State the meaning of (i) plaintext (ii) encryption (iii) decryption

a) Here's the MATLAB program that finds the biggest, smallest number, and average in an array without using any Built-In MATLAB functions:```matlabfunction [max_num, min_num, average] = find_stats()input_arr = [];num = input('Enter a number: ');while num ~= -1    input_arr = [input_arr num];    num = input('Enter a number: ');enddisp("Input Array: ")disp(input_arr)max_num = input_arr(1);min_num = input_arr(1);array_sum = 0;for i = 1:length(input_arr)    array_sum = array_sum + input_arr(i);    if input_arr(i) > max_num        max_num = input_arr(i);    end    if input_arr(i) < min_num        min_num = input_arr(i);    endendaverage = array_sum / length(input_arr);end```

Explanation: The user is prompted to input a number repeatedly until they enter -1 to indicate they're done entering numbers.The input array is then displayed.The maximum number, minimum number, and average of the array are computed using loops and displayed.b) Here's the MATLAB program that sorts an array from smallest to largest for any user without using any Built-In MATLAB functions:```matlabfunction sorted_arr = sort_array()input_arr = [];num = input('Enter a number: ');while num ~= -1    input_arr = [input_arr num];    num = input('Enter a number: ');enddisp("Input Array: ")disp(input_arr)sorted_arr = [];for i = 1:length(input_arr)    min_val = input_arr(1);    min_index = 1;    for j = 1:length(input_arr)        if input_arr(j) < min_val            min_val = input_arr(j);            min_index = j;        end    end    sorted_arr = [sorted_arr min_val];    input_arr(min_index) = 150;endend```

Explanation:The user is prompted to input a number repeatedly until they enter -1 to indicate they're done entering numbers. The input array is then displayed.The array is sorted by repeatedly finding the minimum value in the remaining unsorted portion of the array and appending it to the sorted array.The sorted array is displayed.

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To find the biggest, smallest number, and average in an array without using any built-in MATLAB functions, prompt the user to input an array and use a while loop and conditional statements. To sort an array from smallest to largest, prompt the user to input an array and use nested for loops to compare and swap elements.

To find the biggest, smallest number, and average in an array without using any built-in MATLAB functions, you can prompt the user to input an array of any size and ask them to enter -1 when they are done inputting the array. Here is a step-by-step explanation of how to accomplish this:

To sort an array from smallest to largest without using any built-in MATLAB functions, you can prompt the user to input an array of any size and ask them to enter -1 when they are done inputting the array. Here is a step-by-step explanation of how to accomplish this:

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The expression 2² x 4³ x 8-2 can be simplified to 2^9. Therefore, n equals 9.



To simplify the given expression, we'll start by evaluating each part individually.First, we have 2², which equals 2 × 2 = 4.

Next, we have 4³, which equals 4 × 4 × 4 = 64.

Lastly, we have 8-2, which equals 6.

Now, we can rewrite the expression as 4 × 64 × 6.

To express this in the form 2^n, we need to find the highest power of 2 that divides the number. Let's break down the factors:

4 = 2²

64 = 2^6

6 = 2 × 3

Now, we can rewrite the expression as (2²) × (2^6) × (2 × 3).Simplifying further, we get 2^(2 + 6 + 1), which is equal to 2^9.Therefore, the expression 2² × 4³ × 8-2 can be expressed as 2^9.From this, we can see that n = 9.

Therefore, the correct answer is B. 9.

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The given circle equation is x² + y² = 4 . Now, let W be the set of points on the circle x² + y² = 4.

We need to show that W is not a subspace of R².In order to show that W is not a subspace of R², we need to check whether the following two properties hold or not:

Additivity property: For any vectors u, v in W, u+v is also in W or not.

Multiplication property: For any scalar c and any vector v in W, cv is also in W or not.

Now, let us consider two vectors on the circle x² + y² = 4 and check whether their addition and multiplication by a scalar satisfies the above two properties or not.

Consider the vectors u=(2,0) and v=(-2,0) . These vectors are on the circle x² + y² = 4 .

We have u+v=(2+(-2), 0+0)=(0,0) . Since (0,0) is not on the circle x² + y² = 4, thus the Additivity property is not satisfied for the vectors u and v in W.

Therefore, W is not a subspace of R².

Thus, we have shown that W is not a subspace of R² because it does not satisfy the additivity property and multiplication property.

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The value of (Ā.V)B on the given vector space is x³i + xyzj — x²yk + 2x²yi + 2y²zj — 2xy²k

The given function is 0 = x²y³z⁴. To find the maximum directional derivative, we need to calculate the gradient of the function first.

The gradient of the function is given as, grad(f) = (df/dx) i + (df/dy) j + (df/dz) k

Now, we need to find the partial derivatives of the given function with respect to x, y, and z.  Let's find the partial derivative of f with respect to x.fx = ∂f/∂x = 2xy³z⁴

Here's the partial derivative of f with respect to y.fy = ∂f/∂y = 3x²y²z⁴

And, here's the partial derivative of f with respect to z.fz = ∂f/∂z = 4x²y³z³

Now, the gradient of the function is: grad(f) = (2xy³z⁴) i + (3x²y²z⁴) j + (4x²y³z³) k

Now, we need to find the maximum directional derivative of the given function. We know that the directional derivative is given by the dot product of the gradient and a unit vector in the direction of the maximum derivative.

Therefore, the directional derivative is given as follows: Dᵥ(f) = ∇f . V, where V is the unit vector.

Dᵥ(f) = (2xy³z⁴) i + (3x²y²z⁴) j + (4x²y³z³) k . V

Now, let's find the unit vector in the direction of the maximum derivative. We know that the unit vector is given as: V = (a/|a|) i + (b/|b|) j + (c/|c|) k

where a, b, and c are the directional cosines.

