Carolyn is searching online for tickets to a concert. Two weeks ago, the cost was $50. Now the cost is $59. What is the percent increase

Answer:

12.75

Step-by-step explanation:

The first number, 51, is called the dividend.

The second number, 4 is called the divisor.

Approximately 17.4 inches of fabric were used on a headband for each player, and 9.43 inches of fabric were used on a wristband for each player.

Here we want to find how much fabric was used on a headband and wristband for each player, we need to calculate the total fabric used for headbands and wristbands and divide it by the total number of players.

So the formula we need to use is:

Fabric used on a headband  = Total fabric used for headbands / (Number of players + Number of coaches)

Fabric used on a headband  = 853.58 inches / (46 + 3)

Fabric used on a headband = 853.58 inches / 49 = 17.4 inches

And we also need to use:

Fabric used on a wristband  = Total fabric used for wristbands / Number of players

Fabric used on a wristband = 433.78 inches / 46 = 9.43 inches

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The amount of fabric that was used on a headband and wristband for each player is: 28 inches

The parameters given in the question are:

853.58 inches of fabric used for headbands

433.78 inches of fabric on wristbands

She makes 46 headbands for the players and 3 for the coaches: 46 + 3 = 49 headbands

She only makes wristbands for the players: 46 wristbands

Thus:

Fabric used for each headband = 853.58/49 = 17.42 ≈ 18 inches per headband

Fabric used for wristband = 433.78/46 = 9.43 ≈  10 inches per wrist band.

Total fabric used = 18 + 10 = 28 inches

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Answer:

17 days

Step-by-step explanation:

525-75.50= 449.50

449.50/25 is 17.98, but you can't have a decimal for a day so the answer is rounded down to 17 days

\(\dfrac{(-9-11)/(-4)}{-7^2+(11 \times 2^2)}\\\\=\dfrac{(-20)/(-4)}{-49+44}\\\\=\dfrac{5}{-5}\\\\=-1\)

The category in the Excel Options dialog box that contains the option to change the user name is "General".

In the Excel Options dialog box, the category that contains the option to change the user name is the General category.

Excel graphing methods, Go to Insert > Line after selecting the data. The type of line chart you want may be chosen from a dropdown menu that appears when you click the icon.

We'll use the fourth 2-D line graph (Line with Markers) for this illustration. Your line graph for the chosen data series will be added by Excel.

What are the primary three graphs?

How to Use Graphs in Science

Bar, circle, and line graphs are the three most often utilized graph kinds. Each form of graph may be used to display a certain kind of data.

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Answer:

8$/lb

Step-by-step explanation:

because 8 x 5 is 40 there for each pound would be 8$ if there are 5 pounds

Answer:

the answer is the first one

Step-by-step explanation:

because

Answer

c                                                  hope it helps

Answer:

x is 48 and y is 100

Step-by-step explanation:

Angle y = 100 degree

Angle x = 48 degree

Answer: One abutment

Step-by-step explanation: When a fixed bridge is created, there must be at least one abutment of the bridge.

Answer: Each friend will get 3 gumballs.

Step-by-step explanation:

Given: Pete has 4 packs of gumballs. Each pack has 5 gumballs.

Total gumballs he has = 4 x 5 = 20

Now , if he needs to divide them into 6 friends , then each friend will get (20 ÷ 6)  gumballs

\(20\div 6=\dfrac{20}{6}=\dfrac{18+2}{6}=\dfrac{3\times6+2}{6}=3\dfrac26\)

So each friend will get 3 gumballs.

The three computations covered by the reliability factor table are sample size, index of reliability, and index of precision. Sample size deals with the size of the sample being used in order to achieve a desirable level of reliability.

Index of reliability is used to measure the consistency of results achieved over multiple trials. It does this by calculating the total number of items that contribute significantly to the final result. Finally, the index of precision measures the effect size of the sample, which is determined by comparing the results from the sample with the expected results.

The sample size computation gives the researcher an idea of the number of items that should be included in a sample in order to get the most reliable results. This is done by taking into account a number of factors including the variability of the population, the type of measurements used, and the desired level of accuracy.

The index of reliability is commonly calculated by finding the ratio of the number of items contributing significantly to the total result to the total number of items in the sample. This ratio is then multiplied by 100 in order to get a final score.

