The table below contains information about the distribution of the variables X and Y. Each variable has two levels (categories). The contents of the cells in the table represent the observed frequencies.
Variable X Nivel 1 Nivel 2 Variable y Nivel 1 12 7 19 Nivel 2 7 21 28 19 28 47. Can we say that the variables X and Y are independent?
Yes
No
What did you use to evaluate the independence of the variables? Select the best alternative.
a) Fisher's exact test
b) Binomial distribution
c) Try Chi-Squared

The value of the constant 1/3 can be determined by integrating the joint probability density function (pdf) over its entire domain. The integral should be equal to 1 since the function represents a probability distribution.

Given that fx,y(2,y) = 2x(2− x−y), 0 < x < 1, 0 < y < 1 - x.

To determine the value of the constant 1/3, we integrate the joint probability density function (pdf) over its entire domain:∫∫2x(2−x−y)dydx, 0

. The integral should be equal to 1 since the function represents a probability distribution

Hence, . The integral should be equal to 1 since the function represents a probability distribution

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Let's assume Valley College purchased x boxes of gel pens and y boxes of mechanical pencils.

According to the given information, the cost of each box of gel pens is $17.49 and the cost of each box of mechanical pencils is $16.49.

The total number of boxes purchased is 120, so we can write the equation x + y = 120.

The total cost of the purchase is $2010.80, so we can write the equation 17.49x + 16.49y = 2010.80.

To find the values of x and y, we can solve this system of equations.

Using a method such as substitution or elimination, we can find that x = 48 and y = 72.

Therefore, Valley College purchased 48 boxes of gel pens and 72 boxes of mechanical pencils.

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For Face, the diameter of the workpiece is ØA, so RPM = (900 x 4) / 1.00 = 3600 RPMAnd, for Turn, the diameter of the workpiece is ØB, so RPM = (900 x 4) / 0.80 = 4500 RPM.

The target RPM for each of the following operations are:Face: RPM = (CS x 4) / DWhere,RPM = revolutions per minuteCS = cutting speedD = diameter of the workpiece.The cutting speed is the speed at which the metal is removed by the cutting tool from the workpiece. It is expressed in meters per minute or feet per minute. For aluminum, the cutting speed is 900 SFPM (feet per minute).

Now, let's calculate the target RPM for each of the following operations:Face:RPM = (CS x 4) / DWhere,RPM = revolutions per minuteCS = cutting speedD = diameter of the workpieceFor Face, the diameter of the workpiece is ØA, soRPM = (900 x 4) / 1.00 = 3600 RPMTurn:RPM = (CS x 4) / DWhere,RPM = revolutions per minuteCS = cutting speedD = diameter of the workpieceFor Turn, the diameter of the workpiece is ØB, soRPM = (900 x 4) / 0.80 = 4500 RPM

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To calculate the sample standard deviation, we need to find the variance first. The variance is the average of the squared differences from the mean. Then, we take the square root of the variance to get the standard deviation.

Given data: 153, 156, 168, 138

Step 1: Calculate the mean (average):

Mean = (153 + 156 + 168 + 138) / 4 = 154.75

Step 2: Calculate the variance:

Variance = [(153 - 154.75)^2 + (156 - 154.75)^2 + (168 - 154.75)^2 + (138 - 154.75)^2] / 4

        = (2.5625 + 1.5625 + 157.5625 + 268.5625) / 4

        = 107.5625

Step 3: Calculate the sample standard deviation:

Sample Standard Deviation = √(Variance)

                       = √(107.5625)

                       ≈ 10.37 (rounded to one decimal place)

Step 3: Find the critical value that should be used in constructing the confidence interval.

Since the sample size is small (n = 4) and the population is assumed to be approximately normal, we can use the t-distribution to find the critical value for a 99% confidence level.

Degrees of freedom (df) = n - 1 = 4 - 1 = 3

Using a t-distribution table or a statistical software, the critical value for a 99% confidence level with 3 degrees of freedom is approximately 4.541 (rounded to three decimal places).

Step 4: Construct the 99% confidence interval.

