Practice problem for module 2 A random sample of 50 students GPA reveals that the mean GPA is 2.8 years with a standard deviation of 0.45 years. (a) Construct a 95% Confidence Interval for the mean lifetime of all LED TV. (b) If we want to be 90% confident, and we want to control the maximum error of estimation to be 0.2, how many more students should be added into the given sample?
(c) Would you conclude that the mean GPA more than 2.5 at 5% level of significance?

The general solution of the y" + 2y + y= 7 +75sin2x is given as:

y =  (c₁+c₂x)[tex]e^{-x}[/tex] + 7 - 12cos2x - 9sin2x

The Greek terms trigonon (triangle) and metron (measure) are the origin of the word trigonometry. The connections between the lengths and angles of triangles' sides are the subject of this area of mathematics. An equation with one or more trigonometric ratios of unknown angles is said to as trigonometric. The ratios of sine, cosine, tangent, cotangent, secant, and cosecant angles are used to express it.

y" + 2y' + y = 7 + 75sin2x

Auxlliary equation are (m²+2m+1) = 0

CF = (c₁+c₂x)[tex]e^{-x}[/tex]

PI = [tex]\frac{1}{D^2+2D+1} (7+75sin2x)[/tex]

Now,

[tex]\frac{7}{D^2+2D+1} +\frac{75}{D^2+2D+1} (sin2x)[/tex]

7 -3(2D+3)sin2x

7 - 6D.sin2x - 9sin2x

7 - 6 x 2cos2x - 9sin2x

7 - 12cos2x - 9sin2x

PI = 7 - 12cos2x - 9sin2x

Finally,

y = C.F + P.I

y =  (c₁+c₂x)[tex]e^{-x}[/tex] + 7 - 12cos2x - 9sin2x.

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The cars will be 225 km from point A when they meet and it will take 5 hours for the cars to meet..

Let's denote the distance of the faster car from point A as "x" km. Therefore, the distance of the slower car from point B would be "400-x" km.

We can use the formula for distance, which is:

distance = rate × time

For the faster car, the distance it travels can be expressed as:

x = 45t

where t is the time it takes for the cars to meet.

For the slower car, the distance it travels can be expressed as:

400-x = 35t

Now, we can solve for t by setting these two expressions equal to each other:

45t = 400 - 35t

80t = 400

t = 5

We can then substitute t back into either expression to find the distance from point A:

x = 45t = 45(5) = 225 km

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Probit coefficients are typically estimated using:

A. the method of maximum likelihood.

The method of maximum likelihood is used to estimate the probit coefficients. This method aims to find the coefficients that maximize the likelihood of observing the given sample data. It involves an iterative process to identify the most likely parameter values for the model, making it suitable for nonlinear models like the probit model. Maximum likelihood estimation is a widely used method in econometric analysis due to its desirable properties, such as consistency and asymptotic efficiency.

In summary, probit coefficients are estimated using the method of maximum likelihood, which provides the most accurate and efficient estimates for this type of model.

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To orthogonally diagonalize the matrix A, we will follow these steps:
1. Find the eigenvalues of matrix A.
2. Find the eigenvectors corresponding to each eigenvalue.
3. Normalize the eigenvectors to form an orthogonal basis.
4. Construct the orthogonal matrix Q and the diagonal matrix D.

To orthogonally diagonalize the matrix A, we need to find the eigenvalues and eigenvectors of A.

A = [[3, -1], [-1, 3]]

The characteristic polynomial of A is:

det(A - λI) = det([[3-λ, -1], [-1, 3-λ]]) = (3-λ)² - 1 = λ² - 6λ + 8 = (λ-2)(λ-4)

So the eigenvalues are λ₁ = 2 and λ₂ = 4.

To find the eigenvectors, we need to solve the system of equations:

(A - λ₁I)x = 0 and (A - λ₂I)x = 0

For λ₁ = 2, we have:

(A - 2I)x = [[1, -1], [-1, 1]]x = 0

This system has two linearly independent solutions:

v₁ = [1, 1] and v₂ = [-1, 1]

For λ₂ = 4, we have:

(A - 4I)x = [[-1, -1], [-1, -1]]x = 0

This system has one linearly independent solution:

v₃ = [1, -1]

To orthogonalize the eigenvectors, we need to normalize them and put them as columns of an orthogonal matrix Q.

