Under what circumstances does the sampling distribution of the proportion approximately follow the normal distribution? Choose the correct answer below. A. sampling without replacement when nx and n(1 - x) are each at least 10 B. only for instances of sampling with replacement c. sampling with replacement or without replacement from extremely large populations when nx and n(1-x) are sach at least 5 D. for all instances of sampling with replacement or without replacement from extremely large populations

Answers

Answer 1

The correct answer is C. The sampling distribution of the proportion approximately follows the normal distribution when sampling with replacement or without replacement from extremely large populations when nx and n(1-x) are each at least 5.

The sampling distribution of the proportion is the distribution of proportions obtained from multiple random samples taken from a population. The central limit theorem states that for large sample sizes, the sampling distribution of the proportion will be approximately normally distributed, regardless of whether sampling is done with replacement or without replacement.

In option A, it is mentioned that nx and n(1-x) should be at least 10. This is a more conservative threshold and may not always be necessary for approximation to a normal distribution. Option C, on the other hand, states that nx and n(1-x) should be at least 5. This is a commonly used threshold in statistics and is generally considered sufficient for approximation to a normal distribution for large populations.

Therefore, option C is the correct answer as it includes both sampling with replacement or without replacement and allows for nx and n(1-x) to be at least 5 for approximation to a normal distribution in most cases.

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Related Questions

how long is a parking space ? select the best estimate

Answers

Answer:

So it would not be 6 cm because, that's like the size of a couple paper clips! it's not 6 millimeters either because that's very very small! 6 kilometers is ALOT! Do it would be 6 meters! A meter is about the length of a door frame! I hoped this helped! Good luck on that IXL!

You are going to spend no more than 3 hours hiking. During the 3 hours, you will take a 15 minute break. You can hike at a rate of 2.75 miles per hour. What is the greatest number of miles that you can hike?

Answers

Answer:

3 hours minus 15 minutes = 2 hours 45 minutes (2.75 hours)

(2.75 mph)(2.75 hours) = 7.5625 miles

reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (enter your answer in terms of s.) r(t) = e2t cos(2t) i 6 j e2t sin(2t) k

Answers

To reparametrize the curve with respect to arc length, we need to find the arc length function s(t) and then solve for t in terms of s.

The arc length function is given by:

s(t) = ∫√[r'(t)·r'(t)] dt

where r'(t) is the derivative of r(t) with respect to t.

We can calculate r'(t) as:

r'(t) = (2e^(2t)cos(2t) - 4e^(2t)sin(2t))i + (12e^(2t)sin(2t))j + (2e^(2t)sin(2t) + 6e^(2t)cos(2t))k

Now we can substitute this into the arc length formula and integrate:

s(t) = ∫√[(2e^(2t)cos(2t) - 4e^(2t)sin(2t))^2 + (12e^(2t)sin(2t))^2 + (2e^(2t)sin(2t) + 6e^(2t)cos(2t))^2] dt

This integral looks quite complicated, so we will use a numerical integration method to approximate s(t).

We can use the trapezoidal rule to numerically integrate s(t) between t = 0 and some value t = T:

s(T) ≈ ∑[s(iΔt) + s((i+1)Δt)]/2 * Δt

where Δt = T/n is the step size, and n is the number of intervals we use.

Once we have approximated s(t), we can solve for t in terms of s using numerical methods such as the bisection method or Newton's method.

For example, if we want to find the value of t that corresponds to s = 10, we can solve:

s(t) = 10

for t using numerical methods. Once we have t, we can plug it back into r(t) to get the reparametrized curve in terms of arc length s.

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set up the simplex matrix used to solve the linear programming problem. assume all variables are nonnegative. maximize f = 9x 3y subject to 2x 3y ≤ 300 x 4y ≤ 200.

Answers

The simplex matrix using the linear programming problem gives optimal solution x = 12.5, y = 0, with objective function value f = 9(12.5) + 3(0) = 112.5

To set up the simplex matrix for the given linear programming problem, we need to introduce slack variables for each inequality constraint and form the initial tableau as follows:

Basic Variables x y s1 s2 RHS

s1 2 3 1 0 300

s2 1 4 0 1 200

z -9 -3 0 0 0

In this tableau, x and y are the decision variables, s1 and s2 are the slack variables, and z is the objective function.

