Problem #1: Concrete Dam 3 115-9 A) Draw the flow net for the structure B) Find the uplift pressure at points A, B, and C C) Find the seepage quantity (cu. ft/day/ft of dam) D) Find the total uplift force per foot of dam (1b/ft of dam) E) Find the exit gradient and of safety against piping Я fuctor Soil Parameters: 1) k = 0.0003 in/sec 2) G = 2.65 3) 0 = 0.72 F-731 67 67 100 50 A В Probler 729 --PATO) - - 77/91

if y = 4 x^3 - 5x and dx/dt=4, by Using the chain rule dy/dt = 172 when x = 2.

Given the function y = 4x^3 - 5x and we need to find dy/dt when x = 2 and dx/dt = 4. We can do this using the following steps:

Step 1: Differentiate the function y with respect to x to find dy/dx.
Thus, First, we find f'(x) by taking the derivative of y with respect to x:

dy/dx = d(4x^3 - 5x)/dx = 12x^2 - 5

Step 2: To find dy/dt, we need to find dy/dx and substitute x = 2 into the resulting expression, along with dx/dt = 4. Thus, substitute the given value of x = 2 into the expression for dy/dx.
dy/dx = 12(2)^2 - 5 = 12(4) - 5 = 48 - 5 = 43

Step 3: Use the chain rule to find dy/dt, which states that dy/dt = dy/dx * dx/dt.

Step 4: Finally, we use the chain rule formula to find dy/dt when x = 2:

Substitute the values of dy/dx and dx/dt into the chain rule equation.
dy/dt = 43 * 4 = 172

So, when x = 2 and dx/dt = 4, dy/dt = 172.

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Assume that the linear system [tex]Ax = b[/tex] has more than one solution, and let [tex]x1[/tex] and [tex]x2[/tex] be two distinct solutions. Let [tex]x3 := ux1+7x2[/tex], where µ, [tex]7 E IR[/tex] with the property that [tex]u+n = 1.[/tex]

Then we have: [tex]Ax1 = b and Ax2 = b[/tex] since x1 and x2 are solutions.

Subtracting the second equation from the first, we get: [tex]A(x1 - x2) = 0.[/tex]

Since [tex]x1[/tex] and [tex]x2[/tex]are distinct solutions, we know that [tex]x1 - x2 ≠ 0[/tex].

Therefore,[tex]A(x1 - x2) = 0[/tex] this implies that the columns of A are linearly dependent. That is, there exist scalars [tex]c1, c2, ..., cn[/tex] (not all zero) such that

[tex]c1a1 + c2a2 + ... + cnan = 0,[/tex]

where [tex]a1, a2, ...,[/tex]and an are the columns of A.

Let x be any solution of Ax = b. Then we have:[tex]A(x + tx3) = Ax + tAx3 = b + tAx3[/tex]

where t is any scalar. But we know that [tex]Ax3 = A(ux1 + 7x2) = uAx1 + 7Ax2 = ub + 7b = 8b,[/tex] since [tex]Ax1 = Ax2 = b.[/tex]

Therefore, we have: [tex]A(x + tx3) = b + t(8b) = (1 + 8t)b.[/tex]

Thus, [tex]x + tx3[/tex] is a solution of [tex]Ax = b[/tex] for any scalar t.

In particular, if we take [tex]t = 1/n,[/tex] where n is any nonzero integer, we get:

[tex]x + (1/n)x3 = (1 - 1/n)x + (1/n)ux1 + (7/n)x2.[/tex]

Since [tex]u + 7 = 1,[/tex] we have:[tex](1/n)ux1 + (7/n)x2 = (1/n)((1 - u)x1 + ux1 + 7x2) = (1/n)x1 + (7/n)x2.[/tex]

Therefore, we can write:[tex]x + (1/n)x3 = (1 - 1/n)x + (1/n)x1 + (7/n)x2.[/tex]

This shows that [tex]x + (1/n)x3[/tex] is another solution of Ax = b for any nonzero integer n. Since we can find infinitely many integers n such that 1/n is nonzero, we conclude that there are infinitely many solutions of .

