A graph has 5 vertices: four vertices of degree 3 and a vertex of degree 2 . How many edges are there in the graph? QUESTION 10 A graph has 5 vertices and 10 edges such that two vertices are of degree 3 , a vertex is of degree 2 , and a vertex is of degree 5 . Find the degree of the remaining vertex.

Answer:  not rational

Step-by-step explanation:

It's  not rational because it has no pattern and probably goes on forever

Irrational means goes on forever without pattern.  Ex. [tex]\pi[/tex] or √7

(B) (-1, 2) ordered pair is the best estimate for the solution to the system.

To solve the system {13x + 6y = −30, x − 2y = −4} using a graphing tool, we need to plot the line for each equation and find the point where they intersect.

This point is the solution to the system.

The intersection point appears to be (-1, 2).

Therefore, the best estimate for the solution to the system is the ordered pair (-1, 2).

Therefore, option b. (-1, 2) is the correct option.

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The marginal probability mass function for X is given by P(X = 1) = 6/18 = 1/3P(X = 2) = 5/18P(X = 3) = 5/18.

First, let us compute the marginal probability mass function for X.

p(1,1) + p(2,1) + p(3,1) = 1/9 + 1/3 + 1/9 = 5/9p(1,2) + p(2,2) + p(3,2) = 1/9 + 0 + 1/18 = 1/6p(1,3) + p(2,3) + p(3,3) = 0 + 1/6 + 1/9 = 5/18

Therefore, the marginal probability mass function for X is given by P(X = 1) = 6/18 = 1/3P(X = 2) = 5/18P(X = 3) = 5/18

We are asked to compute E[X|Y = 1], E[X|Y = 2], and E[X|Y = 3]. We know that E[X|Y] = ∑xp(x|y) / p(y)

Therefore, let us compute the conditional probability mass function for X given Y = 1.

p(1|1) = 1/9 / (5/9) = 1/5p(2|1) = 1/3 / (5/9) = 3/5p(3|1) = 1/9 / (5/9) = 1/5

Therefore, the conditional probability mass function for X given Y = 1 is given by P(X = 1|Y = 1) = 1/5P(X = 2|Y = 1) = 3/5P(X = 3|Y = 1) = 1/5

Therefore, E[X|Y = 1] = 1/5 × 1 + 3/5 × 2 + 1/5 × 3 = 1.8

Next, let us compute the conditional probability mass function for X given Y = 2.

p(1|2) = 1/9 / (1/6) = 2/3p(2|2) = 0 / (1/6) = 0p(3|2) = 1/18 / (1/6) = 1/3

Therefore, the conditional probability mass function for X given Y = 2 is given by P(X = 1|Y = 2) = 2/3P(X = 2|Y = 2) = 0P(X = 3|Y = 2) = 1/3

Therefore, E[X|Y = 2] = 2/3 × 1 + 0 + 1/3 × 3 = 2

Finally, let us compute the conditional probability mass function for X given Y = 3.

p(1|3) = 0 / (5/18) = 0p(2|3) = 1/6 / (5/18) = 6/5p(3|3) = 1/9 / (5/18) = 2/5

Therefore, the conditional probability mass function for X given Y = 3 is given by P(X = 1|Y = 3) = 0P(X = 2|Y = 3) = 6/5P(X = 3|Y = 3) = 2/5

Therefore, E[X|Y = 3] = 0 × 1 + 6/5 × 2 + 2/5 × 3 = 2.4

Therefore,E[X|Y=1] = 1.8,E[X|Y=2] = 2,E[X|Y=3] = 2.4.

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The x-coordinate of the point on the parabola where the tangent line has a slope of 18/18 is 5/8.

We are to find the x-coordinate of the point on the parabola y=20x²−12x/13 where the tangent line to the parabola has a slope of 18/18.  

The tangent line to the parabola has a slope of 18/18, so we can find the derivative of the equation y=20x²−12x/13 and set it equal to the given slope.dy/dx = 40x - 12/13

slope = 18/18 = 1

We can set the derivative equal to 1 and solve for x.40x - 12/13 = 1Multiplying both sides of the equation by 13, we have 40x - 12 = 13

Combining like terms, we get

40x = 25Dividing both sides by 40, we obtain x = 25/40 or x = 5/8.

