The stray dog population in a local city is currently estimated to be 1,000. The expected annual rate of increase is predicted to be 0. 7. What will the population be in 4 years? Round your answer to the nearest whole number

The correct expressions which model the value of the phone after t years are given by 300(0.87)t/12 and 300(0.9885)t. Value of a cellular phone depreciates at a rate of 13% every month.

Given a cellular phone's value depreciates at a rate of 13% every month. So, the phone's value will decrease by 13% of its original value every month. Therefore, the equation for the phone's value after t years is given by:

V(t) = $300 × (1 - 0.13)ᵗ, where t is the time in years.

The given expressions, 300(0. 87)/12 and 300(0. 1880)t 300(0. 87)t/12 and 300(0. 9885)t 300(0. 87)124 and 300(0. 1880)t 300(0. 87) 12 and 300(0. 9885)t. Do not model the value of the phone after t years. Therefore, the correct answer is 300(0. 87)t/12 and 300(0. 9885)t.

The value of a cellular phone depreciates at a rate of 13% every month, which means that the remaining value of the phone after one month is 87% of the original value. Therefore, to calculate the value after t years, the equation

V(t) = $300 × (1 - 0.13)ᵗ should be used.

By plugging in the time t in years, we can get the remaining value of the phone. The first option (300(0.87)/12 must be corrected because it only calculates the phone's value after one month, which is not the question asked. Therefore, the correct expression that model the phone's value after t years is given by 300(0.87)t/12 and 300(0.9885)t.

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The best lower bound for the volume is 24 cm³, and the best upper bound is 120 cm³ and the best lower bound for the surface area is 52 cm², and the best upper bound is 148 cm².

a. To determine the best upper and lower bounds for the volume of the rectangular parallelepiped, we can consider the extreme cases by rounding each side to the nearest centimeter.

Lower bound: If we round each side down to the nearest centimeter, we get a rectangular parallelepiped with sides 2 cm, 3 cm, and 4 cm. The volume of this parallelepiped is 2 cm * 3 cm * 4 cm = 24 cm³.

Upper bound: If we round each side up to the nearest centimeter, we get a rectangular parallelepiped with sides 4 cm, 5 cm, and 6 cm. The volume of this parallelepiped is 4 cm * 5 cm * 6 cm = 120 cm³.

Therefore, the best lower bound for the volume is 24 cm³, and the best upper bound is 120 cm³.

b. Similar to the volume, we can determine the best upper and lower bounds for the surface area of the parallelepiped by considering the extreme cases.

Lower bound: If we round each side down to the nearest centimeter, the dimensions of the parallelepiped become 2 cm, 3 cm, and 4 cm. The surface area is calculated as follows:

2 * (2 cm * 3 cm + 3 cm * 4 cm + 4 cm * 2 cm) = 2 * (6 cm² + 12 cm² + 8 cm²) = 2 * 26 cm² = 52 cm².

Upper bound: If we round each side up to the nearest centimeter, the dimensions become 4 cm, 5 cm, and 6 cm. The surface area is calculated as follows:

2 * (4 cm * 5 cm + 5 cm * 6 cm + 6 cm * 4 cm) = 2 * (20 cm² + 30 cm² + 24 cm²) = 2 * 74 cm² = 148 cm².

Therefore, the best lower bound for the surface area is 52 cm², and the best upper bound is 148 cm².

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(a) Cov(X, Y) = 1/2, (b) E(Y|X) = X/2, (c) Cov(X,E(Y|X)) = Cov(X, Y) = 1/2, and the identity Cov(X,E(Y|X)) = Cov(X, Y) holds true for any joint distribution of X and Y.

(a) To compute Cov(X, Y), we need to first find the marginal density of X and the marginal density of Y.

