Put all of these rates of increase in the correct order, from slowest rate of increase to fastest rate of increase: - O(n^2)- O(2^n)- O(n*log(n))- O(n) - O(log(n))- O(n!)

Answer:

C   11.9

Step-by-step explanation:

For a parallelogram,

area = base × height

A = 4.25 cm × 2.8 cm

A = 11.9 cm²

Answer:

C. 11.9

Concept Used:
Area of Parallelogram = b · h

Step-by-step explanation:

Required Area = 2.8 · 4.25

= 11.9 cm²

Note: Please always consider the side which touches the perpendicular as the base unless you are asked to divide the shape into parts and calculate the area separately.

a. The higher-order derivatives are zero, the Taylor series of f(x) centered at x = -1 is simply:

[tex]f(x) = f(-1) + f'(-1)(x+1) + (1/2!)f''(-1)(x+1)^2 + (1/3!)f'''(-1)(x+1)^3 + ...[/tex]

The first four terms are:

[tex]f(x) = 4 - 4(x+1) + 3(x+1)^2 - 9/2(x+1)^3 + ...[/tex]

b.  The expanded form of the Taylor series is [tex]f(x) = -9/2x^3 - 27/2x^2 - 15x + 29 + ...[/tex]

(a) To find the Taylor series of f(x) centered at x = -1, we first need to compute the derivatives of f(x) at x = -1:

[tex]f(x) = -3x^2 + 2x + 5.[/tex]

f'(x) = -6x + 2

f''(x) = -6

f'''(x) = 0

f''''(x) = 0

Since all the higher-order derivatives are zero, the Taylor series of f(x) centered at x = -1 is simply:

[tex]f(x) = f(-1) + f'(-1)(x+1) + (1/2!)f''(-1)(x+1)^2 + (1/3!)f'''(-1)(x+1)^3 + ...[/tex]

Plugging in the values of f(-1), f'(-1), f''(-1), and f'''(-1) gives us the first few terms of the series:

[tex]f(x) = 4 - 4(x+1) + 3(x+1)^2 + ...[/tex]

The first four terms are:

f(x) = 4 - 4(x+1) + 3(x+1)^2 - 9/2(x+1)^3 + ...

(b) To expand the series, we simply need to distribute and simplify each term:

[tex]f(x) = 4 - 4(x+1) + 3(x^2 + 2x + 1) - 9/2(x^3 + 3x^2 + 3x + 1) + ...[/tex]

[tex]f(x) = 4 - 4x - 1 + 3x^2 + 6x + 3 - 9/2x^3 - 27/2x^2 - 27/2x - 9/2 + ...[/tex]

Simplifying further gives:

[tex]f(x) = -9/2x^3 - 27/2x^2 - 15x + 29 + ...[/tex]

So the expanded form of the Taylor series is [tex]f(x) = -9/2x^3 - 27/2x^2 - 15x + 29 + .....[/tex]

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True. If f is a polynomial of degree n and c is a nonzero scalar, then cf is a polynomial of degree n.

A polynomial is an expression consisting of variables raised to non-negative integer powers, multiplied by coefficients. The degree of a polynomial is the highest power of the variable in the polynomial.

If f is a polynomial of degree n, it means that the highest power of the variable in f is n. When we multiply f by a nonzero scalar c, each term in f is multiplied by c, including the term with the highest power of the variable. This means that the highest power of the variable in cf will also be n, since c multiplied by the highest power of the variable in f will result in the same power.

Therefore, cf is a polynomial of degree n, as the highest power of the variable remains unchanged after multiplying f by the nonzero scalar c.

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Mark has left with 44 inches. The correct option is C.

An object or event's attributes are quantified through measurement so that they can be compared to those of other things or occurrences.

To solve this problem, we need to first convert the measurements to a common unit. Let's convert everything to inches:

To find out how much rope Mark had left after cutting off 4 feet 6 inches, we need to subtract 54 inches from 98 inches:

98 inches - 54 inches = 44 inches

Therefore, the answer is C. 44 inches.

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The answer of the given question based on the population is , the correct option is B) a parameter.

In statistics, a population is a group or set of individuals, objects, events, or measurements that share at least one common characteristic. This characteristic is usually a variable or a set of variables that the researcher is interested in studying or measuring. For example, the population might be all the adults living in a particular city, or all the trees in a particular forest.