The formula for the confidence interval is:

Confidence Interval = Mean ± (Critical Value) * (Standard Deviation / √(Sample Size))

Mean = 154.75

Critical Value = 4.541

Standard Deviation = 10.37

Sample Size = 4

Confidence Interval = 154.75 ± (4.541) * (10.37 / √(4))

                  = 154.75 ± (4.541) * (10.37 / 2)

                  = 154.75 ± (4.541) * 5.185

                  = 154.75 ± 23.566

                  ≈ (131.184, 178.316) (rounded to one decimal place)

Therefore, the 99% confidence interval for the mean noise level at such locations is approximately (131.2, 178.3) decibels.

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The global maximum value f(max) of the function ƒ(x, y) = 8x - 5y is 49, and the global minimum value f(min) is -79.

We have,

To find the global extreme values of the function ƒ(x, y) = 8x - 5y subject to the given constraints, consider the critical points and the boundary points of the feasible region.

The feasible region is defined by the inequalities:

y ≥ x − 5

y ≥ −x − 5

y ≤ 3

First, let's find the critical points by finding the gradient of the function and setting it equal to zero.

Gradient of ƒ(x, y) = ∇ƒ(x, y) = (∂ƒ/∂x, ∂ƒ/∂y) = (8, -5)

Setting both partial derivatives equal to zero:

8 = 0 (no solution)

-5 = 0 (no solution)

Since there are no solutions for the gradient, there are no critical points in the interior of the feasible region.

Next, consider the boundary points of the feasible region.

y = x - 5 and y = -x - 5

By setting these two equations equal to each other,

x - 5 = -x - 5

2x = 0

x = 0

Substitute x = 0 into either equation to find the y-coordinate:

y = 0 - 5 = -5

So the point (0, -5) is the intersection of the lines y = x - 5 and y = -x - 5.

y = x - 5 and y = 3

By setting these two equations equal to each other,

x - 5 = 3

x = 8

Substitute x = 8 into either equation to find the y-coordinate:

y = 8 - 5 = 3

So point (8, 3) is the intersection of the lines y = x - 5 and y = 3.

y = -x - 5 and y = 3

By setting these two equations equal to each other,

-x - 5 = 3

x = -8

Substitute x = -8 into either equation to find the y-coordinate:

y = -(-8) - 5 = 3

So the point (-8, 3) is the intersection of the lines y = -x - 5 and y = 3.

Now, evaluate the function ƒ(x, y) = 8x - 5y at these boundary points and compare the values to find the global extreme values.

At (0, -5):

ƒ(0, -5) = 8(0) - 5(-5) = 0 + 25 = 25

At (8, 3):

ƒ(8, 3) = 8(8) - 5(3) = 64 - 15 = 49

At (-8, 3):

ƒ(-8, 3) = 8(-8) - 5(3) = -64 - 15 = -79

To find the global extreme values, we compare these values:

f(max) = 49 (maximum value of the function occurs at point (8, 3))

f(min) = -79 (minimum value of the function occurs at point (-8, 3))

Therefore,

The global maximum value f(max) of the function ƒ(x, y) = 8x - 5y is 49, and the global minimum value f(min) is -79.

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The subset W = {〈x, y, z〉| x = 0 or z = 0} of R3 is not a subspace of R3.

To be a subspace, a subset must satisfy three conditions: it must contain the zero vector, it must be closed under addition, and it must be closed under scalar multiplication.

In the case of W, the zero vector 〈0, 0, 0〉 is not in W because it does not satisfy the conditions x = 0 or z = 0. Therefore, W fails the first condition and cannot be a subspace.

Additionally, W is not closed under addition or scalar multiplication. If we take two vectors 〈0, y1, 0〉 and 〈0, y2, 0〉 from W, their sum 〈0, y1+y2, 0〉 is not in W because the x-component is not zero. Similarly, scalar multiplication of a vector 〈0, y, 0〉 in W by a non-zero scalar would result in a vector with a non-zero x-component.

Hence, W does not satisfy the necessary conditions to be considered a subspace of R3.

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Rounding to the nearest hundredth, the mean of the dataset is approximately 239.86.

To find the mean of the dataset, you can use the formula:

mean = Σx / N

where Σx is the sum of all the values and N is the number of data points.

In this case, Σx = 1679 and N = 7. Plugging these values into the formula:

mean = 1679 / 7 = 239.857142857

Rounding to the nearest hundredth, the mean of the dataset is approximately 239.86.

