divide 3240 into two parts, such that one is 28 percent more than the other​

1. The sample space can be represented as {3, 4, 5, ..., 17, 18}. 2. The sample space can be represented as {(A, B, C), (A, C, B), (B, A, C), (B, C, A), (C, A, B), (C, B, A)}, 3. The sample space can be represented as {(A, B), (A, C), (A, D), (A, E), (B, C), (B, D), (B, E), (C, D), (C, E), (D, E)}

In probability theory, the sample space is the set of all possible outcomes of an experiment. It is denoted by the symbol "Ω" and can be written in set notation.

1. For the first experiment, where a six-sided die is tossed three times and the sum is recorded, the sample space can be represented as follows: Ω = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18}

This is because the minimum sum possible is 3 (by rolling three 1's) and the maximum sum possible is 18 (by rolling three 6's), and all sums in between are possible outcomes.

2. For the second experiment, where a taste tester ranks three varieties of yogurt, A, B, and C, according to preference, the sample space can be represented as follows: Ω = {(A,B,C), (A,C,B), (B,A,C), (B,C,A), (C,A,B), (C,B,A)}

This is because there are six possible orders in which the three varieties of yogurt can be ranked.

3. For the third experiment, where two county supervisors are selected from five supervisors, A, B, C, D, and E, to study a recycling plan, the sample space can be represented as follows: Ω = {(A,B), (A,C), (A,D), (A,E), (B,C), (B,D), (B,E), (C,D), (C,E), (D,E)}

This is because there are 10 possible pairs of supervisors that can be selected from the five available.

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Let X and Y be jointly continuous random variables with joint density function f(x,y)=c(y2−x2)e−2y,−y≤x≤y,0<y<∞. c = 1/4 so that f is a density function.

Throughout the complete range of potential X and Y values, the joint density function must integrate to 1.

The interval [-y, y] for  0 < y < ∞ defines the range of potential values for X and Y. The joint density function integral over this region must therefore be equal to 1.

Integrating the given density function, we have:

∫∫c(y2−x2)e−2y dxdy

= c ∫∫(y2−x2)e−2y dxdy

= c ∫y2e−2y dy − c ∫∫x2e−2y dxdy

= c [−2ye−2y − (−2/2)e−2y]|y=0 to ∞

= c [2 − (−1)]

= c(3)

Therefore, c = 1/4 so that the joint density function integrates to 1.

Complete Question:

Let X and Y be jointly continuous random variables with joint density function f(x,y)=c(y2−x2)e−2y,−y≤x≤y,0<y<∞. Find c so that f is a density function.

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The formula that can be used to describe the sequence is f(x + 1) = Negative three-fourths f(x).

The sequence can be described by the formula f(x + 1) = negative three-fourths f(x), because each term is obtained by multiplying the previous term by negative three-fourths.

To see this, we can start with the first term 64 and apply the formula to find the second term:

f(1) = -3/4 * 64 = -48

Then we can use the same formula to find the third term:

f(2) = -3/4 * (-48) = 36

Continuing in this way, we can use the formula to find each subsequent term in the sequence.

It's worth noting that the sequence is an example of a geometric sequence, where each term is obtained by multiplying the previous term by a constant ratio. In this case, the ratio is negative three-fourths, which means that the sequence alternates between positive and negative values.

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For Eric to get a B in the class, he must score at least 82.5 on the final exam.

Let x be the average score for the final exam. The final exam counts as his two tests, so Eric has a total of five test results.

He has 3 grades before and he has 2 grades for the final exam.

To get a B in the class he must have an average score of 80 or above.

(70 + 82 + 83 + 2x)/5 ≥ 80

Multiplying both sides by 5 gives:

70 + 82 + 83 + 2x ≥ 400

Simplified, it looks like this:

2x ≥ 165

x≧82.5

Therefore, for Eric to get a B in the class, he must score at least 82.5 on the final exam. 

