Consider a population that grows according to the recursive rule Pn=Pn−1+50

, with initial population P0=30

The value of the volume of the cylinder is  706. 5 cm³

The formula for the volume of a cylinder is expressed as;

V = πr²h

Such that the parameters are;

Now, substitute the values into the formula, we get;

Volume, V = 3.14 × 5² × 9

Find the square value and substitute

Volume, V = 3.14 × 25 × 9

Multiply the values

volume, V = 706. 5 cm³

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Answer:

The total amount Joe spent on gasoline and oil can be determined by the expression: 2.85g + 3.15c. This expression represents the cost of g gallons of gasoline at $2.85 per gallon plus the cost of c cans of oil at $3.15 per can.

Step-by-step explanation:

The fraction representing Juan's age relative to Julia's age is -1/10. This means that Juan's age is one-tenth less than Julia's age.

Let's start by translating the given statement into an equation:

[tex]2/3J = 3/5 - 2/3Jl[/tex]

where J is Juan's age and Jl is Julia's age.

Now we can simplify this equation by first multiplying both sides by 15 (the least common multiple of 3 and 5) to get rid of the denominators:

[tex]10J = 9 - 10Jl[/tex]

Next, we can isolate J on one side of the equation by adding 10Jl to both sides:

[tex]10J + 10Jl = 9[/tex]

Finally, we can factor out a 10 from the left-hand side:

[tex]10(J + Jl) = 9[/tex]

Dividing both sides by 10, we get:

[tex]J + Jl = 9/10[/tex]

Now we can express Juan's age as a fraction of Julia's age by dividing both sides of this equation by Jl:

[tex]Jl/Jl + J/Jl = 9/10Jl[/tex]

Simplifying this, we get:

1 + J/Jl = 9/10

Subtracting 1 from both sides, we get:

[tex]J/Jl = 9/10 - 1[/tex]

[tex]J/Jl = -1/10[/tex]

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Answer:

Step-by-step explanation:

as we alderdy have value of y,we can substuite it in the place of y

4x+8(2)=3x+7(2)+14

4x+16=3x+14+14

4x-3x=28-16

x=12

Using the intersection of sets we find that A ∩ B = {1, 3}

A set is a collection of well ordered items.

Givent hat set A = {x| x is an od number greater than 0 and less than 10} and set B = {1, 2, 3, 4}, we desire to find A ∩ B. We proceed as follows.

First since set A = {x| x is an odd number greater than 0 and less than 10}, its elements are A = {1, 3, 5, 7, 9}

Also, set B =  {1, 2, 3, 4}

Since we require A ∩ B which is the intersection of both sets and is  the elements that both sets have in common. Since both sets have element 1 and 3 in common, we have that

A ∩ B = {1, 3}

So, A ∩ B = {1, 3}

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The time till you hit the water, given your height above the water, would be 0.76 seconds.

The maximum height above the pool you would get is 15.39 feet.

We are given h ( t ) = - 16 t ² + 5t + 15

You hit the water at 0 so the formula is:

0 = - 16 t ² + 5t + 15

Using the quadratic equation, we can solve:

t = ( - b ± √ ( b ² - 4  ac ) ) / 2a

t = ( - 5 ± √ ( 5 ² - 4 ( - 16 ) ( 15 ) )) / 2 (- 16)

t = (- 5 ± √ 985 ) / -32

t = 0. 76 seconds

The vertex of the parabolic function would be:

= - b / 2a

= - 5 / ( 2 x - 16 )

=  0. 15625 seconds

Maximum height is therefore:

= -16 ( 0. 15625 ) ² + 5 ( 0. 15625 ) + 15

=  15.39 feet

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If the circumference of a circle is 50. 4 ft, its area is 202.24 sq ft. Correct  option is C: 202.24 sq ft.