The marginal density of X is:

f_X(x) = ∫[0,x] f(x,y) dy

= ∫[0,x] 2e^(-2x) / x dy

= 2e^(-2x)

The marginal density of Y is:

f_Y(y) = ∫[y,∞] f(x,y) dx

= ∫[y,∞] 2e^(-2x) / x dx

= -2e^(-2y)

Next, we can use the formula for covariance:

Cov(X, Y) = E(XY) - E(X)E(Y)

To find E(XY), we can integrate over the joint density:

E(XY) = ∫∫ xyf(x,y) dxdy

= ∫∫ 2xye^(-2x) / x dxdy

= ∫ 2ye^(-2y) dy

= 1

To find E(X), we can integrate over the marginal density of X:

E(X) = ∫ xf_X(x) dx

= ∫ 2xe^(-2x) dx

= 1/2

To find E(Y), we can integrate over the marginal density of Y:

E(Y) = ∫ yf_Y(y) dy

= ∫ -2ye^(-2y) dy

= 1/2

Substituting these values into the formula for covariance, we get:

Cov(X, Y) = E(XY) - E(X)E(Y)

= 1 - (1/2)*(1/2)

= 3/4

Therefore, Cov(X, Y) = 3/4.

(b) To find E(Y | X), we can use the conditional density:

f(y | x) = f(x, y) / f_X(x)

For 0 ≤ y ≤ x, we have:

f(y | x) = (2e^(-2x) / x) / (2e^(-2x))

= 1 / x

Therefore, the conditional density of Y given X is:

f(y | x) = 1 / x, 0 ≤ y ≤ x

To find E(Y | X), we can integrate over the conditional density:

E(Y | X) = ∫ y f(y | x) dy

= ∫[0,x] y (1 / x) dy

= x/2

Therefore, E(Y | X) = x/2.

(c) To compute Cov(X,E(Y | X)), we first need to find E(Y | X) as we have done in part (b):

E(Y | X) = x/2

Next, we can use the formula for covariance:

Cov(X, E(Y | X)) = E(XE(Y | X)) - E(X)E(E(Y | X))

To find E(XE(Y | X)), we can integrate over the joint density:

E(XE(Y | X)) = ∫∫ xyf(x,y) dxdy

= ∫∫ 2xye^(-2x) / x dxdy

= ∫ x^2 e^(-2x) dx

= 1/4

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Given fixed expenses of Rashad for the month of February, which are as follows:Rent = $1,150.00Car Loan = $348.00Internet = $46.14Student Loan = $399.34Total Expenses = $3,291.74.

Rashad can determine his variable expenses by subtracting his fixed expenses from his total expenses for the month.Subtracting the fixed expenses from the total expenses, we get, Variable Expenses = Total Expenses - Fixed Expenses Variable Expenses = $3,291.74 - ($1,150.00 + $348.00 + $46.14 + $399.34)Variable Expenses = $3,291.74 - $1,943.48Variable Expenses = $1,348.26

Therefore, Rashad's variable expenses are $1,348.26.Equation that represents this situation is,Variable Expenses = Total Expenses - Fixed Expenses.Variable Expenses = $3,291.74 - ($1,150.00 + $348.00 + $46.14 + $399.34)Variable Expenses = $3,291.74 - $1,943.48Variable Expenses = $1,348.

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The normal distribution tails never go up again after crossing the horizontal axis. In a normal distribution, the tails of the curve represent the extreme values in either direction.

The tails of the curve extend infinitely in both directions and they get closer and closer to the horizontal axis, but they never touch it.

The curve is symmetrical around the mean and the area under the curve is equal to 1 or 100%.In probability theory, normal distribution is a continuous probability distribution that describes a set of random variables, and is often referred to as the Gaussian distribution. It is a bell-shaped curve and is characterized by the mean and standard deviation. It is an important concept in statistics and is used to describe various natural phenomena, such as heights, weights, IQ scores, etc.

The normal distribution is a bell-shaped curve that describes the distribution of a set of data. The curve is symmetrical around the mean, and the area under the curve is equal to 1 or 100%. The normal distribution is important in statistics because it is used to describe various natural phenomena. It is often used to describe the distribution of heights, weights, IQ scores, etc.

