Let X be a continuous random variable with the density function f(x) = -{/². 1 < x < 2, elsewhere. (a) Define a function that computes the kth moment of X for any k ≥ 1. (b) Use the function in (a)

The required equation in the specified form is f(x) = 3(x - 2)² - 4.

Given that the quadratic function is f(x) = 3x²-12x+8

(a)

Writing the equation in the form f(x) = a(x-h)²+k

Let's first complete the square of the given quadratic equation

            f(x) = 3x²-12x+8,

               f(x) = 3(x² - 4x) + 8

Here, a = 3

         f(x) = 3(x² - 4x + 4 - 4) + 8

                 = 3(x - 2)² - 4

Therefore, the equation in the form f(x) = a(x - h)² + k is given by:

                   f(x) = 3(x - 2)² - 4

The vertex of the graph will be at (h, k) => (2, -4)

Therefore, the required equation in the specified form is f(x) = 3(x - 2)² - 4.

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The replacement times for washing machines follow an exponential distribution, where the probability of a washing machine lasting longer than a certain time t is given by P(X > t) = e^(-λt), and the expected lifetime of a washing machine is E(X) = 1/λ.

1) The correct answer is option C) 1/31. There are 31 days in May, so out of the 365 days in a year, the probability of someone being born on any given day is 31/365. Thus, the probability of someone being born in May is 31/365 or 1/31.

2) The replacement times for washing machines is an example of exponential distribution. Exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently at a constant average rate.

The probability density function for exponential distribution is given by f(x) = λe^(-λx), where λ is the rate parameter and x is the time elapsed. The cumulative distribution function is given by F(x) = 1 - e^(-λx).

To find the probability of a washing machine lasting longer than a certain time t, we can use the complementary cumulative distribution function P(X > t) = 1 - F(t) = e^(-λt).

This means that the probability of a washing machine lasting longer than a certain time t is exponentially decreasing with a rate of λ. The expected lifetime of a washing machine is given by E(X) = 1/λ.

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According to the question The sample provide evidence that the rides are as follows :

a) The appropriate statistical test to conduct in this scenario is the chi-square test for independence.

b) The hypotheses for this test are as follows:

H0: The rides are equally popular.

H1: The rides are not equally popular.

c) Given that the chi-square statistic is 3.144 and the p-value is 0.208, with a significance level of 0.05, we compare the p-value to the significance level to make a conclusion.

Since the p-value (0.208) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Explanation:

Failing to reject the null hypothesis means that we do not have enough evidence to conclude that the rides are not equally popular based on the sample data.

The test does not provide sufficient evidence to suggest that the preferences for the rides are significantly different among the visitors surveyed. Therefore, we cannot conclude that the rides are not equally popular based on this sample.

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The number of flaws in bolts of cloth in textile manufacturing is assumed to be Poisson distributed with a mean of 0.3 flaws per square meter. We are required to calculate the probability that there are at least two flaws in 3.9 square meters of cloth.

Therefore, the probability that there are at least two flaws in 3.9 square meters of cloth is 0.2255 or approximately 0.23.

To solve the given problem, we have to use Poisson probability distribution formula, which is:$$P(X = x) = \frac{{e^{ - \mu } \mu ^x }}{{x!}}$$where $x$ is the number of flaws, $\mu$ is the mean number of flaws, and $e$ is the mathematical constant 2.71828, and $x!$ is the factorial of $x$.

Probability of at least two flaws in 3.9 square meters of cloth can be calculated by using the following formula:$$P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$$We have $3.9$ square meters of cloth, so $0.3 \times 3.9 = 1.17$ flaws are expected. Let $X$ be the random variable representing the number of flaws in 3.9 square meters of cloth.$$P(X = x) = \frac{{e^{ - 1.17} 1.17^x }}{{x!}}$$We have to calculate $P(X \ge 2)$:$$\begin{aligned}P(X \ge 2) &= 1 - P(X = 0) - P(X = 1)\\&= 1 - \frac{{e^{ - 1.17} 1.17^0 }}{{0!}} - \frac{{e^{ - 1.17} 1.17^1 }}{{1!}}\\&= 1 - e^{ - 1.17}  - 1.17e^{ - 1.17}\\&= 0.2255\end{aligned}$$

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The probability that there are at least two flaws in 3.9 square meters of cloth is 0.037, or 3.7%.