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The exact value of sin⁻¹(√2/2) is π/4. The exact value of cos⁻¹(√2/2) is π/4. The exact value of tan⁻¹(-1) is -π/4.

The expression sin⁻¹(√2/2) represents the angle whose sine is √2/2. This angle corresponds to the first quadrant in the unit circle, where both the sine and cosine values are positive. In the first quadrant, the angle π/4 has a sine of √2/2. Therefore, sin⁻¹(√2/2) = π/4.

The expression cos⁻¹(√2/2) represents the angle whose cosine is √2/2. Again, in the first quadrant, the angle π/4 has a cosine of √2/2. Therefore, cos⁻¹(√2/2) = π/4.

The expression tan⁻¹(-1) represents the angle whose tangent is -1. This angle can be found in the fourth quadrant, where the tangent is negative. The angle -π/4 satisfies tan(-π/4) = -1. Therefore, tan⁻¹(-1) = -π/4.

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The critical path for the project is A, C, E, H. Option A is correct.

The critical path for the given project is A, C, E, H. A critical path is a project management technique that identifies the most critical tasks that must be completed on time for the project to finish on schedule. It shows the sequence of activities that, if delayed, would delay the entire project completion time. The expected time (TE) formula is:TE = (a + 4m + b)/6Where,a is the optimistic timeb is the pessimistic timem is the most likely time Using the given data in the table, the expected time for each task and the critical path can be calculated. Activity Activity Predecessor Most Likely Time Pessimistic Time A - 5 9 - B - 15 22 - C A 7 9 - D B 18 24 - E C, D 6 9 - F C, D 12 18 - G E, F 19 20 - H E, F 4 5 -Expected Time:A: TE = (5 + 4(9) + 9)/6 = 7

B: TE = (15 + 4(22) + 22)/6 = 20

C: TE = (7 + 4(9) + 9)/6 = 8

D: TE = (18 + 4(24) + 24)/6 = 22

E: TE = (6 + 4(9) + 9)/6 = 8

F: TE = (12 + 4(18) + 18)/6 = 16

G: TE = (19 + 4(20) + 20)/6 = 21

H: TE = (4 + 4(5) + 5)/6 = 5

Critical path: A - C - E - H = 7 + 8 + 8 + 5 = 28Hence, the critical path for the project is A, C, E, H. Option A is correct.

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The purchase value of transactions that separates the lowest-spending 10% of customers from the remaining customers is R266.80.

The purchase value of transactions, x, at a national clothing store such as Woolworth is normally distributed with a mean of R350 and a standard deviation of R65.

We need to determine the purchase value of transactions that separates the lowest-spending 10% of customers from the remaining customers. It is required to find the z-score for the given probability of 0.10.

The z-score represents the number of standard deviations a given value, x, is from the mean, μ. The formula for z-score is given as

z = (x - μ) / σ

Where,μ = 350

σ = 65

z = z-score

To find the z-score for a probability of 0.10, we use the standard normal distribution table.

From the standard normal distribution table, we find the z-score for a probability of 0.10 as -1.28 (rounded to two decimal places).

-1.28 = (x - 350) / 65

Multiplying both sides by 65, we get

-83.2 = x - 350

Adding 350 on both sides, we get

266.8 = x

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To solve the linear recurrence xₖ₊₃ = -2xₖ₊₂ + xₖ₊₁ + 2xₖ, we can use diagonalization.

First, let's construct the vector vₖ = [xₖ, xₖ₊₁, xₖ₊₂] and the matrices A, P, P⁻¹, and D. We have vₖ₊₁ = Avₖ, and A = PDP⁻¹, where D is a diagonal matrix.

To find the matrices, we can start by setting up the characteristic equation for the recurrence relation: λ³ + 2λ² - λ - 2 = 0. Solving this equation, we find the eigenvalues λ₁ = 1, λ₂ = -2, and λ₃ = -1.

Next, we find the corresponding eigenvectors by solving the equations (A - λI)v = 0, where I is the identity matrix. For each eigenvalue, we obtain a set of eigenvectors. Let's denote these eigenvectors as v₁, v₂, and v₃.

Now, we construct the matrix P using the eigenvectors as its columns. P = [v₁, v₂, v₃]. The matrix P⁻¹ is the inverse of P.

The diagonal matrix D is formed by placing the eigenvalues on its diagonal. D = diag(1, -2, -1).