Therefore, if the linear system [tex]Ax = b[/tex] has more than one solution, then it must have infinitely many solutions.

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The Pigeonhole Principle states that if there are more pigeons than pigeonholes, then there must be at least one pigeonhole with two or more pigeons. In this case, there are 7 days of the week (pigeonholes) and 110 days (pigeons) to choose from.

Therefore, if we divide the 110 days into 7 groups based on the day of the week, the largest group can have at most ⌊110/7⌋ = 15 days. But since we have 7 groups, by the Pigeonhole Principle, at least one group must have more than ⌊110/7⌋ = 15 days. Thus, at least 16 of any 110 days chosen must fall on the same day of the week.

In simpler terms, if you have 110 days to choose from and only 7 days of the week, it is inevitable that some days will have to overlap.

In fact, at least one day of the week must have more than 15 days chosen, which means at least 16 days must fall on that day of the week. This principle can be applied to many situations where there are more items to choose from than categories to put them in.

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The profit is maximum when 40 lawns are mowed.

Given that Kieran is starting a lawn-mowing buisness in her neighborhood. She creates a graph to help her determine what to charge customers per lawn to maximize her profits. She uses {c} to represent the number of lawns she mows and {y} to represent her profit in dollars.

The profit is maximum when 40 lawns are mowed as at this point the the peak of the parabola occurs.

Therefore, the profit is maximum when 40 lawns are mowed.

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To find the volume, you need to multiply all the values together.

1/3 x 5/6 x 2/3 = 5/27

The equal-weighted average, the market capitalization-weighted average, and the volume-weighted average.

The type of average that sums the price of each stock and divides the total by a divisor is called a "price-weighted average". In this type of average, the price of each stock is used as a weight to determine its contribution to the overall index.

For example, suppose we have an index with three stocks: A, B, and C. The price of each stock is $10, $20, and $30, respectively. To calculate the price-weighted average of this index, we would add up the prices of each stock and divide by a divisor, which is usually adjusted for changes in the stock prices or for the addition or removal of stocks from the index. In this case, the calculation would be:

($10 + $20 + $30) / 3 = $20

So the price-weighted average of this index is $20.

One drawback of price-weighted averages is that they are sensitive to changes in the prices of higher-priced stocks, since those stocks have a greater weight in the index. This can lead to distortions in the index if the prices of the higher-priced stocks change significantly. Additionally, price-weighted averages do not take into account the market capitalization or trading volume of each stock, which may not accurately reflect the overall market or sector performance.

Other types of averages that address these limitations include the equal-weighted average, the market capitalization-weighted average, and the volume-weighted average.

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Answer:

The correct answer is d. average.

Step-by-step explanation:

The mean is a measure of central tendency in statistics and is often referred to as the average of a data set. It is calculated by adding up all the values in the data set and dividing by the total number of values. The mean is a common way to summarize a data set and provides a single value that represents the "typical" value of the data. It is not the same as the mode, which is the most frequently occurring value in the data set, or the range, which is the difference between the largest and smallest values in the data set

The chi-square test statistic uses observed frequencies in a contingency table to determine whether an association exists between two nominal random variables by comparing them to expected frequencies.

Here's the step-by-step explanation:
1. Construct a contingency table, showing the observed frequencies of each combination of the two nominal variables.
2. Calculate the expected frequencies for each cell in the table, using the formula: (Row Total * Column Total) / Grand Total.
3. Compute the chi-square test statistic using the formula: Χ² = Σ [(O - E)² / E], where O represents the observed frequencies, E represents the expected frequencies, and Σ indicates the summation of all cells in the table.
4. Determine the degrees of freedom (df) for the chi-square test, using the formula: df = (number of rows - 1) * (number of columns - 1).
5. Compare the calculated chi-square test statistic to the critical value from the chi-square distribution table, using the appropriate degrees of freedom and desired significance level (typically 0.05).
6. If the chi-square test statistic is greater than the critical value, reject the null hypothesis and conclude that there is a significant association between the two nominal variables. If it's less than or equal to the critical value, fail to reject the null hypothesis and conclude that there is no significant association.

