a) Find constants A,B and C such that (x−1)(x−2)(x−3)
3x 2
−12x+11
​
≡ x−1
A
​
+ x−2
B
​
+ x−3
C
​
. (4) b) Express (x+1)(x 2
+1)
−2x
​
as a sum of partial fractions and hence write (x+1)(x 2
+1)
x 4
+x 3
+x 2
−x
​
as a sum of partial fractions. (10)

Given that the number x of bicycle helmets a retail chain is willing to sell per week at a price of p is given by x=81\sqrt{p}+9-350 for 20\le p\le 100. The equilibrium price of bicycle is $43.58 and the equilibrium quantity of helmets is 1179 helmets.

The demand curve for the given equation is:

x = 81√p + 9 - 350

The supply curve is a straight line, and it's given by

y = 50p - 550.

Plot the demand curve in the same graph as the supply curve for 20 ≤ p ≤ 100.

Also, draw horizontal lines representing the equilibrium quantity of helmets.

At the equilibrium price, supply equals demand.800-700-600-500-

Part (a)The equilibrium price for bicycles

The equilibrium price is the one where the quantity demanded and the quantity supplied are the same.

Equating the given supply equation to the demand equation:

50p - 550

= 81√p + 9 - 350

Simplifying this equation:

31p - 191 = 81√p

Taking the square of both sides of the equation:

961p² - 116411p + 36481 = 0

On solving this quadratic equation we get:

p = 43.58 or

p = 87.88

As per the given interval, $20 \le p \le 100$,

so the equilibrium price is

p = 43.58

Part (b)The equilibrium quantity of helmets

Substitute $p

= 43.58$ into the supply equation to get the equilibrium quantity:

q = 50(43.58) - 550q

= 1179

Therefore, the equilibrium quantity of helmets is 1179 helmets.

Hence, the solution is:

The equilibrium price of bicycle is $43.58 and the equilibrium quantity of helmets is 1179 helmets.

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The length of the boundary of the remaining park is 398 meters.

Given,Length of the side of the square park,

ABCD = 100 mSize of each rectangular flower bed = 10 m x 5 mTotal area of each rectangular flower bed = 10 m x 5 m = 50 sq.m

Area of both the rectangular flower beds = 50 sq.m + 50 sq.m = 100 sq.m

Area of the square park = Side x Side= 100 m x 100 m = 10000 sq.m

Area of the remaining park = Total area of the park - Area of both the rectangular flower beds= 10000 sq.m - 100 sq.m

= 9900 sq.m

The remaining park is in the shape of a square with an area of 9900 sq.m.The length of the side of this square can be found as follows:

Area of the square = Side x Side

9900 sq.m = Side x Side Side = √9900 mSide = 99.5 m

Therefore, the length of the boundary of the remaining park is:

Length of the boundary = 4 x Side= 4 x 99.5 m= 398 m

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The equation can be written in the form y = mx + c as y = (3/5)x + 4/5 and the correct option is D, i.e., m = 0.6, c = 0.8.

The linear equation 5y - 3x -4 = 0 can be written in the form y m and c is m = 0.6, c = 0.8. We are given the linear equation 5y - 3x - 4 = 0 which is not in the form y = mx + c. To write this in y = mx + c form, we have to isolate the y variable and rewrite the equation in terms of y.Therefore,

5y - 3x - 4 = 0

⇒ 5y = 3x + 4

⇒ y = (3/5)x + 4/5

Comparing the equation with the standard equation y = mx + c, we get:

m = 3/5 and c = 4/5

Hence, the equation can be written in the form y = mx + c as y = (3/5)x + 4/5 and the correct option is D, i.e., m = 0.6, c = 0.8.

Therefore, we can say that the linear equation 5y - 3x -4 = 0 can be written in the form y m and c as m = 0.6, c = 0.8.

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The applicant should spend $50 on the preparatory course as it doubles his probability of achieving a score above 750, increasing his chances of getting into a top 20 MBA program.

To determine whether the applicant should spend $50 on the preparatory course, we need to evaluate the probability tree.

The tree diagram would consist of two branches: one representing the applicant not taking the course and the other representing the applicant taking the course.

Starting with the branch of not taking the course, the probability is -20% (0.2) of achieving a score above 750.

For the branch of taking the course, the probability of achieving a score above 750 would need to double from 0.2 to 0.4, as stated by the applicant.

Considering the increase in probability and the cost of the course, it would be beneficial for the applicant to spend $50 on the preparatory course since the probability of achieving a score above 750 doubles.

