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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: x=84

Step-by-step explanation:

It should be 84, since the arc is twice the size of angle ADB. Hopefully that makes sense

The ordered pair that describes the location of E'' is (6, -1).

The initial location of Point EE is (-6, 1). To reflect Point EE over the yy-axis, we will change the sign of the x-coordinate while keeping the y-coordinate the same.

Step 1: Reflect Point EE over the yy-axis to create Point E'.
E' = (6, 1)

Next, to reflect Point E' over the xx-axis, you'll change the sign of the y-coordinate while keeping the x-coordinate the same.

Step 2: Reflect Point E' over the xx-axis to create Point E''.
E'' = (6, -1)

Therefore, the ordered pair that describes the location of E'' is (6, -1).

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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

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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The two triangles shows that an angle opposite the longest side of a triangle is the largest angle

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

The two triangles

From the triangles we have the largest angles to be

C = 117.3 and E = 93 degrees

The lennths opposite these sides aere

AB = 6 and DF = 11.94

These lengths are the longest segments on their respective triangles

This means that an angle opposite the longest side of a triangle is the largest angle

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Sydney can row a canoe at a speed of 5 mph in still water.

Let x represent Sydney's speed in still water. When rowing upriver, her effective speed will be (x - 2) mph because she's going against the current, which flows at 2 mph. When rowing downriver, her effective speed will be (x + 2) mph, since she's going with the current.

According to the problem, the time it takes her to row 6 miles upriver is the same as the time it takes her to row 14 miles downriver. We can set up the equation using the formula time = distance / speed:

6 / (x - 2) = 14 / (x + 2)

To solve for x, first cross-multiply:

6(x + 2) = 14(x - 2)

Expand:

6x + 12 = 14x - 28

Now, rearrange and solve for x:

12 + 28 = 14x - 6x

40 = 8x

x = 5

So, Sydney can row a canoe at a speed of 5 mph in still water.

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The area of the preimage is 400units²

Dilation is a transformation, which is used to resize a given object. We can use dilation to either make the objects larger or smaller.

A scale factor shows the relationship between the old shape and new shape.

The scale factor is expressed as ;

scale factor = dimension of the new length / dimension of old length

scale factor = 3/2 = 1.5

old length = x

therefore 1.5 = 30/x

1.5x = 30

divide both sides by 1.5

x = 30/1.5 = 20units

therefore the area of the preimage = l²

= 20²

= 400units²

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This is an unfavorable variance.

To calculate the dollar variance, we subtract the actual amount from the budgeted amount:

Dollar variance = Budgeted amount - Actual amount = $106.00 - $100.00 = $6.00 (favorable)

The dollar variance of $6.00 suggests that the actual expenses were less than the budgeted expenses, which is a favorable variance.

To calculate the percentage variance, we use the following formula:

Percentage variance = (Budgeted amount - Actual amount) / Budgeted amount x 100%

Substituting the values, we get:

Percentage variance = ($106.00 - $100.00) / $106.00 x 100% = 5.66% (rounded to the nearest whole percent)

The percentage variance of 5.66% suggests that the actual expenses exceeded the budgeted expenses by 5.66%, which is an unfavorable variance.

It's important to note that the dollar variance and percentage variance provide different perspectives on the variance, and they should be considered together to fully understand the implications of the variance. In this case, the dollar variance is favorable, indicating that the company spent less than expected.

However, the percentage variance is unfavorable, indicating that the company's expenses were higher than budgeted. The company may use this information to identify areas where they can reduce expenses in the future or adjust their budgeting process to be more accurate.

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You can look the image.

It is very clear.

Answer:

[tex]\sin (Z)=\sf\dfrac{9}{15}[/tex]        [tex]\cos (Z)=\sf\dfrac{12}{15}[/tex]        [tex]\tan(Z)=\sf \dfrac{9}{12}[/tex]

Step-by-step explanation:

To create trigonometric ratios for angle Z in the given right triangle XYZ, we can use the trigonometric ratios.

