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I need help with #13

Answer:

Steps:

tan(50.1°) = x/5

1.19.. = x/5

x = 5 × 1.19 ..

x = 5.97.. ≈ 5.98

1. False: The sum of the probabilities can never exceed one.

2. False: The alpha value indicates the maximum level of error allowed in the investigation.

3. True: The critical value is obtained from the formulas applied to each test.

4. False: H0 is rejected when the test statistic value is greater than the critical value.

5. True: All probability is between zero and one.

6. False: The ANOVA test uses only a portion of the bell curve.

7. True: There are two types of hypotheses.

8. True: The null hypothesis may posit that there is no significant difference between the statistic and the parameter or between two parameters.

9. True: When α = 0.05, it means that the level of significance is 5%, not that you allow 95% error in the study.

10. True: The critical value is the point at which the rejection zone for H0 begins.

11. False: In binomial distribution, r cannot be greater than n.

12. False: The value of H1 depends on the alternative hypothesis.

13. False: In a normal distribution, the probabilities can never exceed one.

14. True: In incompatible events, the sum of probabilities must be one.

15. True: In classical probability, the sample space is always known.

16. False: When n = 30 in a binomial distribution, Pr can never be greater than one.

17. True: All probability values are between 0 and 1.

18. True: In binomial distribution, Pr(0) is one of the probabilities.

19. True: In Poisson distribution, N is known, and its average N is used.

20. True: Incompatible events are mutually exclusive.

21. False: The sum of Pr(Oc) and Pr(Not Oc) can never exceed one.

22. True: Incompatible events can be seen in classical probability.

23. True: The variable "eye color" is a qualitative variable, nominal.

24. True: The "scatter diagram" illustrates the relationship between two variables.

25. True: In the range of grouped data, you subtract the smallest value from the largest value.

26. True: The independent variable is manipulated by the researcher.

27. False: In Poisson distribution, Pr(X > 1) = 1 - Pr(0) - Pr(1).

28. False: In binomial distribution, Pr(r > 1) = 1 - Pr(0) - Pr(1).

29. True: The sum of Pr(Oc) and Pr(Not Oc) is always one.

30. True: In normal distribution, the sum of probabilities equals one.

31. False: In binomial distribution, r cannot be greater than n.

32. True: The bar chart is used to illustrate the relationship of nominal qualitative variables.

33. True: In binomial distribution, the sum of all Pr is 1.

34. False: In the Poisson distribution, Pr(X = 31) is always less than or equal to 1.

35. False: The null and alternative hypotheses are not mutually exclusive.

36. False: The given formula is only true for independent events.

37. True: The given formula is only true for dependent events.

38. True: The correlation coefficient can be negative.

39. True: The scatter plot illustrates the dispersion of the data.

40. True: The sum of all Pr(Oc) is always one.

41. True: When Z is negative, the probability is also negative.

42. False: In normal distribution, Pr(Z ≤ 3) = Pr(Z = 3) + Pr(Z < 3).

43. False: The arithmetic average is found by adding all the data and dividing by n.

44. True: The location of the median is found by looking for (n + 1) / 2.

45. True: Median of ungrouped data does not use outliers.

46. False: In normal distribution, Z is equal to the number of standard deviations away from the mean.

47. False: When Z = 1, the probability is 0.3413, not 0.5 - 0.3413.

48. True: Variance in grouped data is divided by n - 1.

49. False: If r > 1, then the variables are not directly proportional.

50. True: Every probability is a proper fraction.

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Answer:

[tex]2\frac{5}{8} cups = 1\frac{1}{2}\ batches[/tex]

Step-by-step explanation:

Given

[tex]1\frac{3}{4} cups = 1\ batch[/tex]

Required

Number of cups for [tex]1\frac{1}{2}[/tex] batches

[tex]1\frac{3}{4} cups = 1\ batch[/tex]

Multiply both sides by [tex]1\frac{1}{2}[/tex]

[tex]1\frac{1}{2} * 1\frac{3}{4} cups = 1\ batch * 1\frac{1}{2}[/tex]

[tex]1\frac{1}{2} * 1\frac{3}{4} cups = 1\frac{1}{2}\ batches[/tex]

Express fractions as improper fractions

[tex]\frac{3}{2} * \frac{7}{4} cups = 1\frac{1}{2}\ batches[/tex]

[tex]\frac{21}{8} cups = 1\frac{1}{2}\ batches[/tex]

Express fraction as improper fraction

[tex]2\frac{5}{8} cups = 1\frac{1}{2}\ batches[/tex]

Answer: (x+4)² = -8 (y+3)

Step-by-step explanation: I got this correct on Edmentum.