B) a parameter.

A parameter is  value that describes  characteristic of  entire population. It is typically computed from the information obtained from sample of  population, but it is used to describe  entire population. For example, mean income of all households in city is a parameter.

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The positive difference between the two solutions of the quadratic equation [tex]2x^{2}[/tex] + 5x + 12 = 19 -7x is [tex]\frac{\sqrt{200} }{4}[/tex].

We are required to determine the positive difference between the two solutions of the given quadratic equation: [tex]2x^{2}[/tex] + 5x + 12 = 19 -7x

1. Move all terms to the left side of the equation to form a standard quadratic equation:

[tex]2x^{2}[/tex] + 5x + 12 + 7x - 19 = 0

2. Simplify the equation: [tex]2x^{2}[/tex] + 12x - 7=0.

3. Use the quadratic formula to find the solutions for x:

[tex]x = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a}[/tex]

where a=2, b=12, and c=-7.

4. Substitute the values:

[tex]x = \frac{-12 \pm \sqrt{12^{2} -4(2)(-7)}}{2(2)}[/tex]

5. Simplify the expression:

[tex]x = \frac{-12 \pm \sqrt{144 + 56}}{4}[/tex]

6. Calculate the value under the square root:

[tex]x = \frac{-12 \pm \sqrt{200}}{4}[/tex]

7. Now, we have two solutions:

[tex]x_{1} = \frac{-12 + \sqrt{200}}{4}x_{2} = \frac{-12 - \sqrt{200}}{4}[/tex]

8. Find the difference between the solutions:

[tex]x_{1} - x_{2}[/tex] = [tex]\frac{\sqrt{200} }{4}[/tex]

The positive difference between the two solutions is[tex]\frac{\sqrt{200} }{4}[/tex].

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An illustration of comparative research is comparing the population of two distinct states to investigate the prevalence of depression.

Cross-sectional exploration, then again, includes gathering information from a populace at a particular moment, with practically no examination between various gatherings or factors.

Comparative research seeks to identify similarities and differences between two or more groups or variables. This is an example of comparative research because the prevalence of depression is being compared between two distinct states.

Longitudinal research involves collecting data from the same population over an extended period of time, to track changes or patterns over time.

In contrast, archival research entails answering research questions by utilizing existing data sources like historical records or documents. It does not require new data to be gathered from a population.

Because it compares the prevalence of depression in two distinct groups (individuals from two distinct states), comparing the populations of two distinct states to investigate the prevalence of depression is an example of comparative research.

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Answer:

2($3.95 + $4.75)(1.05) = $18.27

The number of men living in the village is 260.

A collection of many linear equations that include the same variables is referred to as a system of linear equations. A linear equation system is often composed of two or more linear equations with two or more variables.A linear equation with two variables, x and y, has the following general form:

                                           [tex]ax + by = c[/tex]

Given:

Total inhabitants in the village: 817

Number of children: 241

There are 56 more women than men in the village

Total adults = Total inhabitants - Number of children

Total adults = 817 - 241

Total adults = 576

Let number of men in the village be 'x' and number of women in the village be 'y',

∴ y=x+56 (given) ..................(1)

Also, x+y=576 .................(2)

From equation (1) and (2),

x + (x + 56) = 576

2x + 56 = 576

2x = 576 - 56

2x = 520

x = 520 / 2

x = 260

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The derivative of the expression y = ln ((15x² + 4x)⁵ (4x² – 2x)) is (120x - 10)/(3x + 1) (15x² + 4x)⁴.

To find the derivative of this expression, we'll need to use the chain rule and product rule. The chain rule allows us to find the derivative of a function inside another function, while the product rule allows us to find the derivative of two functions multiplied together.

Let's start by using the product rule to differentiate the expression inside the natural logarithm:

(15x² + 4x)⁵ (4x² – 2x) = f(x)g(x)

f(x) = (15x² + 4x)⁵ g(x) = (4x² – 2x)

f'(x) = 5(15x² + 4x)⁴ (30x + 4)

Now, we can use the product rule:

y = ln(f(x)g(x))

y' = 1/(f(x)g(x)) (f(x)g'(x) + g(x)f'(x))

y' = 1/((15x² + 4x)⁵ (4x² – 2x)) ((15x² + 4x)⁵(8x-2) + (4x² - 2x)5(15x² + 4x)⁴ (30x + 4))

Simplifying, we get:

y' = (120x - 10)/(3x + 1) (15x² + 4x)⁴

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The integration of the above equation is: [tex]∫e^x/1+e^x dx = ln|1 + e^x| + C[/tex]

For problem 4 using the substitution method, we will substitute a new variable for the part of the integral that is causing difficulty.