To solve the recurrence relation, we can express vₖ as a linear combination of the eigenvectors: vₖ = c₁v₁ + c₂v₂ + c₃v₃, where c₁, c₂, and c₃ are constants.

Finally, we can find the values of c₁, c₂, and c₃ by using the initial conditions: v₀ = [x₀, x₁, x₂] and expressing it in terms of the eigenvectors. Once we have c₁, c₂, and c₃, we can compute Aᵏ = PDᵏP⁻¹ to find the values of xₖ for any k.

Note that the explicit solution for xₖ involves raising D to the power of k, which can be done by raising each diagonal entry to the power of k.

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The probability that the randomly selected ball is a basketball defective is 1.2% and the probability that the randomly selected ball is a good soccer ball is 57%.

a. Probability that the randomly selected ball is a defective basketball:We know that the probability that a basketball is defective is 0.03.

The probability that a randomly selected ball is a basketball is:1 - probability that a ball is a soccer ball = 1 - 0.6 = 0.4

So, the probability that the randomly selected ball is a defective basketball:0.03 x 0.4 = 0.012 or 1.2%.

b. Probability that the randomly selected ball is a good soccer ball:The probability that a soccer ball is good is:1 - probability that a soccer ball is defective = 1 - 0.05 = 0.95

Therefore, the probability that the randomly selected ball is a good soccer ball:0.95 x 0.6 = 0.57 or 57%.

Hence, the probability that the randomly selected ball is a basketball defective is 1.2% and the probability that the randomly selected ball is a good soccer ball is 57%.

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a) The limit as x approaches negative infinity and as x approaches infinity is equal to 1. b) The horizontal asymptote is y = 1. The given function f(x) is as follows; f(x) = (x - 2)(x + 3)Let us first find the x-intercept of the function above;                x-intercept.

When the value of f(x) is zero, that is, f(x) = 0;  (x - 2)(x + 3) = 0, which implies; x - 2 = 0  or x + 3 = 0  => x = 2  or x = -3Therefore, the x-intercepts are (2, 0) and (-3, 0). y-intercept. When x = 0, the value of the function is given by f(0) = (0 - 2)(0 + 3) = -6Therefore, the y-intercept is (0, -6).Vertical asymptotes the vertical asymptotes occur at the values of x where the denominator of the function is equal to zero. Therefore, there is no vertical asymptote as there is no denominator in the given function above.Horizontal asymptotesThe degree of the numerator and denominator is equal; both being 2. Therefore, we can use the following equation to find horizontal asymptotes;y = a_n / b_n = 1/1 = 1.

Domain and Range the domain of a function is all the values of x for which the function is defined; that is, there are no division by zero or square roots of negative numbers in the function. Therefore, the domain of f(x) is all real numbers. The range of a function is all the values that y can take. Since the minimum value of the function is -6 and there is no maximum value of y, the range of f(x) is {y | y ∈ ℝ, y ≥ -6}.c) Sketch of the function the graph of the function f(x) = (x - 2)(x + 3) is given below; d) Evaluation of one-sided limits at the asymptotes. Since there is no vertical asymptote, there is no need to evaluate one-sided limits. The horizontal asymptote is y = 1, which is an equation of a horizontal line.

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The Lemma 1.3 holds true.

The given bases for R are B = < (-1 1), (2 2) >, D = < (0 4), (1 3) >.

a. Change of basis matrix can be found using the following formula: Rep_B(id) = [B]_P^B =[P_B^R]^-1

=[(b1)_R (b2)_R]^-1, where (b1)_R and (b2)_R are the column vectors of the standard basis matrix represented in the basis B.

Hence, Rep_B(id) = [(2 - 1), (2 - 2)], that is, Rep_B(id) = [1, 0; 0, 0].

b. We can find Rep(01) using the matrix representation of the linear transformation 01.

The matrix representation of 01 is given by [01]_R=[01(b1)_R 01(b2)_R] = [b1)_R b2)_R].

Thus, Rep(01) = [2, 2; 2, 2].

c. The representation of 01 in the basis D can be found by the formula, Rep_D(01) = [D]_R^D * Rep_R(01).

Here, [D]_R^D is the change of basis matrix from R to D.

The change of basis matrix from R to D is given by [D]_R^D = [P_D^R]^-1

=[(d1)_R (d2)_R]^-1.

