Assume the results of a an empirical study reveal the following: n= 250, sample mean = 140, sample standard deviation =22. The standard error of the sample mean is closest to 1.39 22 2.72 14

The interpretation of the interval in 1b (95% confidence level) is that we can be 95% confident that the true proportion of the population that hears their cats talking to them falls within the range of 0.1241 to 0.2137.

A.) With an 85% confidence level, the confidence interval that could be used to estimate the proportion of the population that hears their cats talking to them is [0.1459, 0.1919] in set notation, (0.1459, 0.1919) in interval notation, and +/- 0.023 in +/- notation.

B.) With a 95% confidence level, the confidence interval that could be used to estimate the proportion of the population that hears their cats talking to them is [0.1241, 0.2137] in set notation, (0.1241, 0.2137) in interval notation, and +/- 0.045 in +/- notation.

C.) The ranges for 1a and 1b are different because the confidence level affects the width of the interval. A higher confidence level requires a wider interval to provide a more reliable estimate. In this case, the 95% confidence level has a wider range compared to the 85% confidence level.

This means that if we were to repeat the survey multiple times, approximately 95% of the intervals calculated would contain the true proportion.

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a) The 85% confidence interval is given as follows: (0.146, 0.174).

b) The 95% confidence interval is given as follows: (0.142, 0.178).

c) The interval for item a is narrower than the interval for item b, as the lower confidence level leads to a lower critical value and a lower margin of error.

d) We are 95% sure that the true population proportion is between the two bounds found in item b.

A confidence interval of proportions has the bounds given by the rule presented as follows:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which the variables used to calculated these bounds are listed as follows:

For the confidence level of 85%, the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.85}{2} = 0.925[/tex], so the critical value is z = 1.44.

For the confidence level of 95%, the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

The parameters for the confidence interval are given as follows:

[tex]n = 1523, \pi = \frac{243}{1523} = 0.16[/tex]

Hence the bounds of the 85% confidence interval are given as follows:

The bounds of the 95% confidence interval are given as follows:

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Answer:

52

Step-by-step explanation:

Its simple, its just division because she is asking how many for EACH(key word) so 1,248/24 equals 52 giving your answer.

   mark brainliesttt??

Answer:the range is 10 pls gimme brainless plsss I need it

Step-by-step explanation:

The frequency of oranges in the classroom can be calculated as follows:Frequency of oranges in the classroom = Total number of oranges / Total number of fruits= 16/42 = 0.3809 (rounded to 4 decimal places)Thus, the frequency of oranges in the classroom is approximately 0.3809.

To determine the frequency of oranges in the classroom, we need to calculate the proportion of students who brought oranges out of the total number of students.

First, let's calculate the total number of students:

Total students = Number of students with 2 apples + Number of students with 1 apple and 1 orange + Number of students with 2 oranges

Total students = 9 + 4 + 8 = 21

Next, let's calculate the number of students who brought oranges:

Number of students with oranges = Number of students with 1 apple and 1 orange + Number of students with 2 oranges

Number of students with oranges = 4 + 8 = 12

Finally, we can calculate the frequency of oranges:

Frequency of oranges = Number of students with oranges / Total students

Frequency of oranges = 12 / 21 ≈ 0.5714 (rounded to 4 decimal places)

Therefore, the frequency of oranges in the classroom is approximately 0.5714.

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To determine the frequency of oranges in the classroom, it is important to add up the total number of fruits brought to the classroom. The given information is as follows:

2 Apples: 9 students

1 Apple and 1 Orange: 4 students

2 Oranges: 8 students

Thus, the frequency of oranges in the classroom is approximately 0.2105.

Total number of fruits = 2(9) + 1(4) + 2(8)

= 18 + 4 + 16

= 38 fruit in total

So, total number fruits are 38.

Frequency of oranges = Number of oranges / Total number of fruits

There are 8 oranges brought to class, so the frequency of oranges is:

8 / 38 ≈ 0.2105 (rounded to 4 decimal places)

Hence, the frequency of oranges in the classroom is approximately 0.2105.

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Answer:

dunno.

Step-by-step explanation:

duno

Answer:

the answer is 7. :)

Step-by-step explanation:

Simple Proof:

a) In the image, we know that ABCD is a parallelogram and that means opposite angel measures should be the same. We know that angel DCB is made up by angel 1 and 2, and angel DAB and DCB are equal and angel DAB is made up by angel 1 and x. So now we can conclude that angel x is equal to angel 2.

b) According to the definitions of a parallelogram, opposite angel measures have to be the same, while AECF have angle 1 to angel 1 and angel 2 to angel 1. We can conclude that AECF is NOT a parallelogram. (Sorry, you didn't give me the full question so some information remains unclear. )

[tex]area = b \times h \\ = 11 \times 4 \\ = 44[/tex]

Answer:

25

Step-by-step explanation:

(y+2)^2=x

(-7+2)^2=x

(-5)^2=x

(-5)(-5)=x

25=x

Hope that helps :)

Answer:

x=25

Step-by-step explanation:

Plug it innnn plug it innnn

108 inches

hope this helped <3

Answer:

21 packs of gatorade would cost $36.88.

