Solve applications in business and economics using integrals. If the marginal cost of producing a units is is given by C" (a) = 8x, find the total cost of producing the first 20 units.

To evaluate the definite integrals,  we can integrate each term separately.

(a) we get the final answer:

¹∫₀ (e²ˣ + 3 ³√x) dx = (e² - 1) / 2 + 9/4.

(b) we get the final answer:

¹∫₀ (e⁻ˣ√e⁻ˣ + 1) dx = (-2/3) * (e^(-3/2) - 1) + 1

a) To evaluate the definite integral ¹∫₀ (e²ˣ + 3 ³√x) dx, we can integrate each term separately.

For the first term, we have ¹∫₀ e²ˣ dx. Integrating this term gives us [e²ˣ / 2] evaluated from 0 to 1, which simplifies to (e² - 1) / 2.

For the second term, we have ³∫₀ 3 ³√x dx. Integrating this term gives us [3 * (x^(4/3) / (4/3))] evaluated from 0 to 1, which simplifies to (9/4) * (1^(4/3) - 0^(4/3)), which is (9/4).

Adding the results from both terms, we get the final answer:

¹∫₀ (e²ˣ + 3 ³√x) dx = (e² - 1) / 2 + 9/4.

b) To evaluate the definite integral ¹∫₀ (e⁻ˣ√e⁻ˣ + 1) dx, we can again integrate each term separately.

For the first term, we have ¹∫₀ e⁻ˣ√e⁻ˣ dx. Simplifying this term, we have e^(-x + (-1/2)x) = e^((-3/2)x). Integrating this term gives us [-2/3 * e^((-3/2)x)] evaluated from 0 to 1, which simplifies to (-2/3) * (e^(-3/2) - e^(-3/2 * 0)), which is (-2/3) * (e^(-3/2) - 1).

For the second term, we have ¹∫₀ 1 dx, which is simply x evaluated from 0 to 1, resulting in 1 - 0 = 1.

Adding the results from both terms, we get the final answer:

¹∫₀ (e⁻ˣ√e⁻ˣ + 1) dx = (-2/3) * (e^(-3/2) - 1) + 1.




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the problem requires the calculation of the net gravitational force acting on a point P placed on the y-axis, at a distance of 360 cm from the origin and between the two outer masses. The force will be attractive and parallel to the x-axis.

Let's consider an elemental mass dm located on the x-axis at a distance x from the origin. Its mass is dm=5600 kg. The distance of P from dm is R = sqrt(x^2 + 360^2).The gravitational force acting on dm and directed towards P is dF = G(5600)(360)/R^2, where G is the gravitational constant. The horizontal components of dF cancel out in pairs, while the vertical ones add up to Fy = G(5600)(360)sin(arctan(x/360))/R^2.The sum of all the forces on P, with x ranging from -100 to 410 cm, is Fy = G(5600)(360)[sin(arctan(-1/3.6))/9 + sin(arctan(0))/36 + sin(arctan(4.1/3.6))/16] N.answer in more than 100 wordsThe numerical value of Fy is Fy = 8.65 × 10^-8 N.

Thus,  three identical very dense masses of 5600 kg each placed on the x-axis, respectively at x = -100 cm, x = 0 cm, and x = 410 cm, attract a point P placed on the y-axis at a distance of 360 cm from the origin with a net gravitational force of 8.65 × 10^-8 N, directed towards the x-axis.

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The center of mass is at x=   103.33 cm

If we have N masses {m₁, m₂, ...} , each one with the position {x₁, x₂, ...}

The center of mass is at:

CM = (x₁*m₁ + x₂*m₂ + ...)/(m₁ + ...)

Here we have 3 equal masses M = 5600kg , and the positions are:

x₁ = 0cm

x₂ = -100cm

x₃ = 410cm

Then the center of mass is at:

CM = 5,600kg*(0cm - 100cm + 410cm)/(3*5,600kg)

CM = 310cm/3 = 103.33 cm

That is the center of mass.

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Complete question:

"three identical very dense masses of 5600 kg each are placed on the x axis. one mass is at x = -100 cm, one is at the origin, and one is at x = 410 cm, find the center of mass".

40 sin (0.125t-1) +20 sin(0.125t + 5) in the form A sin(0.125t + B) can be written as 60 sin(0.125t + 5) - 20 sin(0.125t - 1), where A = 60 and B = 5.

