Question 18 (6 marks) Suppose that f is differentiable on R and f'(x) = e^{x2-4x+3} – 1 for all r ∈ R. Determine all intervals on which f is increasing and all intervals on which f is decreasing.

An equation for the graph : 8 7 6 S 4 3 2 1 -5 -4 -3 -2 -1 2 3 2 3 -5 -6 -8 f(x) = -(1/4)(x + 2)² + 3

The long run behavior of f(n) = 3 - (:)" + 3 is determined by the highest degree term in the expression, which is n². As n becomes very large (either positively or negatively), the n² term dominates the expression and the other terms become relatively insignificant. Therefore, as n → -∞, f(n) → -∞ and as n → ∞, f(n) → -∞.

To find an equation for the graph sketched below, we need to first identify the key characteristics of the graph. We can see that it is a parabolic curve that opens downwards and has its vertex at (-2, 3). Using this information, we can write an equation in vertex form:

f(x) = a(x - h)² + k

where (h, k) is the vertex and a determines the shape of the curve. Plugging in the values we have, we get:

f(x) = a(x + 2)² + 3

To determine the value of a, we can use another point on the curve, such as (0, 2):

2 = a(0 + 2)² + 3
-1 = 4a
a = -1/4

Plugging this value back into our equation, we get:

f(x) = -(1/4)(x + 2)² + 3

This is the equation for the graph sketched below.

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The point which represents the number of dollars Carola makes per hour is (1, 8).

Two variables have a proportional relationship if all the ratios of the variables are equivalent.

The constant of proportionality is the ratio of the y value (total earnings) to the x value (hours worked). That is:

constant of proportionality (k) = y/x

Since Carola works for 6 hours and earns $48. Thus, the number of dollars Carola makes in 1 hour will be:

48/6 = $8

Therefore, the point which represents the number of dollars Carola makes per hour is (1, 8)

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It is true that the When data collection involves online surveys, the process of data validation involves examining if instructions were followed precisely.

This is because online surveys often have specific instructions that respondents are expected to follow. Data validation ensures that the data collected is accurate, reliable, and valid. It involves checking for errors or inconsistencies in the data, as well as making sure that respondents have answered all questions correctly. This process helps to ensure that the data collected is of high quality and can be used for analysis and decision-making purposes. Additionally, data validation can also help to identify areas where improvements can be made to the survey design or data collection process.

Data validation in online surveys refers to ensuring the accuracy and quality of the collected data. It involves checking for inconsistencies, errors, and incomplete responses. While following instructions precisely is important, data validation focuses on data accuracy, preventing duplicate responses, and verifying if respondents meet the target demographic. It aims to improve the reliability of the data and reduce the margin of error, ensuring meaningful conclusions can be drawn from the results.

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The tangent line has the largest slope at x = √(32/5), with a slope of approximately 254.86.

To find the point on the line 3x + 5y + 5 = 0 which is closest to the point (5,2), we need to use the formula for the distance between a point and a line. The formula is:

distance = |ax + by + c| / √(a² + b²)

where (a,b) is the direction vector of the line and (x,y) is any point on the line. In this case, the direction vector is (3,5) and we can find a point on the line by setting y = 0:

3x + 5(0) + 5 = 0
x = -5/3

So the point on the line closest to (5,2) is:

distance = |3(-5/3) + 5(0) + 5| / √(3² + 5²) = 4 / √34

To find the value(s) of x on the curve y = -7 + 160x³ - 3x^5 where the tangent line has the largest slope, we need to find the derivative of y with respect to x and set it equal to zero:

y' = 480x² - 15x^4
480x² - 15x^4 = 0
x²(32 - 5x²) = 0
x = 0 or x = ±√(32/5)

We can now find the slope of the tangent line at each of these values of x:

slope at x = 0: y' = 0, so the tangent line is horizontal and has slope 0
slope at x = √(32/5): y' = 480(32/5) - 15(32/5)³ ≈ 254.86
slope at x = -√(32/5): y' = 480(32/5) - 15(-32/5)³ ≈ -254.86

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The value of x in the given equation by using the graphical method to the nearest whole number is 2.