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In order to understand the relationships among the complexity classes of these functions, we can categorize them based on their growth rates. Here's a categorization of the given functions:

1. Constant time complexity: 10, 1
2. Logarithmic time complexity: logn, ln n, log(log n)
3. Polynomial time complexity:
  - Linear time complexity: nlogn, n1/k (k > 3)
  - Quadratic time complexity: n2
  - Cubic time complexity: n3
  - Sublinear time complexity: n1/2, n1/3, n1/2 logn, n1/3 logn
4. Exponential time complexity: 2n, 3n
5. Higher-degree polynomial time complexity: nk (k > 3)
6. Super-exponential time complexity: nn
7. Factorial time complexity: n!

These categories are arranged in increasing order of complexity, meaning that functions in higher categories have higher growth rates and belong to more complex complexity classes than those in lower categories.

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Answer:

x = 15

Step-by-step explanation:

The sum of interior angles in a triangle is equal to 180°.

m∠J + m∠K + m∠L = 180

8x - 7 + x + 7 + 2x + 15 = 180

11x + 15 = 180

11x = 165

x = 15

When a single card is randomly drawn from a standard deck of 52 playing cards, the odds in favor of the card being a 3, 4, 5, 6, or 7 can be calculated by considering the total number of these cards and the total number of cards in the deck.

In a standard deck, there are four suits (hearts, diamonds, clubs, and spades), each containing one card of each rank (from Ace to King). Therefore, there is one 3, 4, 5, 6, and 7 in each suit, which results in a total of 4 cards for each rank (3s, 4s, 5s, 6s, and 7s). As there are five ranks in question, there are 4 x 5 = 20 cards that are either a 3, 4, 5, 6, or 7.

To find the odds in favor of drawing one of these cards, we'll divide the number of favorable outcomes (20) by the total number of possible outcomes (52). So, the odds in favor are:

20 (favorable outcomes) / 52 (total outcomes) = 5/13

Therefore, the odds in favor of drawing a 3, 4, 5, 6, or 7 from a standard deck of cards are 5/13.

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Step-by-step explanation:

There are 500 little squares covered....each square is 2 mice

 there are 60 squares that are less than 10 gm

   so that would be 120 mice <10 gm

Answer 1:

This is a compound interest model.

Model:

Value after 28 years = $2600*(1+0.092)^{28} = $22,022.64 (rounded to the nearest cent).

After how many years will the value be $9,734?

$9,734/$2600 = (1+0.092)^t, where t is the number of years.

Solving for t, we get t = 10.14 years (rounded to the nearest hundredth).

Answer 2:

This is an exponential growth model.

Model:

Let P(t) be the population at time t in years.

Then P(t) = 45,000,000*(1+0.057)^t.

To find when the population reaches 100,000,000, we need to solve the equation:

100,000,000 = 45,000,000*(1+0.057)^t

Solving for t, we get t = 21.35 years (rounded to the nearest hundredth).

How many years will it take the population to get to 103,000,000?

103,000,000 = 45,000,000*(1+0.057)^t

Solving for t, we get t = 22.12 years (rounded to the nearest hundredth).

How many years does it take to get to 161,000,000?

161,000,000 = 45,000,000*(1+0.057)^t

Solving for t, we get t = 34.86 years (rounded to the nearest hundredth).

Answer 3:

This is an exponential decay model.

Model:

Let P(t) be the population at time t in years.

Then P(t) = 168,000*(1-0.0347)^t.

To find when the population drops below 508,000, we need to solve the equation:

508,000 = 168,000*(1-0.0347)^t

Solving for t, we get t = 40.36 years (rounded to the nearest hundredth).

Population after 34 years:

P(34) = 168,000*(1-0.0347)^{34} = 76,447.23 (rounded to the nearest hundredth).

Answer 4:

This is an exponential decay model.

Model:

Let A(t) be the amount of substance at time t in years.

Then A(t) = 80*(1-0.03)^t.

Value after 16 years:

A(16) = 80*(1-0.03)^{16} = 37.09 (rounded to the nearest hundredth).

