21 20 15 22 19 23 17 what is the median and range

The two cars must be approximately 177.34 feet apart from each other for Spiderman to have different depression angles to each car.

To find the distance between the two cars, we can use trigonometry and the concept of similar triangles. Let's denote the distance between Spiderman and the nearest car as d1 and the distance between Spiderman and the farther car as d2.

In a right triangle formed by Spiderman, the height of the hotel, and the line of sight to the nearest car, the tangent of the depression angle (52 degrees) can be used:

tan(52) = 600 / d1

Rearranging the equation to solve for d1:

d1 = 600 / tan(52)

Similarly, in the right triangle formed by Spiderman, the height of the hotel, and the line of sight to the farther car, the tangent of the depression angle (46 degrees) can be used:

tan(46) = 600 / d2

Rearranging the equation to solve for d2:

d2 = 600 / tan(46)

Using a calculator, we can compute:

d1 ≈ 504.61 feet

d2 ≈ 681.95 feet

The distance between the two cars is the difference between d2 and d1:

Distance = d2 - d1

Plugging in the values, we have:

Distance ≈ 681.95 - 504.61

Distance ≈ 177.34 feet

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Answer:

B

Step-by-step explanation:

the secant- secant angle LMN is half the difference of the measures of the intercepted arcs , that is

∠ LMN = [tex]\frac{1}{2}[/tex] ( KP - LN)

20° = [tex]\frac{1}{2}[/tex] (96 - LN) ← multiply both sides by 2 to clear the fraction

40° = 96° - LN ( subtract 96° from both sides )

- 56° = - LN ( multiply both sides by - 1 )

56° = LN

Therefore, Heather can make a maximum of 4 biscuits with the given ingredient quantities.

To determine how many biscuits Heather can make, we need to compare the amount of each ingredient required for a single biscuit to the total amount of ingredients she has.

Let's calculate the number of biscuits she can make based on the ingredient quantities provided:

First, we need to find the ratio of each ingredient required per biscuit:

Flour: 100g per biscuit

Butter: 120g per biscuit

Icing sugar: 80g per biscuit

Chocolate: 25g per biscuit

Next, we divide the total amount of each ingredient by the respective ratio to find the maximum number of biscuits she can make:

Flour: 2kg / 100g = 20 biscuits

Butter: 1kg / 120g = 8.33 biscuits (approximately)

Icing sugar: 340g / 80g = 4.25 biscuits (approximately)

Chocolate: 200g / 25g = 8 biscuits

Since we can only make a whole number of biscuits, the limiting factor is the icing sugar. Heather can only make 4 biscuits since she has 4.25 times the required amount of icing sugar.

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Answer: OC. The cost of packaging per weight unit is given by x / 3.

To find the cost of packaging per weight unit, we need to calculate the cost of packaging (given by x^2) divided by the weight of the box (given by x + 2).

Let's substitute x + 2 for the weight in the cost function:

Cost of packaging per weight unit = (Cost of packaging) / (Weight of the box)

= (x^2) / (x + 2)

Now, let's simplify this expression:

Cost of packaging per weight unit = x^2 / (x + 2)

To further simplify, we can divide both the numerator and denominator by x:

Cost of packaging per weight unit = (x * x) / (x * (1 + 2))

= x / (1 + 2)

= x / 3

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The x variable represents the number of tickets sold.

The y variable represents the total amount of money raised from ticket sales.

The equation for the scenario is y = 2.50x.

In the given scenario, the number of tickets sold is represented by the variable x, and the total amount of money raised from ticket sales is represented by the variable y. Since each ticket is priced at $2.50, the equation relating the number of tickets sold (x) and the total amount of money raised (y) is y = 2.50x.

Here's an explanation of each element in the response:

1. The x variable represents the number of tickets sold: This statement correctly identifies the variable x as representing the number of tickets sold. In the equation y = 2.50x, x represents the independent variable, which is the quantity we want to determine.

