what is 28.5 inches in height?

The approximate probability distribution of X−10 has a mean of -9 and a variance of 4. These values are obtained under the assumption of independent and identically distributed observations in the random sample.

1. The approximate probability distribution of X−10 can be characterized by its mean and variance. The mean of X−10 is equal to the mean of X subtracted by 10, which can be calculated as follows. Since X is uniformly distributed between 0 and 2, its mean is (2+0)/2 = 1. Therefore, the mean of X−10 is 1−10 = -9.

2. To calculate the variance of X−10, we make the assumption that the observations in the random sample are independent and identically distributed. Under this assumption, the variance of X−10 can be determined by subtracting 10 from each observation in the sample, calculating the variance of the modified sample, and then summing the individual variances.

3. Since X is uniformly distributed, its variance can be calculated as ((2−0)²)/12 = 1/3. Subtracting 10 from each observation, we obtain a new random variable Y representing the modified sample. The variance of Y is also 1/3. Therefore, the variance of X−10 is (1/3) + (1/3) + ... + (1/3) (12 times) = 4.

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To find the least-squares solution x* of the system Ax=b where `A = ⎣⎡​354​235​⎦⎤​` and `b = ⎣⎡​592​⎦⎤​`, follow these steps:

Step 1: Find the transpose of A and multiply it with A. Then, find the inverse of the product.(A' denotes the transpose of A)A' = ⎣⎡​354​235​⎦⎤​' = ⎣⎡​3​5​4​2​3​5​⎦⎤​A' A = ⎣⎡​3​5​4​2​3​5​⎦⎤​ ⎣⎡​354​235​⎦⎤​ = ⎣⎡​50​30​30​14​10​⎦⎤​A' A is invertible. (The determinant is not equal to zero.) Hence, the inverse of A' A exists. Let's find it.(A' A)^{-1} = 1/260 [⎣⎡​10​−30⎦⎤​ ⎣⎡​−30​50​−5⎦⎤​ ⎣⎡​−5​−5​26​⎦⎤​]

Step 2: Find A'b.A'b = ⎣⎡​3​5​4​2​3​5​⎦⎤​ ⎣⎡​592​⎦⎤​ = ⎣⎡​2176​⎦⎤

​Step 3: Multiply the inverse of A' A with A' b to get the least-squares solution x*.(A' A)^{-1} A' b = 1/260 [⎣⎡​10​−30⎦⎤​ ⎣⎡​−30​50​−5⎦⎤​ ⎣⎡​−5​−5​26​⎦⎤​] ⎣⎡​592​⎦⎤​ = ⎣⎡​8/13​⎦⎤​ ⎣⎡​−18/13​⎦⎤​ ⎣⎡​20/13​⎦⎤​

Therefore, the least-squares solution of the given system is `x* = [8/13, -18/13, 20/13]`.The sketch showing the vector `b`, the image of `A`, the vector `Ax`, and the vector `b−Ax` is as follows: The sketch of the graph has been shown below:

Therefore, the required sketch has been shown above which includes the vector `b`, the image of `A`, the vector `Ax`, and the vector `b−Ax`.

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The expressions A ∪ B and (A ∩ B)' are not equal. The correct answer is false.

The union of two sets A and B, denoted as A ∪ B, represents the set that contains all elements that are in A or in B (or in both). On the other hand, the complement of the intersection of sets A and B, denoted as (A ∩ B)', represents the set of all elements that are not in the intersection of A and B.

If A and B have any elements in common, then the intersection A ∩ B will not be an empty set. In this case, the complement of the intersection, (A ∩ B)', will include all elements that are not in the intersection.

However, the union A ∪ B will contain all elements that are in A, all elements that are in B, and any elements that are in the intersection of A and B.

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The level surface of F(x, y, z) = 19.

Given a function F(x, y, z) = 5n(xy + z)²¹ + 2(yz - x²)² = 19(0, 0, (19 - 5n)/n).

To determine the level surface of F(x, y, z), the partial derivative of the function F(x, y, z) with respect to x, y, and z are computed as follows;

∂F/∂x = -4x2

∂F/∂y = 10n(xy + z)20x + 4(yz - x2)y

∂F/∂z = 10n(xy + z)20z + 2(yz - x2)z

Equating each of these partial derivatives to zero to solve for the critical points of F(x, y, z);

∂F/∂x = -4x² = 0 x² = 0 => x = 0

∂F/∂y = 10n(xy + z)20x + 4(yz - x²)y = 0

Since x = 0 => 0 + 4(yz - 0) y = 0 4yzy = 0 => y = 0 or z = 0

∂F/∂z = 10n(xy + z)20z + 2(yz - x²)z = 0

Since x = 0, y = 0 or z = 0 => 10n(0 + z)20z + 2(0 - 0)z = 0

10nz² + 0z = 0 => z(10nz + 0) = 0

Therefore, the critical points are (0, 0, 0) and (0, 0, 19 - 5n/n).

Now, let's obtain the Hessian matrix of F(x, y, z);

Thus, the determinant of the Hessian matrix is |H| = -640z². From the determinant of H, it can be observed that |H| < 0 when z ≠ 0, thus indicating that (0, 0, 19 - 5n/n) is a saddle point of F(x, y, z). However, when z = 0, |H| = 0, thus indicating that (0, 0, 0) is a degenerate critical point of F(x, y, z).

Therefore, to determine the level surface, we shall evaluate F(x, y, z) at the critical points.

When (x, y, z) = (0, 0, 0), F(0, 0, 0) = 19

When (x, y, z) = (0, 0, 19 - 5n/n), F(0, 0, 19 - 5n/n) = 19.

Thus, the level surface of F(x, y, z) = 19.

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Answer:

  A.  (0, 1)

Step-by-step explanation:

You want to know the y-intercept of f(x) = (1/2)^x.

The y-intercept of a function is the function value when x=0. Put 0 where x is and do the arithmetic.

  f(0) = (1/2)^0

  f(0) = 1

The y-intercept is (x, f(x)) = (0, f(0)) = (0, 1).