[tex]\boxed{\begin{minipage}{9.4 cm}\underline{Trigonometric ratios} \\\\$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}[/tex]

From inspection of the right triangle XYZ:

Substitute these values into the three ratios to create the trigonometric ratios for angle Z:

[tex]\sin (Z)=\sf \dfrac{O}{H}=\dfrac{9}{15}[/tex]

[tex]\cos (Z)=\sf \dfrac{A}{H}=\dfrac{12}{15}[/tex]

[tex]\tan(Z)=\sf \dfrac{O}{A}=\dfrac{9}{12}[/tex]

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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Alternative hypothesis can also be written to reflect an increase in yield if the researchers believed that was a possibility.

In hypothesis testing, the null hypothesis is a statement that assumes there is no difference or no effect between two variables.

The alternative hypothesis, on the other hand, assumes that there is a difference or an effect between the variables being tested.

In this scenario, the null hypothesis would be that the new fertilizer has no effect on the yield of the orange grove. The alternative hypothesis would be that the new fertilizer decreases the yield of the orange grove.

So, the appropriate null and alternative hypotheses for this scenario can be stated as follows:

Null hypothesis (H0): The new fertilizer has no effect on the yield of the orange grove.

Alternative hypothesis (Ha): The new fertilizer decreases the yield of the orange grove.

It is important to note that the alternative hypothesis can also be written to reflect an increase in yield if the researchers believed that was a possibility.

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We solve this problem using the angle of elevation, we can apply the tangent function from trigonometry. The approximate distance between the fishing vessel and the base of the lighthouse is 235.2 meters, which corresponds to option C. 235. 2 meters

Find the approximate distance between the fishing vessel and the base of the 50-meter-tall lighthouse when the angle of elevation is 12 degrees.
Set up the equation using tangent function.
tan(angle of elevation) = (height of lighthouse) / (distance between vessel and lighthouse base)
Plug in the values.
tan(12°) = 50 / distance
Solve for the distance.
distance = 50 / tan(12°)
Calculate the distance using a calculator.
distance ≈ 235.2 meters
So, the approximate distance between the fishing vessel and the base of the lighthouse is 235.2 meters, which corresponds to option C. 235. 2 meters

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Answer:

Step-by-step explanation:

12,787

a. ^p^ represents the proportion of adults who chose chocolate pie, ^q^ represents the proportion of adults who did not choose chocolate pie, n represents the sample size, E represents the margin of error, and p represents the population proportion.

b. The value of α for a 99.99% confidence level is 0.0001.

a. ^p^ represents the sample proportion of adults who chose chocolate pie, ^q^ represents the proportion of adults who did not choose chocolate pie (1 - ^p^), n represents the sample size of 1500, E represents the margin of error of ±3 percentage points (0.03), and p represents the population proportion of adults who choose chocolate pie.

b. To find the value of α for a 99.99% confidence level, we need to subtract the confidence level from 100% to get the level of significance, which is 0.0001. This means that there is a 0.0001 probability of rejecting the null hypothesis when it is actually true.

The level of significance (α) is used to determine the critical value for the test statistic, which is then used to determine whether or not to reject the null hypothesis.

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Step-by-step explanation:

$20-$3-$15= $2

the amount of money spent on the bouncy ball is $2

According to the given condition the two-digit number is 66.

To find the two-digit number that is 11 times its units digit and has a sum of digits equal to 12, we can use the following steps:

1. Let's represent the two-digit number as XY, where X is the tens digit and Y is the units digit.
2. The number is 11 times its units digit, so we can write the equation: 10X + Y = 11Y.
3. The sum of the digits is 12, which means X + Y = 12.
4. Now, we have two equations with two variables:
   - 10X + Y = 11Y
   - X + Y = 12
5. We can solve for X from the second equation: X = 12 - Y.
6. Substitute the value of X in the first equation: 10(12 - Y) + Y = 11Y.
7. Simplify and solve for Y: 120 - 10Y + Y = 11Y.
8. Combine the Y terms: 120 - 9Y = 11Y.
9. Move all the Y terms to one side: 120 = 20Y.
10. Divide by 20 to get Y: Y = 6.
11. Now, substitute the value of Y back into the X equation: X = 12 - 6.
12. Solve for X: X = 6.