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Answer:

C) square root of 28

Step-by-step explanation:

Alright, so the point on the graph is approximately less than 5.5.

[tex]\sqrt5=2.236 \\\sqrt15 =3.87\\\sqrt28 =5.29\\\sqrt63 = 7.93[/tex]

We can see that only the square root of 28 is the correct option.  Hope this helps! :)

(a) Probability of winning first three prizes with multiple wins allowed. (b) Probability of specific individuals winning specific prizes with no multiple wins allowed. (c) Probability of friends winning first four prizes and you winning 56th prize with multiple wins allowed. (d) Probability of friends winning first four prizes and you winning 56th prize with no multiple wins allowed.

(a) The probability of winning the first prize is 1/55 since there are 55 total participants. After winning the first prize, the probability of winning the second prize is 1/54, and for the third prize, it is 1/53. Therefore, the probability of winning the first three prizes is (1/55) * (1/54) * (1/53).

(b) Since no one can win more than one prize, the probability of each person winning their respective prize is 1/55. Therefore, the probability of you winning the fifth prize, Bobby winning the eighth prize, Lindsey winning the tenth prize, and Terry winning the 50th prize is (1/55) * (1/55) * (1/55) * (1/55).

(c) Assuming winning more than one prize is allowed, the probability of each friend winning their respective prize is 1/55. As for you winning the 56th prize, the probability is 1/51 since there are 51 participants left after the first four prizes have been awarded.

Therefore, the probability of your friends winning the first four prizes and you winning the 56th prize is (1/55) * (1/55) * (1/55) * (1/55) * (1/51).

(d) If no one can win more than one prize, the probability of each friend winning their respective prize is 1/55. Since you cannot win any of the first four prizes, the probability of you winning the 56th prize is 1/51.

Therefore, the probability of your friends winning the first four prizes and you winning the 56th prize is (1/55) * (1/55) * (1/55) * (1/55) * (1/51).

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Therefore, the correct option is O t≈ 6.2 seconds.

Given, T(t) = 16+ 44e-0.35t

We need to find how long will it take for the thermometer to reach 23°C. i.e,

we need to find t when T(t) = 23°C.

Substitute T(t) = 23°C in the given equation and solve for

t.23 = 16 + 44e-0.35t

t = -1/0.35

ln [(23-16)/44]≈ 6.2 seconds

Hence, the thermometer will take approximately 6.2 seconds to reach 23°C.

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The graph of the given function f(x) = 2 cos(x) cos (x – 3π/2) is sketched.

Given function is f(x) = 2 cos(x) 2 cos (x – 3π/2)

The given function is a product of two cosines. Let's consider each part of the product separately.

Step 1: 2 cos(x)

First, we will draw the graph of y = cos(x). It is shown below:

For 2 cos(x), we will multiply the amplitude of the cosine by 2.

The graph becomes:

Step 2: cos(x – 3π/2)

We will shift the graph of cos(x) to the right by 3π/2 units to get the graph of cos(x – 3π/2).

It is shown below:

Step 3: 2 cos(x)  cos(x – 3π/2)

To sketch the graph of 2 cos(x) cos(x – 3π/2), we will multiply the ordinates of the two graphs (as it is a product of two functions). It is shown below:

Therefore, the required graph is shown in the figure below:

Conclusion: Thus, the graph of the given function

f(x) = 2 cos(x) cos (x – 3π/2) is sketched.

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It is true that m(A) = n(A) for all A ∈ σ(C). It is given that (X, A) is a measurable space and C is an arbitrary subset of A. And also given that m and n are two σ-finite measures on (X, A) such that m(A)

= n(A) for all A ∈ C.

We are to determine whether m(A)

= n(A) for all A ∈ σ(C) or not.σ(C) is the σ-algebra generated by C. It contains all the countable unions and countable intersections of the sets in C. We are to prove that m(A)

= n(A) for all A ∈ σ (C).If A ∈ C, then m(A)

= n(A) [Given]If A ∉ C, then A is the countable union of disjoint sets of C such that A

= B1 ∪ B2 ∪ .......∞

Then, m(A) = ∑m(Bn)and n(A) = ∑n(Bn)Therefore, m(A)

= n(A) since m(Bn) = n(Bn) for all n ∈ N. Now, if m and n are finite measures, then also we can show that m(A)

= n(A) for all A ∈ σ(C). We can prove this using Monotone class theorem since both m and n are finite.

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As x approaches 0, the limit lim 1-0 sin² (161) 1 evaluates to sin² (161).