__

Additional comment

Any non-zero value to the zero power is 1.

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(a) The standard error of the mean is 1.1672 mpg.

(b) The 95% confidence interval for the sample mean is [22.4744 mpg, 26.5256 mpg].

(a) The standard error of the mean (SE) is calculated by dividing the standard deviation (σ) by the square root of the sample size (n). Therefore, SE = σ / √n. Substituting the given values, we get SE = 3.50 / √9 = 1.1672 mpg (rounded to 4 decimal places).

(b) To determine the interval within which we would expect the sample mean to fall with 95% probability, we use the concept of a confidence interval. Since the population standard deviation (σ) is known, we can use the formula X¯¯¯ ± Z(α/2) * (σ / √n), where X¯¯¯ represents the sample mean, Z(α/2) is the critical value corresponding to the desired confidence level (95% in this case), and n is the sample size. Substituting the given values, we find X¯¯¯ ± 1.96 * (3.50 / √9). Evaluating this expression, we obtain the 95% confidence interval for the sample mean as [22.4744 mpg, 26.5256 mpg] (rounded to 4 decimal places).

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1. The equation of the tangent plane to the surface z = x^2 + xy + y^2 at point P0(-1, 1, 1) is z = -2x + 3y + 4.

2. The equation of the tangent plane to the surface z = x^2y + e^(xy)sin(y) at point P0(1, 0, 0) is z = x - y.

3. The equation of the tangent plane to the surface z = y/x + xy at point P0(4, 1, 8) is z = 3x - 2y - 7.

1. To find the tangent plane to the surface z = x^2 + xy + y^2 at point P0(-1, 1, 1), we need to compute the partial derivatives of z with respect to x and y. The partial derivatives are:

∂z/∂x = 2x + y

∂z/∂y = x + 2y

Evaluating these derivatives at P0, we have ∂z/∂x = -2 + 1 = -1 and ∂z/∂y = -1 + 2 = 1. Thus, the equation of the tangent plane can be written as:

z = z0 + (∂z/∂x)(x - x0) + (∂z/∂y)(y - y0)

  = 1 + (-1)(x + 1) + (1)(y - 1)

  = -2x + 3y + 4

2. For the surface z = x^2y + e^(xy)sin(y), the partial derivatives are:

∂z/∂x = 2xy + y^2e^(xy)sin(y) + e^(xy)cos(y)

∂z/∂y = x^2 + xy^2e^(xy)sin(y) + e^(xy)sin(y)cos(y)

Evaluating these derivatives at P0, we have ∂z/∂x = 0 and ∂z/∂y = 1. Therefore, the equation of the tangent plane is simply z = x - y.

3. For the surface z = y/x + xy, the partial derivatives are:

∂z/∂x = -y/x^2 + y

∂z/∂y = 1/x + x

Evaluating these derivatives at P0, we have ∂z/∂x = -1/16 + 1 = 15/16 and ∂z/∂y = 1/4 + 4 = 17/4. The equation of the tangent plane can be written as:

z = z0 + (∂z/∂x)(x - x0) + (∂z/∂y)(y - y0)

  = 8 + (15/16)(x - 4) + (17/4)(y - 1)

  = 3x - 2y - 7

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(a) In the hypothesis test framework applied to the criminal trial example, the null hypothesis would be that the defendant is innocent, and the alternative hypothesis would be that the defendant is guilty.

In hypothesis testing, the null hypothesis represents the default assumption or the claim that is initially presumed to be true unless there is sufficient evidence to support the alternative hypothesis. In this case, the null hypothesis assumes the defendant's innocence, while the alternative hypothesis asserts their guilt.

The null hypothesis (H0): "The defendant is innocent" and the alternative hypothesis (Ha): "The defendant is guilty."

(b) Rejecting the null hypothesis would mean that the evidence presented by the prosecutor is strong enough to conclude that the defendant is not innocent and, consequently, the defendant is found guilty.

In hypothesis testing, rejecting the null hypothesis implies that the evidence provided is significant enough to support the alternative hypothesis. Therefore, if the null hypothesis is rejected in this context, it would mean that the evidence presented by the prosecutor is convincing, and the defendant is considered guilty.

Rejecting the null hypothesis in this case would lead to the defendant being found guilty based on the evidence presented.

(c) Failing to reject the null hypothesis would mean that the evidence presented by the prosecutor is not strong enough to conclude that the defendant is not innocent. Therefore, the judge would deliver a verdict of "not guilty."

Failing to reject the null hypothesis does not necessarily mean that the null hypothesis is true; it simply means that there is insufficient evidence to support the alternative hypothesis. In this scenario, if the null hypothesis is not rejected, it would mean that the evidence presented by the prosecutor is not convincing enough to rule out the possibility of the defendant's innocence.

Failing to reject the null hypothesis in this context would result in the defendant being declared "not guilty" due to insufficient evidence to prove their guilt.

(d) In the context of this problem, a type II error would occur if the judge fails to reject the null hypothesis (declares the defendant "not guilty") when, in reality, the defendant is guilty.

A type II error in hypothesis testing refers to the situation where the null hypothesis is false, but the test fails to reject it. In this case, a type II error would occur if the judge, despite the defendant being guilty, fails to find sufficient evidence to reject the null hypothesis of the defendant's innocence.

Committing a type II error in this scenario would result in the defendant being declared "not guilty" even though they are actually guilty.

(e) In the context of this problem, a type I error would occur if the judge rejects the null hypothesis (declares the defendant guilty) when, in reality, the defendant is innocent.

A type I error in hypothesis testing refers to the situation where the null hypothesis is true, but the test mistakenly rejects it in favor of the alternative hypothesis. In this case, a type I error would occur if the judge, based on the evidence presented, wrongly concludes that the defendant is guilty and rejects the null hypothesis of the defendant's innocence.

Committing a type I error in this scenario would lead to an incorrect conviction of the defendant, declaring them guilty when they are actually innocent.

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The degree of freedom for a t-statistic with a sample size of 25 individuals is 24.

