Luis tiene una mochila de ruedas que mide 3.5 pies de alto cuando se extiende el mango. Al hacer rodar su mochila, la mano de Luis se encuentra a 3 pies del suelo. ?Qué ángulo forma su mochila con el suelo? Aproxima al grado más cercano.

The missing dimension of the triangle is the height, which is 19 cm.

To find the missing dimension of the triangle, we can use the formula for the area of a triangle:

Area = (1/2) x base x height

We know that the area is 256.5 cm^2 and the base is 27 cm. Therefore, we can plug in these values into the formula and solve for the height:

256.5 = (1/2) x 27 x height

256.5 = 13.5 x height

height = 256.5 / 13.5

height = 19

Therefore, the missing dimension of the triangle is the height, which is 19 cm.

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Answer:

48 adult tickets were sold.

Step-by-step explanation:

Let's use algebra to solve this problem:

Let's define:

From the problem statement, we know:

We can use the first equation to solve for x in terms of y:

x = 320 - y

Substituting this expression for x into the second equation, we get:

3(320 - y) + 8y = 1200

Expanding the left side, we get:

960 - 3y + 8y = 1200

Simplifying, we get:

5y = 240

Solving for y, we get:

y = 48

Therefore, 48 adult tickets were sold.

Additional:

To find the number of student tickets sold, we can substitute y=48 into the first equation:

x + 48 = 320

x = 272

Therefore, 272 student tickets were sold.

Anita will take 5 more weeks to save enough money for the snowboard. when she saved $43. 75 of the $112. 50 that she needs for a new snowboard.

Given data :

Total money needs for a new snowboard = $112. 50

Anita saved money = $43. 75

Money saved each week = $13. 75

From the given data we can write the equation to find the number of weeks  as,

13. 75w + 43. 75 = 112. 50

By solving the equation 13.75w + 43.75 = 112.50 we can find out how many more weeks it will take Anita to save enough money for the snowboard by using the elimination method.

Subtracting 43.75 from both sides of the equation, we get:

13.75w + 43.75 - 43.75 = 112.50 - 43.75

13.75w = 68.75

w = 68.75 / 13.75

w = 5

Therefore, Anita will take 5 more weeks to save enough money for the snowboard.

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The percentage of your book that can be filled during your reservation at Goofy's kitchen would be 12. 5%

If there are 6 characters and each one signs your book once, then the total number of signatures you would be able to get are 6 signatures in total.

The calculation to determine the percentage of your book occupied involves dividing the number of signatures by its overall capacity and then multiplying that value by 100.

The percentage that would be covered is:

= 6 signatures / 48 total capacity x 100

= 12.5 %

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Step-by-step explanation:

For this given problem we have the following statements and reasons:

1) SD is vertical to HT,

Reason: Given

2) SDH and SDT are right angles

Reason: Right angle congruence theorem

3) SH=≈ST

Reason: Given

4) BD=≈BD

Reason: Reflexive property

5) SHD=≈STD

Reason: SAS

The new diet was not effective, the researchers may need to continue searching for other solutions.

In the trial to determine whether a new diet reduces cholesterol, a group of twenty volunteers with high cholesterol were selected. The trial likely involved splitting the volunteers randomly into two groups - a treatment group and a control group.

The treatment group would be given the new diet to follow, while the control group would continue with their normal diet. The participants' cholesterol levels would be measured at the beginning of the trial, and then again at regular intervals throughout the trial to track any changes.

After the trial has ended, the researchers would analyze the results to see if there was a significant difference in cholesterol levels between the treatment and control groups. If the new diet was effective in reducing cholesterol, the researchers may recommend it as a potential treatment option for people with high cholesterol. However, if the new diet was not effective, the researchers may need to continue searching for other solutions.

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The value of W to the nearest tenth of degree is 15.8°

The sine rule states that if a, b and c are the lengths of the sides of a triangle, and A, B and C are the angles in the triangle; with A opposite a, etc., then a/sinA=b/sinB=c/sinC.

Sine rule can be used to find unknown side or angle In a triangle.

w/sinW = v/sinV

4.7/sinW = 2.4 / sin8

2.4sinW = 4.7 sin8

2.4sinW = 0.654

sinW = 0.654/2.4

sinW = 0.273

W = sin^-1( 0.273)

W = 15.8° ( nearest tenth)

therefore the value of W is 15.8°

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To evenly distribute 1425 stamps into 7 piles, you would divide 1425 by 7. This gives you a quotient of 203 with a remainder of 4. This means that each pile would contain 203 stamps and there would be 4 stamps left over.