To find the limit, we must determine what happens to the function as x approaches 0. A limit is a value that a function approaches as x approaches a particular point or infinity. We can calculate a limit by analyzing the function's behavior as x approaches the limiting value from the left or right.

It can be simplified as follows;

Lim 1-0 sin² (161) 1

= lim 1-0 sin² (161) * lim 1-0 1 [using product rule]

= sin² (161) * 1

= sin² (161)

Therefore, this means that as x approaches 0, the limit of the function sin² (161) 1 approaches the value of sin² (161).

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The end behavior of the graph is

- As x approaches positive infinity, the graph decreases without bound.

- As x approaches negative infinity, the graph also decreases without bound.

a) To find the zeros of the function f(x), we set f(x) equal to zero and solve for x:

-(3x - 2)(x + 5)(x - 1)² = 0

From this equation, we can see that the function will be zero when any of the factors are equal to zero. Therefore, the zeros of the function are:

1) x = 2/3 (zero of multiplicity 1)

2) x = -5 (zero of multiplicity 1)

3) x = 1 (zero of multiplicity 2)

To determine whether each zero touches or crosses the x-axis, we can look at the behavior of the function near each zero.

- For x = 2/3, the factor (3x - 2) is positive, so the sign of the function will depend on the signs of the other factors. Since (x + 5) and (x - 1)² are positive for x = 2/3, the function crosses the x-axis at this zero.

- For x = -5, the factor (x + 5) is zero and the other factors are nonzero. Therefore, the function changes sign at this zero, indicating that it crosses the x-axis.

- For x = 1, the factor (x - 1)² is zero, and the other factors are nonzero. Since (x - 1)² is squared, the function touches the x-axis at this zero.

b) To sketch the end behavior of the graph, we examine the leading term of the function as x approaches positive infinity and negative infinity.

The leading term of f(x) is -(3x - 2)(x + 5)(x - 1)². When expanded, the highest power term is x³. Since the coefficient of this term is negative, we know that as x approaches positive infinity, the graph will be decreasing.

Similarly, as x approaches negative infinity, the graph will also be decreasing since the leading term dominates.

Therefore, the end behavior of the graph is as follows:

- As x approaches positive infinity, the graph decreases without bound.

- As x approaches negative infinity, the graph also decreases without bound.

Note that without additional information or the use of a graphing tool, we cannot provide a detailed sketch of the graph of the function.

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Therefore, the range of g(x, y) is [-1, 1].

Let's evaluate g(1, −2):g(1, -2) = sin(6(1) + 8(-2)) = sin(-10) = -0.54402...

Now, let's find the range of g(x, y):

The function g(x, y) = sin(6x + 8y) has a range of [-1, 1] in interval notation.

The sin function has a maximum value of 1 and a minimum value of -1.

Since 6x + 8y can take on any real value, g(x, y) can take on any value between -1 and 1, inclusive.

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The sum of the roots of the indicial equation is -3. Option (e) is correct.

Given differential equation is:  [tex]x^2(x+1)y'' + 4x(x+1)y' - 6y = 0[/tex]

and the domain of the differential equation is x>0. We need to find the sum of the roots of the indicial equation.The standard form of the differential equation is:

[tex]x^2(x+1)y'' + bxy' + cy = 0[/tex]

where,

b and c are constants.In this differential equation,

b = 4x(x+1)

c = -6.

The equation of the indicial roots is obtained by substituting y = x^r in the standard form of the differential equation.

[tex]$$\begin{aligned} x^2(x+1)y'' + 4x(x+1)y' - 6y &= x^2(x+1)r(r-1)x^{r-2} + 4x(x+1)rx^{r-1} - 6x^r \\ &= x^r\left[x^2(x+1)r(r-1) + 4x(x+1)r - 6\right] \end{aligned}$$[/tex]

Let's substitute  [tex]$y=x^r$[/tex] in the given differential equation:

[tex]$$x^2(x+1)y'' + 4x(x+1)y' - 6y = 0$$$$x^2(x+1)r(r-1)x^{r-2} + 4x(x+1)rx^{r-1} - 6x^r = 0$$[/tex]

We can simplify the above equation by factoring out. Now, we have an equation in the form [tex]$p(r)x^r = 0$[/tex] , where [tex]$p(r)$[/tex] is the polynomial in [tex]$r$[/tex].The roots of the polynomial [tex]$p(r)$[/tex] are called the indicial roots. In this equation,

