For the function f(x)= eˣ.¹/² / 9 + 1 find f⁻¹(x)
f⁻¹(x) = ln [9 (x - 1)²] f⁻¹(x) = (9 ln x - 1)²
f⁻¹(x) = [ln [9 (x − 1)]]² f⁻¹(x) = [In (9x - 1)]²

In total there is a whole 89.44% chance that a randomly selected value from the normally distributed variable X will be less than or equal to 14.

The probability of X ≤ 14 can be calculated by standardizing the variable X using the formula z = (X - μ) / σ, where z is the standardized value. In this case, z = (14 - 11.5) / 2 = 1.25.

Next, we look up the cumulative probability corresponding to the standardized value of 1.25 in the standard normal distribution table or use statistical software/tools. The cumulative probability for z = 1.25 is approximately 0.8944.

Therefore, the probability of X ≤ 14 is 0.8944, or approximately 89.44%.

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The approximate value of log base b (45 × 3) is 1.670.

To approximate the value of log base b of 45 times 3, we can use the logarithmic properties to rewrite the expression as the sum of two logarithms:

log base b (45 × 3) = log base b 45 + log base b 3

Now, using the given approximations for log base b of 20.327, log base b of 3, and log base b of 5:

log base b 45 ≈ log base b 20.327 + log base b 5

≈ 0.503 + 0.835

≈ 1.338

log base b 3 ≈ log base b 5 - log base b 2

≈ 0.835 - 0.503

≈ 0.332

Finally, we can substitute these values back into the original expression:

log base b (45 × 3) ≈ 1.338 + 0.332

≈ 1.670

Therefore, the approximate value of log base b (45 × 3) is 1.670.

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The first partial derivatives of z with respect to x and y in the equation x + sin(y + z) = 0 are ∂z/∂x = -1 and ∂z/∂y = -cos(y + z).

To find the first partial derivatives of z with respect to x and y, we need to differentiate the given implicit equation with respect to x and y while treating z as a function of x and y.

Differentiating the equation with respect to x:

∂/∂x (x + sin(y + z)) = 1 + ∂z/∂x

Differentiating the equation with respect to y:

∂/∂y (x + sin(y + z)) = cos(y + z) (1 + ∂z/∂y)

The term ∂z/∂x represents the partial derivative of z with respect to x, and ∂z/∂y represents the partial derivative of z with respect to y.

So, the first partial derivatives of z are:

∂z/∂x = -1

∂z/∂y = -cos(y + z)

These derivatives indicate how the variable z changes with respect to changes in x and y in the given equation x + sin(y + z) = 0. The value of -1 for ∂z/∂x means that for every unit increase in x, z decreases by 1. The value of -cos(y + z) for ∂z/∂y indicates how z changes with respect to changes in y, with the specific relationship determined by the trigonometric function cos(y + z).

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The coefficient of determination is 0.05.Answer: a. Mean square error = 0.25. b. Coefficient of determination (R²) = 0.05.

a. Mean square error is used to measure the goodness of fit of the linear regression model. Mean square error (MSE) is the average squared differences between the predicted value and the actual value. MSE can be calculated using the formula MSE = SSE / (n - k - 1) where SSE is the sum of squared errors, n is the number of observations and k is the number of independent variables.

The given data for two variables x and y are as follows: xi 2yi7Applying the values in the regression equation, we get:ŷ = 1.2 + 2.4x Substituting xi = 2, we get: ŷ = 1.2 + 2.4(2) = 6Therefore, the SSE can be calculated as follows: SSE = ∑(yi - ŷ)² = (7 - 6)² = 1Now, n = 5 and k = 1 (since there is only one independent variable),

Therefore, MSE = SSE / (n - k - 1)= 1 / (5 - 1 - 1)= 0.25Therefore, the mean square error is 0.25.b. The coefficient of determination (R²) is the proportion of the total variation in the dependent variable (y) that can be explained by the variation in the independent variable(s) (x).