By following these steps, the chi-square test uses observed frequencies in a contingency table to determine the presence or absence of an association between two nominal random variables.

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(a) the general solution is [tex]y(x) = c1e^{-5x} + c2e^{-x} + (1/6)x - 1/2[/tex], (b) the probability of getting at least two 5's is [tex](3/216)(5/6) + (1/216) = 1/36[/tex], the general solution of (c) is [tex]y(x) = c1e^{4x} + c2e^{-x}+ (1/10)cos(2x), (d) y(x) = c1e^{4x} + c2e^{-x} + (1/10)cos(2x), (e) y(x) = c1e^{6x} + c2e^{-2x} + 7/2 + (3/2)x[/tex] and (f) The characteristic equation is [tex]r^2 - 2r +[/tex].

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Option (a) k(1 - P) matches the correct equation.

The given circumstance depicts a situation where the pace of progress of the division P of the populace who has heard a letting it be known story is relative to the small portion of the populace who has not yet heard the report.

The differential equation that follows can be used mathematically to represent this situation:

dP/dt = k(1 - P)

where P is the fraction of the population who has heard the news story, and k is the proportionality constant.

Choice (a) k(1 - P) matches the right condition, as it has a similar structure as the given differential condition. This condition expresses that the pace of progress of P is relative to the result of a steady k and the division (1 - P), which addresses the extent of the populace who has not yet heard the report.

On the other hand, Option (b) K(P-1) is incorrect due to its incorrect form. It implies that P's rate of change is inversely proportional to the difference between P and 1, which is not the case in this particular circumstance.

Option c: 1-KP, option d: KP-1, option e: +KP, option g: -KP, option h: None of these, and option i: None of these are incorrect as well.

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Answer:

The hose can run for 7.5 minutes, 9 minutes, and 12 minutes.

The hose can't run for 0 minutes, 7 minutes, 9.75 minutes, or 10.3 minutes.

Here's why:

Vanessa needs to water her plants for at least 7.5 minutes.

Vanessa can't leave the hose running for more than 12 minutes because it will waste water.

Step-by-step explanation:

Answer:

The hose can run for 7.5 minutes, 9 minutes, and 12 minutes.

The hose can't run for 0 minutes, 7 minutes, 9.75 minutes, or 10.3 minutes.

Here's why:

Vanessa needs to water her plants for at least 7.5 minutes.

Vanessa can't leave the hose running for more than 12 minutes because it will waste water.

Step-by-step explanation:

The derivative of y = cos(1 - e / (1 +⁸ˣ 8ˣ)) is y' = [8e⁸ˣsin(1 - e⁸ˣ) × (1 + 8ˣ)] + [(8e⁸ˣln8 + 8ˣln8)cos(1 - e⁸ˣ)] / (1 + 8ˣ)².

To find the derivative of y = cos(1 - e⁸ˣ / (1 + 8ˣ)), we need to use the chain rule and quotient rule.

First, let's find the derivative of the numerator:

y' = -sin(1 - e⁸ˣ) × (-8e⁸ˣ / (1 + 8ˣ)²)

Next, let's find the derivative of the denominator:

y' = (1 + 8ˣ)(-8e⁸ˣln8 - 8ˣln8) / (1 + 8^x)²

Now, using the quotient rule, we can combine these derivatives:

y' = [(-sin(1 - e⁸ˣ) × (-8e⁸ˣ / (1 + 8ˣ)²)) × (1 + 8ˣ)] - [(cos(1 - e⁸ˣ) × (-8e⁸ˣln8 - 8ˣln8)) / (1 + 8ˣ)²]

Simplifying this expression gives:

y' = [8e⁸ˣsin(1 - e⁸ˣ) × (1 + 8eˣ)] + [(8e⁸ˣln8 + 8eˣln8)cos(1 - e⁸ˣ)] / (1 + 8eˣ)²

Therefore, the derivative of y = cos(1 - e / (1 +⁸ˣ 8ˣ)) is y' = [8e⁸ˣsin(1 - e⁸ˣ) × (1 + 8ˣ)] + [(8e⁸ˣln8 + 8ˣln8)cos(1 - e⁸ˣ)] / (1 + 8ˣ)².