[tex]$$x^2(x+1)r(r-1) + 4x(x+1)r - 6 = 0$$[/tex]

Dividing both sides by [tex]$x^2(x+1)$[/tex], we get:

[tex]$$r(r-1) + 4r - \frac{6}{x^2(x+1)} = 0$$$$r^2 + 3r - \frac{6}{x^2(x+1)} = 0$$[/tex]

Using the quadratic formula,

[tex]$$r_1, r_2 = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 1 \cdot \frac{6}{x^2(x+1)}}}{2}$$[/tex]

Simplifying,

[tex]$$r_1, r_2 = \frac{-3 \pm \sqrt{9 + \frac{24}{x^2(x+1)}}}{2}$$$$r_1 + r_2 = \frac{-3 + \sqrt{9 + \frac{24}{x^2(x+1)}}}{2} + \frac{-3 - \sqrt{9 + \frac{24}{x^2(x+1)}}}{2}$$$$= -3$$[/tex]

Therefore, the sum of the roots of the indicial equation is -3. Option (e) is correct.

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The general solution is

[tex]xₜ = 2^t/(1 - 1/2^t) (-2)^t + 2^t.yₜ = 2^t[/tex]

The given linear difference equations are

[tex]:x_{t+1}=-2x_t+y_ty_{t+1}=2y_t[/tex]

with initial values x_0=2 and y_0=1.

To solve the system of linear difference equations, we first solve for x_t in the first equation and substitute it into the second equation as follows:

[tex]xₜ₊₁ = -2xₜ + 2yₜ ; x₀ = 2 x₁ = -2(2) + 2(1) = -2x₂ = -2(-2) + 2(2) = 8x₃ = -2(8) + 2(2) = -12x₄ = -2(-12) + 2(2) = 28x₅ = -2(28) + 2(2) = -54[/tex]

Using the method of undetermined coefficients, we assume a solution of the form [tex]xt = C(-2)^t[/tex] and substitute into the difference equation

[tex](3)xₜ₊₁ = -2xₜ + 2yₜor C(-2)^(t+1) = -2C(-2)^t + 2yₜor C(-2) = -2C + 2yₜor yₜ = C(-2)/2 + C = -C + C(-2)^-1[/tex]

[tex]yₜ = 2^t[/tex]

Substituting, we get

[tex]2^t = -C + C(-2)^-1 or 2^t = C(1 - 1/2^t)or C = 2^t/(1 - 1/2^t)[/tex]

Substituting C gives the general solution

[tex]xt = 2^t/(1 - 1/2^t) (-2)^t + 2^t.[/tex]

The general solution is

[tex]xₜ = 2^t/(1 - 1/2^t) (-2)^t + 2^t.yₜ = 2^t[/tex]

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The graph is attached. One of the feature is the vertex being at (3, -3)

The graph of the equation y = (x - 3)² - 3 is a parabola.

The features are

The term (x - 3)² represents a quadratic term, indicating that the graph will be a parabola.

The vertex of the parabola is located at the point (3, -3)

since the equation is in the form (x - h)² + k,

where (h, k) represents the vertex coordinates.

The coefficient of the quadratic term, (x - 3)², is positive, indicating that the parabola opens upward.

The constant term, -3, shifts the parabola downward by three units.

The graph is attached

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Ario will be at approximately 4.6 meters from the starting point when Miguel starts the race.

From the number line from 0 to 25, and Ario is standing 3 meters from the starting point. Miguel offers Ario a head start, and Miguel will start when the ratio of Ario's completed meters to Ario's remaining meters is 1:4.4.

Assume that Ario has covered x meters when Miguel starts the race. The remaining distance for Ario would be 25 - x meters.

According to the given ratio, set up the equation:

x / (25 - x) = 1 / 4.4

To solve for x, cross-multiply:

4.4x = 25 - x

Combining like terms:

5.4x = 25

Dividing both sides by 5.4:

x ≈ 4.6296

Rounding to the nearest tenth:

x ≈ 4.6

Therefore, Ario will be at approximately 4.6 meters from the starting point when Miguel starts the race.

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1. The \( z \)-score for an IQ test score of 145, when the mean is 100 and the standard deviation is 15, is 3.

The \( z \)-score (or standard score) is a statistical measure that indicates how many standard deviations a data point is above or below the mean of a distribution. It is calculated using the formula:

\[ z = \frac{{X - \mu}}{{\sigma}} \]

where \( X \) is the value of the data point, \( \mu \) is the mean of the distribution, and \( \sigma \) is the standard deviation of the distribution.

