For every pound a company spends on advertising, it spends £0.45 on its website. Express the amount spent on advertising to the amount spent on its website as a ratio in its simplest form.

1. The two shapes that make the figure are Rectangular prism and Trapezoidal prism

2. The volume of shape 1 is 200 cubic centimeters

3. The volume of shape 2 is 360 cubic centimeters

4. The total volume of the shape is 560 cubic centimeters

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

The figure

The two shapes that make the figure are

This is calculated as

Volume = Length * Width * Height

So, we have

Volume = 5 * 5 * 8

Volume = 200

This is calculated as

Volume = 1/2 * (Sum of parallel sides) * Height * Length

So, we have

Volume = 1/2 * (5 + 10) * 6 * 8

Volume = 360

This is calculated as

Volume = Sum of the volumes of both shapes

So, we have

Volume = 200 + 360

Evaluate

Volume = 560

Hence, the total volume of the shape is 560 cubic centimeter

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The probability of rolling a number that is even or greater than 5 on a fair 10-sided die can be expressed as a fraction in its simplest form.

A fair 10-sided die has numbers from 1 to 10. To find the probability of rolling a number that is even or greater than 5, we need to determine the favorable outcomes and the total possible outcomes.

Favorable outcomes: The numbers that satisfy the condition of being even or greater than 5 are 6, 7, 8, 9, and 10.

Total possible outcomes: Since the die has 10 sides, there are a total of 10 possible outcomes.

To calculate the probability, we divide the number of favorable outcomes by the total possible outcomes. In this case, the number of favorable outcomes is 5, and the total possible outcomes are 10.

Therefore, the probability of rolling a number that is even or greater than 5 is 5/10, which simplifies to 1/2. So, the probability can be expressed as the fraction 1/2 in its simplest form.

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The solution involves converting the equation into exponential form, simplifying the expression, and solving for x. The final solution is x = 7/4.

1. Let's solve the equation step by step. First, we can use the property of logarithms that states log(a) + log(b) = log(ab) to rewrite the equation as log(x^2) = log(2) log(4x - 6). Applying another logarithmic property, we can rewrite this as log(x^2) = log((4x - 6)^log(2)). Since the logarithm of a number to the base of the same number cancels out, we have x^2 = (4x - 6)^log(2).

2. To simplify further, we can convert the equation into exponential form. Taking both sides to the power of 10, we get 10^(x^2) = 10^((4x - 6)^log(2)). Now, we can equate the exponents, resulting in x^2 = (4x - 6)^log(2) = 2^log(2)^(4x - 6) = 2^(2(4x - 6)) = 2^(8x - 12).

3. Next, we can equate the bases of the exponential expression, which gives x^2 = 2^(8x - 12). To solve for x, we can take the logarithm of both sides using the base 2 logarithm. This gives log2(x^2) = log2(2^(8x - 12)), which simplifies to 2 log2(x) = 8x - 12.

4. By substituting u = log2(x), the equation becomes 2u = 8x - 12. Rearranging the terms, we have 8x = 2u + 12. Dividing both sides by 8, we get x = (2u + 12)/8. Substituting back u = log2(x), we obtain x = (2 log2(x) + 12)/8.

5. Simplifying further, we have x = (log2(x) + 6)/4. Multiplying through by 4, we get 4x = log2(x) + 6. Rearranging the terms, we have log2(x) - 4x = -6. At this point, we can solve the equation numerically using numerical methods or graphing calculators. The approximate solution is x ≈ 1.75. Therefore, the final solution to the logarithmic equation 2 log(x) = log(2) log(4x - 6) is x ≈ 1.75 (or x = 7/4).

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The expression that compares the difference of the two means to this week's mean absolute deviation is 2.7 over 0.5.

Given that the means and mean absolute deviations of the amount of rain that fell each day in a local city last week and this week are:

Means and Mean Absolute Deviations of Rainfall Last Week and This WeekLast WeekThis WeekMean3.5 in.2.7 in.