Step-by-step explanation:

Mathematically:

8.78 / 5 = 1.756

1.756 * 21 = 36.876 ~= $36.88

Answer:

[tex]\frac{x+20}{x+4}[/tex]

Step-by-step explanation:

Can't cancel out terms like that

need to factor out the top and bottom

[tex]\frac{x^{2} +16x-80}{x^{2} -16}[/tex] = [tex]\frac{(x+20)(x-4)}{(x-4)(x+4)}[/tex]   now cancel out the (x-4) from the top and bottom

= [tex]\frac{x+20}{x+4}[/tex]

Answer & Explanation:

Error: individual terms in an equation in a fraction cannot be directly canceled out.

Correction:

the easy way (solve the quadratic equation in the numerator and complete the square in the denominator):

(x^2 + 16x - 80) / (x^2 - 16)

(x+20)(x-4) / (x+4)(x-4)

x+20 / x+4

the complicated way (manipulate the exponents and common factors):

(x^2 + 16x - 80) / (x^2 - 16)

(x^2 - 4x + 20x - 80) / x^2 - 2^4

x(x^2-1 - 2^2)+4*5(x - 2^4-2) / (x - 2^2)(x + 2^2)

x(x-4)+20(x-4) / (x-4)(x+4)

x(x-4)+(2^2 (5))(x-4) / (x-4)(x+4)

(x-4)(x + 2^2 (5)) / (x-4)(x+4)

(x-4)(x+20) / (x-4)(x+4)

x+20 / x+4

Answer:

x = 3

Step-by-step explanation:

Given that,

The mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3.

We need to find the value of x.

We know that, Mode is the number in a data with Max frequency. x can be 3 or 7. If x = 7, mode becomes 7 and if x = 3, mode equals 3.

Hence, the value of x is equal to 3.

Answer:

"Marty purchased a car. The car cost him $16,500 and it depreciates in value at a rate of 4.3% per year. How much will the car be worth in 12 years?"​

Step-by-step explanation:

Answer:

Area of the lawn = 64π square feet

Step-by-step explanation:

Area of the circular lawn = πd²/4

d is the diameter of the lawn = 16ft

Substitute the given value into the formula

Area = π(16)²/4

Arrea of the lawn = 256π/4

Area of the lawn = 64π square feet

Answer:

9

Step-by-step explanation:

Answer:

cosine = adjacent/hypotenuse

cos A = 20/29   (choice: yellow)

Step-by-step explanation:

Answer:

yellow

Step-by-step explanation:

Answer:

the last answer

Step-by-step explanation:

[tex]V=l*w*h\\l=8\\w=9\\h=1\\V=8*9*1\\V=72[/tex]

Answer:

4

Step-by-step explanation:

12(1/3)

so it is just 12/3 which is 4

Hope that helps :)

Answer:2008

Step-by-step explanation:

Answer:

16.76 inch³

Step-by-step explanation:

volume of cone is 1/3 πr²h

h=4

r=2

then 1/3* 22/7* 2*2 *4

=352/21

=16.76

Answer:

A. Survey

Step-by-step explanation:

Answer:

Option A. Survey

Step-by-step explanation:

What is survey?

Survey is defined as the act of examining a process or questioning a selected sample of individuals to obtain data about a service, product, or process.

Nick collected samples and then concluded his answer which is a survey he conducted.

Correct answer is Option A.

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Answer:

To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.

the yield received by the holder of the bill at the date of maturity is approximately 10.15%.

To calculate the yield received by the holder of the bill at the date of maturity, we need to use the formula for yield to maturity. The formula is:

Yield to Maturity = (Face Value - Purchase Price) / Purchase Price * (365 / Days to Maturity)

In this case:

Face Value = $150,000

Purchase Price = $138,222

Days to Maturity = 60 - 25 = 35

these values in the formula, we can calculate the yield to maturity:

Yield to Maturity = ($150,000 - $138,222) / $138,222 * (365 / 35)

Yield to Maturity ≈ 0.1015 or 10.15%

Therefore, the yield received by the holder of the bill at the date of maturity is approximately 10.15%.

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a) Bisection Method MATLAB code for equation [tex]x^3 + 4x^2 - 10 = 0[/tex] in the interval [0,5]:

function root = bisection_method()

   f = [tex]x^3 + 4*x^2 - 10[/tex];

   a = 0;

   b = 5;

   tol = 1e - 6;

   while (b - a) > tol

       c = (a + b) / 2;

       if f(c) == 0

           break;

       elseif f(a) * f(c) < 0

           b = c;

       else

           a = c;

       end

   end

   root = (a + b) / 2;

end

b) Bisection Method MATLAB code for equation [tex]x^3 - 6x^2 + 10x - 4 = 0[/tex] in the interval [0,4]:

function root = bisection_method()

   f = [tex]x^3 - 6*x^2 + 10*x - 4[/tex];

   a = 0;

   b = 4;

   tol = 1e-6;

   while (b - a) > tol

       c = (a + b) / 2;

       if f(c) == 0

           break;

       elseif f(a) * f(c) < 0

           b = c;

       else

           a = c;

       end

   end

   root = (a + b) / 2;

end

c) Newton's Method MATLAB code for equation [tex]x^3 + 3x + 1 = 0[/tex] in the interval (-2,0]:

function root = newton_method()

   f = [tex]x^3 + 3*x + 1[/tex];

   df =  [tex]3*x^2 + 3[/tex];

