6.95 Suppose that X and Y are positive discrete random variables defined on the same sample space. Use Proposition 6.16 to derive the following formulas for the product and quotient a) pxy(z) = x Px.x(x,z/x) b) PY/x(z) = Ex Px.x(x, xz) c) Specialize the formulas in parts (a) and (b) to the case where X and Y are independent. of X and Y. 606 Tarive the formulae in Brauninn Proposition 6.16 PMF of a Function of Two Discrete Random Variables Let X and Y be two discrete random variables defined on the same sample space and let g be a real-valued function of two variables defined on the range of (X,Y). Then the PMF of the random variable Z = 8(X,Y) is pz(z) = ΣΣ PX, (x, y), (6.26) (x,y)e8-1({z}) forz in the range of Z, and pz(z) = 0 otherwise. In words, ifz is in the range of Z, we ob- tain the probability that Z = z-that is, the probability that 8(X,Y)=z-by summing the joint PMF of X and Y over all points (x, y) in the plane such that g(x, y) = z. Proof Let z be in the range of Z. From the FPF for two discrete random variables, pz(z) = P(Z = z) = P(8(X,Y) = z) = P((X, Y) € 8-'({z})) = 2 px,x(x,y). (x,y)e8-'(z) = as required. Note: We can express Equation (6.26) in the alternate form Pg(x,y)(z) = 2 px,y(x, y), (6.27) 8(x,y)=2 where indicates that the double sum is taken over all x and y such that g(x, y) = z. 8(x,y)=2

The statements that are true for function and graph is the initial value of the function is One-third and the function is a shrink of the function f(x) = 3x. (option a and e).

First, let's define what a function is. A function is a mathematical rule that takes an input value (usually denoted by x) and produces an output value (usually denoted by y or f(x)). In other words, a function is like a machine that takes in a number and spits out another number.

Now, let's talk about the first statement: "The initial value of the function is One-third." The initial value of a function is the value of the output when the input is zero. So, if the initial value of this function is One-third, we can write that as f(0) = One-third.

The fifth and final statement is "The function is a shrink of the function f(x) = 3x." A shrink is a transformation of a function that compresses the graph horizontally. If we replace x in the function f(x) = 3x with a smaller value (such as x/2), we get a new function f(x/2) = 3(x/2) that is a shrink of the original function. So, if the given function is a shrink of f(x) = 3x, then we can write it as f(x) = 3(x/k) for some constant k.

Hence the first and fifth statements are the correct one.

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The number of friends there are in line is 6.

We are given that;

Number of different ways= 720

Now,

The formula for permutations to solve for the number of friends in line:

n! / (n - r)! = 720

We can simplify this equation by noticing that 720 = 6! / (6 - r)!, which means that n! / (n - r)! = 6! / (6 - r)!. Solving for r, we get:

r = n - 6

So, there are n - 6 friends waiting in line to board the roller coaster. To find n, we can substitute r = n - 6 into the original equation:

n! / (n - (n - 6))! = 720

Simplifying this equation, we get:

n! / 6! = 720

n! = 720 * 6!

n = 6

Therefore, by permutation the answer will be 6.

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(a) The fundamental radian frequency of x(t) is 8π.

(b) The Fourier transform of x(t) is X(ω) = 2π[δ(ω) + 3/2π(δ(ω+8π) - δ(ω-8π)) + π(δ(ω+5π) + δ(ω-5π))].

(a) The fundamental radian frequency of x(t) is the highest frequency component with a non-zero coefficient in the Fourier series expansion of x(t). In this case, the coefficient of the term cos(8t) is non-zero, and its corresponding frequency is 8π.

(b) Using the properties of Fourier transforms, we can write x(t) in terms of its Fourier series coefficients and derive X(ω). The final expression is obtained using the delta function representation of the Fourier series coefficients.

(c) The magnitude response |H(ω)| is the absolute value of the Fourier transform of h(t). Plotting the function for 0 ≤ ω ≤ 10π gives a peak at ω = 2π and nulls at ω = 0 and ω = 4π.

(d) The system h(t) is distortionless if the magnitude of its frequency response is constant. Since |H(ω)| varies with ω, the system is not distortionless.

(e) The output y(t) can be obtained by convolving x(t) and h(t) using the time-domain convolution formula. The final expression for y(t) involves integrals of sinc and cosine functions, which can be evaluated numerically.