Let's assume the maximum directional derivative occurs in the direction of the vector V = ai + bj + ck. Therefore, the directional cosines are given as follows:

a/|a| = 2xy³z⁴b/|b| = 3x²y²z⁴c/|c| = 4x²y³z³

Therefore, the vector V is given as:

V = (2xy³z⁴/|2xy³z⁴|) i + (3x²y²z⁴/|3x²y²z⁴|) j + (4x²y³z³/|4x²y³z³|) k

= (2xy³z⁴/√(4x²y⁶z⁸)) i + (3x²y²z⁴/√(9x⁴y⁴z⁸)) j + (4x²y³z³/√(16x⁴y⁶z⁶)) k

= 2xy³z/2xy²z⁴ i + 3x²y²z/3x²y²z³ j + 2xy³/2xy³z³ k= i/z + j/z + k

Therefore, the directional derivative is given as follows:

Dᵥ(f) = (2xy³z⁴) i + (3x²y²z⁴) j + (4x²y³z³) k . (i/z + j/z + k)

= (2xy³z⁴/z) + (3x²y²z⁴/z) + (4x²y³z³/z)

= (2xy²z³) + (3x²yz²) + (4x²y²z)

Now, we need to find the maximum value of Dᵥ(f). For that, we need to find the critical points of Dᵥ(f). Let's find the partial derivatives of Dᵥ(f) with respect to x, y, and z.

Here's the partial derivative of Dᵥ(f) with respect to x.

∂/∂x [(2xy²z³) + (3x²yz²) + (4x²y²z)] = 4xy²z + 6xyz²

Now, the partial derivative of Dᵥ(f) with respect to y.

∂/∂y [(2xy²z³) + (3x²yz²) + (4x²y²z)] = 2xy³z² + 6xyz²

And, the partial derivative of Dᵥ(f) with respect to z.

∂/∂z [(2xy²z³) + (3x²yz²) + (4x²y²z)] = 2xy²z² + 4x²y³

From the above three partial derivatives, we get,

4xy²z + 6xyz² = 0              -----(1)

2xy³z² + 6xyz² = 0         -----(2)

2xy²z² + 4x²y³ = 0      -----(3)

From equation (1), we get, 4yz + 6xz = 06xz = -4yzx = -4yz/6z = -2yz/3

Substitute the value of x in equation (3)

2y(-2yz/3)² + 4(-2yz/3)²y³ = 0

2y(4y²z²/9) + 4y³(4y²z²/9) = 0

2y(4y²z²/9) + 16y⁵z²/9 = 08y³z² = 0

9y²z² = 1y²z² = 1/9y = ± 1/3√z = ± 3√3/9

On substituting the values of x, y, and z, we get the maximum directional derivative as follows:

Dᵥ(f) = (2xy²z³) + (3x²yz²) + (4x²y²z)

= (2)(-2/3)((1/3)²)(3√3)³ + (3)(4/9)((-1/3)²)(3√3)² + (4)((-2/3)²)((1/3)²)(3√3)

= (-16/27)(27√3) + (4/9)(3)(3) + (4/9)√3

= -16√3/9 + 4 + 4√3/9= 4 + 3√3

Therefore, the maximum directional derivative is 4 + 3√3, and it occurs in the direction of the vector V = i/z + j/z + k.Let's find the value of (A bar.V)B. Here are the given vectors.

à = 2yzî — x²yĵ + xz²kB = x²î + yzĵ — xyk

Now, let's calculate Ā.V.Ā.V = Ã . V

Here's the vector V. V = i/z + j/z + k

Now, let's find the dot product of à and

V.Ã.V = (2yzî — x²yĵ + xz²k) . (i/z + j/z + k)= 2yz(i/z) — x²y(j/z) + xz²(k)= 2y — xy + x= x + 2y

Now, we need to find (x + 2y).B.

Here's the vector B. B = x²î + yzĵ — xyk

Now, let's calculate

(Ā.V)B.(Ā.V)B = (x + 2y) B= (x + 2y)(x²i + yzj — xyk)= x³i + xyzj — x²yk + 2x²yi + 2y²zj — 2xy²k

Therefore, the value of (Ā.V)B is x³i + xyzj — x²yk + 2x²yi + 2y²zj — 2xy²k.

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OA. The maximum value of P is 30.

OB. The coordinates of the corner point where the maximum value of P occurs are (4, 6).

To solve the linear programming problem, we can graph the feasible region determined by the constraints and find the corner point that maximizes the objective function P = 3x + 3y.

The constraints are:

2x + y ≤ 20 (equation 1)

x + 2y ≤ 16 (equation 2)

x, y ≥ 0

First, we graph the lines defined by the equations 2x + y = 20 and x + 2y = 16.

By plotting the points where the lines intersect the x and y axes, we can connect them to form the feasible region. The feasible region is the area below or on the lines and within the first quadrant.

Next, we evaluate the objective function P = 3x + 3y at the corner points of the feasible region to find the maximum value.

The corner points of the feasible region are:

A: (0, 0)

B: (0, 8)

C: (6, 0)

D: (4, 6)

Now, we substitute the coordinates of each corner point into the objective function P = 3x + 3y to find the corresponding values of P:

P(A) = 3(0) + 3(0) = 0

P(B) = 3(0) + 3(8) = 24

P(C) = 3(6) + 3(0) = 18

P(D) = 3(4) + 3(6) = 30

The maximum value of P is 30, which occurs at the corner point D: (4, 6).

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We can solve this problem using the method of linear programming, specifically the simplex method.

First, we need to write the constraints in standard form (Ax = b):

2x + 3y - s1 = 6

3x - 2y + s2 = 9

x + 5y = 20

where s1 and s2 are slack variables that allow us to convert the inequality constraints into equality constraints.

Next, we create the initial simplex tableau:

Coefficients x y s1 s2 RHS

4 1 0 0 0 0

3 0 1 0 0 0

0 2 3 -1 0 6

0 3 -2 0 -1 9

0 1 5 0 0 20

The first row represents the objective function coefficients, and the last column represents the right-hand side values of the constraints.

To find the minimum value of C, we need to use the simplex method to pivot until there are no negative values in the bottom row. At each iteration, we select the most negative value in the bottom row as the pivot element and use row operations to eliminate any other non-zero values in the same column.

After several iterations, we arrive at the final simplex tableau:

Coefficients x y s1 s2 RHS

1 0 0 1/7 -3/7 156/7

0 0 1 1/7 2/7 38/7

0 1 0 -3/7 2/7 6/7

0 0 0 5/7 13/7 174/7

0 0 0 4 -1 16

The minimum value of C is found in the last row and last column of the tableau, which is 16. Therefore, the minimum value of C = 4x + 3y subject to the constraints is 16, and it occurs when x = 6/7 and y = 38/7.

