Let f(n)=n 2
and g(n)=n log 3
​
(10)
. Which holds: f(n)=O(g(n))
g(n)=O(f(n))
f(n)=O(g(n)) and g(n)=O(f(n))
​

The determinant of the matrix B is (a) det(A) = -2

from the question, we have the following parameters that can be used in our computation:

det(A) = -2

We understand that

B is the matrix formed when two elementary row operations are performed on A

By definition;

The determinant of a matrix is unaffected by elementary row operations.

using the above as a guide, we have the following:

det(B) = det(A) = -2.

Hence, the determinant of the matrix B is -2

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Domain: {-1, 5, -2, 3, -4, -5}, Range: {-6, -8, 8, -2, -5}. These sets represent the distinct values that appear as inputs and outputs in the given relation.

To determine the values in the domain and range of the given relation, we can examine the set of ordered pairs provided.

The given set of ordered pairs is: {(-1, -6), (5, -8), (-2, 8), (3, -2), (-4, -2), (-5, -5)}

(a) Domain: The domain refers to the set of all possible input values (x-values) in the relation. We can determine the domain by collecting all unique x-values from the given ordered pairs.

From the set of ordered pairs, we have the following x-values: -1, 5, -2, 3, -4, -5

Therefore, the domain of the relation is {-1, 5, -2, 3, -4, -5}.

(b) Range: The range represents the set of all possible output values (y-values) in the relation. Similarly, we need to collect all unique y-values from the given ordered pairs.

From the set of ordered pairs, we have the following y-values: -6, -8, 8, -2, -5

Therefore, the range of the relation is {-6, -8, 8, -2, -5}

It's worth noting that the order in which the elements are listed in the sets does not matter, as sets are typically unordered.

It's important to understand that the domain and range of a relation can vary depending on the specific set of ordered pairs provided. In this case, the given set uniquely determines the domain and range of the relation.

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Distributive property was used incorrectly going from Line 2 to Line 3

The line which used property incorrectly while going from Line 2 to Line 3 is Line 3.

The expressions:

Line 1: -3(m - 3) + 6 = 21

Line 2: -3(m - 3) = 15

Line 3: -3m - 9 = 15

Line 4: -3m = 24

Line 5: m = -8

The distributive property is used incorrectly going from Line 2 to Line 3. Because when we distribute the coefficient -3 to m and -3, we get -3m + 9 instead of -3m - 9 which was incorrectly calculated.

Therefore, -3m - 9 = 15 is incorrect.

In this case, the correct expression for Line 3 should have been as follows:

-3(m - 3) = 15-3m + 9 = 15

Now, we can simplify the above equation as:

-3m = 6 (subtract 9 from both sides)or m = -2 (divide by -3 on both sides)

Therefore, the correct answer is "Distributive property".

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Volume of the solid obtained by rotating the region is 67π/6 .

Given,

Curves:

y=x², y=0, x=1, and x=2 .

The arc of the parabola runs from (1,1) to (2,4) with vertical lines from those points to the x-axis. Rotated around x=4 gives a solid with a missing circular center.

The height of the rectangle is determined by the function, which is x² . The base of the rectangle is the circumference of the circular object that it was wrapped around.

Circumference = 2πr

At first, the distance is from x=1 to x=4, so r=3.

It will diminish until x=2, when r=2.

For any given value of x from 1 to 2, the radius will be 4-x

The circumference at any given value of x,

= 2 * π * (4-x)

The area of the rectangular region is base x height,

= [tex]\int _1^22\pi \left(4-x\right)x^2dx[/tex]

= [tex]2\pi \cdot \int _1^2\left(4-x\right)x^2dx[/tex]

= [tex]2\pi \left(\int _1^24x^2dx-\int _1^2x^3dx\right)[/tex]

= [tex]2\pi \left(\frac{28}{3}-\frac{15}{4}\right)[/tex]

Therefore volume of the solid is,

= 67π/6

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The percent markup based on cost is (P^(6) - 1) x 100%.

To calculate the percent markup based on cost, we need to find the difference between the selling price and the cost, divide that difference by the cost, and then express the result as a percentage.