The formula for the circumference of a circle is C = 2πr, where r is the radius of the circle. We are given that the circumference is 50.4 ft, so we can solve for the radius:

50.4 = 2πr

r = 50.4 / (2π)

r ≈ 8.02 ft

Now we can use the formula for the area of a circle: A = πr²

A = 3.14 * (8.02)²

A ≈ 202.24 sq ft

Therefore, the answer is option C: 202.24 sq ft.

Alternatively, to find the area of a circle with a circumference of 50.4 ft, we will first find the radius using the formula for circumference (C = 2πr) and then use the formula for the area of a circle (A = πr²). Using π = 3.14:

Solve for the radius (r):
C = 2πr
50.4 = 2(3.14)r
r = 50.4 / (2 * 3.14)
r ≈ 8 ft

Calculate the area (A):
A = πr²
A = 3.14 * (8²)
A = 3.14 * 64
A ≈ 201.06 sq ft

The closest answer among the options provided is 202.24 sq ft. Correct  option is C: 202.24 sq ft.

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The total cubic centimeters of space the tissue box requires is 500 cubic centimeters, under the condition that the box is half full.

The volume of a cube is evaluated by multiplying its length by its width by its height. If all sides of a cube are equal,

we can use the formula

V = s³

here

s = length of one side of the cube.

If we let s be the length of one side of the cube and V be its volume,

V = s³

If we know that the tissue box is half full, then let us consider that half of its volume is occupied by tissues. Then

V' = 0.5 × V

Staging V = s³ in the equation

V' = 0.5 × s³

s = 10 cm (given)

V' = 0.5 × 1000 = 500 cubic centimeters

Hence, if Patrick's tissue box has a volume of 1000 cubic centimeters and it is half full, then the tissues occupy 500 cubic centimeters of space.

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The complete question is

Patrick buys a tissue box in the shape of a cube of side 10 centimeters. How many cubic centimeters of space do the tissues occupy if the box is half full? Show all work

To find the elasticity of demand for the function q = D(x) = 460 - x, we can use the formula:

Elasticity = (% change in quantity demanded) / (% change in price)

a) Since the demand function given does not include a price variable, we can assume that price is constant. Therefore, the elasticity of demand for this function is constant and equal to -1.

b) To find the elasticity at x = 103, we need to calculate the percentage change in quantity demanded when x increases from 103 to 104.

At x = 103, quantity demanded is q = D(103) = 460 - 103 = 357.

At x = 104, quantity demanded is q = D(104) = 460 - 104 = 356.

The percentage change in quantity demanded is:

(% change in quantity demanded) = [(new quantity - old quantity) / old quantity] x 100

= [(356 - 357) / 357] x 100 = -0.28%

Since the elasticity of demand is -1, we can say that demand is inelastic at x = 103.

c) To find the elasticity at x = 205, we need to calculate the percentage change in quantity demanded when x increases from 205 to 206.

At x = 205, quantity demanded is q = D(205) = 460 - 205 = 255.

At x = 206, quantity demanded is q = D(206) = 460 - 206 = 254.

The percentage change in quantity demanded is:

(% change in quantity demanded) = [(new quantity - old quantity) / old quantity] x 100

= [(254 - 255) / 255] x 100 = -0.39%

Since the elasticity of demand is -1, we can say that demand is inelastic at x = 205.
Hi! I'd be happy to help you with your question on elasticity of demand.

a) To find the elasticity, we first need the formula for price elasticity of demand (PED), which is:

PED = (% change in quantity demanded) / (% change in price)

Here, we have the demand function D(x) = 460 - x, where x is the price.

b) To find the elasticity at x = 103, we first need to calculate the quantity demanded, which is:

q = D(103) = 460 - 103 = 357

Now, we'll find the derivative of the demand function with respect to price:

dq/dx = -1

Next, we'll use the formula for PED:

PED = (dq/dx * x) / q = (-1 * 103) / 357 = -103/357 ≈ -0.289

Since the absolute value of PED is less than 1, demand is inelastic at x = 103.

c) To find the elasticity at x = 205, we'll follow the same steps:

q = D(205) = 460 - 205 = 255

PED = (dq/dx * x) / q = (-1 * 205) / 255 ≈ -0.804

Again, the absolute value of PED is less than 1, so demand is inelastic at x = 205.