So, the two-digit number is 66.

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The equation has at least one solution in the interval (-1, 1).

To prove that the equation has at least one solution in the interval (-1, 1), we can use the Intermediate Value Theorem.

First, let's simplify the equation by finding a common denominator:

a(x^3+x-2) + b(x^3+2x^2-1) = 0

Now, let's define a new function f(x) = a(x^3+x-2) + b(x^3+2x^2-1). This function is continuous on the interval (-1, 1) because it is a sum of continuous functions.

Next, we will evaluate f(-1) and f(1) to see if the Intermediate Value Theorem can be applied.

f(-1) = a(-1^3-1-2) + b(-1^3+2(-1)^2-1) = -a-b < 0

f(1) = a(1^3+1-2) + b(1^3+2(1)^2-1) = a+3b > 0

Since f(-1) is negative and f(1) is positive, there must be at least one value of x in the interval (-1, 1) such that f(x) = 0, by the Intermediate Value Theorem.
To prove that the given equation has at least one solution in the interval (-1, 1), we can use the Intermediate Value Theorem (IVT). Let's define the function f(x) as follows:

f(x) = a/(x^3 + 2x^2 - 1) + b/(x^3 + x - 2)

Since a and b are positive numbers, we can examine the behavior of f(x) at the endpoints of the interval (-1, 1).

f(-1) = a/((-1)^3 + 2(-1)^2 - 1) + b/((-1)^3 + (-1) - 2)
f(-1) = a/(-1) + b/(-4) < 0

f(1) = a/(1^3 + 2(1)^2 - 1) + b/(1^3 + 1 - 2)
f(1) = a/(2) + b/(0) = a/2 > 0

Since f(-1) < 0 and f(1) > 0, by the Intermediate Value Theorem, there must be at least one point c within the interval (-1, 1) where f(c) = 0. This means that the given equation has at least one solution in the interval (-1, 1).

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The median score, which represents the middle value of the dataset, can be identified by the line inside the box.

The IQR is represented by the length of the box in the box plot.

Based on the provided box plot for the sixth grade math test, we can infer the following information about the distribution's center, variability, and spread:

1. Center: The median score, which represents the middle value of the dataset, can be identified by the line inside the box. This value divides the data into two equal halves and helps to understand the central tendency of the scores.

2. Variability: The Interquartile Range (IQR) represents the variability in the data. It is calculated as the difference between the third quartile (Q3) and the first quartile (Q1).

The IQR is represented by the length of the box in the box plot and indicates how scores are dispersed around the median.

3. Spread: The range of the dataset can be identified by the distance between the minimum and maximum scores, represented by the whiskers in the box plot.

This shows the overall spread of the scores and indicates the extent of variation within the class.

By analyzing these aspects of the box plot, we can better understand the distribution of math test scores in the sixth grade class.

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This box plot shows scores on a recent math test in a sixth grade class. Identify at least three things that you can infer from the box plot about the distribution’s center, variability, and spread.

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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Answer:

Step-by-step explanation:

(for the first question)

1/2 gallons cover 1/6th of the wall then

1-gallon covers 1/3rd of the wall

so 3 gallons cover one wall

(second question)

you have to calculate 85% of $35 because it is 15% percent off.

35*0.85=29.75

The sale price is $29.75.

(third question)

6*3=18

14.4*100/18

1440/18

80

100-80 = 20

The answer is A (20%)

Answer:

3 gallons of paint

$29.75

20%

Step-by-step explanation:

1. Let's break this down:

1/2 gallon of paint covers 1/6 of his wall.

This means that we have to multiply 1/2 by 6, as there would be 6 1/2 gallon sections of his wall.

1/2·6=3

So, Jamel needs 3 gallons of paint.

2. If a pair of shoes has an original price of $35 but it on sale for 15%, we have to first find how much it's now on sale for:

15/100·35

=0.15·35

=5.25

This is the how much it's off, so subtract 5.25 from 35

35-5.25=$29.75

The shoes have a sale price of $29.75

Even though you said this was wrong, you may have to put a dollar sign in front of it.

