The probability that a student who doesn't attend lectures passes the exam is 2/25.
Let's assume that there are 100 students in total. From the given information, we know that 4/5 of the students (80 students) pass the exam, and 1/5 of the students (20 students) fail the exam.
Out of the 80 students who pass the exam, 9/10 of them attend lectures. Therefore, the number of students who pass the exam and attend lectures is (9/10) * 80 = 72.
Out of the 20 students who fail the exam, only 1/2 of them attend lectures. So, the number of students who fail the exam and attend lectures is (1/2) * 20 = 10.
Now, we need to find the probability that a student who doesn't attend lectures passes the exam. This means we need to find the probability of passing the exam among the students who do not attend lectures, which is the complement of attending lectures.
The number of students who do not attend lectures is 100 - 72 - 10 = 18. Out of these 18 students, only 2 students pass the exam.
Therefore, the probability that a student who doesn't attend lectures passes the exam is 2/18, which simplifies to 1/9. In decimal form, this is approximately 0.08, which can be further simplified to 2/25.
Thus, the correct answer is option d) 2/25.
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numbers = (56, 38, 41, 32, 45, 54, 31, 87, 58, 96) Partition(numbers, 0, 6) is called. Assume quicksort always chooses the element at the midpoint as the pivot.
What is the pivot?
What is the low partition?
What is the high partition?
What is numbers after Partition(numbers, 0, 6) completes?
The pivot for the Partition(numbers, 0, 6) operation is 54. The low partition consists of 56, 38, 41, 32, 45, and 31. The high partition consists 87, 58, and 96. After the completion of Partition(numbers, 0, 6), the updated order of the numbers would be (56, 38, 41, 32, 45, 31, 54, 87, 58, 96).
In the given scenario, the quicksort algorithm chooses the element at the midpoint (index 5) of the given range (from 0 to 6) as the pivot. In this case, the element at index 5 is 54. The Partition operation then divides the numbers into two partitions based on their relationship to the pivot. All elements less than the pivot are placed in the low partition, while all elements greater than the pivot are placed in the high partition.
In the given numbers, the elements less than 54 are 56, 38, 41, 32, 45, and 31, which form the low partition. The elements greater than 54 are 87, 58, and 96, which form the high partition. After the completion of the Partition operation, the numbers are rearranged so that the low partition comes before the pivot and the high partition comes after the pivot. Therefore, the updated order of the numbers becomes (56, 38, 41, 32, 45, 31, 54, 87, 58, 96).
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Solve the system by using Gaussian elimination or Gauss-Jordan elimination.
5x−19y+12z=−110
x−4y+3z=−23
2x−3y−7z=−19
we can see that the system is consistent and has a unique solution. The solution to the system is x = 242, y = -132, and z = -1.
To solve the given system of equations using Gaussian elimination or Gauss-Jordan elimination, we will perform row operations to transform the augmented matrix into row-echelon form or reduced row-echelon form.
The augmented matrix representing the system is:
[5 -19 12 | -110]
[1 -4 3 | -23]
[2 -3 -7 | -19]
First, we will perform row operations to create zeros below the first entry in the first column. We can achieve this by multiplying the second row by 5 and adding it to twice the third row. The resulting matrix is:
[5 -19 12 | -110]
[0 1 -2 | -130]
[0 7 -13 | -267]
Next, we will perform row operations to create zeros below the second entry in the second column. We can achieve this by multiplying the second row by 7 and adding it to seven times the third row. The resulting matrix is:
[5 -19 12 | -110]
[0 1 -2 | -130]
[0 0 1 | -1]
Now, we will perform row operations to create zeros above the third entry in the third column. We can achieve this by adding twice the third row to the second row and adding -12 times the third row to the first row. The resulting matrix is:
[5 -19 0 | 2]
[0 1 0 | -132]
[0 0 1 | -1]
Finally, we will perform row operations to create zeros above the second entry in the second column. We can achieve this by adding 19 times the second row to the first row. The resulting matrix is:
[5 0 0 | 242]
[0 1 0 | -132]
[0 0 1 | -1]
From the row-echelon form of the augmented matrix, we can see that the system is consistent and has a unique solution. The solution to the system is x = 242, y = -132, and z = -1.