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The function f(t) is 0 for t < 0, [tex]t^2[/tex] for 0 ≤ t ≤ 1, and t for t > 1.

The Heaviside function H(t) is defined as:

H(t) = 0, if t < 0

H(t) = 1, if t ≥ 0

Using the Heaviside function, we can write the piecewise function f(t) as:

[tex]f(t) = t^2 * H(t) + (t - t^2) * H(t - 1)[/tex]

Here's how the function works:

For t < 0, H(t) = 0, so f(t) = 0

For 0 ≤ t ≤ 1, H(t) = 1, so f(t) = [tex]t^2[/tex]

For t > 1, H(t) = 1 and H(t - 1) = 0, so f(t) = t

Therefore, the function f(t) is 0 for t < 0, [tex]t^2[/tex] for 0 ≤ t ≤ 1, and t for t > 1.

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The area under the normal curve to the right of μ equals 0 . Thus, option C is correct.

Probability is a measure of the likelihood or chance of an event occurring. It is a number between 0 and 1, with 0 representing an impossible event and 1 representing a certain event. The probability of an event is calculated by dividing the number of ways the event can occur by the total number of possible outcomes.

The area under the normal curve to the right of μ equals 0, which means that the entire normal distribution is to the left of μ.

This is because the normal distribution is a symmetric probability distribution, and so half of the area is to the left of the mean and half is to the right. Therefore, if all the area is to the left of μ, then none is to the right.

Option A, σ, represents the standard deviation of the normal distribution and is not related to the area to the right of μ.

Option B, 1/2, is incorrect because it represents the area to the right of the median, which is not necessarily the same as the mean for a normal distribution.

Option D, 1/σ√2π, is incorrect because it represents the height of the normal curve at the mean, not the area to the right of the mean.

hence, The area under the normal curve to the right of μ equals 0.

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Answer:

The total surface area of a cube can be found using the formula 6s^2, where s is the length of an edge.

In this case, s = 3 meters, so the surface area of one face is 3^2 = 9 square meters.

There are 6 faces in a cube, so the total surface area that needs to be painted is:

6 x 9 = 54 square meters

Therefore, the correct answer is 54 m².

For all integers n, 0n 22n – 1 is divisible by 3.

To prove that for all integers n, 0n 22n – 1 is divisible by 3, we can use mathematical induction.
First, we will show that the statement is true for n = 1.
When n = 1, we have 0([tex]2^1[/tex]) - 1 = -1, which is not divisible by 3. However, we can rewrite the expression as 0([tex]2^1[/tex]) - 1 = 2 - 3, which is divisible by 3. Therefore, the statement is true for n = 1.
Next, we assume that the statement is true for some integer k, and we will show that it is also true for k+1.
For k+1, we have:
0([tex]2^(k+1)[/tex]) - 1 = (0[tex](2^k)[/tex] - 1) * [tex]2^1[/tex] + [tex](2^k - 1)[/tex]
We know that 0[tex](2^k)[/tex] - 1 is divisible by 3 since we assumed the statement is true for k.
We also know that 2^k - 1 is divisible by 3 since we can write it as:
[tex]2^k[/tex] - 1 = (2-1) + ([tex]2^2[/tex] - 1) + ([tex]2^3[/tex] - 1) + ... + ([tex]2^k[/tex] - 1)
Each term in the parentheses is divisible by 3 since [tex]2^n[/tex] - 1 is always divisible by 3 for any integer n. Therefore, the sum of all these terms is also divisible by 3.
Combining these two facts, we can conclude that:
[tex]0(2^(k+1))[/tex] - 1 = (0[tex](2^k)[/tex] - 1) * [tex]2^1[/tex] + ([tex]2^k[/tex] - 1)
is divisible by 3.
By mathematical induction, we have shown that the statement is true for all integers n.