(a) For ∫5х/x2 + 1 dx, let u = x2 + 1. Then du/dx = 2x and dx = du/2x. Substituting this in the integral, we get:

∫5х/x2 + 1 dx = ∫5/(2u) du

Now, we can solve this integral easily as:

∫5/(2u) du = (5/2)ln|u| + C

Substituting back u = x2 + 1, we get:

∫5х/x2 + 1 dx = (5/2)ln|x2 + 1| + C

(b)[tex]For ∫(3t^2 – 1)e^t3-t dt, let u = t^3 - t. Then  du/dt = 3t^2 - 1 and dt = du/(3t^2 - 1). Substituting this in the integral, we get:[/tex]

[tex]∫(3t^2 – 1)e^t3-t dt = ∫e^u du/3[/tex]

Solving this integral, we get:

[tex]∫e^u du/3 = (1/3)e^u + C[/tex]

Substituting back u = t^3 - t, we get:

[tex]∫(3t^2 – 1)e^t3-t dt = (1/3)e^(t^3 - t) + C[/tex]

(c) For ∫ln(x)/x dx, let u = ln(x). Then du/dx = 1/x and dx = x du. Substituting this in the integral, we get:

[tex]∫ln(x)/x dx = ∫u du[/tex]

Solving this integral, we get:

∫u du = (1/2)u^2 + C = (1/2)ln^2(x) + C

Substituting back u = ln(x), we get:

∫ln(x)/x dx = (1/2)ln^2(x) + C

(d) For [tex]∫e^x/1+e^x dx, let u = 1 + e^x[/tex]. Then [tex]du/dx = e^x and dx = du/e^x.[/tex] Substituting this in the integral, we get:

[tex]∫e^x/1+e^x dx = ∫du/u[/tex]

Solving this integral, we get:

[tex]∫du/u = ln|u| + C = ln|1 + e^x| + C[/tex]

Substituting back u = 1 + e^x, we get:

[tex]∫e^x/1+e^x dx = ln|1 + e^x| + C[/tex]

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Therefore, the number of common tangents between a pair of circles depends on their relative positions. Without knowing the specific positions of the circles, it is not possible to determine the number of common tangents.

To determine the number of common tangents between a pair of circles, we need to consider their relative positions.

If the circles do not intersect or touch each other, there are 4 common tangents - 2 external tangents and 2 internal tangents.

If the circles touch each other externally, there are 3 common tangents - 1 common external tangent and 2 internal tangents.

If the circles touch each other internally, there are 3 common tangents - 1 common internal tangent and 2 external tangents.

If the circles intersect each other at two distinct points, there are 2 common tangents - each passing through one of the points of intersection.

If the circles coincide, there are infinitely many common tangents.

Therefore, the number of common tangents between a pair of circles depends on their relative positions. Without knowing the specific positions of the circles, it is not possible to determine the number of common tangents.

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The probability that a random sample of 100 bulbs will have an average life of more than 1,825 hours is approximately 0.1056, or 10.56%.

To find the probability that a random sample of 100 bulbs will have an average life of more than 1,825 hours, we need to use the central limit theorem.

First, we need to calculate the standard error of the mean, which is the standard deviation of the population (200 hours) divided by the square root of the sample size (100):

standard error of the mean = 200 / √(100) = 20

Next, we need to standardize the sample mean using the formula:

z = (x - mu) / SE

where x is the sample mean (1,825 hours), mu is the population mean (1,800 hours), and SE is the standard error of the mean (20 hours).

z = (1825 - 1800) / 20 = 1.25

Finally, we need to find the probability that a standard normal distribution is greater than 1.25. We can use a standard normal distribution table or calculator to find this probability, which is approximately 0.1056.

Therefore, the probability that a random sample of 100 bulbs will have an average life of more than 1,825 hours is approximately 0.1056, or 10.56%.