Therefore, the x-coordinate of the point on the parabola where the tangent line has a slope of 18/18 is 5/8.

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The correct answer is d. symmetrical. Therefore, neither a, c, nor b describes the normal distribution accurately.

The normal distribution, also known as the Gaussian distribution or bell curve, is a symmetric probability distribution. It is characterized by a bell-shaped curve, where the data is evenly distributed around the mean. In a normal distribution, the mean, median, and mode are all equal and located at the center of the distribution. A bimodal distribution refers to a distribution with two distinct peaks or modes. An asymmetrical distribution does not exhibit symmetry and can be skewed to one side. Skewness refers to the degree of asymmetry in a distribution, so a skewed distribution is not necessarily a normal distribution.

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The stochastic process which satisfies the measurability condition adapted to σ (B, 0 ≤ s ≤ t) will only be considered.

Adapted process is a stochastic process that depends on time and which is predictable by the available information in a specified probability space.

An adapted stochastic process X(t) is measurable with respect to the given information up to time t. Here, the following stochastic processes X are adapted to σ (B, 0 ≤ s ≤ t):

(i) Xt = f B(t), if and only if f is σ (Bt; 0 ≤ t ≤ T) measurable.

(ii) Xt = max{0, B(t)}, if and only if the event {X(t) ≤ x} is σ (Bt; 0 ≤ t ≤ T) measurable for every x.

As the maximum function is continuous, it is left continuous and thus adapted to the filtration generated by Brownian motion

The stochastic processes adapted to σ (B, 0 ≤ s ≤ t) are as follows:

(i) Xt = f B(t), if and only if f is σ (Bt; 0 ≤ t ≤ T) measurable.

(ii) Xt = max{0, B(t)}, if and only if the event {X(t) ≤ x} is σ (Bt; 0 ≤ t ≤ T) measurable for every x.

The conclusion is that the stochastic process which satisfies the measurability condition adapted to σ (B, 0 ≤ s ≤ t) will only be considered.

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By comparing the z-scores we obtain that the 41-week gestation baby weighs more relative to their gestation period compared to the 32-week gestation baby.

To find the z-score for each baby, we can use the formula:

z = (X - μ) / σ

where X is the baby's weight, μ is the mean weight, and σ is the standard deviation.

For the 32-week gestation baby weighing 2875 grams:

z = (2875 - 2500) / 500 = 0.75

For the 41-week gestation baby weighing 3275 grams:

z = (3275 - 2900) / 415 = 0.9

The z-score measures the number of standard deviations a data point is from the mean.

A positive z-score indicates that the data point is above the mean.

Comparing the z-scores, we see that the 41-week gestation baby (z = 0.9) has a higher z-score than the 32-week gestation baby (z = 0.75).

This means that the 41-week gestation baby weighs more relative to their gestation period compared to the 32-week gestation baby.

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1 solution of a triangle exists because all of the angles are less than 180 degrees and the sides and angles have non-negative values.

Therefore, there exists only one triangle.

The given values are:

a = √2, b = √7, and p = 105

The sine law is applied to determine the angle opposite to a. We know that sin(A)/a = sin(B)/b = sin(C)/c

where A, B, and C are the angles of a triangle, and a, b, and c are the opposite sides to A, B, and C, respectively.

Therefore, sin(A)/√2 = sin(B)/√7

We can now get sin(A) and sin(B) by cross-multiplication:

√7 * sin(A) = √2 * sin(B)sin(A) / sin(B) = √(2/7)

Using the sine law, we can now calculate the angle C:

sin(C)/p = sin(B)/b

Therefore, sin(C) = (105 sin(B))/√7

Using the equation sin²(B) + cos²(B) = 1, we can determine

cos(B) and cos(A)cos(B)

= √(1 - sin²(B)) = √(1 - 2/7)

= √(5/7)cos(A) = (b cos(C))/a

= (√7 cos(C))/√2Since sin(A)/√2

= sin(B)/√7sin(A)

= (√2/√7)sin(B)sin(A)

= (√2/√7) [√(1 - cos²(B))]

We can solve the equations above using substitution to find sin(B) and sin(A).