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Using a standard normal distribution table or the cumulative distribution function (CDF), the area under the standard normal curve to the right of z=−1.5 is approximately 0.0668, rounded to four decimal places.

Explanation:

To find the area under the standard normal curve to the right of z=−1.5, Follow these steps:

Step 1: To find the area under the standard normal curve to the right of z=−1.5, we need to use a standard normal distribution table or calculator.

Using a standard normal distribution table, we can find the area to the right of z=−1.5 is 0.0668 (rounded to four decimal places).

Step 2: Alternatively, we can use a calculator or statistical software to find the area using the cumulative distribution function (CDF) of the standard normal distribution. Using a calculator or software, we get the same result of 0.0668.

Therefore, the area under the standard normal curve to the right of z=−1.5 is approximately 0.0668, rounded to four decimal places.

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Answer:

Step-by-step explanation:

So 20% of the science club students is 10.

I am trying to find 100% as this equals ALL the students on the science club trip.

20% = 10

100% / 20% = 5

This means I need to multiply both sides by 5 to get to 100%
20% = 10

(Multiply both sides by 5)

100% = 50

Therefore there are 50 students in the science club (C)

Answer:

Step-by-step explanation:

To find the x-intercepts of a quadratic function, we set y = 0 and solve for x. So, for the given function:

-1/2x^2 + x + 5/2 = 0

Multiplying both sides by -2 to eliminate the fraction:

x^2 - 2x - 5 = 0

We can solve for x using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = -2, and c = -5:

x = (2 ± sqrt(4 + 20)) / 2

x = (2 ± 2sqrt(6)) / 2

x = 1 ± sqrt(6)

Therefore, the x-intercepts of the quadratic function y = -1/2x^2 + x + 5/2 are

x = 1 + sqrt(6)

and

x = 1 - sqrt(6).

The solution of the given differential equation is :

p(t) = A*e^(k*sin(t)), where A is a constant value.

The differential equation given is:

dp/dt = (k*cos(t))p, where k is a positive constant.

This equation models a population that undergoes yearly fluctuations. To find the solution of this equation, we can use the method of separation of variables.

First, separate the variables by dividing both sides by p and multiplying both sides by dt:

(dp/p) = (k*cos(t))dt

Now, integrate both sides with respect to their respective variables:

∫(1/p)dp = ∫(k*cos(t))dt

Upon integrating, we get:

ln|p| = k*sin(t) + C

To solve for p, take the exponent of both sides:

p(t) = e^(k*sin(t) + C)

Since e^C is also a constant, we can write the solution as:

p(t) = A*e^(k*sin(t))

Here, A is a constant that depends on the initial conditions of the problem. This solution represents the population that undergoes yearly fluctuations based on the given differential equation.

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The best measure of variability for the given data is the range, and it equals 9.

The range is the difference between the maximum and minimum values in a dataset.

As per the question, the maximum value is 4, and the minimum value is 1. Therefore, the range is 4 - 1 = 3.

The interquartile range (IQR) is another measure of variability that is useful for identifying the spread of data.

However, since there are no outliers in the given data, the range is a sufficient measure of variability.

Hence, the best measure of variability for the given data is the range, and it equals 9.

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Answer:

Translation: 3 units right and 7 units down.

Step-by-step explanation:

The mapping rule for a rotation of 90° counter-clockwise about the origin is:

The mapping rule for a dilation of 0.25 about the origin is:

The mapping rule for a translation of 3 units right and 7 units down is:

The mapping rule for a reflection across the y-axis is:

To determine the rule that transforms KLMN to K'L'M'N', take one of the vertices from the pre-image and compare to its corresponding vertex in the image.