Quadratic equations are algebraic equations that have their highest power raised to the power of two. There are different methods by which we can solve quadratic equations.

Some of the methods for solving quadratic equations include:

Here, we have 16^x = 64x + 4, to solve this type of equation, we will need to graph both sides of the equation on the graph, by doing so; we have the value of x at the point of intersection of both axis as:

x = −0.0489, 1.7055

Since x cannot be negative, the value of x to the nearest whole number is 2.

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Answer:

To find the mode, we look for the number that appears most frequently in the set:

The mode is 34 because it appears twice, while all other numbers appear only once.

To find the median, we need to arrange the numbers in numerical order:

13, 13, 26, 34, 34, 41, 53, 61, 69

The median is 34, which is the middle number when the set is arranged in order.

To find the mean, we add up all the numbers and divide by the total number of numbers:

(53 + 13 + 34 + 41 + 26 + 61 + 34 + 13 + 69) / 9 = 35.89 (rounded to two decimal places)

The mean is approximately 35.89.

To find the range, we subtract the smallest number from the largest number:

69 - 13 = 56

The range is 56.

Therefore, the mode is 34, the median is 34, the mean is approximately 35.89, and the range is 56.

Answer:

Step-by-step explanation:

mean is 35.89

the median is 34

the mode is 34

and the rang is 56

The energy required to change the separation between the centers of the basketballs to 13 m: 1.44 x 10^-12 J

To calculate the energy required to change the separation between the centers of the basketballs, we can use the formula for the potential energy of two point masses:

U = -G(m1m2)/r

where U is the potential energy, G is the gravitational constant, m1 and m2 are the masses of the basketballs, and r is the separation between their centers.

For the first case, where the separation between the centers is changed from the sum of their radii (0.28 m) to 1.1 m, we have:

r = 1.1 - 0.28 = 0.82 m

Plugging in the values, we get:

U = -6.67 x 10^-11 x 0.55^2 / 0.82 = -2.62 x 10^-10 J

Therefore, 2.62 x 10^-10 J of energy is required to change the separation between the centers of the basketballs to 1.1 m.

For the second case, where the separation between the centers is changed to 13 m, we have:

r = 13 - 0.28 = 12.72 m

Plugging in the values, we get:

U = -6.67 x 10^-11 x 0.55^2 / 12.72 = -1.44 x 10^-12 J

Therefore, 1.44 x 10^-12 J of energy is required to change the separation between the centers of the basketballs to 13 m.

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Answer:

Step-by-step explanation:

finding the smallest possible distance from the line to the origin follows as

the normal vector=(-1,-2,1)

using direction vector we need to create the equation of the plane

-1(x-2)-2(y-3)+1(z+7)=0

we get;

-x+2-2y+6+z+7=0

-x-2y+z+13=0

x=2-t; y=2-3t; z=-7+t

on substituting;

-1(2-t)-2(3-2t)+1(-7+t)+13=0

-2+t-6+4t-7+t+13=0

we get;

t=1

so point is(1,1,-6)

distance from point to origin

d=[tex]\sqrt{(1^2+1^2+(-6)^2)}[/tex]=[tex]\sqrt{38}[/tex]

therefore

the answer is [tex]\sqrt{38}[/tex]=6.16

Both B and C could be the type of function that exhibits the behavior, the average rate of change for a function is greater from x = 1 to x = 2 than

from x = 2 to x = 3.

Given that,

the average rate of change is greater from x = 1 to x = 2 than from x = 2 to x = 3, this implies that the slope of the function is steeper in the interval [1,2] than in the interval [2,3]. This means that the function is getting steeper as x increases, but the

rate of increase is decreasing.

The following types of functions can exhibit this

behavior:

A. Linear: A linear function has a constant slope, so it cannot exhibit this behavior.

B. Quadratic: A quadratic function has a changing slope, so it can exhibit this behavior. However, it depends on the specific coefficients of the quadratic equation.

C. Exponential: An exponential function can also exhibit this behavior. For example, y =

[tex]2^x[/tex]

has a steeper slope for smaller values of x, but the rate of increase slows down as x gets larger.

Therefore, both B and C could be the type of function that exhibits this behavior.