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The ratio of the corresponding dimensions of the smaller pyramid to the larger pyramid is 1/3.

Dimension is the measure of the distance or length of an obeject.

To calculate the ratio of the corresponding dimensions of the smaller pyramid to the larger pyramid, we use the formula below

Formula:

l/L = √(a/A) ........................ Equation 1

Where:

From the question,

Given:

Substitute these values into equation 1

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The drive-thru is able to estimate their wait time more consistently will be A. Burger Quick, because it has a smaller IQR.

The interquartile range (IQR) is a measure of variability that represents the range of values in the middle 50% of scores in a distribution. It is used when the data is skewed and the median is a better measure of central tendency than the mean.

In this scenario, Burger Quick has an IQR of 14.5, while Fast Chicken has an IQR of 4.5.

The IQR is calculated by subtracting the first quartile (Q1) from the third quartile (Q3).

For Burger Quick, Q1 is 9.5 and Q3 is 24, so the IQR is 24 - 9.5 = 14.5. For

Fast Chicken, Q1 is 10 and Q3 is 14.5, so the IQR is 14.5 - 10 = 4.5.

Therefore, Burger Quick has a larger IQR than Fast Chicken, indicating that its data is more consistent and less spread out. This means that Burger Quick is better able to estimate its wait time consistently.

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The mass of the copper wire is approximately 131.77 lb.

To find the mass of the copper wire, we will first need to calculate its mass per unit length using the given density function[tex]p(t) = 5x^2 + 4x lb/ft,[/tex] and then integrate the function over the length of the wire.
Write down the given density function: [tex]p(t) = 5x^2 + 4x lb/ft[/tex]
2. Write down the limits of integration, which correspond to the length of the wire:

a = 0, b = 3.75 feett.

Set up the integral to find the mass of the wire:

Mass = ∫[p(t) dt] from a to b.

Plug in the density function and limits:

Mass = ∫[tex][5x^2 + 4x dx][/tex]from 0 to 3.75
Integrate the function: Mass = (5/3)x^3 + 2x^2 | from 0 to 3.75
Substitute the upper limit and then subtract the result of the lower limit:
  Mass =[tex][(5/3)(3.75)^3 + 2(3.75)^2] - [(5/3)(0)^3 + 2(0)^2][/tex]
Perform the calculations and round to two decimal places:
  Mass ≈ 131.77 lb.

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To solve this problem, we can use substitution. Let u = 9z - 5, then du/dz = 9 and dz = du/9.

Using this substitution, we can rewrite the integral as:

∫(2/(u+5))(1/9)du

Simplifying this expression:

(2/9) ∫(1/(u+5))du

We can then solve this integral by using the formula for the natural logarithm:

(2/9) ln|u+5| + C

Substituting back in for u:

(2/9) ln|9z| + C
2 zV9z+ - 5 dz is (2/9) ln|9z| + C.


Consider the integral:
∫2z√(9z² - 5) dz

We can use the substitution method. Let's choose the substitution:
u = 9z² - 5

Now, differentiate u with respect to z:
du/dz = 18z

Solve for dz:
dz = du/(18z)

Now, substitute u and dz back into the integral and simplify:
∫(2z√u) * (du/(18z)) = (1/9)∫√u du

Now, integrate with respect to u:
(1/9)(2/3)(u^(3/2))/(3/2) + C = (2/27)u^(3/2) + C

Finally, substitute back the original expression for u:
(2/27)(9z² - 5)^(3/2) + C

So, the indefinite integral is:
(2/27)(9z² - 5)^(3/2) + C

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the restaurant used 754,040 ounces of cream last year.

What is an Equations?