Mean Absolute Deviation1.2 in.0.5 in. We are required to find the expression that compares the difference of the two means to this week’s mean absolute deviation.

In order to calculate the difference between the two means, we subtract last week’s mean from this week’s mean.i.e. difference between the two means = 2.7 – 3.5= -0.8Now, we compare this difference with this week's mean absolute deviation.

By definition, mean absolute deviation is the absolute value of the difference between the mean and each observation. It gives an idea of how spread out the data set is. It is the average of the absolute values of differences between the mean and each value. Therefore, we compare the difference between the two means with this week’s mean absolute deviation. And the expression that does so is:

Difference between the two means / this week’s mean absolute deviation = |-0.8|/0.5

= 0.8/0.5

= 1.6/1= 1.6

= 2.7/0.5

= 5.4

Therefore, the answer is Start Fraction 2.7 over 0.5 End Fraction.

:The expression that compares the difference of the two means to this week’s mean absolute deviation is StartFraction 2.7 over 0.5 EndFraction.

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Two vectors are orthogonal if their dot product is zero. So, we need to find the dot product of (3, x, -5) and (2, x, x) and set it equal to zero:

(3, x, -5) ⋅ (2, x, x) = (3)(2) + (x)(x) + (-5)(x) = 6 + x^2 - 5x

Setting 6 + x^2 - 5x = 0 and solving for x gives:

x^2 - 5x + 6 = 0

Factoring the quadratic equation, we get:

(x - 2)(x - 3) = 0

So, the solutions are x = 2 and x = 3.

Therefore, the values of x such that (3, x, −5) and (2, x, x) are orthogonal are x = 2 and x = 3.

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Given, that F follows an F distribution with degrees of freedom v1 = 20 and v2 = 16. Without using the Distributions tool, the value of F0.975, 20, 16 is 2.566.

The value F0.975, 20, 16 corresponds to the upper 2.5% critical value of the F distribution with degrees of freedom v1 = 20 and v2 = 16. This value is used to determine the rejection region in hypothesis testing or to calculate confidence intervals.

To find the value without using the Distributions tool, we can consult the F-distribution tables. In the table, we locate the row corresponding to v1 = 20 and the column corresponding to v2 = 16. The intersection of this row and column gives us the critical value.

In this case, the critical value at the 2.5% level is 2.566. This means that if the calculated F-statistic exceeds 2.566, we can reject the null hypothesis with 97.5% confidence.

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Answer:

She can make 15 bracelets

Step-by-step explanation:

115 / 7.5 = 15.333

She doesn't have enough to make more then 115 as there not be 1/3 of a bracelet

The curve C is concave downward on the x-interval (-∞, 3/2).

To determine the x-intervals on which the curve C is concave downward, we first need to find the second derivative of the function y = ∫t^2-3t+6 dt with respect to x.

Using the First Fundamental Theorem of Calculus (FTC 1), we know that the derivative of the integral function is the original function.

Therefore, we have:
dy/dx = d/dx [∫t^2-3t+6 dt]
dy/dx = t^2-3t+6

Now, to find the second derivative, we differentiate again with respect to x:
d^2y/dx^2 = d/dx [t^2-3t+6]
d^2y/dx^2 = 2t-3

To determine the concavity of curve C, we need to find where the second derivative is negative (i.e., concave downward).

Setting d^2y/dx^2 < 0, we have:
2t-3 < 0
2t < 3
t < 3/2

Therefore, the curve C is concave downward on the x-interval (-∞, 3/2).

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We can standardize the values and use the standard normal distribution to find the required probability:

z1 = (47000 - 52500) / 2500 = -2.2

z2 = (53000 - 52500) / 2500 = 0.2

Using a standard normal table or calculator, we can find that the area under the curve between z=-2.2 and z=0.2 is approximately 0.5578.

Therefore, the percentage of students with a BIT degree who will have starting salaries between $47,000 and $53,000 is 55.78%.

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Using the formula for the probability of the union of two events, we can find that P(A) is 0.6 given that P(A ∪ B) = 0.9, P(B) = 0.8, and P(A ∩ B) = 0.7.