3. This is worded a little, but assuming:

1 swimming pass is $6, 3 people buy the swimming pass, and the total cost for the 3 passes in total is $14.40, we have to find out the discount.

So, originally, before the discount, the total amount for the 3 swimming passes would've been $18, but there's been a discount and now they only had to pay $14.40.

To solve, we do the following:

subtract the original price by the sale price

18-14.40=3.6

divide by the original price

3.6/18=0.2

multiply by 100 to get into percent

0.2x100=20%

This means that A is the correct choice.

Hope this helps! :)

---------------------------

|           | Popcorn | No Popcorn |

---------------------------

| Drink     |    74   |     15     |

---------------------------

| No Drink  |    10   |     6      |

---------------------------

The completed two-way table is as shown above.

We construct a two-way table summarizing the results. Let's use the information provided:

1. 105 people entered the theatre
2. 84 people had popcorn
3. 10 out of 84 people with popcorn did not have a drink
4. 6 people walked in without popcorn or a drink

Now let's construct the two-way table:

---------------------------
|           | Popcorn | No Popcorn |
---------------------------
| Drink     |    A    |     B      |
---------------------------
| No Drink  |    C    |     D      |
---------------------------

We need to fill in the values for A, B, C, and D:

1. A = People with popcorn and a drink = Total with popcorn - those without a drink = 84 - 10 = 74
2. C = People with popcorn but no drink = 10 (given in the question)
3. D = People with neither popcorn nor a drink = 6 (given in the question)
4. To find B, we need to find the total number of people with a drink. We know 105 people entered, and 6 had neither popcorn nor a drink, so there were 105 - 6 = 99 people with either popcorn or a drink. Since 84 people had popcorn, this means there were 99 - 84 = 15 people with only a drink.
5. B = People with a drink but no popcorn = 15

Now let's fill in the table:

---------------------------
|           | Popcorn | No Popcorn |
---------------------------
| Drink     |    74   |     15     |
---------------------------
| No Drink  |    10   |     6      |
---------------------------

So, the completed two-way table is as shown above.

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The Maximum volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid [tex]9x^2 + y^2 + 36z^2 = 1[/tex]is 4/5.

To find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid [tex]9x^2 + y^2 + 36z^2 = 1:[/tex]

We can use the hint provided.

By symmetry, we can assume that the rectangular box is in the first octant where x, y, and z are all positive.
Let the dimensions of the rectangular box be 2x, 2y, and 2z.

Then the volume of the rectangular box is V = 8xyz.
To maximize V, we need to find the maximum value of xyz that satisfies the equation of the ellipsoid.

Substituting 2x, 2y, and 2z into the equation of the ellipsoid, we get:
[tex](2x/3)^2 + (y/6)^2 + (2z/3)^2 = 1[/tex]
Multiplying both sides by 9/4, we get:
[tex](2x/3)^2 * (9/4) + (y/6)^2 * (9/4) + (2z/3)^2 * (9/4) = 9/4[/tex]
Simplifying, we get:
4x^2/9 + y^2/36 + 4z^2/9 = 1
We can see that this is the equation of an ellipsoid centered at the origin with semi-axes a = 3/2, b = 3, and c = 3/2.
By symmetry, we know that the maximum value of xyz will be achieved when x = y = z. Therefore, we need to find the value of x, y, and z that satisfy the equation of the ellipsoid and maximize xyz.
Substituting x = y = z into the equation of the ellipsoid, we get:
[tex]4x^2/9 + x^2/36 + 4x^2/9 = 1[/tex]
Simplifying, we get:
[tex]x^2 = 9/20[/tex]
Therefore, x = y = z = √(9/20).
Substituting these values into V = 8xyz, we get:
[tex]V = 8(√(9/20))^3 = 4/5[/tex]
Therefore,the Maximum volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid [tex]9x^2 + y^2 + 36z^2 = 1 is 4/5.[/tex]

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Answer:

5.63 degrees  to the nearest hundredth.