Q = [[1/√2, -1/√2, 0], [1/√2, 1/√2, 0], [0, 0, 1]]

The diagonal matrix D has the eigenvalues on the diagonal:

D = [[2, 0], [0, 4]]

Finally, we can check that QTAQ = D:

QTAQ = [[1/√2, -1/√2, 0], [1/√2, 1/√2, 0], [0, 0, 1]]T[[3, -1], [-1, 3]][[1/√2, -1/√2, 0], [1/√2, 1/√2, 0], [0, 0, 1]] = [[2, 0, 0], [0, 4, 0], [0, 0, 4]] = D

Therefore, matrix A is orthogonally diagonalized by Q and D.

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The experimental probability of drawing a diamond is 0.20

From the question, we have the following parameters that can be used in our computation:

Outcome Frequency

Heart 7

Spade 3

Club 6

Diamond 4

For diamond, we have

P(Diamond) = Diamond/Total

So, we have

P(Diamond) = 4/20

Evaluate

P(Diamond) = 0.20

Hence, the value is 0.20

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To determine the coordinates of point P with respect to the initial frame, we first need to find the transformation matrix from frame B to frame A, and then from frame A to the initial frame.

First, let's find the transformation matrix from frame B to frame A. We know that frame A is rotated 90° about x, so its transformation matrix is:

[A] = [1 0 0; 0 0 -1; 0 1 0]

To find the transformation matrix from frame B to frame A, we need to first undo the translation by subtracting the vector (6,-2,10) from point P:

P' = P - (6,-2,10) = (-11,4,-22)

Next, we need to apply the inverse transformation matrix of frame A to P'. The inverse of [A] is its transpose, so:

P'' = [A]T * P' = [1 0 0; 0 0 1; 0 -1 0] * (-11,4,-22) = (-11,-22,-4)

Finally, we need to find the transformation matrix from frame A to the initial frame. Since the initial frame is fixed and not rotated or translated, its transformation matrix is just the identity matrix:

[I] = [1 0 0; 0 1 0; 0 0 1]

So, the coordinates of point P with respect to the initial frame are:

P''' = [I] * P'' = (1*-11, 0*-22, 0*-4) = (-11,0,0)

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If abcde is a regular pentagon, the smallest rotation about e which maps a to d is 144 degrees.

Using the fact that a regular pentagon has rotational symmetry of order 5. This means that if we rotate the pentagon by 72 degrees around its center, it will appear the same as it did before the rotation.

To map a to d, we need to rotate the pentagon by some angle around point e. Since e is the center of rotation, we need to find an angle that is a multiple of 72 degrees.

To determine the smallest angle of rotation, we need to find the number of 72 degree rotations that take us from a to d.

Starting from a, we can count the number of 72 degree rotations we need to make to reach d. We see that we need to make two 72 degree rotations in a counterclockwise direction.

Therefore, smallest rotation about e that maps a to d is a 144 degree counterclockwise rotation.

We can visualize this by drawing a regular pentagon and marking the point a and d on it. Then, we draw a line segment from a to d and a line segment from e to the midpoint of segment ad.

The angle between these two line segments is 144 degrees.

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The probability that the sum is 7 given that neither die showed a 6 is 4/25 or 0.16.

To find the probability that the sum is 7 given that neither die showed a 6, we need to consider the possible outcomes of rolling two dice without any 6s, and then identify the outcomes where the sum is 7.

Determine the total number of possible outcomes without rolling a 6.
Since there are 5 possible outcomes for each die (1, 2, 3, 4, and 5), there are 5 x 5 = 25 possible outcomes for rolling two dice without any 6s.

Identify the outcomes where the sum is 7.
The possible outcomes that result in a sum of 7 are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). However, since neither die can show a 6, we can only consider the following four outcomes: (1, 6), (2, 5), (3, 4), and (4, 3).

Calculate the probability.
The probability that the sum is 7 given that neither die showed a 6 is the number of favorable outcomes divided by the total number of possible outcomes:
P(sum is 7 | no 6s) = (number of outcomes with sum 7) / (total number of outcomes without 6s) = 4 / 25

So, the probability that the sum is 7 given that neither die showed a 6 is 4/25 or 0.16.

