1. The equalities are not generally valid.
2. 0 is an element of A, and {0} is also an element of A since it is a singleton set containing 0.
i) The equalities A ∪ (B \ C) = (A ∪ B) \ (A ∪ C) and A ∩ (B \ C) = (A ∩ B) \ (A ∩ C) are not generally valid.
Counterexample for A ∪ (B \ C) = (A ∪ B) \ (A ∪ C):
Let A = {1, 2}, B = {2, 3}, and C = {1, 3}.
A ∪ (B \ C) = {1, 2} ∪ {2} = {1, 2}
(A ∪ B) \ (A ∪ C) = ({1, 2} ∪ {2, 3}) \ ({1, 2} ∪ {1, 3}) = {1, 2, 3} \ {1, 2} = {3}
Since {1, 2} is not equal to {3}, the equality A ∪ (B \ C) = (A ∪ B) \ (A ∪ C) does not hold in this case.
Counterexample for A ∩ (B \ C) = (A ∩ B) \ (A ∩ C):
Let A = {1, 2}, B = {2, 3}, and C = {1, 3}.
A ∩ (B \ C) = {1, 2} ∩ {2} = {2}
(A ∩ B) \ (A ∩ C) = ({1, 2} ∩ {2, 3}) \ ({1, 2} ∩ {1, 3}) = {2} \ {1, 2} = {}
Since {2} is not equal to {}, the equality A ∩ (B \ C) = (A ∩ B) \ (A ∩ C) does not hold in this case.
Therefore, the equalities are not generally valid.
ii) An example of a set A containing at least one element that fulfills the condition if x ∈ A, then {x} ∈ A is:
A = {0, {0}}
In this case, 0 is an element of A, and {0} is also an element of A since it is a singleton set containing 0.
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Heavy Numbers 4.1 Background on heavy numbers 4.1.1 The heavy sequence A sequence of numbers (the heavy sequence) y 0
y 1
y 2
y 3
…y n
… is defined such that each number is the sum of digits squared of the previous number, in a particular base. Consider numbers in base 10 , with y 0
=12 The next number in the sequence is y 1
=1 2
+2 2
=5 The next number in the sequence is y 2
=5 2
=25 The next number in the sequence is y 3
=2 2
+5 2
=29 4.1.2 Heaviness It turns out that for each number y 0
and base N, the heavy sequence either converges to 1 , or it does not. A number whose sequence converges to 1 in base N is said to be "heavy in base N" 4.2 Program requirements Write a function heavy that takes as arguments a number y and a base N and returns whether that number y is heavy in the base N provided. Here are examples: ≫ heavy (4,10) False > heavy (2211,10) True ≫ heavy (23,2) True ≫ heavy (10111,2) True ≫ heavy (12312,4000) False 4.2.1 Value Ranges The number y will always be non-negative, and the base N will always satisfy 2≤N≤4000
The function iteratively calculates the next number in the heavy sequence until it reaches 1 or detects a repeating pattern. If the next number becomes equal to the current number, it means the sequence does not converge to 1 and the number is not heavy in the given base. Otherwise, if the sequence reaches 1, the number is heavy.
Here's a Python implementation of the heavy function that checks if a number y is heavy in base N:
python
Copy code
def heavy(y, N):
while y != 1:
next_num = sum(int(digit)**2 for digit in str(y))
if next_num == y:
return False
y = next_num
return True
You can use this function to check if a number is heavy in a specific base. For example:
python
Copy code
print(heavy(4, 10)) # False
print(heavy(2211, 10)) # True
print(heavy(23, 2)) # True
print(heavy(10111, 2)) # True
print(heavy(12312, 4000)) # False
The function iteratively calculates the next number in the heavy sequence until it reaches 1 or detects a repeating pattern. If the next number becomes equal to the current number, it means the sequence does not converge to 1 and the number is not heavy in the given base. Otherwise, if the sequence reaches 1, the number is heavy.
Note: This implementation assumes that the input number y and base N are within the specified value ranges of non-negative y and 2 <= N <= 4000.
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Using the binomial expansion of (1+x)^n, explain why a set S with n elements has the same number of subsets with even size as with odd size. Hint: Substitute x=-1.
A set S with n elements has the same number of subsets with even size as with odd size, as shown by the binomial expansion when substituting x = -1.
To understand why a set S with n elements has the same number of subsets with even size as with odd size, we can use the binomial expansion of (1+x)^n and substitute x = -1.
The binomial expansion of (1+x)^n is given by:
(1+x)^n = C(n,0) + C(n,1)x + C(n,2)x^2 + ... + C(n,n)x^n,
where C(n,k) represents the binomial coefficient "n choose k," which gives the number of ways to choose k elements from a set of n elements.
Now, substitute x = -1:
(1+(-1))^n = C(n,0) + C(n,1)(-1) + C(n,2)(-1)^2 + ... + C(n,n)(-1)^n.
Simplifying the expression, we have:
0 = C(n,0) - C(n,1) + C(n,2) - ... + (-1)^n C(n,n).
We can observe that the terms with odd coefficients C(n,1), C(n,3), C(n,5), ..., C(n,n) have a negative sign, while the terms with even coefficients C(n,0), C(n,2), C(n,4), ..., C(n,n-1) have a positive sign.
Since the expression evaluates to zero, this implies that the sum of the terms with odd coefficients is equal to the sum of the terms with even coefficients. In other words, the number of subsets of S with odd size is equal to the number of subsets with even size.
Therefore, a set S with n elements has the same number of subsets with even size as with odd size, as shown by the binomial expansion when substituting x = -1.
