If z is a standard normal variable, find the probability that z lies between -0.55 and 0.55.A. -0.4176B. 0.9000C. -0.9000D. 0.4176

a) p^ = 27/75 = 0.36

The given limit can be expressed as the definite integral:

∫[0 to 1] 3x^8 dx

To express the limit as a definite integral, we can use the definition of a Riemann sum. Let's consider the function f(x) = x^8.

The given limit can be rewritten as:

lim(n→∞) Σ[i=1 to n] (3i^8 / n^9)

Now, let's express this limit as a definite integral. We can approximate the sum using equal subintervals of width Δx = 1/n. The value of i can be replaced with x = iΔx = i/n. The summation then becomes:

lim(n→∞) Σ[i=1 to n] (3(i/n)^8 / n^9)

This can be further simplified as:

lim(n→∞) (1/n) Σ[i=1 to n] (3(i/n)^8 / n)

Taking the limit as n approaches infinity, the sum can be written as:

lim(n→∞) (1/n) ∑[i=1 to n] (3(i/n)^8 / n) ≈ ∫[0 to 1] 3x^8 dx

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The one-sided p value for a z-statistic of 1.72 is approximately 0.0427.

To calculate the one-sided p value for a z-statistic of 1.72:

Step 1: Identify the z-statistic (zstat) given in the question, which is 1.72.

Step 2: Look up the z-statistic in a standard normal (z) table or use an online calculator to find the area to the left of the z-statistic. For a z-statistic of 1.72, the area to the left is approximately 0.9573.

Step 3: Since we want the one-sided p-value, and our z-statistic is positive, we'll calculate the area to the right of the z-statistic. To do this, subtract the area to the left from 1:

P-value (one-sided) = 1 - 0.9573 = 0.0427

The one-sided p-value for a z-statistic of 1.72 is approximately 0.0427.

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If the area of the rectangle is 15, the dimensions of the rectangle are l = √(15) and w = √(15).

The question is referring to a rectangle, we can use the formula for the area of a rectangle, which is A = lw, where A is the area, l is the length, and w is the width.

We are given that the area of the rectangle is 15, so we can set up an equation:

l * w = 15

We are not given any information about the length, so we cannot solve for l and w separately. However, if we assume that the rectangle is a square (i.e., l = w), then we can solve for the dimensions:

l * l = 15

l² = 15

l = √(15)

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Making assumptions about the marginal pmf fx(x) and the conditional pmf h(y|x), probability P(X+Y>4) is 0.35.

To compute P(X+Y>4), we need to consider the possible values of X and Y and calculate the probabilities accordingly.

Let's analyze the scenario step by step:

Randomly choosing X from the interval [0, 4]:

The possible values for X are 0, 1, 2, 3, and 4. We assume a uniform distribution for X, meaning each value has an equal probability of being chosen. Therefore, the marginal pmf fx(x) is given by:

fx(0) = 1/5

fx(1) = 1/5

fx(2) = 1/5

fx(3) = 1/5

fx(4) = 1/5

Choosing Y from the interval [0, x]:

Since the value of X is observed, the range for Y will depend on the chosen value of X. For each value of X, Y can take on values from 0 up to X. We assume a uniform distribution for Y given X, meaning each value of Y in the allowed range has an equal probability. Therefore, the conditional pmf h(y|x) is given by:

For X = 0: h(y|0) = 1/1 = 1

For X = 1: h(y|1) = 1/2

For X = 2: h(y|2) = 1/3

For X = 3: h(y|3) = 1/4

For X = 4: h(y|4) = 1/5

Computing P(X+Y>4):

We want to find the probability that the sum of X and Y is greater than 4. Since X and Y are independent, we can calculate the probability using the law of total probability:

P(X+Y>4) = Σ P(X+Y>4 | X=x) * P(X=x)

= Σ P(Y>4-X | X=x) * P(X=x)

Let's calculate the probabilities for each value of X:

For X = 0: P(Y>4-0 | X=0) * P(X=0) = 0 * 1/5 = 0

For X = 1: P(Y>4-1 | X=1) * P(X=1) = 1/2 * 1/5 = 1/10

For X = 2: P(Y>4-2 | X=2) * P(X=2) = 1/3 * 1/5 = 1/15

For X = 3: P(Y>4-3 | X=3) * P(X=3) = 1/4 * 1/5 = 1/20

For X = 4: P(Y>4-4 | X=4) * P(X=4) = 1/5 * 1/5 = 1/25

Summing up the probabilities:

P(X+Y>4) = 0 + 1/10 + 1/15 + 1/20 + 1/25

= 0.35

Therefore, the probability P(X+Y>4) is 0.35.