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Answer:

y = x + 9

Step-by-step explanation:

We Know

Sarah threw the javelin 9 inches farther than Kimberly.

Let's y represent the total distance Sarah threw and x is the distance Kimberly threw, we have the equation

y = x + 9

a. The null hypothesis is H0: p1 = p2, meaning that there is no difference in the proportion of B2B and B2C marketers who commonly use the business social media tool. The alternative hypothesis is Ha: p1 ≠ p2, indicating that there is a difference in the proportion of B2B and B2C marketers who commonly use the business social media tool.

To determine the test statistic, we can use the formula:

[tex]Z = (\frac{p1-p2}{\sqrt{p(1-p)} } (\frac{1}{n1} +\frac{1}{n2})[/tex]

where p is the pooled sample proportion, n1 and n2 are the sample sizes for B2B and B2C marketers, respectively.

Plugging in the values, we get:

[tex]p = \frac{(x1 + x2)}{(n1 + n2)} = \frac{525+244}{584+417} = 0.677[/tex]

[tex]Z= \frac{ (0.9 - 0.59)}{\sqrt{0.677(1-0.677)(\frac{1}{584}+\frac{1}{417}) } } } = 12.72[/tex]

The rejection region for a two-tailed test with alpha = 0.05 is ±1.96. Since our test statistic (12.72) is outside this range, we reject the null hypothesis and conclude that there is strong evidence of a difference in the proportion of B2B and B2C marketers who commonly use the business social media tool.

b. The p-value is the probability of observing a test statistic as extreme as the one calculated or more extreme, assuming the null hypothesis is true. We can find it using a Z-table or a calculator. Here, the p-value is essentially zero, indicating very strong evidence against the null hypothesis.

Interpreting the p-value, we can say that if the true proportion of B2B marketers who commonly use the business social media tool were equal to that of B2C marketers, the probability of observing a difference as extreme as the one in our sample or more extreme would be very small. Therefore, we can reject the null hypothesis and conclude that the proportion of B2B marketers who commonly use the business social media tool is significantly different from that of B2C marketers.

c. The null hypothesis is H0: p1 = p2, meaning that there is no difference in the proportion of B2B and B2C marketers who commonly use the video social media tool. The alternative hypothesis is Ha: p1 ≠ p2, indicating that there is a difference in the proportion of B2B and B2C marketers who commonly use the video social media tool.

To determine the test statistic, we can use the same formula as in part (a), but with the sample proportions and sizes for the video social media tool:

[tex]p = \frac{(x1 + x2)}{(n1 + n2)} = \frac{309+241}{584+417} = 0.444[/tex]

[tex]Z= \frac{ (0.53 - 0.58)}{\sqrt{0.444(1-0.444)(\frac{1}{584}+\frac{1}{417}) } } } = -1.33[/tex]

The p-value for a two-tailed test with alpha = 0.05 is approximately 0.18. Since this is greater than the significance level, we fail to reject the null hypothesis and conclude that there is not enough evidence to support a difference in the proportion of B2B and B2C marketers who commonly use the video social media tool.

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Answer:

128° + 52° + 38° + x° = 360°

218° + x° = 360°

x = 142

4(3x + 2)4 12 + K (3x + 2) 6 + K is NOT a family of antiderivative of 2(3x + 2). The correct answer is Option b.

To find the antiderivative of 2(3x + 2), follow these steps:

1. Notice the function is 2(3x + 2).
2. Apply the power rule of integration, which states that ∫x^n dx = (x^(n+1))/(n+1) + C, where n ≠ -1.
3. In this case, n = 1, so the antiderivative is (2(3x + 2)^2)/(2) + C.
4. Simplify to obtain (3x + 2)^2 + C.

Option b doesn't match this result, so it is NOT a family of antiderivatives of 2(3x + 2).

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Answer:

1. D

2.C

3. A

4. B

Step-by-step explanation:

times each of the numbers by however many r in the brackets

3×2=6

3×-1=-3

so the answer to 1 will be (6)

(-3)

The confidence interval is in the range of 40.8% ± E.

The confidence interval for a proportion is typically expressed as:
ˆp ± z*SE
where ˆp is the sample proportion, z* is the critical value from the standard normal distribution for the desired level of confidence (e.g. 1.96 for 95% confidence), and SE is the standard error of the proportion.