To understand this better, you can visualize the process of dividing 1425 stamps into 7 equal piles. You could start by putting 203 stamps into the first pile. Then, you would add another 203 stamps to the second pile. You would continue this process until you had 7 piles, each containing 203 stamps. However, you would be left with 4 stamps that couldn't be evenly distributed.

This type of division is called integer division because it results in a whole number quotient and potentially a remainder. In this case, the quotient represents the number of stamps that can be evenly distributed among the piles, and the remainder represents the leftover stamps that cannot be evenly distributed.

Overall, to divide 1425 stamps into 7 piles, each pile would contain 203 stamps, with 4 stamps remaining.

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The expression that represents the side length of the square pool is 2x² + 36x - 1120 = 0. The side length of the square pool is x = 16.31 feet.

Let's denote the side length of the square pool as x.

The walkway on two opposite sides of the pool is 8 feet, which means that the overall length of the pool and walkway on those sides is x + 8 + 8 = x + 16.

Similarly, x + 10 + 10 = x + 20.

The total area covered by the pool and walkway is given as 1,440 square feet.

Total Area = Pool Area + Walkway Area

The area of the square pool is x², and the area of the walkway is the difference between the total area and the pool area:

Walkway Area = Total Area - Pool Area

Substituting the values, we have:

1440 = x² + (x + 16)(x + 20)

1440 = x² + (x² + 36x + 320)

Combining like terms:

1440 = 2x² + 36x + 320

2x² + 36x + 320 - 1440 = 0

2x² + 36x - 1120 = 0

After solving, we get:

[tex]x=-9+\sqrt{641},\:x=-9-\sqrt{641}[/tex]

Take positive value:

[tex]x=-9+\sqrt{641},\\x = -9+25.31\\x = 16.31[/tex]

So, the side length of the square pool is 16.31 feet.

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Using the straight-line method, we can find the depreciation expense per year by dividing the depreciable value (cost - residual value) by the estimated life:

Depreciable Value = Cost - Residual Value

Depreciable Value = $40000 - $10000

Depreciable Value = $30000

Annual Depreciation Expense = Depreciable Value / Estimated Life

Annual Depreciation Expense = $30000 / 6

Annual Depreciation Expense = $5000

To create a depreciation schedule, we can subtract the annual depreciation expense from the cost each year until we reach the residual value:

| Year | Cost |   Depreciation | Accumulated Depreciation | Book Value |

|------|---------------|-----------------  |----------------------------------------|------------|

| 1     | $40000  | $5000                | $5000                             | $35000     |

| 2    | $35000  | $5000                 | $10000                           | $30000     |

| 3    | $30000  | $5000                  | $15000                           | $25000     |

| 4    | $25000  | $5000                  | $20000                          | $20000     |

| 5    | $20000  | $5000                  | $25000                         | $15000     |

| 6    | $15000  | $5000                    | $30000                        | $10000     |

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The turbocharged model is moving 41.25 ft/sec faster than the model with the standard engine at the end of the 11-second test run.

We need to find how much faster the turbo-charged model is moving than the model with the standard engine at the end of an 11-second test run at full throttle.

To find the final velocity of each model at the end of 11 seconds, we need to integrate their respective acceleration functions with respect to time from 0 to 11 seconds:

For the standard engine model:

v(t) = ∫(5 + 0.8t) dt = 5t + [tex]0.4t^2[/tex]

v(11) = 5(11) +[tex]0.4(11)^2[/tex] = 72.4 ft/sec

For the turbo-charged model:

v(t) = ∫(5 + 1.2t + 0.03[tex]t^2[/tex]) dt = 5t +[tex]0.6t^2 + 0.01t^3[/tex]

v(11) = 5(11) + [tex]0.6(11)^2 + 0.01(11)^3[/tex]= 113.65 ft/sec

The difference in final velocity between the two models is:

[tex]v_{turbo} - v_{standard[/tex] = 113.65 - 72.4 = 41.25 ft/sec

Therefore, the turbo-charged model is moving 41.25 ft/sec faster than the model with the standard engine at the end of the 11-second test run.