The Poisson distribution is defined by the parameter λ, which represents the average number of flaws per square meter.

Given that the mean is 0.3 flaws per square meter, we have λ = 0.3.

To find the probability of at least two flaws in 3.9 square meters of cloth, we can calculate the complement of the probability of having zero or one flaw.

P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]

Let's calculate each term step by step:

Probability of zero flaws in 3.9 square meters:

P(X = 0) = e⁻⁰³= 0.7408

Probability of one flaw in 3.9 square meters:

P(X = 1) = 0.3 × e^(-0.3)

= 0.2222

Now, we can calculate the probability of at least two flaws:

P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]

P(X ≥ 2) = 1 - (0.7408 + 0.2222)

P(X ≥ 2)=0.037

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Therefore, the solution to the given system of equations is x = 7 and y = 8.

To solve the system of equations using row operations on an augmented matrix, we first write the system in matrix form:

| 1 1 | | x | | 15 |

| 1 -1 | * | y | = | -1 |

We can apply row operations to transform this matrix into row-echelon form or reduced row-echelon form. Let's use the Gaussian elimination method to solve it:

Step 1: Subtract the first row from the second row:

| 1 1 | | x | | 15 |

| 0 -2 | * | y | = | -16 |

Step 2: Divide the second row by -2 to obtain leading 1:

| 1 1 | | x | | 15 |

| 0 1 | * | y | = | 8 |

Step 3: Subtract the second row from the first row:

| 1 0 | | x | | 7 |

| 0 1 | * | y | = | 8 |

The resulting augmented matrix corresponds to the system of equations:

x = 7

y = 8

Therefore, the solution to the given system of equations is x = 7 and y = 8.

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The area of the lawn watered by the sprinkler is approximately 3.311 square feet.

To determine the area of the lawn watered by the sprinkler, we need to calculate the sector area of the circle covered by the sprinkler's rotation.

First, let's find the radius of the circle. The distance from the sprinkler to the edge of the water projection is 4.1 ft. Since the sprinkler rotates 115°, it covers one-fourth (90°) of the circle.

To find the radius, we can use the trigonometric relationship in a right triangle formed by the radius, half of the water projection (2.05 ft), and the adjacent side (distance from the center to the edge). The adjacent side is found using cosine:

cos(angle) = adjacent / hypotenuse

cos(90°) = 2.05 ft / radius

Solving for the radius:

radius = 2.05 ft / cos(90°) = 2.05 ft

Now that we have the radius, we can calculate the area of the sector covered by the sprinkler:

sector area = (angle / 360°) * π * radius^2

= (115° / 360°) * π * (2.05 ft)^2

Calculating this expression:

sector area ≈ 0.318 * π * (2.05 ft)^2 ≈ 3.311 ft²

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We have shown that the inverse of the two-dimensional orthogonal rotation matrix is equal to its transpose.

In mathematics, an orthogonal rotation matrix is a real matrix that preserves the length of each vector and the angle between any two vectors, including those that are not orthogonal.

In this case, we are to prove that the inverse of the orthogonal rotation matrix is equal to its transpose.

The two-dimensional orthogonal rotation matrix λ is given by

λ = [cos(θ) -sin(θ);

sin(θ) cos(θ)]

where θ is the angle of rotation.

Let's find the inverse of λ:

λ⁻¹ = [cos(θ) sin(θ);-

sin(θ) cos(θ)]/det(λ)

where det(λ) is the determinant of λ, which is

cos²(θ) + sin²(θ) = 1

Therefore,

λ⁻¹ = [cos(θ) sin(θ);-

sin(θ) cos(θ)]

Multiplying both sides by λ, we get

λ⁻¹λ = [cos(θ) sin(θ);-sin(θ) cos(θ)][cos(θ) -sin(θ);

sin(θ) cos(θ)]

λ⁻¹λ = [cos²(θ) + sin²(θ) cos(θ)sin(θ) - cos(θ)sin(θ);

sin(θ)cos(θ) - cos(θ)sin(θ) cos²(θ) + sin²(θ)]

λ⁻¹λ = [1 0;0 1]

This implies thatλ⁻¹ = λ¹And this completes the proof.

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The velocity vector of the particle at any time t is given by v(t) = -2n sin(nt)i + 4n cos(nt)j.