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The equation is written as ; <SVT = 90/2

The measure of the angle SVT = 45 degrees

To determine the value of the angle, we need to take into considerations the properties of a right-angled triangle.

These properties are;

From the information given, we have that;

RVT is a right -angled triangle

Since the diagonal SV bisects the angle 90 degrees into two equal halves, then,

<SVT = 90/2

Divide the angle

<SVT = 45 degrees

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The black population is approximately 13% of the U.S. population because of the census done by the Census Bureau . Option b.

According to the U.S. Census Bureau, the black or African American population in the United States was estimated to be approximately 13.4% of the total population in 2020. This means that out of every 100 people in the U.S., about 13 or 14 are black or African American.

The black population has a long and complex history in the U.S., including periods of slavery, segregation, and discrimination.

Despite ongoing challenges and inequalities, the black community has made significant contributions to American culture, politics, and society. Understanding the demographics of the U.S. population, including the proportion of black individuals, is important for a range of policy and social issues.

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There is about a 33.7% probability that a case of 48 cans will contain at least two cans with a defective ball.

To solve this problem, we can use the binomial distribution. Let's define "success" as getting a can with no defective ball and "failure" as getting a can with at least one defective ball.

The probability of success in one can is:

P(success) = 1 - P(failure) = 1 - 0.025 = 0.975

The probability of failure in one can is:

P(failure) = 0.025

Now, let's define X as the number of cans in a case of 48 that have at least one defective ball. We want to find the probability that X is greater than or equal to 2.

We can use the binomial distribution formula to calculate this probability:

P(X ≥ 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1)

P(X = 0) = (0.975)^48 ≈ 0.223

P(X = 1) = 48C1 (0.975)^47 (0.025)^1 ≈ 0.44

where 48C1 is the number of ways to choose one can out of 48.

Therefore, the probability that a case of 48 cans will contain at least two cans with a defective ball is:

P(X ≥ 2) ≈ 1 - 0.223 - 0.44 ≈ 0.337

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The derivative of f(x) = -4x with respect to x, or dy/dx, is -4.

To find the derivative of f(x) = -4x using first principles.
First, let's recall the definition of the derivative using first principles:
f'(x) = lim (h → 0) [(f(x + h) - f(x))/h]
Now, substitute f(x) = -4x into the definition:
f'(x) = lim (h → 0) [(-4(x + h) - (-4x))/h]
Next, distribute -4 to both x and h in the numerator:
f'(x) = lim (h → 0) [(-4x - 4h + 4x)/h]
Simplify the expression by canceling out -4x and +4x:
f'(x) = lim (h → 0) [(-4h)/h]
Cancel out h in the numerator and denominator:
f'(x) = lim (h → 0) [-4]

Since -4 is a constant and doesn't depend on h, the limit is simply:
f'(x) = -4.

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The vertex of the quadratic equation is (-1, -1) and the axis of symmetry is:

x = -1

For a general quadratic equation:

y = ax² + bx + c

The vertex is at:

x = - b/2a

Here the quadratic is:

y = -9x² - 18x - 1

So the x-value of the vertex is_:

x = 18/(2*-9) = -1

Evaluating the quadratic in that we get:

y = -9*(-1)² - 18*-1 - 1

y = -1

So the vertex is at (-1, -1)

And the axys of symetry is a line:

x = x-value of the vertex = -1

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The probability of getting two heads and one tail is 3/8.
The probability of getting three tails is 1/2 * 1/2 * 1/2 = 1/8.

The probability that Joy's sibling is a brother is 1/2.

(i) To find the probability of getting two heads and one tail, we need to consider the number of possible outcomes where two heads and one tail can occur. The possible outcomes are HHT, HTH, and THH. Each of these outcomes has a probability of 1/2 * 1/2 * 1/2 = 1/8.
(ii) To find the probability of getting three tails, we need to consider the number of possible outcomes where three tails can occur. There is only one outcome where three tails can occur, which is TTT.


If the neighbour has a son named Joy, there are two possibilities for the gender of the other child: it could be a boy or a girl. However, we know that Joy is a boy, so the only possibility for his sibling to be a brother is if the other child is also a boy.

