N a competition for children at a local primary school, a container is filled with six cans of

Cola, five cans of Soda and one can of Fizz. The cans are identical in all respects except for

their contents. If a child is required to draw two cans without replacement from the

container, what is the probability that:

a) One can filled with Cola and one can filled with Fizz will be drawn

b) The cans will have the same contents?

c)The cans will differ with respect to their contents?

Check the picture below.

let's recall, twin sides make twin angles.

There are 91,390,400 different ways to choose president, vice president, treasurer, and secretary from a class of 40 students.

To find the number of different ways to choose president, vice president, treasurer, and secretary from a class of 40 students, we need to use the formula for permutations. This formula is nPr = n! / (n - r)!, where n is the total number of items and r is the number of items being chosen. In this case, n = 40 and r = 4, so we have equations:

40P4 = 40! / (40 - 4)!
40P4 = 40! / 36!
40P4 = (40 x 39 x 38 x 37) / (4 x 3 x 2 x 1)
40P4 = 91,390,400

Therefore, there are 91,390,400 different ways to choose president, vice president, treasurer, and secretary from a class of 40 students. This means that each person in the class has a 1 in 91,390,400 chance of being chosen for all four positions.  

It's important to note that this calculation assumes that each person can only hold one position, and that the order in which the positions are filled matters. If these assumptions are changed, then a different formula would need to be used.

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Answer:

Step-by-step explanation:

In order to solve this, we need to know the definitions of the vocabulary terms and how to identify them in a polynomial expression. Here are the definitions:

To match the vocabulary terms with their “parts” of the given expression, you need to drag and drop the correct numbers from the options. Here are the matches:

Answer:

K = 2cm

Step-by-step explanation:

The figure are similar by AA similarity theorem

A-A similarity theorem was deals with at least the two angles are congrunt then we said that figure is similar.

So in the given figure ∆SRT ~ ∆QPT and if the two triangles are dimilar their coresponding side ratio are equal and their angles are congrunt .

Now our question is to find the scale factor . let it be K

K = QP/SR = PT/RT = QT/ST

K = 4cm/2cm = 6cm/3cm= 6cm /3cm

K = 2cm = 2cm = 2cm ... so the scale factor is 2cm.

The figure above shows two circles, one inside the other. The equation x^2 + y^2 = 2 represents the smaller circle, while the larger circle is not explicitly defined in the question.

However, we can see from the graph that the center of both circles is at the origin (0,0) and that the larger circle has a radius twice as big as the smaller circle.

Therefore, the radius of the larger circle is 2 times the radius of the smaller circle, which is the square root of 2. So the answer is (a) √2.
The given equation of the circle is x^2 + y^2 = 2. To find the radius of this circle, we can rewrite the equation in the standard form of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.

In this case, the equation is already in the standard form with h = 0, k = 0, and r^2 = 2. Therefore, the radius of the larger circle is r = √2. So, the correct answer is a) √2.

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Answer: D

Step-by-step explanation:

To determine the maximum number of stories that can be constructed before friction losses prevent proper valve operation, we need to calculate the pressure loss due to friction as the water flows through the piping to each floor.

Using the Hazen-Williams formula, which is commonly used for sizing water supply systems:

P = (4.52Q1.85L10.67/C1.85)d4.87

where:

P = pressure loss due to friction (psi)

Q = flow rate (gpm)

L = length of pipe (feet)

C = Hazen-Williams coefficient (dimensionless)

d = inside diameter of pipe (inches)

Assuming the flow rate is 10 gpm, the length of pipe from floor to floor is the height of the building divided by the number of stories, and the inside diameter of the pipe is 4 inches (since 4 type-L copper corresponds to a 4-inch nominal diameter), the pressure loss for each story can be calculated using a Hazen-Williams coefficient of 130 for copper piping:

P = (4.52 x 10^1.85 x (1 story height/number of stories)10.67/1301.85)4.87

P = 3.3 x (1/number of stories)^1.85

For example, for a 2-story building, the pressure loss would be:

P = 3.3 x (1/2)^1.85

P = 1.6 psi

To ensure that the minimum working pressure of 25 psi is maintained at each flush valve, the pressure loss for each story cannot exceed 25 - 50 = -25 psi (since lower pressures can cause valve malfunctions).

