find the range of this set of data

All Trigonometric Expressions:

a. ∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx = ∫2sin³θ cos²θ dθ

b. ∫[tex]x^{2} dx/(4x^{2} + 6)[/tex]= ∫tan²θ sec²θ dθ

c. ∫x√(5x + 50)/(x + 118)dx = ∫(5tan²θ – 25)tanθ sec³θ dθ

d. ∫(19 – 50x)/(119 – 5x² + 50x)dx = -2∫dθ/(25tan²θ + 94)

a. The integral ∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx, we can use the substitution x = (2/5)sinθ. This gives dx = (2/5)cosθ dθ and 5x² – 2 = 5(2/5 sinθ)² – 2 = 2cos²θ. Substituting these expressions into the integral, we get:

∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx  

= ∫2sin³θ cos²θ dθ

b. For the integral ∫x²dx/(4x² + 6), we can use the substitution x = tanθ. This gives dx = sec²θ dθ and 4x² + 6 = 4tan²θ + 6 = 2sec²θ. Substituting these expressions into the integral, we get:

∫x²dx/(4x² + 6) = ∫tan²θ sec²θ dθ

c. For the integral ∫x√(5x + 50)/(x + 118)dx, we can use the substitution

x + 25 = 5tan²θ.

This gives x = 5tan²θ – 25 and dx = 10tanθ sec²θ dθ, and

5x + 50 = 25sec²θ. Substituting these expressions into the integral, we get:

∫x√(5x + 50)/(x + 118)dx

= ∫(5tan²θ – 25)tanθ sec³θ dθ

d. For the integral:

∫(19 – 50x)/(119 – 5x² + 50x)dx,

we can use the substitution

5x – 5 = √(50x – 5)tanθ.

This gives x = (1/10)[(tanθ)² + 1] and

dx = (1/5)(tanθ sec²θ) dθ, and 119 – 5x² + 50x

= (25tan²θ + 94)².

Substituting these expressions into the integral, we get:

∫(19 – 50x)/(119 – 5x² + 50x)dx

= -2∫dθ/(25tan²θ + 94)

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Correct Question:

For each of the following integrals find an appropriate trigonometric substitution of the form x = f(t) to simplify the integral.

a. ∫5x * ∫5x * [tex](5x^{2} - 2)^{(3/2)[/tex]dx

b. ∫[tex]x^{2} dx/(4x^{2} + 6)[/tex]

c. ∫x√(5x + 50)/(x + 118)dx

d. ∫(19 – 50x)/(119 – 5x² + 50x)dx

To answer this question, we'll use the exponential distribution and the concept of the probability density function (pdf). Let X be the time until failure of a single device, with a mean lifetime of 20 months. The exponential distribution has the following pdf:

f(x) = (1/μ) * e^(-x/μ),

where μ is the mean lifetime (20 months in this case).

Now, let's find the probability that the first failure occurs at w months among the 5 independent devices. For this, we need to calculate the probability that none of the other 4 devices fail before w months and that the first device fails at w months.

The probability that a single device does not fail before w months is given by the complementary cumulative distribution function (ccdf) of the exponential distribution:

P(X > w) = e^(-w/μ).

Since the devices are independent, the probability that all 4 devices do not fail before w months is:

P(All 4 devices survive > w) = (e^(-w/μ))^4.

Now, the probability that the first device fails at w months is given by the pdf of the exponential distribution:

P(X = w) = (1/μ) * e^(-w/μ).

Finally, we multiply the two probabilities to find the chance that the first failure occurs at w months:

P(First failure at w) = P(All 4 devices survive > w) * P(X = w)
= (e^(-w/μ))^4 * (1/μ) * e^(-w/μ)
= (1/20) * e^(-5w/20).

Thus, the chance that the first failure will occur at w months is given by the expression (1/20) * e^(-5w/20).

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Answer:

29/16, or 1.81 in decimal form.

Step-by-step explanation:

To solve this, we have to find a common denominator between the 2 fractions.

We cannot simplify 15/16 any further without getting a decimal, so let's change 7/8.

A common denominator between the 2 fractions is 16, so multiply 7/8 by 2 to get 14/16.

Now, we have to find out how far she walked in total, so add 14/16 + 15/16:

29/16

Hope this helps! :)

The value of the inflection point is 1/10, and the intervals on which the function is concave up or down are (0, 1/10) and (1/10, ∞).