We start with the coefficients of the decision variables in the objective function, which are -9 and -3. We choose the variable with the most negative coefficient to enter the basis, which is y in this case.

To determine which variable to exit the basis, we calculate the ratio of the right-hand side (RHS) value to the coefficient of the entering variable for each constraint. The smallest nonnegative ratio corresponds to the variable that will exit the basis.

For the y variable, we have the following ratios:

s1: 300/3 = 100

s2: 200/4 = 50

Since the ratio for s2 is smaller, we choose s2 to exit the basis. To pivot, we divide the second row by 4 and perform row operations to eliminate the y variable from the other rows:

Basic Variables x y s1 s2 RHS

s1 2 0 1 -3/4 50

y 1/4 1 0 1/4 50

z -9/4 0 0 9/4 225

The new entering variable is x, with coefficient -9/4 in the objective function. The ratios for x are:

s1: 50/2 = 25

y: 50/(1/4) = 200

Therefore, y exits the basis and we pivot on the (2,1) element:

Basic Variables x y s1 s2 RHS

s1 1/2 0 1/2 -3/8 25

x 1/8 1 -1/8 1/8 12.5

z -9/8 0 9/8 9/8 237.5

The optimal solution is x = 12.5, y = 0, with objective function value f = 9(12.5) + 3(0) = 112.5

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plsssssssssssssssssss help

Answers

Answer:

2210in^3

Step-by-step explanation:

since volume= length *width *height

and area= length * width

u alr know area

so if area=l*w

and area= 130

then 130=l*w

so subsittute 130 for length and width

volume=(130*17)

=2210in^3

Answer:

Step-by-step explanation:

Volume  = Area x Height

Area = 130 in^2

Height = 17 in

Volume = 130 x 17

Volume = 2210

use lagrange multipliers to find the indicated extrema of f subject to two constraints, assuming that x, y, and z are nonnegative. maximize f(x, y, z) = xyzconstraints: x y z = 32, x − y + z = 12f( ___ ) = ____

Answers

The maximum value of f(x, y, z) = xyz subject to the given constraints is f(4, 4, 2) = 32.

To maximize f(x, y, z) = xyz subject to constraints x y z = 32 and x - y + z = 12 using Lagrange multipliers, we need to define a function L(x, y, z, λ, μ) that includes the constraints:

L(x, y, z, λ, μ) = xyz + λ(xyz - 32) + μ(x - y + z - 12)

Now, we need to find the partial derivatives with respect to x, y, z, λ, and μ, and set them to zero:

∂L/∂x = yz + λyz + μ = 0
∂L/∂y = xz + λxz - μ = 0
∂L/∂z = xy + λxy + λ = 0
∂L/∂λ = xyz - 32 = 0
∂L/∂μ = x - y + z - 12 = 0

Solving this system of equations, we get the following:

x = 4
y = 4
z = 2

Now, we can find the maximum value of f:

f(4, 4, 2) = 4*4*2 = 32

Thus, the maximum value of f(x, y, z) = xyz subject to the given constraints is f(4, 4, 2) = 32.

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(1 point) carry out the hexadecimal addition ca10 4f57 directly in hexadecimal. show your work. do not convert the numbers to decimal, add them in decimal and convert the sum back to hexadecimal.

Answers

The hexadecimal addition of CA10 and 4F57 directly in hexadecimal is B197.

How to carry out the hexadecimal addition?

Hi! I'd be happy to help you carry out the hexadecimal addition of CA10 and 4F57 directly in hexadecimal. Here are the steps:

1. Write down the two hexadecimal numbers one below the other, with the least significant digits (rightmost digits) aligned:
  CA10
+ 4F57

2. Perform column-wise addition starting from the rightmost column (least significant digit):
  0 + 7 = 7

3. Move to the next column and add the corresponding digits. If the sum exceeds 15 (F in hexadecimal), carry over the excess to the next column:
  1 + 5 = 6 (no carry over needed)

4. Continue the addition for the remaining columns, remembering to carry over when necessary:
  A + F = 19 (in hexadecimal, this is 13); carry over the 1
  1 + C + 4 = 11 (in hexadecimal, this is B)

5. Combine the results from each column to obtain the final sum: B197

So, the hexadecimal addition of CA10 and 4F57 directly in hexadecimal is B197.