The velocity vector of the particle at any time t can be obtained by taking the derivatives of the position functions with respect to time. Given x(t) = 2cos(nt) and y(t) = 4sin(nt), the velocity vector v(t) is given by v(t) = dx/dt i + dy/dt j.

Taking the derivatives of x(t) and y(t) with respect to t, we get dx/dt = -2n sin(nt) and dy/dt = 4n cos(nt). Therefore, the velocity vector v(t) is:

v(t) = -2n sin(nt)i + 4n cos(nt)j.

This vector represents the instantaneous velocity of the particle at any given time t. The i-component (-2n sin(nt)) represents the velocity in the x-direction, while the j-component (4n cos(nt)) represents the velocity in the y-direction.

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A symbol that completes the inequality 6x - 3y ___ -12 is: C. ≥.

In Mathematics and Geometry, an inequality simply refers to a mathematical relation that is typically used for comparing two (2) or more numerical data and variables in an algebraic equation based on any of the inequality symbols;

Next, we would evaluate the inequality by using specific ordered pairs (x, y) as follows;

(0, 0)

6(0) - 3(0) ? -12

0 ≥ -12

(1, 2)

6(1) - 3(2) ? -12

0 ≥ -12

(-1, 2)

6(-1) - 3(2) ? -12

-12 ≥ -12

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For the first data set representing the number of marbles owned by fifteen different boys:

a) To find the 1st quartile, we arrange the data in ascending order: 13, 20, 33, 51, 55, 58, 64, 69, 70, 80, 86, 88, 93, 94, 99. The 1st quartile is the median of the lower half of the data, which is the median of the first seven numbers. So, the 1st quartile is 58.

b) The 2nd quartile is the median of the entire data set. Since there are 15 data points, the median is the 8th value, which is 69.

c) To find the 3rd quartile, we take the median of the upper half of the data, which is the median of the last seven numbers. So, the 3rd quartile is 93.

d) The box-and-whisker plot represents the minimum value (13), the 1st quartile (58), the median (69), the 3rd quartile (93), and the maximum value (99), with a box indicating the interquartile range (IQR).

For the second data set representing the number of telephone calls made by eighteen executives:

a) The 1st decile is the value below which 10% of the data lies. So, 10% of 18 is 1.8. Since we can't have a fraction of a telephone call, the 1st decile is the second value, which is 10.

b) P34 represents the 34th percentile, which is the value below which 34% of the data lies. So, 34% of 18 is 6.12. Since we can't have a fraction of a telephone call, P34 is the seventh value, which is 15.

c) The median is the value that separates the data into two equal halves. Since there are 18 data points, the median is the average of the ninth and tenth values, which is (18 + 22) / 2 = 20.

d) The third quartile is the value below which 75% of the data lies. So, 75% of 18 is 13.5. Since we can't have a fraction of a telephone call, the third quartile is the fourteenth value, which is 35.

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To maximize the amount of light entering the window, the width should be 2.5 m. The surface area of the window would be approximately 8.07 m².



To find the width that lets in the most light, we can set up an equation using the given perimeter. Let's denote the width of the rectangle as "w" and the radius of the semicircle as "r." The perimeter of the window is the sum of the rectangle's perimeter and half the circumference of the semicircle: 2w + πr = 16 m.

To maximize the amount of light, we need to maximize the surface area of the window. The surface area can be calculated by adding the area of the rectangle to half the area of the semicircle: A = wh + 1/2πr².Now, we can solve for the width that maximizes the surface area. Rearranging the perimeter equation, we have r = (16 - 2w) / π. Substituting this value of r into the surface area equation, we get A = wh + 1/2π[(16 - 2w) / π]².

To find the maximum surface area, we differentiate the equation with respect to w and set it to zero. After simplifying, we find that the width that maximizes the surface area is w = 2.5 m. Substituting this value back into the perimeter equation, we can find r = 1.5 m.Finally, we can calculate the surface area of the window using the obtained values of w and r: A = (2.5)(1.5) + 1/2π(1.5)² ≈ 8.07 m². Therefore, a window with a width of 2.5 m and a surface area of approximately 8.07 m² will let in the most light.

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Given that X and Y are independent random variables, we can use the properties of variance to find Var(X) and Var(Y) based on the given information.

