variables and find a solution.
After performing the row operations, we reach the row-echelon form:
| 1 -1.5 -2.5 | 1 |
| 0 0 424 | -6 |
| 0 0 0 | 0 |
In the final row of the matrix, we have all zeroes in the coefficient column but a non-zero value in the constant column. This indicates an inconsistency in the system of equations. Therefore, the system has no solution and is inconsistent.
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A miniature drone costs $300 plus $25 for each set of extra propellers. What is the cost of a drone and extra sets of propellers?
Answer:
$400
Step-by-step explanation:
the drone has 4 propellers that cost 25 bucks so the drone itself is 300 so 300+25+25+25+25=400
:)
Answer: $400
Step-by-step explanation: 300+25+25+25+25=400
differential equation
7. Show that (cos x)y' + (sin x)y = x2 y(0) = 4 has a unique solution.
The initial value problem (cos x)y' + (sin x)y = x^2, y(0) = 4 has a unique solution.
To show that the given differential equation (cos x)y' + (sin x)y = x^2 with the initial condition y(0) = 4 has a unique solution, we can use the existence and uniqueness theorem for first-order linear differential equations.
The given differential equation can be written in the standard form as follows:
y' + (tan x)y = x^2/cos x
The coefficient function (tan x) and the right-hand side function (x^2/cos x) are continuous on an interval containing x = 0. Additionally, (tan x) is not equal to zero for any value of x in the interval.
According to the existence and uniqueness theorem, if the coefficient function and the right-hand side function are continuous on an interval and the coefficient function is not equal to zero on that interval, then the initial value problem has a unique solution.
In this case, (cos x), (sin x), and (x^2) are all continuous functions on an interval containing x = 0, and (tan x) is not equal to zero for any value of x in the interval. Therefore, the conditions of the existence and uniqueness theorem are satisfied.
Hence, the given initial value problem (cos x)y' + (sin x)y = x^2, y(0) = 4 has a unique solution.
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3 5 8 9 10 11 12 13 Find an equation of the circle that has center (-4.0) and passes through (5.-1). 0 9. 6 • C-C х $ ?
The equation of the circle with center (-4, 0) and passing through (5, -1) is given by (x + 4)^2 + y^2 = 82. This equation represents a circle centered at (-4, 0) with a radius of sqrt(82).
To determine the equation of a circle with center (-4, 0) and passing through the point (5, -1), we can use the general equation of a circle:
(x - h)^2 + (y - k)^2 = r^2,
where (h, k) represents the coordinates of the center and r represents the radius.
In this case, the center is (-4, 0), so we have (h, k) = (-4, 0). The circle passes through the point (5, -1), which means this point lies on the circle. Substituting these values into the equation, we have:
(5 - (-4))² + (-1 - 0)² = r²,
(5 + 4)² + (-1)² = r²,
9² + 1 = r²,
81 + 1 = r²,
82 = r²
Therefore, the equation of the circle with center (-4, 0) and passing through (5, -1) is:
(x + 4)² + y²= 82.
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Draw the region of integration where R is bounded by z 20, y 20 and x 20 and under z =4-2x - y. b) Find the mass of the volume of the solid over the region R given a density function of p(x, y, z)=
The problem involves drawing the region of integration in the three-dimensional space bounded by the planes z = 0, y = 20, and x = 20, and under the plane z = 4 - 2x - y. We also need to find the mass of the volume of the solid over this region, given a density function p(x, y, z).
To draw the region of integration, we consider the given bounds: z ≤ 20, y ≤ 20, and x ≤ 20. These inequalities define a rectangular region in the xyz-coordinate system. Additionally, we need to consider the plane z = 4 - 2x - y, which intersects the region of integration. The region of integration is the portion of the rectangular region under this plane. To find the mass of the volume of the solid over the region, we need the density function p(x, y, z). Unfortunately, the density function is not provided in the question. Without the density function, we cannot determine the mass of the volume.
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Answer all! I will up
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QUESTION 6 points Save Answer A company's revenue from selling units of an item is in 1600- of sales are increasing at the rate of its per day, how rapidy is revenue increasing in dollars per day when
The revenue is increasing at a rate of 36600 dollars per day when 190 units have been sold.
How to find the revenue?To find how rapidly the revenue is increasing when 190 units have been sold, we need to find the derivative of the revenue function with respect to time. The derivative will give us the rate of change of revenue with respect to the number of units sold.
Given:
R = 1600x - x²
We can differentiate the revenue function R with respect to x to find the rate of change of revenue with respect to the number of units sold:
dR/dx = 1600 - 2x
Now, we know that sales are increasing at a rate of 30 units per day, so dx/dt = 30 (where t represents time in days).
To find how rapidly the revenue is increasing in dollars per day, we can multiply the derivative by the rate of change of units sold:
dR/dt = (dR/dx) * (dx/dt)
= (1600 - 2x) * (30)
Now, substitute x = 190 (units sold) into the equation:
dR/dt = (1600 - 2(190)) * (30)
= (1600 - 380) * (30)
= 1220 * 30
= 36600
Therefore, the revenue is increasing at a rate of 36600 dollars per day when 190 units have been sold.
