Write the Central Limit Theorem for sample means. 3. The average time taken to complete a project in a real estate company is 18 months, with a standard deviation of 3 months. Assuming that the project completion time approximately follows a normal distribution, find the probability that the mean completion time of 4 such projects falls between 16 and 19 months.

Answers

Answer 1

The probability that the mean completion time of 4 projects falls between 16 and 19 months is approximately 0.6568 or 65.68%.

The Central Limit Theorem states that for a sufficiently large sample size, the distribution of sample means will approach a normal distribution regardless of the shape of the population distribution.

Specifically, if we have a random sample of n observations drawn from a population with mean μ and standard deviation σ, then the distribution of the sample means will have a mean equal to the population mean μ and a standard deviation equal to the population standard deviation σ divided by the square root of the sample size n.

In this case, the average time taken to complete a project in the real estate company is 18 months, with a standard deviation of 3 months.

Assuming that the project completion time approximately follows a normal distribution, we can use the Central Limit Theorem to find the probability that the mean completion time of 4 such projects falls between 16 and 19 months.

First, we need to calculate the standard deviation of the sample mean. Since we have 4 projects, the sample size is n = 4.

Therefore, the standard deviation of the sample mean is σ/√n = 3/√4 = 3/2 = 1.5 months.

Next, we can standardize the values of 16 and 19 months using the formula z = (x - μ) / (σ/√n), where x is the value, μ is the population mean, σ is the population standard deviation, and n is the sample size.

For 16 months: z1 = (16 - 18) / (1.5) = -2/1.5 = -1.33

For 19 months: z2 = (19 - 18) / (1.5) = 1/1.5 = 0.67

Using a standard normal distribution table, we can look up the probabilities corresponding to the z-scores -1.33 and 0.67.

The table provides the cumulative probabilities for values up to a certain z-score.

For -1.33, the cumulative probability is approximately 0.0918.

For 0.67, the cumulative probability is approximately 0.7486.

To find the probability between these two z-scores, we subtract the cumulative probability associated with -1.33 from the cumulative probability associated with 0.67:

P(-1.33 < Z < 0.67) = 0.7486 - 0.0918 = 0.6568

Therefore, the probability that the mean completion time of 4 projects falls between 16 and 19 months is approximately 0.6568 or 65.68%.

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Related Questions

Evaluate the algebraic expression5m + 4n – 3 whenm=3 and n=4. Show your work.

Answers

Answer:

Given, m = 3 and n = 4

5×3 + 4×4 – 3

15 + 16 - 3

31 - 3

28

don't forget to like and mark me

5. Calculate the area.
9ft
4
11 ft.
ООО
40
44
о
396

Answers

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

Find the volume of the con round you answer to the nearest tenth

Answers

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

Mrs. Canales has 1,248 student pictures to displayaround the school. She plans to put them on 24 poster boards with the same amountof pictures on each poster board. How many student pictures will Mrs.Canales place on each poster board?

Answers

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??

Which shows a correct way to determine the volume of the right rectangular prism? ​

Answers

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]

The fourth one (l x w x h)
8 x 1 x 9= 72

Algebra 2 PLEASE HELP

Answers

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

A rotating lawn sprinkler sprays water in a circular area of grass, as shown in the
picture. The diameter of the circular area of grass is 16 ft. what is the closest measurement to the area in square feet ?

Answers

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

2 Which expressions can be used to find the volume of the rectangular prism?
3 ft
5 ft
7 ft
apply.
5
35 + 5
35 x 3
(7 + 5) x3
7+3+5
35 + 35 + 35

Answers

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.

¿can you help me, please?

Answers

The volume of the prism would be 21

If the mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3. Then what is the value of 'x'.​

Answers

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.

a) give stating reasons five other angles each equal to x
b) prove that AECF is a parallelogram ​

Answers

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. )

multiply
12 times 1/3?

Answers

Answer:

4

Step-by-step explanation:

12(1/3)

so it is just 12/3 which is 4

Hope that helps :)

4
12 / 3 = 4
1.333 x 12 = 4

In 2005, there are 705 cable users in the small town of Whoville. The number of users
increases by 56% each year after 2005. Find the number of users to the nearest whole in 2020.

Answers

Answer:2008

Step-by-step explanation:

"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?"​

Answers

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:

1.13 UNIT TEST GRAPH OF SINUSOIDAL FUNCTION PART 1
What is the equation of the midline for the function f(x)?

f(x) =1/2 sin(x)+6

Answers

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].

Euler's formula, v − e + f = 2, relates the number of vertices (v), the number of edges (e), and the number of faces (f) of a polyhedron. Solve Euler's formula for v.
a) v = e + f + 2
b) v = e + f - 2
c) v = e - f - 2
d) v = e - f + 2

Answers

Euler's formula, v − e + f = 2, relates the number of vertices (v), the number of edges (e), and the number of faces (f) of a polyhedron.

Therefore, the correct option is (d) v = e - f + 2.

To solve Euler's formula for v, we'll have to isolate v on one side of the equation. The first step is to add e to both sides of the equation:

v − e + f + e = 2 + e

v + f = e + 2

Now subtract f from both sides of the equation:

v + f - f = e + 2 - f

v = e + 2 - f

Hence, the solution for Euler's formula for v is:

v = e + 2 - f

Therefore, the correct option is (d) v = e - f + 2.

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A kindergarten teacher asked the students' parents to send their child to school with two fruits (apples and/or oranges). Below are the combinations of fruits brought to class the next day.
Fruit
2 Apples : 9 students
1 Apple and 1 Orange : 4 students
2 Oranges : 8 students

What is the frequency of oranges in the classroom? Round your answer to 4 decimal places.