The cost of a box of Wendy's cupcakes is P^(10). The selling price is P^(16). So the difference between the selling price and the cost is:

P^(16) - P^(10)

We can simplify this expression by factoring out P^(10):

P^(16) - P^(10) = P^(10) (P^(6) - 1)

Now we can divide the difference by the cost:

(P^(16) - P^(10)) / P^(10) = (P^(10) (P^(6) - 1)) / P^(10) = P^(6) - 1

Finally, we can express the result as a percentage by multiplying by 100:

(P^(6) - 1) x 100%

Therefore, the percent markup based on cost is (P^(6) - 1) x 100%.

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The expected amount of dollars to be paid by Miami Co. for the pesos in one year is approximately $1,270,588.24. option e is correct.

We need to consider the inflation rates and the concept of purchasing power parity (PPP).

Purchasing power parity (PPP) states that the exchange rate between two currencies should equal the ratio of their price levels.

Let us assume that PPP holds, meaning that the change in exchange rates will be proportional to the inflation rates.

First, let's calculate the expected exchange rate in one year based on the inflation differentials:

Expected exchange rate = Spot rate × (1 + U.S. inflation rate) / (1 + Mexican inflation rate)

= 0.12× (1 + 0.08) / (1 + 0.02)

= 0.12 × 1.08 / 1.02

= 0.1270588235

Now, we calculate the expected amount of dollars to be paid by Miami Co. for 10 million Mexican pesos in one year:

Expected amount of dollars = Expected exchange rate × Amount of Mexican pesos

Expected amount of dollars = 0.1270588235 × 10,000,000

Expected amount of dollars = $1,270,588.24

Therefore, the expected amount of dollars to be paid by Miami Co. for the pesos in one year is approximately $1,270,588.24.

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The limit limx→[infinity]​f(x) = 36, limx→−[infinity]​f(x) = 36. The function has one horizontal asymptote, y = 36. Option (a) is correct.

Given function is f(x) = 36x + 66x⁻¹We need to evaluate limx→∞​f(x) and limx→-∞​f(x) and find horizontal asymptotes, if any.Evaluate limx→∞​f(x):limx→∞​f(x) = limx→∞​(36x + 66x⁻¹)= limx→∞​(36x/x + 66/x⁻¹)We get  ∞/∞ form and hence we apply L'Hospital's rulelimx→∞​f(x) = limx→∞​(36 - 66/x²) = 36

The limit exists and is finite. Hence the correct choice is A) limx→∞​36x+66x​=36.Evaluate limx→−∞​f(x):limx→-∞​f(x) = limx→-∞​(36x + 66x⁻¹)= limx→-∞​(36x/x + 66/x⁻¹)

We get -∞/∞ form and hence we apply L'Hospital's rulelimx→-∞​f(x) = limx→-∞​(36 + 66/x²) = 36

The limit exists and is finite. Hence the correct choice is A) limx→−∞​36x+66x​=36.  Hence the horizontal asymptote is y = 36. Hence the correct choice is A) The function has one horizontal asymptote, y = 36.

The limit limx→[infinity]​f(x) = 36, limx→−[infinity]​f(x) = 36. The function has one horizontal asymptote, y = 36.

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A Bernoulli distribution represents the probability distribution of a random variable with only two possible outcomes.

Three examples of Bernoulli rv's are as follows:

X = 1 if a randomly selected lightbulb needs to be replaced and X = 0 otherwise X = 1 if a randomly selected shopper purchases a food item at a department store and X = 0 otherwise X = 1 if a randomly selected day has a high temperature of over 100 degrees and X = 0 otherwise. These are the Bernoulli random variables. A Bernoulli trial is a random experiment that has two outcomes: success and failure. These trials are used to create Bernoulli random variables (r.v. ) that follow a Bernoulli distribution.

In Bernoulli's distribution, p denotes the probability of success, and q = 1 - p denotes the probability of failure. It's a type of discrete probability distribution that describes the probability of a single Bernoulli trial. the above three Bernoulli rv's that are different from those given in the text.

A Bernoulli distribution represents the probability distribution of a random variable with only two possible outcomes.

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The order of the given differential equation dy/dx + 2 = -9x is 1.

The differential equations among the given options are:

(a) m dtdx = p

(c) y' = 4x^2 + x + 1

(d) dt^2 d^2z/dx^2 = x + 2

Therefore, options (a), (c), and (d) are differential equations.

Now, let's determine the order of the differential equation dy/dx + 2 = -9x.

The order of a differential equation is determined by the highest order derivative present in the equation. In this case, the highest order derivative is dy/dx, which is a first-order derivative.