Taking the second derivative of y(x), we get:

y''(x) = 240x³ - 24x²

Setting y''(x) equal to zero and solving for x, we get:

x = 0 or x = 1/10

These critical points divide the real line into three intervals:

(-∞, 0), (0, 1/10), and (1/10, ∞)

We evaluate the sign of y''(x) on each of these intervals to determine where the function is concave up or down:

For x < 0: y''(x) < 0, so y(x) is concave down.

For 0 < x < 1/10: y''(x) > 0, so y(x) is concave up.

For x > 1/10: y''(x) > 0, so y(x) is concave up.

Therefore, the function is concave down on the interval (-∞, 0) and concave up on the intervals (0, 1/10) and (1/10, ∞).

To find the inflection point, we set y''(x) equal to zero and solve for x:

240x³ - 24x² = 0

Factor out 24x²:

24x²(10x - 1) = 0

So either x = 0 or x = 1/10.

Since the second derivative changes sign at x = 1/10, this is an inflection point.

Therefore, the inflection point occurs at x = 1/10.

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The question is -

Determine the intervals on which the function is concave up or down and find the value at which the inflection point occurs.

Y = 8x^5 - 3x^4

(Express intervals in interval notation. Use symbols and fractions when needed)

point of influence at x = __________

interval on which function is concave up = ____________

interval on which function is concave down = ___________

Answer: (10.4, 0).

Step-by-step explanation: In order to determine the coordinate where the projectile would land, we need to find the parabolic path followed by the projectile. Since the catapult is launched from the origin (0,0), and has a maximum height at (5.2, 6.3), we can find the vertex of the parabola, which is also (5.2, 6.3).

The standard form of a parabola is y = a(x - h)^2 + k, where (h,k) is the vertex of the parabola. In this case, h = 5.2 and k = 6.3.

We can use the information from the origin to determine the value of 'a'. Substituting x = 0 and y = 0, we get:

0 = a(0 - 5.2)^2 + 6.3

Solving for 'a':

a(5.2)^2 = -6.3

a = -6.3 / (5.2)^2

Now that we have 'a', we can rewrite the equation of the parabola:

y = a(x - 5.2)^2 + 6.3

To find the x-coordinate where the projectile would land, we need to find the other x-intercept (the other point where y = 0). Since the parabola is symmetric, the other x-intercept will be equidistant from the vertex:

x = 5.2 * 2 = 10.4

Now, we can plug in x = 10.4 into the equation to find the y-coordinate:

y = a(10.4 - 5.2)^2 + 6.3

However, since we are looking for the landing coordinate, which is an x-intercept, we know the y-coordinate will be 0.

Thus, the coordinate where the projectile would land is (10.4, 0).

Answer:

Assuming that those are fraction the simplest form would be --> 7/4

Hope this helps

The calculated value of the expression when a = 3 and b = -2 is 1

The expression is given as

(StartFraction 3 a Superscript negative 2 Baseline b Superscript 6 Baseline Over 2 a Superscript negative 1 Baseline b Superscript 5 Baseline EndFraction) squared

Mathematically, this can be expressed as

(3a^-2b^6/2a^-1b^5)^2

The values of a and b are given as

a =3 and b = -2

substitute the known values in the above equation, so, we have the following representation

(3(3)^-2(-2)^6/2(3)^-1(-2)^5)^2

Evaluate

So, we have

1

Hence, the solution is 1

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The first number, 5 x [tex]10^22,[/tex] can be written as 5 followed by 22 zeros:

5,000,000,000,000,000,000,000

To simplify means to make something easier to understand or do by reducing complexity, removing unnecessary details, or using simpler language or concepts.

The second number , 5 × 10², can be written as 5 followed by 2 zeros: 500

The third number, 3.7 x 10⁵³, can be written as 3.7 followed by 53 zeros:

370,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

or in scientific notation as 3.7 x 10⁵³

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Simplify

2. 5 x 10^22. 5 × 10 2 and 3. 7 x 10^53. 7 × 10 5

The sample mean of the population is 3/4 and the variance is 3/80. Using the central limit theorem, P( > 0.8) can be simplified as 0.003.

