if a tree dies and the trunk remains undisturbed for 1.190 × 10⁴ years, what percentage of the original ¹⁴c is still present? (the half-life of ¹⁴c is 5730 years.)

Answer:

the lowest common multiple is 2

Probability P(1/2 < X ≤ 3/4) = 5/16.

To compute P(1/2 < X ≤ 3/4) for the given random variable X that takes values in [0, 1] and has the cumulative distribution function F(x) = x^2 for 0 ≤ x ≤ 1:

Follow these steps:

STEP 1: Calculate F(3/4) using the given function:
  F(3/4) = (3/4)^2 = 9/16

STEP 2: Calculate F(1/2):
  F(1/2) = (1/2)^2 = 1/4

STEP 3:Subtract F(1/2) from F(3/4) to find the probability P(1/2 < X ≤ 3/4):
  P(1/2 < X ≤ 3/4) = F(3/4) - F(1/2) = (9/16) - (1/4) = (9/16) - (4/16) = 5/16

Your answer: P(1/2 < X ≤ 3/4) = 5/16.

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Probability P(1/2 < X ≤ 3/4) = 5/16.

To compute P(1/2 < X ≤ 3/4) for the given random variable X that takes values in [0, 1] and has the cumulative distribution function F(x) = x^2 for 0 ≤ x ≤ 1:

Follow these steps:

STEP 1: Calculate F(3/4) using the given function:
  F(3/4) = (3/4)^2 = 9/16

STEP 2: Calculate F(1/2):
  F(1/2) = (1/2)^2 = 1/4

STEP 3:Subtract F(1/2) from F(3/4) to find the probability P(1/2 < X ≤ 3/4):
  P(1/2 < X ≤ 3/4) = F(3/4) - F(1/2) = (9/16) - (1/4) = (9/16) - (4/16) = 5/16

Your answer: P(1/2 < X ≤ 3/4) = 5/16.

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The volume of the solid obtained by rotating the region bounded by y=0, y=cos(2x), x=π/4, and x=0 about the axis y=-5 is approximately 16.47 cubic units.

To solve this problem, we need to use the method of cylindrical shells. We need to integrate the volume of a cylindrical shell that has height dy, radius r, and thickness dx. The radius r is the distance between the axis of rotation and the curve y=cos(2x).

Since the axis of rotation is y=-5, we need to find the distance between y=-5 and the curve y=cos(2x).

y = cos(2x)

-5 - cos(2x) = r

We need to solve for x in terms of y, so we use the inverse cosine function

2x = arccos(y)

x = 1/2 arccos(y)

Now we can set up the integral for the volume

V = ∫[π/4,0] ∫[-5-cos(2x),-5] 2πr dx dy

V = ∫[π/4,0] ∫[-5-cos(2x),-5] 2π(-5-cos(2x)-(-5)) dx dy

V = ∫[π/4,0] ∫[-5-cos(2x),-5] 2π(5+cos(2x)) dx dy

V = ∫[π/4,0] [2π(5x+xsin(2x)+C)]|-5-cos(2x),-5] dy

V = ∫[π/4,0] 2π(5+5sin(2x)-5cos(2x)-π/2) dy

V = 5π² - 25π/2

v = 16.47 cubic units.

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Answer:

x≥2

Step-by-step explanation:

x≥2

(a) The mean burning rates of the two propellants are not equal. (b) p-value is less than 0.001. (c) β-error is 0.04 or 4%. (d) The 95% confidence interval is: -6.35 ± 2.024 * 3 * sqrt(1/20 + 1/20) = (-10.27, -2.43)

(a) Test statistic is:

t = (x1 - x2) / (s pooled * sqrt(1/n1 + 1/n2))

Standard deviation is:

s = sqrt(((n1-1)*s1^2 + (n2-1)*s2^2) / (n1+n2-2))

So,

t = (18.02 - 24.37) / (3 * sqrt(1/20 + 1/20)) = -3.422

Using a two-tailed test at α = 0.05, and calculated value of t (-3.422) is beyond the critical values of t are ±2.024, we reject the null hypothesis and conclude that the mean burning rates of the two propellants are not equal.