The normal distribution has a unique property that makes it useful in probability theory. The tails of the curve never touch the horizontal axis. The tails represent the extreme values in either direction, and they extend infinitely in both directions. They get closer and closer to the horizontal axis, but they never touch it. This means that the probability of observing an extreme value is very small. The normal distribution is an important concept in statistics, and it is used to make predictions about the future based on past observations.

The normal distribution is a bell-shaped curve that describes the distribution of a set of data. The tails of the curve never touch the horizontal axis. The tails represent the extreme values in either direction, and they extend infinitely in both directions. They get closer and closer to the horizontal axis, but they never touch it.

The normal distribution is important in probability theory and is often used to describe various natural phenomena. It is used to make predictions about the future based on past observations.

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The null and alternative hypotheses are: H0: σ21 = σ22 and Ha: σ21 ≠ σ22(C).

To find the critical value, we need to use the F-distribution with degrees of freedom (df1 = n1 - 1, df2 = n2 - 1) at a significance level of α/2 = 0.005 (since this is a two-tailed test).

Using a calculator or a table, we find that the critical values are F0.005(12,7) = 4.963 (for the left tail) and F0.995(12,7) = 0.202 (for the right tail).

The test statistic is calculated as F = s21/s22, where s21 and s22 are the sample variances and n1 and n2 are the sample sizes. Plugging in the given values, we get F = 5.7^2/5.1^2 = 1.707.

Since this value is not in the rejection region (i.e., it is between the critical values), we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to claim that the population variances are different at the 0.01 level of significance.

So C is correct option.

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The given pairs of points represent coordinates on a graph: (2,1) and (3,1.5), (2,1) and (5,2), (6,2) and (8,2), and (6,2) and (10,1.75). These points indicate different positions in a two-dimensional plane.

In the first pair of points, (2,1) and (3,1.5), the y-coordinate increases from 1 to 1.5 as the x-coordinate increases from 2 to 3. This suggests a positive slope, indicating an upward trend.

The second pair of points, (2,1) and (5,2), shows a similar trend. The y-coordinate increases from 1 to 2 as the x-coordinate increases from 2 to 5, indicating a positive slope and an upward movement.

In the third pair, (6,2) and (8,2), both points have the same y-coordinate of 2. This suggests a horizontal line, indicating no change in the y-coordinate as the x-coordinate increases from 6 to 8.

The fourth pair, (6,2) and (10,1.75), shows a slight decrease in the y-coordinate from 2 to 1.75 as the x-coordinate increases from 6 to 10. This indicates a negative slope, representing a downward trend.

Overall, these pairs of points represent different types of trends on a graph, including upward, horizontal, and downward movements. The relationship between the x and y coordinates can help determine the nature of the trend between the points.

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The equation of the parabola that intersects the x-axis at points (-1, 0) and (3,0) is y = 0.

Given that a parabola intersects the x-axis at points (-1, 0) and (3,0).We know that, when a parabola intersects the x-axis, the y-coordinate of the point on the parabola is 0. Therefore, the two x-intercepts tell us two points that are on the parabola.Thus the vertex is given by:Vertex is the midpoint of these x-intercepts=(x_1+x_2)/2=(-1+3)/2=1The vertex is the point (1,0).Since the vertex is at (1,0) and the parabola intersects the x-axis at (-1,0) and (3,0), the axis of symmetry is the vertical line passing through the vertex, which is x=1.We also know that the parabola opens upwards because it intersects the x-axis at two points.To find the equation of the parabola, we can use the vertex form:y = a(x - h)^2 + kwhere (h, k) is the vertex and a is a constant that determines how quickly the parabola opens up or down.We have h=1 and k=0.Substituting in the x and y values of one of the x-intercepts, we get:0 = a(-1 - 1)^2 + 0Simplifying, we get:4a = 0a = 0Substituting in the x and y values of the other x-intercept, we get:0 = a(3 - 1)^2 + 0Simplifying, we get:4a = 0a = 0Since a = 0, the equation of the parabola is:y = 0(x - 1)^2 + 0Simplifying, we get:y = 0Hence the equation of the parabola that intersects the x-axis at points (-1, 0) and (3,0) is y = 0.