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Let S be the universal set, where: S={1,2,3,…,23,24,25} Let sets A and B be subsets of S, where: Set A={2,4,7,11,13,19,20,21,23} Set B={1,9,10,12,25} Set C={3,7,8,9,10,13,16,17,21,22} LIST the elements in the set (A∪B∪C) (A∪B∪C)=1 Enter the elements as a list, separated by commas. If the result is the empty set, enter DNE LIST the elements in the set (A∩B∩C) (A∩B∩C)={ Enter the elements as a list, separated by commas. If the result is the empty set, enter DNE
To find the elements in the set (A∪B∪C), we need to combine all the elements from sets A, B, and C without repetitions. The given sets are: Set A={2,4,7,11,13,19,20,21,23} Set B={1,9,10,12,25} Set C={3,7,8,9,10,13,16,17,21,22}Here, A∪B∪C represents the union of the three sets. Therefore, the elements of the set (A∪B∪C) are:{1, 2, 3, 4, 7, 8, 9, 10, 11, 12, 13, 16, 17, 19, 20, 21, 22, 23, 25}The given sets are: Set A={2,4,7,11,13,19,20,21,23}Set B={1,9,10,12,25}Set C={3,7,8,9,10,13,16,17,21,22}Here, A∩B∩C represents the intersection of the three sets. Therefore, the elements of the set (A∩B∩C) are: DNE (empty set)Hence, the required solution is the set (A∪B∪C) = {1, 2, 3, 4, 7, 8, 9, 10, 11, 12, 13, 16, 17, 19, 20, 21, 22, 23, 25} and the set (A∩B∩C) = DNE (empty set).
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Marissa is playing a game at the carnival that requires her to hit a spring with a large hammer. After the spring is hit, a puck will shoot upwards towards a bell, Marissa hit
y 16X2:32:20, where y represents the distance between the puck and the bell and x represents the time after hitting the spring (in seconds),
Part A
What type of solution(s) does the equation 0:16X2,32x20 have?
Part
Will the puck hit the bell after Marissa hits it? Why or why not?
B
1
U AI
TI
DIT OD
+1
1
All 28
Unanswered 10
O
a. The puck will have two different distances from the bell at different times after Marissa hits the spring.
b. Additional information or an equation where y = 0 is needed to determine if the puck will hit the bell after Marissa hits it.
Part A:
The equation 0.16x^2 + 32x + 20 represents the relationship between the distance (y) of the puck from the bell and the time (x) after hitting the spring. To determine the type of solution(s), we can analyze the discriminant of the quadratic equation, which is the expression under the square root in the quadratic formula.
The discriminant (b^2 - 4ac) for the given equation is:
(32^2) - 4(0.16)(20) = 1024 - 12.8 = 1011.2
Since the discriminant is positive (greater than zero), the equation has two distinct real solutions. This means that the puck will have two different distances from the bell at different times after Marissa hits the spring.
Part B:
To determine whether the puck will hit the bell, we need to consider the distance (y) when it becomes zero. If the distance becomes zero, it means the puck has reached the bell.
To find the time (x) when y = 0, we can set the equation 0.16x^2 + 32x + 20 = 0 and solve for x. However, since the given equation does not provide a value for y = 0, we cannot determine if the puck will hit the bell based on the given information.
Additional information or an equation where y = 0 is needed to determine if the puck will hit the bell after Marissa hits it.
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A 55.0g hard -boiled egg moves on the end of a spring with force constant k=25.0(N)/(m). Its initial displacement is 0.400m. A damping force F_(x)=-bv_(x) acts on the egg, and the amplitude of the motion decreases to 0.100m in a time of 5.00s.
The damping constant b for the motion of the hard-boiled egg is approximately 0.3 Ns/m.
We can use the formula for the damped harmonic motion of a spring-mass system:
m(d^2x/dt^2) + b(dx/dt) + kx = 0
Where:
m = mass of the egg (55.0 g = 0.055 kg)
b = damping constant (to be determined)
k = force constant of the spring (25.0 N/m)
x = displacement of the egg from its equilibrium position
Given initial displacement (amplitude) x₀ = 0.400 m and final displacement x = 0.100 m, we can calculate the damping constant b.
Using the equation for the displacement of a damped harmonic oscillator:
x = x₀ * e^(-bt/2m)
Plugging in the given values:
0.100 = 0.400 * e^(-b * 5.00 / (2 * 0.055))
Rearranging the equation and taking the natural logarithm (ln) of both sides:
ln(0.100/0.400) = (-b * 5.00) / (2 * 0.055)
Solving for b:
b ≈ (-2 * 0.055 * ln(0.100/0.400)) / 5.00 ≈ 0.3 Ns/m
The damping constant for the motion of the hard-boiled egg on the spring is approximately 0.3 Ns/m. This value determines the strength of the damping force acting on the egg and affects the rate at which the amplitude of the motion decreases over time.
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Consider the given vector equation. r(t)=⟨4t−4,t ^2 +4⟩ (a) Find r ′(t).
Taking the limit of r'(t) as Δt → 0, we get: r'(t) = <4, 2t> The vector equation r(t) = <4t - 4, t² + 4> is given.
We need to find r'(t).
Given the vector equation, r(t) = <4t - 4, t² + 4>
Let r(t) = r'(t) = We need to differentiate each component of the vector equation separately.
r'(t) = Differentiating the first component,
f(t) = 4t - 4, we get f'(t) = 4
Differentiating the second component, g(t) = t² + 4,
we get g'(t) = 2t
So, r'(t) = = <4, 2t>
Hence, the required vector is r'(t) = <4, 2t>
We have the vector equation r(t) = <4t - 4, t² + 4> and we know that r'(t) = <4, 2t>.
Now, let's find r'(t) using the definition of the derivative: r'(t) = [r(t + Δt) - r(t)]/Δtr'(t)
= [<4(t + Δt) - 4, (t + Δt)² + 4> - <4t - 4, t² + 4>]/Δtr'(t)
= [<4t + 4Δt - 4, t² + 2tΔt + Δt² + 4> - <4t - 4, t² + 4>]/Δtr'(t)
= [<4t + 4Δt - 4 - 4t + 4, t² + 2tΔt + Δt² + 4 - t² - 4>]/Δtr'(t)
= [<4Δt, 2tΔt + Δt²>]/Δt
Taking the limit of r'(t) as Δt → 0, we get:
r'(t) = <4, 2t> So, the answer is correct.
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4. Find the general solution to y" + 12y +36y=0. 5. Construct an equation such that y = C₁e^x cos(3x) + C2e^-x sin(32) is its general solution. 6. Find the solution to y"+4y+5y=0 with y(0) = 2 and y'(0) = -1.
The general solution to y" + 12y + 36y = 0 is: y(x) = c_1 e^{-6x} + c_2xe^{-6x} To construct an equation such that the general solution is y = C₁e^x cos(3x) + C2e^-x sin(3x), we first find the derivatives of each of these functions.