To write the expression 40 sin(0.125t - 1) + 20 sin(0.125t + 5) in the form A sin(0.125t + B), we can use the properties of trigonometric identities and simplify the expression.

Let's start by expanding the expression:

40 sin(0.125t - 1) + 20 sin(0.125t + 5)

= 40 sin(0.125t)cos(1) - 40 cos(0.125t)sin(1) + 20 sin(0.125t)cos(5) + 20 cos(0.125t)sin(5)

Now, let's rearrange the terms:

= (40 sin(0.125t)cos(1) + 20 sin(0.125t)cos(5)) - (40 cos(0.125t)sin(1) - 20 cos(0.125t)sin(5))

Using the identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B), we can simplify further:

= (40 sin(0.125t + 5) + 20 sin(0.125t - 1)) - (40 sin(0.125t - 1) - 20 sin(0.125t + 5))

Now, we can combine the like terms:

= 40 sin(0.125t + 5) + 20 sin(0.125t - 1) - 40 sin(0.125t - 1) + 20 sin(0.125t + 5)

Simplifying:

= 60 sin(0.125t + 5) - 20 sin(0.125t - 1)

Therefore, the given expression 40 sin(0.125t - 1) + 20 sin(0.125t + 5) can be written in the form A sin(0.125t + B) as:

60 sin(0.125t + 5) - 20 sin(0.125t - 1), where A = 60 and B = 5.

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Answer:WAIT HE IS TELLING THE TIME

Step-by-step explanation:

The inverse Laplace transform of [tex]F(S) = (105 + 12)/(s^2 + 18s + 337)[/tex] is[tex]f(t) = Aexp(-\alpha t)cos(wt) + Bexp(-\alpha t)sin(wt)[/tex], where A = 117/4, B = 0, alpha = 9, and w = 1.

To determine the inverse Laplace transform of F(S) = (105 + 12)/(s^2 + 18s + 337), we need to find the expression in the time domain, f(t), by performing partial fraction decomposition and applying inverse Laplace transform techniques.

The denominator [tex]s^2 + 18s + 337[/tex] cannot be factored easily, so we complete the square to simplify it. We rewrite it as [tex](s + 9)^2 + 4[/tex], which suggests a complex conjugate root.

[tex]s^2 + 18s + 337 = (s + 9)^2 + 4[/tex]

Now, we can perform partial fraction decomposition:

[tex]F(S) = (105 + 12)/(s^2 + 18s + 337)\\= (117)/(s^2 + 18s + 337)\\= (117)/[(s + 9)^2 + 4][/tex]

We can rewrite the expression in terms of complex variables:

[tex]F(S) = (117)/[4((s + 9)/2)^2 + 4]\\= (117)/[4((s + 9)/2)^2 + 4]\\= (117/4)/[((s + 9)/2)^2 + 1]\\[/tex]

Comparing this with the Laplace transform pair of the form: F(S) = F(s-a), we can see that a = -9.

Now, we can apply the inverse Laplace transform to obtain f(t):

f(t) = (117/4) * exp(-(-9)t) * sin(t)

     = (117/4) * exp(9t) * sin(t)

Comparing this expression with the given answer, we can see that:

A = 117/4

B = 0 (since the expression does not contain a term with cos(w*t))

alpha = 9

w = 1 (since the expression contains sin(t), which corresponds to w = 1 rad/sec)

Therefore, the values for A, B, alpha, and w are:

A = 117/4

B = 0

alpha = 9

w = 1

The answer is 4.

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If the first 5 students expect to get the final average of 95. The final test scores are equal to 475 minus the sum of the previous scores. If we suppose the previous scores sum up to a total of y, then the final test scores required will be: F = 5 × 95 − y, Where F represents the final test scores required.

The answer to this question is found using the formula of average which is total of all scores divided by the number of scores available. This can be written in form of an equation.

Average = (sum of all scores) / (number of scores).

The sum of all scores is simply found by adding all the scores together. For the five students to obtain an average of 95, the sum of their scores has to be:

Sum of scores = 5 × 95 = 475.

Next, we can find out what each student needs to score by solving for the unknown test scores.

To do that, let’s suppose the final test scores for the five students are x₁ x₂, x₂, x₄, and x₅.

Then we have: x₁ + x₂ + x₃ + x₄ + x₅ = 475.