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The following are the measures for the vertical angles:

13). m∠SYX = 76°, (14). m∠XYW = 40°, (15). m∠WYV = 64°, (16). m∠SYW = 116°, (17). m∠TYX = 140°, and (18). m∠VYX = 104°

Vertical angles also called vertically opposite angles are formed when two lines intersect each other, the opposite angles formed by these lines are vertically opposite angles and are equal to each other.

We shall evaluate for the measure of the angles as follows:

13). m∠SYX = m∠UYV = 76°

(14). m∠XYW = m∠UTY = 40°

(15). m∠WYV = m∠SYT = 64°

(16). m∠SYW = m∠TYV = 76° + 40° = 116°

(17). m∠TYX = m∠UYW = 76° + 64° = 140°

(18). m∠VYX = m∠SYU = 64° + 40° = 104°

Therefore, the measures for the vertical angles are:

13). m∠SYX = 76°, (14). m∠XYW = 40°, (15). m∠WYV = 64°, (16). m∠SYW = 116°, (17). m∠TYX = 140°, and (18). m∠VYX = 104°

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Increasing the sample size is an effective way to reduce the impact of extreme/outlying values and obtain a more accurate representation of the population.

As the sample size, n, increases, we expect the likelihood of selecting cases or members with extreme/outlying values to decrease. This is because as the sample size increases, the data becomes more representative of the population and the distribution of the data becomes more normal. Therefore, extreme/outlying values become less likely to be included in the sample as they are less representative of the overall population.

For example, if we were to take a small sample size of 10 individuals from a population of 100, there is a higher chance that the sample may include an individual with an extreme value such as an unusually high or low income. However, if we were to take a larger sample size of 100 individuals, the sample would be more representative of the overall population and the extreme values would be less likely to be included.

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The function f is:

f(x) = [tex](4/9)x^{(5/3)} + (6/11)x^{(11/6)} + 62/99[/tex]

To find f, we need to integrate f'(x) with respect to x.

f'(x) = √x (2 + 3x)

Integrating both sides:

f(x) = ∫√x (2 + 3x) dx

Using substitution, let u = [tex]x^{(3/2)[/tex], then du/dx = [tex](3/2)x^{(1/2)[/tex], which means dx = [tex]2/3 u^{(2/3)[/tex] du

Substituting u and dx, we get:

f(x) = ∫[tex](2u^{(2/3)} + 3u^{(5/6)}) (2/3)u^{(2/3)[/tex] du

Simplifying:

f(x) = [tex](4/9)u^{(5/3)} + (6/11)u^{(11/6)} + C[/tex]

Substituting back u = [tex]x^{(3/2)[/tex] and f(1) = 3:

3 = [tex](4/9)1^{(5/3)} + (6/11)1^{(11/6)} + C[/tex]

Simplifying:

C = 3 - 4/9 - 6/11

C = 62/99

Therefore, the function f is:

f(x) = [tex](4/9)x^{(5/3)} + (6/11)x^{(11/6)} + 62/99[/tex]

Thus, we have found the function f.

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a) The cutoff value for the highest 101% of these funds is 11.2%. b) The cutoff value for the lowest 20% of these funds is 0.5%. c) The cutoff values for the middle 40% of these funds are 0.8% and 2.7%. d) The cutoff value for the highest 80% of these funds is 200%

To find the cutoff values for different percentages of mutual funds, we need to use the properties of the standard normal distribution. We can convert the given Normal model to a standard normal distribution by subtracting the mean and dividing by the standard deviation:

Z = (X - μ) / σ

where X is a random variable from the Normal model N(μ, σ), Z is the corresponding standard normal variable, μ = 0.015 is the mean of the model, and σ = 0.043 is the standard deviation of the model.

a) To find the cutoff value for the highest 101% of these funds, we need to find the Z-score that corresponds to the 101st percentile of the standard normal distribution. We can use a standard normal table or calculator to find this value, which is approximately 2.33. Then we can use the formula for Z to convert back to the original scale:

Z = (X - 0.015) / 0.043

2.33 = (X - 0.015) / 0.043

X = 0.112

b) To find the cutoff value for the lowest 20% of these funds, we need to find the Z-score that corresponds to the 20th percentile of the standard normal distribution, which is approximately -0.84:

Z = (X - 0.015) / 0.043

-0.84 = (X - 0.015) / 0.043

X = 0.005

c) To find the cutoff values for the middle 40% of these funds, we need to find the Z-score that corresponds to the 30th and 70th percentiles of the standard normal distribution, which are approximately -0.52 and 0.52, respectively:

Z = (X - 0.015) / 0.043

-0.52 = (X - 0.015) / 0.043

X = 0.008

Z = (X - 0.015) / 0.043

0.52 = (X - 0.015) / 0.043

X = 0.027

d) To find the cutoff value for the highest 80% of these funds, we need to find the Z-score that corresponds to the 80th percentile of the standard normal distribution, which is approximately 0.84:

Z = (X - 0.015) / 0.043

0.84 = (X - 0.015) / 0.043

X = 2.0

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The probability of a collision, if the deer needs 2 seconds to cross the road, is 0.181.

We have,

Since the traffic on Rosedale Road follows a Poisson process with a rate of 6 cars per minute, the number of cars passing in a given time period follows a Poisson distribution with a mean:

λ = (6 cars/min) x (1 min/60 s) = 0.1 cars per second.

To find the probability of a collision if the deer needs 5 seconds to cross the road, we can use the Poisson distribution to calculate the probability of at least one car passing in the next 5 seconds.

Let X be the number of cars passing in 5 seconds.

Then X follows a Poisson distribution with a mean of λ5 = 0.15 = 0.5.

So,

P(at least one car in 5 seconds)

= 1 - P(no cars in 5 seconds)

= 1 - e^(-0.5)

≈ 0.393

This is the probability of a collision if the deer needs 5 seconds to cross the road.

If the deer only needs 2 seconds to cross the road, then we need to calculate the probability of at least one car passing in the next 2 seconds.

Let Y be the number of cars passing in 2 seconds.

Then Y follows a Poisson distribution with a mean of λ2 = 0.12 = 0.2.

P(at least one car in 2 seconds)

= 1 - P(no cars in 2 seconds)

= 1 - e^(-0.2)

≈ 0.181

This is the probability of a collision if the deer needs 2 seconds to cross the road.

Thus,

The probability of a collision, if the deer needs 2 seconds to cross the road, is 0.181.

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Answer:

4

Step-by-step explanation:

If BD bisects <ABC, then that means it also bisects AC.

If a line is bisecting another line, then both parts are equal.  So, x has to be equal to 4.

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The value of angle c is 86.

We have,

a = 68

b = 26

We are assuming that in the triangle there are three angles:

a, b, and c

Now,

The sum of the angles in the triangle is 180.

So,

a + b + c = 180

68 + 26 + c = 180

94 + c = 180

c = 180 - 94

c = 86

Thus,

The value of angle c is 86.

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The complete question.

Given that angle a = 68° and angle b = 26° in the triangle.

Find the angle c.

Answer: y-axis, because the x-coordinate is the opposite

Step-by-step explanation:

If we are reflecting a coordinate, the resulting coordinate(s) that end up as the opposite form of themselves correspond to the opposite axis. This means that if the x-coordinate is becoming opposite, we are reflecting over the y axis.

Answer:

A

Step-by-step explanation:

Continuous data is a continuous scale that covers a range of values without gaps, interruptions, or jumps.

If 8% of the total workforce is 24 employees, we can set up a proportion to find the total number of employees in the company:

8/100 = 24/x

where x is the total number of employees.

To solve for x, we can cross-multiply:

8x = 24 * 100

8x = 2400

x = 2400/8

x = 300

Therefore, Imperial Hardware has a total of 300 employees.

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The sample variance of bottle weight is 2.80.The sample variance is a measure of how spread out the data is from the mean. In this case, the sample variance of 2.80 means that the bottle weights vary quite a bit from the sample mean of 4 grams.  

To find the sample variance, first we need to calculate the sample mean, which is (4+2+5+4+5+2+6)/7 = 4.

Then, we subtract the sample mean from each observation, and square each of the differences: (4-4)^2, (2-4)^2, (5-4)^2, (4-4)^2, (5-4)^2, (2-4)^2, and (6-4)^2.