When calculating a confidence interval for a population mean using a t-statistic, the degrees of freedom are determined by the sample size minus 1. In this case, the sample size is 25 individuals, so the degrees of freedom would be 25 - 1 = 24.

Degrees of freedom represents the number of independent pieces of information available for estimation. In the context of a t-distribution, it is related to the variability and sample size. With a larger sample size, there is more information available, resulting in higher degrees of freedom.

The t-distribution is used when the population standard deviation is unknown, and the sample size is small. By using the appropriate degrees of freedom, the t-distribution accounts for the additional uncertainty introduced by estimating the population standard deviation from the sample.

In summary, for a sample size of 25 individuals, we would expect 24 degrees of freedom for calculating the t-statistic in order to construct a confidence interval for a population mean.

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The components of the vector v_n are determined by the SVD of X, and specifically, the n-th column of V, denoted as v_n, is orthogonal to the vector b = V^T * a.

In the given problem, we have a data matrix X ∈ R^(m×n) with the property that there exists a vector a = ⟨a_1, ..., a_m⟩ such that for every data point X_i,∗, the entries satisfy X_i,∗^T * a = 0, 1 ≤ i ≤ m. It is also given that the rank of X is n-1

Let X = U * Σ * V^T be the singular value decomposition (SVD) of X, where U ∈ R^(m×m) and V ∈ R^(n×n) are orthogonal matrices, and Σ ∈ R^(m×n) is a diagonal matrix.

Since the rank of X is n-1, we know that there are n-1 non-zero singular values in Σ. Let's assume these non-zero singular values are σ_1, σ_2, ..., σ_{n-1}.

The components of the vector v_n correspond to the last column of the matrix V, denoted as v_n = [v_1, v_2, ..., v_n]^T.

Since V is an orthogonal matrix, its columns are orthogonal to each other and have unit length. Therefore, v_n is a unit vector.

Now, let's consider the equation X_i,∗^T * a = 0, where X_i,∗ represents the i-th row of X.

Using the SVD, we can write X_i,∗^T * a as (U * Σ * V^T)_i,∗^T * a.

Since V is orthogonal, V^T * a is also a vector. Let's denote V^T * a as b = [b_1, b_2, ..., b_n]^T.

Now, the equation X_i,∗^T * a = (U * Σ * V^T)_i,∗^T * a can be written as (U * Σ * b)_i,∗ = 0

Considering that Σ is a diagonal matrix with non-zero singular values σ_1, σ_2, ..., σ_{n-1}, we can see that for (U * Σ * b)_i,∗ to be zero, the i-th row of U must be orthogonal to b.

Since U is an orthogonal matrix, its columns are orthogonal to each other. Therefore, the i-th row of U, denoted as U_i, must be orthogonal to the vector b.

Now, recall that U_i represents the left singular vectors of X. These vectors are orthogonal to each other and correspond to the singular values of X.

Since the rank of X is n-1, there are n-1 non-zero singular values, which means there are n-1 left singular vectors. Therefore, there are n-1 orthogonal vectors in U, and the n-th column of U (U_n) is orthogonal to the vector b.

Consequently, the components of the vector v_n correspond to the n-th column of V, which is orthogonal to the vector b and satisfies the given conditions.

the components of the vector v_n are determined by the SVD of X, and specifically, the n-th column of V, denoted as v_n, is orthogonal to the vector b = V^T * a.

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The probability that one job is created during the first three months of the year in this firm is 0.3033.

The average number of jobs created in a firm is 2 jobs per year.

The probability that one job is created during the first three months of the year in this firm is

We can assume that the number of jobs created follows the Poisson distribution with λ = 2.

We have to find the probability of creating one job during the first three months of the year, which is the probability of creating one job out of the total number of jobs created in the year. As 3 months is 1/4th of the year, the probability of creating one job in the first three months is given by:

P(X = 1) = (λ^x × e^(-λ)) / x!, x = 1, λ = 2

Putting these values in the formula:

P(X = 1) = (2^1 × e^(-2)) / 1!P(X = 1) = 2e^(-2)

Therefore, the probability that one job is created during the first three months of the year in this firm is approximately 0.3033. Hence, the correct option is 0.3033.

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Test Statistic: The formula for calculating the z-score is:$$\frac{(\hat p_1-\hat p_2)-D}{\sqrt{\hat p(1-\hat p)\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}}$$We need to calculate the proportions, sample sizes, and the value of the z-score.the correct answer is option B.

$\hat{p}_{1}=126 / 153=0.824$, $\hat{p}_{2}

=22 / 154

=0.143$

The value of D = 0 (since we want to test if the two proportions are equal)$\hat{p}=\frac{126+22}{153+154}=0.4824$$n_{1}=153, n_{2}=154$$\begin{aligned} z &=\frac{(\hat{p}_{1}-\hat{p}_{2})-D}{\sqrt{\hat{p}(1-\hat{p})\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}} \\ &=\frac{(0.824-0.143)-0}{\sqrt{0.4824(1-0.4824)\left(\frac{1}{153}+\frac{1}{154}\right)}} \\ &=13.04 \end{aligned}$The value of the z-score is 13.04.

Therefore, the test statistic is z = 13.04.b. Confidence Interval: We can use the confidence interval to test the claim. The 98% confidence interval is given by:$\left(\hat{p}_{1}-\hat{p}_{2}-z_{\alpha / 2} \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}, \hat{p}_{1}-\hat{p}_{2}+z_{\alpha / 2} \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\right)$Substituting the values, we get:$$\begin{aligned}\left(\hat{p}_{1}-\hat{p}_{2}-z_{\alpha / 2} \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}, \hat{p}_{1}-\hat{p}_{2}+z_{\alpha / 2} \sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\right)\\=\left(0.681, 0.803\right)\end{aligned}$$The 98% confidence interval is $(0.681,0.803)$.c. Based on the results, is the oxygen treatment effective? The P-value is less than the level of significance of 0.01.