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describe the given set with a single equation or with a pair of equations. the circle of radius centered at and lying in a. the xy-plane b. the yz-plane c. the plane y

Answers

The simplified versions of the given circle equation are a. x^2 + (y-1)^2 = 81, b.(y-1)^2 + z^2 = 81, and c. x^2 + z^2 = 81

a. The circle of radius 9 centered at (0, 1, 0) lying in the xy-plane can be described with the equation:
(x-0)^2 + (y-1)^2 = 9^2
Simplified, this becomes: x^2 + (y-1)^2 = 81

b. The circle of radius 9 centered at (0, 1, 0) lying in the yz- plane can be described with the equation:
(y-1)^2 + (z-0)^2 = 9^2
Simplified, this becomes: (y-1)^2 + z^2 = 81

c. If the circle of radius 9 centered at (0, 1, 0) is lying in the plane y, it means that the y-coordinate is constant throughout the circle. In this case, the equation becomes:
x^2 + z^2 = 9^2
Simplified, this becomes: x^2 + z^2 = 81, with y = 1

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determine the equation of the tangent plane to z = ln(2x y) at (1, 3).

Answers

The equation of the tangent plane to z = ln(2x y) at (1, 3) is x + (1/3)y - z + ln(2) + ln(3) = 0.

To find the equation of the tangent plane to the surface z = ln(2xy) at the point (1, 3), we need to find the partial derivatives of z with respect to x and y at (1, 3), which are:

∂z/∂x = 1/x

∂z/∂y = 1/y

Evaluating these partial derivatives at (1, 3), we get:

∂z/∂x(1, 3) = 1/1 = 1

∂z/∂y(1, 3) = 1/3

So the normal vector to the tangent plane at (1, 3) is given by:

N = <1, 1/3, -1>

Now we need to find the constant term in the equation of the plane. To do this, we substitute the coordinates of the point (1, 3) into the equation of the surface:

z = ln(2xy) = ln(2(1)(3)) = ln(6)

So the equation of the tangent plane is:

x + (1/3)(y - 3) - z + ln(6) = 0

Simplifying, we get:

x + (1/3)y - z + ln(2) + ln(3) = 0

So the equation of the tangent plane to the surface z = ln(2xy) at the point (1, 3) is x + (1/3)y - z + ln(2) + ln(3) = 0.

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P1
A fruit punch that contains 70% truit juice is combined with 100% fruit juice. How many ounces of each should be used to make 54 oz of a mixture that is 90%
Part: 0/2

Answers

Answer:

Let x be the number of ounces of 70% fruit juice. Then 54 - x will be the number of ounces of 100% fruit juice.

.7x + 54 - x = .90(54)

54 - .30x = 48.6

.3x = 5.4, so x = 18

18 oz of 70% fruit juice, 36 ounces of 100% fruit juice.

We need 18 oz of the 70% fruit punch and 36 oz of the 100% fruit juice to make 54 oz of a mixture that is 90% fruit juice.

What is an expression?

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Let's call the amount of the 70% fruit punch we'll use "x" and the amount of 100% fruit juice we'll use "y".

To start, we can create two equations based on the information we have.

The first equation represents the total amount of liquid in the final mixture:

x + y = 54

The second equation represents the percentage of fruit juice in the final mixture:

0.7x + y = 0.9(54)

Simplifying the second equation:

0.7x + y = 48.6

We can now use the first equation to solve for one of the variables in terms of the other.

Let's solve for "y".

y = 54 - x

Substituting this into the second equation:

0.7x + (54 - x) = 48.6

Simplifying as:

0.7x - x = 48.6 - 54

-0.3x = -5.4

x = 18

So we need 18 oz of the 70% fruit punch.

Using the first equation to solve for "y":

y = 54 - x

y = 54 - 18

y = 36

So we need 36 oz of the 100% fruit juice.

Therefore, we need 18 oz of the 70% fruit punch and 36 oz of the 100% fruit juice to make 54 oz of a mixture that is 90% fruit juice.