It ranges from 0 to 1, where 0 means that the independent variable(s) does not explain any of the variation in the dependent variable, and 1 means that the independent variable(s) perfectly explain the variation in the dependent variable.R² is calculated as the ratio of the explained variation to the total variation.

It can be calculated as follows: R² = SSE / SST, where SSE is the sum of squared errors and SST is the total sum of squares. SST is calculated as follows: SST = ∑(y i - ȳ)²where ȳ is the mean of yi

Substituting the given values, we get: SST = ∑(yi - ȳ)²= (7 - 5)² + (7 - 5)² + (7 - 5)² + (7 - 5)² + (7 - 5)²= 2² + 2² + 2² + 2² + 2²= 20Now, SSE = 1 (calculated in part a)Therefore,R² = SSE / SST= 1 / 20= 0.05

Therefore, the coefficient of determination is 0.05.Answer: a. Mean square error = 0.25. b. Coefficient of determination (R²) = 0.05.

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The equation of the word problem is N = 50 + n, where N represents the width of Jack's new TV, n represents the additional width of the new TV compared to the old TV, and 50 represents the width of Jack's old TV.

The equation representing the word problem can be derived as follows:

Let's assume the width of Jack's new TV is N inches. According to the information given, Jack's new TV is n inches wider than his old TV, which was 50 inches wide. This can be expressed as:

N = 50 + n

The equation above represents the relationship between the width of Jack's new TV (N), the width of his old TV (50 inches), and the additional width (n inches) of the new TV.

To further simplify, we can substitute the value of n with the specific number of inches wider Jack's new TV is compared to his old TV. Let's say Jack's new TV is 20 inches wider than his old TV. We can substitute n with 20 in the equation:

N = 50 + 20

Simplifying further, we find:

N = 70

This equation represents the specific case where Jack's new TV is 20 inches wider than his old TV, resulting in a width of 70 inches for the new TV.

In general, the equation can be modified to accommodate any value for n, representing the width difference between the new and old TV:

N = 50 + n

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To construct a confidence interval for the mean amount spent on a child's last birthday gift, we can use the formula:

Confidence Interval = sample mean ± (critical value * standard error)

Given that we have a sample size of 10, a mean of $37, and a standard deviation of $4, we can calculate the standard error as:

Standard Error = standard deviation / sqrt(sample size)

Standard Error = $4 / sqrt(10)

Standard Error ≈ $1.27

Next, we need to determine the critical value corresponding to a 95% confidence level. Since the sample size is small (n < 30), we use a t-distribution instead of a z-distribution. With 10-1 = 9 degrees of freedom, the critical value for a 95% confidence level is approximately 2.262.

Now we can calculate the confidence interval:

Confidence Interval = $37 ± (2.262 * $1.27)

Confidence Interval ≈ $37 ± $2.88

Confidence Interval ≈ ($34.12, $39.88)

Therefore, at a 95% confidence level, the confidence interval for the mean amount spent on a child's last birthday gift is approximately $34.12 to $39.88.

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The quadratic function f(x) = 4x² + 5x - 6 can be rewritten in standard (vertex) form as f(x) = 4(x + 5/8)² - 89/8.

To rewrite the quadratic function in standard form, we complete the square. First, we factor out the leading coefficient of 4 from the quadratic term: f(x) = 4(x² + (5/4)x) - 6. Next, we add and subtract the square of half the coefficient of x, which is (5/8)² = 25/64, inside the parentheses: f(x) = 4(x² + (5/4)x + 25/64 - 25/64) - 6. This allows us to express the quadratic term as a perfect square trinomial.

Simplifying further, we have f(x) = 4((x + 5/8)² - 25/64) - 6. Distributing the 4, we obtain f(x) = 4(x + 5/8)² - 100/64 - 6. Combining the constants, we get f(x) = 4(x + 5/8)² - 100/64 - 384/64, which can be simplified to f(x) = 4(x + 5/8)² - 484/64. Finally, converting the improper fraction to a mixed number, we have f(x) = 4(x + 5/8)² - 7 9/64, which is the quadratic function in standard form.

