. Amy Company sold merchandise of $8,000 to Tory Turnbull with terms 2/10, n/30. Amy Company recorded this transaction using the gross method. If Tory Turnbull paid for all the merchandize within the discount period, the journal entry that Amy Company will make to record the collection of cash would include a: a. Credit to Sales Discount of $160 b. Credit to Account receivable $7,840 c. Debit to Sales Discount of $160 d. Credit to Cash of $160 Select-

24 years and 5 months

add 13 years to ten get 23 add 6 to 11 to get 1 year and 5 months

23 plus 1 years and 5 months = 24 years an 5months

have a nice day please mark brainliest

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

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

Step-by-step explanation:

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

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:

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:

the answer is 7. :)

Step-by-step explanation:

Answer:

cosine = adjacent/hypotenuse

cos A = 20/29   (choice: yellow)

Step-by-step explanation:

Answer:

yellow

Step-by-step explanation:

The unit impulse response of the system described by the equation y[n+2] + 2y[n+1] + y[n] = 2x[n+2] − x[n+1] is h[n] = 2δ[n+2] − δ[n+1] + δ[n], where δ[n] represents the unit impulse function.

To find the unit impulse response, we need to determine the output of the system when an impulse is applied at the input, i.e., x[n] = δ[n].

Substituting x[n] = δ[n] into the given equation, we have:

y[n+2] + 2y[n+1] + y[n] = 2δ[n+2] − δ[n+1].

Since δ[n] = 0 for n ≠ 0 and δ[0] = 1, we can simplify the equation:

y[n+2] + 2y[n+1] + y[n] = 2δ[n+2] − δ[n+1] + δ[n] = 2δ[n+2] − δ[n+1] + δ[n]δ[n].

Now, comparing the equation with the standard form of the unit impulse response:

h[n] = 2δ[n+2] − δ[n+1] + δ[n],

we can conclude that h[n] is the unit impulse response of the given system.

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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. )

The 90% confidence interval for the true population mean textbook weight, based on the given data, is approximately 41.44 ounces to 42.56 ounces.

To construct a confidence interval for the population mean textbook weight, we can use the formula:

Confidence Interval = sample mean ± (critical value * standard error)

Given that the sample mean is 42 ounces and the population standard deviation is 3.8 ounces, we need to determine the critical value and the standard error.

For a 90% confidence interval, the critical value corresponds to a two-tailed z-score of 1.645 (from the standard normal distribution).

The standard error can be calculated as the population standard deviation divided by the square root of the sample size. Since the sample size is not provided, we cannot calculate the exact standard error. However, if we assume a large sample size (usually considered to be greater than 30), we can use the formula for the standard error.

Assuming a large sample size, the standard error would be 3.8 ounces divided by the square root of the sample size.

Using the formula for the confidence interval, we can now calculate the range:

Confidence Interval = 42 ± (1.645 * standard error)

Substituting the values, we get:

Confidence Interval = 42 ± (1.645 * 3.8 / sqrt(sample size))

Since we do not know the sample size, we cannot calculate the exact confidence interval. However, based on the given data, we can conclude that the true population mean textbook weight falls between approximately 41.44 ounces and 42.56 ounces with 90% confidence.

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Evaluating the quadratic function we will get:

g(-2) = -3

g(0) =  -5

g(3) = 31

Here we need to evaluate the quadratic function:

g(x) = -2x² + 3x - 5

To do so, just replace the value of x by the correspondent number.

For example, if x = -2

g(-2) = 2*(-2)² + 3*(-2) - 5 = -3

if x = 0

g(0) = 2*(0)² + 3*(0) - 5 = -5

if x = 3

g(3) = 2*(3)² + 3*(3) - 5 = 31

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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:

"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:

16

Step-by-step explanation:

Alex will run 16 laps in the time it take for Carlos to run 12 laps.

Multiplication of two numbers is defined as the addition of one of the number repeatedly until the times of the other number.

a × b means that a is added to itself b times or b is added to itself a times.

Given are the values related to the number of laps run by Alex and Carlos.

The number of laps run by Alex is forming the multiples of 4 or 4x.

Number of laps run by Carlos is forming the multiples of 3 or 3x.

When x = 1,

Alex : 4 × 1 = 4

Carlos : 3 × 1 = 3

When x = 2,

Alex : 4 × 2 = 8

Carlos : 3 × 2 = 6

When x = 3,

Alex : 4 × 3 = 12

Carlos : 3 × 3 = 9

When x = 4,

Alex : 4 × 4 = 16

Carlos : 3 × 4 = 12

Hence Alex ran 16 laps in the same time for which Carlos ran 12 laps.

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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

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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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:

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:

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:

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:2008

Step-by-step explanation:

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

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:

333 :)

Step-by-step explanation:

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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In the long run, you are expected to win $1.50 when playing the game. Therefore, the correct answer is  :

(B) Yes! In the long run, you are expected to win $1.00.

To determine whether you should play the game based on the long run expected amount you would win, we need to calculate the expected value.

The probability of winning $3 is 2/6 (numbers 1 and 2), the probability of winning $5 is 1/6 (number 3), the probability of winning nothing is 2/6 (numbers 4 and 5), and the probability of losing $2 is 1/6 (number 6).

Now let's calculate the expected value:

Expected Value = (Probability of winning $3 * $3) + (Probability of winning $5 * $5) + (Probability of winning nothing * $0) + (Probability of losing $2 * -$2)

Expected Value = (2/6 * $3) + (1/6 * $5) + (2/6 * $0) + (1/6 * -$2)

Expected Value = ($6/6) + ($5/6) + ($0) + (-$2/6)

Expected Value = $11/6 - $2/6

Expected Value = $9/6

Expected Value = $1.50

Therefore, in the long run, you are expected to win $1.50.

The correct answer is option (B) Yes! In the long run, you are expected to win $1.00.

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Answer:

Given, m = 3 and n = 4

5×3 + 4×4 – 3

15 + 16 - 3

31 - 3

28

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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].