Solving for the maximum number of stories:

3.3 x (1/number of stories)^1.85 <= -25

(1/number of stories)^1.85 <= -25/3.3

1/number of stories <= (-25/3.3)^0.54

number of stories >= 1/(-25/3.3)^0.54

number of stories >= 6.7

Therefore, the maximum number of stories that can be constructed before friction losses prevent proper valve operation is D. This floor only.

The table represents a nonlinear function because both the slope and common difference are not constant.

Based on the pattern in the table, the commission of selling 5 cell phone plans is 600.

In Mathematics, a linear function is a type of function whose equation is graphically represented by a straight line on the cartesian coordinate.

This ultimately implies that, a linear function has the same slope and it is typically used for uniquely mapping an input variable to an output variable, which both increases or decreases simultaneously.

Therefore, this table does not represent a linear function because the slope isn't constant and the common difference is not constant between successive data points.

When x = 5 cell phone plans, y is given by;

y - 400 = 200

y = 600.

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

The differentiation of H(X)=43(X5−3x3+7)⁸ is H′(x) = 344(x^5 - 3x^3 + 7)^7(5x^4 - 9x^2)

To find the derivative of the function H(x) = 43(x^5 - 3x^3 + 7)^8, we need to use the chain rule of differentiation. The chain rule states that if we have a function f(g(x)), then its derivative is given by f'(g(x)) * g'(x). In this case, our function g(x) is (x^5 - 3x^3 + 7), and our function f(x) is (43x^8). So, using the chain rule, we get:

H′(x) = f'(g(x)) * g'(x)

H′(x) = 43(8)(x^5 - 3x^3 + 7)^7 * (5x^4 - 9x^2)

Simplifying the expression, we get:

H′(x) = 344(x^5 - 3x^3 + 7)^7(5x^4 - 9x^2)

Therefore, the derivative of the function H(x) = 43(x^5 - 3x^3 + 7)^8 is H′(x) = 344(x^5 - 3x^3 + 7)^7(5x^4 - 9x^2).

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Option D is correct as it justifies the two systems being equivalent

Multiply by 2 the first equation A

2 (X + 2Y ) = 6

This would give us the first part of the equation B

= 2x + 4y = 12

Then the second equation

Multiply by 5 to get B

5(2/5x - y ) = 3

= 2x - 5y = 15

Hence the last option is the correct answer

Therefore we would say that Option D is correct as it justifies the two systems being equivalent

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B. Mean is not resistant to extreme values in a numeric data set. The mean is affected by outliers or extreme values in a dataset because it takes into account all the values in the dataset, including the extreme values.

Central tendency is a measure that is used to find the central or typical value of a set of data. The most common measures of central tendency are mean, median, and mode. These measures provide different ways of summarizing the data, depending on the nature of the dataset and the research question.

The mean is calculated by adding up all the values in a dataset and dividing by the number of values. The mean is often considered to be the most common measure of central tendency because it takes into account all the values in the dataset. However, the mean is not resistant to extreme values or outliers. An outlier is an observation that is significantly different from other observations in the dataset.

When a dataset contains extreme values or outliers, the mean is often affected, resulting in a value that does not represent the central tendency of the dataset. This is because the mean is influenced by every value in the dataset, including the outliers, and is not a robust measure of central tendency. In contrast, the median and mode are resistant to extreme values and provide better estimates of central tendency when a dataset contains outliers.

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To find the maximum and minimum values of the given function f(θ), we need to find its critical points and endpoints.