2. The y variable represents the total amount of money raised from ticket sales: This statement correctly identifies the variable y as representing the total amount of money raised from ticket sales. In the equation y = 2.50x, y represents the dependent variable, which is determined based on the value of x.

3. The equation for the scenario is y = 2.50x: This equation is derived from the given information that each ticket is priced at $2.50. Multiplying the price per ticket by the number of tickets sold gives the total amount of money raised, which is represented by y in the equation.

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The explicit formula for the sequence 12, 112, 212, 312, 412 is a_n = 100n + 12.

The explicit formula for the given sequence is:

a_n = 100n + 12

In the given sequence, each term is obtained by adding 100 to the previous term. The first term is 12, and each subsequent term is obtained by adding 100 to the previous term.

Using the formula, we can calculate any term in the sequence by substituting the corresponding value of n. For example:

a_1 = 100(1) + 12 = 112

a_2 = 100(2) + 12 = 212

a_3 = 100(3) + 12 = 312

a_4 = 100(4) + 12 = 412

Therefore, the explicit formula for the sequence 12, 112, 212, 312, 412 is a_n = 100n + 12.

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The conclusions that can be drawn from the data shown in the table are:

- Twelve tickets cost $30.00.

- Each additional ticket costs $2.50.

- The table is a partial representation.

- (27.50, 11), (30, 12), and (32.50, 13) are the ordered pairs represented in the table.

The statement "(27.50, 11), (30, 12), and (32.50, 13) are the ordered pairs represented in the table" is correct. From the data shown in the table, we can draw the following conclusions:

1. Twelve tickets cost $30.00: Looking at the "Tickets" column, we can see that the entry "12" corresponds to the "Total Cost" entry of $30.00. Therefore, we can conclude that purchasing twelve tickets would cost $30.00.

2. Each additional ticket costs $2.50: By examining the "Tickets" column, we can observe that for each increase of one ticket (from 11 to 12, and from 12 to 13), the "Total Cost" in the second column increases by $2.50. This consistent pattern suggests that each additional ticket costs $2.50.

3. The table is a partial representation: The table only displays three rows of data, showing the "Tickets" and "Total Cost" for x values of 11, 12, and 13. Since the table does not provide information for all possible values of x, it is a partial representation of the relationship between the number of tickets and the total cost.

The statement "Thirty tickets cost $12.00" is not supported by the given data. The table does not include an entry for 30 tickets, and none of the given entries correspond to that value.

These ordered pairs match the values shown in the table, with the first element representing the number of tickets (x) and the second element representing the total cost (y) in dollars.

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The answer is:

45÷3[90÷2{6+3(19+16)}]

= 45÷3[90÷2*12] (Simplifying inside brackets first)

= 45÷3[90÷24]

=45÷3 *3

= 135

Therefore, the correct answer is: 135

The equation is true when x = 9,x = 9 is a solution to this equation.

To determine which equations have x = 9 as a possible solution, we need to check each equation individually. Here are the equations to consider:

3x - 18 = 15

Substituting x = 9, we have:

3(9) - 18 = 15

27 - 18 = 15

9 = 15

The equation is not true when x = 9. Therefore, x = 9 is not a solution to this equation.

2(x + 4) = 26

Substituting x = 9, we have:

2(9 + 4) = 26

2(13) = 26

26 = 26

The equation is true when x = 9. Therefore, x = 9 is a solution to this equation.

5x + 3 = 2x + 30

Substituting x = 9, we have:

5(9) + 3 = 2(9) + 30

45 + 3 = 18 + 30

48 = 48

The equation is true when x = 9

Based on the analysis, x = 9 is a possible solution for equations 2 and 3.

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The cost of packaging per weight unit is x + 1.

To derive this answer, we first need to understand the given information. The cost of packing a box of chocolates is given by [tex]x^2[/tex], where x is the number of chocolates. However, we also know that a box can never have fewer than 3 chocolates.

Now, let's calculate the weight of a box of chocolates. It is given by x + 2.

To find the cost of packaging per weight unit, we need to divide the cost of packing by the weight of the box. Therefore, the cost of packaging per weight unit can be calculated as ([tex]x^2[/tex]) / (x + 2).