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(a) The equation 1 - x² = ɛe¯x represents a transcendental equation that combines a polynomial function (1 - x²) with an exponential function (ɛe¯x). To sketch the functions, we can start by analyzing each term separately. The polynomial function 1 - x² represents a downward-opening parabola with its vertex at (0, 1) and intersects the x-axis at x = -1 and x = 1. On the other hand, the exponential function ɛe¯x represents a decreasing exponential curve that approaches the x-axis as x increases.

For small values of ε, the exponential term ɛe¯x becomes very small, causing the curve to hug the x-axis closely. As a result, the intersection points between the polynomial and exponential functions occur close to the x-intercepts of the polynomial (x = -1 and x = 1). Since the exponential function is decreasing, there will be two solutions to the equation, one near each x-intercept of the polynomial.

(b) To find a two-term asymptotic expansion for small ε, we assume that ε is a small parameter. We can expand the exponential function using its Maclaurin series:

ɛe¯x = ɛ(1 - x + x²/2 - x³/6 + ...)

Substituting this expansion into the equation 1 - x² = ɛe¯x, we get:

1 - x² = ɛ - ɛx + ɛx²/2 - ɛx³/6 + ...

Ignoring terms of higher order than ε, we obtain a quadratic equation:

x² - εx + (1 - ε/2) = 0.

Solving this quadratic equation gives us the two-term asymptotic expansion for each solution.

(c) To find a three-term asymptotic expansion for small ε, we include one more term from the exponential expansion:

ɛe¯x = ɛ(1 - x + x²/2 - x³/6 + ...)

Substituting this expansion into the equation 1 - x² = ɛe¯x, we get:

1 - x² = ɛ - ɛx + ɛx²/2 - ɛx³/6 + ...

Ignoring terms of higher order than ε, we obtain a cubic equation:

x² - εx + (1 - ε/2) - ɛx³/6 + ...

Solving this cubic equation gives us the three-term asymptotic expansion for each solution.

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The margin of error at a 98% confidence level is approximately 4.26.To find the margin of error (EBM - Error Bound for a Population Mean) at a 98% confidence level.

We need to use the formula:

Margin of Error = Z * (s / sqrt(n))

where Z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.

For a 98% confidence level, the corresponding z-score is 2.33 (obtained from the standard normal distribution table).

Plugging in the values into the formula:

Margin of Error = 2.33 * (10 / sqrt(30))

Calculating the square root and performing the division:

Margin of Error ≈ 2.33 * (10 / 5.477)

Margin of Error ≈ 4.26

Therefore, the margin of error at a 98% confidence level is approximately 4.26.

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The 99% confidence interval for the population proportion (p) is approximately 0.323 to 0.457, and the margin of error is approximately 0.067.

The margin of error and confidence interval can be calculated as follows:

First, we need to find the standard error of the proportion:

SE = sqrt[p(1-p)/n]

where:

p is the sample proportion (0.39 in this case)

n is the sample size (500 in this case)

Substituting the values, we get:

SE = sqrt[(0.39)(1-0.39)/500] ≈ 0.026

Next, we can find the margin of error (ME) using the formula:

ME = z*SE

where:

z is the critical value for the desired confidence level (99% in this case). From a standard normal distribution table or calculator, the z-value corresponding to the 99% confidence level is approximately 2.576.

Substituting the values, we get:

ME = 2.576 * 0.026 ≈ 0.067

This means that we can be 99% confident that the true population proportion falls within a range of 0.39 ± 0.067.

Finally, we can calculate the confidence interval by subtracting and adding the margin of error from the sample proportion:

CI = [p - ME, p + ME]

Substituting the values, we get:

CI = [0.39 - 0.067, 0.39 + 0.067] ≈ [0.323, 0.457]

Therefore, the 99% confidence interval for the population proportion (p) is approximately 0.323 to 0.457, and the margin of error is approximately 0.067.

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Option (C) is correct.

Given:

- Flip a coin that results in Heads with a probability of 1/4 and Tails with a probability of 3/4.

- If the result is Heads, pick X to be Uniform(5,11).

- If the result is Tails, pick X to be Uniform(10,20).

We need to find E(X).

Formula used:

Expected value of a discrete random variable:

X: random variable

p: probability

f(x): probability distribution of X

μ = ∑[x * f(x)]

Case 1: Heads

If the coin flips Heads, then X is Uniform(5,11).

Therefore, f(x) = 1/6, 5 ≤ x ≤ 11, and 0 otherwise.