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solve the given differential equation. (the form of yp is given.) show your work in clear and concise steps. put a box around your final answer. y'' - y = x2 3 (let yp = ax2 bx c.)

Answers

To solve the differential equation y'' - y = x^2/3 using the given form of yp (yp = ax^2 + bx + c), we first need to find the derivatives of yp.

yp = ax^2 + bx + c
yp' = 2ax + b
yp'' = 2a

Substituting yp, yp', and yp'' into the differential equation gives:

2a - (ax^2 + bx + c) = x^2/3

Simplifying this equation and comparing the coefficients of x^2, x, and the constant term gives:

-a = 1/3
b = 0
2a - c = 0

Solving for a, b, and c, we get:

a = -1/3
b = 0
c = -2a = 2/3

Therefore, the particular solution is yp = -x^2/3 + 2/3.

To find the general solution, we need to add the homogeneous solution, which is the solution to the associated homogeneous equation y'' - y = 0. The characteristic equation for this homogeneous equation is r^2 - 1 = 0, which has roots r = ±1.

Therefore, the general solution is:

y = c1e^x + c2e^-x - x^2/3 + 2/3

where c1 and c2 are constants determined by the initial or boundary conditions.

Final answer:

y = c1e^x + c2e^-x - x^2/3 + 2/3

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Section 7.2: Problem 7 (1 point) Previous Problem Problem List Next Problem Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y= x², y = 0, x = 1, about the y-axis

Answers

The volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, x = 1 about the y-axis is π/2 cubic units.

How to find the volume of the solid obtained by rotating the region bounded by the curves?

To find the volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, x = 1 about the y-axis, we can use the method of cylindrical shells.

Consider a thin strip of width dx at a distance x from the y-axis. When this strip is rotated about the y-axis, it generates a cylindrical shell with radius x and height y = x².

The volume of the cylindrical shell is given by:

dV = 2πx(y)dx

= 2πx(x²)dx

= 2πx³dx

To find the total volume, we need to integrate the volume of all such cylindrical shells from x = 0 to x = 1:

V = ∫₀¹ 2πx³ dx

= 2π ∫₀¹ x³ dx

= 2π [x⁴/4]₀¹

= 2π(1/4)

= π/2

Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, x = 1 about the y-axis is π/2 cubic units.

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On Your Own
PAGE 3 in the notes
A propane tank in the shape of a cylinder has a radius of 10 meters and a height of
15 meters. Find the volume Use 3.14 as an approximation of pi. Round your answer
to the nearest tenth.
My drawing:

Answers

The volume of the propane tank that has the shape of a cylinder would be =4,710m³

How to calculate the volume of the cylinder?

The propane tank has the shape of a cylinder with a given radius and height therefore the formula for the volume of a cylinder should be used to calculate it's volume.

To calculate the volume of a cylinder, the formula that should be used would be = πr²h

Where ;

radius = 10m

height = 15 m

Volume = 3.14 × 10×10× 15

= 4,710m³

Therefore,the volume of the propane tank that has the shape of a cylinder would be = 4710m³.

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Sketch the graphs of the following equations y=x+5, y=-(x+5) and y=|x+5|

Answers

Answer:

See below

Step-by-step explanation:

suppose x is χ2 - distribution with degrees of 20. find a point a such that p(x < a) = 0.025

Answers

For X following a χ²-distribution with 20 degrees of freedom, the point 'a' such that P(X < a) = 0.025 is approximately 8.26.

To find the point 'a' such that P(X < a) = 0.025, where X follows a χ²-distribution with 20 degrees of freedom, you need to use the inverse chi-square distribution function (also called the chi-square quantile function).

Here's the step-by-step explanation:

1. Identify the given parameters: X follows a χ²-distribution with 20 degrees of freedom, and we need to find a point 'a' such that P(X < a) = 0.025.

2. Use the inverse chi-square distribution function (quantile function) with the given probability and degrees of freedom. This function will give you the value of 'a' corresponding to the specified probability.

In most statistical software or calculators, you can find this function. For example, in R programming, you can use the "qchisq()" function:

a = qchisq(0.025, df = 20)

3. Calculate the value of 'a'.

In this case, a ≈ 8.26.