We have the following information:

Var(3X - Y) = 12 ...(1)

Var(X + 2Y) = 13 ...(2)

To find Var(X), we can manipulate equation (2) as follows:

Var(X + 2Y) = 13

Var(X) + Var(2Y) = 13 (since X and 2Y are independent)

Var(X) + 4Var(Y) = 13 (applying the property Var(aX) = a^2 * Var(X))

Now, let's substitute equation (1) into the above equation:

12 + 4Var(Y) = 13

4Var(Y) = 13 - 12

4Var(Y) = 1

Var(Y) = 1/4

Therefore, we have found Var(Y) = 1/4.

To find Var(X), we can substitute the value of Var(Y) into equation (2):

Var(X + 2Y) = 13

Var(X) + Var(2Y) = 13 (since X and 2Y are independent)

Var(X) + 4Var(Y) = 13 (applying the property Var(aX) = a^2 * Var(X))

Var(X) + 4 * (1/4) = 13

Var(X) + 1 = 13

Var(X) = 13 - 1

Var(X) = 12

Therefore, we have found Var(X) = 12.

Conclusion:

Var(X) = 12

Var(Y) = 1/4

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The amount of money that will be in the savings account on week 100 is $250.

To find the amount of money that will be in the savings account on week 100, we can use the formula for linear interpolation which is given by:

`(y2 - y1) / (x2 - x1) = (y - y1) / (x - x1)`,

where `y1`, `y2` are the amounts of money in the savings account at week `x1`, `x2` respectively, and we need to find `y` at week `x = 100`.

Given that on week 8, she had $20.00 and on week 12, she had $30.00, we can let

`x1 = 8`,

`y1 = 20`,

`x2 = 12`,

`y2 = 30` and `x = 100`.

Plugging these values into the formula for linear interpolation, we get:(30 - 20) / (12 - 8) = (y - 20) / (100 - 8)

Simplifying, we get:

2.5 = (y - 20) / 92

Multiplying both sides by 92, we get:

230 = y - 20

Adding 20 to both sides, we get:

y = 250

Therefore, the amount of money that will be in the savings account on week 100 is $250.

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The final answer is 4(2x + h) after simplifying the difference quotient f(x+h)-f(x)/h for the function f(x) = 4x² - 4.

To find the difference quotient f(x+h)-f(x)/h for the function

                                   f(x) = 4x² - 4,

we need to substitute the given values into the formula as shown below:

                     f(x+h)-f(x)/h=f((x + h)) - f(x)/h

Substitute

               f(x + h) = 4(x + h)² - 4

             and f(x) = 4x² - 4.

             f(x+h)-f(x)/h= [4(x + h)² - 4] - [4x² - 4]/h

Note: We must expand (x + h)² to simplify the formula.

           f(x+h)-f(x)/h= [4(x² + 2xh + h²) - 4] - [4x² - 4]/h

Now we can solve it step by step:

            f(x+h)-f(x)/h= [(4x² + 8xh + 4h²) - 4 - 4x² + 4]/h

Combine like terms.  

                         f(x+h)-f(x)/h= (8xh + 4h²)/h

Factor out 4h from the numerator.

                      f(x+h)-f(x)/h= (4h(2x + h))/h

Cancel the h in the numerator and denominator.

                           f(x+h)-f(x)/h= 4(2x + h)

The final answer is 4(2x + h) after simplifying the difference quotient f(x+h)-f(x)/h for the function f(x) = 4x² - 4.

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To sketch the curve defined by the equation f(x, y) = c, along with the vector field Vf and the tangent line at a given point. The equation of the tangent line is also provided.  the equation of the tangent line is 9xy = -45.

The curve f(x, y) = c represents a level curve of the function f(x, y), where c is a constant. To sketch the curve, we can choose different values of c and plot the corresponding points on the xy-plane. The vector field Vf represents the gradient vector of the function f(x, y) and can be visualized by drawing arrows indicating the direction and magnitude of the gradient at each point.

In this specific case, the equation is given as 8x² - 3y = 43. To find the tangent line at the point (√√5, −1), we need to determine the gradient of the curve at that point. The gradient vector can be obtained by taking the partial derivatives of the equation with respect to x and y.

Once we have the gradient vector, we can find the equation of the tangent line using the point-slope form. Since the equation of the tangent line is provided as 9xy = -45, we can compare it with the general equation of a line (y - y₁) = m(x - x₁) to identify the slope and the point (x₁, y₁) on the line.