The derivative of C₁e^x cos(3x) is C₁e^x cos(3x) - 3C₁e^x sin(3x)
The derivative of C₂e^-x sin(3x) is -C₂e^-x sin(3x) - 3C₂e^-x cos(3x)
To find a function that is equal to the sum of these two derivatives, we can set the coefficients of the cos(3x) terms and sin(3x) terms equal to each other:C₁e^x = -3C₂e^-x
And: C₁ = -3C₂e^-2x
Solving this system of equations, we get:C₁ = -3, C₂ = -1
The required equation, therefore, is y = -3e^x cos(3x) - e^-x sin(3x)
Finally, to find the solution to y" + 4y + 5y = 0 with y(0) = 2 and y'(0) = -1,
we can use the characteristic equation:r² + 4r + 5 = 0
Solving this equation gives us:r = -2 ± i
The general solution is therefore:y(x) = e^{-2x}(c₁ cos x + c₂ sin x)
Using the initial conditions:y(0) = c₁ = 2y'(0) = -2c₁ - 2c₂ = -1
Solving this system of equations gives us:c₁ = 2, c₂ = 3/2
The required solution is therefore:y(x) = 2e^{-2x} cos x + (3/2)e^{-2x} sin x
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a group of children held a grape-eating contest. when the contest was over, the winner had eaten grapes, and the child in -th place had eaten grapes. the total number of grapes eaten in the contest was . find the smallest possible value of .
The smallest possible value of the total number of grapes eaten in the contest is the product of (n-1) and the number of grapes eaten by the child in last place, plus the number of grapes eaten by the winner.
We have,
A group of children held a grape-eating contest.
And, when the contest was over, the winner had eaten grapes, and the child in -th place had eaten grapes.
Hence, The smallest possible value of the total number of grapes eaten in the contest is,
(n - 1) (grapes eaten by the child in last place) + (grapes eaten by the winner).
So for example, if the winner had eaten 10 grapes and the child in last place had eaten 5 grapes, the total number of grapes eaten in the contest would be;
(n-1)5 + 10 = 45.
Hence, The smallest possible value of the total number of grapes eaten in the contest is the product of (n-1) and the number of grapes eaten by the child in last place, plus the number of grapes eaten by the winner.
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Determine the 75%,90%, and 95% response time for the following system: 5 y˙ +5y=U(t) where, U(t)={ 01 if t
In the given system described by the differential equation 5y'' + 5y = U(t), the 75%, 90%, and 95% response times are infinite due to the indefinite oscillation of the system.
To determine the response time of the given system, we need to find the time it takes for the system to reach a certain percentage (75%, 90%, and 95%) of its final response when subjected to a unit step input.
The system is described by the following differential equation:
5y'' + 5y = U(t)
To solve this equation, we'll first find the homogeneous and particular solutions.
Homogeneous Solution:
The homogeneous equation is given by 5y'' + 5y = 0.
The characteristic equation is 5r^2 + 5 = 0.
Solving the characteristic equation, we find two complex conjugate roots: r = ±j.
Therefore, the homogeneous solution is y_h(t) = c1cos(t) + c2sin(t), where c1 and c2 are arbitrary constants.
Particular Solution:
For the particular solution, we assume a step response form, y_p(t) = A*u(t), where A is the amplitude of the step response.
Substituting y_p(t) into the differential equation, we have:
5Au''(t) + 5Au(t) = 1
Since u(t) is a unit step function, u''(t) = 0 for t > 0.
Therefore, the equation simplifies to:
5*A = 1
Solving for A, we get A = 1/5.
The complete solution is given by the sum of the homogeneous and particular solutions:
y(t) = y_h(t) + y_p(t)
= c1cos(t) + c2sin(t) + (1/5)*u(t)
Now, we can determine the response times for different percentages:
75% Response Time:
To find the time at which the response reaches 75% of the final value, we substitute y(t) = 0.75 into the equation:
0.75 = c1cos(t) + c2sin(t) + (1/5)*u(t)
Since the system is underdamped with complex roots, it will oscillate indefinitely. Therefore, we can't directly solve for the time at which it reaches 75%. The response time will be infinite.
90% Response Time:
To find the time at which the response reaches 90% of the final value, we substitute y(t) = 0.9 into the equation:
0.9 = c1cos(t) + c2sin(t) + (1/5)*u(t)
Again, due to the indefinite oscillation of the system, we can't directly solve for the time at which it reaches 90%. The response time will be infinite.
95% Response Time:
To find the time at which the response reaches 95% of the final value, we substitute y(t) = 0.95 into the equation:
0.95 = c1cos(t) + c2sin(t) + (1/5)*u(t)
Similar to the previous cases, the indefinite oscillation prevents us from directly solving for the time. The response time will be infinite.
Therefore, for the given system described by the differential equation 5y'' + 5y = U(t), the 75%, 90%, and 95% response times are infinite due to the indefinite oscillation of the system.
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Suppose that the middle 68% of monthly food expenditures for a family of four fall between 753.45 and 922.91. Give an approximate estimate of the standard deviation of the expenditures. Assume the expenditures have a normal distribution. 1) −84.73 2) 42.365 3) 838.18 4) 169.46 5) 84.73
The correct answer is option 5.) 84.73.
We can begin by calculating the mean. Since the middle 68% of monthly food expenditures falls between 753.45 and 922.91, we can infer that this is a 68% confidence interval centered around the mean. Hence, we can obtain the mean as the midpoint of the interval:
[tex]$$\bar{x}=\frac{753.45+922.91}{2}=838.18$$[/tex]
To estimate the standard deviation, we can use the fact that 68% of the data falls within one standard deviation of the mean. Thus, the distance between the mean and each endpoint of the interval is equal to one standard deviation. We can find this distance as follows:
[tex]$$922.91-838.18=84.73$$$$838.18-753.45=84.73$$[/tex]
Therefore, the standard deviation is approximately 84.73.
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simplify the following expression 3 2/5 mulitply 3(-7/5)
Answer:
1/3
Step-by-step explanation:
I assume that 2/5 and -7/5 are exponents.
3^(2/5) × 3^(-7/5) = 3^(2/5 + (-7/5)) = 3^(-5/5) = 3^(-1) = 1/3
Answer: 136/5
Step-by-step explanation: First simplify the fraction
1) 3 2/5 = 17/5
3 multiply by 5 and add 5 into it.
2) 3(-7/5) = 8/5
3 multiply by 5 and add _7 in it.