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In the first coordinate system, the graphs of \( y=(x-1)^{1/n} \) for \( n=1, 3, \) and \( 5 \) are shown. In the second coordinate system, the graphs of \( y=(x-1)^{1/n} \) for \( n=2, 4, \) and \( 6 \) are depicted.

To sketch the graph of the equation \( y=(x-1)^{1/n} \) for different values of \( n \), we can follow these steps:

1. For \( n=1, 3, \) and \( 5 \), and in the first coordinate system, we start by choosing some x-values. For each value of x, we calculate the corresponding y-value using the equation \( y=(x-1)^{1/n} \). Plot these points and connect them to form the graph.

2. The graph for \( n=1 \) is a straight line passing through the point (1, 0) and with a slope of 1.

3. For \( n=3 \), the graph has a similar shape to a cube root function. It is symmetric about the y-axis and passes through the point (1, 0).

4. For \( n=5 \), the graph becomes steeper near the point (1, 0) compared to the previous cases. It approaches the x-axis and y-axis but never touches them.

5. Similarly, in the second coordinate system, we repeat the process for \( n=2, 4, \) and \( 6 \). The graphs for these values will have similar characteristics to square root  but with different rates of increase as \( n \) increases.

Remember to label the axes and include appropriate scales to accurately represent the functions on the coordinate systems.

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The exponential equation 3^(1 - 2x) = 5 does not have an exact solution. The solution rounded to three decimal places is x ≈ -0.355

To solve the exponential equation 3^(1 - 2x) = 5, we need to isolate the variable x. We start by taking the logarithm of both sides of the equation.

Applying the logarithm property log(base b) a^c = c*log(base b) a, we have (1 - 2x)log(base 3) 3 = log(base 3) 5. Since log(base 3) 3 = 1, the equation simplifies to 1 - 2x = log(base 3) 5.

Next, we isolate x by subtracting 1 and dividing by -2: -2x = log(base 3) 5 - 1. Dividing by -2, we obtain x = (1 - log(base 3) 5) / 2.

However, this solution cannot be expressed exactly. We can approximate it as a decimal rounded to three decimal places. Using a calculator, we find x ≈ -0.355.

The solution to the equation, rounded to three decimal places, is x ≈ -0.355.

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The value of m that satisfies the condition is m = -3.5. To find the value of m that makes vectors u and v perpendicular, we can use the dot product.

Two vectors are perpendicular if their dot product is zero. The dot product of two vectors u = (5mi + 3j) and v = (2i + 7) can be calculated as follows:

u · v = (5mi)(2i) + (5mi)(7) + (3j)(2i) + (3j)(7)

      = 10m + 35mi + 6j + 21j

      = (10m + 35mi) + (6 + 21)j

For the dot product to be zero, the real and imaginary parts must be zero individually. Therefore, we can equate the real and imaginary parts to zero:

10m + 35mi = 0    -->  (Equation 1)

6 + 21 = 0       -->  (Equation 2)

From Equation 2, we can see that it leads to a contradiction, as 6 + 21 ≠ 0. Therefore, Equation 2 is not satisfied.

Now, let's solve Equation 1:

10m + 35mi = 0

To satisfy this equation, we must have m = -35i/10. Simplifying further, we get:

m = -3.5i

Thus, the value of m that makes vectors u and v perpendicular is m = -3.5.

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The {u, v} is an orthogonal set because the dot product is zero.

Let λ

= ||u|| and μ

= ||v||.

So Z = uu + Av can be expressed as (λ^2 + μ^2)w,

where w

= u/(λ^2+μ^2)1/2 and z

= v/(λ^2+μ^2)1/2.

We need to prove that {w, z} is an orthogonal set.To prove that {w, z} is orthogonal, we need to show that the dot product of the two vectors is zero.

The dot product of w and z is: w.z

= (λ^2+μ^2)(u/(λ^2+μ^2)1/2) .

(v/(λ^2+μ^2)1/2)

= uv/λ^2+μ^2

Therefore, {w, z} is an orthogonal set because the dot product is zero. (b) We need to prove that if ||uv|| = ||u+v||, then {u, v} is an orthogonal set.Using the formula for the dot product, we can say that ||uv||^2

= ||u+v||^2 is equivalent to u.v = 0.

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The internal rate of return (IRR) for the proposed investment with cash flows of $20,000 per year for 5 years, $15,000 per year for 6 years, and a net cash flow of $19,000 in the 12th year, is approximately 12.9%.

In order to calculate the IRR for the given scenario, you need to follow these steps with line breaks after every equal sign:

Step 1: Find the net cash flows for each year.

Year 1 to 5: $20,000,

year 6 to 11: $15,000,

year 12: ($3,000 + $22,000)

                  = $19,000.

Total net cash flow: $240,000.

Step 2: Determine the initial investment which is $90,000.

Step 3: Use a financial calculator or a spreadsheet program to calculate the IRR.

The IRR is approximately 12.9%.

Therefore, the internal rate of return (IRR) for the given proposal is approximately 12.9%.

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The upper bound for a symmetric p^ ± MOE confidence interval for the "None of the Above" category is approximately 0.0184

In the Week 46 Household Pulse Survey (HPS), the table below gives the gender identity responses from people who responded. We need to provide the upper bound for a symmetric p^± MOE confidence interval for the "None of the Above" category. Gender Identity Responses Cisgender Male24440 Cisgender Female 36420 Transgender 227 None of the Above 651 Margin of error: We can use the formula MOE=1/√N, where N is the total sample size, to estimate the margin of error for sample proportions.

In this case, the total sample size is the sum of all categories in the table. N = 24440 + 36420 + 227 + 651 = 61338 people. MOE = 1/√61338 = 0.004029. Symmetric p^ ± MOE confidence interval: We can use the following formula to compute the symmetric p^ ± MOE confidence interval: p^ ± MOE = p^ ± z α/2 MOE, where z α/2 is the z-score of the standard normal distribution for a given level of confidence α/2. We need to find the upper bound for a 95% confidence interval. Therefore,α/2 = 0.05/2 = 0.025.z α/2 = 1.96 (from the z-score table).