Therefore, we can reject the null hypothesis and conclude that there is evidence to support the claim that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. The confidence interval is also entirely above zero, and it suggests that the cure rate for the oxygen treatment is higher than that of the placebo. Hence, the results suggest that the oxygen treatment is effective in curing "cluster" headaches.

Thus, the correct answer is option B.

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The expansion of (a + b)8 = a⁸ + 8a⁷b + 28a⁶b² + 56a⁵b³ + 70a⁴b⁴ + 56a³b⁵ + 28a²b⁶ + 8ab⁷ + b⁸.

Recall the binomial theorem for expansion of powers of (a + b) as follows:

(a + b)⁰ = 1, (a + b)¹ = a + b, (a + b)² = a² + 2ab + b²,

and in general,

(a + b)n = nC₀an + nC₁an-1b + nC₂an-2b² + ... + nCn-1abn-1 + nCnbn,

where nCk = n!/[k!(n - k)!], k = 0, 1, ..., n.

The expansion of (a + b)8 is:

(a + b)⁸ = 8C₀a⁸ + 8C₁a⁷b + 8C₂a⁶b² + 8C₃a⁵b³ + 8C₄a⁴b⁴ + 8C₅a³b⁵ + 8C₆a²b⁶ + 8C₇ab⁷ + 8C₈b⁸.

to find the precise coefficients of (a + b)8, apply the formula given above.

n = 8, and so calculate

nC₀, nC₁, ..., nC₈nC₀ = 8C₀ = 1nC₁ = 8C₁ = 8nC₂ = 8C₂ = 28nC₃ = 8C₃ = 56nC₄ = 8C₄ = 70nC₅ = 8C₅ = 56nC₆ = 8C₆ = 28nC₇ = 8C₇ = 8nC₈ = 8C₈ = 1

Therefore, substitute these values to obtain the precise coefficients of (a + b)8.

(a + b)8 = a⁸ + 8a⁷b + 28a⁶b² + 56a⁵b³ + 70a⁴b⁴ + 56a³b⁵ + 28a²b⁶ + 8ab⁷ + b⁸.

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(a) The graph of y = f(x-2) is the graph of y = f(x) shifted horizontally to the right by 2 units.

(b) The graph of y = f(x) - 4 is the graph of y = f(x) shifted vertically downward by 4 units.

(c) The graph of y = f(x-3) + 1 is the graph of y = f(x) shifted horizontally to the right by 3 units and shifted vertically upward by 1 unit.

(d) The graph of y = f(x+4) + 1 is the graph of y = f(x) shifted horizontally to the left by 4 units and shifted vertically upward by 1 unit.

(a) The graph of y = f(x-2) is the graph of y = f(x) shifted horizontally. When we have a transformation of the form f(x - h), it represents a horizontal shift by h units.

In this case, the function y = f(x-2) indicates a shift of the graph of y = f(x) to the right by 2 units. This means that every point on the original graph is moved 2 units to the right to create the new graph. The general shape and characteristics of the graph remain the same, but it is shifted horizontally to the right.

(b) The graph of y = f(x) - 4 is the graph of y = f(x) shifted vertically. When we have a transformation of the form f(x) + k or f(x) - k, it represents a vertical shift by k units.

For y = f(x) - 4, the graph of y = f(x) is shifted downward by 4 units. Each point on the original graph is moved downward by 4 units to create the new graph. The shape and characteristics of the graph remain unchanged, but it is shifted vertically downward.

(c) The graph of y = f(x-3) + 1 is the graph of y = f(x) shifted horizontally and vertically. Here, the transformation f(x - h) + k represents a horizontal shift by h units and a vertical shift by k units.

In this case, y = f(x-3) + 1 implies a shift of the graph of y = f(x) to the right by 3 units and a shift upward by 1 unit. Each point on the original graph is moved 3 units to the right and 1 unit upward to create the new graph. The general shape and characteristics of the graph remain the same, but it is shifted both horizontally and vertically.

(d) The graph of y = f(x+4) + 1 is the graph of y = f(x) shifted horizontally and vertically. Similarly, the transformation f(x + h) + k represents a horizontal shift by h units and a vertical shift by k units.

For y = f(x+4) + 1, it indicates a shift of the graph of y = f(x) to the left by 4 units and a shift upward by 1 unit. Each point on the original graph is moved 4 units to the left and 1 unit upward to create the new graph. The overall shape and characteristics of the graph remain the same, but it is shifted both horizontally and vertically.

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The solution of the given linear system is [x1, x2] = [ (√2/28-1/98) (7t/2-1/2)].

The given system of linear equations is:

28x1 + 2x2

= (2√2)x1 + (1/2)x204x1 + 16x2t

= x2

Now, let's write the given system of equations in the matrix form [A]x=[B], where x is the column matrix of variables

[x1,x2].28 2 2√2 1/2 28 x1 2x2

= 04 16t 0 1 4 x21 x2

On multiplying the matrices [A] and [x], we get:

28x1 + 2x2

= 2√2x1 + 1/2x204x1 + 16x2t

= x2

Now, we need to solve for x1 and x2 using the Gauss-Jordan method:

[28 2 | 2√2 1/2] [28 2 | 2√2 1/2][04 16t | 0 1]

=> [04 16t | 0 1]R2

= R2 - 4R1/R1

= R1/28      

[1 2/7 | √2/28 1/56][0 16t-4(2/7) | -√2/7 1/7]    [0 16t/7-2/7 | 0 1/7]R2

= R2/(16t/7-2/7)     [1 2/7 | √2/28 1/56][0 1 | 0 7t/2-1/2]R1

= R1-2/7R2     [1 0 | √2/28-1/98 (1/56-2/7(7t/2-1/2))][0 1 | 0 7t/2-1/2]

The solution of the given linear system is:x1

= (√2/28-1/98) x2x2

= 7t/2-1/2

Therefore, the solution of the given linear system is[x1, x2]

= [ (√2/28-1/98) (7t/2-1/2)]

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The 95% two-sided Confidence Interval for the true mean load at specimen failure is (11.107, 16.913) MPa. The 95% Upper Confidence Interval for the true mean load at specimen failure is (13.446, +∞) MPa.