1 solution of a triangle exists because all of the angles are less than 180 degrees and the sides and angles have non-negative values.

Therefore, there exists only one triangle.

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The measure of angle a is:

a = (140° - 96°) / 2 = 44° / 2 = 22°

Therefore, the answer is 22.

1

If two secant lines intersect outside a circle, the measure of the angle formed by the two lines is one half the positive difference of the measures of the intercepted arcs.

In the given diagram, we can see that the intercepted arcs are 96° and 140°. Therefore, the measure of angle a is:

a = (140° - 96°) / 2 = 44° / 2 = 22°

Therefore, the answer is 22.

Answer: 22

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Stem-and-leaf display of the given data is shown below: Stem and Leaf  

1 | 3, 4, 5, 8    

2 | 1, 3, 3    

3 | 2, 4, 5, 5, 6    

4 | 2, 5, 8    

5 | 2, 5, 7

There are two parts of the stem-and-leaf display:  the stem, which is the digits in the greatest place value, and the leaf, which is the digits in the lesser place value. The digits in the least place value of each observation are called leaves and are listed alongside the corresponding stem. This gives a clear picture of the distribution of the data.

The stem-and-leaf display for the given data table is as follows: Stem and Leaf  

1 | 3, 4, 5, 8    

2 | 1, 3, 3    

3 | 2, 4, 5, 5, 6    

4 | 2, 5, 8    

5 | 2, 5, 7

There are two parts of the stem-and-leaf display: the stem, which is the digits in the greatest place value, and the leaf, which is the digits in the lesser place value. The digits in the least place value of each observation are called leaves and are listed alongside the corresponding stem. This gives a clear picture of the distribution of the data. The stem-and-leaf display for the given data table is as follows: Stem and Leaf  

1 | 3, 4, 5, 8    

2 | 1, 3, 3    

3 | 2, 4, 5, 5, 6    

4 | 2, 5, 8    

5 | 2, 5, 7

The stem and leaf plot is a great way to show how data are distributed. It allows you to see the distribution of data in a more meaningful way than just looking at the raw numbers.

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The Taylor polynomial t3(x) for the function f centered at the number a = 1 and [tex]f(x) = ex is e(x-1)^3 + e(x-1)^2 + e(x-1) + e.[/tex]

The Taylor polynomial for f(x) = e^x centered at a = 1, with degree n = 3 is:

[tex]t_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3[/tex]

First, we compute the first three derivatives of f(x) = e^x.

[tex]f(x) = e^x\\f'(x) = e^x\\f''(x) = e^x\\f'''(x) = e^x[/tex]

Substituting a = 1 into each of these yields:

[tex]f(1) = e^1\\= e\\f'(1) = e^1 \\= e\\f''(1) = e^1 \\= e\\f'''(1) = e^1 \\= e[/tex]

Therefore,[tex]t_3(x) = e + e(x-1) + \frac{e}{2!}(x-1)^2 + \frac{e}{3!}(x-1)^3= e(1 + (x-1) + \frac{1}{2!}(x-1)^2 + \frac{1}{3!}(x-1)^3)[/tex]

Simplifying, we get:

[tex]t_3(x) = e(x-1)^3 + e(x-1)^2 + e(x-1) + e[/tex]

Therefore, the Taylor polynomial t3(x) for the function f centered at the number a = 1 and [tex]f(x) = ex is e(x-1)^3 + e(x-1)^2 + e(x-1) + e.[/tex]

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The correlation coefficient must necessarily be equal to -0.9.

The coefficient of determination (R²) is a statistical measure that indicates how well the regression line predicts the data points. It is a proportion of the variance in the dependent variable (y) that can be predicted by the independent variable (x).The value of R² ranges from 0 to 1, with 1 indicating a perfect fit between the regression line and the data points. The closer R² is to 1, the more accurate the regression line is in predicting the data points.

The formula to calculate the correlation coefficient (r) is as follows:r = (n∑xy - ∑x∑y) / √((n∑x² - (∑x)²) (n∑y² - (∑y)²))where x and y are the two variables, n is the number of data points, and ∑ represents the sum of the values.To find the correlation coefficient (r) from the coefficient of determination (R²), we take the square root of R² and assign a positive or negative sign based on the direction of the linear relationship (positive or negative).