Simplifying this expression, we can rewrite it as ([tex]x^2[/tex]) / (x + 2) = (x + 1) - 1 / (x + 2).

Hence, the cost of packaging per weight unit is x + 1, which means that for every unit of weight, the cost of packaging is x + 1.

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Answer:

The 2nd option is correct, a 90 degree counterclockwise rotation about the origin and then a reflection across the y-axis

Step-by-step explanation:

Using chain rule, the derivative of f(x) = 4√(8x³ + 2x⁴) is

f'(x) = (2/√(8x³ + 2x⁴)) * (24x² + 8x³)

To find the derivative of the function f(x) = 4√(8x³ + 2x⁴), we can use the chain rule.

Let's break down the function into its components:

u(x) = 8x³ + 2x⁴ (inside function)

v(u) = 4√u (outer function)

To find the derivative, we apply the chain rule, which states:

(f(g(x)))' = f'(g(x)) * g'(x)

In this case, f(g(x)) = v(u(x)), and g(x) = u(x).

First, let's find the derivative of the inner function u(x):

u'(x) = 24x² + 8x³ (using the power rule for differentiation)

Next, let's find the derivative of the outer function v(u):

v'(u) = 4 * (1/2) * 1/√u = 1/√2u = 2/√u

Now, we can apply the chain rule:

f'(x) = v'(u(x)) * u'(x)

f'(x) = (2/√u) * (24x² + 8x³)

f'(x) = (2/√(8x³ + 2x⁴)) * (24x² + 8x³)

The derivative of the function is  (2/√(8x³ + 2x⁴)) * (24x² + 8x³)

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The 98% confidence interval for the true mean statistics anxiety score (μ) is approximately (39.037, 39.763).

To calculate the 98% confidence interval for the true mean statistics anxiety score (μ) for all undergraduate students in the U.S., we can use the formula:

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

First, we need to find the critical value associated with a 98% confidence level. Since we are assuming a normal distribution, we can use the Z-table or a statistical software to find this value. For a 98% confidence level, the critical value is approximately 2.33.

Next, we calculate the standard error (SE) using the formula:

SE = standard deviation / √sample size

In this case, the standard deviation is 0.9 and the sample size is 33. Plugging these values into the formula, we get: SE = 0.9 / √33 ≈ 0.156

Now, we can calculate the confidence interval:Confidence interval = 39.4 ± (2.33 * 0.156)

Simplifying the expression:Confidence interval ≈ 39.4 ± 0.363

This gives us the interval (39.037, 39.763). This means we are 98% confident that the true mean statistics anxiety score for all undergraduate students in the U.S. falls within this interval.

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Average speed of the humpback whale: 27 kilometers per hour.

To calculate the average speed of the humpback whale, we can use the formula:

Average speed = Total distance / Total time

Given that the humpback whale traveled a distance of 13.5 kilometers in 30 minutes, we need to convert the time to hours. There are 60 minutes in an hour, so 30 minutes is equal to 0.5 hours.

Now, we can substitute the values into the formula:

Average speed = 13.5 kilometers / 0.5 hours

Average speed = 27 kilometers per hour

Therefore, the average speed of the humpback whale is 27 kilometers per hour.

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Rhe point-slope equation of a line with the labeled point (3, -4) is y + 4 = m(x - 3), where 'm' represents the slope of the line.

To find the point-slope equation of a line using the coordinates of a labeled point, you can use the following formula:
y - y₁ = m(x - x₁)
In this formula, (x₁, y₁) represents the coordinates of the labeled point, and m represents the slope of the line.
Given the coordinates (3, -4) of the labeled point, we can substitute these values into the formula:
y - (-4) = m(x - 3)
Simplifying this equation, we get:
y + 4 = m(x - 3)
This is the point-slope equation of the line.
Now, it's important to note that the problem does not provide information about the slope of the line. Therefore, we cannot determine the exact point-slope equation without knowing the slope. The point-slope equation requires the slope value to be defined.
If you have the slope of the line, let's say it is represented by the variable 'm', you can substitute that value into the equation to get the specific point-slope equation. For example, if the slope is 2, the equation becomes:
y + 4 = 2(x - 3)

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The sum diverges to negative infinity.