Here, (d1)_R and (d2)_R are the column vectors of the standard basis matrix represented in the basis D.

Hence, [D]_R^D = [(-3, 1), (2, -1)].

Thus, Rep_D(01) = [(-3, 1), (2, -1)] * [2, 2; 2, 2]

= [(-4, 2), (0, 0)].

d. The proof of Lemma 1.3 is given below:

Rep_Bp(id) * Rep(01)

= [B]_P^Bp * [Bp]_R^B * [R]_R^P * [01]_R * [P]_P^R * [B]_P^B

= [B]_P^Bp * [Bp]_R^B * [01]_P * [B]_P^B

= [B]_P^Bp * [Bp]_B^R * [01]_P

= [B]_P^R * [01]_R

= Rep(01).

Hence, the Lemma 1.3 holds true.

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Step-by-step explanation:

Take 1/2 of the 's' coefficient.....square it and add it to both sides of the equation

1/2 * 10 = 5

5^2 = 25   added to both sides of the equation

The lines intersect at the point (8, 14, -13) or (4, 12, -5).

To find the intersection point of two lines, we need to set their respective parametric equations equal to each other and solve for the values of the parameters.

The given lines are:

L: r = (4, -2, -1) + t(1, 4, -3)

F: r = (-8, 20, 15) + u(-3, 2, 5)

Setting the two equations equal to each other, we have:

(4, -2, -1) + t(1, 4, -3) = (-8, 20, 15) + u(-3, 2, 5)

By comparing the corresponding components, we can write a system of equations:

4 + t = -8 - 3u

-2 + 4t = 20 + 2u

-1 - 3t = 15 + 5u

Simplifying each equation:

t + 3u = -12

4t - 2u = 22

-3t - 5u = 16

We can solve this system of equations to find the values of t and u. Once we have the values, we can substitute them back into the equation for either line to find the corresponding point of intersection.

By solving the system, we find t = 4 and u = -4. Substituting these values into either line equation, we have:

L: r = (4, -2, -1) + 4(1, 4, -3) = (8, 14, -13)

F: r = (-8, 20, 15) + (-4)(-3, 2, 5) = (-8, 20, 15) + (12, -8, -20) = (4, 12, -5)

Therefore, the lines intersect at the point (8, 14, -13) or (4, 12, -5).

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If the equation Ax = b has infinitely many solutions, it implies that the matrix equation Ax = c, where c is a vector, cannot have a unique solution.

Let's assume that the equation Ax = b has infinitely many solutions. This means that there exist multiple vectors x₁, x₂, ..., xn that satisfy Ax = b.

Now, suppose there exists a vector c such that Ax = c has a unique solution.

This would mean that there is only one vector x that satisfies Ax = c. However, since Ax = b has infinitely many solutions, there must be at least two distinct vectors x₁ and x₂ that satisfy Ax = b.

If we substitute x₁ and x₂ into Ax = c, we would obtain two different solutions, c₁ and c₂, respectively. But this contradicts the assumption that Ax = c has a unique solution. Therefore, if Ax = b has infinitely many solutions, it follows that there does not exist a vector c such that Ax = c has a unique solution.

In conclusion, the existence of infinitely many solutions for the equation Ax = b implies the impossibility of finding a vector c that leads to a unique solution in the matrix equation Ax = c.

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The point (20, 195°) can be represented in polar coordinates using two different polar ordered pairs: (20, 195°) and (-20, 15°).

To describe the point (20, 195°) in polar coordinates, we can represent it using two different polar ordered pairs. In polar coordinates, a point is described by its radial distance from the origin (r) and the angle (θ) it makes with the positive x-axis.

(20, 195°):

The first polar ordered pair for the point (20, 195°) is (20, 195°). This representation directly matches the given coordinates, where the radial distance is 20 units and the angle is 195°.

(-20, 15°):

The second polar ordered pair can be obtained by adding or subtracting 180° from the given angle (195°) and changing the sign of the radial distance. In this case, we subtract 180° from 195° and obtain 15°. The negative sign is applied to the radial distance to indicate the opposite direction from the positive x-axis.

Therefore, the second polar ordered pair for the point (20, 195°) is (-20, 15°).