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The Fourier coefficients of the periodic function f(t) on the interval [-1, 1] can be calculated. The coefficient Ao is found to be 1/2, while the coefficient Bn is given by Bn = [tex]\frac{1}{n*\pi }[/tex].

To find the Fourier coefficients of the periodic function f(t), we first calculate the coefficient Ao, which represents the average value of the function over one period. In this case, the function f(t) is defined as 1 on the interval (-1, 1), so the average value over this interval is 1/2. Therefore, Ao = 1/2.

Next, we determine the coefficient Bn, which represents the contribution of the sine component to the function f(t). Bn can be calculated using the formula [tex]B_{n} = \frac{2}{T}[/tex] × [tex]\int\limits^\frac{T}{2} _\frac{-T}{2} \, f(t) * sin(n\omega t)dt[/tex], where T is the period of the function (in this case, T = 2) and ω is the angular frequency (ω = 2π/T = π).

Since f(t) is defined as 1 on (-1, 1) and 0 elsewhere, the integral simplifies to [tex]\int\limits^1_{(-1)} {sin(n\pi t)} \, dt[/tex]. This integral evaluates to [tex]\frac{-1}{n\pi } *cos(n\pi )[/tex], and when evaluated over the interval [-1, 1], we get [tex]\frac{-1}{n\pi } *cos(n\pi )[/tex] - cos(-nπ)) = 0. Therefore, Bn = 0 for all values of n.

However, if we have Bn = [tex]\frac{a}{n^{2} }[/tex], we can set Bn = 1/(nπ) and compare the expressions. This implies a = 1/(π), which is the value of a for the given equation.

In summary, the Fourier coefficient Ao is 1/2, and the coefficient Bn is 0 for all n. However, if we express Bn as [tex]\frac{a}{n^{2} }[/tex], the value of a is 1/(π).

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The inverse of matrix A is computed as A^(-1) = (1/(ad - bc)) * [d -b; -c a], where a, b, c, and d are the elements of matrix A. By substituting the values of matrix A and vector b into the equation x = A^(-1)b, we can find the unique solution for x. In this case, the solution is x = [2; 1].

1. To find the inverse of matrix A = [3 2; 7 5], we first calculate the determinant of A, which is given by ad - bc. In this case, the determinant is (3*5) - (2*7) = 15 - 14 = 1. Since the determinant is nonzero, we can proceed to compute the inverse. The formula for the inverse of a 2x2 matrix is A^(-1) = (1/determinant) * [d -b; -c a]. Substituting the values from matrix A, we have A^(-1) = (1/1) * [5 -2; -7 3] = [5 -2; -7 3].

2. To solve the equation Ax = b, we can multiply both sides by the inverse of A. Here, x = A^(-1)b. Substituting the values, we get x = [5 -2; -7 3] * [10; 23] = [(5*10) + (-2*23); (-7*10) + (3*23)] = [50 -46; -70 + 69] = [4; -1]. Therefore, the unique solution to the equation Ax = [10; 23] is x = [2; 1].

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The values of x in the diagram is as follows:

14. x = 49 degrees

15. x = 58 degrees

Complementary angles are angles that sum up to 90 degrees while supplementary angles are angles that sum up to 180 degrees.

Therefore, let's use the angle relationships to find the angle x in the diagram as follows:

Hence,

14.

x + x - 8 = 90

2x - 8 = 90

2x = 90 + 8

2x = 98

divide both side of the equation by 2

x = 98 / 2

x = 49 degrees

15.

2x + 6 + x = 180

3x + 6 = 180

3x = 180 - 6

3x = 174

divide both sides by 3

x = 174 / 3

x = 58 degrees

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Given that sin(θ) = -5/8, we can determine csc(θ) by finding the reciprocal of sin(θ). In this case, csc(θ) is equal to -8/5.

The sine function (sin) represents the ratio of the length of the side opposite to an angle in a right triangle to the hypotenuse.