We can calculate p^ as follows: p^ = None of the Above/Total = 651/61338 = 0.010605. The upper bound of the confidence interval is: p^ + z α/2 MOE=0.010605 + (1.96) (0.004029)=0.018369. The upper bound for a symmetric p^ ± MOE confidence interval for the "None of the Above" category is approximately 0.0184 (rounded to four decimal places).

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With side lengths a = 16, b = 17, and angle A = 44 degrees, a triangle can be formed. The remaining angles are B ≈ 54.1 degrees and C ≈ 81.9 degrees, with side c ≈ 21.9.



To determine if the given information forms a triangle, we can apply the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Using the given information:

a = 16, b = 17, and A = 44 degrees

First, let's check if the triangle inequality is satisfied for sides a, b, and the third side, c:

a + b > c

16 + 17 > c

33 > c

Now, let's check if the triangle inequality is satisfied for sides b, c, and the third side, a:

b + c > a

17 + c > 16 + c > 16

The triangle inequality holds for both cases, which means that a triangle can be formed with the given information.

To solve the triangle, we can use the Law of Sines. Using the given angle A and side a, we can find the remaining angles and sides. Applying the Law of Sines:

sin(A) / a = sin(B) / b

sin(44) / 16 = sin(B) / 17

Solving for sin(B):

sin(B) = (sin(44) / 16) * 17

B ≈ 54.1 degrees

Now, we can find angle C using the fact that the sum of angles in a triangle is 180 degrees:

C ≈ 180 - A - B

C ≈ 180 - 44 - 54.1

C ≈ 81.9 degrees

Finally, we can find side c using the Law of Sines:

sin(C) / c = sin(A) / a

sin(81.9) / c = sin(44) / 16

Solving for c:

c ≈ (sin(81.9) / sin(44)) * 16

c ≈ 21.9

Therefore, the resulting triangle has side lengths approximately rounded to 1 decimal place: a = 16, b = 17, and c = 21.9. The angles are approximately A = 44 degrees, B = 54.1 degrees, and C = 81.9 degrees.

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1. (ℓ2, ∥⋅∥2) is a normed vector space over R. 2. (ℓ2, ∥⋅∥2) is complete. 3. The closed unit ball B(0,1) is not compact.

To show that (ℓ2, ∥⋅∥2) is a normed vector space over R, we need to verify the following properties:

a) Non-negativity: [tex]||a||_2 \geq 0[/tex] for all a ∈ ℓ2.

b) Definiteness: [tex]||a||_2 = 0[/tex] if and only if a = 0.

c) Homogeneity: [tex]||c. a||_2 = |c| . ||a||_2[/tex] for all c ∈ R and a ∈ ℓ2.

d) Triangle inequality: [tex]||a+b||_2 \leq ||a||_2 + ||b||_2[/tex] for all a, b ∈ ℓ2.

These properties can be easily verified using the definition of the norm [tex]|| \,||_2[/tex] as the square root of the sum of the squares of the elements in the sequence.

To show that (ℓ2, ∥⋅∥2) is complete, we need to demonstrate that every Cauchy sequence in ℓ2 converges to a limit that is also in ℓ2. Let {an} be a Cauchy sequence in ℓ2, meaning that for any ε > 0, there exists N such that for all m, n ≥ N, we have [tex]||a_m - a_n||_2 < \epsilon[/tex].

To show completeness, we need to find a limit point a ∈ ℓ2 such that [tex]\lim_{n \to \infty} ||an - a||_2[/tex]. We can construct this limit point by taking the limit of each component of the sequence. Since each component is a real number and the real numbers are complete, the limit exists for each component. Thus, the limit point a will also be in ℓ2, satisfying the completeness property.

To show that the closed unit ball B(0,1) = {a ∈ ℓ2 : ∥a∥2 ≤ 1} is not compact, we need to demonstrate that there exists an open cover of B(0,1) that does not have a finite sub-cover.

Consider the sequence {en}, where en is the sequence with a 1 in the nth position and 0s elsewhere. Each en is in B(0,1) since ∥en∥2 = 1. Now, for each n, consider the open ball B(en, 1/2). It can be shown that the intersection of any two of these open balls is empty, and hence, no finite sub-cover can cover B(0,1).

Therefore, B(0,1) is not compact.

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Therefore, the correct statement is B. Fail to reject H0 because the P-value is greater than the significance level.

(a) Hypotheses:

The correct statement for the null hypothesis (H0) and alternative hypothesis (Ha) is:

H0: μ ≤ 11.0

Ha: μ > 11.0

We are testing if the mean number of classroom hours per week for full-time faculty (μ) is greater than 11.0.

(b) Calculation of P-value:

To calculate the P-value, we need to find the test statistic. Since we have a small sample size (n = 8) and the population standard deviation is unknown, we can use a t-test.

Using statistical software or a t-distribution table, we can find the test statistic value for a one-sample t-test. Based on the provided data, the test statistic is t = 1.623.

Next, we can calculate the P-value associated with this test statistic. Since we are testing if the mean is greater than the claimed value (11.0), we need to find the area under the t-distribution curve to the right of the test statistic (t = 1.623).

Using software or a t-distribution table, we find that the P-value is approximately 0.070.

Therefore, P ≈ 0.070 (rounded to three decimal places).

(c) Decision:

To make a decision, we compare the P-value to the significance level (α). In this case, α = 0.05.

Since the P-value (0.070) is greater than the significance level (0.05), we fail to reject the null hypothesis.

(d) Conclusion:

Based on the hypothesis test, there is not enough evidence to reject the dean's claim that the mean number of classroom hours per week for full-time faculty is 11.0.

Therefore, the correct statement is:

B. Fail to reject H0 because the P-value is greater than the significance level.

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Given a vector space V of all functions f : R → R and T : V → V be defined as T(f(x)) = x * f(x), where x is a real number.To prove that T is a linear transformation, we need to show that it satisfies two conditions: (i) additivity and (ii) homogeneity.

Condition (i) additivity:T(u + v) = T(u) + T(v)T(u + v) = x * (u + v)T(u + v) = x * u + x * vT(u) + T(v) = x * u + x * vT(u) + T(v) = T(u + v)Hence, the transformation T satisfies the first condition, i.e., additivity.