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The number of drinks that will be sold for 10 games is 1635 drinks.

The basketball concession stand sold 327 drinks in two games

2 games = 327 drinks

using unitary method

Unitary method is a process by which we find the value of a single unit from the value of multiple units and the value of multiple units from the value of a single unit.

1 game = 327/2

1 game = 163.5 drinks

Number of drinks that will be sold for 10 games

10 games = 10 × 1 game

10 games = 10 ×  163.5

10 games = 1635 drinks

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Answer: m = 6

Step-by-step explanation: In a calculator, I put in the following equation:

12.6 = 0.85m + 7.5

and that have me the answer of 6

We can double check this solution by putting the following equation in the calculator:

(0.85 x 6) + 7.5 =

and you will see that is = to 12.6

Answer:

m = 6

Step-by-step explanation:

Isolate the variable, m. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

PEMDAS is the order of operations, and stands for:

Parenthesis

Exponents (& Roots)

Multiplications

Divisions

Additions

Subtractions

~

First, subtract 7.5 from both sides of the equation:

[tex]0.85m + 7.5 = 12.6\\0.85m + 7.5 (-7.5) = 12.6 (-7.5)\\0.85m = 12.6 - 7.5\\0.85m = 5.1[/tex]

Next, isolate the variable, m, by dividing 0.85 from both sides of the equation:

[tex]0.85m = 5.1\\\frac{(0.85m)}{0.85} = \frac{(5.1)}{0.85} \\m = \frac{5.1}{0.85} \\m = 6[/tex]

6 would be your value for m.

~

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The equation of the plane passing through the points (0, 2, 1), (1, 1, 5), and (2, 0, 11) is 12x - 6y - 10z = 0.

To find the equation of the plane passing through three given points,  the point-normal form of the equation. This form uses a point on the plane and the normal vector perpendicular to the plane.

Step 1: Find two vectors on the plane by subtracting the coordinates of one point from the other two points.

Vector 1 = (1, 1, 5) - (0, 2, 1) = (1, -1, 4)

Vector 2 = (2, 0, 11) - (0, 2, 1) = (2, -2, 10)

Step 2: Calculate the cross product of the two vectors to obtain the normal vector to the plane.

Normal vector = Vector 1 × Vector 2

Using the determinant method:

i j k

1 -1 4

2 -2 10

= (1 × 10 - (-1) × (-2))i - (1 × 10 - 4 × (-2))j + (-1 × (-2) - 4 × 2)k

= 12i - 6j - 10k

Therefore, the normal vector is (12, -6, -10).

Step 3: Choose one of the given points as the reference point on the plane. Let's choose (0, 2, 1) as the reference point.

Step 4: Substitute the values into the point-normal form of the equation:

(x - x₁)(A) + (y - y₁)(B) + (z - z₁)(C) = 0

Where (x₁, y₁, z₁) is the reference point, and (A, B, C) are the components of the normal vector.

Substituting the values,

(x - 0)(12) + (y - 2)(-6) + (z - 1)(-10) = 0

Simplifying the equation:

12x - 6y - 10z + 12 - 12 = 0

12x - 6y - 10z = 0.

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To develop Q-Q plots for the exponential and uniform distributions, we first need to order the data in ascending order: 9, 24, 50, 89.

For the exponential distribution, we use the formula F(x) = 1 - e^(-λx) where λ is the rate parameter. We estimate λ using the sample mean, which is 43. We then compute the expected values of F(x) for each observation: 0.001, 0.16, 0.52, 0.83. We plot these expected values against the ordered data on a Q-Q plot.