In summary, the point (20, 195°) can be represented in polar coordinates using two different polar ordered pairs: (20, 195°) and (-20, 15°). The first pair directly matches the given coordinates, while the second pair is obtained by subtracting 180° from the given angle and changing the sign of the radial distance.

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[tex](x+3)^{2} =x^2+6x+9[/tex]. Thus,  the closure property of multiplication is satisfied.

The closure property of multiplication states that when any two elements of a set are multiplied, their product will also be present in that set. The closure property formula for multiplication for a given set S is: ∀ a, b ∈ S ⇒ a × b ∈ S.

[tex](x+3)^2[/tex] is same as product of (x+3) with itself that is (x+3) multiply by (x+3)

(x+3)(x+3)=x(x+3)+3(x+3)

[tex](x+3)(x+3)=x^2+3x+3x+9\\=x^2+6x+9[/tex]

[tex](x+3)(x+3)=x^2+6x+9[/tex]

[tex](x+3)^{2} =x^2+6x+9[/tex]

Thus, any value of x which satisfies [tex]x^2+6x+9[/tex] will also satisfy [tex](x+3)^2[/tex].

Hence, the closure property of multiplication is satisfied.

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(a) Let's solve part (a) step by step.

i) To find the value of k, we need to use the fact that the sum of probabilities in a probability distribution must equal 1. Therefore, we can set up the equation:

5k + 4k + 3k + 2k + k = 1

Combining like terms, we have:

15k = 1

Dividing both sides by 15, we get:

k = 1/15

So the value of k is 1/15.

ii) To find E(X) (the expected value or mean) and Var(X) (the variance), we can use the formulas:

E(X) = Σ(x * P(X = x))

Var(X) = Σ((x - E(X))^2 * P(X = x))

Using the probability distribution given, we can calculate E(X) and Var(X) as follows:

E(X) = (0 * 5/15) + (1 * 4/15) + (2 * 3/15) + (3 * 2/15) + (4 * 1/15)

= 0 + 4/15 + 6/15 + 6/15 + 4/15

= 20/15

= 4/3

Var(X) = (0 - 4/3)^2 * 5/15 + (1 - 4/3)^2 * 4/15 + (2 - 4/3)^2 * 3/15 + (3 - 4/3)^2 * 2/15 + (4 - 4/3)^2 * 1/15

= (0 - 4/3)^2 * 5/15 + (1/3)^2 * 4/15 + (2/3)^2 * 3/15 + (5/3)^2 * 2/15 + (4/3)^2 * 1/15

= (4/3)^2 * (5/15 + 4/15 + 2/15 + 1/15) + (1/9) * (4/15) + (4/9) * (3/15) + (25/9) * (2/15) + (16/9) * (1/15)

= (16/9) * (12/15) + (4/135) + (12/135) + (50/135) + (16/135)

= (16/9) * (16/15)

= 256/135

So, E(X) = 4/3 and Var(X) = 256/135.

To illustrate the probability distribution of X in a graph, we can plot the values of X on the x-axis and the corresponding probabilities on the y-axis.

(b) i) To find the probability that at least 35 faulty chips are produced in one day, we can use the normal approximation. Since the sample size is large (3000), we can assume the distribution of the number of faulty chips follows a normal distribution.

The mean (μ) of the distribution is given by:

μ = (sample size) * (probability of being faulty) = 3000 * 0.008 = 24

The standard deviation (σ) is calculated using the formula:

σ = sqrt((sample size) * (probability of being faulty) * (1 - probability of being faulty))

= sqrt(3000 * 0.008 * (1 - 0.008))

≈ 5.29

To find the probability of at least 35 faulty chips, we calculate the z-score for 35 using the formula:

z = (x - μ) / σ

z = (35 - 24) / 5.29 ≈ 2.08

Using a standard normal distribution table or calculator, we can find the probability associated with z = 2.08, which represents the probability of at least 35 faulty chips.

ii) To find the expected profit made by the factory per day, we need to consider the cost of manufacturing and the revenue from selling non-faulty chips.

The expected profit per chip is given by:

Profit per chip = Revenue per chip - Cost per chip

Revenue per chip = Selling price per chip = RM 2.50

Cost per chip = Manufacturing cost per chip = RM 0.45

The probability of a chip being non-faulty is 1 - probability of being faulty = 1 - 0.008 = 0.992.