Answer:

there are 8 packages left

Step-by-step explanation:

120÷12=10. So there are 10 packages of pencils to begin with. She loses 24 pencils which means that she lost two packages worth of pencils (12+12=24). So, 2 packages-10 packages=8 packages.

Answer:

The Impressionist period was distinctive because composers were focused on creating an impression through building atmospheres, pictures, and sound worlds with music. Composers Debussy and Ravel were at the forefront of this ground-breaking musical style.

Given equation is y''' + 343y = 0 which is a third-order linear differential equation. To find the general solution of the equation, we can use the characteristic equation of the differential equation as follows;

Let y = erx be the trial solution of the differential equation, where e is the exponential function and r is an unknown constant to be determined. Substituting the trial solution into the differential equation, we have; y''' + 343y = 0 y' = rerx, y'' = rerx, y''' = rerx. Substituting into the differential equation, we have;r³erx + 343erx = 0. Factorizing out erx, we have erx(r³ + 343) = 0For erx ≠ 0;r³ + 343 = 0r³ = -343r = (-343)¹/³ = -7 (r = -7, since we are working with real forms of the equation). Therefore, the general solution of the third order linear differential equation is;

y = c₁e^-7x + c₂e^-7x cos(12.124x) + c₃e^-7x sin(12.124x) where c₁, c₂, and c₃ are arbitrary constants.

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The time complexity of the binary search is c) O(NLogN)

An effective searching technique that uses sorted arrays is a binary search. The search space is continually divided in half until the target element is located or the search space is empty, using a divide-and-conquer strategy. It makes the necessary adjustments to the search bounds at each stage by comparing the target element to the centre element of the active search area.

The divide-and-conquer strategy is logarithmic, hence binary search has an O(log n) time complexity. This implies that number of operations needed to discover the target element rises at a logarithmic rate as input size increases. When compared to linear search algorithms with O(n) complexity, where each element must be evaluated sequentially, binary search is significantly more efficient for big arrays due to its logarithmic time complexity.

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Answer:

answer is 24

Step-by-step explanation:

The side length of the rectangular prism with a square base is 13.2 cm if the original container was a cylinder with the height of 18 cm and a diameter of 14 cm.

In geometry, it is defined as the three-dimensional shape having two circular shapes at a distance called the height of the cylinder.

We know the volume of the cylinder is given by:

[tex]\rm V = \pi r^2 h[/tex]

The radius of the cylinder r = 14/2 = 7 cm

The volume of the cylindrical container V = π(7)²(18) = 882π cubic cm

Let's suppose the side length is L:

Then the volume of the rectangular prism is the same as the cylindrical container:

882π = L×L×16       (h = 16 cm)

882π = 16L²  

L = 13.15 ≈ 13.2 cm

Thus, the side length of the rectangular prism with a square base is 13.2 cm if the original container was a cylinder with the height of 18 cm and a diameter of 14 cm.

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Answer:

15 yards amd 3 feet.

Step-by-step explanation:

I wish to get extra three points :)

Answer:

This is 15 yards and 3 feets.

Answer:

-6x + 56

Step-by-step explanation:

-2(3x) + 56

-6x + 56

The critical t-value that corresponds to a 95% confidence level with 15 degrees of freedom is approximately 2.131.

To find the critical t-value that corresponds to a 95% confidence level with 15 degrees of freedom, we can use a t-table or a statistical software.

Using a t-table or a statistical software, we look up the critical t-value for a two-tailed test with a confidence level of 95% and 15 degrees of freedom.