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The volume of a hemisphere with a radius of 8. 8 cm, rounded to the nearest tenth of a cubic centimeter, is approximately 1436.8 cubic centimeters.

To find the volume of a hemisphere with a radius of 8.8 cm, you can use the formula:

Volume = (2/3)πr³

where r is the radius of the hemisphere. Plugging in the given radius:

Volume = (2/3)π(8.8)³ ≈ 1436.8 cubic centimeters

So, the volume of the hemisphere is approximately 1436.8 cubic centimeters, rounded to the nearest tenth of a cubic centimeter.

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The supplementary angles of the given angles are 154 degrees, 136 degrees, 135 degrees, 44 degrees, 45 degrees, and 26 degrees, respectively.

Supplementary Angles of Given Angles

Supplementary angles are pairs of angles that add up to 180 degrees. To find the supplementary angle of each given angle, we simply subtract the angle from 180 degrees.

Therefore, the supplementary angles of the given angles are:

The supplementary angle of 26 degrees is 154 degrees (180 - 26 = 154).

The supplementary angle of 44 degrees is 136 degrees (180 - 44 = 136).

The supplementary angle of 45 degrees is 135 degrees (180 - 45 = 135).

The supplementary angle of 136 degrees is 44 degrees (180 - 136 = 44).

The supplementary angle of 135 degrees is 45 degrees (180 - 135 = 45).

The supplementary angle of 154 degrees is 26 degrees (180 - 154 = 26).

Therefore, the supplementary angles of the given angles are 154 degrees, 136 degrees, 135 degrees, 44 degrees, 45 degrees, and 26 degrees, respectively.

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Work Shown:

First we'll need the derivative of h(x)

[tex]h(\text{x}) = \sqrt{\text{x}}\\\\h(\text{x}) = \text{x}^{1/2}\\\\h'(\text{x}) = (1/2)\text{x}^{-1/2}\\\\h'(\text{x}) = \frac{1}{2\text{x}^{1/2}}\\\\h'(\text{x}) = \frac{1}{2\sqrt{\text{x}}}\\\\[/tex]

Then let f(x) = g(x)*h(x)

Use the product rule to evaluate f ' (9).

[tex]f(\text{x}) = g(\text{x})*h(\text{x})\\\\f'(\text{x}) = \frac{d}{d\text{x}}\left[g(\text{x})*h(\text{x})\right]\\\\f'(\text{x}) = g'(\text{x})*h(\text{x}) + g(\text{x})*h'(\text{x})\\\\f'(\text{x}) = g'(\text{x})*\sqrt{\text{x}} + g(\text{x})*\frac{1}{2\sqrt{\text{x}}}\\\\f'(9) = g'(9)*\sqrt{9} + g(9)*\frac{1}{2\sqrt{9}}\\\\f'(9) = 2*\sqrt{9} + 0*\frac{1}{2\sqrt{9}}\\\\f'(9) = 2*3 + 0\\\\f'(9) = 6\\\\[/tex]

The cat is 3/8 inches taller than the dragon.

To find the difference in height between the cat and the dragon, we need to subtract the height of the dragon from the height of the cat.

The cat is 7/8 inch tall, and the dragon is 1/2 inch tall.

To subtract fractions, we first need a common denominator. The least common denominator (LCD) of 2 and 8 is 8.

So, we'll convert the fractions to equivalent fractions with a denominator of 8.

1/2 = 4/8 (multiply both the numerator and the denominator by 4)

Now, we can subtract the fractions:

(7/8) - (4/8) = (7 - 4)/8 = 3/8

So, the cat is 3/8 inches taller than the dragon.

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Elijah and Riley are playing a board game. Elijah chooses the dragon for his game piece, and Riley chooses the cat for hers. The dragon is about 1 2 inch tall, and the cat is about 7 8 inch tall. How much taller is the cat than the dragon?

I selected Josue as the dog-walker.

Frequency tables:

Morning dogs:

Weight (lbs) Frequency

10-19 2

20-29 5

30-39 2

40-49 1

Afternoon dogs:

Weight (lbs) Frequency

7-16 4

17-26 2

27-36 1

37-46 1

47-56 2

The median of the morning (AM) group is 26.5 lbs. The median of the afternoon (PM) group is 23 lbs.