The formula to find the correlation coefficient (r) from the coefficient of determination (R²) is as follows:

r = ±√R²For example, if R² is 0.81, then r = ±√0.81 = ±0.9

Since the linear regression equation is y = 15- 0.49

x1, this means that the slope of the line is -0.49.

This indicates that there is a negative linear relationship between the two variables, meaning that as x1 increases, y decreases.

Since the correlation coefficient (r) must have a negative sign, we have :r = -0.9

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Therefore,the function that has only one x-intercept at (-6, 0) is f(x) = (x + 6)(x - 6).

In this function, when you set f(x) equal to zero, you get:

(x + 6)(x - 6) = 0

For this equation to be satisfied, either (x + 6) must equal zero or (x - 6) must equal zero. However, since we want only one x-intercept, we need exactly one of these factors to be zero.

If (x + 6) = 0, then x = -6, which gives the x-intercept (-6, 0).

If (x - 6) = 0, then x = 6, but this would give us an additional x-intercept at (6, 0), which we do not want.

Therefore, the function f(x) = (x + 6)(x - 6) has only one x-intercept at (-6, 0).

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Given the cone z² = x²y² and the point (5, 3, 0), we have to find the points on the cone that are closest to the given point.The equation of the cone z² = x²y² can be written in the form z² = k²(x² + y²), where k is a constant.

Hence, the cone is symmetric about the z-axis. Let's try to obtain the constant k.z² = x²y² ⇒ z = ±k√(x² + y²)The distance between the point (x, y, z) on the cone and the point (5, 3, 0) is given byD² = (x - 5)² + (y - 3)² + z²Since the points on the cone have to be closest to the point (5, 3, 0), we need to minimize the distance D. Therefore, we need to find the values of x, y, and z on the cone that minimize D².

Let's substitute the expression for z in terms of x and y into the expression for D².D² = (x - 5)² + (y - 3)² + [k²(x² + y²)]The values of x and y that minimize D² are the solutions of the system of equations obtained by setting the partial derivatives of D² with respect to x and y equal to zero.∂D²/∂x = 2(x - 5) + 2k²x = 0 ⇒ (1 + k²)x = 5∂D²/∂y = 2(y - 3) + 2k²y = 0 ⇒ (1 + k²)y = 3Dividing these equations gives us x/y = 5/3. Substituting this ratio into the equation (1 + k²)x = 5 gives usk² = 16/9 ⇒ k = ±4/3Now that we know the constant k, we can find the corresponding value of z.z = ±k√(x² + y²) = ±(4/3)√(x² + y²)

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To determine how much the volume of the rectangular prism increases when the height is increased by 1 cm, we need to calculate the difference in volumes between the two configurations.

The volume V of a rectangular prism is given by the formula:

V = B * h

where B represents the area of the base and h represents the height.

When the height is increased by 1 cm, the new height becomes (h + 1) cm. The new volume, V', is given by:

V' = B * (h + 1)

To find the increase in volume, we subtract the original volume V from the new volume V':

Increase in volume = V' - V

Substituting the expressions for V and V', we have:

Increase in volume = (B * (h + 1)) - (B * h)

Simplifying, we get:

Increase in volume = B * h + B - B * h

The term B * h cancels out, leaving us with:

Increase in volume = B

Therefore, the increase in volume when the height is increased by 1 cm is equal to the area of the base B.

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The answer is - log x.

Here's the solution for the given problem:

Using the properties of logarithms to evaluate each of the following expressions.

2log 3 + log 4

= 0 5 122(log 3) + log 4

= log (3²) + log 4

= log (3² × 4)

= log 36

= log 6²

Now, we have the expression log 6²

Now, we can write the given expression as log 36

Thus, the final answer is log 36

Ine = In e

(b) 0 - X Ś ?