After how many years will the value be 29.28?

29.28 = 80*(1-0.03)^t

Solving for t, we get t = 24.32 years (rounded to the nearest hundredth).

(a) The subspace of R₃ spanned by the two vectors (1, 1, −1) and (−1, −1, 1) is a plane.

(b) The subspace of R₃ spanned by the three vectors (0, 1, 1), (1, 1, 0), and (0, 0, 0) is a plane.

(c) The subspace spanned by the columns of a 3 by 5 echelon matrix with 2 pivots is a plane in R₃.

(d) The subspace of R₃ consisting of all vectors with positive components is an octant.

(a) The subspace of R₃ spanned by the two vectors (1, 1, −1) and (−1, −1, 1) is a plane. To see this, note that any linear combination of the two vectors is of the form c₁(1, 1, −1) + c₂(−1, −1, 1) = (c₁ − c₂, c₁ − c₂, −c₁ + c₂), where c₁ and c₂ are scalars. Therefore, the subspace is the set of all linear combinations of (1, 1, −1) and (−1, −1, 1) and this set is a plane since any two non-parallel vectors in R3 span a plane.

(b) The subspace of R₃ spanned by the three vectors (0, 1, 1), (1, 1, 0), and (0, 0, 0) is a plane. Note that any linear combination of the three vectors is of the form c₁(0, 1, 1) + c₂(1, 1, 0) + c₃(0, 0, 0) = (c₂, c₁ + c₂, c₁), where c₁, c₂, and c₃ are scalars. Since the third component is always equal to the first component, this subspace lies in a plane parallel to the xz-plane.

(c) Since the echelon matrix has 2 pivots, it has 2 linearly independent columns, and the subspace spanned by these columns is a plane in R₃. Note that the span of the columns of an echelon matrix is always equal to the span of the rows of the matrix.

(d) The subspace of R₃ consisting of all vectors with positive components is an octant. An octant is a region of space that is defined by the positive and negative values of its three coordinates, and in this case, we only allow positive values. Specifically, the octant is given by the set of all vectors (x, y, z) such that x > 0, y > 0, and z > 0.

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The total surface area of the triangular pyramid is: 381 yd²

To find the surface area of the pyramid, we have to find all the surface area of the surfaces and add them together.

Area of base = 11 * 9

Area = 99 yd²

Area of 2 big triangles is:

A = 2(¹/₂ * 11 * 15)

A = 165 yd²

Area of two small triangles is:

A = 2(¹/₂ * 9 * 13)

A = 117 yd²

Thus:

Total surface area = 99 + 165 + 117

Total surface area = 381 yd²

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The correct answer is A) 3.14 is the approximate ratio of the circumference to the diameter of a circle, rounded to the nearest hundredth.

This is due to the fact that the circle's circumference to diameter ratio, often known as pi, is roughly equal to 3.14.

A circle's circumference divided by its diameter is known as pi, and it is a mathematical constant.

Thus, the ratio of circumference to diameter will always be equal to pi, regardless of the size of the circle.

Hence, the circumference to diameter ratio of a circle, rounded to the closest hundredth, is approximately 3.14.

Complete Question:

Liza measures the circumferences and diameters of different circles and records them in a table. What is the approximate ratio of the circumference to the diameter of a circle, rounded to the nearest hundredth?

Options:

A) 3.14

B) 6.28

C) 12.56

D) None of the above

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The lowest price at which this firm would operate in the short run is $7 (option c).

To determine this, we can use the concept of the shutdown point. In the short run, if the price falls below the shutdown point, it would be more profitable for the firm to shut down production temporarily rather than continue operating and incurring losses.

The shutdown point occurs where price is equal to average variable cost (AVC). From the table, we can calculate the AVC for each quantity using the formula AVC = TC/Q.

For example, at Q = 3, AVC = $16.33. This means that if the price falls below $16.33, the firm should shut down.