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The amount of the first powder that the pharmacist uses is -100 milligram and the amount of the second powder that the pharmacist uses is 1500mg

Let the amount of the first powder that the pharmacist uses "x" (in milligrams), and the amount of the second powder "y" (also in milligrams).

0.20x + 0.10y = 130 (equation 1)

0.15x + 0.10y = 180 (equation 2)

Subtract equation 2 from 1

0.20x-0.15x=130-180

0.5x=-50

Divide both sides by 0.5

First powder x=-100

Now plug in -100 to find y

0.20(-100)+0.10y=130

-20+0.10y=130

0.10y=150

Divide both sides by 0.1

y=150/0.1

second  powder y=1500

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The term "daily wage" refers to the amount of money a person earns for their work in one day. In Jodie's case, she earns $15 each day for delivering newspapers to 50 houses. This means that her daily wage is $15.

Similarly, the term "daily earnings" can also be used to describe the money a person makes in one day. In Jodie's case, her daily earnings would also be $15 since she earns that amount each day.

Both terms are commonly used to describe the income earned by individuals who work on a daily wage, such as freelancers, contractors, or hourly workers who are paid on a daily basis. The terms can also be used for individuals who have a fixed salary or hourly rate but work on a daily basis, such as delivery drivers or newspaper carriers like Jodie.

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To evaluate the integral ∫e^(9x+1) dx, we can use a simple substitution. Let's substitute u = 9x + 1. Taking the derivative of both sides with respect to x gives us du/dx = 9, or dx = du/9.

Substituting these values into the integral, we have:

∫e^(9x+1) dx = ∫e^u (du/9)

             = (1/9) ∫e^u du.

Now, we can integrate e^u with respect to u. The integral of e^u is simply e^u. Therefore, we have:

(1/9) ∫e^u du = (1/9) e^u + C,

where C is the constant of integration.

Substituting the original expression for u, we get:

(1/9) e^(9x+1) + C.

So, the result of the integral ∫e^(9x+1) dx is:

(1/9) e^(9x+1) + C.

To check the result, let's differentiate this expression with respect to x:

d/dx [(1/9) e^(9x+1) + C]

= (1/9) d/dx [e^(9x+1)]

= (1/9) e^(9x+1) * d/dx [9x+1]

= (1/9) e^(9x+1) * 9

= e^(9x+1).

The result of differentiating matches the original integrand e^(9x+1), confirming the correctness of our integral evaluation.

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The probability of a flush is: 1,277 / 2,598,960 = 0.0019654, or about 0.2%, In other words, a flush will occur in roughly 1 out of every 510 hands.

we need to determine the number of possible flush hands and divide by the total number of possible hands.

The number of possible flush hands is given by the product of the number of ways to choose 5 cards from a single suit and the number of possible suits (since there are four suits to choose from). Thus, the number of flush hands is: (13 choose 5) * 4 = 1,277

The total number of possible hands is the number of ways to choose 5 cards from a deck of 52: (52 choose 5) = 2,598,960

Therefore, the probability of a flush is: 1,277 / 2,598,960 = 0.0019654, or about 0.2%, In other words, a flush will occur in roughly 1 out of every 510 hands.

It's worth noting that this calculation assumes that the cards are drawn randomly from a well-shuffled deck. In practice,

the probability of a flush (or any other hand) may be affected by various factors, such as the skills of the players, the presence of wild cards, and the rules of the particular game being played.

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The integral that gives the volume of the solid formed by revolving the region about the x-axis is V = [tex]\int\limits^{-2}_{-2}[/tex]π(4−x²)² dx is (8/3)π cubic units.

To find the volume of the solid formed by revolving the region about the x-axis, we can use the disk method.

First, we need to find the limits of integration. The given function y = 4 - x² intersects the x-axis at x = -2 and x = 2. So, the limits of integration will be from -2 to 2.

Next, we need to express the given function in terms of x. Solving for x, we get x = ±√(4-y).

Now, we can set up the integral for the volume using the disk method

V = π [tex]\int\limits^a_b[/tex] (f(x))² dx

where f(x) = √(4-x²), and a = -2, b = 2.