Using this formula, if the sample proportion is 40.8% and we want a 95% confidence interval, we would have:
40.8% ± 1.96*√[(40.8%*(1-40.8%))/n]

where n is the sample size.

Without knowing the sample size, we cannot calculate the exact confidence interval. However, we can express the interval as:
(40.8% ± E)%
where E is the margin of error, which is equal to 1.96*√[(40.8%*(1-40.8%))/n]. This means that we are 95% confident that the true proportion falls within the range of 40.8% ± E.

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The distance between A and B is 5 unit.

We have the coordinates as

A(4, 3) and B(4, -2).

Using Distance formula

d= √(x₂ - x₁)² + (y₂ - y₁)²

So, d = √(-2-3)² + (4-4)²

d= √(-5)² + 0²

d = √25

d= 5

Thus, the distance between A and B is 5 unit.

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a. This is a combination because the order in which the students are chosen does not matter, only the group of four students is selected.
b. There are 4845 different ways that 4 students can be chosen from a group of 20.

a. This is a combination because the order of the chosen students does not matter. In a permutation, the order matters, whereas in a combination, it does not.

b. To find the number of different ways 4 students can be chosen from a group of 20, use the combination formula:

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

Where n = the total number of students (20), k = the number of students to be chosen (4), and ! denotes factorial.
The number of different ways 4 students can be chosen from a group of 20 can be calculated using the combination formula:

nCr = n! / r!(n-r)!

Where n is the total number of students (20) and r is the number of students being chosen (4).

So,

20C4 = 20! / 4!(20-4)!
C(20, 4) = 20! / (4!(20-4)!)
C(20, 4) = 20! / (4! * 16!)
C(20, 4) = 2,432,902,008,176,640,000 / (24 * 20,922,789,888,000)
C(20, 4) = 4845

There are 4,845 different ways to choose 4 students from a group of 20.

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There are 181,440 ways to arrange the letters in "competition" with the constraint that "ie" must appear together. This can be answered by the concept of Permutation.

To arrange the letters in the word "competition" with the constraint that "ie" must appear together, first consider "ie" as a single unit.

There are now 10 distinct elements to arrange: C, O, M, P, T, T, I, O, N, and the combined "ie". There are 9! (9 factorial) ways to arrange these elements. However, we need to account for the repetition of the letters O and T, which appear twice each.

To correct for this, divide the total arrangements by the repetitions:

9! / (2! × 2!) = 181,440 ways

So, there are 181,440 ways to arrange the letters in "competition" with the constraint that "ie" must appear together.

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The first derivative of at x with respect to x is simply the transpose of the matrix a.The first derivative of xt ax with respect to x is 2ax, since taking the derivative of the product of two vectors involves multiplying one of the vectors by the derivative of the other vector, and in this case the derivative of x is the identity matrix (since x is a vector and not a matrix).The second derivative of xt ax with respect to x is simply the matrix 2a, since the second derivative involves taking the derivative of the first derivative.1. Given the vector x and the symmetric matrix A.
2. We need to find the first and second derivatives of the following expressions:
  i. A * x
  ii. x^T * A * x

The question involves vectors, symmetric matrices, and derivatives. Let's break it down step-by-step.

i. First derivative of A * x with respect to x:
To find the derivative of A * x with respect to x, we treat A as a constant matrix. The first derivative is simply the matrix A itself.

Answer: The first derivative of A * x with respect to x is A.

ii. First derivative of x^T * A * x with respect to x:
To find the first derivative of this expression, we'll use the following formula for the derivative of a quadratic form:

d/dx (x^T * A * x) = (A + A^T) * x

Since A is a symmetric matrix, A = A^T. Therefore, the formula becomes:

d/dx (x^T * A * x) = 2 * A * x

Answer: The first derivative of x^T * A * x with respect to x is 2 * A * x.

iii. Second derivative of x^T * A * x with respect to x:
The second derivative of x^T * A * x with respect to x is the derivative of the first derivative (2 * A * x) with respect to x. Since A is a constant matrix, the second derivative is zero.

Answer: The second derivative of x^T * A * x with respect to x is 0.