Using the formula, we have:

μ₁ = ∑[x * f(x)]

Where x varies from 5 to 11 and f(x) = 1/6

μ₁ = (5 * 1/6) + (6 * 1/6) + (7 * 1/6) + (8 * 1/6) + (9 * 1/6) + (10 * 1/6) + (11 * 1/6)

μ₁ = 35/6

Case 2: Tails

If the coin flips Tails, then X is Uniform(10,20).

Therefore, f(x) = 1/10, 10 ≤ x ≤ 20, and 0 otherwise.

Using the formula, we have:

μ₂ = ∑[x * f(x)]

Where x varies from 10 to 20 and f(x) = 1/10

μ₂ = (10 * 1/10) + (11 * 1/10) + (12 * 1/10) + (13 * 1/10) + (14 * 1/10) + (15 * 1/10) + (16 * 1/10) + (17 * 1/10) + (18 * 1/10) + (19 * 1/10) + (20 * 1/10)

μ₂ = 15

Case 3: Both of the above cases occur with probabilities 1/4 and 3/4, respectively.

Using the formula, we have:

E(X) = μ = μ₁ * P(Heads) + μ₂ * P(Tails)

E(X) = (35/6) * (1/4) + 15 * (3/4)

E(X) = (35/6) * (1/4) + (270/4)

E(X) = (35/24) + (270/24)

E(X) = (305/24)

Therefore, E(X) = 305/24.

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Option D, "We cannot say whether the completion of full practice schedules changed because the sample is of only 200 marathon runners," is incorrect.

The participation in full practice schedules demonstrated a statistically significant decrease between March 2018 and March 2019. A random sample of 200 marathon runners was surveyed in March 2018 and March 2019 to determine how often they did a full practice schedule in the week before their scheduled marathon.

In the March 2018 survey, 75%(95%Cl70−77%) of the sample did not complete a full practice schedule in the week before their scheduled marathon.

A year later, in March 2019, the same sample group was surveyed, and 61%(95%Cl57−64%) stated that they did not run a full practice schedule in the week before their competition.

The results suggest that participation in full practice schedules has decreased significantly between March 2018 and March 2019.

The reason why we know that there was a statistically significant decrease is that the confidence interval for the 2019 survey did not overlap with the confidence interval for the 2018 survey.

Because the confidence intervals do not overlap, we can conclude that there was a significant change in the completion of full practice schedules between March 2018 and March 2019.

Therefore, option C, "The participation in full practice schedules demonstrated a statistically significant decrease between March 2018 and March 2019," is the correct answer.

The sample size of 200 marathon runners is adequate to draw a conclusion since the sample was drawn at random. Therefore, option D, "We cannot say whether the completion of full practice schedules changed because the sample is of only 200 marathon runners," is incorrect.

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This equation represents the original ODE after the substitution has been made. dx/du = 2sech(4x)((u + 6)^7 - x^4(u + 6))

To find the ODE in terms of u and x using the given substitution, we start by expressing y in terms of u:

u = y - 6

Rearranging the equation, we get:

y = u + 6

Next, we differentiate both sides of the equation with respect to x:

dy/dx = du/dx

Now, we substitute the expressions for y and dy/dx back into the original ODE:

dx/dy = 2sech(4x)(y^7 - x^4y)

Replacing y with u + 6, we have:

dx/dy = 2sech(4x)((u + 6)^7 - x^4(u + 6))

Finally, we substitute dy/dx = du/dx back into the equation:

dx/du = 2sech(4x)((u + 6)^7 - x^4(u + 6))

Thus, the ODE in terms of u and x is:

dx/du = 2sech(4x)((u + 6)^7 - x^4(u + 6))

This equation represents the original ODE after the substitution has been made.

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The derivative of the function y = 5x - 3x + 9 is 2. The value of the derivative at the point (1, 11) is 2.

To find the derivative of y = 5x - 3x + 9, we take the derivative of each term separately. The derivative of 5x is 5, the derivative of -3x is -3, and the derivative of 9 is 0 (since it is a constant). Therefore, the derivative of the function y = 5x - 3x + 9 is y' = 5 - 3 + 0 = 2.

To evaluate the derivative at the point (1, 11), we substitute x = 1 into the derivative function. So, y'(1) = 2. Hence, the value of the derivative at the point (1, 11) is 2.

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(i) The statement is true. If a divides both b and c, then a also divides any linear combination of b and c with integer coefficients.