Equations are statements in mathematics that consist of two algebraic expressions separated by an equals (=) sign, indicating the equivalence between the expressions on either side. Equations can be solved to determine the value of a variable that represents an unknown quantity. A statement that does not have an "equal to" symbol is not considered an equation and is instead referred to as an expression.

If the restaurant used 50% less cream this year compared to last year, then it means that this year's usage is 50% of last year's usage.

Let x be the amount of cream used last year.

Then we can set up the following equation:

x * 50% = 377,020

To solve for x, we need to isolate it on one side of the equation.

x * 50% = 377,020

x = 377,020 / 50%

To convert 50% to a decimal, we divide it by 100:

x = 377,020 / 0.5

x = 754,040

Therefore, the restaurant used 754,040 ounces of cream last year.

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Answer:

$10.8

Step-by-step explanation:

We know that 3 books are $2.70, so 1 book is 2.70/3 which is $0.9

1 dozen = 12

To find the price of 12 books, we multiply by the cost of 1 book

12 * 0.9 = $10.8

To determine how many times a day we would need to fill the container shown below, we need to calculate its capacity in cubic centimeters. Let's assume that the container is a rectangular prism with dimensions of 30 centimeters (length) x 20 centimeters (width) x 25 centimeters (height). The formula for calculating the volume of a rectangular prism is:

Volume = length x width x height

Plugging in the values we have, we get:

Volume = 30 cm x 20 cm x 25 cm

Volume = 15,000 cubic centimeters

Therefore, the container has a capacity of 15,000 cubic centimeters. To determine how many times we would need to fill it each day, we need to divide the amount of pellets and hay we feed the okapi daily (5,024 cubic centimeters) by the capacity of the container (15,000 cubic centimeters):

5,024 / 15,000 = 0.3356

This means that we would need to fill the container approximately 0.3356 times a day, which is not a practical answer. We need to round this up to the nearest whole number.

The options given to us are 80 times, 40 times, 16 times, and 4 times. Out of these options, the closest whole number to 0.3356 is 1, which means we would need to fill the container once a day.

Therefore, the answer is that we would need to fill the container shown below 1 time a day to feed the okapi their daily amount of pellets and hay.

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Answer:

(0,5.8)

Step-by-step explanation:

Therefore, 32 1/2 can be written as a radical form of 32 + √2.

In mathematics, a fraction is a numerical quantity that represents a part of a whole or a ratio between two numbers. Fractions are written in the form of one integer, called the numerator, written above a line, and another integer, called the denominator, written below the line. The denominator represents the total number of equal parts into which a whole is divided, while the numerator represents the number of those parts being considered.

Here,

We can write 32 1/2 as a mixed number, which is equivalent to the fraction 65/2. To express 65/2 as a radical form, we can simplify the fraction by finding a perfect square that divides evenly into both the numerator and the denominator. In this case, 4 is a perfect square that divides into 64 (the nearest perfect square to 65) and 2, so we can write:

65/2 = (64 + 1)/2

= 64/2 + 1/2

= 32 + 1/2

Now, we can express the fraction 1/2 as a radical by taking the square root of the denominator, which gives:

32 1/2 = 32 + √2

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The sum is equal to 18600.

Let's first simplify the expression inside the brackets:

(i^2 - 21 + 1) - 4 = i^2 - 24

Now we can write the sum using sigma notation:

15 Σ (i^2 - 24), i=1

Using the summation formulas of the first n squares, we have:

Σ i^2 = n(n+1)(2n+1)/6

So, substituting n = 1 to 15, we get:

15 Σ i^2 = 15 Σ n(n+1)(2n+1)/6

Now, let's use the formula for the sum of the first n integers:

Σ i = n(n+1)/2

Substituting n = 1 to 15, we get:

Σ i = 1 + 2 + ... + 15 = 15(15+1)/2 = 120

Substituting these formulas into our original expression, we have:

15 Σ (i^2 - 24), i=1

= 15 [Σ i^2 - 24Σ i], i=1

= 15 [15(15+1)(2(15)+1)/6 - 24(120)]

= 15 [1240]

= 18600

Therefore, the sum of 15 (i^2 - 24) from i = 1 to 15 is equal to 18600.