V = π [tex]\int\limits^{-2}_{-2}[/tex] (√(4-x²))² dx

V = π [tex]\int\limits^{-2}_{-2}[/tex] (4-x²) dx

V = π [4x - (1/3)x³] [tex]|^{-2}_2[/tex]

V = π [(32/3)-(8/3)]

V = (8/3)π

Therefore, the volume of the solid formed by revolving the region about the x-axis is (8/3)π cubic units.

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Answer:

870 meters

Step-by-step explanation:

To find the difference between the distances they walked, we need to calculate the distance each person walked.

We can use the formula distance = speed x time.

Naeem's speed is 1.3 m/s and he took 5 minutes to get to school which is equal to 300 seconds. Therefore, Naeem walked a distance of:

distance = speed x time

distance = 1.3 m/s x 300 s

distance = 390 m

Fina's speed is 1.4 m/s and she took 15 minutes to get to school which is equal to 900 seconds. Therefore, Fina walked a distance of:

distance = speed x time

distance = 1.4 m/s x 900 s

distance = 1260 m

The difference between the distances they walked is:

1260 m - 390 m = 870 meters.

The value of a. P(x > 1,050) = 0.4013, b. P(x < 960) = 0.4207, c. P(x > 1,100) = 0.3085.

a. To find P(x > 1,050), we need to standardize the value of x using the formula:

z = (x - µ) / σ

where µ is the mean of the population and σ is the standard deviation.

In this case, µ = 1000 and σ = 200.

z = (1,050 - 1,000) / 200 = 0.25

Using a standard normal distribution table or calculator, we can find that the probability of z being greater than 0.25 is approximately 0.4013.

Therefore, P(x > 1,050) = 0.4013.

b. To find P(x < 960), we again need to standardize the value of x:

z = (x - µ) / σ

z = (960 - 1000) / 200 = -0.2

Using the same table or calculator, we can find that the probability of z being less than -0.2 is approximately 0.4207.

Therefore, P(x < 960) = 0.4207.

c. To find P(x > 1,100), we standardize x:

z = (x - µ) / σ

z = (1,100 - 1,000) / 200 = 0.5

Using the table or calculator, we can find that the probability of z being greater than 0.5 is approximately 0.3085.

Therefore, P(x > 1,100) = 0.3085.

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The probability that at least 7 of the 10 washing machines will have a lifetime beyond the warranty period is  0.321.

we know that the mean is 8 years and the variance is 16 years^2. Solving these equations for α and β, we get:

α = (Mean / Variance)² = (8 / 16)² = 1/4

β = Variance / Mean = 16 / 8 = 2

Therefore, the pdf of the lifetime of the new model of washing machine is:

f(x) = x^(α-1) e^(-x/β) / (β^α Γ(α))

where Γ(α) is the gamma function.

Substituting the values of α and β, we get:

f(x) = 4 x^(1/4-1) e^(-x/2) / Γ(1/4)

(b) Let X be the number of washing machines out of the 10 that have a lifetime beyond the warranty period.

P(X > 5) = 1 - P(X ≤ 5) = 1 - F(5)

F(x) = Γ(α, x/β) / Γ(α)

where Γ(α, x/β) is the upper incomplete gamma function.

F(x) = Γ(1/4, x/2) / Γ(1/4)

Therefore, the probability that at least 7 of the 10 washing machines will have a lifetime beyond the warranty period is:

P(X > 5) = 1 - F(5) = 1 - Γ(1/4, 5/2) / Γ(1/4) = 0.321

So the probability that at least 7 of the 10 washing machines will have a lifetime beyond the warranty period is  0.321.

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The cost per equivalent unit for conversion costs using weighted average is $4.34.

To determine the cost per equivalent unit for conversion costs using weighted average, we need to consider the costs incurred during a specific period and divide it by the equivalent units of production.

The weighted average method takes into account the costs from the beginning inventory and costs incurred during the current period. It assigns a weight to the beginning inventory costs and a weight to the costs incurred during the period based on the number of units involved.

The formula for calculating the cost per equivalent unit using weighted average is:

Cost per equivalent unit = (Cost from beginning inventory + Cost incurred during the period) / (Equivalent units from beginning inventory + Equivalent units produced during the period)

To determine the specific value, we need the actual cost from beginning inventory, the cost incurred during the period, the equivalent units from beginning inventory, and the equivalent units produced during the period. Without this information, it is not possible to provide an exact answer. However, the correct answer among the options provided will be determined by calculating the cost per equivalent unit using the weighted average method.