By the rule of logarithm for quotient, we have

log (1/x)

= log 1 - log x

= -log x

We can use the same rule for log (0 - x) and write it as

log (0 - x)

= log 0 - log x

= - log x

Now, we have the expression - log x

Thus, the final answer is - log x

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To compute the circulation of the vector field F around the curve C, we need to evaluate the line integral ∮C F · dr, where dr is the differential vector along the curve C. Let's calculate the circulation for each segment of the curve:

(a) Line segment from (0, 0, 0) to (1, 0, 0):

The differential vector dr along this segment is dr = dx i, where i is the unit vector in the x-direction, and dx represents the differential length along the x-axis. Since the vector field F = <y, 0, 0>, we have F · dr = (y)dx = 0, because y = 0 along this line segment. Hence, the circulation along this segment is zero.

(b) Line segment from (1, 0, 0) to (1, 1, 0):

The differential vector dr along this segment is dr = dy j, where j is the unit vector in the y-direction, and dy represents the differential length along the y-axis. Since the vector field F = <y, 0, 0>, we have F · dr = (y)dy = y dy. Integrating y dy from 0 to 1 gives us the circulation along this segment. Evaluating the integral, we get:

∫[0,1] y dy = [y^2/2] from 0 to 1 = (1^2/2) - (0^2/2) = 1/2.

(c) Line segment from (1, 1, 0) to (0, 1, 0):

The differential vector dr along this segment is dr = -dx i, where i is the unit vector in the negative x-direction, and dx represents the differential length along the negative x-axis. Since the vector field F = <y, 0, 0>, we have F · dr = (y)(-dx) = -y dx. Integrating -y dx from 1 to 0 gives us the circulation along this segment. Evaluating the integral, we get:

∫[1,0] -y dx = [-y^2/2] from 1 to 0 = (0^2/2) - (1^2/2) = -1/2.

(d) Line segment from (0, 1, 0) to (0, 0, 0):

The differential vector dr along this segment is dr = -dy j, where j is the unit vector in the negative y-direction, and dy represents the differential length along the negative y-axis. Since the vector field F = <y, 0, 0>, we have F · dr = (y)(-dy) = -y dy. Integrating -y dy from 1 to 0 gives us the circulation along this segment. Evaluating the integral, we get:

∫[1,0] -y dy = [-y^2/2] from 1 to 0 = (0^2/2) - (1^2/2) = -1/2.

To compute the total circulation around the curve C, we sum up the circulations along each segment:

Total Circulation = Circulation(a) + Circulation(b) + Circulation(c) + Circulation(d)

= 0 + 1/2 + (-1/2) + (-1/2)

= 0.

Therefore, the total circulation of the vector field F around the curve C, which is a unit square in the xy-plane, is zero.

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We are given two paraboloids as:z = (-9/2)(x^2 + y^2)andz = 5 - 2(x^2 + y^2)The volume of the solid enclosed between the two paraboloids is given byV = ∫∫R[(5 - 2(x^2 + y^2)) - (-9/2)(x^2 + y^2)] d[tex]z = (-9/2)(x^2 + y^2)andz = 5 - 2(x^2 + y^2)[/tex]A

where R is the region in the xy-plane that is bounded by the circular region of radius a centered at the origin.We can rearrange the equation and simplify it as follows:V = ∫∫R (23/2)x^2 + (23/2)y^2 - 5 dAWe will use polar coordinates (r, θ) to evaluate the integral, and the limits of integration for the radius will be 0 and a, and the limits of integration for the angle will be 0 and 2π.

Hence, we can rewrite the integral as:V = ∫[0, 2π] ∫[0, a] (23/2)r^2 - 5r dr dθEvaluating this integral:V = ∫[0, 2π] [23/6 * a^3 - 5/2 * a^2] dθV = 4π [23/6 * a^3 - 5/2[tex]V = ∫[0, 2π] ∫[0, a] (23/2)r^2 - 5r dr dθ Evaluating this integral:V = ∫[0, 2π] [23/6 * a^3 - 5/2 * a^2] dθV = 4π [23/6 * a^3 - 5/2 * a^2]V = (46/3)πa^3 - 10πa^2[/tex] * a^2]V = (46/3)πa^3 - 10πa^2Hence, the volume of the solid enclosed between the two paraboloids is (46/3)πa^3 - 10πa^2.The explanation has a total of 152 words.