Step-by-step explanation:

Length of the plywood

= area / width

= 10.2 / 2

= 5.1 m

Sin x = 0.8 / 5.1    where x is the agle of elevation

sin x = 0.09804

x = 5.626 degrees

1. (x⁻²y⁵)²*(x⁻³y⁸/x⁻⁶y⁻²), the power and multiplication law is not used correctly.

2. (y⁵/x²)*(y¹⁰/x³), the power and multiplication law is not used correctly.

3.  y⁷/x⁴ * y¹⁰/x³,  the multiplication law is not used correctly.

The exponents are simplified as follows; (using power exponents)

1. (x⁻²y⁵)²*(x⁻³y⁸/x⁻⁶y⁻²)

= (x⁻⁴y¹⁰)*(x⁻⁹y¹⁰)

= x⁻¹³y²⁰

2. (y⁵/x²)*(y¹⁰/x³) (simplify using multiplication and division rule)

(y⁵/x²)*(y¹⁰/x³)

= (y⁵x⁻²)*(y¹⁰x⁻³)

= y¹⁵x⁻⁵

3. y⁷/x⁴ * y¹⁰/x³ (simplify using multiplication and division rule)

y⁷/x⁴ * y¹⁰/x³

= (y⁷x⁻⁴)(y¹⁰x⁻³)

= y¹⁷x⁻⁷

4. y¹⁷/x⁷ (This expression is simplified correctly)

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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 base is decreasing at 1.917 cm/min when altitude is 12cm and  area=84cm².

Let A be the area of the triangle, h be the height of the triangle, and b be the base of the triangle.

Then the formula for the area of a triangle is:

A = (1/2)bh

We are given that dh/dt = 2.5 cm/min (the height is increasing at a rate of 2.5cm/min), and dA/dt = 3 cm²/min (the area is increasing at a rate of 3cm²/min).

We want to find db/dt, the rate of change of the base of the triangle when h = 12 cm and A = 84 cm².

To solve this problem, we need to use the chain rule of differentiation.

We start by differentiating both sides of the formula for the area of a triangle with respect to time t:

dA/dt = (1/2) d/dt (bh)

Next, we can use the product rule of differentiation to find d/dt (bh):

d/dt (bh) = b dh/dt + h db/dt

Substituting this into the previous equation gives:

dA/dt = (1/2) [ b dh/dt + h db/dt ]

Now we can substitute the given values of dh/dt and dA/dt, as well as h = 12 cm and A = 84 cm².

To find db/dt:

3 cm²/min = (1/2) [ b (2.5 cm/min) + 12 cm db/dt ]

Simplifying this expression gives:

6 cm²/min = 2.5 b cm²/min + 12 cm db/dt

Substituting A = 84 cm² and h = 12 cm into the formula for the area of a triangle gives:

84 cm² = (1/2) b (12 cm)

Simplifying this expression gives:

b = 14 cm

Now we can substitute b = 14 cm into the previous equation to find db/dt:

6 cm²/min = 2.5 (14 cm) cm²/min + 12 cm db/dt

Simplifying this expression gives:

db/dt = (6 cm²/min - 35 cm²/min) / (12 cm)

db/dt = -1.917 cm/min (rounded to three decimal places)

Therefore, the base of the triangle is decreasing at a rate of 1.917 cm/min when the height is 12 cm and the area is 84 cm².

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Density of liquid b = 0.4 g/cm³.

Let the volume of liquid B added be V cm³.

The total volume of the mixture = Volume of A + Volume of B = 150 + V cm³

Using the formula:

Density = Mass/Volume

Density of C = (Mass of C) / (Volume of C)

1.1 = 220 / (150 + V)

Solving for V, we get:

V = 100 cm³

Therefore, the volume of liquid B added is 100 cm³.

The total mass of the mixture = Mass of A + Mass of B = (Density of A x Volume of A) + (Density of B x Volume of B)

220 = (1.2 x 150) + (Density of B x 100)

Solving for Density of B, we get:

Density of B = (220 - 180) / 100 = 0.4 g/cm³

Therefore, the density of liquid B is 0.4 g/cm³.

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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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