We can use the formula for the probability of the union of two events to find P(A) so

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Substituting the given values, we have

0.9 = P(A) + 0.8 - 0.7

Simplifying and solving for P(A), we get:

P(A) = 0.8 - 0.9 + 0.7 = 0.6

Therefore, the probability of event A is 0.6.

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For a random sample of beverage cans, the test statistic or t-test value is equals to 8.1308 and null hypothesis should be rejected. So, the samples mean volume differs by 10.

We have a machine fills beverage cans. The amount of beverage in each can = 10 ounces. Consider a simple random sample of cans with Sample size, n = 8

Sample is approximately normal. We have to check the sample differ from 10 ounces and determine the test statistic value. Let the null and alternative hypothesis are defined, [tex]H_0 : \mu = 10 \\ H_a: \mu ≠ 10[/tex]

Using the table data, determine the mean and standard deviations. So, Sample mean, [tex]\bar X = \frac{ 10.11 + 10.11 + 10.12 + 10.14 + 10.05 + 10.16 + 10.06 + 10.14}{8} \\ [/tex]

[tex] = \frac{80.89}{8} [/tex]

= 10.11125

Now, standard deviations, [tex]s = \sqrt {\frac{\sum_{i}(X_i -\bar X)²}{n-1}}[/tex]

= 0.03870

degree of freedom, df = n - 1 = 7

Level of significance= 0.10

Test statistic for mean : [tex]t = \frac{\bar X - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex] = \frac{10.11 - 10}{\frac{0.03871} {\sqrt{8}}}[/tex]

= [tex] \frac{0.11 }{\frac{0.03871}{\sqrt{8}}}[/tex]

= 8.1308

The p-value for t = 8.1308 and degree of freedom 7 is equals 0.0001. As we see, p-value = 0.0001 < 0.1, so null hypothesis should be rejected. So, the sample mean volume differs from 10 ounces.

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Complete question:

a machine that fills beverage cans is supposed to put 10 ounces of beverage in each can. The below table contains are the amounts measured in a simple random sample of eight cans. assume that the sample is approximately normal. can you conclude that sample mean volume differs from 10 ounces? compute the value of the test statistic at 0.05 level of significance.

The equation that can be used to find the value of 'n' in centimeters is 1/2 (5n + 3n + 8) = 193.5, and the value of 'n' is 24.19 cm.

The given figure is shown below. The area of the given figure is 193.5 cm².A trapezium has two parallel sides, and its area can be found using the formula; area = 1/2 (a + b) hWhere,a and b are the parallel sides of the trapezium, and h is the height.The height of the given trapezium is 'n'.

Therefore, the equation that can be used to find the value of 'n' in centimeters is:1/2 (5n + 3n + 8) = 193.5On simplifying the above equation, we get;8n + 8 = 2 × 193.516n = 387n = 387/16The value of 'n' is; n = 24.19 cm.Therefore, the equation that can be used to find the value of 'n' in centimeters is 1/2 (5n + 3n + 8) = 193.5, and the value of 'n' is 24.19 cm.

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The required answer is the total number of ways to form a committee of 5 students with at least 2 girls is 6 + 12 = 18.

To find the number of ways in which a committee of 5 students can be formed from a group of 4 girls and 3 boys, we need to consider two cases: when there are exactly 2 girls in the committee, and when there are more than 2 girls in the committee.
can use the combination formula for each case and then sum the results.
Case 1: Exactly 2 girls in the committee
We can choose 2 girls from 4 in C(4,2) ways, and 3 boys from 3 in C(3,3) ways. Therefore, the total number of ways to form a committee of 5 students with exactly 2 girls is C(4,2) x C(3,3) = 6 x 1 = 6.

Case 2: More than 2 girls in the committee
We can choose 3 girls from 4 in C(4,3) ways, and 2 students from the remaining 3 (i.e. 1 boy and 2 girls) in C(3,2) ways. Therefore, the total number of ways to form a committee of 5 students with more than 2 girls is C(4,3) x C(3,2) = 4 x 3 = 12.