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1. The regression equation to predict Year 12 completion for Indigenous Australians aged 20-24 with 32.1% Internet access:The regression equation for Year 12 completion rate of Indigenous Australians aged 20-24 would be Y = a + bX, where Y is the response variable (Year 12 completion rate).

X is the predictor variable (internet access), a is the intercept, and b is the slope. The equation can be expressed as follows:Y = 12.684 + 0.855 (32.1)Y = 12.684 + 27.453 = 40.137≈ 40.14The predicted Year 12 completion rate for Indigenous Australians aged 20-24 with 32.1% Internet access is 40.14 percent.2. Estimating the increase or decrease in Year 12 completion rate with a 95% confidence interval:From the regression output, we can see that for every one percent increase in internet access, the Year 12 completion rate increases by 0.855 percent on average.

Hence, for a 95% confidence interval, the estimate would be calculated as follows:Lower limit = 0.855 - 1.96 × 0.111 = 0.6394Upper limit = 0.855 + 1.96 × 0.111 = 1.0706Therefore, the 95% confidence interval for an extra one percent increase in internet access would be 0.6394 to 1.0706, or approximately 0.64 to 1.07.This means that we can be 95% confident that the true change in the Year 12 completion rate will be between 0.64% and 1.07% for every one percent increase in internet access. Hence, the answer is option A: increase.

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The curve of intersection between the paraboloid [tex]z = 2x^2 + y^2[/tex]and the parabolic cylinder y = 3[tex]x^2[/tex] can be represented by the vector function

r(t) = ([tex]t, 3t^2, 2t^2 + 9t^4[/tex]).

To find the curve of intersection between the two surfaces, we need to find the values of x, y, and z that satisfy both equations simultaneously. We can start by substituting the equation of the parabolic cylinder, y = 3[tex]x^2[/tex], into the equation of the paraboloid, [tex]z = 2x^2 + y^2[/tex].

Substituting y = 3[tex]x^2[/tex] into z = [tex]2x^2 + y^2[/tex], we get [tex]z = 2x^2 + (3x^2)^2 = 2x^2 + 9x^4[/tex].

Now, we can express the vector function r(t) as (x(t), y(t), z(t)).

Since [tex]y = 3x^2[/tex], we have y(t) = [tex]3t^2[/tex]. And from [tex]z = 2x^2 + 9x^4[/tex], we have [tex]z(t) = 2t^2 + 9t^4[/tex]

For x(t), we can choose x(t) = t, as it simplifies the equations and represents the parameter t directly. Therefore, the vector function representing the curve of intersection is [tex]r(t) = (t, 3t^2, 2t^2 + 9t^4)[/tex].

This vector function traces out the curve of intersection between the paraboloid and the parabolic cylinder as t varies. Each point on the curve is obtained by plugging in a specific value of t into the vector function.

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The completely simplified form of csc(z) cot(z) + tan(z) is 1 / sin²(z) + sin(z) Using the fundamental identities.

Given,

csc(z) cot(z) + tan(z)

We know that:

cot(z) = cos(z) / sin(z) csc(z)

= 1 / sin(z) tan(z)

= sin(z) / cos(z)

Now, csc(z) cot(z) + tan(z)

= 1 / sin(z) × cos(z) / sin(z) + sin(z) / cos(z)

= cos(z) / sin²(z) + sin(z) / cos(z)

The LCM of sin²(z) and cos(z) is sin²(z)cos(z).

Hence, cos(z) / sin²(z) + sin(z) / cos(z)

= cos²(z) / sin²(z) × cos(z) / cos(z) + sin³(z) / sin²(z) × sin²(z) / cos(z)

= cos²(z) / sin²(z) + sin(z) = 1 / sin²(z) + sin(z)

The completely simplified form of csc(z) cot(z) + tan(z) is 1 / sin²(z) + sin(z).

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The probabilities have been computed to be as follows: i) P (S2 = 2, S5 = 1) = 0.0441, ii) P (S2 = 2, S4 = 3, S5 = 1) = 0.0189.

The probabilities can be computed using the formula:

P (Sn = i, Sm = j) = P (Sn = i, Sn - m = j - i) = P (Sn = i)*P (Sn - m = j - i),

where i, j ∈ Z, n > m ≥ 0.