In this case, the equation of the tangent line is 9xy = -45.

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If there are twenty prizes, then the number of prizes that should go to fifth-grade students is 4.

We must distribute the awards proportionally based on the number of pupils in each grade in order to determine how many should go to fifth-graders.

We must first determine the total number of students enrolled in the institution:

Total students = 35 + 38 + 38 + 33 + 36 = 180

Proportion of fifth-grade students = 36 / 180 = 0.2

Number of prizes for fifth-grade students = Proportion of fifth-grade students * Total number of prizes

Number of prizes for fifth-grade students = 0.2 * 20 = 4

Therefore, the number of prizes as per the probability that should go to fifth-grade students is 4.

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Your question seems incomplete, the probable complete question is:

Prizes are to be awarded to the best pupils in each class of an elementary school. The number of students in each grade is shown in the table, and the school principal wants the number of prizes awarded in each grade to be proportional to the number of students. If there are twenty prizes, how many should go to fifth grade students?

Grade 1 2 3 4 5

Students 35 38 38 33 36

A

5

B

4

C

7

D

3

E

2

The given series is 1, 8, 15, 22,...To find the formula for the umpteenth term, an of the progression, we need to use the formula of the general term of an Arithmetic progression (AP), which is given by:an = a1 + (n - 1)da1 is the first term of the APn is the number of terms in the APd is the common difference of the APTaking a1 = 1 and d = 8 - 1 = 7 in the above formula, we get:an = 1 + (n - 1) x 7Simplifying the above equation, we get:an = 7n - 6 Therefore, the formula for the umpteenth term, an of the given arithmetic progression is: an = 7n - 6.

To determine the formula for the umpteenth term, an, of the given progression, we can observe the pattern in the terms.

The given sequence starts with 1 and increases by 7 with each subsequent term

=(8 - 1 = 7, 15 - 8 = 7, 22 - 15 = 7, and so on). We can express this pattern mathematically using the formula: an = a₁ + (n - 1) * d. Where an represents the nth term, a₁ is the first term, n is the term number, and d is the common difference. In this case, the first term is 1 and the common difference is 7. Substituting these values into the formula, we have: an = 1 + (n - 1) * 7

Simplifying further: an = 1 + 7n - 7

an = 7n - 6

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The probability that at least one of five randomly selected 40-year-old males will not live to be 41 years old is approximately 0.01214 or 1.214%.

a) To find the probability that two randomly selected 40-year-old males will live to be 41, we can multiply the individual probabilities together since the events are independent:

P(both live to be 41) = P(live to be 41) * P(live to be 41)

= 0.99757 * 0.99757

≈ 0.99514

Therefore, the probability that two randomly selected 40-year-old males will live to be 41 is approximately 0.99514.

b) Similarly, to find the probability that five randomly selected 40-year-old males will live to be 41, we can multiply the individual probabilities together:

P(all live to be 41) = P(live to be 41) * P(live to be 41) * P(live to be 41) * P(live to be 41) * P(live to be 41) = [tex]0.99757^5[/tex]results to 0.98786.

Therefore, the probability that five randomly selected 40-year-old males will live to be 41 is approximately 0.98786.

c) To find the probability that at least one of five 40-year-old males will not live to be 41, we can use the complement rule. The complement of "at least one" is "none." So, the probability of at least one not living to be 41 is equal to 1 minus the probability that all five live to be 41:

P(at least one does not live to be 41) = 1 - P(all live to be 41)

= 1 - 0.99757^5  which gives value of 0.01214.

Therefore, the probability that at least one of five randomly selected 40-year-old males will not live to be 41 years old is approximately 0.01214 or 1.214%.

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To evaluate the integral ∫(1 to 0) y/e^(3y) dy, we can use integration by substitution.

Let u = 3y. Then, du = 3dy.

When y = 1, u = 3(1) = 3.

When y = 0, u = 3(0) = 0.

The limits of integration can be expressed in terms of u as well.

Now, let's rewrite the integral in terms of u:

∫(1 to 0) y/e^(3y) dy = ∫(3 to 0) (1/3)e^(-u) du.

Next, we can simplify the integral:

∫(3 to 0) (1/3)e^(-u) du = (1/3) ∫(3 to 0) e^(-u) du.