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Based on the crocodile skeleton found with a head length of 62 cm and a body length of 380 cm, it is likely that the species was a Saltwater Crocodile (Crocodylus porosus).

According to the given measurements, it is likely that the species was a Saltwater Crocodile (Crocodylus porosus).  This is because Saltwater Crocodiles are known to have larger sizes compared to other species.

To explain why, let's consider the following steps:

1. Compare the head length and body length to average sizes of different crocodile species.
2. Identify the species whose average size is closest to the given measurements.

Saltwater Crocodiles are the largest living species of crocodiles, with males reaching lengths of over 6 meters (20 feet). The head length of 62 cm and body length of 380 cm (3.8 meters) would likely be within the size range for an adult male Saltwater Crocodile. Other species, such as the Nile Crocodile or the American Alligator, typically do not reach such large sizes, making the Saltwater Crocodile a more plausible candidate based on the given measurements.

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The test statistic for the difference between the means is 2.22.

To determine the test statistic for the difference between the means of two independent populations, use the two-sample t-test:

t = (x₁ - x₂) / √[(σ₁² /n₁) + (σ₂² /n₂)]  

where x₁ and x₂ = sample means, σ₁ and σ₂ = sample standard deviations, and n₁ and n₂ = sample sizes.

Using the given values:

x₁ = $15.80

x₂ = $15.00

σ₁ = $3.00

σ₂ = $2.25

n₁ = 81

n₂ = 64

Calculate the test statistic as:

t = ($15.80 - $15.00) / √[($3.00²/81) + ($2.25²/64)]  

t = 2.22

Therefore, the test statistic for the difference between the means is 2.22.

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Musk is 12 years old, Abu is 18 years old and the sum of their ages is 30.

Let's find out the current ages of Musk and Abu from the given information.

Musk's age is 2/3 of Abu's age.

We can express it as; Musk's age = 2/3 × Abu's age Also, the sum of their age is 30.

So we can express it as: Musk's age + Abu's age = 30

Substitute the first equation into the second one:2/3 × Abu's age + Abu's age = 30

Simplify the equation and solve for Abu's age:5/3 × Abu's age = 30Abu's age = 18

Substitute Abu's age into the first equation to find Musk's age:

Musk's age = 2/3 × 18Musk's age = 12

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Answer:

16 mm

Step-by-step explanation:

P = 2(L + W)

P = 2(2 mm + 6 mm)

P = 2(8 mm)

P = 16 mm

Mr. Baral has to pay Rs 64000 as an annual income tax at an interest of 10% for his stationary shop.

From the question, we have given that if he is unmarried and his income is between Rs 5,00,001 to Rs 7,00,000, he has to pay an annual interest of 10%.

Given annual income in Rs = 640000.

The annual income tax rate he has to pay at = 10%

So, to find out the income tax from the annual income we have to find out the 10% of 640000.

Income tax = 640000/100 * 10 = 64000

From the above analysis, we can conclude that Mr. Baral has to pay 64000 rs of income tax annually.

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Given question is not having enough information, I am writing the complete question below:

Use it to calculate the income taxes. For an individual Income slab Up to Rs 5,00,000 0% Rs 5,00,001 to Rs 7,00,000 10% Rs 7,00,001 to Rs 10,00,000 20% Rs 10,00,001 to Rs 20,00,000 30% Tax rate For couple Tax rate 0% Income slab Up to Rs 6,00,000 Rs 6,00,001 to Rs 8,00,000 Rs 8,00,001 to Rs 11,00,000 20% Rs 11,00,001 to Rs 20,00,000 30%

a) Mr. Baral has a stationery shop. His annual income is Rs 6,40,000. If he is unmarried, how much income tax should he pay? 10%​

The sum of the series Σn=1 to ∞ n(n-1)x^(n) is:

(2x^2(1-x)^3 + 6x^3(1-x)^2)/(1-x)^6

We can differentiate both sides of the equation 1/(1-x)^2 = Σn=1 to ∞ nx^(n-1) with respect to x to obtain:

[1/(1-x)^2]' = [Σn=1 to ∞ nx^(n-1)]'

Then, using the power rule of differentiation, we get:

2/(1-x)^3 = Σn=1 to ∞ n(n-1)x^(n-2)

Multiplying both sides by x, we obtain:

2x/(1-x)^3 = Σn=1 to ∞ n(n-1)x^(n-1)

Differentiating both sides of the equation 2x/(1-x)^3 = Σn=1 to ∞ n(n-1)x^(n-1) with respect to x, we obtain:

[2x/(1-x)^3]' = [Σn=1 to ∞ n(n-1)x^(n-1)]'

Using the power rule of differentiation, we get:

(2(1-x)^3 + 6x(1-x)^2)/(1-x)^6 = Σn=1 to ∞ n(n-1)x^(n-2)

Multiplying both sides by x^2, we obtain:

(2x^2(1-x)^3 + 6x^3(1-x)^2)/(1-x)^6 = Σn=1 to ∞ n(n-1)x^(n)

Therefore, the sum of the series Σn=1 to ∞ n(n-1)x^(n) is:

(2x^2(1-x)^3 + 6x^3(1-x)^2)/(1-x)^6

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The answer as a fraction, decimal, and percent is 3/10, 0.3, and 30%, respectively.

The table on air travel outside of the airport is not provided in the question. However, to answer the question, we can assume that the table contains information about flight arrivals and departure times.In order to determine if a flight arrived on time, we need to know the scheduled arrival time and the actual arrival time. If the actual arrival time is later than the scheduled arrival time, then the flight is considered delayed. If the actual arrival time is earlier than the scheduled arrival time, then the flight is considered early. If the actual arrival time is the same as the scheduled arrival time, then the flight is considered on time.To find the percentage of flights that arrive on time, we need to divide the number of on-time flights by the total number of flights and then multiply by 100. For example, if there are 200 flights and 140 of them arrived on time, then the percentage of flights that arrived on time would be:

(140/200) x 100 = 70%

To find the percentage of flights that did not arrive on time, we need to subtract the percentage of on-time flights from 100. For example, if the percentage of on-time flights is 70%, then the percentage of flights that did not arrive on time would be:

100 - 70 = 30%

Therefore, the answer as a fraction, decimal, and percent is 3/10, 0.3, and 30%, respectively.

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The rejection region is given by: {F(x) ≤ c} ∪ {F(x) ≥ 1 - c} which is of the form {x ≤ x0} ∪ {x ≥ x1}, where x0 and x1 are determined by c.

To show that the rejection region is of the form {x ≤ x0} ∪ {x ≥ x1}, we can use the fact that the critical value c divides the sampling distribution of the test statistic into two parts, the rejection region and the acceptance region.

Let F(x) be the cumulative distribution function (CDF) of the test statistic. By definition, the rejection region consists of all values of the test statistic for which F(x) ≤ c or F(x) ≥ 1 - c.

Since the sampling distribution is symmetric about the mean under the null hypothesis, we have F(-x) = 1 - F(x) for all x. Therefore, if c is the critical value, then the rejection region is given by:

{F(x) ≤ c} ∪ {1 - F(x) ≤ c}

= {F(x) ≤ c} ∪ {F(-x) ≥ 1 - c}

= {F(x) ≤ c} ∪ {F(x) ≥ 1 - c}

This shows that the rejection region is of the form {x ≤ x0} ∪ {x ≥ x1}, where x0 and x1 are determined by c. Specifically, x0 is the value such that F(x0) = c, and x1 is the value such that F(x1) = 1 - c.

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The distance from the plane 10x y z=90 to the plane 10x y z=70, we need to find the distance between a point on one plane and the other plane. The distance from the plane 10x y z=90 to the plane 10x y z=70 is 10sqrt(2) units.