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Graph the following line y= 3/2x-5

Answers

A graph of the linear equation y = 3x/2 - 5 in slope-intercept form is shown in the image attached below.

How to graph the solution to this linear equation?

In order to to graph the solution to the given linear equation on a coordinate plane, we would use an online graphing calculator to plot the given linear equation and then take note of the point that lie on it;

y = 3x/2 - 5

Next, we would use an online graphing calculator to plot the given linear equation as shown in the graph attached below.

Based on the graph (see attachment), we can logically deduce that a possible solution for the linear equation is the ordered pairs (0, -5) and (3.333, 0), which corresponds to the y-intercept and x-intercept respectively.

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I NEED HELP ON THIS ASAP!!!!

Answers

In the two functions as the value of V(x) increases, the value of W(x) also increases.

What is the value of the functions?

The value of functions, V(x) and W(x) is determined as follows;

for h(-2, 1/4); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁻²⁺³ = 2¹ = 2

w(x) = 2ˣ ⁻ ³ = 2⁻²⁻³ = 2⁻⁵ = 1/32

for h(-1, 1/2); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2² = 4

w(x) = 2ˣ ⁻ ³ = 2⁻⁴ = 1/16

for h(0, 1); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2³ = 8

w(x) = 2ˣ ⁻ ³ = 2⁻³ = 1/8

for h(1, 2); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁴ = 16

w(x) = 2ˣ ⁻ ³ = 2⁻² = 1/4

for h(2, 4); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁵ = 32

w(x) = 2ˣ ⁻ ³ = 2⁻¹ = 1/2

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find the volume v of the described solid s. a right circular cone with height 4h and base radius 7r. V=Finding the Volume:The objective is to find the volume of the described solid.The general form of volume of the right circular cone is V=13πr^2hBy applying the given value in the formula to get a resultant part.

Answers

The volume of the described solid is (490/7) π h^3.

The formula for the volume of a right circular cone is:

V = 1/3 πr^2h

In this case, the cone has height 4h and base radius 7r. We need to express the volume in terms of h and r. Since the base radius is 7r, we can write:

r = (1/7) b

where b is the radius of the base of the cone. To find b, we use the fact that the height of the cone is 4h and the base radius is 7r:

h^2 + r^2 = (4h)^2

Substituting r = (1/7) b, we get:

h^2 + (1/49) b^2 = 16h^2

Solving for b, we get:

b^2 = 49(16h^2 - h^2) = 49(15h^2) = 735h^2

Therefore, b = sqrt(735)h, and the volume of the cone is:

V = 1/3 πr^2h = 1/3 π(49/49) b^2 (1/7) 4h = (2/21) π (735h^3)

Simplifying, we get:

V = (490/7) π h^3

Therefore, the volume of the described solid is (490/7) π h^3.

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if a flexible copper cable has 133 strands, each 5.63 mils in diamteter, what is the size of the cable in circular mils

Answers

As a result, the cable is **3317.7 CM⁵ in diameter in round mils.

explain circular mills.

A circular mil (CM) is a measurement of area that is particularly useful for describing the size of a wire or cable's cross section. It is equivalent to the surface area of a circle with a one mil (a thousandth of an inch, or 0.0254 mm²) diameter. It is equivalent to roughly 0.7854 millionths of an inch square.

This formula can be used to get the cable's diameter in circular mils:

- Each strand has a diameter of 5.63 mils.

Each strand has an area of (5.63/2)2 × π, or 24.9 circular mils (CM).

- 133 strands have a total area of 133× 24.9, or 3317.7 CM.

As a result, the cable is **3317.7 CM**⁵ in diameter in round mils.

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Policyholders of a certain insurance company have accidents at times distributed according to a Poisson process with rate λ. The amount of time from when the accident occurs until a claim is made has distribution G.
(a) Find the probability there are exactly n incurred but as yet unreported claims at time t.
(b) Suppose that each claim amount has distribution F, and that the claim amount is independent of the time that it takes to report the claim. Find the expected value of the sum of all incurred but as yet unreported claims at time t.