The mean of the population can be computed as follows:

µ = ∫x f(x) dx from 0 to 1

  = ∫x (3x²) dx from 0 to 1

  = 3/4

The variance of the population can be computed as follows:

σ² = ∫(x-µ)² f(x) dx from 0 to 1

    = ∫(x-(3/4))² (3x²) dx from 0 to 1

    = 3/80

By the Central Limit Theorem, as the sample size n = 80 is large, the distribution of the sample mean  can be approximated by a normal distribution with mean µ and variance σ²/n.

Therefore, P( > 0.8) can be approximated by P(Z >0.8- 0.75)/(sqrt(3/80)/(sqrt(80))), where Z is a standard normal random variable.

Simplifying, we get P( > 0.8) ≈ P(Z > 2.73) ≈ 0.003.

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The given set of ordered pairs represents a function because each output has exactly one corresponding input. So, the correct answer is A) Yes, because for each output there is exactly one input.

A relation between two variables is a set of ordered pairs, where the first element in each pair corresponds to the input or independent variable (usually denoted by x), and the second element corresponds to the output or dependent variable (usually denoted by y).

In the given set of ordered pairs, each output has exactly one corresponding input, and therefore the relation satisfies the definition of a function. For example, the input of 3/4 is associated with only one output of -2, and the output of -7 is associated with only one input of -2. Hence, the relation represents a function.

So, the correct answer is A).

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A simple random sampling of 5 observations from a population containing 400 elements was taken. Then, the point estimate of the population means is 18.

You have a simple random sample of 5 observations from a population containing 400 elements, and the observed values are 12, 18, 19, 20, and 21.

To calculate the point estimate of the population mean, we simply take the average of the sample values.

Point estimate of population mean = (12 + 18 + 19 + 20 + 21)/5 = 18

Therefore, the point estimate of the population means is 18.

To clarify the terms used in the question, a "random sample" is a sample that is selected randomly from the population, meaning that every element in the population has an equal chance of being included in the sample. In this case, a simple random sample of 5 observations was taken. "Elements" refers to the individual units or objects within the population that is being studied. In this case, there were 400 elements in the population.

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The value of the iterated integral is [tex]7/3[/tex] √2 in the given case

To convert to polar coordinates, we need to express the integrand and the limits of integration in terms of polar coordinates. Let's start by finding the limits of integration:

0 ≤ y ≤ √2 - y[tex]^2[/tex]

0 ≤ x ≤ 1

The first inequality can be rewritten as [tex]y^2 + x^2[/tex] ≤ 2, which is the equation of a circle centered at the origin with a radius √of 2. Therefore, the limits of integration in polar coordinates are:

0 ≤ r ≤ √2

0 ≤ θ ≤ π/2

Now, let's express the integrand in polar coordinates:

7xy = 7r cos(θ) sin(θ)

And the differential area element in polar coordinates is:

dA = r dr dθ

Therefore, the integral becomes:

= [tex]7/3[/tex] √2

Therefore, the value of the iterated integral is [tex]7/3[/tex] √2.

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The equation that can be used to find the value of x is 120 = (x + 5) × (120/x - 1/3).

Carmen's first trip was 120 miles, and she traveled at an average of x miles per hour. We can use the formula:

distance = rate × time, which can be written as:
120 miles = x miles/hour × time

where, time is the time for outgoing.

For the return trip, Carmen traveled at a rate that was 5 miles per hour faster, so her speed was (x + 5) miles/hour. The time for the return trip was one-third of an hour less than the time for the outgoing trip, so we can represent the return trip time as (time - 1/3) hours. Using the distance formula again for the return trip:
120 miles = (x + 5) miles/hour × (time - 1/3) hours

Now, let's express both times in terms of x. From the first equation, we can find the time for the outgoing trip as:
time = 120 miles / x miles/hour

Substitute this expression for time in the return trip equation:
120 miles = (x + 5) miles/hour × (120/x - 1/3) hours

Now you have an equation that can be used to find the value of x:
120 = (x + 5) × (120/x - 1/3)

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The number of ways of arrangement is 726, 485, 760 ways, 5040 ways, 210 ways resdpectively

Remember that Permutation is a mathematical technique that determines the number of possible arrangements in a set when the order of the arrangements matters.

How many ways can you choose 5 players from 14

= [tex]14^{} Px_{5}[/tex]

14!/(14-5)!