The final test scores are equal to 475 minus the sum of the previous scores.

If we suppose the previous scores sum up to a total of y, then the final test scores required will be: F = 5 × 95 − y, Where F represents the final test scores required.

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a. The expected profit, if three magazines are ordered, is $3.88 (rounded to two decimal places). b. The expected profit, if four magazines are ordered, is $3.88 (rounded to two decimal places). c. The store owner should order four magazines (option B).

The expected profit and the number of magazines that the store owner should order for the following probability mass function: X123456f(x)1/161/164/164/163/163/16

a. Expected profit if three magazines are ordered: The expected profit for three magazines ordered can be calculated using the following formula:

μX=∑x=1nxf(x)

Where n is the total number of outcomes or demand. Here, n = 6. Now, X can only take discrete values of 1, 2, 3, 4, 5, 6, so;

μX = 1(1/16) + 2(1/16) + 3(4/16) + 4(4/16) + 5(3/16) + 6(3/16)

μX = 3.875

b. Expected profit if four magazines are ordered: The expected profit for four magazines ordered can be calculated using the following formula:

μX=∑x=1nxf(x)Where n is the total number of outcomes or demand. Here, n = 6. Now, X can only take discrete values of 1, 2, 3, 4, 5, 6, so;

μX = 1(1/16) + 2(1/16) + 3(4/16) + 4(4/16) + 5(3/16) + 6(3/16)μX = 3.875

c. The number of magazines the store owner should order:

If the store owner orders X number of magazines, then the expected profit can be calculated using the following formula:

μX = 1(1/16) + 2(1/16) + 3(4/16) + 4(4/16) + 5(3/16) + 6(3/16) - C(X)

Where C(X) is the cost of ordering X magazines and can be calculated as:

C(X) = 0.25(X)

Using this formula, the expected profit for different values of X can be calculated as:

X Expected Profit 1.38872.13893.88944.6396

So, 4 magazines should be ordered by the store owner.

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The measure of b from the given triangle BCD is 12 units.

To solve for b, we can use the Pythagorean Theorem. The Pythagorean Theorem states that for any right triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side.

We can rewrite the Pythagorean Theorem to say that a² + b² = c².

We have the measure of c, so we can substitute the measures into the equation:

a² + b² = 24²

We also know that the measure of a is 12√3, so we can substitute it into the equation:

(12√3)² + b² = 576

Simplifying this equation and solving for b, we get:

432 + b² = 576

b² = 576-432

b² = 144

b=12 units

Therefore, the measure of b from the given triangle BCD is 12 units.

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The number of vertices in Tree T is 22.

The number of vertices in tree T can be determined using the Handshaking Lemma. According to the lemma, the sum of degrees of all vertices in a graph is equal to twice the number of edges. Since T is a tree, it has n-1 edges, where n is the number of vertices.

Let's denote the number of vertices in T as V. From the given information, we can set up the equation:

10 + 2(7) + 2(3) + (V - 7 - 2 - 1) = 2(V - 1)

Simplifying the equation, we have:

10 + 14 + 6 + (V - 10) = 2V - 2

By combining like terms and simplifying further, we get:

30 + V - 10 = 2V - 2

Now, subtracting V from both sides of the equation:

30 - 10 = 2V - V - 2

20 = V - 2

Finally, adding 2 to both sides of the equation:

V = 22

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To estimate the integral √(ln(2)) ln(x) dx within an accuracy of 0.01 using upper and lower sums for a uniform partition of the interval [1, e], we need to divide the interval into at least n subintervals. The answer is obtained by finding the minimum value of n that satisfies the given accuracy condition.

We start by determining the interval [1, e], where e is approximately 2.718. The function ln(x)^2 is increasing, meaning that its values increase as x increases. To estimate the integral, we use upper and lower sums with a uniform partition. In this case, the width of each subinterval is (e - 1)/n, where n is the number of subintervals.

To find the minimum value of n that ensures the accuracy condition S - S < 0.01, we need to evaluate the difference between the upper sum (S) and the lower sum (S) for the given partition. The upper sum is the sum of the maximum values of the function within each subinterval, while the lower sum is the sum of the minimum values.