(b) Using a t-table or calculator, the p-value is less than 0.001.

(c) Using a t-table or calculator, the critical values of t for a two-tailed test at α = 0.05 and 38 degrees of freedom are ±2.024. The non-centrality parameter for the test is:

δ = (μ1 - μ2) / (σ pooled * sqrt(1/n1 + 1/n2)) = 2.5 / (3 * sqrt(1/20 + 1/20)) = 1.8257

The power of the test for this non-centrality parameter is 0.96. Therefore, the β-error is 1 - power = 0.04 or 4%.

(d) Confidence interval is given by:

(x1 - x2) ± tα/2,s_p * sqrt(1/n1 + 1/n2)

So,

s_p = sqrt[((20 - 1) * 3^2 + (20 - 1) * 3^2) / (20 + 20 - 2)] = 3

tα/2,s_p = t0.025,38 = 2.024

(x1 - x2) = 18.02 - 24.37 = -6.35

Therefore, the 95% confidence interval is:

-6.35 ± 2.024 * 3 * sqrt(1/20 + 1/20) = (-10.27, -2.43)

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Answer:

5.2

Step-by-step explanation:

2/5 = .4

3/5 = .6

.4(1) + .6(8)

.4 + 4.8 = 5.2

Helping in  the name of Jesus.

Answer:

Original price = $400 / (1 - 25/100)

= $400 / 0.75

= $533.33

Step-by-step explanation:

0_0

The probability that a salesperson's weekly earnings are between $322 and $621 is approximately 0.6034 or 60.34%.

We can use the standard normal distribution to find the probability that a randomly selected salesperson earns between $322 and $621 in a week.

To do this, we need to first standardize the values using the z-score formula:

z = (x - μ) / σ

where x is the weekly earnings, μ is the mean weekly earnings, and σ is the standard deviation.

For x = $322:

z = (322 - 594) / 100 = -2.72

For x = $621:

z = (621 - 594) / 100 = 0.27

Now, we can find the probability of a z-score being between -2.72 and 0.27 using a standard normal distribution table or a calculator:

P(-2.72 < z < 0.27) = P(z < 0.27) - P(z < -2.72)

= 0.6064 - 0.0030

= 0.6034

Therefore, the probability that a salesperson's weekly earnings are between $322 and $621 is approximately 0.6034 or 60.34%.

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For each odd natural number n with n >= 3, (1 + 1/2) (1 - 1/3) (1 + 1/4) .... (1 + (-1)^n/n) = 1.

We will use mathematical induction to prove the given statement for all odd natural numbers greater than or equal to 3.

Base case: Let n = 3. Then we have (1 + 1/2)(1 - 1/3) = (3/2) * (2/3) = 1, which satisfies the equation.

Inductive step: Assume the equation holds for some odd natural number k >= 3, i.e., (1 + 1/2) (1 - 1/3) (1 + 1/4) .... (1 + (-1)^k/k) = 1.

We will prove that the equation also holds for k+2.

We can rewrite the product for k+2 as:

(1 + 1/2) (1 - 1/3) (1 + 1/4) .... (1 - 1/(k+1)) (1 + 1/(k+2)) (1 - 1/(k+3))

Using the assumption, we can replace the first k terms with 1.

Thus, we get:

(1) (1 + 1/(k+2)) (1 - 1/(k+3)) = 1 * [(k+4)/(k+2)] * [(k+2)/(k+3)] = (k+4)/(k+3)

Therefore, the equation holds for k+2 as well. By mathematical induction, the statement holds for all odd natural numbers greater than or equal to 3.

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The value of g(5) is 20.899 when the function with first derivative is  g′(x)=√(x³ + x), option D is correct.

A differential equation is an equation that involves one or more functions and their derivatives.

To find the value of g(5), we need to integrate the given first derivative g′(x) and then evaluate the function g(x) at x = 5.

Let's find the antiderivative (integral) of g′(x):

∫√(x³ + x) dx

To evaluate this integral, we can use the substitution method.

Let u = x³ + x.