K = (-3, 4)

K' = (0, -3)

As the numerical values of the x and y coordinates have not be swapped or made negative, the transformation cannot be a rotation of 90 degrees about the origin, or a reflection in the y-axis.

As the x and y coordinates of K' are not 0.25 times the x and y coordinates of K, then the transformation cannot be a dilation of 0.25 about the origin.

Therefore, the transformation that transforms KLMN to K'L'M'N' must be:

To check, apply the mapping rule (x, y) → (x+3, y-7) to the vertices of KLMN:

Therefore, this confirms that the transformation is a translation of 3 units right and 7 units down.

The series is convergent because sn can be expressed as a telescoping sum, which means that the series will approach a finite value as n approaches infinity.

To determine whether the series [infinity] cos 4n − cos 4n/2n=1 is convergent or divergent by expressing sn as a telescoping sum, we can rewrite the terms using the identity cos 2x = 2cos²ˣ − 1:

cos 4n − cos 4n/2n=1 = 2cos^24n/2 − 1 − 2cos^24n/2n+1 + 2cos^24n+2/2n+2 − 1

This expression has a telescoping sum because each term cancels with the previous and next terms. So we can simplify it as:

s_n = (2cos² 2n − 1) − (2cos² 2n+1 − 1)

s_n = 2(cos² 2n − cos² 2n+1)

s_n = −2(cos² 2n+1 − cos² 2n)

Therefore, the series is convergent because sn can be expressed as a telescoping sum, which means that the series will approach a finite value as n approaches infinity.

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The points at which the following surface has horizontal tangent planes is x= ±π/6, ±π/2, ±5π/6 and y= ±π/2 or points with x=0, ±π/3, ±2π/3, ±π and y=0, ±π. So, the correct option is option C. Points with x= ±π/6, ±π/2, ±5π/6 and y= ±π/2 or points with x=0, ±π/3, ±2π/3, ±π and y=0, ±π

To find the points at which the surface z = sin(3x)cos(y) has horizontal tangent planes in the region -π to π, we need to find the points where the partial derivatives with respect to x and y are both zero.

1. Find the partial derivative with respect to x: ∂z/∂x = 3cos(3x)cos(y)
2. Find the partial derivative with respect to y: ∂z/∂y = -sin(3x)sin(y)

Now, we need to find the points where both these derivatives are zero.

3. Set ∂z/∂x = 0: 3cos(3x)cos(y) = 0
4. Set ∂z/∂y = 0: -sin(3x)sin(y) = 0

From step 3, we have two cases:
i) cos(3x) = 0, which gives x = ±π/6, ±π/2, ±5π/6
ii) cos(y) = 0, which gives y = ±π/2

From step 4, we also have two cases:
iii) sin(3x) = 0, which gives x = 0, ±π/3, ±2π/3, ±π
iv) sin(y) = 0, which gives y = 0, ±π

Considering all the cases, the points at which the following surface has horizontal tangent planes is Points with x = ±π/6, ±π/2, ±5π/6 and y = ±π/2 or points with x = 0, ±π/3, ±2π/3, ±π and y = 0, ±π.

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The volume of a hemisphere with a diameter of 8.6 cm, is 167 cm^3.

The volume of a hemisphere  can be calculated using the formular below;  (2/3)πr^3 cubic units.

In this case we can see that  

π = constant whose value is equal to 3.14 approximately.

r” = radius of the hemisphere

given that  diameter = 8.6 cm

radius = 8.6 cm/2 = 4.3 cm

(2/3)πr^3  = (2/3) * π * 4.3 ^3

= 166.519 cm^3

Therefore , the volume of a hemisphere with a diameter of 8.6 cm, can be expressed as 166.519 cm^3.

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The probability of drawing at least one red card is (d) 60/90.

The probability of drawing at least one red card can be found by finding the probability of drawing two yellow cards and subtracting that from 1.