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Suppose that Y
1
,Y
2
,…,Y
m
is a random sample of size m from Gamma (α=3,β=θ), where θ is not known. Check whether or not the maximum likelihood estimator
θ
^
is a minimum variance unbiased estimator of the parameter θ.
The maximum likelihood estimator θ-hat is biased, but the bias tends to 0 as the sample size increases. It is not a minimum variance unbiased estimator.
Given, Y1, Y2, …, Ym are the random samples of size m from Gamma (α=3, β=θ), where θ is not known and the aim is to check whether or not the maximum likelihood estimator θ-hat is a minimum variance unbiased estimator of the parameter θ.
Now, let's calculate the maximum likelihood estimator of θ. The density function of Gamma distribution is given by: f(y|θ) = (1/Γ(α) βα) yα-1 e-y/β. The likelihood function of θ is given by: L(θ) = (1/Γ(α) βα)m Πi=1myiα-1 e-yi/β. We can take the natural logarithm of this to obtain: l(θ) = m log (1/Γ(α)) + mα log(β) + (α-1) Σi=1mlog (yi) - (1/β) Σi=1myi.
Now, we differentiate this with respect to θ to obtain: d/dθl(θ) = mα/θ - Σi=1myi/θ = 0. Therefore,θ-hat = (1/m) Σi=1myi = Y-bar. The expected value of θ-hat is given by: E(θ-hat) = E(Y-bar) = (1/m) Σi=1m E(Yi) = (1/m) Σi=1m (αβ) = αβ=3θ. Thus, the maximum likelihood estimator θ-hat is biased but the bias tends to 0 as m tends to infinity.
Now, let's calculate the variance of θ-hat. The variance of θ-hat is given by:
Var(θ-hat) = Var(Y-bar) = (1/m²) Σi=1mVar(Yi) = αβ/m = 3θ/m. Thus, the mean square error (MSE) of θ-hat is given by: MSE(θ-hat) = Var(θ-hat) + Bias²(θ-hat) = (3θ/m) + 0
Thus, the maximum likelihood estimator θ-hat is not a minimum variance unbiased estimator of the parameter θ. Therefore, it can be concluded that the estimator is biased.
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A moving particle starts at an initial position r(0) = (0, 1, 1) with initial velocity v(0) = (0, 0, 1). Its acceleration at any given time t is a(t) = (t, e^t, e^-t). Find its velocity and position vectors at any given time t.
Given that a particle starts at an initial position r(0) = (0,1,1) with initial velocity v(0) = (0,0,1) and its acceleration at any given time t is a(t) = (t, e^t, e^-t). We need to find the velocity and position vectors at any given time t.
Velocity Vector:The acceleration vector is given as a(t) = (t, e^t, e^-t)We know that acceleration is the derivative of velocity with respect to time.So, integrating a(t) w.r.t time we get velocity vector v(t) as: v(t) = (t²/2, e^t, -e^-t) + c1Where c1 is the constant of integration.Using the initial condition given, i.e., v(0) = (0,0,1) we can calculate the value of c1.Substituting v(0) = (0,0,1), we get c1 = (0,0,0)Hence, the velocity vector at any given time t is given as v(t) = (t²/2, e^t, -e^-t)Conclusion: Therefore, the velocity vector at any given time t is v(t) = (t²/2, e^t, -e^-t)Position Vector:The velocity vector is given as v(t) = (t²/2, e^t, -e^-t)We know that velocity is the derivative of position vector with respect to time.So, integrating v(t) w.r.t time we get the position vector r(t) as: r(t) = (t³/6, e^t, e^-t) + c2Where c2 is the constant of integration.
Using the initial condition given, i.e., r(0) = (0,1,1) we can calculate the value of c2.Substituting r(0) = (0,1,1), we get c2 = (0,1,1)Hence, the position vector at any given time t is given as r(t) = (t³/6, e^t+1, e^-t+1)
Therefore, the velocity vector at any given time t is v(t) = (t²/2, e^t, -e^-t) and the position vector at any given time t is r(t) = (t³/6, e^t+1, e^-t+1).
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