For the uniform distribution, we estimate the parameters as a = 9 and b = 89, the minimum and maximum values in the data set. We then compute the expected values of F(x) for each observation using the formula F(x) = (x-a)/(b-a). The expected values for each observation are: 0, 0.167, 0.556, 1.

Looking at the Q-Q plots, we can see that the data points lie closer to the diagonal line for the uniform distribution than the exponential distribution. This suggests that the uniform distribution is a better fit for the data than the exponential distribution.

In summary, based on the Q-Q plots, we can conclude that the uniform distribution appears to be a better fit for the data than the exponential distribution. This may be due to the fact that the data set is relatively small and does not exhibit the exponential decay pattern often seen in larger data sets.

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The probability of observing a sample mean lower than 40.8 is 1.1% assuming the data come from a population that follows a normal model. Therefore, option b. is correct.

The p-value is a measure of the evidence against a null hypothesis. In statistical hypothesis testing, the null hypothesis is typically a statement of "no effect" or "no difference" between two groups or variables. The p-value represents the probability of obtaining a sample statistic (or one more extreme) if the null hypothesis is true.

In this context, the p-value is 1.1%, which means that if the null hypothesis were true (i.e., the population mean is equal to 43.80), the probability of obtaining a sample mean lower than 40.8 is 1.1%. This suggests that the data provide some evidence against the null hypothesis and support the alternative hypothesis that the population mean is less than 43.80.

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The correct answer is a. The P-value represents the probability of observing a sample mean as extreme or more extreme than the one observed, assuming that the data comes from a population that follows a normal model.

In this context, a P-value of 1.1% means that there is a low probability of observing a sample mean lower than 43.80, given that the data comes from a normal distribution. This suggests that the observed sample mean is unlikely to have occurred by chance alone, and provides evidence for a significant difference between the sample mean and the hypothesized population mean.

The P-value represents the probability of observing a sample mean as extreme as, or more extreme than, the one obtained from your data (43.80 mpg) if the true population mean is 40.8 mpg. The P-value of 1.1% indicates that there is a 1.1% chance of obtaining a sample mean of 43.80 or more due to natural sampling variation, assuming the population follows a normal model.

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The mean (expected value) of a binomial distribution is equal to the product of the number of trials and the probability of success on each trial. Therefore, the mean of the binomial distribution for marriages would be 10 multiplied by the probability of divorce within 10 years. The standard deviation of a binomial distribution is equal to the square root of the product of the number of trials, the probability of success on each trial, and the probability of failure on each trial. Since the probability of success (divorce) is already known, we can calculate the probability of failure (not divorcing) by subtracting the probability of success from 1.

The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials. In the case of marriages, the number of trials is 10 years, and the success is divorce within that time period. The probability of divorce within 10 years is not provided in the question, but let's assume it is 50% for the sake of simplicity. Therefore, the mean of the binomial distribution would be 10 multiplied by 0.5, which equals 5. The standard deviation would be the square root of (10 x 0.5 x 0.5), which equals 1.58.

In summary, the mean and standard deviation for the binomial distribution involving marriages depend on the probability of divorce within the specified time period. The mean is equal to the number of years multiplied by the probability of divorce, while the standard deviation is equal to the square root of the product of the number of years, the probability of divorce, and the probability of not divorcing. These calculations can be used to understand the expected number of divorces and the variability around that expectation.

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The length of the side x for the right triangle is equal to be 23.6 to the nearest tenth using the Pythagoras rule.

The Pythagoras rule states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 

For the right triangle;

x² = 14² + 19²

x² = 196 + 361

x² = 557

x = √557 {take square root of both sides}

x = 23.6008

Therefore, the length of the hypotenuse side x is equal to be 23.6 to the nearest tenth using the Pythagoras rule.

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The total area of the figure in this problem is given as follows:

41.3 ft².

The area of the composite figure is given by the sum of the areas of all the parts that compose the figure.