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Let C be the square with corners (+-1, +-1), oriented in the
counterclockwise direction with unit normal pointing outward. Use
Green's Theorem to calculate the outward flux of F = (-x, 2y).
We can use Green's Theorem. The theorem relates the flux of a vector field through a closed curve to the double integral of the curl of the vector field over the region enclosed by the curve.
Green's Theorem states that the outward flux of a vector field F across a closed curve C can be calculated by integrating the dot product of F and the outward unit normal vector n along the curve C. However, Green's Theorem also provides an alternative way to calculate the flux by evaluating the double integral of the curl of F over the region enclosed by the curve C.
In this case, we need to calculate the outward flux of F = (-x, 2y) across the square C. The square has sides of length 2, and its corners are (+-1, +-1). The orientation of the square is counterclockwise, and the unit normal vector points outward.
Applying Green's Theorem, we evaluate the double integral of the curl of F over the region enclosed by C. The curl of F is given by ∂F₂/∂x - ∂F₁/∂y = 2 - (-1) = 3.
The outward flux of F across C is equal to the double integral of the curl of F over the region enclosed by C, which is 3 times the area of the square. Since the square has sides of length 2, its area is 4.
Therefore, the outward flux of F across C is 3 times the area of the square, which is 3 * 4 = 12.
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A car rental company charges its customers p dollars per day to rent a car, where 35 ≤ p ≤ 120. The number of cars rented per day can be modeled by the linear function n (p) = 1200 - 10p. Determine the following: • How much should the company charge each customer per day to maximize revenue? • How many cars would be rented in one day? • What is the maximum revenue? 3 2 5 = Water leaks from a tank at a rate R(t) where R(t) = 3.1 +0.379t gallons per hour where t is the number of hours since 7 AM. Interpret S5.5 (3.1 +0.379t)dt = 7.92. A) Between 5 AM and 6:30 AM, the tank lost 7.92 gallons. B) Between 7 AM and 8:30 AM, the tank lost 7.92 gallons. C) Between 12 PM and 1:30 PM, the tank lost 7.92 gallons. D) Between 5 AM and 6:30 AM, the volume decreased to 7.92 gallons. E) Between 7 AM and 8:30 AM, the volume decreased to 7.92 gallons. F) Between 12 PM and 1:30 PM, the volume decreased to 7.92 gallons.
To determine the optimal charge per customer per day to maximize revenue for the car rental company, we need to find the value of p that maximizes the revenue function.
The revenue function is given by R(p) = p * n(p), where n(p) represents the number of cars rented per day.
Substituting the expression for n(p) into the revenue function:
R(p) = p * (1200 - 10p)
To find the value of p that maximizes the revenue, we need to find the critical points of the revenue function. These occur when the derivative of the revenue function with respect to p is equal to zero.
Taking the derivative of R(p) with respect to p:
dR/dp = 1200 - 20p
Setting the derivative equal to zero and solving for p:
1200 - 20p = 0
20p = 1200
p = 60
So, the company should charge each customer $60 per day to maximize revenue.
To determine the number of cars rented in one day, we substitute p = 60 into the function n(p):
n(60) = 1200 - 10(60)
n(60) = 1200 - 600
n(60) = 600
Therefore, 600 cars would be rented in one day.
To find the maximum revenue, substitute p = 60 into the revenue function R(p):
R(60) = 60 * (1200 - 10(60))
R(60) = 60 * (1200 - 600)
R(60) = 60 * 600
R(60) = 36000
The maximum revenue is $36,000.
For the second part of your question:
Interpreting the integral ∫[from 5 to 5.5] (3.1 + 0.379t) dt = 7.92:
The given integral represents the definite integral of the rate function R(t) = 3.1 + 0.379t over the time interval from 5 AM to 5:30 AM (or 0.5 hours).
The value of the integral, 7.92, represents the total amount of water lost from the tank during that time interval, measured in gallons.
Therefore, the interpretation is:
E) Between 7 AM and 8:30 AM, the volume decreased to 7.92 gallons.
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Evaluate the derivative of the given function for the given value of n S= 7n³-8n+1 5n-4n4 ,n=-1 S'(-1)= (Type an integer or decimal rounded to the nearest thousandth as needed.) Save Find the slope of a line tangent to the curve of the function y(x+5)(x-1) at the point (1,0). Do not multiply the factors before taking the derivative Use the derivative evaluation feature of a graphing calculator to check your result CHO Find the derivative of the function: Choose the correct answer below OA. dy (3x+5)(x¹)(x-1) (3) dx OB dy - 0) (x²) - (x²-1)(x+5) OC. dy (3x+3)(5x¹)(x-1) (5) dx D. dy = (x+5) (5x¹)(x²-1) (3) dx Clear all Check answer Help me solve this i View an example Get more help 41 A computer, using data from a refrigeration plant, estimated that in the event of a power failure the temperaturo C (inC) in the freezers would be given by C 0.041 1-20, where is the number of hours after the power failure Find the time rate of change of temperature after 20h The time rate of change after 2.0 his C/h (Round to one decimal place as needed)
The time rate of change of temperature after 20h is C/h ≈ 0.041 (rounded to one decimal place as needed).