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The statement ''the shape of a sampling distribution of sample means that follows the requirements of the central limit theorem will be approximately bell-shaped.'' is true. The statement ''A sampling distribution of sample means has a mean, μ /√n 1to.'' is false.

(a) According to the central limit theorem, when the sample size is sufficiently large, the sampling distribution of sample means will approximate a bell-shaped distribution, regardless of the shape of the population from which the samples are drawn. This is one of the key properties of the central limit theorem.

(b) The correct formula for the mean of a sampling distribution of sample means is μ, not μ/√n. The mean of the sampling distribution of sample means is equal to the population mean (μ). The formula μ/√n is used to calculate the standard deviation (σ) of the sampling distribution, not the mean.

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The Laplace transform of the function \(f(t) = 5t^3 - 5\sin(4t)\) is given by: \[F(s) = \frac{120}{s^4} - \frac{20}{s^2+16}\]

To find the Laplace transform of the given function \(f(t) = 5t^3 - 5\sin(4t)\), we can apply the properties and formulas of Laplace transforms.

The Laplace transform of a function \(f(t)\) is defined as:

\[

F(s) = \mathcal{L}\{f(t)\} = \int_0^\infty f(t)e^{-st}\,dt

\]

where \(s\) is the complex frequency variable.

Let's find the Laplace transform of each term separately:

1. Laplace transform of \(5t^3\):

Using the power rule of Laplace transforms, we have:

\[

\mathcal{L}\{5t^3\} = \frac{3!}{s^{4+1}} = \frac{5\cdot3!}{s^4}

\]

2. Laplace transform of \(-5\sin(4t)\):

Using the Laplace transform of the sine function, we have:

\[

\mathcal{L}\{-5\sin(4t)\} = -\frac{5\cdot4}{s^2+4^2} = -\frac{20}{s^2+16}

\]

Now, we can combine the Laplace transforms of the individual terms to obtain the Laplace transform of the entire function:

\[

\mathcal{L}\{f(t)\} = \mathcal{L}\{5t^3 - 5\sin(4t)\} = \frac{5\cdot3!}{s^4} - \frac{20}{s^2+16} = \frac{120}{s^4} - \frac{20}{s^2+16}

\]

This is the Laplace transform representation of the function \(f(t)\) in the frequency domain. The Laplace transform allows us to analyze the function's behavior in the complex frequency domain, making it easier to solve differential equations and study the system's response to different inputs.

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These real-world applications help young learners see the practical applications of geometry and develop a deeper understanding of geometric concepts while making learning more engaging and meaningful.

Relevance: Connecting math to real-world applications helps students see the practical value and relevance of the concepts they are learning. It provides a meaningful context and motivation for learning.

Engagement: Real-world applications make math more interesting and engaging for students. It brings concepts to life and helps students see how math is used in everyday life.

Deep understanding: By applying math concepts to real-world situations, students develop a deeper understanding of the concepts and their connections. It promotes critical thinking, problem-solving skills, and the ability to apply mathematical knowledge in different contexts.

Transferability: Real-world applications help students see how math concepts can be transferred and applied to various situations. It promotes the ability to apply learned concepts to new and unfamiliar problems.

Some real-world applications of geometry that would be appropriate for young learners include:

Measurement: Young learners can apply geometric concepts to measure and compare the lengths, areas, and volumes of objects in their environment. For example, measuring the length of a room, comparing the sizes of different shapes, or estimating the volume of a container.

Navigation and Maps: Young learners can use geometry to understand maps, directions, and spatial relationships. They can learn about reading maps, understanding coordinates, and finding distances between locations.

Architecture and Construction: Exploring geometric shapes, angles, and symmetry can help young learners understand the principles of architecture and construction. They can design and build simple structures using different shapes and understand the importance of stability and balance.

Art and Design: Geometry plays a significant role in art and design. Young learners can explore symmetry, patterns, and shapes in various art forms. They can create tessellations, explore rotational symmetry, or design patterns using geometric shapes.

Everyday Objects: Geometry is present in everyday objects around us. Young learners can identify and classify shapes in their environment, such as identifying spheres, cubes, cylinders, and cones in objects like balls, boxes, cups, and ice cream cones.