Condition (ii) homogeneity:T(cu) = cT(u)T(cu) = cx * uT(cu) = c(x * u)T(u) = x * uT(cu) = cT(u).

Hence, the transformation T satisfies the second condition, i.e., homogeneity. Therefore, T is a linear transformation.b) To prove that every real number is an eigenvalue of T, we need to show that there exists a non-zero vector v in V such that Tv = λv for some scalar λ. Let f(x) = c be a constant function such that c is a real number. Then, we haveT(f(x)) = x * f(x)T(f(x)) = x * c = c * xNow, if we take v = f(x), then we haveTv = T(f(x)) = c * x = λvwhere λ = c. Hence, c is an eigenvalue of T for every real number c.Therefore, every real number is an eigenvalue of T.

Thus, we can conclude that T is a linear transformation, and every real number is an eigenvalue of T.

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Given, the function[tex]$f(x,y,z)=x−y+z$[/tex] subjected to the constraint[tex]$x^2 + y^2 + z^2 = 2$[/tex].To find the extreme values of the function using Lagrange multipliers, we need to consider the following equation.

[tex]$$ L(x,y,z,\lambda) = f(x,y,z) - \lambda(g(x,y,z)-c)$$Where, $g(x,y,z) = x^2 + y^2 + z^2$ and $c = 2$[/tex] is the constant.We have to differentiate L w.r.t x, y, z and $\lambda$ respectively and equate each to zero.

[tex]$$ \begin{aligned}\frac{\partial L}{\partial x} &= 1 - 2x\lambda = 0\\\frac{\partial L}{\partial y} &= -1 - 2y\lambda = 0\\\frac{\partial L}{\partial z} &= 1 - 2z\lambda = 0\\\frac{\partial L}{\partial \lambda} &= x^2 + y^2 + z^2 - 2 = 0\end{aligned} $$[/tex]

Now, from first three equations above, we get,[tex]$$ x = \frac{1}{2\lambda}, \: y = -\frac{1}{2\lambda}, \: z = \frac{1}{2\lambda} $$[/tex].

Substituting the value of[tex]$\lambda$[/tex] in the values of [tex]$x$, $y$ and $z$[/tex], we get $$ \begin{aligned}\textbf

[tex]Case 1: }\lambda &= \frac{\sqrt{3}}{2} \\x = \frac{1}{2\lambda}, \: y = -\frac{1}{2\lambda}, \: z = \frac{1}{2\lambda} \\x &= \frac{\sqrt{3}}{3}, \: y = -\frac{\sqrt{3}}{3}, \: z = \frac{\sqrt{3}}{3} \\\textbf[/tex]

[tex]{Case 2: } \lambda &= -\frac{\sqrt{3}}{2} \\x = \frac{1}{2\lambda}, \: y = -\frac{1}{2\lambda}, \: z = \frac{1}{2\lambda} \\x &= -\frac{\sqrt{3}}{3}, \: y = \frac{\sqrt{3}}{3}, \: z = -\frac{\sqrt{3}}{3}\end{aligned} $$[/tex]

Now, to find the extreme values of the function, we can substitute the values of [tex]$x$, $y$ and $z$[/tex] in the function [tex]$f(x,y,z)$[/tex] for each case:

[tex]$$\textbf{Case 1: }f\left(\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\right) = \frac{\sqrt{3}}{3} - \left(-\frac{\sqrt{3}}{3}\right) + \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}$$[/tex]

Hence, the maximum value is [tex]$f(x,y,z) = \frac{2\sqrt{3}}{3}$ which occurs at $x = \frac{\sqrt{3}}{3}, \: y = -\frac{\sqrt{3}}{3}, \: z = \frac{\sqrt{3}}{3}$ .[/tex]

[tex]$$\textbf{Case 2: }f\left(-\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},-\frac{\sqrt{3}}{3}\right) = -\frac{\sqrt{3}}{3} - \left(\frac{\sqrt{3}}{3}\right) - \frac{\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}$$[/tex]

Hence, the minimum value is [tex]$f(x,y,z) = -\frac{2\sqrt{3}}{3}$ which occurs at $x = -\frac{\sqrt{3}}{3}, \: y = \frac{\sqrt{3}}{3}, \: z = -\frac{\sqrt{3}}{3}$.[/tex]

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The value of 2f(x₃) by using Trapezoidal Rule is: 4.32.

To approximate the definite integral using the Trapezoidal Rule, we divide the interval [0, 1] into subintervals and approximate the area under the curve by summing the areas of trapezoids formed by adjacent points.

In this case, the integral of ∫0¹ sin²(x)+1 dx will be approximated using the Trapezoidal Rule with n = 5, meaning we will divide the interval into 5 subintervals.

Step 1: Determine the width of each subinterval.

The width, denoted as Δx, is calculated by dividing the interval length by the number of subintervals:

Δx = (b - a) / n = (1 - 0) / 5 = 1/5 = 0.2

Step 2: Calculate the sum of the function values at the endpoints of each subinterval.

Evaluate the function at the endpoints of each subinterval (including the endpoints of the interval) and sum the values:

f(x0) = f(0) = sin^2(0) + 1 = 0^2 + 1 = 1

f(x1) = f(0.2) = sin^2(0.2) + 1

f(x2) = f(0.4) = sin^2(0.4) + 1

f(x3) = f(0.6) = sin^2(0.6) + 1

f(x4) = f(0.8) = sin^2(0.8) + 1

f(x5) = f(1) = sin^2(1) + 1 = 1

Step 3: Calculate the approximate value of the integral using the Trapezoidal Rule.