The probability of drawing a yellow card on the first draw is 6/10. The probability of drawing a yellow card on the second draw, without replacement, is 5/9 (since there are only 9 cards left). So the probability of drawing two yellow cards in a row is:

(6/10) * (5/9) = 30/90

To find the probability of drawing at least one red card, we can subtract this from 1:

1 - 30/90 = 60/90

So the answer is (d) 60/90.

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The total number of fish caught over the first 14 years is1,554,393 fish over the first 14 years.

The number of fish caught each year decreases by 3%, which means the company catches 97% of the previous year's total.

Therefore, the number of fish caught each year can be calculated as follows:

Year 1: 130,000

Year 2: 130,000 x 0.97 = 126,100

Year 3: 126,100 x 0.97 = 122,243

Year 4: 122,243 x 0.97 = 118,419

Year 5: 118,419 x 0.97 = 114,627

Year 6: 114,627 x 0.97 = 110,867

Year 7: 110,867 x 0.97 = 107,138

Year 8: 107,138 x 0.97 = 103,441

Year 9: 103,441 x 0.97 = 99,775

Year 10: 99,775 x 0.97 = 96,140

Year 11: 96,140 x 0.97 = 92,535

Year 12: 92,535 x 0.97 = 88,960

Year 13: 88,960 x 0.97 = 85,416

Year 14: 85,416 x 0.97 = 81,902

Therefore, the total number of fish caught over the first 14 years is:

130,000 + 126,100 + 122,243 + 118,419 + 114,627 + 110,867 + 107,138 + 103,441 + 99,775 + 96,140 + 92,535 + 88,960 + 85,416 + 81,902 = 1,554,393

Rounded to the nearest whole number, the company caught a total of 1,554,393 fish over the first 14 years.

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Answer:

Step-by-step explanation:

To calculate the y-component, we need to determine the y-component for each of these masses:

If m1 is at the origin, it is at (0,0). This means that it is at y=0.

If m2 is at (4.0,5.4), it is at y = 5.4.
If m3 is at (1.0, 2.8), it is at y = 2.8.

Thus, we can use the equation for finding equilibrium, which is each mass x position, divided by all the masses:

(m1 * 0 + m2 * 5.4 + m3 * 2.8) / (m1+m2+m3) = 2.53 (3 sig figs)

To find the net change in velocity from time t=1 to time t=5, we need to integrate the acceleration function a(t) from t=1 to t=5. The net change in velocity from time t=1 to time t=5 is approximately 37.539 units (rounded to three decimal places).

To find the net change in velocity from time t=1 to time t=5, we need to find the definite integral of the acceleration function a(t) = ln(3t^4) with respect to time over the interval [1, 5]. To do this, we integrate a(t) with respect to t:∫[1 to 5] ln(3t^4) dtLet's call the antiderivative of a(t) as v(t), which represents the velocity function:v(t) = ∫ln(3t^4) dtNow, to find the net change in velocity, we evaluate v(t) at t=5 and t=1, and subtract the results:Net change in velocity = v(5) - v(1)Once you compute this, you will have the net change in velocity from time t=1 to time t=5.

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To solve for n in the given equation:

2 - (1/2n) = 3n + 16/(2 - 1/n)

First, we can simplify the right-hand side of the equation by finding a common denominator for the fraction:

2 - (1/2n) = (3n(2n - 1) + 16n)/(2n - 1)

Next, we can simplify the left-hand side of the equation by combining like terms:

(4n - 1)/2n = (3n(2n - 1) + 16n)/(2n - 1)

We can then cross-multiply and simplify:

(4n - 1)(2n - 1) = 3n(2n - 1) + 16n

8n^2 - 6n + 1 = 6n^2 + 11n

2n^2 - 17n + 1 = 0

Using the quadratic formula, we can solve for n:

n = (17 ± sqrt(17^2 - 4(2)(1)))/(2(2))

n = (17 ± sqrt(281))/4

Therefore, the two solutions for n are:

n = (17 + sqrt(281))/4 or n = (17 - sqrt(281))/4

Both solutions are real numbers, but they are not integers.

Mark Brainleist