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For fx=ln(x^4x): f(x) = ln(x^4x) = 5ln(x), f'(x) = 5/x, f''(x) = -5/x^2. For fx=2e^-x^2: f(x) = 2e^-x^2, f'(x) = -4xe^-x^2, f''(x) = (8x^2 - 4)e^-x^2. Using chain rule and product rule, we can find the solutions.

To find the second derivative f''(x) for the given functions, we'll first find the first derivative f'(x) and then derive it again.

For the first function, f(x) = ln(x^4/x):
1. Simplify: f(x) = ln(x^3)
2. Find f'(x) using the chain rule: f'(x) = (1/x^3) * 3x^2 = 3/x
3. Find f''(x): f''(x) = -3/x^2

For the second function, f(x) = 2e^(-x^2):
1. Find f'(x) using the chain rule: f'(x) = 2(-2x)e^(-x^2) = -4xe^(-x^2)
2. Find f''(x) using the product rule: f''(x) = -4e^(-x^2) - 4x(-2x)e^(-x^2) = -4e^(-x^2) + 8x^2e^(-x^2)

So, the second derivatives are f''(x) = -3/x^2 for the first function and f''(x) = -4e^(-x^2) + 8x^2e^(-x^2) for the second function.

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The value of C is calculated to be -6.

If F(x) = x + 9x + C and F(1) = 4, we can use the given information to solve for C by using Green's theorem;

F(1) = 1 + 9(1) + C = 4

Simplifying, we get:

C = 4 - 1 - 9 = -6

Therefore, the value of C is -6.

If F(x) = x + 9x + C and F(1) = 4, we can still use the given information to solve for C:

F(1) = 1 + 9(1) + C = 4

Simplifying, we get:

C = 4 - 1 - 9 = -6

Now, we can use the value of C to simplify the expression for F(x):

F(x) = x + 9x - 6x^2

Simplifying further, we get:

F(x) = -6x^2 + 10x

Therefore, FC(x) = x² FC = -6x² + 10x² = 4x².

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The given statement "The random sample of 500 park users indicated that the average number of visits per month was 4.56. This is a statistic by the city council" is true because it is a numerical measure.

In statistics, a statistic is a numerical measure that summarizes a sample of data. It is used to estimate or infer the characteristics of the population from which the sample was drawn. The value of a statistic is calculated from the sample data and is subject to random variability due to sampling error.

In this case, the Parks and Recreation manager for the city of Detroit has reported that a random sample of 500 park users indicated an average of 4.56 visits per month. This value is calculated from the sample data and represents a statistic, as it is based on a sample and is subject to sampling variability.

The city council should view this value as a statistic and not as a parameter, which is a numerical measure that describes a characteristic of a population.

While the sample statistic can be used to make inferences about the population parameter, it is important to recognize that the sample statistic is subject to random variability and may not perfectly represent the population parameter.

Therefore, the statement that the value of 4.56 visits per month should be viewed as a statistic by the city council is true.

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Answer: We can start by using algebra to solve the problem. Let x and y be the other two numbers.

Step-by-step explanation: average = (sum of numbers) / (number of numbers)

Substituting the given values, we get:

18 = (21 + x + y) / 3

Multiplying both sides by 3, we get:

54 = 21 + x + y

Subtracting 21 from both sides, we get:

x + y = 33

Therefore, the sum of the three numbers is:

21 + x + y = 21 + 33 = 54

So the sum of the three numbers is 54.

Ans:

Sure! Let's solve the problem step by step:

step-1

We are given that the average of three numbers is 18. Let's call these three numbers a, b, and c. Then, we can write:

(a + b + c)/3 = 18

step-2

We want to find the sum of the remaining two numbers if one of them is 21. Let's assume that a = 21. Then, we have:

(21 + b + c)/3 = 18

step-3

We can simplify this equation by multiplying both sides by 3:

21 + b + c = 54

step-4

Now, we can solve for b + c by subtracting 21 from both sides:

b + c = 33

step-5

Therefore, the sum of the remaining two numbers is 33.

So, the steps to solve this problem are:

Write the equation for the average of the three numbers.

Assume one of the numbers is 21.

Rewrite the equation using this assumption.

Simplify the equation.

Solve for the sum of the remaining two numbers by isolating them on one side of the equation.

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The local minimum of the function f(x, y) is (-1.957, 0.391) with a value of -241.427.

To find the local maxima, minima and saddle points of the function f(x, y), we need to follow these steps:

Find the partial derivatives of f(x, y) with respect to x and y.