The lowest price at which the firm would operate is where AVC is equal to the lowest cost in the table. In this case, the lowest cost is $14 at Q = 1. However, we need to check that this is not below the shutdown point.

At Q = 1, AVC = $41. Since $14 is above $41, it is safe to conclude that the firm would continue operating at a price of $7 or higher.

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The appropriate CPT code for the transcatheter retrieval, percutaneous, of intravascular foreign body, including radiological supervision and interpretation and imaging guidance is 37187.

This code is used to report the percutaneous retrieval of an intravascular foreign body using transcatheter techniques, including radiological supervision and interpretation as well as imaging guidance.

The procedure involves the use of specialized tools and catheters that are inserted through a small incision in the skin and guided to the location of the foreign body using fluoroscopy or other imaging modalities. Once the foreign body is located, it can be removed using a variety of techniques, including suction or grasping devices.

It is important to note that CPT codes are constantly updated, and the correct code may vary based on the specific details of the procedure and the individual circumstances of the patient. It is essential to consult the latest coding resources and guidelines and to document the procedure thoroughly to ensure accurate coding and billing.

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The weight of the dog is calculated by multiplying its mass of 40.0 kg with the acceleration due to gravity of 9.80 m/s^2, resulting in a weight of 392 N.

The weight of an object is the force exerted on it due to gravity. The formula to calculate weight is

weight = mass x acceleration due to gravity

In this case, the mass of the dog is 40.0 kg, and the acceleration due to gravity on Earth is 9.80 m/s^2. Plugging these values into the formula, we get

weight = 40.0 kg x 9.80 m/s^2

weight = 392 N (Newtons)

Therefore, the weight of the dog is 392 Newtons.

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Answer:

The slope is -2/3, so B is correct.

Step-by-step explanation:

Start at (0, 1). Go down 2 units, then right 3 units. You will be at (3, -1). So the slope of this line is -2/3. B is the correct answer.

where [tex]$0\leq u\leq 2\pi$[/tex] and [tex]$0\leq v\leq 1$[/tex].

We have the equation of the paraboloid as [tex]$\$ 64 z=4 x^{\wedge} 2+16 y^{\wedge} 2 \$$[/tex], or equivalently[tex], $\$ z=\backslash$ frac $\{1\}$ $\{16\}\left(x^{\wedge} 2+4 y^{\wedge} 2\right) \$$[/tex]

To parameterize the surface, we can use the following parameterization:

[tex]$\mathbf{r}(u, v)=4 v \cos u \mathbf{i}+2 v \sin u \mathbf{j}+\frac{1}{16}\left(4 v^2 \cos ^2 u+16 v^2 \sin ^2 u\right) \mathbf{k}$[/tex]

where [tex]$\$[/tex] 0 [tex]$\backslash$[/tex] leq u [tex]$\backslash$[/tex] leq [tex]$2 \backslash$[/tex] pi [tex]$\$$[/tex] and [tex]$\$ 0 \backslash$[/tex] leq [tex]$\backslash \backslash$[/tex] leq [tex]$1 \$$[/tex].

To verify that this parameterization indeed gives us the desired surface, we substitute [tex]$\$ x=4 v \backslash \cos u \$$[/tex] and [tex]$\$ y=2 v \backslash \sin u \$$[/tex] into the equation for [tex]$\$ z \$$[/tex] :

[tex]$$z=\frac{1}{16}\left(4 v^2 \cos ^2 u+4 v^2 \sin ^2 u\right)=\frac{1}{4} v^2$$[/tex]

Thus, the parameterization of the paraboloid is:

[tex]$$\mathbf{r}(u, v)=4 v \cos u \mathbf{i}+2 v \sin u \mathbf{j}+\frac{1}{4} v^2 \mathbf{k}$$[/tex]

where [tex]$0\leq u\leq 2\pi$[/tex] and [tex]$0\leq v\leq 1$[/tex].

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Using percentages we know that blue is 18.52% of the total percentage.

A percentage is a figure or ratio stated as a fraction of 100 in mathematics.  