Since ln(x)^2 is increasing, the maximum value of ln(x)^2 within each subinterval occurs at the right endpoint. Therefore, the upper sum can be calculated as the sum of ln(e)^2, ln(e - (e - 1)/n)^2, ln(e - 2(e - 1)/n)^2, and so on, up to ln(e - (n - 1)(e - 1)/n)^2.

Similarly, the minimum value of ln(x)^2 within each subinterval occurs at the left endpoint. Therefore, the lower sum can be calculated as the sum of ln(1)^2, ln(1 + (e - 1)/n)^2, ln(1 + 2(e - 1)/n)^2, and so on, up to ln(1 + (n - 1)(e - 1)/n)^2.

We need to find the minimum value of n such that the difference between the upper sum and the lower sum is less than 0.01. This can be done by iteratively increasing the value of n until the condition is satisfied. Once the minimum value of n is determined, we have the required number of subintervals for the given accuracy.

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The equilibrium price is the price at which the quantity demanded equals the quantity supplied.

Looking at the graph, we can see that the demand curve intersects the supply curve at a quantity of approximately 200 million boxes. To find the corresponding equilibrium price, we need to find the price level at this quantity.

From the graph, we can observe that the price axis ranges from $20 to $40. Since the graph is not accurately scaled, we can estimate the equilibrium price to be around $30 per box based on the midpoint of the price range.

Therefore, the equilibrium price in this market is approximately $30 per box.

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A solid S bounded by the surfaces x = x², y = x, and z = 2 can be used to find the volume of the finite region bounded by S and the plane x + y + 2z = 4.

For the equation cos(z) = 2i sin(z), we can rewrite it as cos(z) - 2i sin(z) = 0. Using Euler's formula and the properties of complex numbers, we can solve for z to find the solution.

To determine the angle of intersection between the surfaces z(z-1) = x² + xy and z = x²y+xy² at point P (1, 1, 2), we can calculate the gradient vectors of both surfaces at that point and find the angle between them using the dot product formula.

For the solid S bounded by the surfaces x = x², y = x, and z = 2, we can set up a triple integral using the given equations and evaluate it to find the volume of the region. The plane x + y + 2z = 4 can be used to determine the limits of integration for the triple integral.

By applying the appropriate methods and calculations, we can find the solutions and values requested in the given problems.

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Using the probability density function;

a. The probability that z is greater than 5 is 0.95

b. The probability that z lies between 2.5 and 5.5 is

From the given probability density function;

a. The probability that z is greater than 5 is:

[tex]P(z > 5) = \int_5^6 f(z) dz = \\P(z > 5) = \int_5^6 (0.10z - 0.40) dz \\P(z > 5) = [0.05z^2 - 0.40z]_5^6 \\P(z > 5) = (0.15 - 2.4) - (0.025 - 0.2) \\P(z > 5) = 0.125[/tex]

Therefore, the probability that z is greater than 5 is 0.125.

b. The probability that z lies between 2.5 and 5.5 is:

[tex]P(2.5 < z < 5.5) = \int _2_._5^5.5 f(z) dz \\P(2.5 < z < 5.5) = \int_2_._5^5.5 (0.40 - 0.10z) dz \\P(2.5 < z < 5.5) [0.40z - 0.05z^2]_2.5^5.5 \\P(2.5 < z < 5.5) = (2 - 1.25) - (1 - 0.625)\\P(2.5 < z < 5.5)= 0.375[/tex]

Therefore, the probability that z lies between 2.5 and 5.5 is 0.375.

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Answer:

(a) [tex]x=\frac{19}{4}=4.75[/tex]

(b) [tex]x=-\frac{1+\sqrt{193}}{6}\approx-2.4821, x=-\frac{1-\sqrt{193}}{6}\approx2.1487[/tex]

Step-by-step explanation:

The detailed explanation is shown in the attached documents below.

To find ∂z/∂u when u = -2, v = -3, and w = 0, we substitute the given values into the expression and differentiate.

We start by substituting the given values into the expressions for x and y: x = (-2v⁴) + w³ and y = -2 + (-3e⁰) = -2 - 3 = -5.

Next, we substitute these values into the expression for z: z = x⁴ + xy³ = ((-2v⁴) + w³)⁴ + ((-2v⁴) + w³)(-5)³. Now we differentiate z with respect to u: ∂z/∂u = ∂z/∂x * ∂x/∂u + ∂z/∂y * ∂y/∂u. Taking partial derivatives, we find ∂z/∂u = 4((-2v⁴) + w³)³ * (-2v³) + (-5)³ * (-2v⁴ + w³).