To construct the Confidence Intervals, we need to calculate the mean load at specimen failure and the standard error of the mean. For a 95% two-sided Confidence Interval, the critical value is obtained from the t-distribution with (n-1) degrees of freedom, where n is the sample size.

Step 1: Calculate the mean and standard deviation

The sample mean load at specimen failure is calculated by summing up all the values and dividing by the sample size (n = 22). In this case, the sample mean is 13.382 MPa. The standard deviation is a measure of the spread of the data around the mean. For this sample, the standard deviation is 3.256 MPa.

Step 2: Calculate the standard error of the mean

The standard error of the mean is calculated by dividing the standard deviation by the square root of the sample size. In this case, the standard error of the mean is 0.693 MPa.

Step 3: Calculate the Confidence Intervals

For a two-sided Confidence Interval, we need to consider the critical value at α/2, where α is the significance level (5% or 0.05). With (n-1) degrees of freedom, we consult the t-distribution table or use statistical software to find the critical value. For a sample size of 22 and α/2 = 0.025, the critical value is approximately 2.074.

To construct the 95% two-sided Confidence Interval, we use the formula:

Mean ± (Critical Value * Standard Error)

For the true mean load at specimen failure, the 95% two-sided Confidence Interval is calculated as:

13.382 ± (2.074 * 0.693) = (11.107, 16.913) MPa

For the 95% Upper Confidence Interval, we only need to calculate the upper bound. The formula is:

Mean + (Critical Value * Standard Error)

For the true mean load at specimen failure, the 95% Upper Confidence Interval is calculated as:

13.382 + (2.074 * 0.693) = (13.446, +∞) MPa

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The values of all sub-parts have been obtained.

(i). The equation of the circle in standard form is (x - 9)² + (y - 6)² = 8².

(ii). The center of the given circle is (9, 6) and the radius is 8.

(b). The value of (f∘g)(x) is x² - 4x + 8 and the domain is [2, ∞).

(i). Equation of the circle in standard form:

To write the equation of the circle in standard form, first, bring all the constant terms to one side of the equation and complete the square for x and y terms.

x² − 18x + y² − 12y + 17 = 0

x² - 18x + y² - 12y = -17

Completing the square for x terms:

(x - 9)² - 81 + y² - 12y = -17

(x - 9)² + y² - 12y = 64

Completing the square for y terms:

(x - 9)² + (y - 6)² = 8²

This is the equation of the circle in standard form.

(ii). Center and Radius:

The standard form of the circle equation is, (x - a)² + (y - b)² = r².

The center is (a, b) and radius is r.

The center of the given circle is (9, 6) and the radius is 8.

(b). (f∘g)(x) and domain:

(f∘g)(x) means f(g(x)).

First, we need to find g(x).g(x) = x - 2

Now, substitute g(x) in place of x in f(x) equation to get

(f∘g)(x).(f∘g)(x) = f(g(x))

                      = f(x-2)

                      = (x-2)² + 4

                      = x² - 4x + 8

The domain of the function (f∘g)(x) is the set of all values of x for which the function is defined.

Since the domain of g(x) is all real numbers, we need to find the domain of (f∘g)(x) that makes the expression under the square root to be non-negative.

Domain of (f∘g)(x) = {x | x-2≥0}

Domain of (f∘g)(x) = [2, ∞).

Therefore, the domain of (f∘g)(x) is [2, ∞).

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To determine the number of 200-milligram tablets that should be given daily to a patient who weighs 169 pounds, we need to convert the weight from pounds to kilograms and then calculate the dosage based on the prescribed dosage of 10 milligrams per kilogram of body weight.

Given that 1 pound is approximately equal to 0.45 kilograms, we convert the weight of the patient, which is 169 pounds, to kilograms by multiplying it by 0.45. Thus, the weight of the patient is approximately 76.05 kilograms.

Next, we calculate the total dosage by multiplying the weight of the patient in kilograms by the prescribed dosage of 10 milligrams per kilogram. Therefore, the total dosage is approximately 760.5 milligrams.

To find the number of 200-milligram tablets needed, we divide the total dosage by the dosage per tablet. Hence, the number of tablets required daily is approximately 4 tablets.

In conclusion, a patient who weighs 169 pounds should receive approximately 4 200-milligram tablets each day according to the prescribed dosage of 10 milligrams per kilogram of body weight.

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Simplified form of both sides of equation is same:LHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t). RHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t). Identity is verified.

To verify the identity (1 - sin²(t) + 8 cos²(t))² + 81 sin²(t) cos²(t) = 81 cos²(t)(1 - sin²(t) + 8 cos² (t))² + 81 sin²(t) cos²(t), we need to simplify both sides of the equation and show that they are equal.

Let's simplify each side step by step:

Left-hand side (LHS):

(1 - sin²(t) + 8 cos²(t))² + 81 sin²(t) cos²(t)

= (1 - 2sin²(t) + 8 cos²(t))² + 81 sin²(t) cos²(t)

= (1 - 2sin²(t) + 8 cos²(t))(1 - 2sin²(t) + 8 cos²(t)) + 81 sin²(t) cos²(t)

= (1 - 2sin²(t) + 8 cos²(t) + 81 sin²(t) cos²(t)) + (16sin⁴(t) - 32sin²(t)cos²(t) + 64cos⁴(t)) + 81 sin²(t) cos²(t)

= 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

Right-hand side (RHS):

81 cos²(t)(1 - sin²(t) + 8 cos² (t))² + 81 sin²(t) cos²(t)

= 81 cos²(t)(1 - 2sin²(t) + 16 cos⁴(t) - 2sin²(t) + 16sin⁴(t) + 64 cos⁴(t) + 16sin²(t) - 32sin²(t)cos²(t) + 128cos⁴(t)) + 81 sin²(t) cos²(t)

= 81 cos²(t)(1 + 16sin⁴(t) + 64 cos⁴(t) + 16sin²(t) - 32sin²(t)cos²(t) + 128cos⁴(t)) + 81 sin²(t) cos²(t)

= 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

As we can see, the simplified form of both sides of the equation is the same:

LHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

RHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

Therefore, the identity is verified.