Then,

P (Sn = i) = (p/q) ^ (n+i)/2√πn, and

P (Sn - m = j - i) = (p/q) ^ ((n-m+ (j-i))/2) √ ((n+m- (j-i))/π(n-m))

For, P (S2 = 2, S5 = 1), i.e., i = 2, j = 1, n = 5, m = 2.

Then,

P (S2 = 2, S5 = 1) = P (S2 = 2) * P (S3 = -1) * P (S4 = -2) * P (S5 - 2 = -1) = (0.3) * (0.7) * (0.7) * (0.3) = 0.0441

For, P (S2 = 2, S4 = 3, S5 = 1), i.e., i = 2, j = 1, n = 5, m = 4.

Then,

P (S2 = 2, S4 = 3, S5 = 1) = P (S2 = 2) * P (S2 = 3 - 4) * P (S5 - 4 = 1 - 2) = (0.3) * (0.7) * (0.3) = 0.0189

Thus, we have computed the required probabilities as follows:

P (S2 = 2, S5 = 1) = 0.0441

P (S2 = 2, S4 = 3, S5 = 1) = 0.0189

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The quadratic function d = 0.0515r² + 1.1r equations graphed and attached

The equation plotted is

d = 0.0515r² + 1.1r

The key features includes

a. The graph open upwards: This is so since the quadratic term"r²" of the equation does not have a negative sign

b. The vertex of the quadratic equation is the lowest part of the graph which is (-10.7, 5.9)

c. the intercepts

the x-intercept or the roots are (-21.4, 0) and (0, 0)

the y-intercepts is (0, 0)

d. axis of symmetry

The symmetry is at x = -b/2x = -1.1(2 * 0.0515) = -10.68

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The interest on a principal of $1,245 at an interest rate of 5% for a period of 2 years is $124.50.

To calculate the interest, we can use the formula:

Interest = Principal × Rate × Time

Given:

Principal (P) = $1,245

Rate (R) = 5% = 0.05 (in decimal form)

Time (T) = 2 years

Plugging these values into the formula, we have:

Interest = $1,245 × 0.05 × 2

Calculating the expression, we get:

Interest = $1245 × 0.1

Interest = $124.50

It's important to note that the interest calculated here is simple interest. Simple interest is calculated based on the initial principal amount without considering any compounding over time. If the interest were compounded, the calculation would be different.

In simple interest, the interest remains constant throughout the period, and it is calculated based on the principal, rate, and time. In this case, the interest is calculated as a percentage of the principal for the given time period.

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The cross-correlation function of the random processes v(t) and w(t) is:

R_vw(tau) = -X^2 sin(wo(t+tau))cos(wot).

The cross-correlation function of the random processes v(t) and w(t) can be found by taking the expected value of their product. Since X and Y are independent random variables with zero mean and variance ², their cross-correlation function simplifies as follows:

R_vw(tau) = E[v(t)w(t+tau)]

= E[(X cos(wot) + Y sin(wot))(Y cos(wo(t+tau)) - X sin(wo(t+tau)))]

= E[XY cos(wot)cos(wo(t+tau)) - XY sin(wot)sin(wo(t+tau)) + Y^2 cos(wo(t+tau))sin(wot) - X^2 sin(wo(t+tau))cos(wot))]

Since X and Y are independent, their expected product E[XY] is zero. Additionally, the expected value of sine and cosine terms over a full period is zero. Therefore, the cross-correlation function simplifies further:

R_vw(tau) = -X^2 sin(wo(t+tau))cos(wot)

Thus, the cross-correlation function is given by:

R_vw(tau) = -X^2 sin(wo(t+tau))cos(wot).

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The scatterplot suggests that the relationship between x and y is moderately scattered.

The given measurements are

x 3 8 5 4D -1 0y 5 4 9 4

The scatterplot suggests that the relationship between x and y is moderately scattered.

A scatterplot is a plot where a dependent and an independent variable are plotted to observe the relationship between them.

The correlation coefficient and regression lines are used to describe the correlation between the two variables.

When a scatterplot is moderately scattered, the points in the plot are not concentrated at a certain point and they do not follow a strict trend.

Instead, the plot will form a pattern of points that are spread out in a random way. It suggests that there is a weak to moderate correlation between x and y.