For a two-tailed test, we divide the desired confidence level by 2 to account for both tails of the t-distribution. In this case, we divide 95% by 2, which gives us 0.475.

Looking up the critical t-value for a confidence level of 0.475 and 15 degrees of freedom, we find that the critical t-value is approximately 2.131.

Therefore, the critical t-value that corresponds to a 95% confidence level with 15 degrees of freedom is approximately 2.131.

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The system x = tx is Lyapunov stable.

To determine if the system x = tx is Lyapunov stable, we need to check if it satisfies the Lyapunov stability criterion. The criterion states that for every ε > 0, there exists a δ > 0 such that if ||x(0) - xe|| < δ, then ||x(t) - xe|| < ε for all t > 0.

Let's analyze the given system x = tx:

Taking the derivative of x with respect to t, we get:

dx/dt = x

This is a first-order linear ordinary differential equation.

The general solution to this equation is:

x(t) = C×[tex]e^{t}[/tex]

where C is the constant of integration.

Now, let's consider the initial condition x(0) = x0, where x0 is some constant.

Using this initial condition, we can solve for the constant C:

x(0) = C × e⁰

x0 = C

So, the specific solution to the initial value problem is:

x(t) = x₀ × [tex]e^{t}[/tex]

Now, we can check if the system is Lyapunov stable.

For any ε > 0, let's consider the case where δ = ε. Then, if ||x(0) - xe|| < δ, we have:

||x(0) - xe|| = ||x₀ - x₀ × [tex]e^{t}[/tex]|| = ||x₀(1 - [tex]e^{t}[/tex])||

Since e^t > 1 for t > 0, we can rewrite the expression as

||x₀(1 - [tex]e^{t}[/tex])|| = x₀ × ||[tex]e^{t}[/tex] - 1||

Now, we want to show that ||x(t) - xe|| < ε for all t > 0. We have

||x(t) - xe|| = ||x₀ × [tex]e^{t}[/tex] - x₀ × [tex]e^{t}[/tex]|| = 0

Since ||x(t) - xe|| = 0 for all t > 0, it satisfies the condition ||x(t) - xe|| < ε.

Therefore, the given system x = tx is Lyapunov stable.

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The question is incomplete the complete question is :

Determine whether the origin of the following system is Lyapunov stable: x= tx. Use Lyapunov's stability criterion .

For the given equation x⁸ + y⁸ + z⁵ = 6xyz, the value of partial derivatives are ∂z/∂x = (6yz - 8x⁷) / (5z⁴ - 6xz) and ∂z/∂y = (6xz) / (8y⁷ - 6xy)

To find the partial derivatives ∂z/∂x and ∂z/∂y for the equation x⁸ + y⁸ + z⁵ = 6xyz, we need to differentiate the equation with respect to x and y while treating y and z as constants.

Taking the partial derivative with respect to x (∂z/∂x), we differentiate each term separately:

8x⁷ + 0 + 5z⁴ (∂z/∂x) = 6yz + 6xz(∂z/∂x)

Simplifying the equation, we get:

8x⁷ + 5z⁴ (∂z/∂x) = 6yz + 6xz(∂z/∂x)

Now, let's take the partial derivative with respect to y (∂z/∂y):

0 + 8y⁷ + 0 (∂z/∂y) = 6xz + 6xy(∂z/∂y)

Simplifying further:

8y⁷ (∂z/∂y) = 6xz + 6xy(∂z/∂y)

To find the values of ∂z/∂x and ∂z/∂y, we need to isolate the partial derivatives:

∂z/∂x = (6yz - 8x⁷) / (5z⁴ - 6xz)

∂z/∂y = (6xz) / (8y⁷ - 6xy)

These equations give the partial derivatives of z with respect to x and y for the given equation x⁸ + y⁸ + z⁵ = 6xyz.

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Answer:

y = 2/3 x - 3

Step-by-step explanation:

substitute the given slope and the coordinate into the slope-intercpet form:

y = mx + b  to solve for b,

-1 = 2/3(3) + b

=> b = -3

y = 2/3 x - 3

Answer:

Plot attached

Step-by-step explanation:

We want to locate √36 and ∛36 on the given number line

Now, √36 = 6.