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Simplified form of both sides of equation is same:LHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t). RHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t). Identity is verified.

To verify the identity (1 - sin²(t) + 8 cos²(t))² + 81 sin²(t) cos²(t) = 81 cos²(t)(1 - sin²(t) + 8 cos² (t))² + 81 sin²(t) cos²(t), we need to simplify both sides of the equation and show that they are equal.

Let's simplify each side step by step:

Left-hand side (LHS):

(1 - sin²(t) + 8 cos²(t))² + 81 sin²(t) cos²(t)

= (1 - 2sin²(t) + 8 cos²(t))² + 81 sin²(t) cos²(t)

= (1 - 2sin²(t) + 8 cos²(t))(1 - 2sin²(t) + 8 cos²(t)) + 81 sin²(t) cos²(t)

= (1 - 2sin²(t) + 8 cos²(t) + 81 sin²(t) cos²(t)) + (16sin⁴(t) - 32sin²(t)cos²(t) + 64cos⁴(t)) + 81 sin²(t) cos²(t)

= 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

Right-hand side (RHS):

81 cos²(t)(1 - sin²(t) + 8 cos² (t))² + 81 sin²(t) cos²(t)

= 81 cos²(t)(1 - 2sin²(t) + 16 cos⁴(t) - 2sin²(t) + 16sin⁴(t) + 64 cos⁴(t) + 16sin²(t) - 32sin²(t)cos²(t) + 128cos⁴(t)) + 81 sin²(t) cos²(t)

= 81 cos²(t)(1 + 16sin⁴(t) + 64 cos⁴(t) + 16sin²(t) - 32sin²(t)cos²(t) + 128cos⁴(t)) + 81 sin²(t) cos²(t)

= 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

As we can see, the simplified form of both sides of the equation is the same:

LHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

RHS = 81 cos²(t) + 16sin⁴(t) + 64cos⁴(t) + 81 sin²(t) cos²(t) - 32sin²(t)cos²(t)

Therefore, the identity is verified.

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a. The probability of selecting a jury of all students is 1.

b. The probability of selecting a jury of two students and six faculty is 0

c. The probability of selecting a jury of two students and six faculty is given by 0

We need to find the probability of selecting a jury of all students and all faculty, and probability of selecting a jury of two students and six faculty.

(a) Probability of selecting a jury of all students can be found by selecting 8 students out of the given 11 students.

Hence, the probability of selecting a jury of all students is given by:

`P(All students)

= C(11, 8)/C(11, 8)

= 1`

Thus, the probability of selecting a jury of all students is 1.

(b) Probability of selecting a jury of all faculty can be found by selecting 8 faculty members out of 3.

Thus, the probability of selecting a jury of all faculty is given by: `

P(All faculty) = C(3, 8)/C(11, 8)

= 0`

Thus, the probability of selecting a jury of all faculty is 0.

(c) Probability of selecting 2 students and 6 faculty members out of the given 11 students and faculty members can be found as follows:

We need to select 2 students from 11 students and 6 faculty members from 3 faculty members available.

Therefore, probability of selecting a jury of two students and six faculty is given by:

`P(2 students and 6 faculty) = (C(11, 2) × C(3, 6))/C(11, 8)

= 0`

Thus, the probability of selecting a jury of two students and six faculty is 0.

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Option b is correct. this option best describes the significance level in the context of this scenario.

The statement that best describes the significance level in the context of this scenario is:

b. The probability of concluding that the defect rate is equal to 0.09 when in fact it is greater than 0.09.

The significance level, also known as the alpha level, represents the threshold at which we reject the null hypothesis. In hypothesis testing, we set a significance level to determine how much evidence we need against the null hypothesis in order to reject it.

In this scenario, if the significance level is set at, for example, 0.05, it means we are willing to tolerate a 5% chance of making a Type I error, which is the probability of concluding that the defect rate is equal to 0.09 (null hypothesis) when it is actually greater than 0.09 (alternative hypothesis).

Therefore, option b correctly describes the significance level as the probability of concluding that the defect rate is equal to 0.09 when in fact it is greater than 0.09.

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The function f(x) = ±cos(x) satisfies the hypotheses and consequence of the Mean Value Theorem.

The function f(x) = x² - 4x + 3 satisfies the hypotheses and consequence of Rolle's Theorem.

The Highway Patrol Officer at Town B is justified in citing the driver of the sports car based on the Mean Value Theorem.

For BOOMSound Corp., the Revenue Function R(q) is determined using the Price-Demand Function. The Profit Function P(q) is then found by subtracting the Daily Cost Function from the Revenue Function.

To find the extrema of the Profit Function over the interval [20, 90], the process used for finding extrema is applied.

For the functions f(x) = 2cos(x) + sin(2x) and f(x) = -x, the process is used to find the extrema over the given intervals, and the results can be verified using a graphing utility.

To verify the Mean Value Theorem, we need to check if the function is continuous on the given interval and differentiable on the open interval. For f(x) = ±cos(x), it satisfies these conditions. The Mean Value Theorem states that there exists at least one point c in the interval where the derivative of the function is equal to the average rate of change of the function over the interval.

For Rolle's Theorem, we need to check if the function is continuous on the closed interval and differentiable on the open interval, and if the function values at the endpoints are equal. For f(x) = x² - 4x + 3, it satisfies these conditions. Rolle's Theorem states that there exists at least one point c in the interval where the derivative of the function is zero.

In the scenario with the sports car, the Mean Value Theorem can be applied. Since the car traveled from Town A to Town B in one hour at a constant speed of 60mph, its average velocity over that interval is 60mph. The Mean Value Theorem guarantees that at some point during the journey, the car must have been traveling at exactly 60mph.