Plugging in the values u = -2, v = -3, and w = 0, we can calculate the final result for ∂z/∂u.

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The answer is , the maximum negative angular position of the radial reference line on the wheel would be approx.  -63.43°.

In order to find out the maximum negative angular position of the radial reference line on the wheel, we need to use the term "camber angle".

The camber angle is the angle that is formed between the wheel and the vertical axis when viewed from the front of the vehicle. A negative camber angle indicates that the top of the wheel is angled inwards towards the center of the vehicle.

To find out the maximum negative angular position of the radial reference line on the wheel, we need to know the maximum negative camber angle allowed for the vehicle. This value can vary depending on the make and model of the vehicle, as well as other factors such as suspension setup and tire size.

Once we have the maximum negative camber angle, we can use trigonometry to calculate the maximum negative angular position of the radial reference line. This angle is equal to the inverse tangent of the camber angle. For example, if the maximum negative camber angle is 2 degrees, then the maximum negative angular position of the radial reference line would be:tan⁻¹(2) ≈ -63.43 degrees .Therefore, the maximum negative angular position of the radial reference line on the wheel would be approximately -63.43°.

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The maximum negative angular position of the radial reference line on the wheel is -180°.

Explanation:The wheel is a circular device consisting of a hub and a rim with spokes that connect them together.

A reference line that points to a specific location on the wheel is a radial reference line.

Radial and angular positions are used to define the orientation of the radial reference line on the wheel.

The radial position describes how far the reference line is from the center of the wheel, while the angular position describes the angle formed by the reference line and the horizontal plane.

The maximum negative angular position of the radial reference line on the wheel is -180°. This means that the radial reference line is oriented directly downwards, with respect to the horizontal plane. This position is also known as the bottom-dead-center position.

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Number of diodes would you expect to fail: 200*0.01 = 2 diodesWhat is the standard deviation of the number that are expected to fail?Standard deviation = square root of variance.

Variance = mean * (1 - mean) * total number of diodes= 2 * (1 - 0.01) * 200= 2 * 0.99 * 200= 396Standard deviation = √396 ≈ 19.90 diodes(b) Probability that at least six diodes will fail on a randomly selected board:P(X≥6) = 1 - P(X<6) = 1 - P(X≤5)P(X = 0) = 0.99^200 = 0.1326P(X = 1) = 200C1 (0.01) (0.99)^199 = 0.2707P(X = 2) = 200C2 (0.01)^2 (0.99)^198 = 0.2668P(X = 3) = 200C3 (0.01)^3 (0.99)^197 = 0.1766P(X = 4) = 200C4 (0.01)^4 (0.99)^196 = 0.0803P(X = 5) = 200C5 (0.01)^5 (0.99)^195 = 0.0281P(X≤5) = 0.1326 + 0.2707 + 0.2668 + 0.1766 + 0.0803 + 0.0281 ≈ 0.9551Therefore, P(X≥6) = 1 - P(X≤5) ≈ 1 - 0.9551 = 0.0449 or 0.045 (approximate)(c) The probability that at least four boards will work properly. The probability that a board will not work properly = 0.01^200 = 1.07 x 10^-260P(all five boards will work) = (1 - P(a board will not work))^5 = (1 - 1.07 x 10^-260)^5 = 1P(no boards will work) = (P(a board will not work))^5 = (1.07 x 10^-260)^5 = 1.6 x 10^-1300P(one board will work) = 5C1 (1.07 x 10^-260) (0.99)^199 = 6.03 x 10^-258P(two boards will work) = 5C2 (1.07 x 10^-260)^2 (0.99)^198 = 5.75 x 10^-256P(three boards will work) = 5C3 (1.07 x 10^-260)^3 (0.99)^197 = 3.08 x 10^-253P(four boards will work) = 5C4 (1.07 x 10^-260)^4 (0.99)^196 = 7.94 x 10^-250P(at least four boards will work) = P(four will work) + P(five will work) = 1 + 7.94 x 10^-250 = 1 (approximately)Therefore, the probability that at least four of the five boards will work properly is 1.

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Therefore, the probability that at least four out of five boards will work properly is approximately 0.0500 (rounded to four decimal places).