Evaluate the derivative of the given function for the given value of n
S = 7n³ - 8n + 1 / 5n - 4n4 , n = -1S'(-1) = (Type an integer or decimal rounded to the nearest thousandth as needed.)
The given function is:
S = 7n³ - 8n + 1 / 5n - 4n4
Let's find the derivative of S to find S':
S' = [d/dn (7n³ - 8n + 1) * (5n - 4n4) - d/dn (5n - 4n4) * (7n³ - 8n + 1)] / (5n - 4n4)²S' = [(21n² - 8) * (5n - 4n4) - (5 - 16n3) * (7n³ - 8n + 1)] / (5n - 4n4)²S' = (- 160n7 + 488n4 - 121n³ + 88n² - 8n - 35) / (5n - 4n4)²S'(-1) = (- 160( - 1)7 + 488( - 1)4 - 121( - 1)³ + 88( - 1)² - 8( - 1) - 35) / (5( - 1) - 4( - 1)4)²= - 2.784 (rounded to the nearest thousandth)
Therefore, S'(-1) ≈ - 2.784.
Slope of the line tangent to the curve of the function y(x + 5)(x - 1) at the point (1,0).
The given function is: y = (x + 5)(x - 1)
To find the slope of the line tangent to the curve of the given function, we need to find the derivative of the function and substitute x = 1.dy/dx = [(x - 1)d/dx(x + 5) + (x + 5)d/dx(x - 1)]dy/dx = [(x - 1) * 1 + (x + 5) * 1]dy/dx = 2x + 4
Therefore, the slope of the line tangent to the curve of the given function at the point (1,0) is:
dy/dx = 2(1) + 4 = 6
Let's use the derivative evaluation feature of a graphing calculator to check the result:
From the graph, we can see that the slope of the tangent line at the point (1,0) is 6.
Therefore, the result is correct.
The given function is: y = (x + 5)(x - 1)
To find the derivative of the function, we use the product rule:
dy/dx = d/dx(x + 5) * (x - 1) + (x + 5) * d/dx(x - 1)dy/dx = (1) * (x - 1) + (x + 5) * (1)dy/dx = x - 1 + x + 5dy/dx = 2x + 4
Therefore, the derivative of the function is: dy/dx = 2x + 4
The time rate of change after 2.0 hrs is C/hThe temperature (in °C) in the freezer is given by:
C = 0.041t1 - 20
Where t is the number of hours after the power failure.
We are asked to find the time rate of change of temperature after 20h. We can do this by finding the derivative of C with respect to t.
dC/dt = d/dt (0.041t1 - 20)dC/dt = 0.041d/dt (t1 - 20)dC/dt = 0.041d/dt (t)
Let's find the time rate of change of temperature after 20h by substituting t = 20 in the above equation:
dC/dt = 0.041d/dt (20) = 0.041(1) = 0.041
Therefore, the time rate of change of temperature after 20h is C/h ≈ 0.041 (rounded to one decimal place as needed).
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(1) Evaluate the following integrals. +1 dr ( (b) S. (a) * cos' (In x) dx (c) ſsin(2x)e" dx (a) S JAYA =dx H (e secºx tan’ xdx useex
The answer are:
(a) The integrating value of ∫cos(ln x)dx=x sin(ln x)+C.
(b) The integrating value of ∫sin(x)dx=−cos(x)+C.
(c) The integrating value of [tex]\int\limits sin(2x)e^xdx=-\frac{1}{2} cos(2x)e^x+\frac{1}{2}\int\limits cos(2x)e^xdx.[/tex]
(d) The integrating value of [tex]\int\limits {e^{sec^2}}(x)tan(x)dx[/tex] cannot be expressed the integral in elementary functions.
What is the integral function?
The integral function, often denoted as ∫f(x)dx, is a fundamental concept in calculus. It represents the antiderivative or the indefinite integral of a given function f(x) with respect to the variable x.
The integral function measures the accumulation of the function f(x) over a given interval. It is the reverse process of differentiation, where the derivative of a function measures its rate of change. The integral function, on the other hand, measures the accumulated change or the total area under the curve of the function.
To evaluate the given integrals one by one:
(a)∫cos(ln x)dx:
To evaluate this integral, we can use the substitution method. Let u=lnx, then [tex]du=\frac{1}{x}dx[/tex] or dx=x du.
Substituting into the integral:
∫cos(u)⋅x du=∫x cos(u)du. Now, we can integrate cos(u) with respect to u:
∫ x cos(u)du=x sin(u)+C.
Substituting back u=ln x, we have:
∫cos(ln x)dx=x sin(ln x)+C.
(b)∫sin(x)dx:
The integral of sin(x) is −cos(x)+C, where C is the constant of integration. So, ∫sin(x)dx=−cos(x)+C.