In this problem, sin(θ) is given as -5/8. To find csc(θ), we need to calculate the reciprocal of sin(θ). The reciprocal of a number is obtained by dividing 1 by that number.

Since sin(θ) = -5/8, we can write csc(θ) as 1/sin(θ). By substituting the value of sin(θ) as -5/8, we get csc(θ) = 1/(-5/8).

To divide by a fraction, we invert the divisor and multiply. Therefore, csc(θ) = 1 * (8/-5) = -8/5.

In conclusion, if sin(θ) is given as -5/8, then csc(θ) is equal to -8/5. The cosecant function (csc) represents the reciprocal of the sine function, and by applying the appropriate calculations, we can determine the value of csc(θ) based on the given information.

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According to the Central Limit Theorem, the distribution of the sample means will be approximately normal.

In particular, if we take random samples of size n from a distribution with mean μ and variance σ², then as the sample size n increases, the distribution of the sample means approaches a normal distribution with mean μ and variance σ²/n.What is the CLT?The central limit theorem (CLT) describes the behavior of the sample means from any population (not necessarily normal) as the sample size increases.

When the sample size is large enough, the distribution of the sample means is approximately normal, regardless of the shape of the original population distribution.I n summary, according to the Central Limit Theorem, the distribution of the sample means from a random sample of size 25 drawn from a distribution with mean 5 and variance 16 is approximately normal with a mean of 5 and a variance of 16/25.

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In language pertaining to the context of the problem, we can say:

Using methods that are correct 90% of the time, we estimate that the average weight of turkeys is between 14.0498 lbs and 16.6036 lbs.

To form a 90% confidence interval for the average weight of a turkey using the given data, we can use the following steps:

1. Calculate the sample mean:

Sum up all the turkey weights and divide by the total number of turkeys:

Mean = (19 + 21 + 15 + 14 + 12 + 20 + 10 + 18 + 12.5 + 15 + 13 + 12 + 15.4 + 18 + 16) / 15 ≈ 15.3267

2. Calculate the sample standard deviation:

Find the square root of the sum of squared deviations from the mean divided by (n-1):

Standard deviation = sqrt(((19-15.3267)^2 + (21-15.3267)^2 + ... + (16-15.3267)^2) / (15-1)) ≈ 2.9561

3. Calculate the margin of error:

The margin of error is determined by multiplying the critical value (z-score) by the standard deviation and dividing by the square root of the sample size. For a 90% confidence level, the critical value is approximately 1.645:

Margin of error = 1.645 * (2.9561 / sqrt(15)) ≈ 1.2769

4. Calculate the confidence interval:

The confidence interval is obtained by subtracting the margin of error from the sample mean and adding it to the sample mean:

Lower bound = Mean - Margin of error = 15.3267 - 1.2769 ≈ 14.0498

Upper bound = Mean + Margin of error = 15.3267 + 1.2769 ≈ 16.6036

with 90% confidence, we estimate that the mean weight of turkeys is between approximately 14.0498 lbs and 16.6036 lbs.

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Forecast accuracy measures how accurately the forecast aligns with the actual outcome of a future event. It is an essential measure in production planning and control (PPC) to analyze the forecasting performance of the system.

PPC or production planning and control is a tool that helps in managing resources in the production process. It includes a set of functions that assists in maintaining inventory levels, scheduling of production, and managing workloads in the manufacturing process.Forecasting is one of the primary functions of PPC, which helps to estimate the future demand for a product or service.

Accurate forecasting is essential in PPC as it helps in avoiding overproduction, underproduction, and stockouts. Therefore, it is crucial to measure the accuracy of the forecast to determine the effectiveness of the PPC system in place.There are various methods to measure the forecast accuracy, such as Mean Absolute Deviation (MAD), Mean Squared Error (MSE), Mean Absolute Percentage Error (MAPE), Symmetric Mean Absolute Percentage Error (SMAPE), and Tracking Signal. All these methods give a value to the difference between the forecasted demand and the actual demand.Therefore, forecast accuracy the measurement of forecast accuracy is an essential tool in PPC to estimate the effectiveness of the forecasting system.