Here we want to find the value of the sum:

[tex]\sum_{m=1}^{ \infty}} (-11/2)*(3/2)^{m + 1}[/tex]

So, that sum goes for infinite values of m, that is bad because you can see that the term with an exponent is larger than 1.

So when m is a really large value, then the term will also be a really large value, which means that the fuction eventually diverges to negative inifnity.

The usual rule that we need to check is that, for large values of m, as m increases, the absolute value of each term decreases.

Here this cleraly does not happen, so the sum diverges.

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Rounding to the nearest whole number, the work required to pump the oil out of the spout is approximately 5,068,032π J.

To find the work required to pump the oil out of the spout, we need to consider the potential energy of the oil. The work done is equal to the change in potential energy.The potential energy of an object is given by the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

Given that the density of the oil is 900 kg/m^3 and the tank is half full, we can determine the mass of the oil. The volume of the tank is calculated using the formula for the volume of a cylinder: V = πr^2h, where r is the radius and h is the height.

The volume of the tank is (1/2)π(12^2)(4) = 288π m^3.

Since the oil is half full, the volume of the oil is (1/2)(288π) m^3.

The mass of the oil is the density multiplied by the volume:

m = (900 kg/m^3)(1/2)(288π m^3) = 129,600π kg.

The height of the oil is 4 m.

Now, we can calculate the potential energy:

PE = mgh = (129,600π kg)(9.8 m/s^2)(4 m) = 5,068,032π J.

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Although the chef used fresh ingredients, Karen knew the pasta dish was tasty but not healthy.Type of Parallel Structure: Parallelism using correlative conjunctions

Every year, I go on a long hiking trip where I like to take a break away from the hustle of the city and enjoy the peacefulness within nature.

Type of Parallel Structure: Parallelism in a series of items

Derek enjoys playing baseball with his friends, going on camping trips with his dad, and traveling to different cities throughout the year.

Type of Parallel Structure: Parallelism in a series of items

I like playing hockey more than I like to play soccer.

Type of Parallel Structure: Parallelism in a comparative sentence

Jonathan enjoys watching comedy at the movie theater more than he likes watching horror films at the movie theater.

Type of Parallel Structure: Parallelism in a comparative sentence

When I go to the park, I like bringing a blanket and to pack a picnic basket full of sandwiches and fruit.

Type of Parallel Structure: Parallelism using the same verb tense

In these sentences, the type of parallel structure used in each sentence has been matched correctly.

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Answer:

[tex]\textsf{8.} \quad \textsf{(A)}\;\;\overline{XB}[/tex]

[tex]\textsf{9.} \quad \textsf{(D)}\;\;\overleftrightarrow{BD}[/tex]

Step-by-step explanation:

The radius is the distance from the center of a circle to any point on its circumference.

The center of the given circle is point X.

Therefore, the radii in the given circle are line segments XB, XA and XC.

[tex]\hrulefill[/tex]

A tangent is a straight line that touches a circle at only one point.

The line BD touches the circle at point B.

Therefore, the tangent of the given circle is line BD.

Answer:

8. A

9. D

Step-by-step explanation:

The radius is a straight line from the midpoint to the circle's circumference.

A Tangent is a line going through the circumference of the circle.

Statement 1: In triangle ADC and BCD, AD = BC (opposite sides of a rectangle are congruent).

Statement 2: Angle ADC = Angle BCD (angles of a rectangle are 90°).

Statement 3: DC = DC (reflexive property of equality).

Statement 4: Triangle ADC and BCD are congruent (by SAS postulate).

Statement 5: AC = BD (by CPCTC).

The statement that completes Jimmy's proof is:

DC = DC (reflexive property of equality)

The reflexive property of equality states that any quantity is equal to itself. In this case, statement 3 is stating that the diagonal DC is equal to itself, which is true by the reflexive property of equality.