(ii) The statement is false. There exist integers for which 3 divides 2x but does not divide x.

(iii) The statement is true. For any integer x, choosing y = x satisfies the divisibility conditions.

(i) Statement: For all integers a, b, and c, if a divides b and a divides c, then for all integers m and n, a divides (mb + nc).

To prove this statement, we can use the property of divisibility. If a divides b, it means there exists an integer k such that b = ak. Similarly, if a divides c, there exists an integer l such that c = al.

Now, let's consider the expression mb + nc. We can write it as mb + nc = mak + nal, where m and n are integers. Rearranging, we have mb + nc = a(mk + nl).

Since mk + nl is also an integer, let's say it is represented by the integer p. Therefore, mb + nc = ap.

This shows that a divides (mb + nc), as it can be expressed as a multiplied by an integer p. Hence, the statement is true.

(ii) Statement: For all integers x, if 3 divides 2x, then 3 divides x.

To disprove this statement, we need to provide a counterexample where the statement is false.

Let's consider x = 4. If we substitute x = 4 into the statement, we get: if 3 divides 2(4), then 3 divides 4.

2(4) = 8, and 3 does not divide 8 evenly. Therefore, the statement is false because there exists an integer (x = 4) for which 3 divides 2x, but 3 does not divide x.

(iii) Statement: For all integers x, there exists an integer y such that 3 divides (x + y) and 3 divides (x - y).

To prove this statement, we can provide a general construction for y that satisfies the divisibility conditions.

Let's consider y = x. If we substitute y = x into the statement, we have: 3 divides (x + x) and 3 divides (x - x).

(x + x) = 2x and (x - x) = 0. It is clear that 3 divides 2x (as it is an even number), and 3 divides 0.

Therefore, by choosing y = x, we can always find an integer y that satisfies the divisibility conditions for any given integer x. Hence, the statement is true.

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If f(x)=2x²−3x+1 and g(x)=3x−1, the rules of the functions:(i) 2f−3g= 4x² - 21x + 5, (ii) fg= 6x³ - 12x² + 6x - 1, (iii) g/f= 9x² - 5x, (iv) f∘g= 18x² - 21x + 2, (v) g∘f= 6x² - 9x + 2, (vi) f∘f= 8x⁴ - 24x³ + 16x² + 3x + 1, (vii) g∘g= 9x - 4

To find the rules of the function, follow these steps:

(i) 2f − 3g= 2(2x²−3x+1) − 3(3x−1) = 4x² - 12x + 2 - 9x + 3 = 4x² - 21x + 5. Rule is 4x² - 21x + 5

(ii) fg= (2x²−3x+1)(3x−1) = 6x³ - 9x² + 3x - 3x² + 3x - 1 = 6x³ - 12x² + 6x - 1. Rule is 6x³ - 12x² + 6x - 1

(iii) g/f= (3x-1) / (2x² - 3x + 1)(g/f)(2x² - 3x + 1) = 3x-1(g/f)(2x²) - (g/f)(3x) + (g/f) = 3x - 1(g/f)(2x²) - (g/f)(3x) + (g/f) = (2x² - 3x + 1)(3x - 1)(2x) - (g/f)(3x)(2x² - 3x + 1) + (g/f)(2x²) = 6x³ - 2x - 3x(2x²) + 9x² - 3x - 2x² = 6x³ - 2x - 6x³ + 9x² - 3x - 2x² = 9x² - 5x. Rule is 9x² - 5x

(iv)Composite function f ∘ g= f(g(x))= f(3x-1)= 2(3x-1)² - 3(3x-1) + 1= 2(9x² - 6x + 1) - 9x + 2= 18x² - 21x + 2. Rule is 18x² - 21x + 2

(v) Composite function g ∘ f= g(f(x))= g(2x²−3x+1)= 3(2x²−3x+1)−1= 6x² - 9x + 2. Rule is 6x² - 9x + 2

(vi)Composite function f ∘ f= f(f(x))= f(2x²−3x+1)= 2(2x²−3x+1)²−3(2x²−3x+1)+1= 2(4x⁴ - 12x³ + 13x² - 6x + 1) - 6x² + 9x + 1= 8x⁴ - 24x³ + 16x² + 3x + 1. Rule is 8x⁴ - 24x³ + 16x² + 3x + 1

(vii)Composite function g ∘ g= g(g(x))= g(3x-1)= 3(3x-1)-1= 9x - 4. Rule is 9x - 4

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The probability of playing the number 1, 5, and 6 is 1/6, and the probability of playing the number 8 is 0.