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A continuous random variable is a variable that can take on any value within a certain range. A continuous random variable is defined as a random variable whose value is a real number. It has a range of possible values. Since the variables can take on a continuum of possible values, they cannot be counted.

Continuous random variables are numerical variables that may take on any value between two points. An example of a continuous random variable is the time it takes for a leaf to fall from a tree. The time it takes for a leaf to fall can take on any value between zero and infinity. The probability distribution of a continuous random variable is described using a probability density function (pdf).Continuous random variables are typically measured using an infinite number of decimal points. This is in contrast to discrete random variables, which are typically measured using whole numbers. Since continuous random variables can take on an infinite number of values, the probability of any one value occurring is typically zero. Instead, we describe the probability distribution using a probability density function (pdf).

Continuous random variables are numerical variables that may take on any value between two points. An example of a continuous random variable is the time it takes for a leaf to fall from a tree. The time it takes for a leaf to fall can take on any value between zero and infinity. The probability distribution of a continuous random variable is described using a probability density function (pdf).A probability density function is a mathematical function that describes the likelihood of a continuous random variable falling within a particular range of values. The pdf is often represented graphically as a curve. The total area under the curve is equal to one. The probability of a continuous random variable falling within a particular range of values is equal to the area under the curve that corresponds to that range of values.The expected value of a continuous random variable is calculated using an integral. The integral is the sum of the product of each possible value of the random variable and its probability density. The variance of a continuous random variable is calculated using a similar formula, but the sum is squared.This is in contrast to discrete random variables, which are typically measured using whole numbers. Since continuous random variables can take on an infinite number of values, the probability of any one value occurring is typically zero. Instead, we describe the probability distribution using a probability density function (pdf).

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We can use the observable data to estimate the parameters consistently.

(a) Simultaneity issue in SEM: The main issue with SEM (simultaneous equation model) is that the endogenous variable is not independent of the error term. Here, the variables (x1, 72, 73, 4) are exogenous variables and (1.2) are endogenous variables.

When an endogenous variable is a function of another endogenous variable, we refer to this as simultaneity.

One example of simultaneity is when the price of a good and the demand for that good are mutually dependent.

If demand for a good is high, the price increases and vice versa, which leads to the issue of simultaneity in the equation.(b) Reduced form equations for 1 and 2:

To get reduced form equations for 1 and 2, we need to eliminate endogenous variables.

x3 = 2x2 is given, let us put it in the equation to get:3/1 = 013/2 + 0271 + 03x2 + 1 + U1

=> 3/1 = (0.13 + 0.27(2x2)) + 03x2 + 1 + U1

=> 3/1 = (0.13 + 0.54x2) + 1 + U1

=> 3/1 = 1.13 + 0.54x2 + U1

=> 1 = -3/1.13 - 0.54x2 - U1/1

Where the coefficients on x2 and U1 are identifiable.

1/2 = 3₁3₁ + 8₂x1 + 3x3 + ₁x₁ + U2₁ =>

1/2 = 3(0.13 + 0.27(2x2)) + 8x1 + 3x3 + 1 + U2

=> 1/2 = 0.39 + 0.81x2 + 8x1 + 3x3 + 1 + U2

=> 0.5 = 0.39 + 0.81x2 + 8x1 + 3x3 + 1 + U2

=> 0.11 = 0.81x2 + 8x1 + 3x3 + U2/2

Where the coefficients on x2, x1, x3, and U2 are identifiable.

(c) Identification of the two structural equations:

We can use the observable data to estimate the parameters consistently when we have a set of equations that are just identified or overidentified.

However, if we have an under-identified model, we cannot estimate the parameters consistently.

To check for identification, we need to check the rank of the matrix.

When we have a linear relationship between 2 and 3, the matrix rank is 2, which means we have two equations and two endogenous variables, and hence the model is just identified.

Therefore, we can use the observable data to estimate the parameters consistently.