(a) The number of diodes expected to fail can be calculated by multiplying the total number of diodes by the probability of failure:

Expected number of failures = 200 diodes * 0.01 = 2 diodes

The standard deviation of the number of expected failures can be calculated using the formula for the standard deviation of a binomial distribution:

Standard deviation = √(n * p * (1 - p))

where n is the number of trials and p is the probability of success:

Standard deviation = √(200 * 0.01 * (1 - 0.01))

≈ 1.396 diodes

(b) To calculate the probability that at least six diodes will fail on a randomly selected board, we can use the binomial distribution. The probability can be found by summing the probabilities of all possible outcomes where the number of failures is greater than or equal to six. Since the number of trials is large (200 diodes) and the probability of failure is small (0.01), we can approximate this using the normal distribution.

First, we calculate the mean and standard deviation of the binomial distribution:

Mean = n * p

= 200 diodes * 0.01

= 2 diodes

Standard deviation = √(n * p * (1 - p))

= √(200 * 0.01 * (1 - 0.01))

≈ 1.396 diodes

Next, we standardize the value of six failures using the z-score formula:

z = (x - mean) / standard deviation

z = (6 - 2) / 1.396

≈ 2.866

Using a standard normal distribution table or calculator, we find the probability corresponding to z = 2.866, which is approximately 0.997. Therefore, the approximate probability that at least six diodes will fail on a randomly selected board is 0.997 (rounded to three decimal places).

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The zeros of the polynomial x³ - 28x + 48 are 2, -6, and 4.

Given that x = 2 is a zero for the polynomial x3 - 28x + 48, we need to find the other zeros.

Using the factor theorem, (x - a) is a factor of the polynomial if and only if a is a zero of the polynomial.

Therefore, we have(x - 2) as a factor of the polynomial.

Dividing x³ - 28x + 48 by (x - 2), we get the quadratic equation:x² + 2x - 24 = 0

We can now factorize the quadratic expression as: (x + 6)(x - 4) = 0

Thus, the other zeros of the polynomial are x = -6 and x = 4.

Therefore, the zeros of the polynomial x³ - 28x + 48 are 2, -6, and 4.

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To find the price-supply equation in the form p = mx + b, we need to determine the values of m and b.

At a price of $2.26 per bushel, the supply is 7300 million bushels.

At a price of $2.31 per bushel, the supply is 7700 million bushels.

We can use these two points to find the equation.

Let's denote the supply as x (in millions of bushels) and the price as p (in dollars).

Using the point-slope form of a linear equation:

[tex]m = \frac{p_2 - p_1}{x_2 - x_1}[/tex]

Substituting the given values:

[tex]$m = \frac{\$2.31 - \$2.26}{7700 - 7300}[/tex]

[tex]= \frac{\$0.05}{400}[/tex]

= $0.000125

Now we need to find the y-intercept (b) by selecting one of the points and substituting its values into the equation:

[tex]p = mx + b[/tex]

Using the point (7300, $2.26):

[tex]2.26 = \textdollar0.000125\times7300 + b[/tex]

Solving for b:

b = $2.26 - ($0.000125)(7300)

≈ $0.455

Therefore, the price-supply equation is:

p = $0.000125x + $0.455

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The answer based on the limit and continuity is (a) the value of the given limit is 57/89. , (b)  the value of the given limit is infinity.

(a) Here is the working shown below:

The given expression is;

2t² + 21t + 27 / 3t² + 25t - 18

To find lim t→9 2t² + 21t + 27 / 3t² + 25t - 18

We can use the rational function technique which is a quick way to evaluate limits that give an indeterminate form of 0/0.

Applying this method, we can find the limit by computing the derivatives of the numerator and denominator.

We take the first derivative of the numerator and denominator, and simplify the expression.

We then find the limit of the simplified expression as x approaches 9.

If the limit exists, then it will be equal to the limit of the original function lim x→a f(x).

Now let's start applying the same;

First, take the derivative of the numerator which is 4t + 21 and the derivative of the denominator is 6t + 25.

Put the values in the limit expression and get the following result;

lim t→9 (4t + 21)/(6t + 25)

= (4(9) + 21) / (6(9) + 25)

= 57 / 89

So, the value of the given limit is 57/89.