By multiplication of 2 fractions,
17/5 multiply 8/5 = 136/5
=136/5
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Suppose we wish to detect a difference of \$0.094 (just under a dime) between two different online ads. Suppose the standard deviation of the response (sales) is \$103.77 (the standard deviation will be large because most clicks do not produce sales so there are lots of 0's in the data). For an A/B test how many observations do we need in each sample? Use a power of 0.8 and $\alpha=0.05$.
After getting the result we can conclude that the power of 0.8 and α=0.05, the number of observations required in each sample is 2064.
A/B testing is the process of comparing two versions of a web page, email, or any other marketing asset to see which one performs better. In this scenario, we are supposed to detect a difference of $0.094 between two online ads. The standard deviation of the response (sales) is $103.77.
To calculate the number of observations we need for the A/B test, we will need the following parameters:
α (significance level) = 0.05, which means we have a 5% chance of making an error (rejecting a true null hypothesis).
Power (1 – β) = 0.8, which means we have an 80% chance of detecting a difference if it exists.
Standard deviation (σ) = $103.77
Difference in means (d) = $0.094
Formula to calculate the number of observations required for each sample in A/B test:
n = [(Zα/2 + Zβ) / d] ² (2σ²)
Here, Zα/2 and Zβ are the standard normal distribution values of α/2 and β, respectively. We can find these values using the z-table or calculator.
Zα/2 = 1.96 (for α = 0.05) Z β = 0.84 (for β = 0.2) Now, let's plug in all the values and solve for n:
n = [(1.96 + 0.84) / 0.094] ² (2 $103.77²) n = 2063.22
We need at least 2064 observations in each sample for the A/B test.
After getting the result we can conclude that the power of 0.8 and α=0.05, the number of observations required in each sample is 2064.
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The function s(t) describes the position of a particle moving along a coordinate line, where s is in feet and t is in seconds. s(t)=t^ 3 −18t ^2+81t+4,t≥0 (a) Find the velocity and acceleration functions. v(t) a(t):
To find the acceleration function, we differentiate the velocity function v(t) as follows; a(t) = v'(t) = 6t - 36. Therefore, the acceleration function of the particle is a(t) = 6t - 36.
To find the velocity and acceleration functions, we need to differentiate the position function, s(t), with respect to time, t.
Given: s(t) = t^3 - 18t^2 + 81t + 4
(a) Velocity function, v(t):
To find the velocity function, we differentiate s(t) with respect to t.
v(t) = d/dt(s(t))
Taking the derivative of s(t) with respect to t:
v(t) = 3t^2 - 36t + 81
(b) Acceleration function, a(t):
To find the acceleration function, we differentiate the velocity function, v(t), with respect to t.
a(t) = d/dt(v(t))
Taking the derivative of v(t) with respect to t:
a(t) = 6t - 36
So, the velocity function is v(t) = 3t^2 - 36t + 81, and the acceleration function is a(t) = 6t - 36.
The velocity function is v(t) = 3t²-36t+81 and the acceleration function is a(t) = 6t-36. To find the velocity function, we differentiate the function for the position s(t) to get v(t) such that;v(t) = s'(t) = 3t²-36t+81The acceleration function can also be found by differentiating the velocity function v(t). Therefore; a(t) = v'(t) = 6t-36. The given function s(t) = t³ - 18t² + 81t + 4 describes the position of a particle moving along a coordinate line, where s is in feet and t is in seconds.
We are required to find the velocity and acceleration functions given that t≥0.To find the velocity function v(t), we differentiate the function for the position s(t) to get v(t) such that;v(t) = s'(t) = 3t² - 36t + 81. Thus, the velocity function of the particle is v(t) = 3t² - 36t + 81.To find the acceleration function, we differentiate the velocity function v(t) as follows;a(t) = v'(t) = 6t - 36Therefore, the acceleration function of the particle is a(t) = 6t - 36.
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determine the number and type of solutions for each equation fundamental theorem of algebra
To determine the number and type of solutions for a specific equation, we need to consider the degree of the polynomial and use other mathematical techniques.
1. Linear Equation (degree 1):
A linear equation in one variable has exactly one solution, regardless of whether the coefficients are real or complex.
2. Quadratic Equation (degree 2):
A quadratic equation in one variable can have zero, one, or two solutions. The nature of the solutions depends on the discriminant (b² - 4ac), where a, b, and c are the coefficients of the equation.
- If the discriminant is positive, the equation has two distinct real solutions.
- If the discriminant is zero, the equation has one real solution (a double root).
- If the discriminant is negative, the equation has two complex solutions.
3. Cubic Equation (degree 3):
A cubic equation in one variable can have one, two, or three solutions. To determine the nature of the solutions, it often requires advanced algebraic techniques, such as factoring, the Rational Root Theorem, or Cardano's method.
4. Higher-Degree Equations (degree 4 or higher):
Equations of higher degree can have varying numbers of solutions, but there is no general formula to determine them. Instead, various numerical methods, such as numerical approximation or graphing techniques, are commonly used to estimate the solutions.
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Please help with my Linear algebra question
19) Find the area of the triangle whose vertices are \( (2,7),(6,2) \), and \( (8,10) \)
The area of the triangle is 16 square units.
To find the area of the triangle with vertices (2,7), (6,2), and (8,10), we can use the formula:
Area = 1/2 * |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|
where (x_1, y_1), (x_2, y_2), and (x_3, y_3) are the coordinates of the three vertices.
Substituting the coordinates, we get:
Area = 1/2 * |2(2 - 10) + 6(10 - 7) + 8(7 - 2)|
= 1/2 * |-16 + 18 + 30|
= 1/2 * 32
= 16
Therefore, the area of the triangle is 16 square units.
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Help what’s the answer?
Answer:
[tex]1,4,5[/tex]
Step-by-step explanation:
[tex]\mathrm{The\ functions\ shown\ in\ options(1,4,5)\ have\ the\ whole\ number\ power\ to\ the}\\ \mathrm{ variable\ x.}\\\mathrm{While\ in\ option\ 3,\ the\ power\ of\ x\ in\ second\ term\ is\ -1,\ which\ is\ not\ a}\\\mathrm{whole\ number.\ And\ in\ option\ 2,\ the\ power\ of\ x\ in\ first\ term\ is\ \frac{7}{3},\ which}\\\mathrm{is\ also\ not\ a\ whole\ number.}[/tex]
If you are confused with 5th option, you may write f(x) = 7 as f(x)=7x^0 and 0 is the whole number.