Set these partial derivatives equal to zero and solve for x and y to find the critical points.

Find the second partial derivatives of f(x, y) with respect to x and y.

Evaluate these second partial derivatives at each critical point.

Use the second partial derivatives to determine the nature of each critical point (whether it is a local maximum, minimum, or saddle point).

Let's follow these steps:

Find the partial derivatives of f(x, y) with respect to x and y.

[tex]f_x = 26x + 5y + 99[/tex]

[tex]f_y = 10y + 5x + 4[/tex]

Set these partial derivatives equal to zero and solve for x and y to find the critical points.

26x + 5y + 99 = 0

10y + 5x + 4 = 0

Solving these equations simultaneously, we get:

x = -1.957

y = 0.391

Find the second partial derivatives of f(x, y) with respect to x and y.

[tex]f_xx = 26[/tex]

[tex]f_xy = 5[/tex]

[tex]f_yy = 10[/tex]

Evaluate these second partial derivatives at each critical point.

At (-1.957, 0.391), we have:

[tex]f_xx = 26[/tex]

[tex]f_xy = 5[/tex]

[tex]f_yy = 10[/tex]

Use the second partial derivatives to determine the nature of each critical point.

Let's compute the discriminant[tex]D = f_xx * f_yy - (f_xy)^2[/tex] at the critical point:

[tex]D = (26 * 10) - (5^2) = 255[/tex]

Since D > 0 and[tex]f_xx[/tex]  > 0 at the critical point, we conclude that (-1.957, 0.391) is a local minimum of f(x, y).

Therefore, the function f(x, y) has only one critical point which is a local minimum at (-1.957, 0.391), and there are no saddle points.

The value of the function at the critical point is:

[tex]f(-1.957, 0.391) = 13(-1.957)^2 + 5(-1.957)(0.391) + 8(0.391)^2 + 99(-1.957) + 4(0.391) + 17 = -241.427[/tex]

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Using Kepler's Third Law, we can find the distance of the exo-planet from its star. The formula is P^2 = a^3, where P is the orbital period in years and a is the average distance from the planet to the star in astronomical units (AU).

the average distance of Le Grosse Homme from its star is approximately 2 AU, which is answer choice b).

To determine the distance of the exo-planet Le Grosse Homme to the star, we will use Kepler's Third Law, which states that the square of the orbital period (P) of a planet is proportional to the cube of the semi-major axis (a) of its orbit. The formula is given as:

P² = a³

In this case, the orbital period P is 2.8284 years. We can now solve for the semi-major axis (a), which represents the distance from the planet to the star in astronomical units (AU).

Step 1: Square the orbital period
(2.8284)² = 7.99968336

Step 2: Find the cube root of the squared orbital period to get the distance (a)
a = ∛(7.99968336) ≈ 2.0 AU

So, the distance of the exo-planet Le Grosse Homme to the star is approximately 2.0 AU. The correct answer is option (b) ~2.0 AU.

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The function  sin x + cos y = 2y does not have critical points and there are there are no relative maxima or minima.

To find the critical points, we need to find the values of x and y such that the partial derivatives are equal to zero:

∂/∂x(sin x + cos y) = cos x = 0

∂/∂y(sin x + cos y) = -sin y = 2

From the first equation, we get cos x = 0, which means x = π/2 + nπ, where n is an integer.

From the second equation, we get -sin y = 2, which has no real solutions. Therefore, there are no critical points.

Since there are no critical points, there are no relative maxima or minima.

Hence, the function  sin x + cos y = 2y does not have critical points and there are there are no relative maxima or minima.

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The probability that exactly four of them are blue ones is 38.073%.

To solve this problem, we can use the formula for calculating the probability of an event:
P(event) = (number of ways the event can occur) / (total number of possible outcomes)

In this case, we want to calculate the probability of launching exactly four blue fireworks out of seven. We can use the combination formula to find the number of ways this can occur:

C(8,4) = 8! / (4! * (8-4)!) = 70

This means there are 70 ways to choose four blue fireworks out of the remaining eight.