(14*13*12*11*10*9*8*7*6*5*4*3*2*1)/9*8*7*6*5*4*3*2*1

= 726, 485, 760 ways

3. How many ways can you line up 7 books on a shelf?

= 7!

= 5040 ways

4.  10!/(10-6)!6!

10*9*8*7*6*5*4*3*2*1/6*5*4*3*2*1*4*3*2*1

= 5040/24

= 210 ways

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The inequality that shows how many questions, q, will be on Mr. Pham's quiz is: 0 ≤ q ≤ 15

What is inequalities?

In mathematics, an inequality is a mathematical statement that indicates that two expressions are not equal. It is a statement that compares two values, usually using one of the following symbols: "<" (less than), ">" (greater than), "≤" (less than or equal to), or "≥" (greater than or equal to).

This inequality states that the number of questions, q, must be greater than or equal to zero (since there cannot be a negative number of questions), but less than or equal to 15 (since Mr. Pham's quiz will have at most 15 questions).

Therefore, the inequality that shows how many questions, q, will be on Mr. Pham's quiz is: 0 ≤ q ≤ 15

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a. The test is a left-tailed test. b. What is the test statistic is -1.73.

a. The test is a left-tailed test because we are testing the claim that less than 10% of treated subjects experienced headaches.

b. To find the test statistic, we'll use the normal distribution as an approximation to the binomial distribution. Here are the steps:

Step 1: Determine the null and alternative hypotheses.
H0: p = 0.10 (The proportion of treated subjects experiencing headaches is equal to 10%.)
H1: p < 0.10 (The proportion of treated subjects experiencing headaches is less than 10%.)

Step 2: Calculate the sample proportion (p-hat).
p-hat = number of subjects with headaches / total subjects = 19 / 276 ≈ 0.0688

Step 3: Determine the standard error.
SE = sqrt((p * (1 - p)) / n) = sqrt((0.10 * (1 - 0.10)) / 276) ≈ 0.0180

Step 4: Calculate the test statistic (z-score).
z = (p-hat - p) / SE = (0.0688 - 0.10) / 0.0180 ≈ -1.73

So, the test statistic is -1.73 (rounded to two decimal places).

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The quantity that maximizes total revenue is 40. The correct option is d. 40.

To maximize total revenue, we need to find the quantity where marginal revenue (MR) equals zero.

Given the MR equation:

MR = 40 - Q

Set MR to zero and solve for Q:

0 = 40 - Q

Q = 40

So, the quantity that maximizes total revenue is 40. The correct option is d. 40.

Maximizing total revenue requires optimizing prices, bundling products, cross-selling/upselling, improving customer retention, expanding the customer base, and reducing costs. Make sure to regularly check on revenue performance and modify strategies accordingly.

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Suppose a monopolist has the following equations: P=40-0.5Q MR=40-Q What is the quantity that maximizes total revenue? a. 25 10 30 d.40 e. 20 T1e MC=10

The answer to the first question is Full Factorial Design, and the answer to the second question is Fractional Factorial Designs.

Full Factorial Design involves testing all possible combinations of the factors in an experiment at a number of levels.

Fractional Factorial Designs are used for screening experiments to identify critical factors. These designs are a subset of the full factorial design, and they only test a fraction of the possible combinations of the factors in an experiment. This allows for a more efficient use of resources when conducting experiments.

Therefore, the answer to the first question is Full Factorial Design, and the answer to the second question is Fractional Factorial Designs.

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The sample space for choosing one card is, S={I, L, L, M, P, T, U, Y}

Since, We know that;

A sample space is a set of potential results from a random experiment. The letter "S" is used to denote the sample space. Events are the subset of possible experiment results. Depending on the experiment, a sample area could contain a variety of results.

Given that,

A set of eight cards were labeled with M, U, L, T, I, P, L, Y.

Here, the sample space is;

{ S, U, B, T, R, A, C, T}

Now, Elements in order is,

⇒ S = {I, L, L, M, P, T, U, Y}

Therefore, the sample space for the given cards is,

S = {I, L, L, M, P, T, U, Y}

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There are two free variables, so the dimension of the eigenspace is 2. So the dimensions of the associated eigenspaces are 2 for all three eigenspace.