Expected profit per chip = (Revenue per chip) * (Probability of non-faulty chip) - (Cost per chip) * (Probability of non-faulty chip)

= RM 2.50 * 0.992 - RM 0.45 * 0.992

= RM 2.48

Since the factory manufactures 3000 chips per day, the expected profit made by the factory per day is:

Expected profit per day = Expected profit per chip * Number of chips per day

= RM 2.48 * 3000

= RM 7440

Therefore, the expected profit made by the factory per day is RM 7440.

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The probability that all Kathy's uniforms have odd numbers would be = 1/2.

To calculate the probability of having only odd numbers the formula that should be used would be given below as follows:

Probability = possible outcome/sample space

Where possible outcome = all odd numbers between 0-99 = 50

The sample space = 100

The probability = 50/100 = 1/2

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The Area of Hexagon is 384√3 unit².

We have,

Side of Hexagon = 16 unit

We know the Formula of area of Hexagon

= 3√3/2 (a)²

where is the length of side of Hexagon                                                                

Now, substituting the value of side length as

= 3√3/2 (16)²

= 16 x 8 x 3 x √3

= 384√3 unit²

Thus, the Area of Hexagon is 384√3 unit².

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The total surface area of the apparatus is approximately 16454.87 mm².

To calculate the total surface area of the apparatus, we need to find the surface area of the two metal cubes and the surface area of the cylinder.

Each metal cube has six faces, and since they are identical, we only need to calculate the surface area of one cube. The surface area of a cube can be found by multiplying the length of one edge by itself and then multiplying by 6. In this case, the length of each edge is 30 mm, so the surface area of one cube is:

Surface area of cube = 6 * (30 mm * 30 mm) = 6 * 900 mm² = 5400 mm²

Now, let's calculate the surface area of the cylinder. The surface area of a cylinder can be found by adding the areas of the two circular bases and the lateral surface area. The formula for the lateral surface area of a cylinder is given by 2 * π * radius * height.

In this case, the radius of the cylinder is equal to the length of one edge of the cube, which is 30 mm, and the height of the cylinder is also 30 mm (since the cubes are fixed to the cylinder). Therefore, the lateral surface area of the cylinder is:

Lateral surface area of cylinder = 2 * π * 30 mm * 30 mm = 1800π mm²

The total surface area of the apparatus is the sum of the surface area of the two cubes and the surface area of the cylinder:

Total surface area = 2 * surface area of cube + surface area of cylinder

= 2 * 5400 mm² + 1800π mm²

≈ 10800 mm² + 5654.87 mm²

≈ 16454.87 mm²

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A firm's variable costs are costs that increase as quantity produced increases. These costs are often illustrated by the increasingly steeper slope of the total cost curve.

As production increases, variable costs such as raw materials, labor, and energy expenses increase in proportion to the level of output. This is due to factors like the need for additional resources or increased utilization of existing resources.

On the other hand, a firm's fixed costs are costs that are incurred even if there is no output. These costs do not change in the short run as production increases. Examples of fixed costs include rent, depreciation of equipment, and insurance. Regardless of the level of production, fixed costs remain constant, representing the expenses that the firm must cover to maintain its operations and infrastructure.

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The equivalent expression for 1.5 cubed over 1.3 raised to the fourth power all raised to the power of negative six is 1.5 to the eighteenth power over 1.3 to the twenty-fourth power.

Here are the steps to simplify the expression:

Apply the negative power rule: (1.5 cubed over 1.3 raised to the fourth power) raised to the power of negative six is equal to (1.3 raised to the fourth power over 1.5 cubed) raised to the power of six.

Apply the power of a quotient rule: (1.3 raised to the fourth power over 1.5 cubed) raised to the power of six is equal to (1.3 raised to the fourth power)⁶ / (1.5 cubed)⁶.

Apply the power of a power rule: (1.3 raised to the fourth power)⁶ is equal to 1.3(⁴*⁶) = 1.3²⁴.

Apply the power of a power rule: (1.5 cubed)⁶ is equal to 1.5(³*⁶) = 1.5¹⁸.

Therefore, the equivalent expression is 1.5¹⁸ / 1.3²⁴.

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For f(x) = x² - x, we know that it is a parabolic function. When it's written as x(x-1), it tells us that it's a parabolic function that intersects the x-axis at 0 and 1.