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The marching of figure images with their formulas are:

Figure 1:

P = a + b + c

P = 5 + 5 + 8

Figure 2:

A = lw

A = 5 * 8

Figure 3:

V = lwh

V = 8 * 5 * 5

The formula for the area of a triangle is:

Area = ¹/₂ * base * height

Formula for area of a rectangle is:

Area = length * width

Formula for volume of a cuboid is:

V = length * width * height

1) For the triangle, we are not give the height and so we can only find the perimeter which is:

P = a + b + c

P = 5 + 5 + 8

2) The area of this rectangle is:

A = lw

A = 5 * 8

3) Volume of this cuboid is:

V = lwh

V = 8 * 5 * 5

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Answer:

Step-by-step explanation:

You want q and the probability of 3 successes in 8 trials if the probability of success is 0.45.

The value designated q is the complement of p, the probability of success.

  q = 1 -p

  q = 1 -0.45

  q = 0.55

The probability of 3 successes is ...

  P(3 of 8) = 8C3·p^3·q^(8-3) = 56·0.45^3·0.55^5

  P(3 of 8) ≈ 56°0.091125·0.050328 ≈ 0.256826

The probability of exactly 3 successes in 8 trials is about 0.2568.

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a) To calculate the missing values, we first need to find the 2018 turnover value. That is, 290.62 M Ft.

b) Average Relative Change =  5.18% and Average Absolute Change = 13.54 M Ft.

c) The ratios for 2016 indicate that the firm experienced a 4% increase in turnover compared to the previous year (2015).

a) To calculate the missing values, we first need to find the 2018 turnover value. We know that the 2015 turnover is 100% of the previous year, so 2015 and 2014 turnover values are the same (250 M Ft). Now we can use the ratios given for the subsequent years:

2016 Turnover = 2015 Turnover * (100% + Ratio)
260 M Ft = 250 M Ft * (100% + Ratio)
Ratio = (260 / 250) - 1 = 0.04 or 4%

2017 Turnover = 2016 Turnover * (100% + Ratio)
275 M Ft = 260 M Ft * (100% + Ratio)
Ratio = (275 / 260) - 1 ≈ 0.0577 or 5.77%

2018 Turnover = 2017 Turnover * (100% + Ratio)
2018 Turnover = 275 M Ft * (100% + 0.0577) ≈ 290.62 M Ft

b) Now, we can calculate the average relative and absolute change:
Average Relative Change = (4% + 5.77% + 5.77%)/3 ≈ 5.18%

Absolute Change (2016) = 260 - 250 = 10 M Ft
Absolute Change (2017) = 275 - 260 = 15 M Ft
Absolute Change (2018) = 290.62 - 275 ≈ 15.62 M Ft
Average Absolute Change = (10 + 15 + 15.62) / 3 ≈ 13.54 M Ft

c) The ratios for 2016 indicate that the firm experienced a 4% increase in turnover compared to the previous year (2015). This means the firm was successful in generating more revenue in 2016 as compared to 2015, which could be attributed to various factors such as improved marketing strategies, expansion in the market, or better product offerings.

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a. At the 0.05 level of​ significance, there is evidence of a linear relationship between the plate gap of the​ bag-sealing machine and the tear rating of a bag of​ coffee.

b. A​ 95% confidence interval estimate of the population​ slope, betaβ1 is (0.5590, 0.8606).

a. To test for the linear relationship between plate gap and tear rating, we can use the null and alternative hypotheses:

H0: β1 = 0 (there is no linear relationship)
Ha: β1 ≠ 0 (there is a linear relationship)

We can use the t-test to test this hypothesis. The test statistic is calculated as:

t = b1 / (S1 / [tex]\sqrt(n)[/tex])

where b1 is the sample slope, S1 is the standard error of the slope, and n is the sample size. Substituting the values given in the question, we get:

t = 0.7098 / (0.2146 / sqrt(30)) = 5.05

Using a t-distribution with n-2 = 28 degrees of freedom and a significance level of 0.05, we can find the critical values as ±2.048. Since the calculated t-value of 5.05 is greater than the critical value of 2.048, we reject the null hypothesis and conclude that there is evidence of a linear relationship between plate gap and tear rating at the 0.05 level of significance.

b. To construct a 95% confidence interval estimate of the population slope β1, we can use the formula:

b1 ± tα/2(S1 / [tex]\sqrt(n)[/tex])

where tα/2 is the critical value from the t-distribution with n-2 degrees of freedom and a confidence level of 95%. Substituting the values given in the question, we get:

b1 ± 2.048(0.2146 / [tex]\sqrt(30)[/tex]) = 0.7098 ± 0.1508

Therefore, the 95% confidence interval for the population slope β1 is (0.5590, 0.8606).