The figure in this problem is composed as follows:

Then the area of the triangle is given as follows:

At = 0.5 x 6 x 3 = 9 ft².

The area of the semicircle is given as follows:

Ac = π x 3² = 28.3 ft².

The area of the square is given as follows:

As = 2² = 4 ft².

Then the total area of the figure is given as follows:

9 + 28.3 + 4 = 41.3 ft².

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Therefore, the first partial derivatives of the function f(x, y) are:

∂/∂x [f(x,y)] = cos(e^x) - y*sin(y)

∂/∂y [f(x,y)] = xcos(x) - ysin(e^y)

To find the partial derivatives of the function f(x, y) = ∫yx cos(e^t) dt with respect to x and y, we can use the Leibniz rule for differentiating under the integral sign.

First, we'll find the partial derivative with respect to x:

∂/∂x [f(x,y)]

= ∂/∂x [∫yx cos(e^t) dt]

= d/dx [∫yx cos(e^t) dt] evaluated at the limits of integration

Using the chain rule of differentiation, we have:

d/dx [∫yx cos(e^t) dt] = d/dx [cos(e^x)*x - cos(y)*y]

Evaluating this derivative gives:

∂/∂x [f(x,y)] = cos(e^x) - y*sin(y)

Now, we'll find the partial derivative with respect to y:

∂/∂y [f(x,y)]

= ∂/∂y [∫yx cos(e^t) dt]

= d/dy [∫yx cos(e^t) dt] evaluated at the limits of integration

Using the Leibniz rule again, we have:

d/dy [∫yx cos(e^t) dt] = d/dy [sin(e^y)*y - sin(x)*x]

Evaluating this derivative gives:

∂/∂y [f(x,y)] = xcos(x) - ysin(e^y)

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We would need a sample size of approximately 167 grade point averages to ensure that the sample mean is within 0.01 of the population mean with a 99% confidence level using the range rule of thumb.

To ensure that the sample mean is within 0.01 of the population mean with a 99% confidence level, the number of grade point averages needed depends on the standard deviation of the population. The answer can be obtained using the range rule of thumb.

The range rule of thumb states that for a normal distribution, the range is typically about four times the standard deviation. Since we want the sample mean to be within 0.01 of the population mean, we can consider this as the range.

A 99% confidence level corresponds to a z-score of approximately 2.58. To estimate the standard deviation of the population, we need to assume a sample size. Let's assume a conservative estimate of 30, which is generally considered sufficient for the Central Limit Theorem to apply.

Using the range rule of thumb, the range would be approximately 4 times the standard deviation. So, 0.01 is equivalent to 4 times the standard deviation.

To find the standard deviation, we divide 0.01 by 4. So, the estimated standard deviation is 0.0025.

To determine the number of grade point averages needed, we can use the formula for the margin of error in a confidence interval: Margin of Error = (z-score) * (standard deviation / √n).

Rearranging the formula to solve for n, we have n = ((z-score) * standard deviation / Margin of Error)².

Plugging in the values, n = ((2.58) * (0.0025) / 0.01)² = 166.41.

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The experimental probability of the next menu being from a Chinese restaurant is 1/10.

To find the experimental probability, we need to calculate the ratio of the number of menus from Chinese restaurants to the total number of menus.

In this case, the number of menus from Chinese restaurants is 2, and the total number of menus is the sum of all the types of menus:

Total number of menus = 2 + 9 + 1 + 2 + 6 = 20

Therefore, the experimental probability of the next menu being from a Chinese restaurant is:

P(Chinese) = Number of menus from Chinese restaurants / Total number of menus

= 2 / 20

= 1/10

So, the experimental probability is 1/10.

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Answer:

1/10

Step-by-step explanation:

i have no explanation

The confidence interval for the population mean, μ which confidence level, 95% or 99%, will result in the confidence interval giving a more accurate estimate of μ is :

D. The 99% confidence level will give a more accurate estimate of μ since the margin of error will be smaller for this confidence level.

When constructing a confidence interval for the population mean, increasing the confidence level will result in a wider interval, as there is a higher level of certainty that the true population mean falls within the interval. However, a wider interval means that the margin of error, or the amount by which the interval is likely to differ from the true population mean, will also be larger.