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The probability that Tom eats one gummy and one fish is 1/6.

If he grabs a red gummy on the first try, there are now three equally likely outcomes for the second try: he can either grab a green gummy, a red fish, or a green fish. If he grabs a red fish on the second try, then he has eaten one of each and we're done. So the probability of this happening is:

P(red gummy on first try and red fish on second try) = P(red gummy on first try) × P(red fish on second try | red gummy on first try)

= 1/4 × 1/3

= 1/12

Similarly, if he grabs a green gummy on the first try, there are again three equally likely outcomes for the second try, and if he grabs a green fish on the second try, then he has eaten one of each. So the probability of this happening is:

P(green gummy on first try and green fish on second try) = P(green gummy on first try) × P(green fish on second try | green gummy on first try)

= 1/4 × 1/3

= 1/12

Therefore, the probability that Tom eats one gummy and one fish is the sum of these two probabilities:

P(one gummy and one fish) = P(red gummy on first try and red fish on second try) + P(green gummy on first try and green fish on second try)

= 1/12 + 1/12

= 1/6

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36.74098197 kg in 81 lbs. Likewise, the question how many pounds in 81 kilogram has the answer of 178.57443237 lbs in 81 kg.

81 kg * 2.2046226218 lbs / 1 kg= 178.57443237lbs

The weight of 81 kilograms is equivalent to 178.57443237 pounds (81kg = 178.57443237lbs). It is simple to convert 81 kg to lb. Simply use our calculator above or the method to convert 81 kg to pounds. To convert 81 kg to pounds, multiply the kilogram mass by 2.2046226218. The formula for converting 81 kilograms to pounds is [lb] = 81 * 2.2046226218. Thus, 81 kilos in pounds are 178.57443237 pounds.

One pound is about equivalent to 0.45359237 kilos (kg). The pound to kilogram conversion is accomplished by multiplying the provided pound value by 0.45359237. In addition, one kilogram is about equal to 2.2046226218 pounds.

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The distribution of X is a binomial distribution with n = 80 and p = 0.36. Using the distribution of X, the probability that Mark has between 27 and 32 (including 27 and 32) hits is 0.1919. The distribution of p is a normal distribution with mean μ = 0.36 and standard deviation σ = 0.05367. Using the distribution of p, the probability that Mark has between 27 and 32 hits is 0.4344.

a. The distribution of X is a binomial distribution with n = 80 and p = 0.36.

Since we are dealing with a large number of trials (80 at-bats) and a binary outcome (hit or no hit), we can model X using a binomial distribution. The distribution of X is B(n=80, p=0.36), where n is the number of trials, and p is the probability of success (getting a hit).

b. Using the binomial distribution, the probability that Mark has between 27 and 32 (including 27 and 32) hits is:
P(27 ≤ X ≤ 32) = \(\sum_{k=27}^{k=32} P(X=k)\)

= \(\sum_{k=27}^{k=32}(80 choose k) \times 0.36^k \times (1-0.36)^{(80-k)}\)

= 0.1919 (rounded to 4 decimal places)

c. The distribution of p is a normal distribution with mean μ = p = 0.36 and standard deviation

\(\sigma = \sqrt{((p\times(1-p))/n)}\)

\(= \sqrt{((0.36(1-0.36))/80)}\)

= 0.05367.

d. Using the normal distribution, we can standardize the range of 27 to 32 hits to the corresponding range of sample proportions using the formula:

z = (x - μ) / σ
where x is the number of hits, μ is the mean proportion (0.36), and σ is the standard deviation of the proportion (0.05367).

So, for 27 hits:
z = (27/80 - 0.36) / 0.05367 = -0.4192

For 32 hits:
z = (32/80 - 0.36) / 0.05367 = 0.7453

Then, we can use the standard normal distribution table or calculator to find the probability that z is between -0.4192 and 0.7453:
P(-0.4192 ≤ z ≤ 0.7453) = 0.4344

Therefore, the probability that Mark has between 27 and 32 hits is approximately 0.4344.

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The 95% confidence interval to estimate the variance of the weights of the packages prepared by the machine is [0.16, 0.32].