First, we take the derivative of f(θ) with respect to θ:

f'(θ) = -2 sin(θ) - 2 cos(θ) sin(θ)

Setting f'(θ) = 0, we get:

-2 sin(θ) - 2 cos(θ) sin(θ) = 0

Simplifying and factoring out -2 sin(θ), we get:

-2 sin(θ) (1 + cos(θ)) = 0

This gives us two critical points: θ = 0 and θ = π.

Next, we need to evaluate the function at the endpoints of the given interval [0, 2π]:

f(0) = 2 cos(0) + cos2(0) = 3

f(2π) = 2 cos(2π) + cos2(2π) = 3

Now we compare the values of f(θ) at the critical points and endpoints to find the maximum and minimum values:

f(0) = 3 (endpoint)

f(π) = 1 (critical point)

f(2π) = 3 (endpoint)

Therefore, the maximum value of f(θ) is 3 and the minimum value is 1.

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Answer:

Step-by-step explanation:

15.3

+1.285

Answer = 16.585

Answer:

False

Step-by-step explanation:

In order to compare fractional terms both sides, you have to make sure that both sides are equal in denominator. In this case, our denominators are 3 and 12. To equalize the denominator of both sides, find the LCM of 3 and 12 which is 12.

Therefore, we multiply 2/3 by 4/4. Therefore:

[tex]\displaystyle{\dfrac{2\cdot 4}{3\cdot 4} = \dfrac{9}{12}}\\\\\displaystyle{\dfrac{8}{12} = \dfrac{9}{12}}[/tex]

The equation is false since 8/12 does not actually equal 9/12. You can also try cutting the pieces into 12 pieces total and compare both shaded regions; you'll see that both are not equivalent at all.

The range of values in the upper 15.85% of the distribution is given as follows:

x > 19.4.

The z-score of a measure X of a normally distributed variable that has mean represented by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is obtained by the equation presented as follows:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The mean and the standard deviation are given as follows:

[tex]\mu = 16, \sigma = 3.4[/tex]

The value of interest is Z with a p-value of 1 - 0.1585 = 0.8415 = 1, hence the value of x is obtained as follows:

1 = (x - 16)/3.4

x - 16 = 3.4

x = 19.4.

Hence the interval is:

x > 19.4.

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The option that gives chart elements a realistic, three-dimensional look is D. Plot Area.

The option that gives chart elements a realistic, three-dimensional look is the plot area. The plot area is the portion of the chart that displays the data points and is typically the main focus of the chart. Adding shading and depth to the plot area, can give the chart a more lifelike appearance and make the data more visually engaging. Additionally, three-dimensional charts can also be created by using specialized charting software that allows for the creation of 3D visualizations. These types of charts can be useful for highlighting patterns and trends in the data, especially when dealing with large datasets.
In a chart, the plot area is the space where the actual data points, bars, lines, or pie slices are displayed. By applying three-dimensional effects to the plot area, you can make the chart elements appear more realistic and visually appealing. This can enhance the viewer's understanding of the data being represented and make the chart more engaging.
In summary, to give chart elements a realistic, three-dimensional look, you should focus on the plot area (Option D). The axis (Option A), chart area (Option B), and legend (Option C) do not directly contribute to creating a three-dimensional appearance for the chart elements.

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When rolling a single six-sided die, there are three odd numbers: 1, 3, and 5, and three even numbers: 2, 4, and 6. Since we are rolling two dice, each with six possible outcomes, there are a total of $6 \times 6 = 36$ possible outcomes.

To find the sample space corresponding to the event that both dice are odd, we need to determine all possible outcomes in which both dice land on odd numbers. There are three odd numbers on each die, so there are three possibilities for the first die and three possibilities for the second die, for a total of $3 \times 3 = 9$ outcomes where both dice are odd.

Therefore, the size of the sample space that corresponds to the event that both dice are odd is 9.

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The probability P(X < 14) when μ₁ =.... =μ= 15 and o₁2 = ... = o2/7 = 7 is 0.

Let X1, ..., X7 be independent normal random variables, each distributed as N(μ1, o₁²) for i = 1,...,7. Given that μ₁ = ... = μ = 15 and o₁² = ... = σ²/7 = 7, you want to find P(X < 14) rounded to the second decimal place.