The scatterplot suggests that the relationship between x and y is moderately scattered.

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The probability that more than 20 customers will arrive at the drive-thru is 0.025.

Poisson distribution is used to find the probability of a certain number of events occurring in a specified time interval. It is used when the event is rare, and the sample size is large. In Poisson distribution, we have an average arrival rate (λ) and the number of arrivals (x).

The probability of x events occurring during a given time period is :

P(x;λ) = λx e−λ/x!, where e is the Euler's number (e = 2.71828182846) and x! is the factorial of x.

To find the probability that more than 20 customers will arrive at the drive-thru during a given hour, we use the Poisson distribution as follows:

P(x > 20) = 1 - P(x ≤ 20)

P(x ≤ 20) = ∑ (λ^x * e^-λ)/x!, x = 0 to 20

Let's find the probability P(x ≤ 20).

P(x ≤ 20) = ∑ (λ^x * e^-λ)/x!; x = 0 to 20

P(x ≤ 20) = ∑ (17^x * e^-17)/x!; x = 0 to 20

P(x ≤ 20) = 0.975

Therefore, P(x > 20) = 1 - P(x ≤ 20)

P(x > 20) = 1 - 0.975

P(x > 20) = 0.025

This means that the probability that more than 20 customers will arrive at the drive-thru is 0.025.

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Answer: 9 s

Step-by-step explanation:

Given:

h(t) = -16t² + 144t

Find:

time when h=0    They want to know when the rocket hits the ground.  This happens when h=0

Solution:

0 = -16t² + 144t                     >Take out Greatest Common Factor

0 = (-16t )(t - 9)                      >Set each parenthesis = 0

(-16t ) = 0   and    (t - 9)=0           >Solve for t

t = 0          and      t =  9              >When the rocket launches t=0 and h=0

                                                      Also,    t=9 when h=0

t=9 s

1 gallon of paint will cover approximately 32.14 square feet when applied with a coat that is 0.05 inches thick.

To determine the surface area that 1 gallon of paint will cover, we need to convert the given thickness of 0.05 inches to feet.

Since 1 foot is equal to 12 inches, we have 0.05 inches/12 = 0.004167 feet as the thickness.

The coverage area of paint can be calculated by dividing the volume of paint (in cubic feet) by the thickness (in feet).

Since 1 gallon is equal to 231 cubic inches, and there are [tex]12^3 = 1728[/tex] cubic inches in 1 cubic foot, we have:

1 gallon = 231 cubic inches / 1728 = 0.133681 cubic feet.

Now, to calculate the surface area covered by 1 gallon of paint with a thickness of 0.004167 feet, we divide the volume by the thickness:

Coverage area = 0.133681 cubic feet / 0.004167 feet ≈ 32.14 square feet.

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Given a continuous random variable X that takes values between 0 and 2 with probability density function p(x) = 0.5, we are to find the expected value E(X) and the variance Var(X).

Expected value E(X)The expected value of a continuous random variable is defined as the integral of the product of the random variable and its probability density function over its range. That is,E(X) = ∫x p(x) dxIn this case, p(x) = 0.5 for 0 ≤ x ≤ 2. Hence,E(X) = ∫x p(x) dx= ∫x(0.5) dx = 0.5(x²/2)|0²= 0.5(2)²/2= 0.5(2)= 1Answer: E(X) = 1Var(X)The variance of a continuous random variable is defined as the expected value of the square of the deviation of the variable from its expected value. That is,Var(X) = E((X - E(X))²)We have already calculated E(X) as 1. Hence,Var(X) = E((X - 1)²)The squared deviation (X - 1)² takes values between 0 and 1 for 0 ≤ X ≤ 2. Hence,Var(X) = ∫(X - 1)² p(x) dx= ∫(X - 1)²(0.5) dx= 0.5 ∫(X² - 2X + 1) dx= 0.5 (X³/3 - X² + X)|0²= 0.5 [(2³/3 - 2² + 2) - (0)] = 0.5 (8/3 - 4 + 2)= 0.5 (2/3)Answer: Var(X) = 1/3

Therefore, E(X) = 1 and Var(X) = 1/3.

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