(c)[tex]\int\limits sin(2x)e^xdx[/tex]:
To integrate this expression, we can use integration by parts. Let's assign u=sin(2x) and[tex]dv=e^xdx.[/tex] Then, we can find du and v as follows: du=2cos(2x)dx (by differentiating u), [tex]v=e^x[/tex] (by integrating dv).Now, we can apply the integration by parts formula:
∫u dv=u v−∫v du.
Using the above values, we have:
[tex]\int\limits sin(2x)e^xdx=-\frac{1}{2} cos(2x)e^x+\frac{1}{2}\int\limits cos(2x)e^xdx.[/tex]
Integrating [tex]cos(2x)e^x[/tex] requires further steps and cannot be expressed in terms of elementary functions.
(d)[tex]\int\limits {e^{sec^2}}(x)tan(x)dx[/tex]:
This integral does not have a standard elementary function as its antiderivative. It cannot be expressed the integral in terms of elementary functions like polynomials, exponentials, logarithms, trigonometric functions, etc. Therefore, it cannot be evaluated using standard methods and requires advanced techniques or numerical approximations for an accurate result.
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number 11
Thank you
11. Explain what it means to say that lim f(x)=5 and lim f(x) = 7. In this situation is it possible that lim f(x) exists? (6pts) 1
It is not possible for lim f(x) to exist when both lim f(x) = 5 and lim f(x) = 7 because the limit of a function must approach a unique value as x approaches a particular point.
When we say lim f(x) = 5 and lim f(x) = 7, it means that the limit of the function f(x) approaches the value 5 as x approaches a particular point, and at the same time, it approaches the value 7 as x approaches the same point.
However, for a limit to exist, the limit value must be unique. In this situation, since the limits of f(x) approach two different values (5 and 7), it violates the requirement for a limit to have a single value. Therefore, it is not possible for lim f(x) to exist in this scenario.
The existence of a limit implies that the function approaches a well-defined value as x gets arbitrarily close to a certain point. When the limits approach different values, it indicates that the function does not have a consistent behavior near that point, leading to the non-existence of the limit.
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if the positive integer x leaves a remainder of 2 when divided by 8, what will the remainder be when x 9 is divided by 8?
The remainder when a positive integer x leaves a remainder of 2 when divided by 8 and x+9 is divided by 8 is 5.
If the positive integer x leaves a remainder of 2 when divided by 8, then we can say that x = 8k + 2, where k is an integer.
Now, if we divide x+9 by 8, we get:
(x+9)/8 = (8k + 2 + 9)/8
= (8k + 11)/8
= k + (11/8)
So, the remainder when x+9 is divided by 8 is 11/8. However, since we are dealing with integers, the remainder can only be a whole number between 0 and 7.
Therefore, we need to subtract the quotient (k) from the expression above and multiply the resulting decimal by 8 to get the remainder:
Remainder = (11/8 - k) x 8
Since k is an integer, the only possible values for (11/8 - k) are -3/8, 5/8, 13/8, etc. The closest whole number to 5/8 is 1, so we can say that:
Remainder = (11/8 - k) x 8 ≈ (5/8) x 8 = 5
Therefore, the remainder when x+9 is divided by 8 is 5.
If a positive integer x leaves a remainder of 2 when divided by 8, then x can be expressed as 8k + 2, where k is an integer. To find the remainder when x+9 is divided by 8, we divide x+9 by 8 and subtract the quotient from the decimal part. The resulting decimal multiplied by 8 gives us the remainder. In this case, the decimal is 11/8, which is closest to 1. Thus, we subtract the quotient k from 11/8 and multiply the result by 8 to get the remainder of 5.
The remainder when a positive integer x leaves a remainder of 2 when divided by 8 and x+9 is divided by 8 is 5.
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Determine the root of. f(x) = 9 ⅇ^(-x) sin (x) - 0.8 Using the Newton-Raphson method (starting point is, Xo = 0.3). Perform just two iterations A. x F(x)
0.4000 0.9078
0.6000 -0.0806
B. x F(x)
0.034 -0.50456
0.094 -0.03073
C. x F (x)
0.5078 0.1731
0.7435 -0.1343
D. x F(x) 0.5731 0.0515 0.4658 -0.0358
Using the Newton-Raphson method with a starting point of X₀ = 0.3, the root of the equation f(x) = 9e^(-x)sin(x) - 0.8 was approximated in two iterations. The calculations showed that the root of the equation lies around x = 0.7435.
The Newton-Raphson method is an iterative numerical method used to find the roots of a given equation. It involves updating the current approximation of the root based on the tangent line to the curve at that point. In each iteration, the formula x₁ = x₀ - f(x₀)/f'(x₀) is used, where x₀ is the current approximation and f'(x₀) is the derivative of the function.
In the given problem, the function f(x) = 9e^(-x)sin(x) - 0.8 is given, and we need to find its root using the Newton-Raphson method. Starting with X₀ = 0.3, we perform two iterations to approximate the root.
In the first iteration, plugging X₀ = 0.3 into the function, we calculate f(X₀) = 0.9078. Using the derivative of the function, we find f'(X₀) = -8.9469. Applying the Newton-Raphson formula, we get X₁ = X₀ - f(X₀)/f'(X₀) = 0.3 - 0.9078/(-8.9469) = 0.4000. Evaluating the function at X₁, we find f(X₁) = 0.9078.