Similarly, we can find the number of ways to choose the remaining three fireworks from the seven red ones:

C(7,3) = 7! / (3! * (7-3)!) = 35

Therefore, the total number of ways to choose seven fireworks out of the remaining fifteen is:

C(15,7) = 15! / (7! * (15-7)!) = 6435

To find the probability of launching exactly four blue fireworks out of seven, we can plug in these values into the formula:

P(4 blue out of 7) = (number of ways to choose 4 blue and 3 red) / (total number of ways to choose 7)

P(4 blue out of 7) = (70 * 35) / 6435 = 490/1287

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a. The inequality is satisfied when 2 < x < 4, so the interval on which f is decreasing is: (-∞, 2) U (4, ∞)

b. The inequality is satisfied when 2 < x < 4, so the interval on which f is decreasing is: (2, 4)

c. The local minimum value of f is f(4) = 9, and the local maximum value of f is f(2) = 23.

(a) To find the intervals on which f is increasing, we need to find where the derivative of f is positive.

So we first find the derivative:

[tex]f'(x) = 6x^2 - 36x + 48[/tex]

Now we solve for where f'(x) > 0:

[tex]6x^2 - 36x + 48[/tex] > 0

[tex]x^2 - 6x + 8[/tex] > 0

(x-2)(x-4) > 0

The inequality is satisfied when x < 2 or x > 4, but since the sign of f'(x) changes at x=2 and x=4,

we have two separate intervals on which f is increasing:

(-∞, 2) U (4, ∞)

(b) To find the intervals on which f is decreasing, we need to find where the derivative of f is negative.

So we look for where f'(x) < 0:

[tex]6x^2 - 36x + 48[/tex] < 0

[tex]x^2 - 6x + 8[/tex] < 0

(x-2)(x-4) < 0

The inequality is satisfied when 2 < x < 4, so the interval on which f is decreasing is: (2, 4)

(c) To find the local maximum and minimum values of f, we need to find the critical points of f, which are the values of x where f'(x) = 0 or where f'(x) does not exist.

[tex]f'(x) = 6x^2 - 36x + 48 = 6(x-2)(x-4)[/tex]

So f'(x) = 0 when x = 2 or x = 4.

We also need to check the endpoints of the intervals where f is increasing or decreasing.

At x = 2, f''(x) = 12x - 36 = -12 < 0, so x = 2 is a local maximum.

At x = 4, f''(x) = 12x - 36 = 12 > 0, so x = 4 is a local minimum.

Finally, we check the endpoints of the intervals where f is increasing or decreasing.

When x approaches negative infinity, f(x) approaches infinity, so there is no local minimum.

When x approaches positive infinity, f(x) approaches infinity, so there is no local maximum.

Therefore, the local minimum value of f is f(4) = 9, and the local maximum value of f is f(2) = 23.

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The change in net worth of the company between 2020 and 2030 is $16,000

The net worth of the company in 2030 is $56,000.

The net worth f(t) of a company is growing at a rate f'(t) = [tex]2000 - 12t^2[/tex]dollars per year, where t is in years since 2020.

To determine how the net worth of the company is expected to change between 2020 and 2030, we can integrate the rate of growth function over the interval [0, 10], which gives us:

∫[0,10] f'(t) dt = ∫[0,10] [tex](2000 - 12t^2)[/tex] dt = [tex][2000t - 4t^3][/tex] from 0 to 10

= [tex](200010 - 4\times10^3)[/tex] - (0 - 0) = 20000 - 40000 = -20000

This negative result indicates that the net worth of the company is expected to decrease between 2020 and 2030.

If the company is worth $40,000 in 2020, then its net worth in 2030 can be found by adding the change in net worth to the initial value of $40,000.

Therefore:

Net worth in 2030 = $40,000 + (-$20,000) = $20,000

This means that the net worth of the company is expected to be $20,000 in 2030, which is a significant decrease from its initial value of $40,000 in 2020.

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As sample size decreases, the degrees of freedom decrease.

This affects both the group and error degrees of freedom. Specifically:

The group degrees of freedom (df between) decrease as the number of groups decreases. This is because there are fewer groups to estimate the population means from, resulting in fewer degrees of freedom for group differences.

The error degrees of freedom (df within) decrease as the sample size within each group decreases. This is because there is less information available to estimate the variation within each group, resulting in fewer degrees of freedom for residual variation.

The total degrees of freedom (df total) also decrease as the number of observations decreases, since the total degrees of freedom is equal to the number of observations minus 1.

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