Answers

a. The expected value of the sum of all incurred but as yet

b. Unreported claims at time t is λt times the expected value of a single claim amount.

What is probability?

Probability is a branch of mathematics that deals with measuring the likelihood of an event occurring. It involves quantifying the chances of different outcomes of a random experiment, such as flipping a coin, rolling a die, or drawing a card from a deck.

According to the given information:

(a) Let N(t) be the number of claims incurred up to time t, and let S be the set of times when claims are incurred but not yet reported. Then, the probability that there are exactly n incurred but as yet unreported claims at time t is given by:

P(N(t) - |S| = n) = P(N(t) = n + |S|) × P(|S|)

Since the occurrence of claims follows a Poisson process with rate λ, the probability of n + |S| claims in time t is:

P(N(t) = n + |S|) = ( + |S| / (n + |S|)!)

The distribution of the time until a claim is reported, G, gives the probability that a claim is reported within some time interval after it is incurred. The probability that a claim is incurred but not reported by time t is given by:

P(|S|) = P(G > t)

Putting all these pieces together, we get:

P(N(t) - |S| = n) = ( + |S| / (n + |S|)!) ×) × P(G > t)

(b) Let X_i denote the claim amount for the i-th incurred but as yet unreported claim. Then, the total claim amount for all incurred but as yet unreported claims at time t is:

Y(t) = Σ_i= - |S| X_i

We can find the expected value of Y(t) by using the law of total expectation:

E(Y(t)) = E[E(Y(t) | N(t), S)]

Given N(t) and S, the expected value of Y(t) is just the sum of the expected values of the claim amounts for the unreported claims:

E(Y(t) | N(t), S) = Σ_i= E(X_i)

Since the claim amounts are independent and identically distributed according to F, we have:

E(X_i) = E(F)

Thus, we get:

E(Y(t)) = E[E(Y(t) | N(t), S)] = E[(n + |S|)E(F)] = λt × E(F)

Therefore, the expected value of the sum of all incurred but as yet unreported claims at time t is λt times the expected value of a single claim amount.

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invent a paired data set, consisting of five pairs of observations, for which y¯1 and y¯2 are not equal, and sey¯1 > 0 and sey¯2 > 0, but sed¯ = 0.'

Answers

Pair 1: (2, 4), Pair 2: (4, 6), Pair 3: (6, 8), Pair 4: (8, 10), Pair 5: (10, 12)

How to create a paired data set?

To create a paired data set with the given conditions, we need to have five pairs of observations where the average of y1 and y2 are not equal, and the standard errors for y1 and y2 are greater than 0, but the standard error of the differences (sed) is 0.

Step 1: Invent five pairs of observations
Here's a paired data set that meets these requirements:

Pair 1: (2, 4)
Pair 2: (4, 6)
Pair 3: (6, 8)
Pair 4: (8, 10)
Pair 5: (10, 12)

Step 2: Verify that y¯1 and y¯2 are not equal
Calculate the average of y1 values and y2 values:

y¯1 = (2+4+6+8+10)/5 = 30/5 = 6
y¯2 = (4+6+8+10+12)/5 = 40/5 = 8

Since y¯1 ≠ y¯2 (6 ≠ 8), this condition is met.

Step 3: Verify that sey¯1 > 0 and sey¯2 > 0
As the data sets have different values, it's evident that the standard errors for y1 and y2 are greater than 0.

Step 4: Verify that sed = 0
Calculate the differences (d) for each pair:

d1 = 4 - 2 = 2
d2 = 6 - 4 = 2
d3 = 8 - 6 = 2
d4 = 10 - 8 = 2
d5 = 12 - 10 = 2

Since all differences are the same (2), the standard error of the differences (sed) is 0.
So, the paired data set provided meets all the given conditions.

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An item is regularly priced at $30. It is now priced at a discount of 80% off the regular price. What is the price now?

Answers

Answer: $24

Step-by-step explanation: (30*80)/100

The price now is $6.

use the big-θ theorems (not bounding with big-o and big-ω!) to find good reference functions for each of the following:i. 3n2+5n(2n+7)ii. n(n+1)/2iii. 3n+3n6+5logn2(n)iv. n(n-1)/(6n2+log2(n))

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We can see that this function is of the same order of magnitude as 1/n. Therefore, we can say that (n-1)/(6n) = Θ(1/n), and 1/n is a good reference function for this function.