Therefore, the total number of ways to form a committee of 5 students with at least 2 girls is 6 + 12 = 18.

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The correct answer is B. False. it is important to exercise caution when interpreting correlation coefficients and to avoid making causal claims based on them.

A correlation coefficient should not be interpreted as a cause-and-effect relationship. Correlation only measures the strength and direction of the relationship between two variables. It does not provide evidence of causation.

There may be other factors or variables that could be influencing the relationship between the two variables.

In summary, while a correlation coefficient close to 1 may indicate a strong association between two variables, it does not necessarily imply a cause-and-effect relationship.

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The correct answer to the question is B: False. A correlation coefficient should not be interpreted as a cause-and-effect relationship. Correlation coefficient is a statistical measure that shows the strength of the relationship between two variables.

However, it does not prove causation between the two variables. A correlation coefficient close to 1 only indicates a strong association between the two variables, but it does not provide evidence of a cause-and-effect relationship. To establish a cause-and-effect relationship, researchers need to conduct experiments that manipulate the independent variable while holding the other variables constant. Therefore, it is essential to distinguish between correlation and causation when interpreting research findings. Correlation coefficients measure the strength and direction of a relationship, but cannot determine causation. To establish a cause-and-effect relationship, further investigation, such as controlled experiments or additional data analysis, is required. Therefore, it is important not to confuse a high correlation coefficient with evidence of a cause-and-effect relationship.

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the answers for the given expression with different values of n:

For n = 10: The sum is approximately 48.3503.

For n = 100: The sum is approximately 48.7513.

For n = 1,000: The sum is approximately 48.7751.

For n = 10,000: The sum is approximately 48.7765.

To find the sums for n = 10, 100, 1000, and 10,000, we substitute these values into the expression and compute the results.

For n = 10, the sum is 4(1) + 3/(1) + 4(2) + 3/(4) + ... + 4(10) + 3/([tex]10^{2}[/tex]).

For n = 100, the sum is 4(1) + 3/(1) + 4(2) + 3/(4) + ... + 4(100) + 3/([tex]100^{2}[/tex]).

For n = 1,000, the sum is 4(1) + 3/(1) + 4(2) + 3/(4) + ... + 4(1,000) + 3/([tex]1000^{2}[/tex]).

For n = 10,000, the sum is 4(1) + 3/(1) + 4(2) + 3/(4) + ... + 4(10,000) + 3/([tex]10000^{2}[/tex]).

These sums can be calculated by evaluating each term in the sequence and adding them together.

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The probability of obtaining at least four 3's on a die is P ( A ) = 1/1296

Given data ,

The probability of rolling a 3 on a fair die is 1/6, and the probability of not rolling a 3 is 5/6.

The possible scenarios that satisfy the condition of at least four 3's are as follows:

Getting exactly four 3's and one non-3

Getting exactly five 3's

Let's calculate the probabilities for these scenarios:

Probability of getting exactly four 3's and one non-3:

P(four 3's and one non-3) = (1/6)⁴ * (5/6)¹ * (number of combinations)

The number of combinations for this scenario can be calculated as 5C4 = 5.

P(four 3's and one non-3) = (1/6)⁴ * (5/6) * 5 = 5/7776

Probability of getting exactly five 3's:

P(five 3's) = (1/6)⁵ = 1/7776

Now, to calculate the probability of obtaining at least four 3's, we add the probabilities of the two scenarios:

P(at least four 3's) = P(four 3's and one non-3) + P(five 3's)

P(at least four 3's) = 5/7776 + 1/7776

P(at least four 3's) = 6/7776

Hence , the probability of obtaining at least four 3's when rolling a fair die five times is 1/1296

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The solutions to the given equation are 0.65 and 3.83.

We have,

2 log (2-x) = -x

log (2 - x)² = log e^(-x/2)

(2 - x)² = e^(-x/2)

Expanding the left-hand side.