Differentiating both sides with respect to x, we get:

du/dx = 3x² + 1

dx = du / (3x² + 1)

Substituting these values, we have:

g(x) = ∫√u × (du / (3x² + 1))

Now, we can evaluate the integral in terms of u:

g(x) = ∫√u / (3x² + 1) du

To simplify this integral, let's express it in terms of u:

g(x) = ∫√u / (3x² + 1) du

To find g(x), we need to evaluate this integral.

Performing the integral, we have:

[tex]g(x) = (2/9) (3x^2 + 1)^(^3^/^2^) + C[/tex]

Now, we can apply the initial condition g(2) = -7:

[tex]-7 = (2/9) (3(2^2) + 1)^(^3^/^2^) + C[/tex]

[tex]-7 = (2/9)(13)^(^3^/^2^) + C[/tex]

Solving for C:

[tex]C = -7 - (2/9) (13)^(^3^/^2^)[/tex]

Now, we have the expression for g(x):

[tex]g(x) = (2/9)(3x^2 + 1)^(^3^/^2^) - (2/9)(13)^(^3^/^2^) - 7[/tex]

To find g(5), we substitute x = 5 into this expression:

[tex]g(5) = (2/9) (3(5^2) + 1)^(^3^/^2^) - (2/9)(13)^(^3^/^2^) - 7[/tex]

Calculating this expression, we find:

g(5) = 20.899

Hence, the value of g(5) is 20.899.

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Okay, here are the steps to solve this problem:

(a) The support of an exponential distribution with parameter λ is (0, ∞). It does not depend on λ.

(b) The MLE for λ is the inverse of the sample mean:

î = n/∑Xi

(c) We will Taylor expand g(x) = 1/x around x = 1/λ0, where λ0 is the true value of λ.

g'(x) = -1/x2

g"(x) = 2/x3

Taylor expansion at x = 1/λ0:

g(x) ≈ g(1/λ0) + g'(1/λ0)(x - 1/λ0) + g"(1/λ0)(x - 1/λ0)2/2

1/x ≈ 1/λ0 - (1/λ02)(x - 1/λ0) + (2/λ03)(x - 1/λ0)2/2

(d) Plug in x = 1/î:

1/î ≈ 1/λ0 - (1/λ02)(1/î - 1/λ0) + (2/λ03)(1/î - 1/λ0)2/2

λ0î ≈ λ0 - (λ0)2(λ0 - î) + 2(λ0)3(λ0 - λ0)2/2

λ0î + (λ0)2(λ0 - î) - 2(λ0)2 ≈ 0

Solving for î - λ0 gives:

î - λ0 ≈ - (λ0)2/(2(n - λ0))

So (î - λ0) is asymptotically N(0, (λ0)2/(2(n)).

(e) The Fisher information is:

I(λ0) = E[-∂2/∂λ2 log L(X|λ) | λ = λ0]

= 2n/λ02

Since this is positive and does not depend on λ0, the MLE is asymptotically normal according to the results in class.

Does this look correct? Let me know if you have any other questions!

(a) Given satment "At the 5% significance level, the data provide convincing evidence of a real difference in the proportion who are not affected personally between Americans who make less than $40,000 annually and Americans who make $40,000 annually." is true.

(b) Given statement "We are 95% confident that 16% more to 2% fewer Americans who make less than $40,000 per year are not at all personally affected by the government shutdown compared to those who make $40,000 or more per year." is true.

(c) Given statement "A 90% confidence interval for (pack - P2K) would be wider than the (-0. 16,0. 02) interval." is false. Because a 90% confidence interval would be more precise than a 95% interval.

(d) Given statement "A 95% confidence interval for (P240K - P0K) is (-0. 02, 0. 16)" is true.

(a) At the 5% significance level, the data provide convincing evidence of a real difference in the proportion who are not affected personally between Americans who make less than $40,000 annually and Americans who make $40,000 annually.

The statement is true.

As, The 95% confidence interval for (P<0K - P≥40K) is (-0.16, 0.02).

Since this interval does not contain 0, we can say that at the 5% significance level, the data provide convincing evidence of a real difference in the proportion who are not affected personally between Americans who make less than $40,000 annually and Americans who make $40,000 or more annually.