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Therefore, the cross product is (7, 16, 7) in the same direction as the thumb in the right-hand rule.

To calculate the cross product of the vectors 2K(2, -1, 3) and (3, -1, 2), you can use the following formula where i, j, and k are the unit vectors in the x, y, and z directions respectively and a = 2K(2,-1, 3) and b = (3,-1,2) are the two vectors. The cross product can be calculated as: So, the cross product is (7, 16, 7) in the same direction as the thumb in the right-hand rule. Therefore, the main answer is: (7, 16, 7).

Therefore, the cross product is (7, 16, 7) in the same direction as the thumb in the right-hand rule.

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The probability that the marathon runners finish in less than 150 minutes is approximately 0.1587.

The given distribution is a normal distribution with a mean of 180 minutes and a standard deviation of 30 minutes.

Let x be a random variable representing the finishing times of a marathon.

Thus, x ~ N (180, 30²).

To find the probability that the marathon runners finish in less than 150 minutes, we need to find P(x < 150).

Here's how we can find it:

z = (x - μ) / σ,

where μ = 180,

σ = 30.z

= (150 - 180) / 30

= -1.p(z < -1)

= 0.1587, using a standard normal distribution table.

Thus,P(x < 150) = P(z < -1) = 0.1587 (approx).

Therefore, the probability that the marathon runners finish in less than 150 minutes is approximately 0.1587.

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Using the given sin θ = 4/8, we can calculate the value of tan θ to determine the correct option. The correct option is option (d) .

To find the value of tan θ, we can use the identity tan θ = sin θ / cos θ. Given sin θ as 4/8, we need to find cos θ in order to calculate tan θ. Using the Pythagorean identity sin² θ + cos² θ = 1, we can solve for cos θ by substituting the value of sin θ: (4/8)² + cos² θ = 1.

Simplifying, we get 16/64 + cos² θ = 1, which further simplifies to 1/4 + cos² θ = 1. Solving for cos θ, we find cos θ = √3/2.

Now we can calculate tan θ using tan θ = sin θ / cos θ, which gives us (4/8) / (√3/2) = 4/(8√3/2) = 4√3/8 = √3/2. Therefore, option (d) is the correct answer.


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a) The percentage of steers weighing over 1150 pounds is 31.46%

b) The percentage of steers weighing under 900 pounds is  1.43%

c) The percentage of steers weighing between 1200 and 1250 pounds is 5.82%.

The given problem is about the normal distribution of the weights of steers, with mean µ = 1104 pounds and standard deviation σ = 94 pounds.

This problem is solvable using the normal distribution table and the z-score formula. The z-score of a random variable x is given by:z = (x - µ) / σ where x is the observed value of the variable.

The z-score measures the number of standard deviations away from the mean that a value is located. Let's solve the problem part by part:

a) To find the percentage of steers weighing over 1150 pounds, we need to calculate the area under the normal distribution curve to the right of 1150.

The z-score for this value is given by:z = (x - µ) / σ = (1150 - 1104) / 94 = 0.489

The area to the right of this z-score can be found from the normal distribution table.Using the table, we find that the area to the right of z = 0.49 is 0.3146.

So, the percentage of steers weighing over 1150 pounds is:P(x > 1150) = 31.46%

b) To find the percentage of steers weighing under 900 pounds, we need to calculate the area under the normal distribution curve to the left of 900.

The z-score for this value is given by:z = (x - µ) / σ = (900 - 1104) / 94 = -2.170

The area to the left of this z-score can be found from the normal distribution table.

Using the table, we find that the area to the left of z = -2.17 is 0.0143.

So, the percentage of steers weighing under 900 pounds is:P(x < 900) = 1.43%

c) To find the percentage of steers weighing between 1200 and 1250 pounds, we need to calculate the area under the normal distribution curve between these two values.

We need to find the z-scores for these values first.

z1 = (x1 - µ) / σ = (1200 - 1104) / 94 = 1.02z2 = (x2 - µ) / σ = (1250 - 1104) / 94 = 1.54

The area between these z-scores can be found from the normal distribution table.