In a fair die, since there are six faces numbered 1 to 6, the probability of rolling a specific number is given by:

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

(a) Probability of rolling the number 1:

There is only one face with the number 1, so the number of favorable outcomes is 1. The total number of possible outcomes is 6.

Probability of playing the number 1 = 1/6

(b) Probability of rolling the number 5:

There is only one face with the number 5, so the number of favorable outcomes is 1. The total number of possible outcomes is 6.

Probability of playing the number 5 = 1/6

(c) Probability of rolling the number 6:

There is only one face with the number 6, so the number of favorable outcomes is 1. The total number of possible outcomes is 6.

Probability of playing the number 6 = 1/6

(d) Probability of rolling the number 8:

Since the die has only six faces numbered 1 to 6, there is no face with the number 8. Therefore, the number of favorable outcomes is 0.

Probability of playing the number 8 = 0/6 = 0

So, the probability of playing the number 1, 5, and 6 is 1/6, and the probability of playing the number 8 is 0.

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The solution to the equation is -1.5 or -3/2.

We have the equation:

x² + 3-2x= 1+ x² +5

Combine Terms and subtract x² from both sides:

x² - x² + 3 -2x = 1 + 5 + x² - x²

3 -2x = 1 + 5

Add:

3 -2x = 6

Combine Terms and subtract 3 from both sides:

-2x + 3 -3 = 6 - 3

-2x = 3

Dividing by -2 we get:

x = 3/(-2)

x = -3/2

x = -1.5

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Thus, the expression for Average Profit (in dollars per unit) when q million units are produced is given as

P(q)/q = 20/q + 70

The given model of profit isP(q) = 20 + 70 ln(q)thousand dollars

Where q is the number of million units produced.

Therefore, Total profit (in thousand dollars) earned by producing 'q' million units

P(q) = 20 + 70 ln(q)thousand dollars

Average Profit is defined as the profit per unit produced.

We can calculate it by dividing the total profit with the number of units produced.

The total number of units produced is 'q' million units.

Therefore, the Average Profit per unit produced is

P(q)/q = (20 + 70 ln(q))/q thousand dollars/units

P(q)/q = 20/q + 70 ln(q)/q

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To solve the equation (x + y - 1)² dx + 9dy = 0 using substitution techniques, we can substitute u = x + y - 1. This will help us simplify the equation and solve for u.

Let's start by substituting u = x + y - 1 into the equation:

(u)² dx + 9dy = 0

To solve for dx and dy, we differentiate u = x + y - 1 with respect to x:

du = dx + dy

Rearranging this equation, we have:

dx = du - dy

Substituting dx and dy into the equation (u)² dx + 9dy = 0:

(u)² (du - dy) + 9dy = 0

Expanding and rearranging the terms:

u² du - u² dy + 9dy = 0

Now, we can separate the variables by moving all terms involving du to one side and terms involving dy to the other side:

u² du = (u² - 9) dy

Dividing both sides by (u² - 9):

du/dy = (u²)/(u² - 9)

Now, we have a separable differential equation that can be solved by integrating both sides:

∫(1/(u² - 9)) du = ∫dy

Integrating the left side gives us:

(1/6) ln|u + 3| - (1/6) ln|u - 3| = y + C

Simplifying further:

ln|u + 3| - ln|u - 3| = 6y + 6C

Using the properties of logarithms:

ln| (u + 3)/(u - 3) | = 6y + 6C

Exponentiating both sides:

| (u + 3)/(u - 3) | = e^(6y + 6C)

Taking the absolute value, we have two cases to consider:

(u + 3)/(u - 3) = e^(6y + 6C) or (u + 3)/(u - 3) = -e^(6y + 6C)

Solving each case for u in terms of x and y will give us the solution to the original differential equation.

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The probability that X=1 for condition

λ=3.7 is 0.0134.

Assuming a Poisson distribution, to find the probability of a random variable X, that can take values from 0 to infinity, for a given parameter λ of the Poisson distribution, we use the formula

P(X=x) = ((e^-λ) * (λ^x))/x!

where x is the random variable value, e is the Euler's number which is approximately equal to 2.718, and x! is the factorial of x.

Using these formulas, we can calculate the probabilities of the given values of x for the given values of λ.

a. Given λ=2.5, we need to find P(X=3).