The approximation of the definite integral is given by:

[f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + 2f(x4) + f(x5)] * Δx / 2

Using the values calculated in Step 2 and Δx = 0.2, we can plug them into the formula and solve for the approximation:

Approximation = [1 + 2f(0.2) + 2f(0.4) + 2f(0.6) + 2f(0.8) + 1] * 0.2 / 2

Now we need to find the values of f(0.2), f(0.4), f(0.6), and f(0.8). Calculate these values by evaluating sin^2(x) + 1 at the respective points:

f(0.2) = sin^2(0.2) + 1

f(0.4) = sin^2(0.4) + 1

f(0.6) = sin^2(0.6) + 1

f(0.8) = sin^2(0.8) + 1

After evaluating these expressions, substitute the results back into the approximation formula:

Approximation = [1 + 2f(0.2) + 2f(0.4) + 2f(0.6) + 2f(0.8) + 1] * 0.2 / 2

Finally, calculate the value of 2f(x3):

2f(x3) = 2 * f(0.6)

After evaluating f(0.6), substitute the result into the expression:

2f(x3) = 2 * f(0.6)

= 2*(\sin^{2}(0.6) + 1)

= 4.32

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a) The average rate of change in profit is calculated for different intervals of sweatshirt sales.

b) The average rate of change in profit increases as the number of sweatshirts sold increases, indicating a higher profit per sweatshirt.

c) The rate of change in profit is predicted to stay positive as long as the coefficient of the quadratic term in the profit function remains negative, meaning profit will continue to grow with increasing sweatshirt sales.

a) The average rate of change in profit for the given intervals are as follows:

i) 1 ≤ s ≤ 2: The average rate of change in profit is $2.85 thousand per sweatshirt.

ii) 2 ≤ s ≤ 3: The average rate of change in profit is $3.7 thousand per sweatshirt.

iii) 3 ≤ s ≤ 4: The average rate of change in profit is $4.55 thousand per sweatshirt.

iv) 4 ≤ s ≤ 5: The average rate of change in profit is $5.4 thousand per sweatshirt.

b) As the number of sweatshirts sold increases, the average rate of change in profit on each sweatshirt also increases. This means that for each additional sweatshirt sold, the profit generated increases at a faster rate.

c) Based on the given profit function, the rate of change in profit will stay positive as long as the coefficient of the quadratic term (-0.30s^2) remains negative. This means that as the number of sweatshirts sold continues to increase, the profit will continue to grow. However, it's important to note that there may be a point where the profit growth slows down or reaches a maximum due to factors such as market saturation or diminishing returns.

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10. A company that sells sweatshirts finds that the profit can be modelled by P(s)=−0.30s^2 +3.5s+11.15, where P(s) is the profit, in thousands of dollars, and s is the number of sweatshirts sold (expressed in thousands). a) Calculate the average rate of change in profit for the following intervals. i) 1≤s≤2 ii) 2≤s≤3 iii) 3≤s≤4 iv) 4≤s≤5 b) As the number of sweatshirts sold increases, what do you notice about the average rate of change in profit on each sweatshirt? What does this mean? c) Predict if the rate of change in profit will stay positive. Explain what this means.

Given High School Competency Test A mandatory competency test for high school sophomores has a normal distribution with a mean of 405 and a standard deviation of 111, Round the final answers to the nearest whole number and intermediate z-value calculations to 2 decimal places. The minimum score needed to receive the award is 576.

To find the minimum score needed to receive the award, we need to find the z-score corresponding to the top 6% of students and then convert it back to the raw score.

First, let's find the z-score using the standard normal distribution table or a calculator. Since the top 6% is considered, we look for the z-score that corresponds to a cumulative probability of 1 - 0.06 = 0.94.

Looking up the z-score for a cumulative probability of 0.94, we find it to be approximately 1.55.

Now, we can use the z-score formula to find the raw score:

z = (x - mean) / standard deviation

Rearranging the formula to solve for x, we have:

x = z * standard deviation + mean

Substituting the values we have:

x = 1.55 * 111 + 405

x ≈ 171.05 + 405

x ≈ 576.05

Rounding to the nearest whole number, the minimum score needed to receive the award is 576.

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Thus, n² - 2 is not divisible by 4.

In each case, n² - 2 is not divisible by 4. this completes the proof by cases.

For every even integer n,  n² - 2  is not divisible by 4.

This can be proven by cases as follows:

When n is even: Let n = 2k for some integer k. Then, n² - 2 = (2k)² - 2 = 4k² - 2.

Now we need to consider two cases separately:

Case 1: k is even If k is even, then k² is even too. So, k² = 2m for some integer m.

Then,

4k² - 2 = 8m - 2 = 2(4m - 1). Thus, n² - 2 is not divisible by 4.

Case 2: k is odd If k is odd, then k² is also odd.

So, k² = 2m + 1 for some integer m. Then,4k² - 2 = 8m + 2 = 2(4m + 1).

Thus, n² - 2 is not divisible by 4.

When n is odd:

Let n = 2k + 1 for some integer k.

Then, n² - 2 = (2k + 1)² - 2 = 4k² + 4k - 1 - 2 = 4k² + 4k - 3.

Now we need to consider two cases separately:

Case 1: k is even If k is even, then k² is even too.

So, k² = 2m for some integer m.

Thus,4k² + 4k - 3 = 8m + 4k - 3 = 2(4m + 2k - 1) + 1. Thus, n² - 2 is not divisible by 4.

Case 2: k is odd If k is odd, then k² is also odd. So, k² = 2m + 1 for some integer m.

Thus,4k² + 4k - 3 = 8m + 8k + 1 = 2(4m + 4k) + 1.

Thus, n² - 2 is not divisible by 4.In each case, n² - 2 is not divisible by 4.

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The 99% confidence interval for the proportion of defective chips is (0.0054, 0.0546).

Part A:

Given that a random sample of 12 wafers were drawn from a slider fabrication process, which gives the following photoresist thickness in micrometers: 10, 11, 9, 8, 10, 10, 11, 8, 9, 10, 11, 12. We need to construct a 95% confidence interval for the mean of all wafers thickness produced by this factory.

To construct a 95% confidence interval for the mean, we need to determine the sample mean, sample standard deviation, and sample size.

Sample mean `x¯` can be calculated as:

`x¯ = (10 + 11 + 9 + 8 + 10 + 10 + 11 + 8 + 9 + 10 + 11 + 12) / 12 = 10`

Sample standard deviation s can be calculated as:

`s = sqrt[((10 - 10)² + (11 - 10)² + (9 - 10)² + (8 - 10)² + (10 - 10)² + (10 - 10)² + (11 - 10)² + (8 - 10)² + (9 - 10)² + (10 - 10)² + (11 - 10)² + (12 - 10)²) / (12 - 1)] = sqrt(2.727) = 1.6507`

The sample size n is 12.