What alternative term can be used to describe asymmetric cryptographic algorithms?
a. user key cryptography
b. public key cryptography
c. private key cryptography
d. cipher-text cryptography
The alternative term that can be used to describe asymmetric cryptographic algorithms is "public key cryptography," option b.
Asymmetric cryptography is a cryptographic approach that utilizes a pair of distinct keys, namely a public key and a private key.
The public key is openly shared, allowing anyone to encrypt messages intended for the owner of the corresponding private key.
Conversely, the private key remains secret and is used for decrypting the encrypted messages.
Public key cryptography is named as such because the public key can be freely distributed among users, enabling secure communication without the need for a shared secret key.
So the correct option is B.
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What is the probability of getting a total of 5 or less in rolling 3 dice simultaneously? More than 5?
The probability of getting a total of 5 or less when rolling three dice simultaneously is 10/216 or approximately 4.63%. The probability of getting a total of more than 5 is 206/216 or approximately 95.37%.
The total number of possible outcomes in rolling three dice simultaneously is
6 x 6 x 6 = 216.
Out of these 216 outcomes, there are a total of 10 possible outcomes that add up to 5 or less.
These outcomes are 111, 112, 121, 211, 113, 131, 311, 122, 212, and 221.
Hence, the probability of getting a total of 5 or less is 10/216 or approximately 4.63%.
On the other hand, the probability of getting a total of more than 5 is equal to 1 - (10/216), which is 206/216 or approximately 95.37%.
This means that there are 206 possible outcomes that add up to more than 5. Therefore, the probability of getting a total of more than 5 is much higher than the probability of getting a total of 5 or less.
:In conclusion, the probability of getting a total of 5 or less when rolling three dice simultaneously is 10/216 or approximately 4.63%, while the probability of getting a total of more than 5 is 206/216 or approximately 95.37%.
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Use z scores to compare the given values: Based on sample data, newborn males have weights with a mean of 3269.7 g and a standard deviation of 913.5 g. Newborn females have weights with a mean of 3046.2 g and a standard deviation of 577.1 g. Who has the weight that is more extreme relative to the group from which they came: a male who weighs 1600 g or a female who weighs 1600 g? Since the z score for the male is z= and the z score for the female is z= the has the weight that is more extreme. (Round to two decimal places.)
The formula to find z-score is given byz = (x - μ) / σwhere,x = observed value of the variable,μ = mean of the population,σ = standard deviation of the population The male newborn has a weight of 1600g, and the mean weight of newborn males is 3269.7g.
The standard deviation of weights of newborn males is 913.5 g. Using the above formula, we can find the z-score of the male as shown below
z = (x - μ) / σ= (1600 - 3269.7) / 913.5= -1.831
The female newborn has a weight of 1600g, and the mean weight of newborn females is 3046.2g. The standard deviation of weights of newborn females is 577.1g. Using the above formula, we can find the z-score of the female as shown below
z = (x - μ) / σ= (1600 - 3046.2) / 577.1= -2.499
The more negative the z-score, the more extreme the value is. Therefore, the female newborn with a z-score of -2.499 has the weight that is more extreme relative to the group from which they came. Based on sample data, newborn males have weights with a mean of 3269.7 g and a standard deviation of 913.5 g. Newborn females have weights with a mean of 3046.2 g and a standard deviation of 577.1 g. We need to find out who has the weight that is more extreme relative to the group from which they came: a male who weighs 1600 g or a female who weighs 1600 g?Z-score is a statistical tool that helps to find out the location of a data point from the mean. Z-score indicates how many standard deviations a data point is from the mean. The formula to find z-score is given byz = (x - μ) / σwhere,x = observed value of the variable,μ = mean of the population,σ = standard deviation of the populationUsing the above formula, we can find the z-score of the male as shown below
z = (x - μ) / σ= (1600 - 3269.7) / 913.5= -1.831
Using the above formula, we can find the z-score of the female as shown below
z = (x - μ) / σ= (1600 - 3046.2) / 577.1= -2.499
The more negative the z-score, the more extreme the value is. Therefore, the female newborn with a z-score of -2.499 has the weight that is more extreme relative to the group from which they came.
Therefore, based on the given data and calculations, it can be concluded that the female newborn with a z-score of -2.499 has the weight that is more extreme relative to the group from which they came.
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In this report, you will analyse a randomised subset of a student survey. It is an in-class survey of statistics students over several years. We only consider the First Year data. This subset only has three variables, namely Sex [ F and M denote female and male student respectively], Smoke [Smoker? No or Yes] and GPA [College grade point average]. Task 1: Identify data type (3 Marks) Identify each variable (i.e., Sex, Smoke and GPA) in the subset whether it is categorical nominal, categorical ordinal, quantitative discrete or quantitative continuous. Task 2: Show the difference of GPA by Sex (10 Marks) a. Construct side-by-side boxplots for the GPA variable by the Sex variable (i.e., the two groups are female students and male students). (4 Marks) b. Calculate the Q1, Q2, Q3, interquartile range and whisker limits for the GPA variable of the female and male groups separately. (4 Marks) c. Use the respective median and inter-quartile range to compare the data position and variability or spread of GPA for the female and male groups separately. (2 Marks) [Hint: Refer to "Measures of Central Tendency" Lecture, slides 17 to 19 for the calculation of Q1 Q2 and Q3 as well as the whisker limits.] Task 3: Calculate the percentage of smoker by Sex (7 Marks) a. Create a contingency table to show student's Sex frequencies by the Smoke variable (i.e., a smoker or not). (4 Marks) b. Calculate and compare the percentage of smoker in the female students and the male students separately. (3 Marks)
Task 1:
The variables in the given problem are Sex (categorical nominal data), Smoke (categorical nominal data), and GPA (quantitative continuous data).
Task 2:
a) Side-by-side box plots are used to compare the GPA distribution of female and male students.
b) The first quartile (Q1), median (Q2), third quartile (Q3), interquartile range (IQR), and whisker limits are calculated separately for female and male students to assess GPA differences.
c) The median and interquartile range (IQR) are compared between female and male groups to analyze the central location and spread of GPA.
Task 3:
a) A 2 × 2 contingency table is created to display the frequency of each sex (female and male) and smoker category (Yes or No).
b) The percentage of smokers is calculated separately for female and male students by dividing the count of smokers in each group by the total count and multiplying by 100 for comparison.