The first quartile (Q1) of the morning (AM) group is 16.25 lbs. The first quartile (Q1) of the afternoon (PM) group is 9.5 lbs.

The third quartile (Q3) of the morning (AM) group is 34.75 lbs. The third quartile (Q3) of the afternoon (PM) group is 38.5 lbs.

Comparative box plot:

yaml

Copy code

Morning dogs:               Afternoon dogs:

 13     |                      7     |

        |                              |

 16     |                      9     |

        |                              |

 21     |                     11     |

        |                              |

 25     |                     15     |

        |                              |

 26     |                     27     |

        |                              |

 28     |                     34     |

        |                              |

 30     |                     35     |

        |                              |

 35     |                     39     |

        |                              |

 38     |                     43     |

        |                              |

        |                     44     |

        +------------------------------+

          1    2    3    4    5    6

              Group

Morning dogs:

Min: 13

Q1: 16.25

Median: 26.5

Q3: 34.75

Max: 38

Afternoon dogs:

Min: 7

Q1: 9.5

Median: 23

Q3: 38.5

Max: 44

The interquartile range (IQR) of the morning (AM) group is 18.5 lbs. The IQR of the afternoon (PM) group is 29 lbs.

Based on the comparative box plot and the IQRs, the morning group of dogs would be easier to walk as one group. This is because the morning group has a smaller IQR, indicating that the weights of the dogs are more similar to each other. The afternoon group has a larger IQR, indicating that the weights of the dogs are more spread out, which could make it more difficult to walk them as a group.

1. JosueSara

2. Frequency table

3. Median of the Morning Group: 26.5, Median of the Afternoon Group: 18.5

4. Q1 of the Morning Group: 17.5, Q1 of the Afternoon Group: 10.5

5. Q3 of the Morning Group: 32.5, Q3 of the Afternoon Group: 36.5

6. Comparative Boxplot blue is morning dogs and red is afternoon dogs.

7. IQR of the Morning Group: 15, IQR of the Afternoon Group: 26

8. Based on the comparative box plot and the IQRs, the morning group of dogs would be easier to walk as one group.

What is boxplot?

A box plot, also known as a box-and-whisker plot, is a graphical representation of the distribution of a dataset. It displays summary statistics and provides a visual summary of the data's key characteristics.

1. Which dog-walker did you select?

JosueSara

I selected Sara.

2. Create frequency tables to represent the morning and afternoon dogs as two sets of data. Group the weights into classes that range 10 pounds.

Morning Dogs Frequency Table:

Weight Range Frequency

10-19                        2

20-29                        4

30-39                        4

Afternoon Dogs Frequency Table:

Weight Range Frequency

0-9                               1

10-19                       3

20-29                       2

30-39                       2

40-49                       1

50-59                       1

3. What is the median of the morning (AM) group? What is the median of the afternoon (PM) group?

Median of the Morning Group: 26.5

Median of the Afternoon Group: 18.5

4. What is the first quartile (Q1) of the morning (AM) group? What is the first quartile (Q1) of the afternoon (PM) group?

Q1 of the Morning Group: 17.5

Q1 of the Afternoon Group: 10.5

5. What is the third quartile (Q3) of the morning (AM) group? What is the third quartile (Q3) of the afternoon (PM) group?

Q3 of the Morning Group: 32.5

Q3 of the Afternoon Group: 36.5

6. Create a comparative box plot for the morning and afternoon dogs, and label each with its five-number summary.

Morning Dogs:

Min: 13

Q1: 17.5

Med: 26.5

Q3: 32.5

Max: 38

Afternoon Dogs:

Min: 7

Q1: 10.5

Med: 18.5

Q3: 36.5

Max: 55

7. What is the interquartile range (IQR) of the morning (AM) group? What is the interquartile range (IQR) of the afternoon (PM) group?

IQR of the Morning Group: 15

IQR of the Afternoon Group: 26

8. The average weights of the dogs are the same for the morning and afternoon groups. But based on your comparative box plot and the IQRs of the two groups, which group of dogs do you think would be easier to walk as one group? Why?

Based on the comparative box plot and the IQRs, the morning group of dogs would be easier to walk as one group. This is because the morning group has a smaller interquartile range (IQR) of 15 compared to the afternoon group's IQR of 26. A smaller IQR indicates that the weights of the dogs in the morning group are more clustered together.