   [tex]x_0[/tex] = -1;

   tol = 1e-6;

   while abs(f([tex]x_0[/tex])) > tol

       [tex]x_0 = x_0 - f(x_0) / df(x_0)[/tex];

   end

   root = [tex]x_0[/tex];

end

d) Fixed-Point Method MATLAB code for equation [tex]x^3 - 2x - 1 = 0[/tex] in the interval (1.5,2]:

function root = fixed_point_method()

   g = [tex](x^3 - 1) / 2[/tex];

   [tex]x_0 = 2[/tex];

   tol = 1e-6;

   while abs([tex]g(x_0) - x_0[/tex]) > tol

       [tex]x_0 = g(x_0)[/tex];

   end

   root = [tex]x_0[/tex];

end

e) Secant Method MATLAB code for equation 1 - 2*exp(-x) - sin(x) = 0 in the interval (0,4]:

function root = secant_method()

   f = 1 - 2*exp(-x) - sin(x);

   [tex]x_0[/tex] = 0;

   [tex]x_1[/tex] = 1;

   tol = 1e-6;

   while abs(f([tex]x_1[/tex])) > tol

       [tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];

       [tex]x_0 = x_1[/tex];

       [tex]x_1 = x_2[/tex];

   end

   root = [tex]x_1[/tex];

end

f) Secant Method MATLAB code for equation [tex]2 - x^3 + 4*x^2 - 10 = 0[/tex] in the interval [0,4]:

function root = secant_method()

   f = [tex]2 - x^3 + 4*x^2 - 10[/tex];

   [tex]x_0 = 0[/tex];

   [tex]x_1 = 1[/tex];

   tol = 1e-6;

   while abs(f([tex]x_1[/tex])) > tol

       [tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];

       [tex]x_0 = x_1[/tex];

       [tex]x_1 = x_2[/tex];

   end

   root = [tex]x_1[/tex];

end

MATLAB provides several numerical methods for solving equations. In this case, we have used the Bisection Method, Newton's Method, Fixed-Point Method, and Secant Method to solve different equations.

The Bisection Method starts with an interval and iteratively narrows it down until the root is found within a specified tolerance. It relies on the intermediate value theorem.

Newton's Method, also known as Newton-Raphson Method, approximates the root using the tangent line at an initial guess. It iteratively refines the guess until the desired accuracy is achieved.

The Fixed-Point Method transforms the equation into an equivalent fixed-point iteration form. It repeatedly applies a function to an initial guess until convergence.

The Secant Method is a modification of the Newton's Method that uses a numerical approximation of the derivative. It does not require the derivative function explicitly.

By implementing these methods in MATLAB, we can numerically solve various equations and find their roots within specified intervals.

These numerical methods are powerful tools for solving equations when analytical solutions are not feasible or not known.

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Answer:

The equation of the midline for the function [tex]f(x)[/tex] is [tex]y = 6[/tex].

Step-by-step explanation:

The sinusoidal function of the form [tex]y = A_{o}+A\cdot \sin x[/tex] is a periodic function whose range is bounded between [tex]A_{o}-A[/tex] (minimum) and [tex]A_{o}+A[/tex] (maximum). The equation of the midline is a line paralel to the x-axis, that is:

[tex]y = c,\forall\, c\in \mathbb{R}[/tex] (1)

Where [tex]c[/tex] is mean of the upper and lower bounds of the sinusoidal function, that is:

[tex]c = \frac{(A_{o}+A+A_{o}-A)}{2}[/tex]

[tex]c = A_{o}[/tex] (2)

If we know that [tex]y = \frac{1}{2}\cdot \sin x + 6[/tex], then the equation of the midline for the function [tex]y[/tex] is:

[tex]c = A_{0} = 6[/tex]

[tex]y = 6[/tex]

The equation of the midline for the function [tex]f(x)[/tex] is [tex]y = 6[/tex].

Answer:

Given, m = 3 and n = 4

5×3 + 4×4 – 3

15 + 16 - 3

31 - 3

28

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Real-World Application: Image Compression: Image compression is a fundamental concept in the field of computer graphics and image processing.

Linear algebra plays a crucial role in various image compression techniques. Let's consider a complete solution for image compression using concepts of linear algebra.

Linear algebra concepts, such as matrix operations and transformations, are fundamental to every step of the image compression process outlined above. By applying these techniques, we can achieve efficient storage and transmission of images while balancing the trade-off between image quality and compression ratio.

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Step-by-step explanation:

a^2 +b^2 = c^2

b^2 = c^2 - a^2

3 - solving for h

A = bh

h = A ÷h

4 - solving for r

I = prt

r = I÷ pt

5- solving for b

A= 1/2 bh

can rewrite as

A = bh ÷2

b = 2A ÷ h