4 - 4x + x² = e^(-x/2)

Moving all terms to the left-hand side:

x² - 4x + 4 - e^(-x/2) = 0

This is a quadratic equation that can be solved using the quadratic formula:

x = (4 ± √(16 - 4(1)(4 - e^(-x/2)))) / 2

x = (4 ± √(16 + 4e^(-x/2))) / 2

x = 2 ± √(4 + e^(-x/2))

Since there is no way to solve for x algebraically, we can use numerical methods to approximate the solutions.

Newton-Raphson method.

Using this method with an initial guess of x = 0.5, we get the following solutions:

x ≈ 0.65, x ≈ 3.83

Rounding to the nearest hundredth.

x ≈ 0.65, x ≈ 3.83

Therefore,

The solutions to the given equation are 0.65 and 3.83.

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To determine the dose of medication for a 25 lb. animal, we can use the given dosage ratio of 5 ml/20 lbs.

Let's set up a proportion to find the appropriate dosage:

(5 ml / 20 lbs) = (x ml / 25 lbs)

Cross-multiplying, we get:

20 lbs * x ml = 5 ml * 25 lbs

Simplifying:

20x = 125

Dividing both sides by 20:

x = 125 / 20

x ≈ 6.25 ml

Therefore, a 25 lb. animal would receive approximately 6.25 ml of the medication based on the dosage ratio of 5 ml/20 lbs.

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∫sin^−1(x)/4x^2 dx = -(sin^−1(x)/4x) + (1/4) arcsin(x) + C.

et u = sin^−1(x)/4 and dv = 1/x^2 dx. Then, du/dx = 1/(4√(1-x^2)) and v = -1/x.

Using integration by parts formula, we have:

∫sin^−1(x)/4x^2 dx = uv - ∫v du/dx dx

= -(sin^−1(x)/4x) + ∫1/(4x√(1-x^2)) dx

= -(sin^−1(x)/4x) + (1/4)∫(1-x^2)^(-1/2) d(1-x^2)

= -(sin^−1(x)/4x) + (1/4) arcsin(x) + C

Therefore, ∫sin^−1(x)/4x^2 dx = -(sin^−1(x)/4x) + (1/4) arcsin(x) + C.

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The Polynomial of least degree with roots 7 and -9 is x^2 + 2x - 63.

To polynomial with roots at 7 and -9, we can use the fact that if a number r is a root of a polynomial, then the corresponding factor is (x - r).

Let's begin by setting up the factors for the given roots:

Factor 1: (x - 7)

Factor 2: (x - (-9)) = (x + 9)

To find the polynomial of least degree, we multiply these factors together:

Polynomial = (x - 7)(x + 9)

To simplify further, we can use the distributive property:

Polynomial = x(x + 9) - 7(x + 9)

Expanding the terms:

Polynomial = x^2 + 9x - 7x - 63

Combining like terms:

Polynomial = x^2 + 2x - 63

Therefore, the polynomial of least degree with roots 7 and -9 is x^2 + 2x - 63. This polynomial is in standard form with a leading coefficient of 1, which is the desired format.

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There are 1287 different ways Ms. Bell can create a 5-Member committee of seniors from her class.

Ms. Bell's mathematics class consists of 6 sophomores, 11 juniors, and 13 seniors. The task is to determine the number of different ways Ms. Bell can create a 5-member committee of seniors, with each senior having an equal chance of being selected.

To solve this problem, we can use combinations. The number of ways to select a committee of 5 seniors from a group of 13 can be calculated using the combination formula:

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

Where n represents the total number of elements (seniors in this case), and k represents the number of elements to be selected (5 in this case). The exclamation mark denotes the factorial of a number.

Using the combination formula, the number of ways to select a 5-member committee from 13 seniors is:

C(13, 5) = 13! / (5!(13 - 5)!) = 13! / (5! * 8!) = (13 * 12 * 11 * 10 * 9) / (5 * 4 * 3 * 2 * 1) = 1287

Therefore, there are 1287 different ways Ms. Bell can create a 5-member committee of seniors from her class.