To calculate this interval, we can use the chi-squared distribution. Since the sample size is relatively large (n=93), we can assume that the sampling distribution of the sample variance is approximately normal. Using the chi-squared distribution with 92 degrees of freedom (df = n - 1), we can find the values of chi-squared that correspond to the lower and upper 2.5% tails of the distribution.

For the lower tail, we find the chi-squared value that corresponds to the 2.5th percentile with 92 degrees of freedom, which is 69.18. For the upper tail, we find the chi-squared value that corresponds to the 97.5th percentile with 92 degrees of freedom, which is 115.14.

We can then use these chi-squared values and the sample variance to construct the confidence interval as follows:

(lower chi-squared value / (n-1)) * sample variance ≤ population variance ≤ (upper chi-squared value / (n-1)) * sample variance

(69.18 / 92) * 0.23 ≤ population variance ≤ (115.14 / 92) * 0.23

0.16 ≤ population variance ≤ 0.32

Therefore, we can be 95% confident that the true variance of the weights of the packages prepared by the machine is between 0.16 and 0.32.

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Answer/Step-by-step explanation:

C = 82

Pythagorean Theorem: a^2 + b^2 = c^2

\(80^{2} = 6400\\18^{2} = 324\\6400 + 324 = 6724\\\sqrt{6724} = 82\)

The true statement about the function of area of by rectangle and width of rectangle are:

The correct answer option is options A and D

A rectangle is a quadrilateral having opposing sides parallel and four right angles.

At point (x, y) = (10, 100) the area is maximum.

This means that the area of the rectangle is maximum when width is 10m.

If the width of the rectangle is more than 10m, the area reduces or decreases.

Therefore, the statement Greater width relates to greater area as long as the width is more than 10 m is false.

The statement "To get the greatest area, the width should be 20" is false.

Complete question:

Simon has a certain length of fencing to enclose a rectangular area. The function A, models give the rectangles area (in square meters) as a function of its width (in meters).

Wlhich of these statements are true?

A. Greater width relates to greater area as long as the width is less than 10.

B. Greater width relates to greater area as long as the width is more than 10.

C. To get the greatest area, the width should be 10

D. To get the greatest area, the width should be 20

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The probability of exactly 4 successes in 5 trials with a probability of success of 60% is 0.2592 or about 25.9%.

It is expressed as a number between 0 and 1, where 0 represents an impossible event (i.e., it will never happen), and 1 represents a certain event (i.e., it will always happen).

The probability of exactly 4 successes in 5 trials of a binomial experiment with a probability of success of 60% can be calculated using the binomial distribution formula:

P(X = k) = (n choose k) * \(p^k\) * \((1-p)^(n-k)\)

where:

The likelihood that X will result in k successes is P(X = k).

The total number of trials is n, which in this instance is 5.

The number of successes for which we want to determine the probability, in this case k = 4, is k.

P is the likelihood that a trial would be successful (p in this instance is 0.6).

The binomial coefficient, (n choose k), is equivalent to n! / (k! * (n-k)!)

Plugging in the values, we get:

P(X = 4) = (5 choose 4) * \(0.6^4\) * \((1-0.6)^(5-4)\)

= 5 * \(0.6^4\) * \(0.4^1\)

= 0.2592

Therefore, the probability of exactly 4 successes in 5 trials with a probability of success of 60% is 0.2592 or about 25.9% when rounded to the nearest tenth of a percent.

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Answer:

\( \sf \: a) \: 27 = 9p \: ; p = 3\)

Step-by-step explanation:

Given information,

→ Simon have 27 pieces of candy.

→ Each person will get 9 pieces.

Now we have to,

→ Find the required equation.

The equation will be,

→ 27 = 9p

→ 9p = 27

=> As each person (p) gets 9 pieces.

Then the value of p will be,

→ 9p = 27

→ p = 27 ÷ 9

→ [ p = 3 ]

Hence, option (a) is correct.

Answer:the graph shows a straight line going up. The line starts at the bottom left corner and goes up towards the upper right corner. If it was non-ohmic the line would have a curve in it.

Step-by-step explanation: I hope I’m right

The probability is approximately 0.000183, or about 0.0183%. It's a very low probability, highlighting the challenge of randomly guessing the correct combination.