And ∛36 = 3.3

I have attached a plot number line both values as we are told in the question

Answer:

1.2

Step-by-step explanation:

You have to subtract 7.2 - 6 = 1.2

East and south direction motion are perpendicular to each other. The hiker went approx 4 miles in south.

East or west are perpendicular (forming 90° ) to north or south directions.

North is 180° to south, and east is 180 degrees to west.

If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:

[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]

where |AB| = length of line segment AB.

Considering the diagram attached below, we deduce that the triangle ABC is right angled triangles, and thus, its sides' lengths will follow Pythagoras theorem.

The movement the hiker did in the south direction is the length of the side BC, denoted by |BC| = x miles (assume)

Then, by applying Pythagoras theorem, we get:

[tex]|AC|^2 = |AB|^2 + |BC|^2\\(7.2)^2 = 6^2 + x^2\\51.84 = 36 + x^2\\\\\text{Subtracting 36 from both the sides and taking root, we get}\\\\x = \sqrt{15.86} \approx 3.98 \approx 4 \: \rm miles[/tex]

(took positive root since distance is a non-negative quantity)

Thus, the hiker went approx 4 miles in south.

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Answer:

in terms of degree it is linear equation

in terms of number of terms it is binomial

[tex]y = (e^{(-2t)} + C . e^{(-t)})^{(-1)}[/tex].  is the explicit solution to the given Bernoulli differential equation. we used an appropriate substitution to transform it into a linear differential equation.

The given Bernoulli differential equation is in the form [tex]dy/dt - y = e^{(-2t)} . y^2[/tex]. Bernoulli equations can be transformed into linear equations by using a substitution. In this case, we can let [tex]v = y^{(-1)[/tex]be our substitution.

Differentiating v with respect to t gives dv/dt = -y^(-2) * dy/dt. Substituting this into the original equation, we have:

[tex]-dv/dt - v = e^{(-2t)}.[/tex]

Now, we have transformed the Bernoulli equation into a linear first-order differential equation. We can solve this equation using standard methods. Multiplying through by -1, we get [tex]dv/dt + v = -e^{(-2t)}[/tex].

The integrating factor for this linear equation is given by [tex]e^{ \[ \int_{}^{}1\,dt \] = e^t[/tex]. Multiplying both sides of the equation by e^t, we have:

[tex]e^t. (dv/dt + v) = -e^{(t-2t)[/tex].

Simplifying further, we get [tex]d/dt (e^t.v) = -e^{(-t)[/tex].

Integrating both sides with respect to t, we obtain:

[tex]e^t . v = \[ \int_{}^{} -e^{(-t)} \,dt \][/tex].

Evaluating the integral, we get:

[tex]e^t . v = e^{(-t) }+ C[/tex],

where C is the constant of integration.

Finally, solving for v, we have:

[tex]v = e^{(-2t)} + C .e^{(-t)[/tex].

Since [tex]v = y^{(-1)[/tex], we can take the reciprocal of both sides to find the explicit solution for y:

[tex]y = (e^{(-2t) }+ C .e^{(-t)})^{(-1)}[/tex].

This is the explicit solution to the given Bernoulli differential equation.

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The function v = x + 1, the norm is v = sqrt(integral of (x+1)^2 dx from 0 to 1), which evaluates to sqrt(5/3). The angle between u and v is arccos(1/√5).

a) The norm f of a function f in the vector space C[0, 1] with the given inner product is defined as the square root of the inner product of the function with itself. In this case, f = sqrt((f, f)) = sqrt(integral of f(x) * f(x) dx over the interval [0, 1]). For the function u = x, the norm is u = sqrt(integral of x^2 dx from 0 to 1), which evaluates to sqrt(1/3). For the function v = x + 1, the norm is v = sqrt(integral of (x+1)^2 dx from 0 to 1), which evaluates to sqrt(5/3).

b) The inner product space C[0, 1] induces a norm on the set of functions, and we can define the distance between two functions as the norm of their difference. In this case, the norm of the difference between the functions u and v is |u - v| = sqrt((u - v, u - v)) = sqrt(integral of (x - (x+1))^2 dx from 0 to 1), which simplifies to sqrt(1/3). Therefore, |u - v| = sqrt(1/3).