To find good reference functions for each of the given functions using the big-Theta notation, we need to find a lower and an upper bound that is both of the same order of magnitude as the function.

i. 3n² + 5n(2n+7)

Expanding the expression inside the parentheses and simplifying, we get:

3n² + 10n² + 35n

= 13n² + 35n

We can see that this function is of the same order of magnitude as n^2. Therefore, we can say that 13n² + 35n = Θ(n²), and n² is a good reference function for this function.

ii. n(n+1)/2

Expanding and simplifying, we get:

n(n+1)/2 = (n² + n)/2

We can see that this function is of the same order of magnitude as n². Therefore, we can say that (n² + n)/2 = Θ(n²), and n² is a good reference function for this function.

iii. 3n + [tex]3n^6[/tex] + 5logn(2n)

We can see that the term [tex]3n^6[/tex] dominates the other terms for large values of n. Therefore, we can say that 3n + [tex]3n^6[/tex] + 5logn(2n) = Θ([tex]n^6[/tex]), and [tex]n^6[/tex] is a good reference function for this function.

iv. n(n-1)/(6n² + log2(n))

For large values of n, the logarithmic term becomes negligible compared to the quadratic term in the denominator. Therefore, we can approximate the function as:

n(n-1)/(6n² + log2(n)) ≈ n(n-1)/(6n²)

= (n-1)/(6n)

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A group of researchers found that the probability of completing a given optional assignment is 0.16. They then took a random sample of n-26 students. What is the expected number of students that will complete the assignment?

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We can, probability of completing a given optional assignment is 0.16, expect about 4.16 students out of the 26 to complete the optional assignment.

What is the total expected number of students with complete the assignment?

The expected number of students that will complete the assignment can be found using the formula:

E(X) = np

where X is the random variable representing the number of students who complete the assignment, n is the sample size, and p is the probability of completing the assignment.

Substituting the given values, we get:

E(X) = 26 * 0.16

E(X) = 4.16

Therefore, we can expect about 4.16 students out of the 26 to complete the optional assignment. Since we can't have a fractional part of a student, we can round this up or down to the nearest whole number as needed.

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I really need help im bad at Algebra

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The line of best fit for the scatter plot is given as follows:

y = 2x.

How to define a linear function?

The slope-intercept representation of a linear function is given by the equation presented as follows:

y = mx + b

The coefficients of the function and their meaning are described as follows:

m is the slope of the function, representing the change in the output variable y when the input variable x is increased by one.b is the y-intercept of the function, which is the initial value of the function, i.e., the numeric value of the function when the input variable x assumes a value of 0. On a graph, it is the value of y when the graph of the function crosses the y-axis.

The graph touches the y-axis at the origin, hence the intercept b is given as follows:

b = 0.

Hence:

y = mx.

When x = 20, y = 40, hence the slope m is given as follows:

20m = 40

m = 40/20

m = 2.

Hence the equation is:

y = 2x.

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Question is in the picture.

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Based on the options provided, the roots of the quadratic equation associated with the graph are -6 and 3, as they are the values where the graph intersects the x-axis. Thus, option A is correct.

What is the quadratic equation?

In a quadratic equation in the form of [tex]"ax^2 + bx + c = 0"[/tex]  , the roots (or solutions) can be determined from the x-intercepts of the associated graph of the quadratic equation.

which represent the points where the graph intersects the x-axis. These points are also known as the "zeros" or "roots" of the quadratic equation.

In the case of the values -6 and 3 that you mentioned, they are the x-intercepts or zeros of the graph. This means that when you plug in -6 and 3 into the equation or function that corresponds to the graph, the result is zero.

Based on the options provided, the roots of the quadratic equation associated with the graph are -6 and 3, as they are the values where the graph intersects the x-axis.

Therefore, option a.  [tex]-6[/tex] and  [tex]3 i[/tex]s the correct answer.

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Where do I get the solutions manual of "Linear Algebra and its Applications" by Gilbert Strang?