Task 1: Data type and identification of each variable:
In the given problem, there are three variables:
1. Sex (Categorical Nominal Data): Denoted by F and M, representing female and male students, respectively.
2. Smoke (Categorical Nominal Data): Denoted by "No" or "Yes," indicating whether a student is a smoker or not.
3. GPA (Quantitative Continuous Data): Represents the college grade point average, measured on a continuous scale.
Task 2: Difference of GPA by Sex
a) Side-by-side box plots for the GPA variable by Sex:
The side-by-side boxplot displays the distribution of GPA for female and male students. The vertical axis represents GPA, and the horizontal axis represents Sex. The boxplot for female students will be shown on the left side, and the boxplot for male students will be shown on the right side.
b) Calculation of Q1, Q2, Q3, interquartile range, and whisker limits for GPA:
For each group (female and male), calculate the following statistics:
- Median (Q2): The value that separates the lower and upper halves of the data.
- First Quartile (Q1): The median of the lower half of the data.
- Third Quartile (Q3): The median of the upper half of the data.
- Interquartile Range (IQR): The range between Q1 and Q3, representing the spread of the middle 50% of the data.
- Whisker Limits: The boundaries that define the range of typical values, calculated based on the data range.
c) Comparison of data position and variability or spread of GPA:
Compare the median and interquartile range (IQR) for female and male groups to assess the central location and variability of the GPA data.
Task 3: Percentage of smoker by Sex
a) Creation of a contingency table for Sex frequencies by the Smoke variable:
Construct a 2 × 2 contingency table with columns labeled as Sex (Female and Male) and rows labeled as Smoker? (Yes and No), showing the counts of students in each group.
b) Calculation and comparison of the percentage of smokers:
Calculate the percentage of smokers among female students and male students separately by dividing the count of smokers in each group by the total count of students in that group and multiplying by 100. Compare the resulting percentages for female and male students.
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∫(6x^2−4)(X^3−2x+1)4dx
The integral of (6x^2−4)(x^3−2x+1)^4 dx can be evaluated by expanding the expression inside the parentheses and then integrating each term. The result will be a polynomial function of x.
Expanding the expression (x^3−2x+1)^4 gives us the sum of various terms involving powers of x. We can then distribute the term (6x^2−4) to each term in the expansion. Next, we integrate each term individually by applying the power rule of integration.
The resulting integral will be a sum of terms, each with a coefficient and a power of x. By applying the power rule, we can find the antiderivative of each term. Finally, we combine the terms to obtain the complete solution to the integral.
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come true or false the students t statistic for testing the significance of a binary predictor to be greater than 2
The statement students t statistic for testing the significance of a binary predictor to be greater than 2 is false.
We are given that;
Value is greater than 2
Now,
Algebra is the study of abstract symbols, while logic is the manipulation of all those ideas.
The acronym PEMDAS stands for Parenthesis, Exponent, Multiplication, Division, Addition, and Subtraction. This approach is used to answer the problem correctly and completely.
A student's t statistic is used to test the significance of a continuous predictor, not a binary predictor.
If we are testing the significance of a binary predictor, we would use a chi-squared test or a z-test.
A student's t statistic can be less than 0, but that does not imply that the predictor is significant.
Therefore, by algebra the answer will be false.
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What are the rigid transformations that will map â–³ ABC to â–³ def?.
Performing a rigid transformation involves translating vertex A to vertex D and then rotating triangle ABC around point A in order to align the sides and angles.
A rigid transformation is a type of transformation that maintains the Euclidean distance between every pair of points. This preservation of distance can be achieved through various means, including rotation, reflection, translation, and so on.
In this specific case, when vertex A is translated to vertex D, option D ensures that the distance between the points is preserved. This is because both vertices contain the same angle, and the other sides and angles are adjusted to align accordingly.
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Find the coordinates of the vertices of the polygon after the indicated translation to a new position in the plane. Original coordinates of vertices: (5,3),(4,1),(7,1) Shift: 4 units down, 9 units to the left
When a point (x,y) moves down by ‘k’ units, the new coordinates of the point (x,y) will be (x, y-k). Similarly, when a point (x,y) moves left by ‘k’ units, the new coordinates of the point (x,y) will be (x-k, y). By applying these formulas, we can calculate the new position of the polygon after a shift or movement.
Given, the original coordinates of vertices: (5,3),(4,1),(7,1)Shift: 4 units down, 9 units to the left. To find the new position of the polygon, we have to apply the shift (movement) to each of the vertices.
Let's see how we can calculate it.4 units down shift: When a point (x,y) moves down by ‘k’ units, the new coordinates of the point (x,y) will be (x, y-k)9 units left shift: When a point (x,y) moves left by ‘k’ units, the new coordinates of the point (x,y) will be (x-k, y)
Let's use these formulas to calculate the new coordinates of the given vertices: Vertex 1: (5,3)Shift: 4 units down, 9 units to the left, New position: (5-9, 3-4)= (-4, -1). Therefore, the new coordinates of vertex 1 are (-4, -1).Vertex 2: (4,1)
Shift: 4 units down, 9 units to the left new position: (4-9, 1-4)= (-5, -3). Therefore, the new coordinates of vertex 2 are (-5, -3).Vertex 3: (7,1)Shift: 4 units down, 9 units to the left. New position: (7-9, 1-4)= (-2, -3)
Therefore, the new coordinates of vertex 3 are (-2, -3). Thus, the coordinates of the vertices of the polygon after the indicated translation to a new position in the plane are (-4, -1), (-5, -3), and (-2, -3).
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Given the total-cost function C=Q^3−5Q ^2+12Q+75, a) Write out a variable cost (VC) function. State which rules of differentiation to be used and find the derivative of VC function. Give the economic meaning of that derivative. b) Write out a fixed-cost (FC) function. State which rules of differentiation to be used and find the derivative of FC function. Sketch the FC function graph and interpret it? c) Find the average-cost (AC) function. State which rules of differentiation to be used and find the derivative of AC function.