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To find the marginal average profit function, we first need to find the average profit function. The average profit function is given by:

A(x) = P(x)/x

where P(x) is the total profit function.

So, we have:

A(x) = (2x - 4)(9x - 7)/x

Now, to find the marginal average profit function, we take the derivative of the average profit function with respect to x:

A'(x) = [2(9x - 7) + (2x - 4)(9)]/x^2

Simplifying this expression, we get:

A'(x) = (20x - 50)/x^2

Therefore, the marginal average profit function is:

A'(x) = 20/x - 50/x^2

To find the marginal average profit function, we first need to find the derivative of the total profit function P(x) = (2x - 4)(9x - 7).

Using the product rule, we get:

P'(x) = (2x - 4)(9) + (2)(9x - 7)

P'(x) = 18x - 36 + 18x - 14

Now, simplify the expression:

P'(x) = 36x - 50

So, the marginal average profit function is P'(x) = 36x - 50.

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For part a, the confidence interval for the population proportion of college students who work to pay for tuition and living expenses is 0.42 to 0.52, with a certain level of confidence (usually 95% or 99%).
For part b, the confidence interval is slightly wider and ranges from 0.41 to 0.53. This could be due to a larger sample size or a lower level of confidence.
For part c, as the confidence level increases from 95% to 99%, the margin of error becomes larger.

a. To calculate the confidence interval for the population proportion of college students who work to pay for tuition and living expenses, we use the given range of 0.42 to 0.52. This interval indicates that we can be confident that the true population proportion falls between 42% and 52% of college students. This means that we are 95% confident that the true population proportion falls within this interval based on the sample data.

b. Similarly, for the second provided confidence interval, we use the given range of 0.41 to 0.53. This interval indicates that we can be confident that the true population proportion falls between 41% and 53% of college students.

c. When the confidence level is increased, the margin of error becomes larger. This is because a higher confidence level requires a wider interval to ensure that the true population proportion falls within the specified range with greater certainty. This is because a higher level of confidence requires a wider interval to capture the true population proportion. As a result, the precision of the estimate decreases as the margin of error increases.

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The integral of ∫∫x dA = 16/3.

To evaluate the given integral ∫∫x dA over the region D, we can change to polar coordinates.

In polar coordinates, x = r cos(θ) and y = r sin(θ), where r is the distance from the origin to the point (x, y), and θ is the angle between the positive x-axis and the line connecting the origin to the point (x, y).

The region D is bounded by the circles x^2 + y^2 = 16 and x^2 + y^2 = 4x, which can be rewritten in polar coordinates as r^2 = 16 and r^2 = 4r cos(θ), respectively. Solving for r, we get r = 4 cos(θ) for the inner circle and r = 4 for the outer circle.

Thus, the integral can be written as:

∫∫x dA = ∫(θ=0 to π/2) ∫(r=4cosθ to 4) r cos(θ) r dr dθ

Simplifying this expression, we get:

∫∫x dA = ∫(θ=0 to π/2) ∫(r=4cosθ to 4) r^2 cos(θ) dr dθ

Integrating with respect to r first, we get:

∫∫x dA = ∫(θ=0 to π/2) [cos(θ) (64/3 - 16cos^3(θ))] dθ

Finally, integrating with respect to θ, we get:

∫∫x dA = 16/3

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Jesse will pay $2,975 in income tax and Jesse will have $72,025 after paying his income tax if the local income tax rate is 8. 5%.

Amount earned = $75000

Income tax rate = 8. 5%

Income = $40,000

A. To calculate the income tax, Jesse, we need to estimate the taxable income of Jesse.

Taxable income = $75,000 - $40,000

Taxable income = $35,000

Income tax = $35,000 x 0.085

Income tax = $2,975

Therefore, we can conclude that Jesse will pay $2,975 in income tax.

B. To find out the remaining amount Jesse has after paying income tax,

Remaining amount = $75,000 - $2,975

Remaining amount  = $72,025

Therefore, we can conclude that Jesse will have $72,025 after paying his income tax.

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The critical value of t for testing the significance of each of the independent variable's coefficients will have 130 degrees of freedom.