Moving on to the second iteration, we repeat the same process with the new approximation X₁ = 0.4000. Calculating f(X₁) = -0.0806 and f'(X₁) = -9.2269, we can determine the next approximation. Applying the Newton-Raphson formula, we find X₂ = X₁ - f(X₁)/f'(X₁) = 0.4000 - (-0.0806)/(-9.2269) = 0.6000. Evaluating the function at X₂, we obtain f(X₂) = -0.0806.
Therefore, after two iterations, we find that the root of the equation f(x) = 9e^(-x)sin(x) - 0.8 is approximately x = 0.6000. However, it's worth noting that the exact root is not given, so this is an approximation based on the provided data.
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which of the following is the binary equivalent to the decimal number 218?
O 1101 O 10101110 O 110110 O 11111100 O 1110
The binary equivalent to the decimal number 218 is 1101 1010.
To convert decimal to binary, we need to continuously divide the decimal number by 2 until the quotient is 0. The remainder of each division will give us the binary digits from right to left. In this case, 218 divided by 2 gives a quotient of 109 with a remainder of 0 (LSB). We then divide 109 by 2, which gives a quotient of 54 with a remainder of 1. We continue this process until we reach 0. The binary digits are read from the remainder column in reverse order, which gives us 1101 1010. This is the correct binary equivalent to the decimal number 218.
The binary equivalent of the decimal number 218 is 11011010. Here's a breakdown of the conversion process:
218 ÷ 2 = 109, remainder = 0 (2^1)
109 ÷ 2 = 54, remainder = 1 (2^3)
54 ÷ 2 = 27, remainder = 0 (2^2)
27 ÷ 2 = 13, remainder = 1 (2^4)
13 ÷ 2 = 6, remainder = 1 (2^5)
6 ÷ 2 = 3, remainder = 0 (2^3)
3 ÷ 2 = 1, remainder = 1 (2^1)
1 ÷ 2 = 0, remainder = 1 (2^0)
Putting the remainders together from top to bottom: 11011010
Therefore, the binary equivalent of 218 is 11011010.
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Consider the function f(x) = fax² +a x ≥ 0 \bx-6 x < 0 (a) Find the value of a such that f(x) is continuous at x = 0. a= (b) Given that f is continuous at x = 0 (that is, using your value of a), id
Substituting x = 0 into the expression, we have: f(0) = a(0)^2
f(0) = 0. So, regardless of the value of "a," when x = 0, f(0) will always be equal to 0.
(a) To find the value of "a" such that the function f(x) is continuous at x = 0, we need to ensure that the left-hand limit and right-hand limit of f(x) as x approaches 0 are equal.
First, let's find the left-hand limit:
[tex]lim(x→0-) f(x) = lim(x→0-) (bx - 6)[/tex]
Since x approaches 0 from the left side, we use the definition of f(x) for x < 0, which is bx - 6.
Now, let's find the right-hand limit:
[tex]lim(x→0+) f(x) = lim(x→0+) (ax^2)[/tex]
Since x approaches 0 from the right side, we use the definition of f(x) for x ≥ 0, which is ax^2.
For f(x) to be continuous at x = 0, the left-hand limit and right-hand limit must be equal.
Therefore, equating the left-hand and right-hand limits, we have:
[tex]bx - 6 = a(0)^2bx - 6 = 0bx = 6x = 6/b[/tex]
To ensure f(x) is continuous at x = 0, the value of "a" should be such that x = 6/b.
(b) Given that f is continuous at x = 0 (using the value of a obtained in part (a)), we need to find the value of f(0).
Since x = 0 falls into the range x ≥ 0, we use the definition of f(x) for x ≥ 0, which is ax^2.
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(1 point) Take the Laplace transform of the following initial value problem and solve for Y(s) = ({y(t)} y" + 4y' +13y = {, t, 0
Inverse laplace transform of Y(s) is: [tex]y(t) = [(t/3)e^(-2t) + (1/3)cos(3t)] u(t)[/tex] for the differential equation.
The given differential equation is y'' + 4y' + 13y = 0, with initial conditions y(0) = 0 and y'(0) = t.
In mathematics and engineering, the Laplace transform is an integral transform that is used to solve differential equations and examine dynamic systems. In order to represent the frequency domain, it transforms a function of time into a function of the complex variable s. An exponential term, e(-st), multiplied by the function's integral yields the Laplace transform, where s is a complex number.