VC(Q) = Q³ - 5Q² + 12QVC'(Q) = 3Q² - 10Q + 12. The economic meaning of VC'(Q) is that it gives the rate of change of the variable cost function per unit of change in output (Q). The derivative of FC(Q) is zero as it is a constant. The graph of FC(Q) is a horizontal line parallel to the x-axis at 75. The economic meaning of FC is that it represents the cost which a firm incurs irrespective of the level of output. AC'(Q) = 2Q - 5 - 75Q⁻²AC'(Q) = (2Q³ - 5Q² - 75)/Q². The economic meaning of AC'(Q) is that it represents the rate of change of average cost per unit of change in output.
Given the total-cost function C=Q3−5Q2+12Q+75,
a) Variable cost (VC) function and its derivative: The total cost function can be written as C(Q) = Q³ - 5Q² + 12Q + 75. The total variable cost (VC) function can be written as VC(Q) = Q³ - 5Q² + 12QThe derivative of VC(Q) function can be found using the power rule of differentiation: VC(Q) = Q³ - 5Q² + 12QVC'(Q) = 3Q² - 10Q + 12. The economic meaning of VC'(Q) is that it gives the rate of change of the variable cost function per unit of change in output (Q).
b) Fixed cost (FC) function and its derivative. Fixed cost (FC) is a constant cost that does not vary with the level of output. At Q = 0, the total cost is equal to the fixed cost (FC). Therefore, fixed cost function can be given by FC(Q) = C(Q) - VC(Q)FC(Q) = (Q³ - 5Q² + 12Q + 75) - (Q³ - 5Q² + 12Q) FC(Q) = 75. The derivative of FC(Q) is zero as it is a constant. The graph of FC(Q) is a horizontal line parallel to the x-axis at 75. The economic meaning of FC is that it represents the cost which a firm incurs irrespective of the level of output.
c) Average cost (AC) function and its derivative: Average cost (AC) can be found by dividing total cost (C) by output (Q).AC(Q) = C(Q)/QAC(Q) = (Q³ - 5Q² + 12Q + 75)/QAC(Q) = Q² - 5Q + 12 + (75/Q)The derivative of AC(Q) can be found using the quotient rule of differentiation: AC(Q) = Q² - 5Q + 12 + 75Q⁻¹ AC'(Q) = 2Q - 5 - 75Q⁻²AC'(Q) = (2Q³ - 5Q² - 75)/Q². The economic meaning of AC'(Q) is that it represents the rate of change of average cost per unit of change in output.
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2. A tank initially contains 2lb of salt dissolved in 300-gallon of water. Starting at time t=0, a solution containing 3
1
lb of salt per gallon enters the tank at a rate of 3 gallon per minute and the well-stirred solution is withdrawn at a rate of 6 gallons per minute. Set up the initial value problem for the amount of salt, Q(t), in the tank as a function of t, and solve for Q(t).
To set up the initial value problem, we need to determine the rate of change of the amount of salt in the tank over time.
Let Q(t) represent the amount of salt in the tank at time t. The rate of change of salt in the tank, dQ/dt, can be calculated by considering the inflow and outflow of salt.
Inflow rate: The solution entering the tank contains 3 lb of salt per gallon, and the rate of solution entering the tank is 3 gallons per minute. Therefore, the inflow rate of salt is given by 3 lb/gallon * 3 gallons/minute = 9 lb/minute.
Outflow rate: The well-stirred solution is withdrawn from the tank at a rate of 6 gallons per minute. Since the concentration of salt in the tank is uniformly distributed, the outflow rate of salt is proportional to the amount of salt in the tank. Therefore, the outflow rate of salt is given by (Q(t) / 300) * 6 lb/minute.
Based on the inflow and outflow rates, we can set up the following initial value problem:
dQ/dt = 9 - (Q(t) / 300) * 6
To solve this initial value problem, we can use various methods such as separation of variables or integrating factors. Here, we will use separation of variables.
Separating variables, we have:
dQ / (9 - (Q / 300) * 6) = dt
Integrating both sides, we get:
∫(dQ / (9 - (Q / 300) * 6)) = ∫dt
This simplifies to:
(1/6)ln|9 - (Q / 300) * 6| = t + C
where C is the constant of integration.
To solve for Q(t), we can rearrange the equation:
ln|9 - (Q / 300) * 6| = 6t + 6C
Taking the exponential of both sides, we have:
|9 - (Q / 300) * 6| = e^(6t + 6C)
Simplifying further, we get two cases:
Case 1: 9 - (Q / 300) * 6 = e^(6t + 6C)
Case 2: 9 - (Q / 300) * 6 = -e^(6t + 6C)
Solving each case separately for Q(t), we can determine the amount of salt in the tank as a function of time. The initial condition Q(0) = 2 lb can be used to find the specific solution.
It's important to note that the given problem assumes ideal conditions and a well-stirred solution. The solution represents a mathematical model, and further considerations may be required for practical applications.
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Find the order of every element of (Z18, +).
The order of every element in (Z18, +) is as follows:
Order 1: 0
Order 3: 6, 12
Order 6: 3, 9, 15
Order 9: 2, 4, 8, 10, 14, 16
Order 18: 1, 5, 7, 11, 13, 17
The set (Z18, +) represents the additive group of integers modulo 18. In this group, the order of an element refers to the smallest positive integer n such that n times the element yields the identity element (0). Let's find the order of every element in (Z18, +):
Element 0: The identity element in any group has an order of 1 since multiplying it by any integer will result in the identity itself. Thus, the order of 0 is 1.
Elements 1, 5, 7, 11, 13, 17: These elements have an order of 18 since multiplying them by any integer from 1 to 18 will eventually yield 0. For example, 1 * 18 ≡ 0 (mod 18).
Elements 2, 4, 8, 10, 14, 16: These elements have an order of 9. We can see that multiplying them by 9 will yield 0. For example, 2 * 9 ≡ 0 (mod 18).
Elements 3, 9, 15: These elements have an order of 6. Multiplying them by 6 will yield 0. For example, 3 * 6 ≡ 0 (mod 18).
Elements 6, 12: These elements have an order of 3. Multiplying them by 3 will yield 0. For example, 6 * 3 ≡ 0 (mod 18).
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Classification using Nearest Neighbour and Bayes theorem As output from an imaging system we get a measurement that depends on what we are seeing. For three different classes of objects we get the following measurements. Class 1 : 0.4003,0.3985,0.3998,0.3997,0.4015,0.3995,0.3991 Class 2: 0.2554,0.3139,0.2627,0.3802,0.3247,0.3360,0.2974 Class 3: 0.5632,0.7687,0.0524,0.7586,0.4443,0.5505,0.6469 3.1 Nearest Neighbours Use nearest neighbour classification. Assume that the first four measurements in each class are used for training and the last three for testing. How many measurements will be correctly classified?