This is because the degrees of freedom for a t-test in a regression model with 5 independent variables and 136 observations is calculated as (n - k - 1) where n is the number of observations and k is the number of independent variables.

Therefore, (136 - 5 - 1) = 130 degrees of freedom.

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Julio needs to dig 3/4 ft of hole.

Given that, Julio needs to dig a hole that is 2 feet deep. He has already dug a hole that is 1 ¼ feet deep.

We need to find that how many more inches does Julio need to dig,

Here we will subtract the already dug hole from the total depth of the hole,

2 - 1 ¼

= 3/4

Hence, Julio needs to dig 3/4 ft of hole.

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The circulation of F counterclockwise around the square is 2/3, and the outward flux of F across C is -2/3.

First, we need to find the circulation of the vector field F counterclockwise around the square bounded by y = 0, x = 0, and y = x.

Using Green's Theorem, we have:

circulation = ∮CF · dr = ∬R (∂Q/∂x - ∂P/∂y) dA

where R is the region enclosed by C and F = P i + Q j.

Here, P = 6y^2 - x^2 and Q = -x^2 - 6y^2, so

∂Q/∂x = -2x, and ∂P/∂y = 12y.

Therefore,

circulation = ∫0^1 ( ∫x^2^0 (-2x) dy) dx + ∫1^0 ( ∫0^(1-x) 12y dx) dy

Simplifying and evaluating the integrals, we get:

circulation = 2/3

Next, we need to find the outward flux of F across C.

Using Green's Theorem again, we have:

flux = ∫∫R (∂P/∂x + ∂Q/∂y) dA

Here,

∂P/∂x = -2x and ∂Q/∂y = -12y.

Therefore,

flux = ∫0^1 ( ∫x^2^0 (-2x) dy) dx + ∫1^0 ( ∫0^(1-x) (-12y) dx) dy

Simplifying and evaluating the integrals, we get:

flux = -2/3

So the circulation of F counterclockwise around the square is 2/3, and the outward flux of F across C is -2/3.

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The Cartesian equation is: 16y^2 - 64y + x^2 + y^2 = 64. Since the equation contains both x^2 and y^2 terms and there are no cross terms (xy), the curve is a conic section.

To convert the polar equation r = -8cos(theta) + 4sin(theta) to Cartesian coordinates, we use the identities x = rcos(theta) and y = rsin(theta), giving us:

x = -8cos(theta)cos(theta) + 4sin(theta)cos(theta)
y = -8cos(theta)sin(theta) + 4sin(theta)sin(theta)

Simplifying these equations using the identities cos^2(theta) + sin^2(theta) = 1 and 2sin(theta)cos(theta) = sin(2theta), we get:

x = -8cos^2(theta) + 4sin(theta)cos(theta)
y = -4sin(2theta)

To describe the resulting curve, we notice that the x-coordinate is a combination of cos^2(theta) and sin(theta)cos(theta), which suggests a horizontal shift of a cosine function with amplitude 8 and period pi/2. The y-coordinate is -4sin(2theta), which is a scaled and reflected sine function with amplitude 4 and period pi. Therefore, the curve is a cardioid with symmetry about the y-axis, option C.

To convert the polar equation r = -8 cos θ + 4 sin θ to Cartesian coordinates, we can use the following relationships: x = r cos θ and y = r sin θ.

First, substitute r into x and y equations:

x = (-8 cos θ + 4 sin θ) cos θ
y = (-8 cos θ + 4 sin θ) sin θ

Now, use the Pythagorean identity sin^2 θ + cos^2 θ = 1 to eliminate θ:

cos θ = x / (-8 + 4y)
sin θ = y / (-8 + 4y)

Square both equations and add them together:

x^2 / (-8 + 4y)^2 + y^2 / (-8 + 4y)^2 = 1

Simplify the equation:

x^2 + y^2 = (-8 + 4y)^2

Expand the equation:

x^2 + y^2 = 64 - 64y + 16y^2

Rearrange the terms:

16y^2 - 64y + x^2 + y^2 - 64 = 0

Finally, the Cartesian equation is:

16y^2 - 64y + x^2 + y^2 = 64

Now let's analyze the curve. Since the equation contains both x^2 and y^2 terms and there are no cross terms (xy), the curve is a conic section. Comparing it with the general equation of a conic section, we can conclude that the curve is a cardioid with symmetry about the x-axis (Option E).