A % is a quantity or ratio that, in mathematics, represents a portion of one hundred.

A dimensionless relationship between two numbers can be represented in a variety of ways, such as through ratios, fractions, and decimals.

The symbol "%" is frequently written after the number to indicate percentages.

So, find the percentage of blue as follows:

= 105 + 220 + 50 + 75 = 405

The blue percentage would be:

= 75/405 * 100

= 18.52

Therefore, using percentages we know that blue is 18.52% of the total percentage.

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a) The measure of the angle in radians is approximately 1.92 radians.

b) The formula for converting degree measure to radian measure is radians = degrees x π / 180.

a. To convert degrees to radians, we use the formula:

radians = degrees x π / 180

where π (pi) is a mathematical constant that represents the ratio of the circumference of a circle to its diameter (approximately equal to 3.14159).

Substituting the given value, we get:

radians = 110 x π / 180 ≈ 1.92

b. The formula to convert degree measure to radian measure is given by:

radians = degrees x π / 180

This formula is derived from the fact that a full rotation of an angle measures 360 degrees or 2π radians. Therefore, we can set up a proportion to relate degrees and radians:

degrees / 360 = radians / 2π

Solving for radians, we get:

radians = degrees x 2π / 360

Simplifying this expression, we get:

radians = degrees x π / 180

So, the formula for converting degree measure to radian measure is simply the ratio of the angle in degrees to 180, multiplied by π.

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Here are the probabilities you're looking for:

a) p(-.67 ≤ z ≤ -.23): This is the probability that the z-score falls between -0.67 and -0.23. You can find this by using a standard normal table or calculator to look up the values and subtracting the smaller value from the larger one.

b) p(0 ≤ z ≤ 1.96): This is the probability that the z-score falls between 0 and 1.96. Again, use a standard normal table or calculator to find the values and subtract accordingly.

c) p(-1.28 ≤ z ≤ 0): This is the probability that the z-score falls between -1.28 and 0. To find this, look up the values on a standard normal table or calculator and subtract as before.

d) p(z > 4.2): This is the probability that the z-score is greater than 4.2. To find this, look up the value for 4.2 on a standard normal table or calculator and subtract it from 1, as the total probability under the curve is equal to 1.

Remember to use a standard normal table or calculator to obtain the exact probabilities for each scenario.

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A)   The t-value for this case is 2.179.B) The t-value for this case is -1.676.C)  The t-value for this case is 2.750.D) The t-values for this case are -1.708 and 1.708. E) The t-values for this case are -2.021 and 2.021.

To find the t-values for each of the given cases, we can use a t-distribution table or a calculator. Here are the solutions to each case:

(a) Upper tail area of 0.025 with 12 degrees of freedom:

Using a t-distribution table, we can find the t-value for the upper tail area of 0.025 with 12 degrees of freedom. The value is 2.179. Therefore, the t-value for this case is 2.179.

(b) Lower tail area of 0.05 with 50 degrees of freedom:

Using a t-distribution table or a calculator, we can find the t-value for the lower tail area of 0.05 with 50 degrees of freedom. The value is -1.676. Therefore, the t-value for this case is -1.676.

(c) Upper tail area of 0.01 with 30 degrees of freedom:

Using a t-distribution table or a calculator, we can find the t-value for the upper tail area of 0.01 with 30 degrees of freedom. The value is 2.750. Therefore, the t-value for this case is 2.750.

(d) Where 90% of the area falls between these two t values with 25 degrees of freedom:

We can use a t-distribution table or a calculator to find the t-values that correspond to the 5th and 95th percentiles of the t-distribution with 25 degrees of freedom. The values are -1.708 and 1.708, respectively. Therefore, the t-values for this case are -1.708 and 1.708.

(e) Where 95% of the area falls between these two t values with 45 degrees of freedom:

We can use a t-distribution table or a calculator to find the t-values that correspond to the 2.5th and 97.5th percentiles of the t-distribution with 45 degrees of freedom. The values are -2.021 and 2.021, respectively. Therefore, the t-values for this case are -2.021 and 2.021.