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The vertex is the highest or lowest point on the parabola, the line of symmetry is a vertical line passing through the vertex and dividing the parabola into two equal halves, and the number of x-intercepts depends on the discriminant of the quadratic equation.

The vertex is a point on a parabola where the curve changes direction. It is the highest or lowest point on the curve, depending on whether it opens up or down. The vertex is represented as (h, k), where h is the x-coordinate and k is the y-coordinate.

The line of symmetry is a vertical line that divides the parabola into two equal halves. It passes through the vertex and is equidistant from the two arms of the parabola. The equation of the line of symmetry is x = h.

The number of x-intercepts is determined by the equation of the parabola. If the discriminant of the quadratic equation is positive, then the parabola intersects the x-axis at two distinct points, and there are two x-intercepts. If the discriminant is zero, then the parabola touches the x-axis at one point, and there is one x-intercept. If the discriminant is negative, then the parabola does not intersect the x-axis, and there are no x-intercepts.

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The critical point x = 0 does not yield a local minimum or maximum. The function is always decreasing.

To find the critical points and intervals for the function y = x^3 / (x^2 + 1), we'll first find the derivative using the Quotient Rule:

y'(x) = [(x^2 + 1)(3x^2) - x^3(2x)] / (x^2 + 1)^2

y'(x) = (3x^4 + 3x^2 - 2x^4) / (x^2 + 1)^2

y'(x) = (x^2 - 2x^2) / (x^2 + 1)^2

Now, we'll find the critical points by setting the derivative equal to zero:

0 = (x^2 - 2x^2) / (x^2 + 1)^2

0 = x^2(1 - 2) / (x^2 + 1)^2

0 = -x^2 / (x^2 + 1)^2

This equation is equal to zero only when x = 0. So, the critical point is x = 0.

Next, we'll use the First Derivative Test to determine if the critical point yields a local min or max. To do this, we'll evaluate the sign of y'(x) to the left and right of x = 0.

1. Left of x = 0 (for example, x = -1):
y'(-1) = (-1)^2(1 - 2) / (-1^2 + 1)^2 = -1 / 1^2 = -1 (negative)

2. Right of x = 0 (for example, x = 1):
y'(1) = (1)^2(1 - 2) / (1^2 + 1)^2 = -1 / 2^2 = -1/4 (negative)

Since the derivative is negative on both sides of the critical point, the function is decreasing for all x. Thus, the critical point x = 0 does not yield a local minimum or maximum. The function is always decreasing.

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(a) The greatest amount of snowfall in any of the cities = 50 inches.

(b) The data is most concentrated in the second quarter (Q₂), ranging from 6 inches to 20 inches.

(c) The data is most spread out in the fourth quarter (Q₄), ranging from 32 inches to 50 inches.

A box plot is a type of graphical representation that summarizes the distribution of a dataset based on the five-number summary: the minimum value, the first quartile (Q1), the median, the third quartile (Q3), and the maximum value.

A box plot consists of a rectangular box, which spans from Q1 to Q3, with a vertical line inside it representing the median. The length of the box represents the interquartile range (IQR), which is the range between Q1 and Q3.

Whiskers, which are lines extending from the top and bottom of the box, indicate the range of the dataset outside of the IQR.  

Here we have

Jyllina created this box plot representing the number of inches of snow that fell this winter in different nearby cities

The data ranges from 4 to 50 inches

4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50

(a) The greatest amount of snowfall in any of the cities will equal the highest value which is 50 inches.

(b) The data is most concentrated in the second quarter (Q₂), which ranges from 6 inches to 20 inches.

This can be inferred from the fact that the box plot shows the smallest range between the minimum and maximum values, as well as the smallest size of the box, which represents the interquartile range (IQR).