Using the table, we find that the area between z = 1.02 and z = 1.54 is 0.0582.

So, the percentage of steers weighing between 1200 and 1250 pounds is:P(1200 < x < 1250) = 5.82%

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The joint probability density function (p .d .f) of X and Y is given by: f(x ,y) = {x y for 0 < x < y < 1,0 otherwise}

In order to determine marginal density functions, we integrate the joint density function over the limits of the variables we want to remove. Here we need to find marginal density functions of X and Y.

To do so, we will integrate the joint pdf with respect to y and x to obtain the marginal pdf of X and Y respectively.

Summary: The marginal density functions of X and Y are as follows :f x (x ) = ∫f( x ,y) d y, limits of 0 to 1, which is= ∫x^1(x)(y)dy= x/2fy(y) = ∫f(x, y)dx, limits of 0 to y, which is= ∫0^y(x)(y)dx= y^2/2

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1. The probability that the 4th surviving patients is the 6th patient is 0.9. ; 2.  The probability that the 1st surviving patient is the 4th patient is 0.9 * 0.9 * 0.9 * 0.1 ; 3.   The probability of taking 2 out of 4 selected patients 0.33, ; 4. The probability that the clinic will have 4 patients 0.168.

1. The probability that the 4th surviving patients is the 6th patient is 0.9, as the probability of a patient recovering from the delicate heart operation is given as 0.9.

2. The probability that the 1st surviving patient is the 4th patient is 0.9 * 0.9 * 0.9 * 0.1, since the patient should recover for the first three times and fail to recover on the fourth attempt, which has a probability of 0.1.

3. The probability of taking 2 out of 4 selected patients that have heart disease when there are 5 patients with the disease is given by:

C(5,2) * C(10,2) / C(15,4) = (10 * 45) / 1365 = 0.33, where C stands for combinations.

4. The probability that the clinic will have 4 patients in the next hour is given by:

P(X = 4) = (e^-3 * 3^4) / 4! = 0.168, where e is the mathematical constant e and the Poisson distribution formula is used to calculate the probability that an event will occur a certain number of times during a specified time period.

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The integral of the two terms as shown below:[tex]∫(1 - x²)^(1/2)dx = 1/2(θ + 1/2sin(2θ)[/tex] + C)where C is the constant of integration.

To integrate (1-x²)^(1/2) using substitution method, we use the following steps:

Step 1: We let x

= sin(θ)dx = cos(θ)dθ1-x²

= cos²(θ)

Step 2: We substitute the expression derived from Step 1 into the original function to obtain∫(1 - x²)^(1/2)dx=∫cos²(θ)dθ

Step 3: We then apply the double angle formula to obtain:cos²(θ) = (1 + cos(2θ))/2Step 4: We substitute this expression back into the integral to obtain:

∫(1 - x²)^(1/2)dx = ∫(1 + cos(2θ))/2dθ∫(1 - x²)^(1/2)dx

= 1/2 ∫(1 + cos(2θ))dθ

Step 5: Evaluate the integral of the two terms as shown below:∫(1 - x²)^(1/2)dx = 1/2(θ + 1/2sin(2θ) + C)where C is the constant of integration.

Finally, we substitute x = sin(θ) back into the expression above to obtain the final solution.

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Answer:

This is a geometric sequence with common ratio 1/4.

The correct answer is d.