That is because, for a quadratic function f(x) = ax² + bx + c, the roots can be found using the quadratic formula and the discriminant D, which is b² - 4ac > 0, allowing the function to cross the x-axis at two different points.

For g(x) = log(2x + 1), the expression 2x + 1 must be positive for the function to be defined. That means that x has to be greater than -1/2.

The graph of this function is always increasing, meaning it does not intersect the x-axis.

Because it is continuous and increasing on the interval (-1/2, ∞), the function is always positive on this interval.

Therefore, the interval during which both functions are positive is (−1/2, ∞).

Therefore, the correct answer is option (C) (1, [infinity]).

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The correct answer is:-The data are randomly assigned to one of ten folds.

There are 10 iterations. For each iteration, there are 90 observations in the training set and 10 in the validation set.

In 10-Fold Cross Validation, the data is divided into 10 equally sized folds. Each iteration of the cross-validation process involves using 9 folds for training and 1 fold for validation. The process is repeated 10 times, with each fold serving as the validation set once. This ensures that every observation in the dataset is used for both training and validation.

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a) The probability that a student is a female or spends less than 10 hours studying is 0.683.

b) The probability that a student is male and spends less than 5 hours studying is 0.183.

c) The probability that a student spends more than 10 hours studying given that the student is a male is 0.255.

a) The probability that a student is female or spends less than 10 hours studying can be calculated using the formula:

P(Female or <10 hours) = P(Female) + P(<10 hours) - P(Female and <10 hours)

We have the following probabilities:P(Female) = 80/150 = 0.53P(<10 hours) = 44/150 = 0.293

P(Female and <10 hours) = 21/150 = 0.14

Substituting the values in the formula:P(Female or <10 hours) = 0.53 + 0.293 - 0.14 = 0.683

So, the probability that a student is a female or spends less than 10 hours studying is 0.683.

b) The probability that a student is male and spends less than 5 hours studying can be calculated using the formula:P(Male and <5 hours) = P(Male) × P(<5 hours|Male)

We have the following probabilities:

P(Male) = 70/150 = 0.47P(<5 hours|Male) = 27/70 = 0.39

Substituting the values in the formula:P(Male and <5 hours) = 0.47 × 0.39 = 0.183

So, the probability that a student is male and spends less than 5 hours studying is 0.183.

c) The probability that a student spends more than 10 hours studying given that the student is male can be calculated using the formula:

P(More than 10 hours|Male) = P(More than 10 hours and Male) / P(Male)

We have the following probabilities:P(More than 10 hours and Male) = 18/150 = 0.12P(Male) = 70/150 = 0.47

Substituting the values in the formula:P(More than 10 hours|Male) = 0.12 / 0.47 ≈ 0.255

So, the probability that a student spends more than 10 hours studying given that the student is a male is approximately 0.255.

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Answer:

B. 2x - 8.

Step-by-step explanation:

2(1) - 8 = -6

2(2) - 9 = -4

2(3) - 8 = -2

2(4) - 8 = 0

(a) The probability of the system being down in the next hour, given that it is initially running, is 0.30. (b) The steady-state probabilities of the system being in the running state and the down state are approximately 0.60 and 0.40, respectively.

(a) If the system is initially running, the probability of the system being down in the next hour can be found using the transition probabilities. From the given data, the transition probability from Running to Down is 0.30. Therefore, the probability of the system being down in the next hour is 0.30.

(b) To find the steady-state probabilities of the system being in the running state and in the down state, we need to find the probabilities that remain constant in the long run. This can be done by solving the system of equations:

[tex]P_{running}[/tex] = 0.30 * [tex]P_{running}[/tex]+ 0.70 * [tex]P_{down}[/tex]

[tex]P_{down}[/tex] = 0.20 * [tex]P_{running}[/tex] + 0.80 *[tex]P_{down}[/tex]

Solving these equations, we can find the steady-state probabilities:

[tex]P_{running}[/tex] = 0.30 / (0.30 + 0.20) ≈ 0.60

[tex]P_{down}[/tex] = 0.20 / (0.30 + 0.20) ≈ 0.40

Therefore, the steady-state probability of the system being in the running state is approximately 0.60, and the steady-state probability of the system being in the down state is approximately 0.40.

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