The Revenue Function R(q) is obtained by multiplying the Price-Demand Function p(q) by the quantity q. Using the given information, R(q) = q(550 - 4.59q). The Profit Function P(q) is then found by subtracting the Daily Cost Function C(q) = 8100 + 55q from the Revenue Function. Simplifying R(q) and P(q) yields the final expressions.

To find the extrema of the Profit Function over the interval [20, 90], we can take the derivative of P(q) and set it equal to zero. Solving for q gives the critical points, and by evaluating the second derivative at these points, we can determine if they correspond to a maximum or minimum.

For the functions f(x) = 2cos(x) + sin(2x) and f(x) = -x, the process is repeated. The derivatives are calculated, and critical points are found by setting the derivatives equal to zero. By evaluating the second derivative at these points, we can determine if they correspond to a maximum or minimum. The results can be confirmed using a graphing utility.

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The sum of the draws can take any value between 0 and 294 (6 * 49), the probability of the sum being less than 100 or more than 200 is essentially 100%.

a) The chance that the sum of 49 random draws will be 160 or more, estimated using a normal approximation, is approximately 99.97%.

b) The chance that the sum of 49 random draws will be 125 or less, estimated using a normal approximation, is approximately 0.21%.

c) The chance that the sum of 49 random draws will be either less than 100 or more than 200, estimated using a normal approximation, is approximately 100%.

Let's go through each question and the corresponding explanations:

a) The chance that the sum of 49 random draws will be 160 or more, estimated using a normal approximation, is approximately 99.97%.

ich states that the distribution of the sum of a large number of independent and identically distributed random variables will approach a normal distribution. Since we are drawing with replacement, each draw is considered independent and has the same probability distribution.

b) The chance that the sum of 49 random draws will be 125 or less, estimated using a normal approximation, is approximately 0.21%.

Similarly, we can use the central limit theorem to estimate this probability. We calculate the mean and standard deviation of the individual draws, and then use the normal distribution to estimate the probability of the sum being 125 or less.

c) The chance that the sum of 49 random draws will be either less than 100 or more than 200, estimated using a normal approximation, is approximately 100%.

Since the sum of the draws can take any value between 0 and 294 (6 * 49), the probability of the sum being less than 100 or more than 200 is essentially 100%. This is because the normal distribution is continuous and extends to both positive and negative infinity.

It's important to note that these estimates are based on the assumption of a normal distribution approximation and may not be exact. However, for a large number of random draws, the normal approximation tends to be quite accurate.

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the general solution of the augmented matrix is x1 = -0.111 - 8.9475x3x2 = -8.95x3x3 is free. Therefore, the correct option is A. x1 = -0.111 - 8.9475x3 x2 = -8.95x3

the augmented matrix is

⎣⎡​10 00 −30 0100 00 01−1089000−45800⎦⎤​

can be written as [A/B], where A and B are the coefficient matrix and the constant matrix, respectively. So, the system of equation represented by the given augmented matrix is Ax = B. Hence,

[A/B] = ⎣⎡​10 00 −30 0100 00 01−1089000−45800⎦⎤​

can be written as (A/B) = ⎣⎡​10 00 −30 0100 00 01−1089000−45800⎦⎤​

find the general solution of the given system using the Gauss-Jordan elimination process. Perform elementary row operations on (A/B) to convert A into an identity matrix.

Interchange R1 and R3:

⎣⎡​0 00 01−10810 00 −30 0000−45800⎦⎤​

Multiply R2 by (-3) and add it to R1:

⎣⎡​0 00 01−10810 00 00−339040⎦⎤​

Divide R2 by -10:

⎣⎡​0 00 01−10810 00 00−339040⎦⎤​

Next, multiply R3 by (-1) and add it to R2:

⎣⎡​0 00 01−10810 00 00−339040⎦⎤​

Divide R3 by 40:

⎣⎡​0 00 01−10810 00 000−8.95⎦⎤​

Write the row reduced matrix as [I/F], where I is the identity matrix and F is the transformed constant matrix. Therefore, [I/F] = ⎣⎡​10 00 00 00−0.111−8.94750−8.95⎦⎤​

So, the solution of the system is given by x = F. Hence, the general solution of the given system is x1 = -0.111 - 8.9475x3x2 = -8.95x3x3 is free. Therefore, the correct option is A. x1 = -0.111 - 8.9475x3 x2 = -8.95x3 The system has infinite solutions.

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The statement is false. The series (a){a} ΣRDE does not converge.

The series (a){a} ΣRDE converges, we need to analyze its terms and their behavior. Let's break down the series step by step:

1. The series starts with (a){a}, which indicates a product of the variable 'a' with itself. However, we don't have any information about the value or properties of 'a', so we cannot make any assumptions about this product.

2. The series then continues with ΣRDE, which suggests a summation involving the variables R, D, and E. Again, without any specific information about these variables, we cannot determine the behavior or convergence of this summation.

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a. The critical value(s) is/are z = -1.28.

b. The correct conclusion is A. Reject H0. There is sufficient evidence to warrant rejection of the claim that p = 1/2.

a. To find the critical value(s), we need to refer to the standard normal distribution table.

Using a significance level (α) of 0.10, we are conducting a one-tailed test (since we're only interested in one direction of the distribution, either greater than or less than). Since the test statistic is negative (-1.91), we're looking for the critical value in the left tail of the standard normal distribution.

From the standard normal distribution table, the critical value for a significance level of 0.10 in the left tail is approximately -1.28.

Therefore, the critical value is z = -1.28.

b. To determine whether we should reject or fail to reject H0 (the null hypothesis), we compare the test statistic (z = -1.91) with the critical value (-1.28).

Since the test statistic is smaller (more negative) than the critical value, it falls in the critical region. This means we reject the null hypothesis.

Thus, the correct conclusion is:

A. Reject H0. There is sufficient evidence to warrant rejection of the claim that p = 1/2.

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The shape of the sampling distribution of sample proportions can be approximated by a normal distribution if certain conditions are met. One of these conditions is that the sample size is sufficiently large, typically considered to be at least 30.

In this case, the sample size is 500, which meets the requirement for a normal approximation.