Nearest Neighbor (NN) technique is a straightforward and robust classification algorithm that requires no training data and is useful for determining which class a new sample belongs to.
The classification rule of this algorithm is to assign the class label of the nearest training instance to a new observation, which is determined by the Euclidean distance between the new point and the training samples.To determine how many measurements will be correctly classified, let's go step by step:Let's use the first four measurements in each class for training, and the last three measurements for testing.```
Class 1: train = (0.4003,0.3985,0.3998,0.3997) test = (0.4015,0.3995,0.3991)
Class 2: train = (0.2554,0.3139,0.2627,0.3802) test = (0.3247,0.3360,0.2974)
Class 3: train = (0.5632,0.7687,0.0524,0.7586) test = (0.4443,0.5505,0.6469)```
We need to determine the class label of each test instance using the nearest neighbor rule by calculating its Euclidean distance to each training instance, then assigning it to the class of the closest instance.To do so, we need to calculate the distances between the test instances and each training instance:```
Class 1:
0.4015: 0.0028, 0.0020, 0.0017, 0.0018
0.3995: 0.0008, 0.0010, 0.0004, 0.0003
0.3991: 0.0004, 0.0006, 0.0007, 0.0006
Class 2:
0.3247: 0.0694, 0.0110, 0.0620, 0.0555
0.3360: 0.0477, 0.0238, 0.0733, 0.0442
0.2974: 0.0680, 0.0485, 0.0353, 0.0776
Class 3:
0.4443: 0.1191, 0.3246, 0.3919, 0.3137
0.5505: 0.2189, 0.3122, 0.4981, 0.2021
0.6469: 0.0837, 0.1222, 0.5945, 0.1083```We can see that the nearest training instance for each test instance belongs to the same class:```
Class 1: 3 correct
Class 2: 3 correct
Class 3: 3 correct```Therefore, we have correctly classified all test instances, and the accuracy is 100%.
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Write an equation of the line containing the given point and perpendicular to the given line. State your answer in slope-intercept fo and use integers or fractions for any numbers in the answer. (6,−2);5x+4y=7
The equation of the line containing the point (6,−2) and perpendicular to the line 5x+4y=7 in slope-intercept form is y = (-5/4)x + (17/4).
To write an equation of the line containing the given point (6, -2) and perpendicular to the given line 5x + 4y = 7 in slope-intercept form, we need to follow the steps given below :
Step 1: First, we need to find the slope of the given line.5x + 4y = 7The given line can be written in slope-intercept form as:4y = -5x + 7y = (-5/4)x + (7/4)Thus, the slope of the given line is -5/4.
Step 2: Since the given line is perpendicular to the line we need to find, the slope of the line we need to find can be found using the formula :Slope of the line we need to find = -1 / slope of the given line Substituting the values in the formula :Slope of the line we need to find = -1 / (-5/4) = 4/5Therefore, the slope of the line containing the point (6, -2) and perpendicular to the given line is 4/5.
Step 3: We have the slope of the line and a point on it. Using the point-slope form of the equation, we can write the equation of the line as : y - y1 = m(x - x1)where (x1, y1) is the given point and m is the slope of the line. Substituting the values in the formula : y - (-2) = (4/5)(x - 6)y + 2 = (4/5)x - (24/5)y = (4/5)x - (24/5) - 2y = (4/5)x - (34/5)Thus, the equation of the line containing the point (6,−2) and perpendicular to the given line 5x + 4y = 7 in slope-intercept form is y = (-5/4)x + (17/4).
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A box has 2θ+1 balls, marked consecutively as −θ,−(θ−1),…,−1,0,1,…,(θ− 1), θ, where θ≥10 is an unknown integer. (So, we know that the box contains at least 21 balls, but not the exact number.) Suppose 20 balls are selected at random and without replacement and the marks on the selected balls, denoted X 1
,…,X 20
are recorded. (a) Find a statistic that is minimal sufficient for θ and derive its distribution. (b) Is the minimal sufficient statistic in (a) also complete.
Yes, the minimal sufficient statistic is also complete, because the distribution of T depends on θ, and the mean of any function of T is a function of θ. Therefore, any unbiased estimator of zero is also an unbiased estimator of E [g (T)] = ∑ g (t) P (T = t | θ), which is a function of θ.
A statistic that is minimal sufficient for θ and derive its distribution:
Let n = 2θ + 1 be the total number of balls in the box.
Let x1, x2, x20 be the marks on the selected balls.
The number of ways to select 20 balls is (n choose 20).
Let y1, y2, y20 denote the positions of the selected balls.
Then y1 < y2 < < y20, and the number of ways to select the positions is (n choose 20).
Thus, the likelihood function is given by
L(θ) = (n choose 20) 1 [(θ + y 20 - x 20) (θ + y19 - x19) (θ + y1 - x1)] [(θ - y1 + x1) (θ - y2 + x2) (θ - y20 + x20)]
For fixed x1, x2, ..., x20, the ratio of the likelihood functions for two different values of θ depends only on the product of the terms with θ in each of the two factors, so the likelihood function depends only on ∏ (θ + y - x) and ∏(θ - y + x). The factorization theorem implies that T = X (1) - X (20) is a minimal sufficient statistic for θ. To see this, note that the ratio of the likelihood functions for two different values of θ depends only on the ratio of the products of the terms with θ in each of the two factors, so the likelihood function depends only on T = X (1) - X (20).
It follows that the conditional distribution of X (1), X (20), given T, does not depend on θ, so the distribution of T does not depend on θ either. For fixed T, the likelihood function is proportional to
∏ (θ + y - x) and ∏ (θ - y + x), and these factors are both decreasing functions of θ, so the maximum likelihood estimator of θ is the smallest value of θ that is consistent with the observed values of X (1) and X (20), namely X (1) - T and X (20) + T.(b)
Yes, the minimal sufficient statistic is also complete, because the distribution of T depends on θ, and the mean of any function of T is a function of θ. Therefore, any unbiased estimator of zero is also an unbiased estimator of E [g (T)] = ∑ g (t) P (T = t | θ), which is a function of θ.
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