(c) The data is most spread out in the fourth quarter (Q₄), which ranges from 32 inches to 50 inches.

This can be inferred from the fact that the box plot shows the largest range between the minimum and maximum values, as well as the largest size of the box, which represents the IQR.

Additionally, the whiskers, which represent the range of values outside the IQR, are also the longest in this quarter, indicating that there are more extreme values in this range compared to the other quarters.

Therefore,

(a) The greatest amount of snowfall in any of the cities = 50 inches.

(b) The data is most concentrated in the second quarter (Q₂), ranging from 6 inches to 20 inches.

(c) The data is most spread out in the fourth quarter (Q₄), ranging from 32 inches to 50 inches.

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Complete Question:

[tex]~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-3}~,~\stackrel{y_1}{-6})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{0})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(~~5 - (-3)~~)^2 + (~~0 - (-6)~~)^2} \implies d=\sqrt{(5 +3)^2 + (0 +6)^2} \\\\\\ d=\sqrt{( 8 )^2 + ( 6 )^2} \implies d=\sqrt{ 64 + 36 } \implies d=\sqrt{ 100 }\implies d=10[/tex]

The cost of the stock for the investor would be calculated as follows:

Cost of 1 share = $98.93
Number of shares purchased = 500

Total cost of the stock = Cost per share x Number of shares
Total cost of the stock = $98.93 x 500
Total cost of the stock = $49,465

The broker charges 2% of the cost of the stock, so the broker's fee would be calculated as:

Broker's fee = 2% x Total cost of the stock
Broker's fee = 2% x $49,465
Broker's fee = $989.30

Therefore, the total cost of the stock for the investor including the broker's fee would be:

Total cost of the stock + Broker's fee = $49,465 + $989.30
Total cost of the stock + Broker's fee = $50,454.30

So the cost of the stock for the investor including the broker's fee is $50,454.30.

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One way Olivia might have achieved this is by performing a translation followed by a rotation.

One way Olivia might have achieved the transformation of ABCD onto A" B" C" D" is by performing a translation followed by a rotation. Here's a possible approach:

By combining these two transformations, Olivia could have successfully moved ABCD onto A" B" C" D" and achieved the desired configuration. It's important to note that there could be other valid combinations of transformations as well.

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A linear function would be the  best fit for the data.

A function that would be the best for this data is: D. y = -4/25(x) + 10

The amount of snow that would be on the ground when the temperature reaches 55° is 1.2 inches.

In this scenario, the temperature would be plotted on the x-axis (x-coordinate) of the scatter plot while the snow (inches) would be plotted on the y-axis (y-coordinate) of the scatter plot through the use of Microsoft Excel.

On the Microsoft Excel worksheet, you should right click on any data point on the scatter plot, select format trend line, and then tick the box to display an equation for the line of best fit (trend line) on the scatter plot.

From the scatter plot (see attachment) which models the relationship between the temperature and the snow (inches), an equation for the line of best fit is given by:

y = -0.16x + 10

y = -4/25(x) + 10

When x = 55, the amount of snow is given by;

y = -4/25(55) + 10

y = 1.2 inches.

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Therefore, (70+72)/2 = **71 minutes** is the median time. B)42.5 minutes as a result. C,D)The range is determined by deducting the dataset's smallest value from highest value.

A)When the data are organized in order of magnitude, the median time is the middle value. The median in this situation is the average of the 14th and 15th values, which are 70 and 72, respectively. There are 28 data points in this situation. Therefore, (70+72)/2 = **71 minutes** is the median time.

b) The median of the lowest half of the data constitutes the lower quartile (Q1). We must arrange the data in descending order of magnitude before determining the median of the first half of the data in order to determine Q1.  is the average of the seventh and eighth values, which are 40 and 45, respectively, in the first half of the data, which consists of 14 values. Q1 = (40+45)/2 = **42.5 minutes as a result.

b) The median of the upper half of the data constitutes the upper quartile (C). We must first organise the data in descending order of magnitude before determining the median of the remaining data in order to determine Q3. Q3 is the average of the seventh and eighth values from the last, which are 90 and 105, respectively, in the second half of the data, which consists of 14 values. Q3 = (90+105)/2 = **97.5 minutes**3 as a result.

d) The range is determined by deducting the dataset's smallest value from highest value.