Take the point (0,0,9) on the plane 10x y z=90.
The distance between a point and a plane can be found using the formula:
distance = | ax + by + cz - d | / sqrt(a^2 + b^2 + c^2)
where a, b, and c are the coefficients of the x, y, and z variables in the plane equation, d is the constant term, and (x, y, z) is the coordinates of the point.
For the plane 10x y z=70, the coefficients are the same, but the constant term is different, so we have:
distance = | 10(0) + 0(0) + 10(9) - 70 | / sqrt(10^2 + 0^2 + 10^2)
distance = | 20 | / sqrt(200)
distance = 20 / 10sqrt(2)
distance = 10sqrt(2)
Therefore, the distance from the plane 10x y z=90 to the plane 10x y z=70 is 10sqrt(2) units.

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Answer:

k = [tex]\frac{12}{5}[/tex]

Step-by-step explanation:

[tex]\frac{12}{5}[/tex] = [tex]\frac{1}{6k}[/tex] ( cross- multiply )

72k = 5 ( divide both sides by 72 )

k = [tex]\frac{5}{72}[/tex]

Answer: k=8.4 or 42/5

Step-by-step explanation: to find k you take 1 2/5 and divide it by 1/6. When I did it I got 8.4. To check my work I replaced the variable in the equation and it was correct.

Gerry's decision will depend on a variety of factors, including the recommendations of his friends, the course syllabus, and his own personal preferences. It is important for him to carefully consider all of these factors before making his final decision.

Gerry is registering for classes next semester and he is deciding between two teachers, Dr. Anderson and Dr. Bean. In order to make an informed decision, Gerry speaks to 17 friends that previously took the course from Dr. Anderson and 17 friends that took it from Dr. Bean. Out of the 17 friends that took Dr. Anderson's course, 8 highly recommend him. Out of the 17 friends that took Dr. Bean's course, 11 highly recommend him.
Based on the recommendation of his friends, Gerry may be inclined to choose Dr. Bean, as he received more highly positive recommendations than Dr. Anderson. However, there are other factors that Gerry may want to consider before making his final decision. For example, Gerry may want to look at the syllabus for each course and compare them to see which one would be a better fit for his academic goals. He may also want to look at the times that each course is offered to see which one fits best with his schedule. Additionally, he may want to read reviews of both professors on websites such as Rate My Professor to see what other students have said about their teaching styles.
Ultimately, Gerry's decision will depend on a variety of factors, including the recommendations of his friends, the course syllabus, and his own personal preferences. It is important for him to carefully consider all of these factors before making his final decision.

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In both the cases when n is even and when n is odd, the expression is odd, we can conclude that if n ∈ Z, then [tex]5n^2 + 3n + 7[/tex]is odd.

To prove the proposition "if n ∈ Z, then[tex]5n^2 + 3n + 7[/tex]is odd" using a direct proof with cases, we consider two cases: when n is even and when n is odd.

Case 1: n is even.

Assume n = 2k, where k ∈ Z. Substituting this into the expression, we have [tex]5(2k)^2 + 3(2k) + 7 = 20k^2 + 6k + 7[/tex]. Notice that [tex]20k^2[/tex] and 6k are both even since they can be factored by 2. Adding an odd number (7) to an even number results in an odd number. Hence, the expression is odd when n is even.

Case 2: n is odd.

Assume n = 2k + 1, where k ∈ Z. Substituting this into the expression, we have [tex]5(2k + 1)^2 + 3(2k + 1) + 7 = 20k^2 + 16k + 15[/tex]. Again, notice that [tex]20k^2[/tex]and 16k are even. Adding an odd number (15) to an even number results in an odd number. Therefore, the expression is odd when n is odd.

Since we have covered all possible cases and in each case, the expression is odd, we can conclude that if n ∈ Z, then 5n^2 + 3n + 7 is odd.

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Answer:

the answer is B 25 calories/1 ounce

explanation:

6 ounce/150 calories = X/ 1 calories

= 25/1

The derivative of h(x) with respect to x, evaluated at x = 5, is h'(5) = 13.