Using the fundamental theorem of calculus, we can integrate e^(-u):

(1/3) ∫(3 to 0) e^(-u) du = (1/3) [-e^(-u)] from 3 to 0.

Now, let's substitute the limits of integration:

(1/3) [-e^(-0) - (-e^(-3))].

Simplifying further:

(1/3) [-1 + e^(-3)].

Therefore, the value of the integral ∫(1 to 0) y/e^(3y) dy is (1/3)[-1 + e^(-3)].

To evaluate the integral ∫(1 to 0) y/e^(3y) dy, we can use integration by substitution.

Let u = 3y. Then, du = 3dy.

When y = 1, u = 3(1) = 3.

When y = 0, u = 3(0) = 0.

The limits of integration can be expressed in terms of u as well.

Now, let's rewrite the integral in terms of u:

∫(1 to 0) y/e^(3y) dy = ∫(3 to 0) (1/3)e^(-u) du.

Next, we can simplify the integral:

∫(3 to 0) (1/3)e^(-u) du = (1/3) ∫(3 to 0) e^(-u) du.

Using the fundamental theorem of calculus, we can integrate e^(-u):

(1/3) ∫(3 to 0) e^(-u) du = (1/3) [-e^(-u)] from 3 to 0.

Now, let's substitute the limits of integration:

(1/3) [-e^(-0) - (-e^(-3))].

Simplifying further:

(1/3) [-1 + e^(-3)].

Therefore, the value of the integral ∫(1 to 0) y/e^(3y) dy is (1/3)[-1 + e^(-3)].

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The polar coordinates (r, θ) for the point (-2, 2) are approximately (2√2, -π/4).

To find the polar coordinates (r, θ) of a point given its Cartesian coordinates (x, y), you can use the following formulas:

r = √(x² + y²)

θ = atan2(y, x)

Let's calculate the polar coordinates for the given Cartesian coordinates (-2, 2):

Calculate the value of r:

r = √((-2)² + 2²)

r = √(4 + 4)

r = √8

r = 2√2

Calculate the value of θ:

θ = atan2(2, -2)

θ = atan2(1, -1) (simplifying the fraction)

θ = -π/4 (approximately -0.7854 radians or -45 degrees)

Therefore, the polar coordinates (r, θ) for the point (-2, 2) are approximately (2√2, -π/4).

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Using Pythagoras Theorem, the distance between two ships after 1.5 hours is approximately 47 nautical miles.

Given the bearing of the first ship = 32 at 26 knots The bearing of the second ship = 122 at 18 knots Time = 1.5 hours We need to calculate the distance between two ships after 1.5 hours. We can find the distance using the formula: Distance = Speed × Time

Distance of the first ship = 26 knots × 1.5 hours = 39 nautical miles Distance of the second ship = 18 knots × 1.5 hours = 27 nautical miles

The angle between the bearings of the two ships = 122 - 32 = 90°

Use Pythagoras Theorem to find the distance between the two ships, we have:

Distance² = 39² + 27²

Distance² = 1521 + 729

Distance² = 2250

Distance = √2250

Distance ≈ 47.43

So, the distance between two ships after 1.5 hours is approximately 47 nautical miles.

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To evaluate the expression (-1 + 2i) * (2 + 2i), we can use the distributive property of complex numbers.

The distributive property of complex numbers is a fundamental property that allows us to multiply a complex number by a sum or difference of complex numbers. It states that for any complex numbers a, b, and c, the following property holds:

a * (b + c) = a * b + a * c

In other words, when multiplying a complex number, a by the sum or difference of two complex numbers (b + c), we can distribute the multiplication to each term within the parentheses.

(-1 + 2i) * (2 + 2i) = -1 * 2 + (-1) * 2i + 2i * 2 + 2i * 2i

= -2 - 2i + 4i + 4i^2

= -2 - 2i + 4i + 4(-1)

= -2 - 2i + 4i - 4

= -6 + 2i

Therefore, the expression (-1 + 2i) * (2 + 2i) simplifies to -6 + 2i in the form a + bi.

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a. A solution to the quadratic inequality x² - 25 > -2x - 10 is x < -5 or x > 3.

b. The solution is shown on the number line attached below.

c. The solution as an interval is (-∞, -5) ∪ (3, ∞).

In Mathematics and Geometry, the standard form of a quadratic equation is represented by the following equation;

ax² + bx + c = 0

Part a.