In this case, we have an IQ test score of 145, a mean of 100, and a standard deviation of 15. To find the \( z \)-score, we substitute these values into the formula:

\[ z = \frac{{145 - 100}}{{15}} = \frac{{45}}{{15}} = 3 \]

The \( z \)-score of 3 indicates that the IQ test score of 145 is three standard deviations above the mean. This means that the score is relatively high compared to the average performance on the IQ test.

By knowing the \( z \)-score, we can compare IQ test scores from different distributions and determine the relative position of a score within a distribution. A positive \( z \)-score indicates a score above the mean, while a negative \( z \)-score indicates a score below the mean. In this case, a \( z \)-score of 3 shows that the IQ test score of 145 is significantly above the mean.

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Option A is correct, the most general antiderivative is [tex]\frac{7}{3}. e^{\:3x}\:dx+6e^-^x+C[/tex].

To find the antiderivative of the given function, we can apply the power rule for integration.

The power rule states that if f(x) is of the form f(x)=axⁿ.

where a is a constant and  n is a real number (except -1), then the antiderivative of f(x) with respect to x is given by [tex]\frac{a}{n+1}x^{n+1}+C[/tex].

Applying the power rule to each term in the given function, we have:

[tex]\int \:\left(7e^{3x}\:-6e^{-x}\:\right)dx=\int \:\:7e^{\:3x}\:dx-\int 6e^{\:-x}\:dx[/tex]

Integrating each term individually:

[tex]\:\int \:7e^{\:3x}\:dx=\:\frac{7}{3}e^{\:3x}+C_1[/tex]

[tex]\:\int \:6e^{\:-x}\:dx=\:-6e^{\:-x}+C_2\\[/tex]

Combining the results, we get:

[tex]\int \:\left(7e^{3x}\:-6e^{-x}\:\right)dx=\frac{7}{3}. e^{\:3x}\:dx+6e^-^x+C[/tex]

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3 3 157 - 20157 + 757 - H Sy Sy Sy X Ś

In the given expression, we have a series of additions and subtractions with various terms. To simplify this expression, we need to evaluate the arithmetic operations step by step.

Add 3 and 3

The sum of 3 and 3 is 6.

Subtract 20157 from 157

When we subtract a larger number from a smaller number, we get a negative result. Here, the difference between 157 and 20157 is -20000.

Add 757 to -20000

Adding 757 to -20000 gives us -19243.

Subtract H from -19243

Since the variable H represents a positive real number, we cannot determine the exact value of this subtraction without additional information. Therefore, the result is expressed as -19243 - H.

Add Sy, Sy, Sy, X, and Ś to -19243 - H

Similarly, without any specific values assigned to these variables, we cannot simplify this part further. Hence, the final result is -19243 - H + Sy + Sy + Sy + X + Ś.

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When considering the expression "Add or subtract as indicated. Assume that the variable represents a positive real number," it implies that we need to perform addition or subtraction operations on the given numbers or variables.

The input you provided consists of a combination of numbers, symbols, and letters that don't follow a clear pattern.

It is essential to provide specific instructions or clarify how the operations should be applied.

Without further information, it is not possible to determine the correct interpretation or calculation.

Please provide additional details or clarify the desired operation so that I can assist you further.

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If ANFO has a cost of $1500 per tonne, the cost of explosives per tonne of rock is $3240.

To determine the blasthole spacing, we need to consider the burden and spacing relationship, as well as the desired powder factor. The burden (B) and spacing (S) are related by the equation B = 1.2S. We can calculate the burden using this equation.

Given that the hole diameter is 250 mm, the bench height is 15 m, and the stemming column length is 4 m, the effective depth for the explosive is the bench height minus the stemming column length and subdrill, which is 15 m - 4 m - 1.5 m = 9.5 m.

To calculate the blasthole spacing, we use the formula:

Spacing (S) = Burden (B) + Hole Diameter (D) = B + D.

From the given burden and spacing relationship (B = 1.2S), we substitute B in terms of S:

S = 1.2S + 250 mm.

Simplifying the equation, we have:

0.2S = 250 mm.

Converting the units, we get:

0.2S = 0.25 m.

Solving for S:

S = 0.25 m / 0.2 = 1.25 m.

Therefore, the blasthole spacing is 1.25 meters.

To calculate the cost of explosives per tonne of rock, we use the powder factor, which is given as 0.8 kg/m³. The density of the rock is 2.7 g/cc.