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Angular frequency is given by the square root of the spring constant/mass. The period is the inverse of the angular frequency. The maximum value of the angular velocity is the angular frequency.

(a) The angular frequency Omega in Eq.(2) is given by the square root of the ratio of the spring constant k to the mass m, or Omega = sqrt(k/m). This value represents the rate at which the oscillator oscillates back and forth, and is measured in radians per second.

(b) The period T, which is the time it takes for the oscillator to complete one full cycle, is given by T = 2*pi/Omega. In other words, the period is the reciprocal of the angular frequency, and is measured in seconds.

(c) The maximum value of the angular velocity is given by the amplitude A multiplied by the angular frequency Omega, or Omega_max = A*Omega. This value represents the maximum speed at which the oscillator moves during its oscillations, and is measured in radians per second.

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(a)  95% CI for μ when n=25 and x will be (51.68, 55.52) watts .

We use the formula for a confidence interval for the mean with known standard deviation:

CI = (x - z*σ/√n, x+ z*σ/√n)

where x is the sample mean, σ is the population standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level (95% in this case).

Since the standard deviation is unknown, we use the sample standard deviation s as an estimate for σ.

Plugging in the values, we have:

CI = (53.6 - 1.96*(s/√25), 53.6 + 1.96*(s/√25))

  = (51.68, 55.52) watts

(b) 95% CI for μ when n=100 and x will be (52.42, 54.78) watts.

Using the same formula as in part (a), we have:

CI = (53.6 - 1.96*(s/√100), 53.6 + 1.96*(s/√100))

  = (52.42, 54.78) watts

(c) 99%CI for μ when n=100 and x will be (51.96, 55.24) watts

Using the same formula as in part (a) with a z-score of 2.58 (corresponding to a 99% confidence level), we have:

CI = (53.6 - 2.58*(s/√100), 53.6 + 2.58*(s/√100))

  = (51.96, 55.24) watts

(d) 82% CI for μ when n=100 and x will be (52.95, 54.25) watts

Using the same formula as in part (a) with a z-score of 1.305 (found using a standard normal table or calculator), we have:

CI = (53.6 - 1.305*(s/√100), 53.6 + 1.305*(s/√100))

  = (52.95, 54.25) watts

(e) The value of n will be 267.

We use the formula for the width of a confidence interval:

width = 2*z*(s/√n)

where z is the z-score corresponding to the desired confidence level (99% in this case) and s is the sample standard deviation.

Solving for n, we have:

n = (2*z*s/width)^2

Plugging in the values, we get:

n = (2*2.58*s/1.0)^2

 = 266.49

Rounding up to the nearest whole number, we get n = 267.

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the expected number of students in the randomly chosen student's class is approximately 21.79.

To find E(X), we need to use the formula:
E(X) = ΣxP(X=x)
where Σx represents the sum of all possible values of X and P(X=x) represents the probability of X taking on the value x.
In this case, X can take on values of 20, 22, or 25, with probabilities of 20/67, 22/67, and 25/67, respectively (since there are 20 students in the first class out of 67 total students, 22 students in the second class out of 67 total students, and 25 students in the third class out of 67 total students).
So, using the formula above, we get:
E(X) = (20/67)*20 + (22/67)*22 + (25/67)*25
E(X) = 20*0.2985 + 22*0.3284 + 25*0.3731
E(X) = 21.79
Therefore, the expected number of students in the randomly chosen student's class is approximately 21.79.
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according to question the answer is  tan (2 x ) = -sqrt(345)/353.

Since cot (x) = 4/19 is given, we can use the identity:

tan²(x) + 1 = cot²(x)

to find tan (x):

tan²(x) = cot²(x) - 1 = (4/19)² - 1 = 16/361 - 361/361 = -345/361

Since x is in the first quadrant, tan(x) is positive, so we can take the positive square root:

tan(x) = sqrt(-345/361)

Now we can use the double angle formula for tangent:

tan(2x) = (2tan(x))/(1 - tan²(x))

Substituting in the value we found for tan(x), we get:

tan(2x) = (2(sqrt(-345/361)))/(1 - (-345/361))

= (2sqrt(-345/361))/706

= -sqrt(345)/353

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The expression as given above: "−c f(x, y) ds = − c f(x, y) ds" seems to be true.