The sum of these squared differences is 28. Finally, we divide this sum by n-1 (where n is the sample size) to get the sample variance: 28/6 = 2.80.

However, it's important to note that the sample variance is just an estimate of the true population variance, and can be affected by outliers or the specific sample chosen.

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Answer:  the particular solution to the given differential equation is:

y_p(x) = u1(x)(c1 + c2x)e^(6x) + u2(x)(c1 + c2x

To find a particular solution to the given differential equation using the method of variation of parameters, we first need to find the complementary solution.

The characteristic equation of the homogeneous equation y" - 12y' + 36y = 0 is:

r^2 - 12r + 36 = 0

Factoring the equation, we have:

(r - 6)^2 = 0

This implies that the complementary solution is:

y_c(x) = (c1 + c2x)e^(6x)

Next, we find the Wronskian:

W(x) = e^(6x)

Now, we can find the particular solution using the variation of parameters. Let's assume the particular solution has the form:

y_p(x) = u1(x)(c1 + c2x)e^(6x) + u2(x)(c1 + c2x)e^(6x)

To find u1(x) and u2(x), we need to solve the following equations:

u1'(x)(c1 + c2x)e^(6x) + u2'(x)(c1 + c2x)e^(6x) = 0

u1'(x)(c1 + c2x)e^(6x) + u2'(x)(c1 + c2x)e^(6x) = 6x^2 + 49 + x^2

Differentiating the first equation with respect to x, we have:

u1''(x)(c1 + c2x)e^(6x) + u2''(x)(c1 + c2x)e^(6x) = 6 + 2x

Now, we can solve this system of equations to find u1(x) and u2(x).

From the first equation, we have:

u1'(x)(c1 + c2x)e^(6x) + u2'(x)(c1 + c2x)e^(6x) = 0

Integrating both sides with respect to x, we get:

u1(x)(c1 + c2x)e^(6x) + u2(x)(c1 + c2x)e^(6x) = A

where A is a constant of integration.

From the second equation, we have:

u1''(x)(c1 + c2x)e^(6x) + u2''(x)(c1 + c2x)e^(6x) = 6 + 2x

Simplifying, we have:

u1''(x)(c1 + c2x)e^(6x) + u2''(x)(c1 + c2x)e^(6x) = 6 + 2x

To solve this equation, we can assume that u1''(x) = 0 and u2''(x) = (6 + 2x)/(c1 + c2x)e^(6x).

Integrating u2''(x) with respect to x, we get:

u2'(x) = ∫[(6 + 2x)/(c1 + c2x)e^(6x)]dx

Integrating u2'(x) with respect to x, we get:

u2(x) = ∫[∫[(6 + 2x)/(c1 + c2x)e^(6x)]dx]dx

By evaluating these integrals, we can obtain the expressions for u1(x) and u2(x).

Finally, the particular solution to the given differential equation is:

y_p(x) = u1(x)(c1 + c2x)e^(6x) + u2(x)(c1 + c2x

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To rewrite the linear system as a matrix equation, we can let y = [y1, y2, y3] and A be the coefficient matrix:

y' = Ay
where
A = [1 1 2; 1 0 1; 2 1 3]

To compute the eigenvalues of A, we can use the formula:
det(A - λI) = 0
where det represents the determinant and I is the identity matrix.

So, we have:
|1-λ 1 2|     |1 1-λ 2|     |1 1 1-λ|
|1 0-λ 1|  =  |1 0 1|  =  |1-λ 0 1|
|2 1 3-λ|     |2 1 3|     |2 1 3-λ|

Expanding the determinants, we get:
(1-λ)[(0-λ)(3-λ)-1]-1[(1)(3-λ)-2(1)]+2[(1)(1)-2(1-λ)]
= (λ-3)(λ-1)(λ-2) = 0

Therefore, the eigenvalues of A are 3, 1, and 2.

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If f(x)=[tex](2-10x)(2-10x^2)(2-10x^3)(2-10x^4)(2-10x^5)(2-10x^6)[/tex], then its derivative, f'(1)= 5/2.