To solve this problem, follow these steps:

1. Calculate the mean and variance of the sum of the random variables.
Since X1, ..., X7 are independent and identically distributed, the mean of their sum is the sum of their means, and the variance of their sum is the sum of their variances. Thus, for the sum S = X1 + ... + X7, we have:

Mean of S = 7 * μ = 7 * 15 = 105
Variance of S = 7 * σ²/7 = 7 * 7 = 49

2. Calculate the standard deviation of the sum.
The standard deviation is the square root of the variance, so:

Standard deviation of S = √49 = 7

3. Calculate the z-score for X < 14.
The z-score is calculated as:

z = (X - Mean) / Standard deviation

Since we want to find probability P(X < 14), we need to find the z-score for X = 14:

z = (14 - 105) / 7 = -91 / 7 = -13

4. Find the probability P(X < 14) using the z-score.
Using the z-score of -13, you can look up the corresponding probability in a standard normal distribution table or use a calculator that provides this functionality. A z-score of -13 is extremely far from the mean, so the probability P(X < 14) is very close to 0.

To the second decimal place, the probability P(X < 14) is approximately 0.00.

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There are 187 students in the sixth grade.

We can start by using the information that 40% of the students bought a pass to ride an unlimited number of rides. This means that 60% of the students did not buy a pass and paid for each ride individually.

Let's assume that the total number of students in the sixth grade is "x". Then, 40% of the students bought a pass, which is 0.4x. The remaining 60% paid for each ride individually, which is 0.6x.

We also know that 75 students paid for each ride individually. Therefore, we can set up an equation:

0.6x = 75

Solving for x, we get:

x = 75 / 0.6

x = 125 / 0.4

x = 187.5

Since we can't have half a student, we round up to the nearest whole number. Therefore, there are 187 students in the sixth grade.

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The approximate value of the car in 3 years is $1648.37.

We have,

To calculate the approximate value of the car in 3 years, we can use the formula for exponential decay:

V = Vo e^(-kt)

where V is the value of the car after t years, Vo is the initial value of the car, k is the decay rate (in this case, 0.3), and e is the mathematical constant e (approximately equal to 2.718).

Substituting the given values.

V = 6000 x e^(-0.33)

V = 6000 x e^(-0.9)

V ≈ $1648.37

Therefore,

The approximate value of the car in 3 years is $1648.37.

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The critical value tc for the confidence level c=0.99 and sample size n=21 is 2.845.

To find the critical value tc for the confidence level c = 0.99 and sample size n = 21, we can use the t-distribution table or a statistical software package.

t-distribution critical value formula is: tc = t(α/2, n-1)

For a confidence level of 0.99, the significance level α is:

α = 1 - c = 1 - 0.99 = 0.01

Using a t-distribution table with 20 degrees of freedom (n-1), we can find the t-value with a cumulative probability of 0.005 in the upper tail, since we are interested in the critical value at the two-sided 99% confidence level.

The t-value with a cumulative probability of 0.005 and 20 degrees of freedom is 2.845.

Therefore, the critical value tc for the confidence level c = 0.99 and sample size n = 21 is:

tc = t(α/2, n-1) = t(0.005, 20) = 2.845

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The circumference of the circle is given as follows: C = 109.9 in.

The relationship between the diameter of a circle and it's circumference is presented in this problem.

We have,

The circumference of a circle of radius r is given by the equation presented as follows:

C = 2πr

The radius of a circle represents the distance between the center of the circle and a point on the circumference of the circle, while the diameter of the circle is the distance between two points on the circumference of the circle that pass through the center. Hence, the diameter’s length is twice the radius length.

The diameter is twice the radius, hence, for a circle with diameter 35 in, the radius is given as follows:

r = 0.5 x 35

r = 17.5 in.

Then the circumference of the circle is given as follows:

C = 2 x pi x 17.5

C = 109.9 in.