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The solutions manual for "Linear Algebra and its Applications" by Gilbert Strang can typically be found on online bookstores, such as Amazon or Barnes & Noble.

Alternatively, you may also be able to find a digital copy of the solutions manual through online forums or academic websites. It is important to note that obtaining unauthorized copies of copyrighted materials is illegal, so be sure to obtain the solutions manual through legitimate sources.


You can find the solutions manual for "Linear Algebra and its Applications" by Gilbert Strang on various online platforms, such as Amazon or the publisher's website. You can also check your local bookstore or academic library for a copy of the solutions manual.

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Consider the random variable X having pdf fX (x) = .6 (x^2/β), for 0 < x ≤1. (a) Find the value of β that makes fX (x) a valid pdf. (b) Find the cdf for the random variable X. (c) Find the probability that the random variable X is greater than 2.

Answers

(a) The value of β that makes fX(x) a valid pdf is β = 0.2.

(b) The cdf for the random variable X is FX(x) = (3[tex]x^3[/tex]/10), for 0 < x ≤ 1.

(c) The probability that the random variable P(X > 2) = 0, since the range of possible values for X is from 0 to 1.

How to find the value of β?

To find the value of β that makes fX(x) a valid pdf, we need to ensure that the area under the pdf from 0 to 1 is equal to 1.

∫0¹ fX(x) dx = ∫0¹ 0.6(x²/β) dx = 0.6/β ∫0¹ x² dx = 0.6/β [[tex]x^3[/tex]/3]0¹ = 0.6/β * (1/3) = 1

Solving for β, we get:

0.6/β * (1/3) = 1

β = 0.6/3 = 0.2

Therefore, β = 0.2 makes fX(x) a valid pdf.

How to find the cdf for random variable?

To find the cdf for the random variable X, we integrate the pdf from 0 to x:

FX(x) = ∫[tex]0^x[/tex] fX(t) dt = ∫[tex]0^x[/tex] 0.6(t²/0.2) dt = 3[tex]t^3[/tex]/10|[tex]0^x[/tex] = (3[tex]x^3[/tex]/10) - 0

Therefore, the cdf for X is:

FX(x) = (3[tex]x^3[/tex]/10), for 0 < x ≤ 1

How to find the probability?

To find the probability that X is greater than 2, we need to use the fact that the probability of an event outside the sample space is 0. Since the range of possible values for X is from 0 to 1, the probability that X is greater than 2 is 0.

P(X > 2) = 0

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Let f be the function given by f(x)=(x^2+x)cos(5x). What is the average value of f on the closed interval 2≤x≤6 ?a. -7.392b. -1.848c. 0.722

Answers

Average value of f on the closed interval 2≤x≤6 ≈ -1.848

How to find average value of f?

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

1/(b-a) × integral from a to b of f(x) dx

So, in this case, we need to find:

1/(6-2) × integral from 2 to 6 of f(x) dx

First, let's find the integral of f(x):

integral of (x²+x)cos(5x) dx

= (1/5) × integral of (x²+x) d(sin(5x))   (integration by parts)

= (1/5) × [(x²+x)sin(5x) - integral of (2x+1)sin(5x) dx]

= (1/5) × [(x²+x)sin(5x) + (2x+1)(cos(5x))/5] + C

So, the average value of f on [2,6] is:

1/(6-2) * integral from 2 to 6 of f(x) dx

= 1/4 × [(6²+6)sin(30) + (2×6+1)(cos(30))/5 - (2²+2)sin(10) - (2×2+1)(cos(10))/5]

≈ -1.848

Therefore, the answer is (b) -1.848 (rounded to three decimal places)

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A cylinder has a base diameter of 6 centimeters and a height of 18 centimeters. What is its volume in cubic centimeters,

Answers

Answer: 508.9380099cm³ or 508.94cm³ (2 decimal places)

Step-by-step explanation:

Volume of a cylinder = [tex]V = \pi r^{2} h[/tex]   (Volume = pi x radius squared x height)

We have the diameter so we half it to get the radius

6 ÷ 2 =3

Then we do [tex]\pi[/tex] x 3² x 18 = 162[tex]\pi[/tex] or 508.9380099

1 decimal places = 508.9

2 decimal places = 508.94

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