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Answer:

[tex] \frac{4}{3} \pi {r}^{3} = 2881[/tex]

[tex]r = \sqrt[3]{ \frac{2881}{ \frac{4}{3} \pi} } = 8.83[/tex]

The radius of the beach ball is about 8.83 cm.

The geometric solid suggested by a typical American house is:

d. Prism

A typical American house often has a rectangular base and parallel, congruent faces.

This shape is best represented by a rectangular prism.

The geometric solid suggested by a typical American house is a prism, specifically a rectangular prism.

A prism is a three-dimensional solid that has two congruent and parallel bases that are connected by a set of parallelograms.

A rectangular prism has two rectangular bases and rectangular faces that are perpendicular to the bases.

Most American houses are rectangular in shape and have a flat roof, which suggests that they are in the form of a rectangular prism.

The walls of the house form the rectangular faces of the prism, and the roof forms the top face of the prism.

The rectangular shape of the house provides a practical and functional design that allows for efficient use of interior space.

It is also an aesthetically pleasing design that has become a standard for American homes.

d. Prism.

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When using the approximation sin θ ≈ θ for a pendulum, it is important to specify the angle θ in radians only (option a).

This approximation is derived from the small-angle approximation, which states that for small angles, the sine of the angle is approximately equal to the angle itself when expressed in radians. This approximation becomes more accurate as the angle decreases, and is generally valid for angles less than about 10 degrees (0.174 radians).

The reason for using radians in this approximation is that radians are a more natural unit for angles in mathematical calculations, as they are dimensionless and relate directly to the arc length on a circle. Degrees and revolutions are more convenient for everyday use but can introduce scaling factors in mathematical expressions, complicating calculations.

To ensure accuracy and proper application of the small-angle approximation for pendulums, always express the angle θ in radians when using sin θ ≈ θ.

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Rounding to two decimal places, the upper and lower limits of the confidence interval are: Upper limit = 2,130.78 square feet, Lower limit = 1,631.22 square feet

To find the 90% confidence interval for the mean size of all houses in this city, we need to use the formula:

CI = X ± (Zα/2) * (σ/√n)

Where X is the sample mean (1,881.0 square feet), σ is the population standard deviation (328.00 square feet), n is the sample size (8), and Zα/2 is the critical value for the 90% confidence level (1.645).

Plugging in the values, we get:

CI = 1,881.0 ± (1.645) * (328.00/√8)

Simplifying the equation, we get:

CI = 1,881.0 ± 249.78

Rounding to two decimal places, the upper and lower limits of the confidence interval are:

Upper limit = 2,130.78 square feet
Lower limit = 1,631.22 square feet

Therefore, we can be 90% confident that the mean size of all houses in this city is between 1,631.22 and 2,130.78 square feet.

To calculate the 90% confidence interval for the mean size of all houses in this city, we need to use the given information:

Sample size (n) = 8
Sample mean (x) = 1,881.0 square feet
Sample standard deviation (s) = 328.00 square feet

We also need the t-distribution critical value for a 90% confidence interval and 7 degrees of freedom (n-1 = 8-1 = 7). Using a t-table or calculator, the t-value is approximately 1.895.

Next, calculate the standard error:
Standard Error (SE) = s / √n = 328 / √8 ≈ 115.99

Now, calculate the margin of error:
Margin of Error (ME) = t-value * SE = 1.895 * 115.99 ≈ 219.84

Finally, calculate the lower and upper limits of the 90% confidence interval:
Lower Limit = x - ME = 1881 - 219.84 ≈ 1661.16
Upper Limit = x + ME = 1881 + 219.84 ≈ 2100.84

So, the 90% confidence interval for the mean size of all houses in this city, rounded to two decimal places, is (1661.16, 2100.84) square feet.

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Answer:

x2 + 4x +4

Step-by-step explanation:

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To achieve the least area, create two separate square enclosures, each with a side length of 25 m. For the greatest area, make one enclosure with a side length of 49 m and another with a side length of 1 m.

To determine the plans for the least and greatest areas using 200 m of fencing, we'll consider two cases: one square enclosure and two square enclosures.