To find h'(5) if h(x) = r(x) s(x), we need to differentiate the function h(x) with respect to x and evaluate it at x = 5.

Using the product rule, we differentiate h(x) as follows:

h'(x) = r'(x) s(x) + r(x) s'(x)

Now, let's substitute the given values into the equation:

r(5) = 4, s(5) = 3, r'(5) = -1, and s'(5) = 4.

h'(x) = r'(x) s(x) + r(x) s'(x)

h'(5) = r'(5) s(5) + r(5) s'(5)

Plugging in the values, we get:

h'(5) = (-1)(3) + (4)(4)

h'(5) = -3 + 16

h'(5) = 13

Therefore, the derivative of h(x) with respect to x, evaluated at x = 5, is h'(5) = 13.

In simpler terms, h'(5) represents the rate of change of the function h(x) at x = 5. In this case, h(x) is the product of two functions, r(x) and s(x). By applying the product rule, we differentiate each function and multiply them together. Substituting the given values, we find that h'(5) equals 13. This means that at x = 5, the function h(x) is changing at a rate of 13 units per unit change in x.

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The number of students that are science majors can be thought of as a binomial random variable because:

1. There are a fixed number of trials (students) in the sample.
2. Each trial (student) has only two possible outcomes: being a science major or not being a science major.
3. The probability of success (being a science major) remains constant for each trial (student).
4. The trials (students) are independent of each other, meaning the outcome for one student does not affect the outcomes of the other students.

These four characteristics satisfy the conditions of a binomial random variable, which is why the number of science majors among a group of students can be modeled using a binomial distribution.

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Given, the rectangular pool of 4m in width and 10m in length. A grass border of width x is to be placed around the pool as shown below.

[tex]\overline{A'B'}=\overline{CD}=10+x\;\;\;\;

and

\;\;\;\;\overline{A'D'}=\overline{CB}=4+x[/tex]

So, the length of the rectangular pool along with the grass border on either side becomes

10 + x + 10 + x = 20 + 2x

and the width becomes

4 + x + 4 + x = 8 + 2x.

Total Area of the rectangular pool with grass border

= 112m²

Thus, we get an equation as;

Area of the rectangular pool with grass border = Area of pool + Area of grass border[tex](20+2x)(8+2x)=40+20x+16x+4x^2=112[/tex][tex]\

Rightarrow 4x^2 + 36x - 72 = 0[/tex]

Now, we have to solve the above quadratic equation to find the value of x.

On solving we get;

x = 3m or x = -6m

Since x cannot be negative, the only valid solution is x = 3m.

Hence, option (D) x² + 7x = 18 allows us to determine the value of x.

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The value of x is -sqrt(55)/8.

Let's use the Pythagorean theorem to find the value of x.

Since P is on the unit circle, we know that the distance from the origin to P is 1. Let's call the x-coordinate of P "x".

We can use the Pythagorean theorem to write:

x^2 + (-3/8)^2 = 1^2

Simplifying, we get:

x^2 + 9/64 = 1

Subtracting 9/64 from both sides, we get:

x^2 = 55/64

Taking the square root of both sides, we get:

x = ±sqrt(55)/8

Since P is in quadrant III, we know that x is negative. Therefore,

x = -sqrt(55)/8

So the value of x is -sqrt(55)/8.

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The composition R2(R1(x)) is a rotation about the center C with angle of rotation given by the angle between the vectors P-Q and R2(R1(P))-C.

Lemma 10.3.3 states that any rigid motion of the plane is either a translation a rotation about a fixed point or a reflection across a line.

To prove that the composition of two rotations with the same center is a rotation can use the following argument:

Let R1 and R2 be two rotations with the same center C and let theta1 and theta2 be their respective angles of rotation.

Without loss of generality can assume that R1 is applied before R2.

By Lemma 10.3.3 know that any rotation about a fixed point is a rigid motion of the plane.

R1 and R2 are both rigid motions of the plane and their composition R2(R1(x)) is also a rigid motion of the plane.