Using the formula for Poisson distribution

P(X=3) = ((e^-2.5) * (2.5^3))/3!

P(X=3) = ((e^-2.5) * (15.625))/6

P(X=3) = 0.0667 (rounded to 4 decimal places)

Therefore, the probability that X=3 when

λ=2.5 is 0.0667.

b. Given λ=8.0,

we need to find P(X=9).

Using the formula for Poisson distribution

P(X=9) = ((e^-8.0) * (8.0^9))/9!

P(X=9) = ((e^-8.0) * 262144.0))/362880

P(X=9) = 0.1054 (rounded to 4 decimal places)

Therefore, the probability that X=9 when

λ=8.0 is 0.1054.

c. Given λ=0.5, we need to find P(X=4).

Using the formula for Poisson distribution

P(X=4) = ((e^-0.5) * (0.5^4))/4!

P(X=4) = ((e^-0.5) * 0.0625))/24

P(X=4) = 0.0111 (rounded to 4 decimal places)

Therefore, the probability that X=4 when

λ=0.5 is 0.0111.

d. Given λ=3.7, we need to find P(X=1).

Using the formula for Poisson distribution

P(X=1) = ((e^-3.7) * (3.7^1))/1!

P(X=1) = ((e^-3.7) * 3.7))/1

P(X=1) = 0.0134 (rounded to 4 decimal places)

Therefore, the probability that X=1 when

λ=3.7 is 0.0134.

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We can create 19 words using the letters U, S, and C where each letter is used at least once and the total length is 6, and at least as many Us as Ss and at least as many Ss as Cs

The given letters are U, S, and C. There are 4 different cases we can create words using the letters U, S, and C.

All letters are distinct: In this case, we have 3 letters to choose from for the first letter, 2 letters to choose from for the second letter, and only 1 letter to choose from for the last letter.

So the total number of ways to create words using the letters U, S, and C is 3 x 2 x 1 = 6.

Two letters are the same and one letter is different: In this case, there are 3 ways to choose the letter that is different from the other two letters.

There are 3C2 = 3 ways to choose the positions of the two identical letters. The total number of ways to create words using the letters U, S, and C is 3 x 3 = 9.

Two letters are the same and the third letter is also the same: In this case, there are only 3 ways to create the word USC, USU, and USS.

All three letters are the same: In this case, we can only create one word, USC.So, the total number of ways to create words using the letters U, S, and C is 6 + 9 + 3 + 1 = 19

Therefore, we can create 19 words using the letters U, S, and C where each letter is used at least once and the total length is 6, and at least as many Us as Ss and at least as many Ss as Cs, and the word is lexicographically first among all of its rearrangements.

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The probability of selecting a blue and a white marble is approximately 15.79%.

The total number of marbles in the bag is:

4 + 5 + 5 + 6 = 20

To calculate the probability of selecting a blue marble followed by a white marble, we can use the formula:

Probability = (Number of ways to select a blue marble) x (Number of ways to select a white marble) / (Total number of ways to select 2 marbles)

The number of ways to select a blue marble is 6, and the number of ways to select a white marble is 5. The total number of ways to select 2 marbles from 20 is:

20 choose 2 = (20!)/(2!(20-2)!) = 190

Substituting these values into the formula, we get:

Probability = (6 x 5) / 190 = 0.15789473684

Rounding this to the nearest hundredth gives us a probability of 15.79%.

Therefore, the probability of selecting a blue and a white marble is approximately 15.79%.

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The z-score for the Boston Red Sox, with 69 wins, is approximately -1.08.

To find the z-score for the Boston Red Sox, we can use the formula:

z = (x - μ) / σ

Where:

x is the value we want to convert to a z-score (69 wins for the Red Sox),

μ is the mean of the dataset (79),

σ is the standard deviation of the dataset (9.3).

Substituting the given values into the formula:

z = (69 - 79) / 9.3

Calculating the numerator:

z = -10 / 9.3

Dividing:

z ≈ -1.08

Rounding the z-score to the nearest hundredth, we get approximately z = -1.08.

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the answers are: - P(H and W) = 0.25

- P(W|H) ≈ 0.4167

- P(H|W) ≈ 0.7143

- P(H) = 0.60

- P(W) = 0.35

let's break down the given information:

P(H) represents the probability of an at-home game.

P(W) represents the probability of a win.

P(H and W) represents the probability of an at-home game and a win.

P(W|H) represents the conditional probability of a win given that it is an at-home game.