The center of the sampling distribution of sample proportions is equal to the population proportion. In this case, the population proportion is known to be 42%, so the center of the sampling distribution of sample proportions is also 42%.

The spread of the sampling distribution of sample proportions can be measured by the standard deviation, which is determined by the population proportion and the sample size. The formula for the standard deviation of the sampling distribution of sample proportions is:

Standard Deviation = sqrt((p * (1-p)) / n)

where p is the population proportion and n is the sample size.

In this case, the population proportion is 42% (0.42) and the sample size is 500, so we can calculate the standard deviation as follows:

Standard Deviation = sqrt((0.42 * (1-0.42)) / 500)

Calculating this, the standard deviation is approximately 0.0246 (rounded to four decimal places).

Therefore, the shape of the sampling distribution of sample proportions is approximately normal, the center is 42%, and the spread, as measured by the standard deviation, is approximately 0.0246.

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The value of b that gives the largest average value of the function g(x) on the interval [0, b] is 9b² - 18b + 9b⁹ - (9/10)b¹⁰ = 0.

Differentiation in mathematics is the process of finding the derivative of a function. The derivative measures how a function changes as its independent variable (usually denoted as 'x') varies. It provides information about the rate of change of the function at any given point, as well as the slope of the tangent line to the graph of the function at that point.

Combining like terms and rearranging the equation:

9b² - 18b + 9b⁹ - (9/10)b¹⁰ = 0

we can consider the average value as a function of b.

The average value of a function f(x) on an interval [a, b] is given by the formula:

Avg = (1/(b-a)) * ∫[a,b] f(x) dx

In this case, the function g(x) = 18x - 9x⁹, and we want to maximize the average value on the interval [0, b]. So, we need to maximize the following expression:

Avg(b) = (1/b) * ∫[0,b] (18x - 9x⁹) dx

To find the maximum value of Avg(b), we need to differentiate Avg(b) with respect to b and find the value of b where the derivative is equal to zero.

d(Avg)/db = -1/b² * ∫[0,b] (18x - 9x⁹) dx + (1/b) * (18b - 9b⁹)

Setting the derivative equal to zero:

0 = -1/b² * ∫[0,b] (18x - 9x⁹) dx + (1/b) * (18b - 9b⁹)

Multiplying through by b^2:

0 = -∫[0,b] (18x - 9x⁹) dx + (18b - 9b⁹)

Rearranging the equation:

∫[0,b] (18x - 9x⁹) dx = 18b - 9b⁹

To evaluate the integral, we need to find the antiderivative of the function inside the integral:

∫[0,b] (18x - 9x⁹) dx = 9x^2 - (9/10)x¹⁰ |_0^b = 9b² - (9/10)b¹⁰

Substituting this result back into the equation:

9b² - (9/10)b¹⁰ = 18b - 9b⁹

Combining like terms and rearranging the equation:

9b² - 18b + 9b⁹ - (9/10)b¹⁰ = 0

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The given data shows the counts for party affiliation (Republican, Democrat, None) and opinions on the US economy (Better, Same, Worse) from a sample of 651 adults. The counts for each category are provided, indicating the distribution of responses.

In the given CBS newspoll, the results of party affiliation and opinions on the US economy were collected from a random sample of 651 adults. The counts for each category are as follows: Republican: 104, Democrat: 44, None: 38, Better: 12, Same: 21, Worse: 87. Republican: 90, Democrat: 137, None: 118.

The first paragraph summarizes the given data, providing the counts for each category of party affiliation (Republican, Democrat, None) and opinions on the US economy (Better, Same, Worse).

In the given data, there are different counts for each category, representing the number of individuals who fall into each group. For party affiliation, there were 104 Republicans, 44 Democrats, and 38 individuals who identified as None. Regarding the opinion on the US economy, there were 12 individuals who believed it was getting better, 21 who thought it was staying the same, and 87 who believed it was getting worse. Additionally, there were 90 Republicans, 137 Democrats, and 118 individuals who identified as distribution . These counts provide a snapshot of the respondents' party affiliation and their opinions on the state of the US economy during the time of the survey.

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(a) The average credit score for mortgages purchased by the company has increased since 2017.

(b) The correct option for the critical value is B. 1.676.

(c) The correct options are: B. The population is normally distributed. A. The population standard deviation is small.

a) The null and alternative hypotheses are as follows:

H0: μ2017 = μ2021 (average credit score for mortgages purchased in 2017 is equal to the average credit score for mortgages purchased in 2021)

H1: μ2021 > μ2017 (average credit score for mortgages purchased in 2021 is greater than the average credit score for mortgages purchased in 2017)

Based on this, we can use a one-tailed t-test with a significance level of 0.05 to determine if there is enough evidence to conclude that the average credit score for mortgages purchased by the company has increased since 2017.

b) To determine the appropriate critical value, we need to use a one-tailed t-test with 47 degrees of freedom (assuming a sample size of 50 and a normally distributed population).

At a significance level of 0.05 and 47 degrees of freedom,

The critical value is 1.676.

To calculate the appropriate test statistic, we need to find the t-value associated with the sample mean and standard deviation, as well as the hypothesized population mean of 2017 (μ2017 = 700).

Assuming the sample mean and standard deviation are 720 and 50, respectively, the test statistic can be calculated as:

t = (720 - 700) / (50 / √(50)) = 3.18

With a calculated test statistic of 3.18 and a critical value of 1.676, we can reject the null hypothesis and conclude that there is sufficient evidence to suggest that the average credit score for mortgages purchased by the company has increased since 2017.

c) Using Excel, the precise p-value for this test can be calculated using the formula "=T.DIST.RT(3.18, 47)" (assuming the sample mean is 720, the sample standard deviation is 50, and the hypothesized population mean is 700). The resulting p-value is 0.0017.

If the sample size were 13, the assumptions that would need to be made to perform this analysis are:

B. The population is normally distributed.

A. The population standard deviation is small.

These assumptions are necessary to ensure that the t-statistic follows a t-distribution and that the calculated confidence intervals and p-values are accurate.

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