c) The inner product (f, g) between the functions f and g is defined as the integral of their pointwise product over the interval [0, 1]. For the functions u = x and v = x + 1, (u, v) = integral of (x * (x + 1)) dx from 0 to 1, which evaluates to 7/6.

d) The angle between two functions u and v in an inner product space can be computed using the definition of the inner product and the norms of the functions. The angle theta between u and v is given by the equation cos(theta) = (u, v) / (u * v). In this case, cos(theta) = (7/6) / (sqrt(1/3) * sqrt(5/3)), which simplifies to 1/√5. Therefore, the angle between u and v is arccos(1/√5).

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If Harriet Hanson Robinson (Larcom) had started her career in the mill as a machine worker instead of as a doffer, her initial impression of the mill may have been significantly different.

Here are a few ways her perspective might have been altered:

Physical Experience: As a machine worker, Larcom would have been directly involved in operating the machinery. She would have experienced the physical demands and potentially dangerous conditions associated with working with heavy machinery. This hands-on experience might have given her a greater appreciation for the risks and challenges faced by the workers. Social Interaction: Machine workers often worked in close proximity to each other, tending to the machines and coordinating their tasks. By working alongside her fellow machine workers, Larcom would have had more direct contact and interaction with her peers. This could have provided her with a deeper understanding of the camaraderie, unity, and challenges faced by the workers as a community.

Skill Development: Operating the machines required a certain level of technical skill and knowledge. If Larcom had started as a machine worker, she would have gained expertise in machine operation and maintenance. This technical knowledge could have given her a different perspective on the machinery, its intricacies, and its impact on the workers. Perspective on Management: As a machine worker, Larcom might have had more direct interactions with mill managers and overseers. This could have given her insights into the management practices, decision-making processes, and power dynamics within the mill. Such firsthand experiences might have influenced her initial impression of the mill's management and their treatment of workers.

Overall, starting as a machine worker would have provided Larcom with a different vantage point and firsthand experience of the mill's operations. This could have shaped her understanding of the work environment, the challenges faced by the workers, and the dynamics between management and labor.

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Answer:

(2 - a) (x² - y)

Step-by-step explanation:

2x² + ay - ax² - 2y

=> 2x² - 2y + ay - ax²

=> 2 (x² - y) + -a (-y + x²)

=> 2 (x² - y) + -a (x² - y)

=> (2 - a) (x² - y)

Answer:

[tex] x=\sqrt{10} [/tex]

[tex] y = \sqrt {14} [/tex]

[tex] z = \sqrt {35} [/tex]

Step-by-step explanation:

By geometric mean theorem:

[tex] x=\sqrt{2\times 5} [/tex]

[tex] x=\sqrt{10} [/tex]

By Pythagoras Theorem:

[tex] y^2 = x^2 + 2^2 [/tex]

[tex] y^2 = (\sqrt{10}) ^2 + 2^2 [/tex]

[tex] y^2 = 10 + 4 [/tex]

[tex] y^2 = 14 [/tex]

[tex] y = \sqrt {14} [/tex]

Again by Pythagoras Theorem:

[tex] z^2 = x^2 + 5^2 [/tex]

[tex] z^2 = (\sqrt{10}) ^2 + 5^2 [/tex]

[tex] z^2 = 10 + 25 [/tex]

[tex] z^2 = 35 [/tex]

[tex] z = \sqrt {35} [/tex]

Answer:

18

Step-by-step explanation:

Formula

P = 2 * (L + W)

Solution

L = 6

W = 3

P = 2 (6 + 3)

P = 2 (9)

P = 18

the approximate values of y at x = X₀ + 0.1 and x = X₀ + 0.2 are:

y ≈ 0.9282 at x = 0.1

y ≈ 0.8281 at x = 0.2

To solve the given differential equation, we will use an appropriate method, such as the Euler's method, to approximate the values of y at specific points.