Therefore, the completed proof is:

Statement 1: In triangle ADC and BCD, AD = BC (opposite sides of a rectangle are congruent).

Statement 2: Angle ADC = Angle BCD (angles of a rectangle are 90°).

Statement 3: DC = DC (reflexive property of equality).

Statement 4: Triangle ADC and BCD are congruent (by SAS postulate).

Statement 5: AC = BD (by CPCTC).

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The amount in the account right after the last deposit, rounded to the nearest dollar, is approximately $3,982.

To calculate the amount in the account right after the last deposit, we need to use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:

A = the final amount

P = the principal (initial deposit)

r = annual interest rate (in decimal form)

n = number of times interest is compounded per year

t = number of years

In this case, the principal is $2,580, the annual interest rate is 3% (0.03 in decimal form), the interest is compounded biannually (twice per year), and the total duration is 15 years.

First, let's calculate the number of compounding periods:

Since interest is compounded twice a year, and there are 15 years, the total number of compounding periods is 15 * 2 = 30.

Now, we can plug the values into the formula:

A = $2,580(1 + 0.03/2)^(2*30)

A = $2,580(1 + 0.015)^60

A = $2,580(1.015)^60

Using a calculator, we find that (1.015)^60 ≈ 1.545.

A = $2,580 * 1.545

A ≈ $3,982.10

Therefore, the amount in the account right after the last deposit, rounded to the nearest dollar, is approximately $3,982.

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The whole number that has no predecessor is 1.

499 is to the left of 500 on a number line.

The successor of the greatest 5-digit number is 100,000.

The additive identity is 0.

The number of whole numbers is infinite.

1 is called the multiplicative identity.

The result of (77) × 99 is 7,623.

The smallest whole number is 0.

The greatest two-digit number exactly divisible by 18 is 90.

The greatest 7-digit number using the digits 4, 6, and 9 with repetition is 999,9999.

Rearranging the numbers using the property of addition:

7326 + 139 + 674 + 861 = (7326 + 861) + 674 + 139 = 8187 + 674 + 139 = 9,000 + 674 + 139 = 9,813.

Finding the product using the distributive property:

798 x 998 = (700 + 90 + 8) x (900 + 90 + 8) = 700 x 900 + 700 x 90 + 700 x 8 + 90 x 900 + 90 x 90 + 90 x 8 + 8 x 900 + 8 x 90 + 8 x 8 = 630,000 + 63,000 + 5,600 + 81,000 + 8,100 + 720 + 7,200 + 720 + 64

= 1,448,504.

The largest 6-digit number exactly divisible by 45 is 999,990.

Simplifying the expression:

75 - [30 + (3 x (18 ÷ 6))] = 75 - [30 + (3 x 3)] = 75 - [30 + 9] = 75 - 39 = 36.

Ramesh buys 15 computers and 15 printers.

Cost of one computer = Rs. 75,326

Cost of one printer = Rs. 8,265

Using the distributive property of multiplication:

Total cost = (Cost of one computer x Number of computers) + (Cost of one printer x Number of printers)

Total cost = (Rs. 75,326 x 15) + (Rs. 8,265 x 15)

Total cost = Rs. 1,129,890 + Rs. 123,975

Total cost = Rs. 1,253,865.

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Using simultaneous equation, the solution to the system of linear equations are 1223 $10 tickets, 1332 $20 tickets, and 763 $30 tickets were sold.

Let's solve the problem step by step.

Let:

x = number of $10 tickets sold

y = number of $20 tickets sold

z = number of $30 tickets sold

From the given information, we can form the following equations:

Equation 1: x + y + z = 3318 (Total number of tickets sold)

Equation 2: y = x + 109 (109 more $20 tickets than $10 tickets were sold)

Equation 3: 10x + 20y + 30z = 61760 (Total sales from ticket sales)

We can use these three equations to solve for the values of x, y, and z.