To convert the density of rock from g/cc to kg/m³:

Rock density = 2.7 g/cc = 2700 kg/m³.

The amount of explosive required per cubic meter of rock is:

Explosive amount = Powder factor * Rock density = 0.8 kg/m³ * 2700 kg/m³ = 2160 kg/1000 m³.

Since the explosive density is given as 0.9 g/cc, which is equivalent to 900 kg/m³, the cost of explosives per tonne of rock can be calculated as:

Cost of explosives = (Explosive amount / 1000) * Cost of ANFO per tonne = (2160 / 1000) * $1500 = $3240.

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In order to solve the given triangle, we will use the Sine law and Cosine law.  Let's start with Sine Law:

According to Sine Law, [tex]a/sin A = b/sin B \\= c/sin C[/tex]

where A, B, and C are the angles opposite to sides a, b, and c respectively. So, we have[tex]a/sin A = b/sin 54°\\ = 5/sin C     ........(i)[/tex]

Now, let's find A using Cosine Law Cosine law states that:

[tex]c² = a² + b² - 2ab cos C[/tex] On substituting the values, we get

[tex]5² = 8² + b² - 2×8×b cos 54° b² - 16b cos 54° + 39 \\= 0[/tex]

On solving the above quadratic equation, we get b = 6.02 (approx.)

[tex]C = 180° - A - BC \\= 180° - 66.96° - 54°C \\= 59.04°[/tex]

Hence, the required sides and angles are given as:

[tex]b = 6.02 (approx.)\\C = 59.04°A\\ = 66.96°[/tex]

[tex](f - g)(x) = f(x) - g(x)  \\= (3x² - 7) - (x - 6)(f - g)(3) \\= f(3) - g(3)  \\= (3*3² - 7) - (3 - 6)(f - g)(3) \\= 14[/tex]

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Given: y'=8x²-3xTo find: All possible functions with the given derivative.So we need to find the integral of the given function y', which will give us the possible functions. Let us solve for it using integration;

∫y' dx = ∫8x²-3x dx= 8 ∫x²dx - 3 ∫xdx

= 8 [ x³/3] - 3 [ x²/2 ] + C

= (8/3) x³ - (3/2) x² + C

where C is a constant of integration.

So, the possible functions with the given derivative y' are;A. (8/3) x³ - (3/2) x² + CB. (8/3) x³ - (3/2) x² - (2/3) x + CC. (8/3) x³ - (3/2) x² + (2/3) x + CD. (8/3) x³ - (3/2) x² - (2/3) x + C

The answer is, A. (8/3) x³ - (3/2) x² + C.

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(a) Find the expected counts for the table The formula for calculating expected counts is: E = (Row total * Column total) / Grand total The calculation of expected counts is given in the following table below.  Wife's Parents Divorced? Couple Separated Couple Intact Total Yes (24 + 118) = 142 (296 * 142) / 510 = 82.99 (296 - 82.99)

= 213.01 296 No (340 - 142) = 198 (296 * 198) / 510 = 114.01 (296 - 114.01) = 181.99 340 Total 342 296 638 (b) Calculate the chi-square test statistic The formula for calculating the chi-square test statistic is:χ2 = Σ(O − E)2 / EWhere:Σ = Summation signO = Observed frequency E = Expected frequency.

The calculation of the chi-square test statistic is given in the following table below.  Wife's Parents Divorced? Couple Separated Couple Intact Total Yes 24 118 142 (24 − 82.99)2 / 82.99 = 36.18 (118 − 213.01)2 / 213.01 = 32.69 69.87 No 296 − 142 = 154 1542 / 114.01 = 202.38 142 − 296 + 154 = 0 202.38 Total 320 174 494 36.18 + 32.69 + 202.38 = 271.25Therefore, the chi-square test statistic is 271.25.

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1. The total number of students who took the exam is 40 is option c.

2. The probability that the five cards picked are suited is approximately is option d. 0.0020.

Probability is a branch of mathematics that deals with the likelihood of events occurring. In this response, we will provide detailed solutions to two probability problems. We will explain the steps involved in solving each problem using mathematical terms.

Solution to Problem 1:

Let's denote the number of students who took the exam as 'x.' We are given that the probability of passing Chemistry is 70% and Physics is 50%. None of the students failed in both subjects, and 8 students passed both subjects.

To solve this problem, we can use the principle of inclusion-exclusion. The principle states that to find the total number of students who passed at least one subject, we need to sum the number of students who passed Chemistry, the number of students who passed Physics, and then subtract the number of students who passed both subjects.