Therefore, if we want a more precise estimate of the population mean, we should choose a confidence level that results in a smaller margin of error. Since the margin of error decreases as the confidence level decreases, the 99% confidence level will give a more accurate estimate of μ than the 95% confidence level.

Therefore, the correct answer is : (D)

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Including the sampling of the (b) parameter in the Gibbs sampler can lead to issues with convergence.

In the Gibbs sampler, as it progresses, the samples are assumed to be drawn from the prior conditional distributions given the observed data. However, if the sampling of a particular variable is included, such as the (b) parameter, the Gibbs sampler may not converge.

The Gibbs sampler is a Markov chain Monte Carlo (MCMC) algorithm used for drawing samples from a joint distribution when the conditional distributions are easier to sample from individually. In each iteration of the Gibbs sampler, the values of variables are updated one at a time based on their conditional distributions.

Ideally, the Gibbs sampler aims to converge to the target distribution, allowing for efficient estimation and inference. However, the inclusion of certain variables in the sampling process can affect the convergence properties of the sampler. Specifically, if the (b) parameter is sampled in the Gibbs sampler, it may prevent convergence.

The convergence of the Gibbs sampler relies on the Markov chain satisfying certain conditions, such as irreducibility, aperiodicity, and ergodicity. When a parameter like (b) is included, it may introduce dependencies or correlations that violate these conditions, preventing the sampler from reaching a stationary distribution.

Therefore, including the sampling of the (b) parameter in the Gibbs sampler can lead to issues with convergence. It is important to carefully consider the impact of including or excluding variables in the sampling process and assess the convergence properties of the Gibbs sampler in each specific case.

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To determine whether the series converges or diverges, we need to analyze the given series. The series is:

Σ (5^n / (4n - 2)), from n = 1 to infinity.

To check for convergence, we can apply the Ratio Test, which involves finding the limit of the ratio between consecutive terms. Let's denote the term a_n as (5^n / (4n - 2)). Then, we'll compute the limit as n approaches infinity:

lim (n→∞) (a_(n+1) / a_n) = lim (n→∞) ((5^(n+1) / (4(n+1) - 2)) / (5^n / (4n - 2)))

Simplifying this expression, we get:

lim (n→∞) (5^(n+1) / 5^n) * ((4n - 2) / (4(n+1) - 2))

The first part of the limit simplifies to:

lim (n→∞) 5 = 5

The second part of the limit becomes:

lim (n→∞) ((4n - 2) / (4n + 2)) = 1

Multiplying both limits, we get:

5 * 1 = 5

Since the limit is greater than 1, the Ratio Test indicates that the series diverges.

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A null hypothesis makes a claim about a population parameter.

So, the correct is A

In statistical hypothesis testing, the null hypothesis is a statement that there is no significant difference between two or more variables or groups. It assumes that any observed difference is due to chance or sampling error.

The alternative hypothesis, on the other hand, is the opposite of the null hypothesis and states that there is a significant difference between the variables or groups being compared.

It is important to test the null hypothesis because it helps to determine whether the observed results are due to chance or a real effect.

Failing to reject a null hypothesis when it is false is known as a type II error, which can have serious consequences in some fields.

Hence the answer of the question is A.

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Therefore, ∫2^0 x·f'(x) dx = 0.

Using the integration by parts formula ∫u dv = uv - ∫v du, we have

∫2^0 x·f'(x) dx = [-x·f(x)]_0^2 + ∫0^2 f(x) dx

Since f(0) = 1 and f(2) = 5, we can apply the mean value theorem for integrals to get a value c in (0,2) such that

∫0^2 f(x) dx = f(c)·(2-0) = 2f(c)

Also, we know that ∫2^0 f(x) dx = -∫0^2 f(x) dx = -2f(c).

Thus, we have

∫2^0 x·f'(x) dx = [-x·f(x)]_0^2 + ∫0^2 f(x) dx

= -2f(c) + 2f(c)

= 0

Therefore, ∫2^0 x·f'(x) dx = 0.