To solve the initial value problem, first we have to take the Laplace transform of the differential equation and the initial conditions. Laplace transform of y'' is given as [tex]s^2Y(s) - sy(0) - y'(0)[/tex]
Laplace transform of y' is given as sY(s) - y(0)
We get: Laplace transform of y'' + 4 Laplace transform of y' + 13Laplace transform of y = Laplace transform of (0)
We get: [tex]s^2Y(s) - st - 1 + 4(sY(s) - 0) + 13Y(s) = 0=>\\\\ s^2Y(s) + 4sY(s) + 13Y(s) = st + 1Y(s)(s^2 + 4s + 13) = \\\\st + 1Y(s) = (st + 1) / (s^2 + 4s + 13)[/tex]
Now we need to take the inverse Laplace transform of Y(s) to get the solution of the initial value problem. For that, we need to factorize the denominator as [tex]s^2 + 4s + 13 = (s + 2)^2 + 9[/tex]
By partial fraction method, we can write the equation asY(s) = [tex](st + 1) / (s^2 + 4s + 13) = \\(st + 1) / [(s + 2)^2 + 9]=\\ [(t/3)(s + 2) + (1/3)] / [(s + 2)^2 + 9][/tex]
Taking inverse Laplace transform of Y(s), we get: [tex]y(t) = [(t/3)e^(-2t) + (1/3)cos(3t)][/tex] u(t)Where u(t) is the unit step function.
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1. What is the farthest point on the sphere 2² + y2 + z2 = 16 from the point (2, 2, 1) ? ) 8 (a) 8 3 4 3 3 (b) ( 8 8 4 3'3'3 8 (c) 8 4 3'3 3 8 (d) 8 3 3) 3 (e) ) 8 8 4 3'3'3
The farthest point on the sphere 2² + y² + z² = 16 from the point (2, 2, 1) is (8/3, 8/3, 4/3). Among the given options, the closest match to the coordinates (8/3, 8/3, 4/3) is option (c) 8 4 3'3 3 8.
To find the farthest point on the sphere 2² + y² + z² = 16 from the point (2, 2, 1), we can use the distance formula. The farthest point will have the maximum distance from the given point.
The distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) in 3D space is given by the formula:
distance = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)
In this case, the given point is (2, 2, 1), and we need to find the farthest point on the sphere. Let's assume the coordinates of the farthest point are (x, y, z).
Substituting the values into the distance formula, we have:
distance = √((x - 2)² + (y - 2)² + (z - 1)²)
To find the farthest point, we want to maximize the distance. However, since the equation of the sphere 2² + y² + z² = 16 represents a spherical surface, the maximum distance will be along the radius of the sphere.
The equation of the sphere can be rewritten as:
x² + y² + z² = 4
Since the center of the sphere is at (0, 0, 0), the point (2, 2, 1) is not on the surface of the sphere.
Therefore, the farthest point on the sphere from (2, 2, 1) will lie on the line connecting the center of the sphere to the point (2, 2, 1).
The coordinates of the farthest point can be found by scaling the direction vector of the line connecting the center to (2, 2, 1) to have a length of 4 (radius of the sphere).
Scaling the direction vector (2, 2, 1) gives us:
(2, 2, 1) * (4/√(2² + 2² + 1²))
Simplifying, we get:
(2, 2, 1) * (4/√9) = (2, 2, 1) * (4/3)
Multiplying the scalars with the vector components, we get:
(8/3, 8/3, 4/3)
The sphere's farthest point from the point (2, 2, 1) is (8/3, 8/3, 4/3), which is determined by the formula 22 + y2 + z2 = 16.
Option (c) 8 4 3'3 3 8 is the option that matches the coordinates (8/3, 8/3, and 3/3) the most closely.
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If two individuals in the same population have identical X scores, they also will have identical z-scores.
TRUE or FALSE
TRUE. If two individuals in the same population have identical X scores, they also will have identical z-scores.
The z-score of an individual in a population is calculated using the formula:
z = (X - μ) / σ
where X is the individual's score, μ is the population mean, and σ is the population standard deviation.
If two individuals in the same population have identical X scores, it means they have the same value for X. Therefore, when calculating the z-score for each individual using the same population mean and standard deviation, the numerator (X - μ) will be the same for both individuals.
Since the numerator is the same, the z-score for both individuals will also be the same. Therefore, if two individuals have identical X scores in a population, they will have identical z-scores. Hence, the statement is TRUE.
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please
cas moil law gagang d bila In Exercises 1-4, find the work done by the force of F(x) newtons along the x-axis from x = a meters to x = b meters. w odt 1.F(x) = xe-x/3, a = 0, b=5 01 21 19th 30 are to
The work done by the force of F(x) newtons along the x-axis from x = a meters to x = b meters is :
-3xe^(-x/3) - 27e^(-x/3) + C, where C is a constant.
The work done by the force of F(x) newtons along the x-axis from x = a meters to x = b meters is to be found given :
F(x) = xe^(-x/3),
a = 0, b = 5.
We know that,
Work done = Integration of F(x) with respect to x from a to b
Using the above formula, we get:
W = Integration of xe^(-x/3) with respect to x from 0 to 5
Let u = -x/3.
Then,
du/dx = -1/3
or dx = -3 du
When x = 0, u = 0.
When x = 5, u = -5/3.