To calculate the probability that Sofie correctly identifies the set of 3 out of 33 ingredients used in the cake by randomly guessing, we can use the concept of combinations.

The total number of possible combinations of 3 ingredients chosen from a set of 33 ingredients can be calculated using the combination formula:

C(n, r) = n! / (r!(n - r)!)

where n is the total number of ingredients (33 in this case), and r is the number of ingredients chosen (3 in this case).

Plugging in the values:

C(33, 3) = 33! / (3!(33 - 3)!)

        = 33! / (3! * 30!)

        = (33 * 32 * 31) / (3 * 2 * 1)

        = 5456

There are 5456 possible combinations of 3 ingredients that Sofie can choose from.

Since Sofie is randomly guessing, there is only one correct combination out of the total possible combinations. Therefore, the probability of Sofie correctly identifying the set of 3 ingredients is:

Probability = 1 / 5456 ≈ 0.000183

So, the probability is approximately 0.000183, or about 0.0183%. It's a very low probability, highlighting the challenge of randomly guessing the correct combination.

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Answer:

Step-by-step explanation:

OK so i have already answered it but B was incorrect i think but i did it differently and i got -20000000.

The answers in standard form are

(a) 3.45 × 10² .

(b) 6.48 × 10⁻³ .

The exponent of a number indicates how long it will take to multiply that number in its entirety. A number's power or exponent, in other words, tells how many times it must be multiplied by itself.

Any integer, fraction, or decimal can serve as the basis in this situation. Additionally, the exponent can have any value, whether positive or negative.

Given :

(a) (2.3 × 10⁴) × (1.5×10⁻²)

    =\($ 2.3\times1.5\times10^{(4-2)\)

    = 3.45 × 10² .

(b) 3.6 × 10⁻⁵× 1.8 × 10²

\(= 3.6\times1.8\times10^{(-5+2)\)

= 6.48 × 10⁻³

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hi! im chimken and i have your answers!!!

1. polygon: is a closed figure where the sides are all line segments.

2. lateral face: a side of three-dimensional figure that is not a base.

3. pyramid: a solid figure that has a polygon for a base and triangles for sides. it is named for the shape of its base.

4. polyhedron: formed by two parallel, congruent, polygonal bases connected by lateral faces that are parallelograms.

5. i don't know this one :(

6. a net: a two dimensional representation of a three-dimensional figure that is unfolded along it's edges so that each face of the figure is shown in two dimensions.

7. face: is a plane figure that serves as one side of a solid figure.

8. i don't know this one :(

9. there are 4 possible answers. rectangular prism, sphere, cone, cylinder: 3D geometric shapes with the basic three-dimensional shapes.

10. triangular prism: with a known base and height of its face.

i hope this helped!!!

braise bingus!

The questions for grade 7 math worksheets are typically prepared by math teachers, curriculum specialists, and educational publishers.

These professionals use their knowledge of math concepts and pedagogy to create questions that are appropriate for the grade level and aligned with educational standards.

Additionally, they may use feedback from students and other teachers to refine the questions and ensure they are effective for teaching and learning. It is important for the questions to be clear, accurate, and challenging in order to help students develop their math skills and prepare for more advanced concepts.

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Answer:

y = 2x + 1

Step-by-step explanation:

You can see a pattern in the group of x's that the numbers go up by 1 Also in the y-set that the numbers go up by 2. So this pattern is linear, that means the "rule" you are looking for does not have exponents or square roots or any very complicated stuff. You can use a guess and check method. Say to yourself how can I get a 7 out, when I put a 3 in? "times by2 and plus 1" works.

3 times 2, and plus1

gives you 7.

Test it on the other numbers.

-1 times2, and plus1

2(-1)+1 = -1

2(0)+1 = 1

2(1)+1 = 3

2(2)+1 = 5

It works for all the numbers.

You can calculate it also, using any two pairs of (x,y) from the data set. Put y-y on top of a fraction and x-x on the bottom. You will get the slope and that is the 2 in the "rule"

(3,7) and (2,5) for example. 7-5 so put 2 on top and 3-2 so put 1 on the bottom. 2/1 is just 2. From the point (0,1) we know the y-intercept is 1. This also gives the equation y=2x+1.

If you are just starting to learn this, probably just guess and check a rule. The rule has to work for all the points.