What’s the answer???

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:

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

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:

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

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:

cosine = adjacent/hypotenuse

cos A = 20/29   (choice: yellow)

Step-by-step explanation:

Answer:

yellow

Step-by-step explanation:

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

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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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. )

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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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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Answer:

4

Step-by-step explanation:

12(1/3)

so it is just 12/3 which is 4

Hope that helps :)

Answer:

22= 6(1x+5x)

Step-by-step explanation:

He could have 4 $5 bills and 2 $1 bills.

Answer:

he has 4 $5 bills

Step-by-step explanation:

22 - 2 = 20

20 divided by 4 = 5

so 4 $5 bills and 2 $1 bills

The derivative of f(x) = -x² + 4x - 9 is f'(x) = -2x + 4.  Evaluating f'(x) at x = 1, 2, and 3 gives f'(1) = 2, f'(2) = 0, and f'(3) = -2. To find the derivative of the function f(x) = -x² + 4x - 9, we will use the four-step process.

After applying the process, we obtain the derivative f'(x) = -2x + 4. Evaluating this derivative at x = 1, x = 2, and x = 3 gives us f'(1) = 2, f'(2) = 0, and f'(3) = -2.

The four-step process involves the following steps:

1. Begin by applying the power rule, which states that the derivative of [tex]x^n[/tex] is [tex]nx^{(n-1)[/tex], where n is a constant. In this case, we have -x², so the derivative becomes -2x.

2. Apply the power rule to the next term, which is 4x. The derivative of 4x is 4.

3. Since -9 is a constant term, its derivative is zero.

4. Combine the derivatives obtained in steps 1, 2, and 3 to find the overall derivative of the function f(x). In this case, f'(x) = -2x + 4.

To find the values of f'(1), f'(2), and f'(3), we substitute the corresponding values of x into the derivative function.

When x = 1, f'(1) = -2(1) + 4 = 2.

When x = 2, f'(2) = -2(2) + 4 = 0.

When x = 3, f'(3) = -2(3) + 4 = -2.

Therefore, the derivative of f(x) is f'(x) = -2x + 4, and the values of f'(1), f'(2), and f'(3) are 2, 0, and -2, respectively.

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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:

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

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:

SA = 48 cm²

Step-by-step explanation:

SA = (3x12) + (4x3) = 48 cm²

108 inches

hope this helped <3

Answer:2008

Step-by-step explanation:

The first, second, and third quartiles from the set of data (3,4,5,6,7,37,100) include the following:

Q₁ = 4.

Q₂ = 6.

Q₃ = 37

Based on the data set, the first quartile (Q₁) can be calculated as follows;

Q₁ = [(n + 1)/4]th term

Q₁ = (7 + 1)/4

Q₁ = 2nd term

Q₁ = 4.

For the second quartile (Q₂), median, or 50th percentile, we have the following:

second quartile (Q₂) = 4th term

second quartile (Q₂) = 6.

For the third quartile (Q₃), we have:

Q₃ = [3(n + 1)/4]th term

Q₃ = 3 × 2nd term

Q₃ = 6th term

Q₃ = 37

In conclusion, a box plot for the given data set is shown in the image attached below.

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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) 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:

3/6

Step-by-step explanation:

1 2 and 3 are less than 4 so only 3 numbers

Answer:

9

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:

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:

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