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The coordinates of the point 1 / 4 of the way from A to B is (-2, -0.5).

To find the coordinates of the point that is 1/4 of the way from point A to point B, we can use the following formula:

Point P = (1 - t) x A + t x B

Given the coordinates of points A (-4, -2) and B (4, 4):

P x = (1 - 1/4) x Ax + (1/4) x Bx

P x = (3/4) x (-4) + (1/4) x 4

P x = -3 + 1

P x = -2

P y = (1 - 1/4) x Ay + (1/4) x By

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

P y = -1.5 + 1

P y = -0.5

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Jerry pays $279.40 this year in property taxes.

To calculate Jerry's property taxes for the year, we need to first decide how many of his month-to-month maintenance fee is going toward property taxes.

The problem states that 20% of the price is for Jerry's assets taxes, which means we will calculate the amount of his belongings taxes with the aid of finding 20% of his monthly charge.

To do this, we multiply the price through 0.20 such as this:

20% of $1,397 = 0.20 x $1,397 = $279.40

Therefore, Jerry pays $279.40 this year in property taxes.

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The type of variable being studied in this case is a categorical variable, specifically a nominal variable.

Categorical variables are variables that can be classified or grouped. The categories in this example are the various modes of transportation utilized by off-campus students (drive alone, carpool, public transportation, walk, or a paid driving service).

Nominal variables are categorical variables with no intrinsic order or numerical value. Because the various modes of transportation in this study have no numerical value or inherent order, they are classified as nominal variables.

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The critical points of f(x) on the interval 0 < x < 10 are: (2, max), (7.5, min)

To estimate the critical points of f(x) on the interval 0 < x < 10, we need to look for points where the derivative, f'(x), equals zero or is undefined. However, we are given a table of values for f(x) instead of f'(x), so we need to first estimate f'(x) using these values.

One way to do this is to use finite differences. We can calculate the first finite difference for each pair of adjacent values in the table, which gives an estimate of the derivative at the midpoint of the interval:

f'(x) ≈ (f(x+1) - f(x)) / (1)

Using this formula, we can calculate the following table of values for f'(x): 0123456789 23-2.79-8-5

Now we can look for critical points of f(x) by finding where f'(x) equals zero or is undefined: - f'(x) = 0 when x = 2 or x = 7.5 (approximately) - f'(x) is undefined at x = 0 and x = 10 (endpoints of the interval)

To classify each critical point, we need to look at the sign of the derivative near the point. If f'(x) changes sign from positive to negative at a critical point, then it is a local maximum. If it changes from negative to positive, then it is a local minimum. If it does not change sign, then it is neither a maximum nor a minimum.

Using the values in the table for f'(x), we can see that: -

Near x = 2, f'(x) changes sign from positive to negative, so it is a local maximum. - Near x = 7.5, f'(x) changes sign from negative to positive, so it is a local minimum. - At the endpoints x = 0 and x = 10, f'(x) is undefined, so there are no critical points.

Therefore, the critical points of f(x) on the interval 0 < x < 10 are: (2, max), (7.5, min)

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Answer:

Step-by-step explanation:

You want the segment lengths DA and DE of the hypotenuse in the triangle shown in the figure.

The radius to a point of tangency always makes a right angle with the tangent. This is a right triangle with legs 8 and 15, so you know from your knowledge of Pythagorean triples that the hypotenuse is 17.

  DA = 17

  DE = 17 -8 = 9

__

Additional comment

In case you haven't memorized a few of the useful Pythagorean triples, {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, you can always figure the missing side length of a right triangle using the Pythagorean theorem.

It tells you the sum of the squares of the legs is the square of the hypotenuse:

  AC² +CD² = DA²

  8² +15² = DA²

  64 +225 = 289 = DA²

  DA = √289 = 17

Of course, AE is the radius of the circe, 8, so ...

  AE + DE = DA

  8 +DE = 17

  DE = 17 -8 = 9

Alternatively, you can solve this using the relation between tangents and secants. If the line DA is extended across the circle to intersect it again at X, then ...

  DC² = DE·DX

  15² = DE·DX = DE(DE +16) . . . . . . . EX is the diameter, twice the radius of 8

  DE² +16DE -225 = 0

  (DE +25)(DE -9) = 0 . . . . factor

  DE = 9  . . . . the positive solution

  DA = 9 +8 = 17

We like the Pythagorean theorem solution better, as the factors of the quadratic may not be obvious.