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The lengths of the sides of triangle PQR are as follows:

Side PQ: 3 units

Side QR: approximately 6.71 units

Side RP: 6 units

To find the lengths of the sides of triangle PQR, we can utilize the distance formula, which states that the distance between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) in 3D space is given by:

d = √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)

Now, let's proceed to find the lengths of the sides of triangle PQR.

Side PQ:

The coordinates of points P and Q are P(3, 6, 5) and Q(5, 4, 4) respectively. Applying the distance formula, we have:

PQ = √((5 - 3)² + (4 - 6)² + (4 - 5)²)

= √(2² + (-2)² + (-1)²)

= √(4 + 4 + 1)

= √9

= 3

Therefore, the length of side PQ is 3 units.

Side QR:

The coordinates of points Q and R are Q(5, 4, 4) and R(5, 10, 1) respectively. Using the distance formula, we can calculate the length of side QR:

QR = √((5 - 5)² + (10 - 4)² + (1 - 4)²)

= √(0² + 6² + (-3)²)

= √(0 + 36 + 9)

= √45

≈ 6.71

Hence, the length of side QR is approximately 6.71 units.

Side RP:

To find the length of side RP, we need to calculate the distance between points R(5, 10, 1) and P(3, 6, 5). By applying the distance formula, we get:

RP = √((3 - 5)² + (6 - 10)² + (5 - 1)²)

= √((-2)² + (-4)² + 4²)

= √(4 + 16 + 16)

= √36

= 6

Therefore, the length of side RP is 6 units.

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The endpoints of the latus rectum of the parabola with equation [tex](x-2)^{2}[/tex] = -8(y-7) are (2, 7) and (2, -9).

The given equation of the parabola is in the form [tex](x-h)^{2}[/tex] = 4p(y-k), where (h, k) represents the vertex and 4p represents the length of the latus rectum. Comparing this with the given equation [tex](x-2)^{2}[/tex] = -8(y-7), we can see that the vertex is (2, 7) since (h, k) = (2, 7). The coefficient of (y-7) is -8, so 4p = -8, which implies p = -2. Since the latus rectum is a line passing through the focus and perpendicular to the axis of symmetry, its length is equal to 4p. Thus, the length of the latus rectum is 4(-2) = -8. The latus rectum is parallel to the x-axis, and its endpoints can be found by adding and subtracting the length of the latus rectum to the y-coordinate of the vertex. Hence, the endpoints of the latus rectum are (2, 7 + (-8)) = (2, -1) and (2, 7 - (-8)) = (2, 15), or in simplified form, (2, -9) and (2, 7).

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The current age of Katie is 1 year and the current age of Elena is 5 years

Statement 1

Let Katie's age be x

Let Elena's age be y

x - 9 = 2(y - 9)

x - 9 = 2y - 18

x - 2y = -18 + 9

x - 2y = - 9

Statement 2;

x + 4 = y

x - y = -4

We then have that;

x - 2y = - 9 ---- (1)

x - y = -4 ----- (2)

x = -4 + y -----(3)

Substitute (3) into (1)

-4 + y - 2y = -9

-y = -9 + 4

y = 5

The substitute y = 4 into (1)

x - 2(5) = -9

x = -9 + 10

x = 1

We can see that we have used the equations to show that the current ages of Katie and Elena are 5 years and 1 year respectively.

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The fraction of this week's allowance spent on gummy bears is 56/x. The money spent on candy will be 3/4x. Now, out of the total amount spent on candy, 56 were spent on gummy bears.

Given that,

56 was spent on gummy bears.
I spent 3/4 of this week's allowance on candy.
Let the week's allowance be x
Therefore, money spent on candy = 3/4 of x = (3/4)x
To find:

A fraction of this week's allowance is spent on gummy bears.
Now, we know that 56 was spent on gummy bears.

Therefore, the fraction of this week's allowance spent on gummy bears is 56/x.

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The integral diverges, the series Σ11/n also diverges by the Integral Test. Therefore, the answer is DIVERGES.