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Missing Information

The problem is incomplete, hence the relation between the diameter of a circle and it's circumference is presented.

the discrepancy between the sample mean and the claimed protein content is due to chance error in the measurements

To determine whether the discrepancy between the sample mean and the claimed protein content of 4 grams can be due to chance error in the measurements, we can perform a hypothesis test.

The null hypothesis, H0, is that the true mean protein content is equal to 4 grams. The alternative hypothesis, Ha, is that the true mean protein content is less than 4 grams.

Using a one-tailed t-test with a significance level of 0.05 and four degrees of freedom (n-1=5-1=4), we calculate the t-value to be -1.386. The critical t-value for a one-tailed test at a significance level of 0.05 and four degrees of freedom is -2.132.

Since our calculated t-value of -1.386 is greater than the critical t-value of -2.132, we fail to reject the null hypothesis. This means that we do not have enough evidence to conclude that the true mean protein content is less than 4 grams. Therefore, it is possible that the discrepancy between the sample mean and the claimed protein content is due to chance error in the measurements.

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The value of the missing leg using Pythagoras theorem is 12.5.

Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“

To calculate the missing leg, we use Pythagoras theorem.

Where:

Substitute these values into equation 1 and solve for c

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There are 252 different teams that can be selected from the 10 available engineers for the upcoming project.

To solve this problem, we can use the combination formula. The number of combinations of choosing r items from n items is given by nCr = n!/(r!(n-r)!). In this case, n = 10 (the total number of available engineers) and r = 5 (the number of engineers to be selected for the team). Plugging these values into the formula, we get:

10C5 = 10!/(5!(10-5)!) = 252

Therefore, there are 252 different teams that can be selected from the 10 available engineers for the upcoming project.

It's important for the engineering manager to carefully consider the skills and strengths of each engineer when selecting the team. By choosing a diverse group with complementary skills, the team will be better equipped to tackle the project and achieve success. The manager should also consider the individual personalities and work styles of each engineer to ensure that they can work together effectively and collaboratively. Overall, the team selection process should be thoughtful and strategic to maximize the project's potential for success.

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Answer:3xf

Step-by-step explanation:

you just reverse the 40∞•§ with the 589€¢¨•§•ª¶¶ which gives you 3xf

Answer: 527

Step-by-step explanation:

As applied to a polygon, a diagonal is a line segment joining any two non-consecutive vertices. Therefore, a quadrilateral has two diagonals, joining opposite pairs of vertices.

The sampling method used in this study is a simple random sample. This is because each person's name has an equal chance of being selected when the darts are blindly thrown at the dartboard.

The sample is also random because there is no predetermined criteria or grouping used in selecting individuals.

The other sampling methods such as systematic, stratified, and cluster involve pre-determined groups or criteria that the individuals must meet, which is not the case with the dartboard method.

Additionally, convenience sampling involves selecting individuals based on their availability or accessibility, which is not applicable in this scenario as the individuals are selected randomly without any biases or preferences.

Overall, the study is using a simple random sampling method to select individuals for the sample.

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In the 1984 domestic violence study conducted by Sherman and Berk, the ethical concern was that they potentially withheld a beneficial treatment.

The domestic violence study conducted by Sherman and Berk in 1984 raised an ethical concern in that they financially profited from the research. This raises the question of whether their motives were purely altruistic or whether they were driven by financial gain. Additionally, the study did not adhere to special protections for vulnerable populations such as women and children who may have been victims of domestic violence. This raises concerns about the validity and generalizability of the study's findings. Furthermore, the study potentially withheld a beneficial treatment, which raises questions about the ethical responsibility of researchers to ensure that their subjects receive the best possible care. Finally, there are also concerns that the researchers may have deceived their subjects, which raises questions about the integrity and transparency of the research process. In conclusion, the ethical concerns raised by this study highlight the need for researchers to carefully consider the impact of their research on vulnerable populations and to ensure that they adhere to the highest ethical standards.
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