First, let's substitute Equation 2 into Equation 1:

x + (x + 109) + z = 3318

2x + 109 + z = 3318

2x + z = 3209 (Equation 4)

Now, let's substitute the value of y from Equation 2 into Equation 3:

10x + 20(x + 109) + 30z = 61760

10x + 20x + 2180 + 30z = 61760

30x + 30z = 59580

x + z = 1986 (Equation 5)

We now have a system of equations (Equations 4 and 5) with two variables (x and z). We can solve this system to find the values of x and z.

Multiplying Equation 4 by 30, and Equation 5 by 2, we get:

60x + 30z = 96270 (Equation 6)

2x + 2z = 3972 (Equation 7)

Now, subtract Equation 7 from Equation 6:

(60x + 30z) - (2x + 2z) = 96270 - 3972

58x + 28z = 92298

Simplifying, we have:

29x + 14z = 46149 (Equation 8)

Now, we can solve Equations 5 and 8 simultaneously:

x + z = 1986 (Equation 5)

29x + 14z = 46149 (Equation 8)

Multiplying Equation 5 by 14, and Equation 8 by 1, we get:

14x + 14z = 27804 (Equation 9)

29x + 14z = 46149 (Equation 8)

Now, subtract Equation 9 from Equation 8:

(29x + 14z) - (14x + 14z) = 46149 - 27804

15x = 18345

Divide both sides of the equation by 15:

x = 18345 / 15

x = 1223

Substituting the value of x into Equation 5, we can find z:

1223 + z = 1986

z = 1986 - 1223

z = 763

Now that we have the values of x and z, we can substitute them back into Equation 1 to find y:

1223 + y + 763 = 3318

y + 1986 = 3318

y = 3318 - 1986

y = 1332

Therefore, the solution to the problem is:

x = 1223 (number of $10 tickets sold)

y = 1332 (number of $20 tickets sold)

z = 763 (number of $30 tickets sold)

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The density of Liquid A is equal to the density of Liquid C when they are mixed in the specified ratio.

To determine the density of Liquid C, we need to find the mass and volume of Liquid C and then calculate the density by dividing the mass by the volume.

Given that the density of Liquid A is in a 4:3 ratio with the density of Liquid B, and the mass of Liquid A is in a 5:2 ratio with the mass of Liquid B, we can assume that the ratio of their volumes is also 5:2. This is because density is the ratio of mass to volume.

When 3 cans of Liquid B are mixed with 1 can of Liquid A to make Liquid C, the volume ratio remains the same. So, the volume of Liquid C would be 5 + 3 = 8 units.

Since the density is the mass divided by the volume, the density of Liquid A would remain the same. Therefore, the density of Liquid A is equal to the density of Liquid C.

In conclusion, the density of Liquid A is equal to the density of Liquid C.

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An example of a data set where the mean is equal to one of the numbers in the set is 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, with a mean of 11.

Here's an example of a data set where the mean is equal to one of the numbers in the set:

Data set: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20

In this data set, the mean (average) value is calculated by summing up all the numbers in the set and dividing by the total number of values. In this case, the sum of the numbers is 110, and since there are 10 numbers in the set, the mean is 110/10 = 11.

As we can see, the number 11 is present in the data set itself and coincidentally, it is also the mean value of the set. This happens because the other numbers are symmetrically distributed around the mean, balancing out to yield the same value.

It's important to note that this is just one example, and there can be various data sets where the mean matches one of the numbers. The occurrence of such a scenario depends on the values within the data set and their distribution.

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Question: Data set [2, 4, 6, 8, 10, 12, 14, 16, 18, 20]

Consider the data set provided above. Is there any number in the data set that is equal to the mean of the data set?

Tom's percentage score is above 75%, which is the passing threshold, we can conclude that Tom has indeed passed the assessment.

To determine if Tom has passed the assessment, we need to calculate his percentage score out of the total marks.

Percentage Score = (Tom's Score / Total Marks) * 100

Given that Tom's score is 98 marks and the total marks are 128:

Percentage Score = (98 / 128) * 100 ≈ 76.5625%

Since Tom's percentage score is above 75%, which is the passing threshold, we can conclude that Tom has indeed passed the assessment.