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Answer:24

Step-by-step explanation:

LOOK I KNOW IT IS THE ANSWER TRUST

The given statement "Compared to a z-score, a hypothesis test with a t statistic requires more information from the sample." is True because hypothesis test need more information than z-score.

A z-score is used when the population standard deviation is known, and the sample size is large. In contrast, a t-statistic is used when the population standard deviation is unknown, and the sample size is small.

When conducting a hypothesis test with a t-statistic, more information from the sample is required to estimate the population standard deviation. This is because the sample standard deviation is used to estimate the population standard deviation, and it is not always an accurate estimate, especially when the sample size is small.

In addition, the use of a t-distribution introduces more uncertainty into the test, as the distribution is wider and flatter than the normal distribution used for z-scores.

Therefore, when conducting a hypothesis test with a t-statistic, more information from the sample is required to account for the increased uncertainty introduced by the smaller sample size and the estimation of the population standard deviation.

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Answer:

So, after 5 years, the desk would be worth approximately $1,003.65.

Step-by-step explanation:

The correct exponential function that models the value, V(t), after t years is:

V(t) = 750(1.06)^t

where t is the number of years since the purchase.

This formula is based on the fact that the value of the desk increases by 6% each year, so we multiply the initial value ($750) by (1 + 0.06) raised to the power of the number of years (t).

For example, after 5 years, the value of the desk would be:

V(5) = 750(1.06)^5

V(5) = 750(1.3382)

V(5) = 1003.65

So, after 5 years, the desk would be worth approximately $1,003.65.

The probability mass function (PMF) of x1 is given by P(x1) = (1/4) if x1 = -1, (1/2) if x1 = 0, and (1/4) if x1 = 1.

To determine the PMF of x1, first recognize that x1, x2, and x3 are identically distributed, which means they all share the same probability mass function. The moment generating function (mgf) of x1 is given as m(t) = (1/4) (1/2) exp(-t) (1/4) exp(t). You can rewrite this as m(t) = (1/4) exp(-t) + (1/2) exp(0) + (1/4) exp(t).

Now, recall that the mgf is defined as m(t) = E[exp(t * x1)] for a discrete random variable x1, where E represents the expected value. To find the PMF, simply compare the coefficients of the mgf with the general form of the mgf for a discrete random variable:

m(t) = Σ P(x1) exp(t * x1),

where the summation is over all possible values of x1. From the given mgf, we can identify the coefficients and find the PMF as follows:

P(x1 = -1) = 1/4,
P(x1 = 0) = 1/2,
P(x1 = 1) = 1/4.

Hence, the PMF of x1 is P(x1) = (1/4) if x1 = -1, (1/2) if x1 = 0, and (1/4) if x1 = 1.

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Answer: Both Green: 29%; Red then Green: 25%

Step-by-step explanation: Both Green: 32/59 * 31/58 = 0.28988 = 29%

Red then Green: 27/59 * 32/58 = 0.2524 = 25%

Your answer: A) Flextime is designed to give an employee greater control of their schedule, allowing them to have more flexibility in their work arrangements and making the work environment more motivating.

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Quantitative data may be either numeric or nonnumeric.

Quantitative data refers to data that can be measured or expressed numerically. This type of data can be either numeric, such as numerical values or counts, or nonnumeric, such as codes or labels that can be assigned numerical values for analysis.

Numeric quantitative data includes measurements like height, weight, temperature, age, and scores on a test, while nonnumeric quantitative data may include categories or labels such as gender (coded as 0 for male and 1 for female), ethnicity (coded as 1 for Asian, 2 for African American, 3 for Caucasian, etc.), or education level (coded as 1 for high school, 2 for college, 3 for postgraduate, etc.).

Therefore, quantitative data can be either numeric or nonnumeric, depending on how it is represented and used for analysis.