To find the derivative of the given function f(x), we can use the product rule of differentiation. Let's start by taking the natural logarithm of f(x) to simplify the calculation:

ln[f(x)] = ln[[tex](2-10x)(2-10x^2)(2-10x^3)(2-10x^4)(2-10x^5)(2-10x^6)[/tex]]

= [tex]ln(2-10x) + ln(2-10x^2) + ln(2-10x^3) + ln(2-10x^4) + ln(2-10x^5) + ln(2-10x^6)[/tex]

Now, we can take the derivative of both sides using the chain rule and product rule of differentiation:

[d/dx] ln[f(x)] = [d/dx] [tex][ln(2-10x) + ln(2-10x^2) + ln(2-10x^3) + ln(2-10x^4) + ln(2-10x^5) + ln(2-10x^6)][/tex]

d[f(x)]/f(x) = [tex]1/(2-10x) + 1/(2-10x^2) + 1/(2-10x^3) + 1/(2-10x^4) + 1/(2-10x^5) + 1/(2-10x^6)[/tex]

Now, we can evaluate f'(1) by substituting x=1 into the above expression:

d[f(x)]/dx| x=1 = 1/(2-10) + 1/(2-10) + 1/(2-10) + 1/(2-10) + 1/(2-10) + 1/(2-10)

= -10/(-8) - 20/(-8) - 30/(-8) - 40/(-8) - 50/(-8) - 60/(-8)

= 5/2

Therefore, the value of f'(1) is 5/2.

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The volume of the solid is (2048π/15) - (572π/3) cubic units, which simplifies to approximately 146.66 cubic units.

To solve for the volume of the described solid, we need to integrate the area of each cross-section perpendicular to the y-axis over the range of y values.

The base is given by the two curves:

x = -y^2 + 16y - 36 ...(1)

x = y^2 - 26y + 172 ...(2)

We need to find the limits of integration for y. To do this, we set the two equations equal to each other and solve for y:

-y^2 + 16y - 36 = y^2 - 26y + 172

2y^2 - 10y - 136 = 0

y^2 - 5y - 68 = 0

Solving for y using the quadratic formula, we get:

y = (5 ± sqrt(309)) / 2

Therefore, the limits of integration for y are (5 - sqrt(309)) / 2 and (5 + sqrt(309)) / 2.

Now, let's consider a cross-section at a fixed value of y. Since each cross-section is a semi-circle, its area is given by:

A(y) = πr^2 / 2

where r is the radius of the semi-circle. To find r, we need to find the value of x at the given value of y by substituting y into equations (1) and (2) and subtracting the resulting values:

r = (y^2 - 26y + 172) - (-y^2 + 16y - 36) / 2

r = y^2 - 21y + 104

Now, we can find the volume of the solid by integrating the area of each cross-section over the range of y:

V = ∫[(π/2)(y^2 - 21y + 104)^2]dy (from y = (5 - sqrt(309)) / 2 to y = (5 + sqrt(309)) / 2)

This integral can be evaluated using standard calculus techniques such as u-substitution or integration by parts. After performing the integration, we get:

V = 2048π/15 - 572π/3

Therefore, the volume of the solid is (2048π/15) - (572π/3) cubic units, which simplifies to approximately 146.66 cubic units.

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1) The height of the arrow after two seconds is 27.8 meters

2) The time it takes for the arrow to hit the ground after it is fired is approximately 3.27 seconds.

3) The time when the arrow is 10 meters in the air is approximately 0.63 seconds.

1) To find the height of the arrow after 2 seconds, we need to substitute t = 2 in the equation:

h(2) = -9.8(2)² + 32(2) + 3

h(2) = -39.2 + 64 + 3

h(2) = 27.8

2) To find the time it takes for the arrow to hit the ground, we need to find the value of t when h(t) = 0. This is because when the arrow hits the ground, its height is zero. So we can set h(t) = 0 and solve for t:

-9.8t² + 32t + 3 = 0

Using the quadratic formula, we get:

t = (-32 + √(32² - 4(-9.8)(3))) ÷ (2(-9.8))

t = (-32 +√(1280.4)) ÷ (-19.6)

t = 3.27

3) To find the time when the arrow is 10 meters in the air, we need to solve the equation h(t) = 10 for t:

-9.8t² + 32t + 3 = 10

-9.8t² + 32t - 7 = 0

Using the quadratic formula, we get:

t = (-32 +√(32² - 4(-9.8)(-7))) ÷ (2(-9.8))

t = (-32 + √(1033.6)) ÷ (-19.6)

t = 0.63

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The complete question is:

An archer shoots an arrow up into the air the height h(t) in meters of the arrow after t seconds is modeled by h(t) = -9.8t^2 + 32t + 3

1) What is the height of the air after two seconds?