To determine which of the given values are complete eigenvalues, we need to find the characteristic polynomial of the matrix. This is done by finding the determinant of (A - λI), where A is the matrix and λ is the eigenvalue:

| 3-λ   2    0   -4    1 |
| 0     1-λ  3    1    0 |
| 1     -1   -1-λ 4    0 |
| -1    0    2    -4-λ 0 |
| 1     1    -1   0    1-λ|

Expanding along the first row, we get:

(3-λ) | 1-λ  3   1   0 |
    |-1   2-λ 4   0 |
    |1    -1  -4-λ 0 |
    |1    -1  0    1-λ |

= (3-λ)[(2-λ)(1-λ)(1-λ) + 4(-1)(1-λ) + 0(4-λ)] - (-1)[(1-λ)(1-λ)(4-λ) + 0(1-λ) + 0(-1)] + (1)[(1-λ)(4-λ)(0) - (2-λ)(1-λ)(-1)] - (1)[(1-λ)(-1)(-1) - (2-λ)(-1)(0)]

= (3-λ)[λ^3 - 6λ^2 + 9λ - 4] + (λ-1)[4λ^2 - 10λ + 6] + (λ-1)(λ-4) - (λ-2)

= λ^5 - 11λ^4 + 44λ^3 - 78λ^2 + 60λ - 16

Now we can check which of the given values satisfy the characteristic polynomial:

(a) 3, († 2), 0, 1
Substituting each value into the polynomial, we get:
3^5 - 11(3^4) + 44(3^3) - 78(3^2) + 60(3) - 16 = 0
2^5 - 11(2^4) + 44(2^3) - 78(2^2) + 60(2) - 16 ≠ 0
0^5 - 11(0^4) + 44(0^3) - 78(0^2) + 60(0) - 16 ≠ 0
1^5 - 11(1^4) + 44(1^3) - 78(1^2) + 60(1) - 16 = 0

So the complete eigenvalues for this matrix are 3, 0, 1.

To find the dimension of the associated eigenspace for each eigenvalue, we need to find the nullspace of (A - λI). For each eigenvalue, we can do this by row reducing the matrix (A - λI) and finding the number of free variables. The dimension of the associated eigenspace is then equal to the number of free variables.

(a) λ = 3:

| 0 -1  1  1 -1 |
| 0 -2  4  0  1 |
| 1 -1 -4  2  1 |
|-1  0  2 -7  1 |
| 1  1 -1  0 -2 |

RREF:

| 1  0 -2  0  0 |
| 0  1 -2  0  0 |
| 0  0  0  1  0 |
| 0  0  0  0  1 |
| 0  0  0  0  0 |

There are two free variables, so the dimension of the eigenspace is 2.

(a) λ = 0:

| 3  2  0 -4  1 |
| 0  1  3  1  0 |
| 1 -1 -1  4  0 |
|-1  0  2 -4  0 |
| 1  1 -1  0  1 |

RREF:

| 1  0 -2  0  0 |
| 0  1 -2  0  0 |
| 0  0  0  1  0 |
| 0  0  0  0  0 |
| 0  0  0  0  0 |

There are two free variables, so the dimension of the eigenspace is 2.

(a) λ = 1:

| 2  2  0 -4  1 |
| 0  0  3  1  0 |
| 1 -1 -2  4  0 |
|-1  0  2 -5  1 |
| 1  1 -1  0  0 |

RREF:

| 1  0 -1  0  0 |
| 0  1 -1  0  0 |
| 0  0  0  1 -1 |
| 0  0  0  0  0 |
| 0  0  0  0  0 |

There are two free variables, so the dimension of the eigenspace is 2.

So, the dimensions of the associated eigenspaces are 2 for all three eigenvalues.

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The negative critical point is approximately -1.812.

The critical values for a two-tailed hypothesis test with a significance level of α = 0.10 and 10 degrees of freedom (sample size - 1) can be found using a t-distribution table or a statistical software.

a) The positive critical point can be found by looking up the t-distribution table or using a statistical software to find the t-value that corresponds to a cumulative probability of 0.95 with 10 degrees of freedom. The value is approximately 1.812.

b) The negative critical point can be found by finding the t-value that corresponds to a cumulative probability of 0.05 with 10 degrees of freedom. Since the t-distribution is symmetric, this value is the negative of the positive critical point. Therefore, the negative critical point is approximately -1.812.