The effect of R1 followed by R2 on a point P in the plane. Let P' be the image of P under R1 and let P'' be the image of P' under R2.

Then, we have:

P'' = R2(R1(P))

= R2(P')

Let theta be the angle of rotation of the composition R2(R1(x)).

We want to show that theta is also a rotation about the center C.

To find a point Q in the plane that is fixed by the composition R2(R1(x)).

The angle of rotation theta must be the angle between the line segment CQ and its image under the composition R2(R1(x)).

Let Q be the image of C under R1, i.e., Q = R1(C).

Then, we have:

R2(Q) = R2(R1(C)) = C

This means that the center C is fixed by the composition R2(R1(x)). Moreover, for any point P in the plane, we have:

R2(R1(P)) - C = R2(R1(P) - Q)

The right-hand side of this equation is the image of the vector P-Q under the composition R2(R1(x)).

The composition R2(R1(x)) is a rotation about the center C angle of rotation given by the angle between the vectors P-Q and R2(R1(P))-C.

The composition of two rotations with the same center is a rotation about that center.

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The 95% confidence interval is 1.23 to 1.51

From the question, we have the following parameters that can be used in our computation:

Sample, n = 22

Mean, x = 1.37 oz

Standard deviation, s = 0.33 oz

Start by calculating the margin of error using

E = s/√n

So, we have

E = 0.33/√22

E = 0.07

The 95% confidence interval is

CI = x ± zE

Where

z = 1.96 i.e. z-score at 95% CI

So, we have

CI = 1.37 ± 1.96 * 0.07

Evaluate

CI = 1.37 ± 0.14

This gives

CI = 1.23 to 1.51

Hence, the 95% confidence interval is 1.23 to 1.51

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If one score in a correlational study is numerical and the other is non-numerical, the non-numerical variable can be used to organize the scores into separate groups which can then be compared with a (d) chi-square hypothesis test.

A chi-square hypothesis test can be used to analyze the relationship between a numerical and a non-numerical variable in a correlational study where the non-numerical variable is used to group the scores.

This test is used to determine whether there is a significant association between the two variables.

The other options, t-test, mixed-design analysis of variance, and single factor analysis of variance, are statistical tests that are used for different types of research designs and are not appropriate for analyzing the relationship between a numerical and non-numerical variable in a correlational study.

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Given the stock is purchased for $72.50 and it is expected that each day the stock will decrease by $0.05.

Let x = time (in days) and

P(x) = stock price (in $).

To find how many days it will take for the stock price to be equal to $65, we need to solve for x such that P(x) = 65.So, the equation of the stock price is

: P(x) = 72.50 - 0.05x

We have to solve the equation P(x) = 65. We have;72.50 - 0.05

x = 65

Subtract 72.50 from both sides;-0.05

x = 65 - 72.50

Simplify;-0.05

x = -7.50

Divide by -0.05 on both sides;

X = 150

Therefore, it will take 150 days for the stock price to be equal to $65

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The average answer time for the entire day on Monday was 15.9 seconds.

For the average answer time for the entire day on Monday, we need to calculate the total time spent answering calls and divide by the total number of calls answered.

Let's first calculate the total time spent answering calls:

- For the 500 calls before noon, the total time spent answering calls is: 500 x 16 = 8,000 seconds

- For the 350 calls from noon to 6pm, the total time spent answering calls is: 350 x 14 = 4,900 seconds

- For the 150 calls on Monday night, the total time spent answering calls is: 150 x 20 = 3,000 seconds

So the total time spent answering calls on Monday is:

8,000 + 4,900 + 3,000 = 15,900 seconds

Now, let's calculate the total number of calls answered:

500 + 350 + 150 = 1,000 calls

Finally, we can calculate the average answer time for the entire day on Monday:

Average answer time = Total time spent answering calls / Total number of calls answered

= 15,900 / 1,000

= 15.9 seconds

Therefore, the average answer time for the entire day on Monday was 15.9 seconds.

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