Next, we would determine the solution for the given quadratic inequality as follows;

x² - 25 > -2x - 10

By rearranging and collecting like-terms, we have the following:

x² + 2x + 10 - 25 > 0

x² + 2x - 15 > 0

x² + 5x - 3x - 15 > 0

x(x + 5) -3(x + 5) > 0

(x + 5)(x - 3) > 0

x + 5 > 0

x < -5

x - 3 > 0

x > 3.

Therefore, the solution for the given quadratic inequality is x < -5 or x > 3.

Part b.

In this exercise, we would use an online graphing calculator to plot the given solution x < -5 or x > 3 as shown on the number line attached below.

Part c.

The solution for the given quadratic inequality x² - 25 > -2x - 10 as an interval should be written as follows;

(-∞, -5) ∪ (3, ∞).

As an inequality, the solution for the given quadratic inequality x² - 25 > -2x - 10 should be written as follows;

-5 > x > 3

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The dot product of ū - v = (3ī + 2j - 8k) · (ī - 25 - 3k) is equal to -83.

To evaluate the dot product, we multiply the corresponding components of the two vectors and sum them up.

The given vectors are:

ū = 3ī + 2j - 8k

v = ī - 25 - 3k

Now, let's calculate the dot product:

(3ī + 2j - 8k) · (ī - 25 - 3k)

= (3 * 1) + (2 * 0) + (-8 * (-3))

(3 * 0) + (2 * (-25)) + (-8 * (-1))

(3 * (-3)) + (2 * (-0)) + (-8 * (-0))

= 3 + 0 + 24

0 - 50 + 8

9 + 0 + 0

= -83

Therefore, the dot product of ū - v is -83.

Explanation (additional details):

The dot product, also known as the scalar product, is a mathematical operation between two vectors that results in a scalar quantity. It is calculated by multiplying the corresponding components of the vectors and then summing them up.

In this case, we have two vectors: ū = 3ī + 2j - 8k and v = ī - 25 - 3k. To find their dot product, we multiply the coefficients of the same variables in each vector and add them together.

For the first component, we have (3 * 1) = 3.

For the second component, we have (2 * 0) = 0.

For the third component, we have (-8 * (-3)) = 24.

Similarly, for the remaining components:

(3 * 0) = 0, (2 * (-25)) = -50, (-8 * (-1)) = 8,

(3 * (-3)) = -9, (2 * (-0)) = 0, and (-8 * (-0)) = 0.

Adding all these products together, we get:

3 + 0 + 24 + 0 - 50 + 8 - 9 + 0 + 0 = -83.

Hence, the dot product of ū - v is -83, indicating that the two vectors are not orthogonal and have a negative scalar relationship.

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The concentration of salt in the tank approaches 0.025 kg/L as time approaches infinity. The amount of salt in the tank after 3.5 hours is 50 kg. The amount of sugar in the tank at the beginning is 0 kg.

The amount of sugar after t minutes is 0.09t kg. The limit of y(t) as t approaches infinity is 205.2 kg.

The concentration of salt in the tank approaches 0.025 kg/L as time approaches infinity because the rate of salt entering the tank is equal to the rate of salt leaving the tank. The amount of salt in the tank after 3.5 hours is 50 kg because the rate of salt entering the tank is equal to the rate of salt leaving the tank.

The amount of sugar in the tank at the beginning is 0 kg because the tank contains pure water. The amount of sugar after t minutes is 0.09t kg because the rate of sugar entering the tank is equal to the rate of sugar leaving the tank. The limit of y(t) as t approaches infinity is 205.2 kg because the rate of sugar entering the tank is greater than the rate of sugar leaving the tank.

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To find the value of z corresponding to a cumulative standardized normal distribution of 0.2090, we can use a standard normal distribution table or a calculator. The value of z is approximately -0.82 when rounded to two decimal places.

In a standard normal distribution, the cumulative standardized normal distribution represents the area under the curve to the left of a given z-score. In this case, we are given a cumulative probability of 0.2090, which indicates that 20.90% of the area under the curve lies to the left of the corresponding z-score.

By referring to a standard normal distribution table or using a calculator that provides the cumulative distribution function (CDF) for the standard normal distribution, we can find the closest corresponding z-score. In this case, the value of z that corresponds to a cumulative probability of 0.2090 is approximately -0.82 when rounded to two decimal places.