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Since the arrival times are independent and uniformly distributed, the probability that a single person arrives in the first hour is 1/10^6. Therefore, the number of people N that arrive in the first hour follows a binomial distribution with parameters n=10^6 and p=1/10^6.

The probability that exactly i people arrive in the first hour is then given by the binomial probability mass function:

P{N=i} = (10^6 choose i) * (1/10^6)^i * (1 - 1/10^6)^(10^6 - i)

Using the normal approximation to the binomial distribution, we can approximate this probability as:

P{N=i} ≈ φ((i+0.5 - np) / sqrt(np(1-p)))

where φ is the standard normal probability density function. Plugging in the values of n=10^6 and p=1/10^6, we get:

P{N=i} ≈ φ((i+0.5 - 1) / sqrt(1*0.999999)) = φ(i - 0.5)

Therefore, an approximation for P{N=i} is given by the standard normal density function evaluated at i-0.5.

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A graph that represents the inequality y > x² - 3 include the following: A. graph A.

In Mathematics and Geometry, the general form of a quadratic function can be modeled and represented by using the following quadratic equation;

y = ax² + bx + c

Where:

Since the leading coefficient (value of a) in the given quadratic function  y > x² - 3 is positive 1, we can logically deduce that the parabola would open upward. Also, the value of the quadratic function f(x) would be minimum at -3.

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The probability that she will get a white button and a blue button is 18/30 * 3/29 = 9/145 or approximately 0.062.

The total number of buttons is 18 + 9 + 3 = 30. The probability of getting a white button on the first draw is 18/30. After drawing a white button, there are 29 buttons left, including 3 blue buttons, so the probability of getting a blue button on the second draw is 3/29.

To find the probability of both events happening, we multiply the probabilities:

18/30 * 3/29 = 9/145

Therefore, the probability that she will get a white button and a blue button is 9/145 or approximately 0.062.

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The domain of this relation is the set of all non-negative real numbers that can be expressed as: A weight from 0 to 5 ounces A weight from 5 to 20 ounces that is a multiple of 0.2 ounces A weight greater than 20 ounces is a multiple of 0.15 ounces.

The domain of this relation is the set of all possible weights of packages that can be shipped using the postal service.

Since the postal service charges $2 for packages up to 5 ounces, the domain includes all weights from 0 to 5 ounces. For packages weighing between 5 and 20 ounces, the domain includes all weights from 5 to 20 ounces, with each weight being a multiple of 0.2 ounces.

For packages weighing more than 20 ounces, the domain includes all weights greater than 20 ounces, with each weight being a multiple of 0.15 ounces.

The domain does not include negative numbers or numbers that are not expressible using the above criteria.

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The amount of money in the account after 16 years is approximately $17,526.64.

Using the formula for continuous compounding, we have:

V = Pe[tex]^(rt)[/tex]

where V is the value of the account after t years, P is the principal initially invested, e is the base of the natural logarithm, r is the annual interest rate, and t is the time in years.

Substituting the given values, we get:

V = 8430e[tex]^(0.038*16)[/tex]

Simplifying this expression, we have:

V = 8430[tex]e^0.608[/tex]

Using a calculator, we get:

V ≈ $17,526.64

Therefore, the amount of money in the account after 16 years is approximately $17,526.64.

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The number of people that are expected to go to the youth wing on Sunday would be = 543.

For Saturday;

The number of people that went to the youth wing = 497

The number of people that went to social issues = 369

The number of people that went to Fiction and Literature = 416

The total number of people that went to the central library = 497+369+416 = 1,282

For Sunday, the number of people that would visit the youth wing ;

= 497/1282× 1400/1

= 695800/1282

= 543 (to the nearest whole number)

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The time taken for the smokestack A is 22.5 hours.

The time taken for the smokestack B is 45 hours.

The time taken for the smokestack is calculated as follows;

Let's the rate at which smokestack B emits pollution = r

Then smokestack A = 2r

Their total rate of pollution combined;

= r + 2r

= 3r

The total amount of pollution they emitted after 15 hours;

= 3r x 15

= 45r pollution

The time taken for each to emit the same amount;

rate of B = r pollution/hr

time of B = 45r/r = 45 hours

rate of A = 2r

time of A = 45r/2r = 22.5 hours

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Answer:

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