Answers

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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Please help I’ll give brainliest if you answer all :)

Answers

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

From a survey taken by Survey 'R Us, 243 respondents out of the 1523 cat owners surveyed claim that their cats speak to them.
A.) With an 85% confidence level, provide the confidence interval that could be used to estimate the proportion of the population that hears their cats talking to them: use all three notations Set notation, interval notation, +/- notation
B.) Do the same as you did for 1a, but use a 95% confidence level instead Set Notation, Interval Notation, +/- notation
C.) Describe the differences between the ranges given for 1a and 1b. Why are the ranges different D.) Provide an interpretation for the interval given in 1b.

Answers

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.

What is a confidence interval of proportions?

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:

[tex]\pi[/tex] is the sample proportion, which is also the estimate of the parameter.z is the critical value.n is the sample size.

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:

[tex]0.16 - 1.44\sqrt{\frac{0.16(0.84)}{1523}} = 0.146[/tex][tex]0.16 + 1.44\sqrt{\frac{0.16(0.84)}{1523}} = 0.174[/tex]

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

[tex]0.16 - 1.96\sqrt{\frac{0.16(0.84)}{1523}} = 0.142[/tex][tex]0.16 + 1.96\sqrt{\frac{0.16(0.84)}{1523}} = 0.178[/tex]

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What is the perimeter of AOJL?
3
Ρ 2
K
M
9

Answers

Answer:

dunno.

Step-by-step explanation:

duno

answer: ummm letter?


explanation: letters belong in english not math i hate math

Nick is curious about which cell phone provider is most used by his neighbors. He asks several neighbors about their provider and draws a conclusion based on the answers he received. What kind of statistical study did Nick conduct? A. survey B. experiment C. observational study D. theoretical study

Answers

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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9 ft =_________in. How many inches????

Answers

108 inches

hope this helped <3

Write a Matlab code to solve the following problems. 1-use Bisection Method x3 + 4x2 - 10 = 0 for x = [0,5] x3 - 6x2 + 10x - 4 = 0 for xe [0,4] 2-Use Newton Method x3 + 3x + 1 = 0 for x = (-2,0] 3-Use fixed point Method x3 - 2x - 1 = 0 for x E (1.5,2] 4-Use secant Method 1-2e -* - sin(x) = 0 for x € (0,4] 2-x3 + 4x2 - 10 = 0 for x € [0,4]

Answers

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

How to find the MATLAB code be used to solve different equations numerically?

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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What is the value of 3^2? (3^2 means 3 raised to the second power.)

Answers

Answer:

9

Step-by-step explanation:

at traget a 5 pack of gaterade cost 8.78 how much would 21 packs of gatorade cost

Answers

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

How far apart are - 7 and |-7| on a number
line?

Answers

Answer:

the answer is 7. :)

Step-by-step explanation:

If x=(y+2)^2 and y= -7 then what is the value of x

Answers

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

Create a real-world application and its complete solution that requires concepts of linear algebra.

Answers

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.

Image Representation:

To begin, we need to represent the image as a matrix. Each pixel in the image can be represented as a vector of color intensities. For example, in an RGB image, each pixel can be represented as a vector [R, G, B]. The entire image can then be represented as a matrix, where each row corresponds to a pixel vector.

Discrete Cosine Transform (DCT):

DCT is a widely used technique in image compression. It converts the spatial information of the image into frequency information. The DCT operation can be represented using a matrix multiplication. Let's assume our image matrix is A. We can calculate the DCT coefficients by performing the following operation:
D = T * A * T^T,
where T is the DCT transform matrix and D is the matrix of DCT coefficients.

Quantization:

Quantization is the process of reducing the number of unique values in the DCT coefficient matrix. It involves dividing the matrix elements by a quantization matrix. The quantization matrix determines the level of compression. By adjusting the values of the quantization matrix, we can control the trade-off between image quality and compression ratio.

Entropy Coding:

After quantization, we can apply entropy coding techniques to further reduce the size of the compressed image. Entropy coding is based on the statistical properties of the data. Techniques like Huffman coding or arithmetic coding can be used to assign shorter codes to frequently occurring DCT coefficients and longer codes to less frequent ones.

Reconstruction:

To reconstruct the compressed image, we reverse the compression steps. First, we perform the inverse entropy coding to obtain the quantized DCT coefficients. Then, we perform the inverse quantization by multiplying the quantized coefficients with the quantization matrix. Finally, we apply the inverse DCT operation to obtain the reconstructed image matrix.

Lossless/Lossy Compression:

The level of compression can be adjusted based on the application requirements. If lossless compression is desired, the quantization step is skipped, and the original image can be exactly reconstructed. However, if lossy compression is acceptable, quantization is performed, resulting in some loss of image quality.

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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plssssssss help solve

Answers

Answer:

cosine = adjacent/hypotenuse

cos A = 20/29   (choice: yellow)

Step-by-step explanation:

Answer:

yellow

Step-by-step explanation:

An electrician bent a section of copper wire into a partial circle as shown. The dimensions are


given in feet (ft).


2.5 f


880


2.5


ft


What is the length of the section of wire to the nearest hundredth of a foot?

Answers

Answer:

Length = 3.84 feets (nearest hundredth)

Step-by-step explanation:

The length of section of wire can be obtyaiejd using the length of of a arc formular :

Length of arc = θ/360° * 2πr

Radius, r = 2.5 feets

Length of arc = (88/360) *2πr

Length of arc = (88/360) * 2π*2.5

Length of arc = 0.244444 * 15.707963

Length of arc = 3.8397

Length = 3.84 feets (nearest hundredth)

Answer:

Step-by-step explanation:

3.84

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