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The distance between the two straight lines in twisted position can be found by determining the shortest distance between any two points on the lines.

To find the distance, we can choose a point on one line and find its shortest distance to the other line. Let's consider a point P on the first line with coordinates (x, y, z) = (2 - t, 3 + 4t, 2t). Now, we need to find the value of parameter t that minimizes the distance between P and the second line.

Substituting the coordinates of P into the equation of the second line, we get the coordinates of the closest point Q on the second line. Then, we can calculate the distance between P and Q using the Euclidean distance formula: d = √[(x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)²].

By simplifying the expression, we obtain the equation for the distance between the two lines in terms of the parameter t.

To find the twisted position, we can set the derivative of the distance equation with respect to t equal to zero and solve for t. The value of t obtained will give us the twisted position at which the two lines are closest to each other.



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Therefore, the probability that the first failure occurs after 15 firings is approximately 0.085 rounded to the nearest ten-thousandth.

Given that the first failure of a type of missile occurs on the last firing per every 20 successive independent firings. We need to find the probability that the first failure occurs after 15 firings.

Given, The number of firings before the first failure follows geometric distribution with probability of success, p = 1/20 (Since it occurs on the last firing per every 20 successive independent firings)

Let X be the number of firings before the first failure, then X ~ Geometric(p) ⇒ X ~ Geometric(1/20)

Now, we need to find P(X > 15 | X > 10)

Probability of the first failure occurs after at least 10 firings:

[tex](X > 10) = (1 - p)^{(10 - 1)} * p[/tex]

[tex]= (19/20)^9 * 1/20[/tex]

= 0.382

For a geometric distribution, P(X > n + k | X > k) = P(X > n), for all n ≥ 0

P(X > 15 | X > 10) = P(X > 5)

[tex]= (1 - p)^{(5 - 1) }* p[/tex]

[tex]= (19/20)^4 * 1/20[/tex]

= 0.085

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1. The solution to the equation is x = 19/4.

2. The solutions to the equation are x = -4 and x = 3.

1. To solve the equation 3/(x+2) = 1/(7-x), we can cross-multiply:

3(7-x) = 1(x+2)

21 - 3x = x + 2

21 - 2 = x + 3x

19 = 4x

x = 19/4

Therefore, the solution to the equation is x = 19/4.

2. To solve the equation (3-x)(x-5) - 2x² / (x²-3x-10) = 2/(x+2), we can simplify and rearrange the equation:

[(3-x)(x-5) - 2x²] / (x²-3x-10) = 2/(x+2)

Expanding the numerator and simplifying the denominator:

[(3x - 8 - x²) - 2x²] / (x² - 3x - 10) = 2/(x+2)

Combining like terms in the numerator:

[-3x² + 3x - 8] / (x² - 3x - 10) = 2/(x+2)

Multiplying both sides by (x² - 3x - 10) and simplifying:

-3x² + 3x - 8 = 2(x² - 3x - 10)

-3x² + 3x - 8 = 2x² - 6x - 20

Rearranging the equation to form a quadratic equation:

2x² - 3x² + 3x - 6x - 8 + 20 = 0

-x² - 3x + 12 = 0

-(x+4)(x-3) = 0

Setting each factor equal to zero and solving for x:

x+4 = 0 -> x = -4

x-3 = 0 -> x = 3

Therefore, the solutions to the equation are x = -4 and x = 3.

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The series (-1)-¹√n n=2 (n-3)² converges absolutely.

Here's how we can solve the problem. We need to use the Limit Comparison Test, as it is the most straightforward method to determine the convergence of this type of series.

Let us use the Limit Comparison Test:

We can say that we need to select the series such that the ratio tends to a finite, nonzero limit as n approaches infinity. We are going to compare the series with the test series:

`1/n²`.∑`|aₙ|`=∑ | (-1)-¹√n n=2 (n-3)² |

For `n>=2, (-1)-¹√n>=0` and `(n-3)²>=0`,

we can conclude that `|(-1)-¹√n| (n-3)² <= n²`∑ `|aₙ| <=∑ 1/n² where the latter series is convergent by the p-series test

∑`|aₙ|` is convergent by the Comparison Test, and it follows that it is absolutely convergent.

Therefore, the series converges absolutely.

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