If f(x) + g(x), f(x) = g(x), f(x) · g(x), X f(x): x + 7 g(x) = x² (a) f(x) + g(x) (b) f(x) - g(x) (c) f(x) · g(x) . f(x) (d) g(x) (e) f(g(x)) (f) g(f(x)) = f(x) g(x) f(g(x)), and g(f(x)), if define then- he expression is: f(x) · g(x) = x³ + 7x²

To find the expressions requested, we will substitute the given functions into the respective equations. Let's solve each part one by one:

Given:

f(x) = x + 7

g(x) = x²

(a) f(x) + g(x):

Substituting the functions:

f(x) + g(x) = (x + 7) + (x²)

Combining like terms:

f(x) + g(x) = x + 7 + x²

(b) f(x) - g(x):

Substituting the functions:

f(x) - g(x) = (x + 7) - (x²)

Expanding the expression:

f(x) - g(x) = x + 7 - x²

(c) f(x) · g(x):

Substituting the functions:

f(x) · g(x) = (x + 7) · (x²)

Expanding the expression:

f(x) · g(x) = x³ + 7x²

(d) g(x):

Substituting the function:

g(x) = x²

(e) f(g(x)):

Substituting the functions:

f(g(x)) = f(x²)

Substituting f(x) = x + 7 into f(g(x)):

f(g(x)) = x² + 7

(f) g(f(x)):

Substituting the functions:

g(f(x)) = g(x + 7)

Substituting g(x) = x² into g(f(x)):

g(f(x)) = (x + 7)²

Expanding the expression:

g(f(x)) = x² + 14x + 49

(g) f(x) · g(x), if defined:

We already solved this in part (c), and the expression is:

f(x) · g(x) = x³ + 7x²

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Answer:

Step-by-step explanation:

The volume of the rectangular box will be maximum when the length of the box is 7.14 cm and the height of the box is 238.10 cm³.

Let's consider the given sheet of metal.

Let the width of the rectangular box to be x.

So, the length of the box = 20 - 2x (as we have to remove width on both sides)

The height of the box = We have the formula of volume of a rectangular box as,

Volume of the rectangular box = length × width × heightV =

x(20 - 2x)yV = (20x - 2x²)yV = 20xy - 2x²y

We need to maximize the volume of the rectangular box by finding the values of x and y. We know that,

Area of metal sheet = Area of rectangular box + Area of waste metal sheet

50 × 20 = xy + 2xy + x(20 - 2x)50 × 20 =

3xy + 20x50 × 20 - 20x = 3xy50(20 - x)

= 3xySo, xy = 50(20 - x)/3Putting this value in the above equation, we get:V = 20x(50 - x)/3 - (2x²) maximizing V, dV/dx = 0dV/dx = 20(50 - 2x)/3 - 4x = 0(100 - 2x)/3 = 4x/3x = 100/14. ≈ 7.14 cm Putting this value in the above equation,  we get:y = 50(20 - 7.14)/3y ≈ 238.10 cm³

Therefore, the dimensions that will give the box with the largest volume are: x = 7.14 cm = 238.10 cm³

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The length of side BC is approximately 8.72 km.

To find the length of side BC using the cosine rule, we can use the following formula:

BC² = AB² + AC² - 2 AB AC Cos(A)

where BC represents the length of side BC, AB represents the length of side AB, AC represents the length of side AC, and A represents the angle opposite to side BC.

Plugging in the given values:

BC² = (25.3 km)² + (16.7 km)² - 2 (25.3 km) (16.7 km) Cos(68.5°)

BC² = 640.09 km² + 278.89 km² - 2 × 25.3 km × 16.7 km × cos(68.5°)

BC² = 919.98 km² - 843.91 km²

BC² = 76.07 km²

Taking the square root of both sides:

BC = √76.07 km

BC ≈ 8.72 km

Therefore, the length of side BC is approximately 8.72 km.

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Answer:

Yes, the lines on the graph at 3 and -3 a part  of the solution,

Step-by-step explanation:

The inequality [tex]|y| \leq 3[/tex] contains all the values of [tex]y[/tex] 3 units from the origin including the values 3 and -3.

Thus, the lines on the graph y =-3 and y = 3 are the part of the solution.

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Based on the provided expression, it seems you are trying to evaluate the limit:

lim(x→2) g(x)

where it is given that lim(x→2) g(x) = 9.

Using the given information, we can directly substitute the limit value into the expression:

lim(x→2) g(x) = 9

Therefore, the limit of g(x) as x approaches 2 is equal to 9.

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