In this instance, Ben's practise time can be anywhere from **10 minutes** to **180 minutes**. Range then equals maximum value - minimum value, which in this case is 180 - 10 = **170 minutes**

e) Extreme values that are beyond the typical range of a dataset's values are known as outliers. We can use a criterion that states that any value that sits more than 1.5 times the interquartile range (IQR) below Q1 or above Q3 is regarded as an outlier to ascertain whether there are outliers in this dataset. When Q1 is subtracted from Q3, the result is the IQR: Q3 - Q1 = 97.5 - 42.5 = **55 minutes**3. By using this rule, we can see that the dataset contains the outliers **180** and **120** minutes.

The term "interquartile range" is IQR. It is a measure of variability that is based on quartilizing a dataset. The first quartile (Q1) is subtracted from the third quartile to determine the IQR. (Q3). It is a representation of the middle 50% of the data's range.

f) A box-and-whisker plot illustrates a dataset's quartiles, outliers, and range1. For Ben's practice, here's how to create a box-and-whisker plot:

- Create a number line with all the values Ben practised with.

- Draw a box spanning Q1 through Q3.

- Inside the box, at the location of Q2, draw a vertical line. (the median).

Draw whiskers from the box's two ends to all values that are not outliers.

- Place every outlier on the graph as a separate point, outside of any whiskers.

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The rate of change of revenue per unit is $8 when the unit cost is changing by $8 and the revenue is $4,000.

We are given that the unit cost (C) and unit revenue (R) are related by the equation C = R + 10621.

We want to find the rate of change of revenue per unit, dR/dx, when the unit cost (C) is changing by $8 per unit and the revenue (R) is $4,000.

1. Differentiate the given equation with respect to x: dC/dx = dR/dx

2. Plug in the given values: dC/dx = $8 (cost per unit is changing by $8) R = $4,000 (revenue per unit)

3. Solve for the cost per unit (C) using the equation

C = R + 10621

C = $4,000 + 10621

C = $14,621

4. Since dC/dx = dR/dx, and we know dC/dx = $8,

we can find the rate of change of revenue per unit (dR/dx) when the cost per unit is changing by $8: dR/dx = $8

Thus, the rate of change of revenue per unit is $8 when the unit cost is changing by $8 and the revenue is $4,000.

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At point (-43,0), f has a maximum. At point (0,4), f has a minimum. At point (463,0), f has a maximum. At point (0,0), f may have a critical point. none of the given points are critical points of the function f (x, y) = xy + 33 - 48y.

Hi! To analyze the critical points of the function f(x, y) = xy + 33 - 48y, we first need to find the partial derivatives with respect to x and y:

fx = ∂f/∂x = y
fy = ∂f/∂y = x - 48

Now, we can analyze the given points:

1. (-43, 0)
At this point, fx = 0 and fy = -48. Since both partial derivatives are not equal to 0, this point is not a critical point.

2. (0, 4)
At this point, fx = 4 and fy = 0. Again, both partial derivatives are not equal to 0, so this is not a critical point.

3. (463, 0)
At this point, fx = 0 and fy = 463 - 48 = 415. Since both partial derivatives are not equal to 0, this is not a critical point.

4. (0, 0)
At this point, fx = 0 and fy = -48. Since both partial derivatives are not equal to 0, this is not a critical point.

5. (0, -4)
At this point, fx = -4 and fy = -48. Again, both partial derivatives are not equal to 0, so this is not a critical point.

In summary, none of the given points are critical points of the function f(x, y) = xy + 33 - 48y.

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