Let's calculate the number of students who passed at least one subject:

Number of students who passed Chemistry = 0.7x

Number of students who passed Physics = 0.5x

Number of students who passed both subjects = 8

Total number of students who passed at least one subject = (Number of students who passed Chemistry) + (Number of students who passed Physics) - (Number of students who passed both subjects)

Substituting in the values, we have:

Total number of students who passed at least one subject = 0.7x + 0.5x - 8

Since none of the students failed in both subjects, the number of students who passed at least one subject is equal to the total number of students. Therefore, we can set the equation equal to 'x' and solve for it:

0.7x + 0.5x - 8 = x

Simplifying the equation:

1.2x - 8 = x

0.2x = 8

x = 8 / 0.2

x = 40

Therefore, the total number of students who took the exam is 40.

Hence, the answer to Problem 1 is option c. 40.

Solution to Problem 2:

We are given that we are picking five cards from a standard deck of 52 cards. We need to find the probability that all five cards picked are suited, meaning they all belong to the same suit.

To solve this problem, we can use the concept of combinations. The number of ways to choose five cards from a particular suit is denoted as C(13, 5), as there are 13 cards in each suit (hearts, diamonds, clubs, spades) and we need to choose 5 cards. Similarly, the total number of ways to choose any five cards from the deck is C(52, 5).

The probability of picking five suited cards can be calculated as:

Probability = (Number of favorable outcomes) / (Total number of possible outcomes)

Number of favorable outcomes = Number of ways to choose 5 cards from a single suit = C(13, 5)

Total number of possible outcomes = Number of ways to choose any 5 cards from the deck = C(52, 5)

Using the formula for combinations, we have:

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

Substituting in the values, we get:

Number of favorable outcomes = C(13, 5) = 13! / (5!(13-5)!)

Total number of possible outcomes = C(52, 5) = 52! / (5!(52-5)!)

Calculating the values:

Number of favorable outcomes = 1,287

Total number of possible outcomes = 2,598,960

Now, we can calculate the probability:

Probability = Number of favorable outcomes / Total number of possible outcomes

Probability = 1,287 / 2,598,960

Simplifying the fraction:

Probability ≈ 0.000495

Therefore, the probability that the five cards picked are suited is approximately 0.000495.

Hence, the answer to Problem 2 is option d. 0.0020.

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The dot product of vectors v and w is 0. The angle between v and w is 90 degrees. The vectors v and w are orthogonal.

To find the dot product of two vectors, v and w, we multiply the corresponding components of the vectors and then sum them up. In this case, v = 7i + 5j and w = 5i - 7j. To calculate the dot product vw, we multiply the corresponding components:

vw = (7i)(5i) + (5j)(-7j) = 35i^2 - 35j^2

Since i^2 = j^2 = -1, the dot product simplifies to:

vw = 35(-1) - 35(-1) = -35 - (-35) = 0

Therefore, the dot product of v and w is 0.

To find the angle between v and w, we can use the formula:

cosθ = (v · w) / (|v| |w|)

where θ is the angle between the vectors, v · w is the dot product of v and w, and |v| and |w| are the magnitudes of v and w, respectively.

Since the dot product vw is 0 and both v and w have nonzero magnitudes, the numerator of the formula is 0. Therefore, cosθ = 0, which implies that θ is 90 degrees.

Hence, the angle between v and w is 90 degrees, indicating that the vectors are orthogonal.

In summary, the dot product of v and w is 0, the angle between v and w is 90 degrees, and the vectors v and w are orthogonal.

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The cabin pressure is lower than the outside pressure, the air in the bag will expand.

To calculate the new volume of air in the bag, we can use the ideal gas law equation: PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the ideal gas constant, and T is temperature.

Since we are assuming the temperature is constant, we can rearrange this equation to solve for V.P1V1 = P2V2, where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, and V2 is the final volume.

Substituting the given values:P1 = 105.5 kPa,V1 = 140.1 mL,P2 = 70.0 kPa,

V2  = 105.5 kPa x 140.1 mL

V2  = 70.0 kPa x V2

V2 = 221.14 mL

The new volume of air in the bag is 221.14 mL.

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The polar coordinates for the point (-7, 2) are approximately (7.28, -0.29).