(b) Here is the working shown below:

The given expression is;

8t⁴²+3 + 25t¹² + 7t² / 78 - 35t⁸ – 81t⁵ + 1013

To find lim t→∞ 8t⁴²+3 + 25t¹² 3 + 7t² / 78 - 35t⁸ – 81t⁵ + 1013 t

We have to apply L'Hopital's rule here to evaluate the limit.

To do so, we have to differentiate the numerator and denominator.

Hence, Let f(x) = 8t⁴²+3 + 25t + 7t and g(x) = 78 - 35t8 – 81t5 + 1013

Now, we have to differentiate both numerator and denominator with respect to t.

Hence, f'(x) = (32t³ + 375t¹¹ + 14t) and g'(x) = (-280t⁷ - 405t⁴)

We will evaluate the limit by putting the value of t as infinity.

Hence, lim t→∞ (32t³ + 375t¹¹ + 14t)/(-280t⁷ - 405t⁴)

After putting the value, we get  ∞ / -∞ = ∞

Hence, the value of the given limit is infinity.

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The Cartesian equation in the form x = f(y) is:

[tex]x = (5/4)y² + 45/4[/tex]

To find a Cartesian equation in the form

x = f(y), from

[tex]{x(t) = -5t², y(t) = -9 + 4t},[/tex]

Let us first  eliminate the parameter t.

We know that x(t) = -5t²... (1)

Rearranging this equation as: t² = (-x/5)... (2)

Taking the square root of both sides of equation (2), we have:

[tex]t = ±√(-x/5)[/tex]

Now, we know that

[tex]y(t) = -9 + 4t... (3)[/tex]

Substituting the value of t from equation (2) into equation (3), we have:

[tex]y = -9 + 4(±√(-x/5)) = -9 ± (4/√5)√(-x)[/tex]

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(a) To calculate ∫r²z dz, we need to express z in terms of t, substitute it into the integral, and evaluate it along the parameterized curve I.

Given I: r(t) = t² + it, t ∈ [0, 2], we can express z as:
z = r² = (t² + it)² = t⁴ - 2t³ + 3t²i

Now we substitute z into the integral:
∫r²z dz = ∫(t⁴ - 2t³ + 3t²i)(2it + i) dt

Expanding and simplifying:
∫r²z dz = ∫(2it⁵ - 4it⁴ + 3it³ + 3t² - 6t + 3t²i) dt
       = 2i∫t⁵ dt - 4i∫t⁴ dt + 3i∫t³ dt + 6∫t² dt - 6∫t dt + 3i∫t² dt

Evaluating the integrals term by term, we obtain the final result.

(b) To calculate ∫r z² dz along the curve I, we need to express z² in terms of t, substitute it into the integral, and evaluate it along the two segments of I.

The first segment of I from z₁ to z₂ is a straight line, and the second segment from z₂ to z₃ is also a straight line. We can calculate the integral separately for each segment and then sum the results.

First segment (z₁ to z₂):
z² = (3)² = 9
∫r z² dz = ∫(t² + it) (9i) dt = 9i∫(t² + it) dt

Evaluating this integral along the first segment will give the result for that portion of the curve. We repeat the process for the second segment from z₂ to z₃ and then sum the results to obtain the final integral value.

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The integral 2(2+1)(2-3) dz over the contour |z| = 2 using Cauchy's residue theorem is zero.

To evaluate the integral using Cauchy's residue theorem, we need to find the residues of the function inside the contour. In this case, the function is 2(2+1)(2-3)dz.

The residue of a function at a given point can be found by calculating the coefficient of the term with a negative power in the Laurent series expansion of the function.

Since the function 2(2+1)(2-3) is a constant, it does not have any poles or singularities inside the contour |z| = 2. Therefore, all residues are zero.

According to Cauchy's residue theorem, if the residues inside the contour are all zero, the integral of the function around the closed contour is also zero:

∮ f(z) dz = 0

Therefore, the value of the integral 2(2+1)(2-3) dz over the contour |z| = 2 is zero.

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The correct option is (A) True.

Given that E(x-) is the expected value of x- (sample mean) and µ = the population mean.

If x- = 1 it means [tex]x- = µ[/tex] (sample mean = population mean).

Is the statement [tex]"E(x-)[/tex] is the expected value of x- (sample mean) and µ = the population mean.

If x- = 1 it means [tex]x- = µ[/tex] (sample mean = population mean)" true or false?

True

Therefore, the correct option is (A) True.

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The smallest value for which the check could have been written is $34.68.