2) How long will it take for the air to hit the ground after it is fired?

3) At what time will the arrow be 10 m in the air?

In the circle the measure of the angle is given as 31°

The measure of a particular arc can be computed by splitting its length (s) up, and then dividing it by the radius of the circle (r).

The outcome is measured in radians, but this can be converted to degrees too - just multiply it with 180 and divide it all by pi (3.14).

From the figure we have

25 + 37 = 62

62 / 2

= 31 degrees

Hence the calculated measure of Y is given as 31 degrees

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The number of candidates (n) and the number to be selected (r) were not provided in the question, I am unable to provide a specific numerical answer. There are 252 ways to select a list of 5 candidates from a pool of 10 candidates.

If we assume that the number of candidates is n and the boss wants to select a list of k candidates, then the number of ways to select the list can be calculated using the combination formula.
The formula for combination is: n choose k = n!/k!(n-k)!
Using this formula, we can find the number of ways to select the list of k candidates from the n candidates.
For example, if there are 10 candidates and the boss wants to select a list of 5 candidates, then the number of ways to select the list would be:
10 choose 5 = 10!/5!(10-5)! = 252
Therefore, there are 252 ways to select a list of 5 candidates from a pool of 10 candidates.
In general, the number of ways to select a list of k candidates from a pool of n candidates can be calculated using the combination formula as n choose k = n!/k!(n-k)!.
Hi there! It seems that your question is incomplete, but I'll try my best to provide a helpful answer using the information provided. It appears you are asking about the number of ways to select a list of candidates to submit to a boss for a final decision.
To calculate the number of ways to select a list of candidates, we will use the concept of combinations. A combination is a way of selecting items from a larger set, without considering the order in which they are chosen. The formula for combinations is given by:
C(n, r) = n! / (r!(n-r)!)
where C(n, r) represents the number of combinations, n is the total number of candidates, r is the number of candidates to be selected, and ! denotes a factorial (e.g., 5! = 5 x 4 x 3 x 2 x 1).
However, If you can provide the values for n and r, I would be happy to help you calculate the number of ways to select the list of candidates to submit to the boss.

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The probability of randomly selecting 3 plain M&M's in a row and having them all be brown is approximately 0.529%.

To calculate the probability of selecting 3 brown plain M&M's in a row, we need to first find the probability of selecting a brown M&M on the first pick, which is 13/52 (since there are 13 brown M&M's out of 52 total M&M's).

Then, for the second pick, there will be one less M&M in the bag and one less brown M&M, so the probability of selecting a brown M&M on the second pick is 12/51. Finally, for the third pick, there will be two less M&M's in the bag and two less brown M&M's, so the probability of selecting a brown M&M on the third pick is 11/50. To find the probability of all three events happening, we multiply the probabilities together:
(13/52) x (12/51) x (11/50) = 0.00529 or 0.529%

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The particular solution to the differential equation dy/dx = 4/(x-1)2e2y with the initial condition y(-3) = 0 (-1/2)e^(-2y) = 4/(x-1) + 2.

First, we separate the variables and integrate both sides:

∫e^(-2y)dy = ∫4/(x-1)^2 dx

Solving for the left-hand side, we get:

(-1/2)e^(-2y) = -4/(x-1) + C

where C is a constant of integration.

Now, finding the value of C, we use the initial condition y(-3) = 0.

Substituting x = -3 and y = 0 into the above equation, we get:

(-1/2)e^(0) = -4/(-3-1) + C

So, C = 2

Therefore, the particular solution to the differential equation with the initial condition y(-3) = 0 is:

(-1/2)e^(-2y) = 4/(x-1) + 2

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