To convert the rectangular coordinates (-7, 2) to polar coordinates, we can use the following formulas:

r = sqrt(x^2 + y^2)

θ = arctan(y / x)

In this case, x = -7 and y = 2. Substituting these values into the formulas:

r = sqrt((-7)^2 + 2^2)

= sqrt(49 + 4)

= sqrt(53)

≈ 7.28

θ = arctan(2 / -7)

≈ -0.29 radians

Therefore, the polar coordinates for the point (-7, 2) are approximately (7.28, -0.29).

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The effectiveness of a maintenance program is affected most by the reliability of installed, replaceable items. This is because the reliability of installed replaceable items plays a key role in the effectiveness of a maintenance program.

The effectiveness of a maintenance program depends upon various factors like the availability of spare parts, skill level of the maintenance personnel, reliability of installed replaceable items, etc.

The four options provided in the question are Availability of spare parts, Designed ease of installation and removal of components, Reliability of installed replaceable items, and Skill level of the maintenance personnel.

Out of these options, the reliability of installed replaceable items has the most significant effect on the effectiveness of a maintenance program. This is because if the installed replaceable items are not reliable, then the maintenance program will not be able to function properly.

In such a case, the maintenance personnel will have to spend more time and effort to maintain the equipment, leading to increased downtime and decreased efficiency.

In contrast, if the installed replaceable items are reliable, then the maintenance program will be able to function smoothly.

The maintenance personnel will be able to quickly and easily replace the faulty components, minimizing downtime and increasing efficiency.

Therefore, the reliability of installed replaceable items is the most critical factor affecting the effectiveness of a maintenance program.

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a. sin y = -tan y * cos y = (-3/4) * (-7/8) = 21/32

b. sin (2x) = 2 * (-5/7) * (-2/7) = -5/7

c. cos (x-y) = (-2/7) * (-7/8) + (-5/7) * (21/32) = 21/28

a) Since both angles lie in quadrant IV, we know that both sin x and tan y are negative. We can use the Pythagorean identity, sin^2 x + cos^2 x = 1, to solve for cos x.

cos^2 x = 1 - sin^2 x = 1 - (-5/7)^2 = 1 - (25/49) = 24/49

cos x = sqrt(24/49) = -2/7

We can also use the identity tan^2 y = 1 - cos^2 y to solve for cos y.

cos^2 y = 1 - tan^2 y = 1 - (-3/4)^2 = 1 - 9/16 = 7/16

cos y = sqrt(7/16) = -7/8

sin y = -tan y * cos y = (-3/4) * (-7/8) = 21/32

b) Since x is in quadrant IV, we know that sin (2x) is negative. We can use the double angle formula, sin (2x) = 2 * sin x * cos x, to solve for sin (2x).

sin (2x) = 2 * (-5/7) * (-2/7) = -5/7

c) Since x and y are in quadrant IV, we know that cos (x-y) is positive. We can use the difference of angle formula, cos (x-y) = cos x * cos y + sin x * sin y, to solve for cos (x-y).

cos (x-y) = (-2/7) * (-7/8) + (-5/7) * (21/32) = 21/28

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The required solution fax ax dx = (5/3)x³ - 7.5x + C.

Given: √6x²dx=0

To calculate: 6x dx. We have to differentiate and simplify the above equation.

√6x²dx=0

=> ∫6x²^(1/2) dx=0

Using the power rule of integration, we have

∫xn dx= xn+1/(n+1)

Hence, ∫6x²^(1/2) dx

= 6x^(2+1)/(2+1) + C

= 2x³+C

As ∫6x²^(1/2) dx=0,

2x³=-C

Now, 6x dx

= d(2x³)/dx= 6x²

Using x²0x = ¹27 dx, we have:

2x³ = 27

=> x³=27/2

=> x=3∛2

∴ 6x dx = 6* (3∛2)²

= 54∛2

Hence, 6x dx = 54∛2.

Given: S₁x²ax- 6

To calculate: fax ax dx.

We have,

S₁x²ax- 6= (2.5)

S₁x²ax- (2.5)(6/2)

Using the formula S₁x²ax = x³/3, we have:

S₁x²ax- 6

= (2.5) x³/3 - 7.5

= 5x³/6 - 7.5

∴ fax ax dx= ∫5x³/6 - 7.5 dx

Using the sum rule of integration, we get: fax ax dx

= 5 ∫x³/6 dx - ∫7.5 dx

Now, ∫x³/6 dx= 6(x³/6)/3 + C

= x³/3 + C

On integrating -7.5, we get, -7.5x.

Now, fax ax dx= 5(x³/3) - 7.5x + C

= (5/3)x³ - 7.5x + C

Thus, fax ax dx = (5/3)x³ - 7.5x + C.

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