The formula to calculate the confidence interval is:

`CI = x¯ ± (t_(alpha/2)(n-1) * (s/sqrt(n)))`

Where `t_(alpha/2)(n-1)` is the t-distribution value at α/2 (alpha/2) level of significance with (n-1) degrees of freedom. Here, α is the level of significance, which is 1 - confidence level. So, α = 1 - 0.95 = 0.05. Therefore, α/2 = 0.025.

From the t-distribution table with 11 degrees of freedom, we can find the value of t_(alpha/2)(n-1) = t_(0.025)(11) = 2.201.

So, the confidence interval is:

`CI = 10 ± (2.201 * (1.6507 / sqrt(12))) = (8.607, 11.393)`

Hence, the 95% confidence interval for the mean of all wafers thickness produced by this factory is (8.607, 11.393).

Part B:

Given that a quality inspector inspected a random sample of 300 memory chips from a production line and found 9 are defectives. We need to construct a 99% confidence interval for the proportion of defective chips.

To construct a 99% confidence interval for the proportion of defective chips, we need to determine the sample proportion, sample size, and z-value.

Sample proportion `p` can be calculated as:

`p = 9 / 300 = 0.03`

Sample size n is 300.

The z-value at 99% confidence level is obtained from the z-table and is approximately 2.576.

The formula to calculate the confidence interval is:

`CI = p ± z_(alpha/2) * sqrt((p * (1 - p)) / n)`

Where `z_(alpha/2)` is the z-distribution value at α/2 (alpha/2) level of significance. Here, α is the level of significance, which is 1 - confidence level. So, α = 1 - 0.99 = 0.01. Therefore, α/2 = 0.005.

So, the confidence interval is:

`CI = 0.03 ± 2.576 * sqrt((0.03 * 0.97) / 300) = (0.0054, 0.0546)`

Hence, the 99% confidence interval for the proportion of defective chips is (0.0054, 0.0546).

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To determine the probability of drawing a poker hand consisting of two pairs, we need to calculate the number of favorable outcomes (hands with two pairs) and divide it by the total number of possible outcomes (all possible poker hands).

First, let's calculate the number of favorable outcomes:

Choose two denominations out of 13: C(13, 2) = 13! / (2!(13-2)!) = 78

For each chosen pair, choose two suits out of four: C(4, 2) = 4! / (2!(4-2)!) = 6

Choose one card of a different denomination: C(11, 1) = 11! / (1!(11-1)!) = 11

Now, let's calculate the total number of possible outcomes (all possible poker hands):

Choose any five cards out of 52: C(52, 5) = 52! / (5!(52-5)!) = 2,598,960

Finally, we can calculate the probability of drawing a poker hand consisting of two pairs:

P(Two pairs) = (number of favorable outcomes) / (total number of possible outcomes)

           = (78 * 6 * 11) / 2,598,960

           = 6,084 / 2,598,960

           ≈ 0.00235

Therefore, the probability of drawing a poker hand consisting of two pairs is approximately 0.00235, or about 0.235%.

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The problem described involves a combination.

In combinatorics, a permutation refers to the arrangement or ordering of objects, while a combination refers to the selection of objects without regard to their order.

In this particular problem, the medical researcher needs to select 27 people out of 95 volunteers.

The order in which the 27 people are selected is not important; what matters is the combination of people chosen.

Therefore, we are dealing with a combination problem.

To determine the number of ways to select 27 people from a group of 95, we can use the formula for combinations:

C(n, r) = n! / (r!(n-r)!)

In this case, n represents the total number of volunteers (95), and r represents the number of people to be selected (27).

Therefore, the problem involves a combination rather than a permutation.

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(f) The statement is true.

Let's consider the statement: "For all p > 0, there exists q, n > 0, such that for all s, t € R, |s — t| < q implies |s" – t¹| < p."

To prove this statement, we need to show that for any positive value of p, there exist positive values of q and n such that if the absolute difference between two real numbers s and t is less than q, then the absolute difference between the s-th power of s and the t-th power of t is less than p.

Let's take any positive value of p. We can choose q = p/2 and n = 1. Now, consider any two real numbers s and t such that |s - t| < q.

We know that if |s - t| < q, then |s - t| < p/2. Now, let's consider the function f(x) = x^n. Since n = 1, f(x) = x.

We have |s - t| < p/2, which implies |s^n - t^n| = |s - t| < p/2. Since f(x) = x, this implies |s - t| = |s - t|^n < p/2. Therefore, we have shown that |s - t| < q implies |s^n - t^n| < p, satisfying the condition of the statement.

Since we have successfully shown that for any positive value of p, there exist positive values of q and n such that |s - t| < q implies |s^n - t^n| < p, the statement is true.

(g) The statement is false.

Let's consider the statement: "For all s, t € R, there exists p, q > 0, such that for all n € Z, |s − t| ≤ q implies (tn ≤ |p| and s^n ≤ |p|)."

To disprove this statement, we need to show that there exists at least one pair of real numbers s and t for which no values of p and q can satisfy the given conditions.

Let's take s = t = 0. Now, consider any positive values of p and q. We have |s - t| = |0 - 0| = 0, which is less than or equal to q for any positive value of q.

Now, for any integer n, we have tn = 0, which is always less than any positive value of p. However, s^n = 0^0, which is undefined.

Since there exist values of s and t for which the condition of the statement cannot be satisfied by any values of p and q, the statement is false.

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Therefore, 80 gallons of the 45% fruit punch should be mixed with 10 gallons of the 90% fruit punch to create a fruit punch that is 50% fruit juice.

Let's assume we need to mix x gallons of the 45% fruit punch with 10 gallons of the 90% fruit punch to create a fruit punch that is 50% fruit juice.

To find the amount of fruit juice in the final mixture, we can calculate it by adding the amounts of fruit juice from each punch.

The amount of fruit juice in the 45% punch is 45% of x gallons, which is 0.45x gallons.

The amount of fruit juice in the 90% punch is 90% of 10 gallons, which is 0.9 * 10 = 9 gallons.

In the final mixture, the total amount of fruit juice will be the sum of the fruit juice amounts from both punches, which should be equal to 50% of the total volume of the mixture.