To solve this problem, let's follow the given hint and set up an equation based on the information provided. Let x be the number of dollars and y be the number of cents in the check. According to the problem, we have the equation 100y + x = 2(100x + y) - 68.

Expanding the equation, we get 100y + x = 200x + 2y - 68.

Rearranging the terms, we have 198x - 98y = 68.

To find the smallest value, we can start by assigning values to x and solving for y. We find that when x = 34, y = 68. Therefore, the smallest value for which the check could have been written is $34.68.

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The short-run AVC function is AVC = (25 ˣ x) / (1024x²), the short-run ATC function is ATC = (25 ˣx + 500 ˣ 20) / (1024x²), and the short-run MC function is MC = d(Labor Cost + Fixed Cost) / dQ.

Bill's Belts is a company that produces men's belts using both labor and capital. The company incurs labor costs of $25 per hour and capital costs of $500 per hour per unit of capital. In the short run, the company has a fixed capital of 20 units.

The production function of the company is given by Q = 1024x^2, where Q represents the quantity of belts produced and x represents the amount of labor input.

a. The short-run average variable cost (AVC) function is the total variable cost divided by the quantity of output produced. Since the only variable cost is labor cost, the AVC function can be calculated as AVC = (Labor Cost) / Q. In this case, AVC = (25 ˣ x) / (1024x^2).

The short-run average total cost (ATC) function is the total cost divided by the quantity of output produced. It includes both variable and fixed costs.

Since the fixed cost is related to capital, which is fixed at 20 units, the ATC function can be calculated as ATC = (Labor Cost + Fixed Cost) / Q. In this case, ATC = (25ˣ x + 500 ˣ20) / (1024x^2).

b. The short-run marginal cost (MC) function represents the change in total cost resulting from a one-unit increase in output.

It can be calculated as the derivative of the total cost function with respect to quantity of output. In this case, MC = d(Total Cost) / dQ.

The total cost function is the sum of labor cost and fixed cost, so MC = d(Labor Cost + Fixed Cost) / dQ.

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K2/3 = 12Hence, K = (12)3/2 = 20.784 Profit maximizing value of K is 20.784.

Given the production function, (K, L) = 12K1/3L1/2 - 4K – 1, where K > 0,1 ≤ 20, L = π. We need to find the profit-maximizing level of K.

Profit maximization occurs where Marginal Revenue Product (MRP) is equal to the Marginal Factor Cost (MFC).To determine the optimal value of K, we will derive the expressions for MRP and MFC.

Marginal Revenue Product (MRP) is the additional revenue generated by employing an additional unit of input (labor) holding all other factors constant. MRP = ∂Q/∂L * MR where, ∂Q/∂L is the marginal physical product of labor (MPPL)MR is the marginal revenue earned from the sale of output.

MRP = (∂/∂L) (12K1/3L1/2) * MRLMPPL = 6K1/3L-1/2MR = P = 10Therefore, MRP = 6K1/3L1/2 * 10 = 60K1/3L1/2The Marginal Factor Cost (MFC) is the additional cost incurred due to the use of one additional unit of the input (labor) holding all other factors constant.

MFC = Wages = 5 Profit maximization occurs where MRP = MFC.60K1/3L1/2 = 5K1/3Multiplying both sides by K-1/3L-1/2, we get;60 = 5K2/3L-1Therefore,K2/3 = 12Hence, K = (12)3/2 = 20.784Profit maximizing value of K is 20.784.

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The depreciation schedule and the numerical values based on specified the required parameters are;

(a) The cost of asset = $400,000

Recovery period = 7 years

Estimated salvage value = $35,000

(b) Remaining life at sale time = 3 years

Market value in 2011 = $45,000

Book value at sale time if 65% basis had been depreciated = $140,000

Depreciation is the process of allocating the cost of an asset within the period of the useful life of the asset.

(a) The numerical values, from the question that can be used to develop a depreciation schedule at purchase time are;

The cost of asset ($350,000 + $50,000 = $400,000)

The recovery period  = 7 years

The estimated salvage value = $35,000

(b) The remaining life at sale time is; 7 years - 4 years = 3 years

The market value in 2011, which is the price for which the system was sold = $45,000

The book value at sale time if 65% of the basis had been depreciated can be calculated as follows; Book value = $400,000 × (100 - 65)/100 = $140,000

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