Substituting these values, we get:
W = Integration of xe^(-x/3) with respect to x from 0 to 5=
W = -Integration of 3u(e^u)(-3du)
(substituting x = -3u and dx = -3 du)
W = 9
Integration of ue^u du
Using Integration by Parts with u = u and dv = e^u du, we get:
W = 9[(u)(e^u) - Integration of e^u du]
W = 9[(u)(e^u) - e^u] + C
Now, substituting u = -x/3, we get:
W = 9[(-x/3)(e^(-x/3)) - e^(-x/3)] + C
W = -3xe^(-x/3) - 27e^(-x/3) + C
Thus, the work done -3xe^(-x/3) - 27e^(-x/3) plus a constant.
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(5 points) l|v|| = 3 ||0|| = 1 The angle between v and w is 2 radians. Given this information, calculate the following: (a) v- w = 2.9981 (b) ||10 + 2w|| 4.99 (c) ||2v – 1w| 5.00
To calculate the values requested, we'll use the given information and apply the properties of vector operations.
(a) Vector subtraction: To calculate v - w, we subtract the components of w from the corresponding components of v.
[tex]v - w = |v| * |w| * cos(2) ≈ 3 * 1 * cos(2) ≈ 2.9981[/tex]Therefore, v - w is approximately equal to 2.9981.(b) Magnitude of the sum: To calculate ||10 + 2w||, we substitute the given values into the formula ||A + B|| = √(A · A + B · B + 2A · B).[tex]||10 + 2w|| = √(10 · 10 + 2 · 2 + 2 · 10 · 1) = √(100 + 4 + 20) = √124 ≈ 11.1355[/tex]Therefore, the magnitude of the sum 10 + 2w is approximately 11.1355.
(c) Magnitude of the difference: To calculate ||2v - w||, we substitute the given values into the formula ||A - B|| = √(A · A + B · B - 2A · B).
[tex]||2v - w|| = √(2 · 2 · 2 + 1 · 1 - 2 · 2 · 1) = √(8 + 1 - 4) = √5 ≈ 2.2361[/tex]
Therefore, the magnitude of the difference 2v - w is approximately 2.2361.
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før 16x3 + 732 + 125x + 100 Consider the indefinite integral dx 24 + 25x2 Then the integrand has partial fractions decomposition a 6 cx + d + x2 х X2 + 25 where + a = b = = C = d = = Integrating term by term, we obtain that 16x3 + 7x2 + 125x + 100 da x4 + 25x2 f6z" = +C
∫ (16x^3 + 7x^2 + 125x + 100) dx = (2/3)ln|6x + 25| + C1 + C2 where C1 and C2 are constants of integration.
To solve the given problem, let's break it down step by step.
We are given the expression:
∫ (24 + 25x^2) dx
Next, we need to perform the partial fraction decomposition on the integrand.
Let the decomposition be:
(24 + 25x^2) = (a/(6x + d)) + ((bx + c)/(x^2 + 25))
We need to find the values of a, b, c, and d.
Multiplying both sides by the denominator (6x + d)(x^2 + 25), we get:
(24 + 25x^2) = a(x^2 + 25) + (bx + c)(6x + d)
Expanding the right side, we have:
24 + 25x^2 = ax^2 + 25a + (6bx^2 + dx + 6cx^3 + cx^2)
Comparing the coefficients of like terms on both sides, we get the following equations:
a + 6c = 0 (coefficient of x^3 terms)
25a + d = 0 (coefficient of x^2 terms)
6b = 0 (coefficient of x^2 terms)
25a + 6c = 24 (constant term)
d = 25 (constant term)
Solving these equations, we find:
c = 0
b = 0
a = 4
d = 25
Therefore, the partial fractions decomposition is:
(24 + 25x^2) = (4/(6x + 25)) + (0/(x^2 + 25))
Now, we can integrate term by term:
∫ (16x^3 + 7x^2 + 125x + 100) dx = ∫ (4/(6x + 25)) dx + ∫ (0/(x^2 + 25)) dx
Evaluating the integrals, we get:
∫ (4/(6x + 25)) dx = (2/3)ln|6x + 25| + C1
∫ (0/(x^2 + 25)) dx = C2
Finally, combining the results, we have:
∫ (16x^3 + 7x^2 + 125x + 100) dx = (2/3)ln|6x + 25| + C1 + C2
Note: C1 and C2 are constants of integration.
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(1 point) A rectangle is inscribed with its base on the I-axis and its upper corners on the parabola y = 8 - x? What are the dimensions of such a rectangle with the greatest possible area? Width = Hei
The dimensions of the rectangle with the greatest possible area are a width of 8 units and a height of 4 units.
To find the dimensions of the rectangle with the greatest area, we can use optimization techniques. Since the base of the rectangle is on the x-axis, its width is equal to the x-coordinate of the upper corners. Let's denote this width as x.
The height of the rectangle is determined by the y-coordinate of the upper corners. Since the upper corners lie on the parabola y = 8 - x, the height of the rectangle can be expressed as y = 8 - x.
The area of the rectangle is given by the formula A = width × height. Substituting the expressions for width and height, we have A = x(8 - x) = 8x - x².
To find the maximum area, we need to find the critical points of the area function A(x) = 8x - x². Taking the derivative of A(x) with respect to x and setting it equal to zero, we get dA/dx = 8 - 2x = 0. Solving for x, we find x = 4.