Case 1: One square enclosure
Perimeter = 200 m
Since the perimeter of a square is 4 * side length (s), we have:
200 = 4 * s
s = 50 m

Area of one square enclosure = s^2 = 50^2 = 2500 m^2

Case 2: Two square enclosures
Let s1 and s2 be the side lengths of the two square enclosures.
Perimeter = 200 m
4 * (s1 + s2) = 200
s1 + s2 = 50
Since the area of a square is side length squared, we have:
Area = s1^2 + s2^2

To minimize the area, make the side lengths equal:
s1 = s2 = 25 m
Minimum area = 2 * (25^2) = 2 * 625 = 1250 m^2

To maximize the area, make one side length as large as possible while keeping the perimeter constraint:
s1 = 49 m, s2 = 1 m
Maximum area = 49^2 + 1^2 = 2401 + 1 = 2402 m^2

Therefore, to achieve the least area, create two separate square enclosures, each with a side length of 25 m. For the greatest area, make one enclosure with a side length of 49 m and another with a side length of 1 m.

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The position function given the acceleration function and an initial value is s(t) = 2t^2 - (t^3)/6 + 8t.

To find the position function s(t) given the acceleration function a(t) = 4 - t, and initial values v(0) = 8 and s(0) = 0, follow these steps:

1. Integrate the acceleration function a(t) to find the velocity function v(t).

∫(4 - t) dt = 4t - (t^2)/2 + C1

2. Use the initial value v(0) = 8 to find the constant C1.

8 = 4(0) - (0^2)/2 + C1 => C1 = 8

So, v(t) = 4t - (t^2)/2 + 8

3. Integrate the velocity function v(t) to find the position function s(t).

∫(4t - (t^2)/2 + 8) dt = 2t^2 - (t^3)/6 + 8t + C2

4. Use the initial value s(0) = 0 to find the constant C2.

0 = 2(0)^2 - (0^3)/6 + 8(0) + C2 => C2 = 0

So, the position function s(t) = 2t^2 - (t^3)/6 + 8t.

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The value of the function y at x = 2 is approximately 4.5504.

Let's start by finding the general solution to the homogeneous equation xy'' - 4xy' + 6y = 0. We can assume a solution of the form y = [tex]x^r[/tex] and substitute it into the equation to get:

xy'' - 4xy' + 6y = r*(r-1)[tex]x^r[/tex] - [tex]4rx^r + 6x^r = (r^2 - 4r + 6)*x^r[/tex]

So, we want to find the values of r that make the above expression equal to 0. This gives us the characteristic equation:

[tex]r^2 - 4r + 6 = 0[/tex]

Using the quadratic formula, we get:

r = (4 ± [tex]\sqrt(16[/tex] - 4*6))/2 = 2 ± i

Therefore, the general solution to the homogeneous equation is:

[tex]y_h(x) = c1x^2cos(ln(x)) + c2x^2sin(ln(x))[/tex]

Now, we need to find a particular solution to the non-homogeneous equation xy'' - 4xy' + 6y = [tex]6*x^5[/tex]. We can guess a solution of the form [tex]y_p = Ax^5[/tex] and substitute it into the equation to get:

xy'' - 4xy' + 6y = [tex]60Ax^3 - 120Ax^3 + 6Ax^5 = 6*x^5[/tex]

So, we need to choose A = 1/6 to make the equation hold. Therefore, the general solution to the non-homogeneous equation is:

[tex]y(x) = y_h(x) + y_p(x) = c1x^2cos(ln(x)) + c2x^2sin(ln(x)) + x^{5/6[/tex]

Using the initial conditions y(1) = 2 and y'(1) = 7, we get:

c1 + c2 + 1/6 = 2

-2c1ln(1) + 2c2ln(1) + 5/6 = 7

The second equation simplifies to:

2*c2 + 5/6 = 7

Therefore, c2 = 31/12. Using this value and the first equation, we get:

c1 = 13/12

So, the solution to the non-homogeneous equation is:

[tex]y(x) = 13/12x^2cos(ln(x)) + 31/12x^2sin(ln(x)) + x^{5/6[/tex]

Finally, we can find the value of y(2):

y(2) = [tex]13/122^2cos(ln(2)) + 31/122^2sin(ln(2)) + 2^{5/6[/tex]

y(2) = 4.5504

Therefore, the value of the function y at x = 2 is approximately 4.5504.

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