P(H|W) represents the conditional probability of an at-home game given that it is a win.

Given the information provided:

P(H) = 0.60 (60% of games were at-home games)

P(W) = 0.35 (35% of games were wins)

P(H and W) = 0.25 (25% of games were at-home wins)

To find the desired proportions:

1. P(W|H) = P(H and W) / P(H) = 0.25 / 0.60 ≈ 0.4167 (approximately 41.67% of at-home games were wins)

2. P(H|W) = P(H and W) / P(W) = 0.25 / 0.35 ≈ 0.7143 (approximately 71.43% of wins were at-home games)

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Answer: proper number of sig figs. are :

              a) 6.22 x 10⁷ g/Lb

              b) 0.312

              c) 1.33270

              d)  12500.210

a) Given: 12500. g and 0.201 mL

Let's convert the units of mL to L.= 0.000201 L (since 1 mL = 0.001 L)

Therefore,12500. g / 0.201 mL = 12500 g/0.000201 L = 6.2189055 × 10⁷ g/L

Now, since there are three significant figures in the number 0.201, there should also be three significant figures in our answer.

So the answer should be: 6.22 x 10⁷ g/Lb

b) Given: (9.38 - 3.16) / (3.71 + 16.2)

Therefore, (9.38 - 3.16) / (3.71 + 16.2) = 6.22 / 19.91

Now, since there are three significant figures in the number 9.38, there should also be three significant figures in our answer.

So, the answer should be: 0.312

c) Given: (0.000738 + 1.05874) x (1.258)

Therefore, (0.000738 + 1.05874) x (1.258) = 1.33269532

Now, since there are six significant figures in the numbers 0.000738, 1.05874, and 1.258, the answer should also have six significant figures.

So, the answer should be: 1.33270

d) Given: 12500. g + 0.210

Therefore, 12500. g + 0.210 = 12500.210

Now, since there are five significant figures in the number 12500, and three in 0.210, the answer should have three significant figures.So, the answer should be: 1.25 x 10⁴ g

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The original number is 10t + o = 10(10) + 7 = 107.

Let's call the tens digit of the original number "t" and the ones digit "o".

From the problem statement, we know that:

t + o = 17   (Equation 1)

And we also know that the number with the digits reversed is thirty more than 5 times the tens' digit of the original number. We can express this as an equation:

10o + t = 5t + 30   (Equation 2)

We can simplify Equation 2 by subtracting t from both sides:

10o = 4t + 30

Now we can substitute Equation 1 into this equation to eliminate o:

10(17-t) = 4t + 30

Simplifying this equation gives us:

170 - 10t = 4t + 30

Combining like terms gives us:

140 = 14t

Dividing both sides by 14 gives us:

t = 10

Now we can use Equation 1 to solve for o:

10 + o = 17

o = 7

So the original number is 10t + o = 10(10) + 7 = 107.

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It is evident from the above truth table that the statement X + 1 = 1 is true since the sum of X and 1 is always equal to 1.

A truth table is a table used in mathematical logic to represent logical expressions. It depicts the relationship between the input values and the resulting output values of each function. Here is the truth table proof for each of the following expressions. A + A’ = 1Truth Table for A + A’A A’ A + A’ 0 1 1 1 0 1 0 1 1 0 0 1 1 1 1 0It is evident from the above truth table that the statement A + A’ = 1 is true since the sum of A and A’ results in 1. (A + B)’ = A’B’ Truth Table for (A + B)’ A B A+B (A + B)’ 0 0 0 1 0 1 1 0 1 1 1 0 1 1 0 1. It is evident from the above truth table that the statement (A + B)’ = A’B’ is true since the complement of A + B is equal to the product of the complements of A and B.

(AB)’ = A’ + B’ Truth Table for (AB)’ A B AB (AB)’ 0 0 0 1 0 1 0 1 1 0 0 1 1 1 0 0It is evident from the above truth table that the statement (AB)’ = A’ + B’ is true since the complement of AB is equal to the sum of the complements of A and B. XX’ = 0. Truth Table for XX’X X’ XX’ 0 1 0 1 0 0. It is evident from the above truth table that the statement XX’ = 0 is true since the product of X and X’ is equal to 0. X + 1 = 1. Truth Table for X + 1 X X + 1 0 1 1 1. It is evident from the above truth table that the statement X + 1 = 1 is true since the sum of X and 1 is always equal to 1.

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