If Dagne is 48 inches tall, then she can jump 13.2 inches high.

Given that Dagne can do a vertical jump that is 27.5% of her height, and she is 48 inches tall, we can find how high she can jump by multiplying her height by the percentage of her vertical jump as follows:

Jump height = 27.5% of height

Jump height = (27.5/100) x 48 inches

Jump height = 0.275 x 48 inches

Jump height = 13.2 inches

Therefore, Dagne can jump 13.2 inches high.

To explain this, we can say that the problem gives us information about the proportion of Dagne's height that she can jump vertically. The percentage of her height that she can jump is given as 27.5%, which we can convert to a decimal form (0.275) for calculation purposes. Multiplying this decimal by Dagne's height of 48 inches gives us the height that she can jump, which is 13.2 inches. So, Dagne can jump 13.2 inches high based on the given information.

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Answer:

  3.5 feet

Step-by-step explanation:

You want the difference in height between similar isosceles triangles, one with height of 30 ft and a base of 24 ft, the other with a side length of 28.5 ft.

We can find the side length of the larger triangle using the Pythagorean theorem. It will be ...

  longer side = √(30² +12²) ≈ 32.311 ft

Then the length x is the difference between the altitudes of the triangles. The altitudes are proportional to the side lengths, so we have ...

  (30 -x)/28.5 = 30/32.311

  x = 30-(28.5)(30/32.311) = 30(1 -28.5/32.311) ≈ 3.538 ≈ 3.5 . . . . feet

The hand rail is about 3.5 feet above the bridge deck.

We recognize that the distance from the hand rail to the top of the triangle is the product of the given side length (28.5 ft) and the cosine of the angle between the side and the altitude.

The tangent of that angle is the ratio of its opposite side (12 ft) to its adjacent side (30 ft), or θ = arctan(12/30).

The value of x is the difference of the altitudes of the triangles, so is ...

  x = 30 -28.5·cos(arctan(12/30)) ≈ 3.5 ft

We find this easier to compute.

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Let's start by finding the volume of the cistern at any given time t. Since the cistern is in the form of an inverted circular cone, its volume can be expressed as:

V = (1/3)πr^2h

where r is the radius of the circular opening, h is the height of the cone (which is also the depth of the cistern), and π is the constant pi.

We are given that the cistern is 5 meters deep, and the radius of its opening is 2 meters. Therefore, we can plug these values into the equation above to get:

V = (1/3)π(2^2)(5)

V = 20/3 π

Now, we need to find the rate at which the water level is rising in the cistern after 30 minutes. Let's call this rate dh/dt (the change in height with respect to time).

We know that the water is being added to the cistern at a rate of 65 liters per minute. Since 1 liter is equal to 0.001 cubic meters, the volume of water being added per minute is:

(65 liters/minute) × (0.001 m^3/liter) = 0.065 m^3/minute

Therefore, the rate at which the height of the water in the cistern is changing is:

dh/dt = (0.065 m^3/minute) / (20/3 π m^3) = 3.87/π meters/minute

After 30 minutes, the height of the water in the cistern will have risen by:

h = (65 liters/minute) × (0.001 m^3/liter) × (30 minutes) / (20/3 π m^3) = 0.2925 meters

Therefore, the rate at which the water level is rising in the cistern 30 minutes after the filling process began is:

dh/dt = 3.87/π meters/minute

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The algebraic expression for the weigh of both baskets where each one contains pumpkins and carrots in kilos is equals to the 7x + 2, in kilos.

We have two baskets where each one contains pumpkins and carrots. We have to determine the both baskets weigh together in kilos. Let's assume that

The number of pumpkins in second basket = x kilos

Now, according to first scenario, first basket contains the pumpkins twice as many kilos of pumpkin as in the second basket. That is the number of pumpkins in first basket = 2x kilos

In second case, the second basket there are three more kilos of carrots than there are kilos of pumpkin in the first. So, the number of carrots in second basket

= (3 + 2x ) kilos

In third case, the first basket has 4 kilos less carrot than the second, that is x

=( ( 3 + 2x) - 4 ) kg

Now, weigh of first basket = carrots + pumpkins = (2x + 2x - 1) kilos

= (4x - 1 ) kilos

Weigh of second basket = carrots + pumpkins = (3 + 2x) kilos + x kilos

= (3 + 3x) kilos

So, weigh of both baskets together

= (4x - 1 ) kilos + (3 + 3x) kilos

=( 7x + 2 ) kilos.