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(a) The following code creates a row vector M1 containing all values in the range [1,100) that are also multiples of 3 or 5:

M1 = [];

for i = 1:99

   if mod(i,3) == 0 || mod(i,5) == 0

       M1(end+1) = i;

   end

end

(b) The following code creates a row vector M2 containing all values in the range [1,100) that are multiples of 3 or 5, but not multiples of both:

M2 = [];

for i = 1:99

   if (mod(i,3) == 0 && mod(i,5) ~= 0) || (mod(i,3) ~= 0 && mod(i,5) == 0)

       M2(end+1) = i;

   end

end

For both parts a and b, a for loop is used to iterate through the range [1,100). Within the loop, the mod() function is used to check if the current value is divisible by 3 or 5.

In part a, the || (or) operator is used to check if either condition is true, and if so, the value is appended to the M1 vector using the end+1 syntax.

In part b, the && (and) operator is used to check if the value is divisible by only one of the numbers, and not both. If so, the value is appended to the M2 vector using the end+1 syntax.

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(a) The differential equation be,

⇒ dy/dt = 550 - (y(t)/12200)  67.1

(b) The required solution is,

⇒y(t) = √[200000 exp(550t + (33.55/24400) 200000²)]

According to the given information,

We know that the rate of change of pollutant amount in the lake is equal to 550 tons/yr, and that the amount of water entering the lake is equal to the amount of water leaving the lake, which is 67.1 km³/yr.

Using this information,

we can write a differential equation for y(t) as follows,

⇒ dy/dt = 550 - (y(t)/12200)  67.1

This equation expresses the rate of change of pollutant amount in the lake at time t as the difference between the rate at which pollutants are added to the lake (550 tons/yr) and the rate at which pollutants are diluted by the incoming and outgoing water flow.

The term (y(t)/12200 tons/km³) represents the pollutant concentration in the lake at time t,

And the factor of 67.1 km³/yr converts this concentration to a rate of pollutant removal from the lake.

The initial condition is y(0) = 200000,

we can use the method of separation of variables.

First, we rearrange the equation to isolate y and t on opposite side,

⇒ (1/y) dy = [550 - (67.1/12200) y] dt

Integrating both sides with respect to their respective variables, we get,

⇒ ln|y| = 550t - (67.1/12200) (1/2) y²+ C

where C is the constant of integration. Solving for y, we get:

⇒ y(t) = ±sqrt[exp(550t - (33.55/12200) * y²+ C)]

Using the initial condition y(0) = 200000, we can solve for C:

⇒ ln|200000| = - (33.55/24400) x 200000²+ C

⇒ C = ln|200000| + (33.55/24400) x 200000²

Substituting this value of C back into the general solution for y(t), we get,

⇒ y(t) = ±√[exp(550t - (33.55/12200) y² + ln|200000| + (33.55/24400) 200000²)]

We can simplify this expression by only considering the positive square root, since the amount of pollutant in the lake cannot be negative:

⇒ y(t) = √[exp(550t - (33.55/12200) y² + ln|200000| + (33.55/24400) 200000²)]

This is the expression for y(t) that satisfies the differential equation and the initial condition y(0) = 200000.

Now, to describe what happens to the amount of pollutants in the lake as t goes from 0 to infinity, we can analyze the behavior of the solution. The exponential term in the square root grows very quickly with time, so after a certain point, the other terms become negligible in comparison.

As t approaches infinity, the term (33.55/12200) y² in the exponent can be ignored, and we get:

y(t) ≈ √[exp(550t + ln|200000| + (33.55/24400) * 200000²)]

= √[200000 exp(550t + (33.55/24400) 200000²)]

This expression suggests that the amount of pollutants in the lake will continue to increase over time, approaching a value that grows exponentially with time.

However, the rate of increase will slow down over time, since the exponential term approaches a finite limit. Eventually, the amount of pollutants in the lake will stabilize, but this will likely be at a very high level that could be harmful to the ecosystem and human health.

Therefore, it is important to take measures to reduce pollutant inputs to the lake and prevent further contamination.

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