A 3 ox serving of roasted skinless chicken breast contain 140 cal, 24 g of protein, 2 g of fat, 11 mg of calcium, and 61 mg of sodium. One half cup of potato salad contains 160 cal, 4 g of protein, 13 g of fat, 21 mg of calcium, and 656 mg of sodium. One brooccoli spear contains 40 cal, 5 g of protein, 1 g of fat, 81 mg of calcium, and 23 mg of sodim. Use this information to complete following parts.
a) Write a 1×5 ​matrices, C,​ P, and B that represent the nutritional values of the​ chicken, potato​ salad, and​ broccoli, respectively. Give the nutritional values in the following​ order: Cal, g of​ protein, g of​ fat, mg of​ calcium, and mg of sodium.
C=
P=
B=

The height of the cone is 30 cm

The first step is to write out the parameters given in the question

Volume of the cone= 2560π cm³

Diameter of the circular base is 32 cm

radius= diameter/2

= 32/2

= 16

The formula for calculating the height of the cone is

Height= volume ÷ 1/3 πr²

height= 2560 πcm³ ÷ 1/3 πr²

height= 2560 ÷ 0.333(16²)

height= 2560 ÷ 0.333(256)

height= 2560÷ 85.248

height= 30

Hence the height of the cone is 30 cm

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Answer:

height = 9.54cm

Step-by-step explanation:

Volume of a cone = 2560cm³

Radius = diameter/ 2 = 32/2 = 16

height = ?

volume of a cone =

[tex]volume \: of \: a \: cone = \frac{1}{3} \pi {r}^{2} h \\ 2560 = \frac{1}{3} \times \frac{22}{7} \times 16 \times 16 \times h \\ 2560 = \frac{22}{21} \times 256 \times h \\ 2560 = \frac{5,632h}{21} \\ 21 \times 2560 = 21 \times \frac{5632h}{21} \\ 53,760 = 5632h \\ \frac{53760}{5632} = \frac{5632h}{5632} \\ 9.54cm = h[/tex]

height = 9.54cm

To produce an A note on a C string (5 notes higher) that is 70 cm long we should place a fret at a distance of 74.92cm from bridge.

To find where to place the fret, we use the formula:

distance from bridge = (string length) x [tex]2^{(n/12)[/tex]

In this case, the string length is 70 cm and we want to produce an A note on a C string, which is 5 notes higher.         So n = 5.

distance from bridge = [tex]70 * 2^{(5/12)[/tex]

Using a calculator, we get:

distance from bridge ≈ 74.92 cm

Therefore, the answer is d. 74.92 cm, rounded to the nearest hundredth.

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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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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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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.

volume =πr^2h

=π(9)^2*12

=π81*12

=972πcm^3

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]

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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a. The slope and intercept of the least-squares line are -0.752 and 88.761, respectively.

b. The linear model may be useful for predicting ozone levels from relative humidity, but only to a limited extent. The R-squared value of 0.22 indicates that only 22% of the variation in ozone levels can be explained by relative humidity.

Additionally, the 95% prediction interval for a new observation (20.86, 66.37) is relatively wide, which suggests that the model may not be very precise in its predictions.

c. To predict the ozone level for a day when the relative humidity is 50%, we plug in x = 50 into the regression equation: Ozone = 88.8 - 0.752(50) = 53.8 ppb.

d. The correlation between relative humidity and ozone level can be found by taking the square root of the R-squared value, which gives us a correlation coefficient of 0.47.

This indicates a moderate positive correlation between relative humidity and ozone levels.

e. To find a 90% confidence interval for the mean ozone level for days where the relative humidity is 60%, we use the formula:

Mean ozone level ± (t-value)*(SE Fit)/sqrt(n), where the t-value is obtained from a t-distribution with n-2 degrees of freedom and a 90% confidence level.

For n = 120 and a 90% confidence level, the t-value is approximately 1.66. Plugging in the values, we get: 43.62 ± 1.66*(1.2)/sqrt(120), which simplifies to (42.39, 44.85).

f. To determine if a predicted ozone level of 80 ppb is reasonable when the relative humidity is 60%, we can calculate the 95% prediction interval for a new observation with x = 60: 43.62 ± 2.064*(11.43) = (21.18, 66.06).

Since the predicted value of 80 ppb falls outside of this range, it is not a reasonable prediction.

A more reasonable range of predicted values would be the 95% prediction interval, which gives us a range of (21.18, 66.06).

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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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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]

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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Answer:

1+3x

Step-by-step explanation:

[tex] \sqrt{1 } = 1 \\ 1 + 3x = 1 + 3x[/tex]

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 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 minimum sample size required is approximately 2507 individuals to accurately poll the proportion of Americans favoring a government-run, single-payer healthcare system within a 2% margin of error and with 99% confidence.

To determine the minimum sample size required for your poll on the proportion of Americans favoring a government-run, single-payer healthcare system, you need to consider the desired margin of error (2%), the confidence level (99%), and the estimated proportion from a previous poll (21%).

Step 1: Identify the critical value (Z-score) for a 99% confidence level. You can use a Z-score table or calculator for this. The critical value is approximately 2.576.

Step 2: Determine the margin of error (E). In this case, the margin of error is 2%, or 0.02.

Step 3: Use the estimated proportion (p) from the previous poll, which is 21%, or 0.21. Calculate the estimated proportion for not favoring the single-payer system (q), which is 1 - p, or 0.79.