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Answer: Answer:

3. For some x-values, the y-value of the exponential function is smaller.

4. For some x-values, the y-value of the exponential function is greater.

5. For any x-value greater than 7, the y-value of the exponential function is greater.

Step-by-step explanation:

We are given the functions,

It is required to find the true statements of the functions.

From the graphs of the functions below, we have that the graphs intersect at the point (6.32,79.878).

To the left of the point, we have that the exponential function have smaller y-values than the parabola.

To the right of the point, we have that the exponential function have greater y-values than the parabola.

Moreover, after x= 7, the y-values of the exponential function are always greater than parabola.

Thus, the correct options are 3, 4 and 5.

Step-by-step explanation:

Answer:

- For any x-value, the y-value of the exponential function is always greater. (False)

- For any x-value, the y-value of the exponential function is always smaller. (False)

- For some x-values, the y-value of the exponential function is smaller. (True)

- For some x-values, the y-value of the exponential function is greater. (True)

- For any x-value greater than 7, the y-value of the exponential function is greater. (Not enough information provided to determine)

- For equal intervals, the y-values of both functions have a common ratio. (False)

Answer:

1,560 miles / (3 days × 8 hours/day) =

1,560 miles / 24 hours = 65 miles per hour

The equation is y = x - 3

and the points (x, y) is (0,-3) , (-1, -4) and (1, -2)

The equation is as follows:

y = x - 3

Substituting 'x = 0' in the given equation, we get

y = 0 - 3

y = -3

Substituting 'x = - 1' in the given equation, we get

y = - 1 - 3

y = -4

Substituting 'x = 1' in the given equation, we get

y = 1 - 3

y = -2

➢ Pair of points of the given equation are shown in the below table.

x             y

0            -3

-1            -4

1             -2

➢ Now draw a graph using the points.

➢ See the attachment graph.

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The answer is not provided in the options given. The closest option is 0.172, but the correct answer is 0.155 (rounded to three decimal places).

To calculate the standard deviation of X, we first need to find the mean or expected value of X. We can do this by integrating the given pdf over the range 0.90 to 1.80:

E(X) = ∫[0.90,1.80] x*f(x) dx

= ∫[0.90,1.80] x*(265/652)*x^3 dx

= (265/652) * ∫[0.90,1.80] x^4 dx

= (265/652) * [x^5/5] from x=0.90 to x=1.80

≈ 1.315

Next, we can calculate the variance of X using the formula:

Var(X) = E(X^2) - [E(X)]^2

To find E(X^2), we integrate the pdf squared over the same range:

E(X^2) = ∫[0.90,1.80] x^2*f(x) dx

= ∫[0.90,1.80] x^2*(265/652)*x^3 dx

= (265/652) * ∫[0.90,1.80] x^5 dx

= (265/652) * [x^6/6] from x=0.90 to x=1.80

≈ 1.464

Var(X) = E(X^2) - [E(X)]^2

≈ 1.464 - 1.315^2

≈ 0.024

Finally, we take the square root of the variance to obtain the standard deviation:

sigma = sqrt(Var(X))

≈ sqrt(0.024)

≈ 0.155

Therefore, the answer is not provided in the options given. The closest option is 0.172, but the correct answer is 0.155 (rounded to three decimal places).

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The possible values of the coefficient (a) and an exponent (b) are CThe missing coefficient (a) can be any multiple of 3. The missing exponent (b) must be 5.

The two monomials are given as

ax³ 6xᵇ

Such that we have the LCM to be

LCM = 18x⁵

Since the coefficient of the LCM is 18, then the following is possible

a * 6 = multiples of 18

Divide both sides by 6

a = multiples of 3

Next, we have

LCM of x³ * xᵇ = x⁵

So, we have

b = 5 (bigger exponent)

Hence, the true statement is (c)

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