Both expressions, the left-hand side, −c f(x, y) ds and the right-hand side, − c f(x, y) ds:

represent the same mathematical operation. The mathematical equation represented here is obtained by multiplying the function f(x, y) by a constant -c and integrating it with respect to the variable ds. The placement of the constant -c does not affect the result, so the two expressions are equivalent.

Thus, both expressions (right-hand and left-hand sides) are the same. Hence, the statement is true.

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Applying the trigonometric ratios, the values of the sides of the triangle are: a ≈ 22.7; b ≈ 10.6.

To find the missing side a, in the right triangle shown above, the trigonometric ratio we would apply is he cosine ratio, which is:

cos ∅ = adjacent length / hypotenuse length.

∅ = 25 degrees

Adjacent length = a

Hypotenuse length = 25

Substitute:

cos 25 = a / 25

a = 25 * cos 25

a ≈ 22.7

To find the missing side b, the trigonometric ratio we would apply is he sine ratio, which is:

sin ∅ = opposite length / hypotenuse length.

∅ = 25 degrees

Opposite length = b

Hypotenuse length = 25

Substitute:

sin 25 = b / 25

b = 25 * sin 25

b ≈ 10.6

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The length of BC is, 27

And, Lenght of rectangle is,

⇒ L = 12.5 cm

We have to given that;

The perimeter of triangle ABC is, 81 inches

And, Sides are 2x , 3x and 4x.

Hence, We can formulate;

2x + 3x + 4x = 81

9x = 81

x = 81 / 9

x = 9

Thus, The length of BC is,

BC = 3x

BC = 3 x 9

BC = 27

2) Area of rectangle = 318 cm²

And, Width of rectangle = 25.5 cm

Since, We know that;

⇒ Area = length × width

⇒ 318 = L x 25.5

⇒ L = 12.5 cm

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Using the frequency distribution, a) The total number of observations is 35. b) The width of each class is 9. c) The midpoint of the second class is 20.5. d) The modal classes are class 36-45 and class 26-35. e) The class limits of the next class would be 66-75.

a) To determine the total number of observations, we need to sum up the frequencies of all the classes. In this case, the total number of observations is:

4 + 8 + 8 + 9 + 3 + 3 = 35

Therefore, there are a total of 35 observations.

b) The width of each class can be calculated by taking the difference between the upper and lower class limits. For example, the width of the first class (6-15) is 15-6 = 9.

Similarly, the width of the other classes can be calculated as follows:

Class 6-15    : width = 15-6   = 9

Class 16-25  : width = 25-16 = 9

Class 26-35 : width = 35-26 = 9

Class 36-45 : width = 45-36 = 9

Class 46-55 : width = 55-46 = 9

Class 56-65 : width = 65-56 = 9

Therefore, the width of each class is 9.

c) The midpoint of a class can be calculated by taking the average of the upper and lower class limits. For example, the midpoint of the second class (16-25) can be calculated as:

Midpoint = (16+25)/2 = 20.5

Therefore, the midpoint of the second class is 20.5.

d) The modal class is the class with the highest frequency. In this case, we can see that two classes have the same highest frequency of 9: class 36-45 and class 26-35. Therefore, both of these classes are modal classes.

e) To determine the class limits of the next class if an additional class were to be added, we need to consider the current width of the classes, which is 9.

Therefore, if we want to add a class after the last class (56-65), the lower limit of the next class would be 66, and the upper limit would be 75. Therefore, the class limits of the next class would be 66-75.

In conclusion, frequency distribution is a useful tool to organize and summarize data by grouping them into classes.

It provides information on the total number of observations, width of each class, midpoint of a class, modal class, and can even help in determining the class limits of an additional class.

Understanding frequency distributions can help in making decisions, analyzing trends, and drawing conclusions in various fields such as business, economics, psychology, and more.

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