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Answer:

Refer to the step-by-step explanation, please follow along very carefully. Answers are encased in two boxes.

Step-by-step explanation:

Given the following function, find it's derivative using the definition of derivatives. Evaluate the function when θ=1, 11, and 3/11

[tex]p(\theta)=\sqrt{11\theta}[/tex]

[tex]\hrulefill[/tex]

The definition of derivatives states that the derivative of a function at a specific point measures the rate of change of the function at that point. It is defined as the limit of the difference quotient as the change in the input variable approaches zero.

[tex]f'(x) = \lim_{{h \to 0}} \dfrac{{f(x+h) - f(x)}}{{h}}[/tex][tex]\hrulefill[/tex]

To apply the definition of derivatives to this problem, follow these step-by-step instructions:

Step 1: Identify the function: Determine the function for which you want to find the derivative. In out case the function is denoted as p(θ).

[tex]p(\theta)=\sqrt{11\theta}[/tex]

Step 2: Write the difference quotient: Using the definition of derivatives, write down the difference quotient. The general form of the difference quotient is (f(x+h) - f(x))/h, where "x" is the point at which you want to find the derivative, and "h" represents a small change in the input variable. In our case:

[tex]p'(\theta) = \lim_{{h \to 0}} \dfrac{{p(\theta+h) - p(\theta)}}{{h}}\\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11(\theta + h)} - \sqrt{11\theta} }{h}[/tex]

Step 3: Take the limit:

We need to rationalize the numerator. Rewriting using radical rules.

[tex]p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11(\theta + h)} - \sqrt{11\theta} }{h} \\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11\theta + 11h} - \sqrt{11\theta} }{h}\\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11}\sqrt{\theta+h} - \sqrt{11}\sqrt{\theta} }{h}[/tex]

Now multiply by the conjugate.

[tex]p'(\theta)= \lim_{h \to 0} \dfrac{\sqrt{11}\sqrt{\theta+h} - \sqrt{11}\sqrt{\theta} }{h} \cdot \dfrac{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} }{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} } \\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{(\sqrt{11}\sqrt{\theta+h} - \sqrt{11}\sqrt{\theta} )(\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} )}{h(\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} )} \\\\\\[/tex]

[tex]\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{11h}{h(\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} )}\\\\\\\Longrightarrow p'(\theta)= \lim_{h \to 0} \dfrac{11}{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} }[/tex]

Step 4: Simplify the expression: Evaluate the limit by substituting the value of h=0 into the difference quotient. Simplify the expression as much as possible.

[tex]p'(\theta)= \lim_{h \to 0} \dfrac{11}{\sqrt{11}\sqrt{\theta+h} + \sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{\sqrt{11}\sqrt{\theta+(0)} + \sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{\sqrt{11}\sqrt{\theta} + \sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{2\sqrt{11}\sqrt{\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{2\sqrt{11\theta} }\\\\\\\Longrightarrow p'(\theta)= \dfrac{11}{2\sqrt{11\theta} }[/tex]

[tex]\therefore \boxed{\boxed{p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta} }}}[/tex]

Thus, we have found the derivative on the function using the definition.

It's important to note that in practice, finding derivatives using the definition can be a tedious process, especially for more complex functions. However, the definition lays the foundation for understanding the concept of derivatives and its applications. In practice, there are various rules and techniques, such as the power rule, product rule, and chain rule, that can be applied to find derivatives more efficiently.[tex]\hrulefill[/tex]

Now evaluating the function at the given points.  