Step 4: Apply the formula for the minimum sample size (n):

n = (Z^2 * p * q) / E^2

n = (2.576^2 * 0.21 * 0.79) / 0.02^2

n ≈ 2506.73

Since you cannot have a fraction of a person, round up to the nearest whole number. The minimum sample size required is approximately 2507 individuals to accurately poll the proportion of Americans favoring a government-run, single-payer healthcare system within a 2% margin of error and with 99% confidence.

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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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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 3rd Maclaurin polynomial for f(x) = tan x is P(x) = x + (1/3)x^3.

Recall that the Maclaurin series expansion for tan x is given by:

tan x = x + (1/3)x^3 + (2/15)x^5 + ...

To find the 3rd Maclaurin polynomial, we only need to take the first three terms of the series expansion, since n = 3.

Thus, the 3rd Maclaurin polynomial for f(x) = tan x is given by:

P(x) = x + (1/3)x^3.

Note that the first two terms of the polynomial, x and (1/3)x^3, correspond to the first two terms of the Maclaurin series expansion for tan x.

The third term of the polynomial would correspond to the next term in the series expansion, which is (2/15)x^5. However, since we are only finding the 3rd Maclaurin polynomial, we do not need to include this term.

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Jhon bought a rectangular door mat that was 1/2 meter long and 3/10 meter wide. The area of the rectangular door mat is 3/20 square meters.

Find the area of John's rectangular door mat that is 1/2 meter long and 3/10 meter wide, you'll need to multiply the length by the width.
Identify the length and width.
Length = 1/2 meter
Width = 3/10 meter
Multiply the length and width to find the area.
Area = Length × Width
Area = (1/2) × (3/10)
Calculate the multiplication.
Area = 3/20 square meters
The area of the rectangular door mat is 3/20 square meters.

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In the given equation, -7 + 4c = 7c + 6, the solution is c = -13/3

From the question, we are to solve the one-variable linear equation

From the given information,

The given equation is

-7 + 4c = 7c + 6

To solve the equation, we will determine the value of c

Solving the equation

-7 + 4c = 7c + 6

Subtract 4c from both sides of the equation

-7 + 4c - 4c = 7c - 4c + 6

-7 = = 3c + 6

Subtract 6 from both sides of the equation

-7 - 6 = 3c + 6 - 6

-13 = 3c

This can be wroitten as

3c = -13

Divide both sides by 3

3c/3 = -13/3

c = -13/3

Hence, the value of c is -13/3

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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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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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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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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 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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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.

The number of horizontal rails for 10 sections is

The rule for the post is to multiply the number of section by 3

The rule for the rail is to multiply the post by 2

The rules is calculated using the unit value and comparing with other values

the rule for the number of post

1 section requires 3 posts hence multiplication by 3, comparing shows that multiplying by 3 gives the number of post

the rule for the number of rails

1 section requires 6 rails hence multiplication by 6, comparing shows that multiplying each section by 6 gives the correponding number of rails

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To solve this problem, we can use the multiplication rule of probability, which states that the probability of two or more independent events occurring together is the product of their individual probabilities.

Let L, R, and S be the events that a vehicle turns left, turns right, and goes straight on, respectively. We want to find the probability of the following event:

E: One vehicle goes straight on and the other two either both turn left or both turn right.

We can break down event E into two sub-events: one vehicle goes straight on and the other two vehicles both turn left or both turn right. Let's calculate the probabilities of these sub-events separately:

- Probability that one vehicle goes straight on: P(S) = 0.3

- Probability that the other two vehicles both turn left or both turn right: P((LL) or (RR)) = P(LL) + P(RR)

To find P(LL) and P(RR), we can use the multiplication rule again. Since the events are independent, we can multiply their individual probabilities:

- Probability that two vehicles both turn left: P(LL) = P(L) × P(L) = 0.55 × 0.55 = 0.3025

- Probability that two vehicles both turn right: P(RR) = P(R) × P(R) = 0.15 × 0.15 = 0.0225

Therefore, the probability of event E is:

P(E) = P(S) × P((LL) or (RR))

    = 0.3 × (P(LL) + P(RR))

    = 0.3 × (0.3025 + 0.0225)

    = 0.0975

So the probability that, of the next three vehicles approaching the junction from town A, one goes straight on and the other two either both turn left or both turn right is 0.0975 or approximately 0.1 (rounded to one decimal place).

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The sum of the series is 1/16. The given series is: ∑ [infinity] 5(−1)nπ2n + 1 / 32n + 1(2n + 1)!

To find the sum of the series, we can use the ratio test to check the convergence of the series. First, let's take the ratio of the (n+1)th term to the nth term: | a(n+1) / a(n) | = 5π2 / 32(2n + 3)(2n + 2)(2n + 1)

As n approaches infinity, the denominator of the ratio tends to infinity, making the ratio go to zero. Therefore, by the ratio test, the series converges.

Now, we need to find the sum of the series. To do this, we can use the formula for the sum of an infinite series: S = lim [n → ∞] Sn, where Sn is the nth partial sum of the series.

Using partial fractions, we can write the series as: 5π2 / 32n + 1 (2n + 1)! = 1 / 64 [ 1 / (n!) - 1 / (2n + 1)! ] - 5π2 / 32(2n + 3)(2n + 2)(2n + 1)

Substituting this expression into Sn and simplifying, we get: Sn = (1 - cos(π/4n+1)) / 32

Taking the limit as n approaches infinity, we get: S = lim [n → ∞] Sn = 1 / 16 Therefore, the sum of the series is 1/16.

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