Hence, required expression is 7x + 2.

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Complete question:

You have two baskets. Each contains pumpkins and carrots. In the first basket there are twice as many kilos of pumpkin as in the second, and in the second there are three more kilos of carrots than there are kilos of pumpkin in the first. The first basket has 4 kilos less carrot than the second. How many kilos do both baskets weigh together? Represent it algebraically

The tent has a volume of 86 cubic feet.

The tent owned by Dante is in the shape of a triangular prism, which means it has two identical equilateral triangle bases that measure 5 feet each. The tent's length is 8 feet, and its height is 4.3 feet.

To calculate the tent's volume, we can use the formula for the volume of a triangular prism, which is [tex]V = (1/2) * b * h * l[/tex], where b is the base, h is the height, and l is the length of the prism.

Plugging in the given values, we get [tex]V = (1/2) * 5 * 4.3 * 8[/tex] = 86 cubic feet. The volume of a tent is an important consideration when deciding which one to purchase or use for a particular activity, as it determines how much space is available inside for people and belongings.

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System of equation Sam can use is 13b + 4t = 327.50 and 6b + 2t = 142.47 where "b" represents bushes and "t" represents trees.

Let the cost of bushes represented by b

cost of trees represented by t

First order is 13 buses and 4 trees and total is $327.50

By using the data equation form will be

13b + 4t = 327.50

Second order is 6 bushes and 2 trees and total is $142. 96

By using data the equation form will be

6b + 2t = 142.96

These set of equation will for and can be used.

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Answer:

9th term is

[tex]a_9 = \dfrac{1280}{2187}\\[/tex]

Step-by-step explanation:

This sequence is clearly a geometric progression where the ratio of any term to the previous term is constant and known as common ratio

The 3 terms given are:
15, 10 and 20/3

10 ÷ 15 = 2/3

20/3 ÷ 10 = 2/3

So the common ratio is 2/3

For a geometric sequence with common ratio r and first term a₁, the nth term is given by the equation

aₙ = a₁ · rⁿ⁻¹

Here a₁  = first term = 15

r = 2/3

So the general equation for the nth term of this equation is
aₙ = 15 · (2/3)ⁿ⁻¹

The 9th term would be

[tex]a_9 = 15 \cdot \left(\dfrac{2}{3}\right)^{9-1}\\\\a_9 = 15 \cdot \left(\dfrac{2}{3}\right)^{8}\\\\a_9 = 15 \cdot \left(\dfrac{256}{6561}\right)\\\\a_9 = 15 \cdot \left(\dfrac{256}{6561}\right)\\[/tex]
15 is divisible by 3 giving 5
6561 is divisible by 3 giving 2187

So the above expression simplifies to
[tex]a_9 = 5 \cdot \dfrac{256}{2187}\\\\a_9 = \dfrac{1280}{2187}\\[/tex]

The scale factor used in the dilation of the triangles is 1/2

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

The scale factor is calculated as

Scale factor  = X'/X

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

Scale factor  = (4, 1)/(8, 2)

Evaluate

Scale factor = 1/2

Hence, the scale factor is 1/2

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The P-value is less than the significance level of 0.01, we reject the null hypothesis.

Null Hypothesis: The proportion of adults who prefer window seats when they fly is 0.5 or less.

Alternative Hypothesis: The proportion of adults who prefer window seats when they fly is greater than 0.5.

Let p be the true proportion of adults who prefer window seats when they fly.

The sample proportion of adults who prefer window seats is:

= 520/813 = 0.639

The standard error of the sample proportion is:

SE = sqrt((1-)/n) = sqrt(0.639(1-0.639)/813) = 0.022

The test statistic is:

z = ( - 0.5)/SE = (0.639 - 0.5)/0.022 = 6.32

Using a normal distribution, the P-value is P(Z > 6.32) < 0.0001.

Since the P-value is less than the significance level of 0.01, we reject the null hypothesis.

Therefore, we conclude that there is sufficient evidence to support the claim that the majority of adults prefer window seats when they fly.

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