Plugging this value back into the equation for the height, we find y = 8 - x = 8 - 4 = 4.
Therefore, the rectangle with the greatest possible area has a width of 4 units and a height of 4 units.
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1. Evaluate the following limits, if they exist, using an appropriate strategy. a) lim Vx+7-v11+2x b) [35-x-2 lim X-3 x2-9 x+4 15-x-Vx+13
a) The limit of (7-v)/(11+2x) as x approaches infinity does not exist.
b) The limit of (35-x-2)/(x^2-9)/(x+4)/(15-x-√(x+13)) as x approaches 3 is 2.
a) To evaluate the limit of (7-v)/(11+2x) as x approaches infinity, we consider the behavior of the expression as x becomes very large. As x approaches infinity, the denominator grows without bound, while the numerator remains constant. In this case, the limit does not exist because the expression becomes undefined (division by infinity). There is no specific value to which the expression tends as x approaches infinity.
b) To evaluate the limit of (35-x-2)/(x^2-9)/(x+4)/(15-x-√(x+13)) as x approaches 3, we substitute x = 3 into the expression and simplify. Plugging in x = 3, we get (35-3-2)/(3^2-9)/(3+4)/(15-3-√(3+13)). This simplifies to (30)/(0)/(7)/(12-√16), which further simplifies to 0/0/7/12-4. To proceed, we need to simplify the remaining division. The denominator 12-4 evaluates to 8. Thus, the limit becomes 0/0/7/8, which is equivalent to 0/0. This indeterminate form requires further analysis. We can apply L'Hôpital's rule by differentiating the numerator and the denominator separately, or factor and simplify the expression to resolve the indeterminate form and find the final limit.
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Although a line has infinitely points (solutions), what are the two intercept points of the line below? (The Importance is that we use intercept points to graph in standard form.)
The two intercept points of the line is (3, 0) and (0, -2).
We have a graph from a line.
Now, take two points from the graph as (3, 0) and (0, -2)
Now, we know that slope is the ratio of vertical change (Rise) to the Horizontal change (run)
So, slope= (change in y)/ Change in c)
slope = (-2-0)/ (0-3)
slope= -2 / (-3)
slope=2/3
Now, the equation of line is
y - 0 = 2/3 (x-3)
y= 2/3x - 3
Now, to find y intercept put x= 0
y= -3
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Find the radius of convergence, R, of the series. 00 Σ n!x" 2.5.8.... · (3n - 1) n=1 R= Find the interval, I, of convergence of the series. (Enter your answer using interval notation.) I =
The given series is:00 Σ n!x^(2.5.8.... · (3n - 1))n=1. To find the radius of convergence, R, of the given series, we use the ratio test.
Apply the ratio test.Using the ratio test:lim | a_(n+1)/a_n | = lim (n+1)!|x|^(2.5.8.... · (3(n+1) - 1))/n!|x|^(2.5.8.... · (3n - 1))= lim (n+1)|x|^(3n+2)|x|^(2.5.8.... · (-2))= |x|^(3n+2)lim (n+1) = ∞, as n → ∞n∴ lim | a_(n+1)/a_n | = ∞ > 1.
Therefore, the series diverges for all values of x.
Hence, the radius of convergence, R, of the given series is 0.
Now, let's determine the interval of convergence, I, of the given series.
The series diverges for all values of x, so there is no interval of convergence.
Therefore, I = Ø (empty set) is the interval of convergence.
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Let X₁, X, be a random sample from a normal distribution with unknown mean and known variance o². Find the maximum likelihood estimator of μ and show that it is a function of a minimal sufficient statistic.
The maximum likelihood estimator (MLE) of the unknown mean μ for a random sample X₁, X₂ from a normal distribution with known variance σ² is obtained by maximizing the likelihood function. In this case, we will show that the MLE of μ is a function of a minimal sufficient statistic.
To find the MLE of μ, we need to maximize the likelihood function. The likelihood function for a normal distribution is given by L(μ, σ² | X₁, X₂) = f(X₁, X₂ | μ, σ²), where f is the probability density function of the normal distribution.
Taking the natural logarithm of the likelihood function, we get the log-likelihood function: log L(μ, σ² | X₁, X₂) = log f(X₁, X₂ | μ, σ²).
To find the MLE of μ, we differentiate the log-likelihood function with respect to μ and set it equal to zero. Solving this equation gives us the MLE of μ, denoted as ȳ, which is simply the sample mean.
Now, to show that the MLE of μ is a function of a minimal sufficient statistic, we can use the factorization theorem. The joint probability density function of X₁, X₂ given μ and σ² can be factorized as f(X₁, X₂ | μ, σ²) = g(T(X₁, X₂) | μ, σ²)h(X₁, X₂), where T(X₁, X₂) is a minimal sufficient statistic and h(X₁, X₂) does not depend on μ.
Since the MLE ȳ is a function of T(X₁, X₂), which is a minimal sufficient statistic, it follows that the MLE of μ is a function of a minimal sufficient statistic.
Therefore, the MLE of μ is ȳ, the sample mean, and it is a function of a minimal sufficient statistic.
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