So, we can set up the equation: 0.45x + 9 = 0.5 * (x + 10)

Simplifying the equation, we get: 0.45x + 9 = 0.5x + 5

Rearranging and solving for x, we find: 0.05x = 4

x = 4 / 0.05

x = 80

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The result of the differential equation y" - 4y = xe* by variation of parameters is:y = c1e2t + c2e-2t - 1/16 * xe2t + 1/16 * xe-2t

The differential equation is y'' - 4y = xe*To solve the differential equation y'' - 4y = xe* by variation of parameters, we first have to find the general solution of the associated homogeneous differential equation y'' - 4y = 0. The characteristic equation of the homogeneous differential equation is:

r2 - 4 = 0

On solving the above equation, we get:

r1 = 2 and r2 = -2

Hence, the general solution of the associated homogeneous differential equation is:

yh = c1e2t + c2e-2t

Now, we assume the particular result of the differential equation as:

yp = u1(t)e2t + u2(t)e-2t

Then, y'p = 2u1(t)e2t - 2u2(t)e-2t

and

y''p = 4u1(t)e2t + 4u2(t)e-2t

Substituting the above values of yp, y'p and y''p in the differential equation y'' - 4y = xe*, we get:

4u1(t)e2t + 4u2(t)e-2t - 4(u1(t)e2t + u2(t)e-2t) = xe*

On simplifying the above equation, we get:

u1(t)e2t = -1/16 * x and u2(t)e-2t = 1/16 * x

On solving the above two equations, we get:u1(t) = -1/16 * x * e-2t and u2(t) = 1/16 * x * e2t

Therefore, the particular result of the differential equation y'' - 4y = xe* is:yp = u1(t)e2t + u2(t)e-2t = -1/16 * x * e2t + 1/16 * x * e-2t

Hence, the general result of the differential equation y'' - 4y = xe* is:y = yh + yp = c1e2t + c2e-2t - 1/16 * x * e2t + 1/16 * x * e-2t

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The 99% confidence interval for the population mean is (15.816, 26.184).

i. The number z α/2 needed in the construction of a confidence interval when the level of confidence is 99% is 2.576.ii. To construct a 99% confidence interval for the population mean, we can use the following formula:  $\overline{X}±\frac{z_{\frac{\alpha}{2}}\sigma}{\sqrt{n}}$Given that,Sample size (n) = 36Sample mean ($\overline{X}$) = 21Sample standard deviation ($\sigma$) = 9Level of confidence = 99%To find the confidence interval, first we need to find the value of z α/2 as follows:  z α/2 = 2.576 (using z-table for 99% confidence level)Now, substituting the given values in the formula, we have:  $\overline{X}±\frac{z_{\frac{\alpha}{2}}\sigma}{\sqrt{n}}$= 21 ± $\frac{2.576 \times 9}{\sqrt{36}}$= 21 ± 5.184= (21 - 5.184, 21 + 5.184)So, the 99% confidence interval for the population mean is (15.816, 26.184).

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The elementary matrix that replaces row 2 of a 5 × n matrix with row 2 plus 1.3 times row 4 is E = [1 0 0 0 0;0 1 0 1.3 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1] and the inverse of E is E^-1= [1 0 0 0 -0.52;0 0.77 0 -1 0.39;0 -0.62 1 0 0;0 -0.23 0 1 -0.3;0 -0.31 0 0 0.77].

The Elementary Operations of Type Two are considered. The row operation that replaces row s with row s plus a non-zero scalar, λ, times row t is an elementary matrix of the second type. It is given that E kk = 1 for all k,E st = λ, and E ij = 0 for all other entries i, j.The given matrix is of the form 5 × n and the elementary matrix of the second type that replaces row 2 of the matrix with row 2 plus 1.3 times row 4 is to be determined. Here, row 2 of the matrix is replaced with row 2 + 1.3 * row 4.

Hence, λ = 1.3.The elementary matrix of the second type that replaces row 2 of a 5 × n matrix with row 2 plus 1.3 times row 4 is as follows:E= [1 0 0 0 0;0 1 0 1.3 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1]The inverse of E has to be determined. It can be found by performing the row operations on E to get the identity matrix. Let I be the identity matrix. Then [E|I] = [I|E^-1].The following row operations are performed on [E|I].1. Add -1.3 times row 2 to row 4.2. Swap row 2 and row 3.3. Swap row 3 and row 4.4. Swap row 2 and row 3.5. Add -1 times row 5 to row 4.6. Add -1 times row 4 to row 3.7. Add -1 times row 3 to row 2.8.

Add -1 times row 2 to row 1.The final matrix is obtained as follows: [I|E^-1] = [E|I] = [1 0 0 0 0.52;0 0.77 0 -1 0.39;0 -0.62 1 0 0;0 -0.23 0 1 -0.3;0 -0.31 0 0 0.77]Hence, the inverse of E is as follows:E^-1= [1 0 0 0 -0.52;0 0.77 0 -1 0.39;0 -0.62 1 0 0;0 -0.23 0 1 -0.3;0 -0.31 0 0 0.77]Therefore, the elementary matrix that replaces row 2 of a 5 × n matrix with row 2 plus 1.3 times row 4 is E = [1 0 0 0 0;0 1 0 1.3 0;0 0 1 0 0;0 0 0 1 0;0 0 0 0 1] and the inverse of E is E^-1= [1 0 0 0 -0.52;0 0.77 0 -1 0.39;0 -0.62 1 0 0;0 -0.23 0 1 -0.3;0 -0.31 0 0 0.77].

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The p-value of the test is given as follows:

0.3974.

Before obtaining the p-value of the test, we must obtain the test statistic, which is given by the equation presented as follows:

[tex]z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]

In which:

The parameters for this problem are given as follows:

[tex]p = 0.2, n = 90, \overline{p} = \frac{19}{90} = 0.2111[/tex]

Hence the test statistic is given as follows:

[tex]z = \frac{0.2111 - 0.2}{\sqrt{\frac{0.2(0.8)}{90}}}[/tex]

z = 0.26.

Looking at the z-table, for a right-tailed test with z = 0.26, the p-value is given as follows:

1 - 0.6026 = 0.3974.

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