To apply the Integral Test, the function f(x) must be continuous, positive, and decreasing on the interval [a, ∞).

In this case, we are considering the series Σ11/n. We can define the function f(x) = 1/x, which is continuous, positive, and decreasing on the interval [1, ∞).

Now, we can apply the Integral Test:

∫1∞ 1/x dx = lim t→∞ ln(t) - ln(1) = ∞

Since the integral diverges, the series Σ11/n also diverges by the Integral Test. Therefore, the answer is DIVERGES.

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Let x be the number of student tickets and y be the number of adult tickets. There are 12 tickets total. Therefore: `x + y = 12`The cost of student tickets is $9 and the cost of adult tickets is $12.

We know that the cost of all 12 tickets is $120. Therefore: `9x + 12y = 120`We can solve this system of equations by substitution or elimination.

Let's use substitution: Solve the first equation for `x`: `x = 12 - y`Substitute that into the second equation: `9(12 - y) + 12y = 120`Simplify and solve for `y`: `108 - 9y + 12y = 120` `3y = 12` `y = 4`Now we know that 4 adult tickets were bought. We can substitute that back into the first equation to find the number of student tickets: `x + 4 = 12` `x = 8`Therefore, 8 student tickets were bought.

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This proves that the expected value of a Poisson distribution with parameter r is equal to r.

To prove that E(X) = r for X ~ Poisson(r), we use the definition of expected value:

E(X) = ∑x P(X = x) x

where the sum is taken over all possible values of X, and P(X = x) is the probability that X takes on the value x.

For the Poisson distribution, the probability mass function is:

P(X = x) = (e^-r * r^x) / x!

where r is the parameter of the Poisson distribution, representing the expected number of events per unit time (or space or other interval), and x is a non-negative integer.

Substituting this expression into the definition of expected value, we get:

E(X) = ∑x P(X = x) x

= ∑x (e^-r * r^x) / x! * x

= ∑x (e^-r * r^x) / (x-1)! (using x! = x(x-1)!)

= e^-r ∑x (r^x / (x-1)!)

= e^-r r ∑(x-1) [(r^(x-1)) / ((x-1)!)] (using the substitution k = x-1)

= e^-r r ∑k [(r^k) / k!]

= e^-r r e^r (using the power series expansion of e^r)

Therefore, we have:

E(X) = r

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To find the probability of "no bridge collapse from strong earthquakes" during the next 20 years, we need to calculate the probability of no bridge collapses during the first 20 years, and then multiply it by the probability that no bridge collapses occur during the next 20 years.

The probability of no bridge collapses during the first 20 years is equal to the probability of no bridge collapses during the first 20 years given that no bridge collapses have occurred during the first 20 years, multiplied by the probability that no bridge collapses have occurred during the first 20 years.

The probability of no bridge collapses given that no bridge collapses have occurred during the first 20 years is equal to 1 - the probability of a bridge collapse during the first 20 years, which is 0.7.

The probability that no bridge collapses have occurred during the first 20 years is equal to 1 - the probability of a bridge collapse during the first 20 years, which is 0.7.

Therefore, the probability of "no bridge collapse from strong earthquakes" during the next 20 years is:

1 - 0.7 * 0.7 = 0.27

So the probability of "no bridge collapse from strong earthquakes" during the next 20 years is 0.27

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a) No, R is not symmetric. b) No, R is not reflexive. c) Yes, R is transitive.

To see this, consider the subsets {2, 4} and {3, 6}. These subsets are disjoint, so {2, 4}R{3, 6}. However, {3, 6} is also disjoint from {2, 4}, so {3, 6}R{2, 4} is not true. For any subset X of A, X and the empty set are disjoint, so XRX cannot be true. To see this, suppose that XRY and YRZ, where X, Y, and Z are subsets of A. Then X and Y are disjoint, and Y and Z are disjoint. Since the empty set is disjoint from any set, we have that X and Z are disjoint as well. Therefore, X and Z satisfy the definition of the relation, so XRZ is true. A binary relation R across a set X is reflexive if each element of set X is related or linked to itself.

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