The Euler's method uses the equation:

y₍ₙ₊₁₎ = yₙ + h * f(xₙ, yₙ),

where:

yₙ is the value of y at xₙ

h is the step size,

f(x, y) is the derivative of y with respect to x (i.e., dy/dx), and

y₍ₙ₊₁₎ is the approximation of y at the next point x₍ₙ₊₁₎ = xₙ + h.

Let's apply Euler's method to find the values of y at x = X₀ + 0.1 and x = X₀ + 0.2, where X₀ is the initial condition x = 0 and y(X₀) = 1.

Given the differential equation:

dy/dx + xy = 3x² - [tex]e^y[/tex] + 2

Rewriting the equation in the form:

dy/dx = -xy + 3x² - [tex]e^y[/tex] + 2

We have the initial condition:

y(0) = 1

Using Euler's method with a step size of h = 0.1:

1. At x = X₀ + 0.1:

  y₁ = y₀ + h * [ -x₀ * y₀ + 3 * x₀² - [tex]e^y_0[/tex] + 2 ]

      = 1 + 0.1 * [ -(0) * (1) + 3 * (0)² - e¹ + 2 ]

      = 1 + 0.1 * (0 - e + 2)

      = 1 + 0.1 * (-e + 2)

      = 1 + 0.1 * (-2.7183 + 2)

      = 1 + 0.1 * (-0.7183)

      = 1 - 0.07183

      ≈ 0.9282

2. At x = X₀ + 0.2:

  y₂ = y₁ + h * [ -x₁ * y₁ + 3 * x₁² - [tex]e^y_1[/tex] + 2 ]

      = 0.9282 + 0.1 * [ -(0.1) * (0.9282) + 3 * (0.1)₂ - [tex]e^{0.9282}[/tex] + 2 ]

      = 0.9282 + 0.1 * (-0.009282 + 0.003 - [tex]e^{0.9282}[/tex] + 2)

      ≈ 0.8281

Therefore, the approximate values of y at x = X₀ + 0.1 and x = X₀ + 0.2 are:

y ≈ 0.9282 (correct to four decimal places) at x = 0.1

y ≈ 0.8281 (correct to four decimal places) at x = 0.2

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(a) The sample size needed is 1,498 if the researcher uses the prior proportion of 0.669.

n = 1.497.87 or 1,498

(b)  The researcher should use a sample size of 1,692 if the proportion is unknown.

n = 1,691.06 or 1,692

When a researcher is designing a sampling study to measure a population parameter, like the proportion, the sample size must be determined. The key to calculating sample size is the margin of error the researcher desires. The smaller the margin of error, the larger the sample size needed.

(a) The sample size needed is 1,498 if the researcher uses the prior proportion of 0.669.

To find sample size for the proportion use the equation:

[tex]n=\frac{Z^2\pi (1-\pi )}{e^2}[/tex]

Where:

Substituting values:

[tex]n=\frac{1.6449^2*0.50*0.331}{0.02^2}[/tex]

n = 1.497.87 or 1,498

(b) The researcher should use a sample size of 1,692 if the proportion is unknown.

Use the same equation as in part (a). However, to maximize [tex]\pi (1-\pi )[/tex] use 0.5 for [tex]\pi[/tex].

Substituting the values:

[tex]n=\frac{1.6449^2*0.50*0.500}{0.02^2}[/tex]

n = 1,691.06 or 1,692

Always round up because it is impossible to have a fraction of an observation.

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Answer: A. 3560 chickens

Step-by-step explanation:

Plugged in the 6 days into the equation provided

Birds will be infected with the Flu after 6 days will be 3560 chickens

An exponential function is a mathematical function of the following form: f ( x ) = a^x. where x is a variable, and a is a constant called the base of the function.

Given function: y = 6000(1 - [tex]e^{-0.15t}[/tex])

Total birds are =6000

Now, take t= 6 days.

So, y = 6000(1 - [tex]e^{-0.15t}[/tex])

y= 6000(1- [tex]e^{-0.15 * 6}[/tex])

 = 6000 (1-0.40656)

 = 6000*0.59343

 = 3560.58204

 = 3560 birds

Hence, 3560 birds get infected in 6 days.

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