[tex]p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}; \ p'(1)=??, \ p'(11)=??, \ p'(\frac{3}{11} )=??[/tex]

When θ=1:

[tex]p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}\\\\\\\Longrightarrow p'(1)= \dfrac{\sqrt{11} }{2\sqrt{1}}\\\\\\\therefore \boxed{\boxed{p'(1)= \dfrac{\sqrt{11} }{2}}}[/tex]

When θ=11:

[tex]p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}\\\\\\\Longrightarrow p'(11)= \dfrac{\sqrt{11} }{2\sqrt{11}}\\\\\\\therefore \boxed{\boxed{p'(11)= \dfrac{1}{2}}}[/tex]

When θ=3/11:

[tex]p'(\theta)= \dfrac{\sqrt{11} }{2\sqrt{\theta}}\\\\\\\Longrightarrow p'(\frac{3}{11} )= \dfrac{\sqrt{11} }{2\sqrt{\frac{3}{11} }}\\\\\\\therefore \boxed{\boxed{p'(\frac{3}{11} )= \dfrac{11\sqrt{3} }{6}}}[/tex]

Thus, all parts are solved.

The width of the cooling tower at the base is approximately 20 meters

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

[tex]\frac{x^2}{400} - \frac{(y - 110)^2}{2304} = 1[/tex]

The above equation is an equation of a hyperbols

The general equation for a hyperbola that has a center at (h, k) is

[tex]\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1[/tex]

Using the above as a guide, we have the following:

a² = 400

So, we have

a = 20

Hence, the width is 20 meters

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Question

Cooling towers are used to remove or expel heat from a process. A cooling towers walls are modeled by. x^2/400 - (y - 110)^2/2304 = 1 where the measurements are in meters

What is the width of the cooling tower at the base of the structure? Round your answer to the nearest whole number

The relationship shown in the diagram can be described as follows: For each input value, there is a corresponding output value. The output value is obtained by performing certain operations on the input value according to the rules specified in the diagram.

2.2.1: The relationship shown in the diagram represents a function where each input value corresponds to a specific output value. The diagram may include various operations or rules to transform the input values into their respective output values.

2.2.2: Using the information provided in the flow diagram, we can complete the table as follows:

  - For input 0, the output is 1.

  - For input 1, the output is 2.

  - For input 2, the output is 4.

  - For input 4, the output is LO.

  - For input 5, the output is 182.

  - For input 182, the output is 2.

  - For input 2, the output is 4.

  - For input -1, the output is 12-10.

  - For input 12-10, the output is 2.

2.2.3: To calculate the input value for the given output value of -21, we follow these steps:

  - Start with the output value -21.

  - Reverse the operations or rules specified in the diagram to transform the output back into the input.

  - Apply the reverse operations in the opposite order to obtain the input value.

Please note that without a specific diagram or additional information, it is challenging to provide precise steps for reversing the operations or rules. The steps may vary depending on the complexity and specifics of the diagram.

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The correct step that comes after constructing a circle when constructing an inscribed square is to use a straightedge to draw a diameter of the circle (option C).

When constructing an inscribed square by hand, the step that comes after constructing a circle is to use a straightedge to draw a diameter of the circle. Therefore, the correct answer is C.

To understand why drawing a diameter comes after constructing a circle, let's review the steps involved in constructing an inscribed square:

1. Start by constructing a circle: To do this, you would use a compass and a fixed point as the center to draw a circle.

2. Draw a diameter of the circle: A diameter is a line segment that passes through the center of the circle and divides it into two equal parts. Using a straightedge, you can draw a straight line that passes through the center of the circle.

3. Find the midpoint of the diameter: The midpoint is the point on the diameter that divides it into two equal parts. You can use a compass or measure the distance from each end of the diameter to find the midpoint.

4. Draw a perpendicular bisector: With the midpoint as the center, use a compass to draw an arc that intersects the diameter on both sides. This arc will create two points on the diameter.

5. Connect the points: Use a straightedge to connect the two points on the diameter. This line segment will be one side of the inscribed square.

6. Repeat the process: Repeat steps 2 to 5 to draw the other three sides of the square, using the circle as a guide.

By drawing a diameter of the circle, you establish a reference line that will be the base for constructing the sides of the inscribed square. It allows you to accurately position the square within the circle and ensure that its vertices lie on the circumference.

Therefore, the correct step that comes after constructing a circle when constructing an inscribed square is to use a straightedge to draw a diameter of the circle (option C).

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