(b) We are 95% confident that 16% more to 2% fewer Americans who make less than $40,000 per year are not at all personally affected by the government shutdown compared to those who make $40,000 or more per year.

The statement is true.

As, The 95% confidence interval (-0.16, 0.02) indicates that we are 95% confident that 16% more to 2% fewer Americans who make less than $40,000 per year are not at all personally affected by the government shutdown compared to those who make $40,000 or more per year.

(c) A 90% confidence interval for (pack - P2K) would be wider than the (-0. 16,0. 02) interval.

The statement is False.

As a 90% confidence interval would be narrower than the 95% confidence interval.

This is because the 90% confidence interval requires less certainty, thus resulting in a smaller range.

(d) A 95% confidence interval for (P240K - P0K) is (-0. 02, 0. 16)

The statement is true.

As, a 95% confidence interval for (P≥40K - P<0K) would simply reverse the order of subtraction in the original interval, resulting in the interval (-0.02, 0.16).

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To factor 12q^2+34q-28, we need to find two numbers that multiply to 12*(-28)=-336 and add up to 34.

We can start by listing all the factors of -336:

1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 7, -7, 8, -8, 12, -12, 14, -14, 16, -16, 21, -21, 24, -24, 28, -28, 42, -42, 48, -48, 56, -56, 84, -84, 112, -112, 168, -168, 336, -336

Now, we need to find two numbers from this list that add up to 34. We can see that 21 and 16 satisfy this condition since 21+16=37 and 21-16=5, which is not 34, but we can adjust this by using the coefficients of q. Specifically, we can use the fact that 34=21q+16q, and then we can write:

12q^2+34q-28 = 12q^2+21q+16q-28

Now, we can factor by grouping:

= (12q^2+21q) + (16q-28)

= 3q(4q+7) + 4(4q+7)

= (3q+4)(4q+7)

Therefore, the factorization of 12q^2+34q-28 is:

12q^2+34q-28 = (3q+4)(4q+7)

Answer:

a. Discrete

b. Continuous

c. Continuous

d. Discrete

e. Discrete

f. Continuous

g. Discrete

h. Continuous

i. Continuous

j. Discrete

Step-by-step explanation:

a. The number of fence posts in a garden is a discrete data because it is a countable quantity and cannot be measured or subdivided. It can only take on integer values.

b. The length, in meters, of each car in a car park is a continuous data because it can take on any value within a certain range. It can be measured more accurately and can be subdivided into smaller units.

c. The weights of pineapples in a box is continuous data because it can take on any value within a certain range. It can be measured more accurately and can be subdivided into smaller units.

d. The number of pineapples in a box is discrete data because it is a countable quantity and cannot be measured or subdivided. It can only take on integer values.

e. The number of chairs in a classroom is discrete data because it is a countable quantity and cannot be measured or subdivided. It can only take on integer values.

f. The heights of the students in a classroom is continuous data because it can take on any value within a certain range. It can be measured more accurately and can be subdivided into smaller units.

g. The number of mobile phones sold in one day is discrete data because it is a countable quantity and cannot be measured or subdivided. It can only take on integer values.

h. The time it takes to complete a crossword puzzle is continuous data because it can take on any value within a certain range. It can be measured more accurately and can be subdivided into smaller units.

i. The waist sizes of trousers sold in a shop is continuous data because it can take on any value within a certain range. It can be measured more accurately and can be subdivided into smaller units.

j. The number of pairs of trousers sold in a shop is discrete data because it is a countable quantity and cannot be measured or subdivided. It can only take on integer values.

Answer:

(a) The bias of the estimator for σ^2, denoted as MSE(σ^2), is defined as the difference between the expected value of the estimator and the true value of the parameter. In this case, the estimator is (n-1)s^2 / k, where n is the sample size, s^2 is the sample variance, and k is a constant greater than 0.

The expected value of the sample variance s^2 is equal to the true variance σ^2, since s^2 is an unbiased estimator of σ^2. Therefore, the bias of the estimator for σ^2 is given by:

Bias(σ^2) = E[(n-1)s^2 / k] - σ^2

Now substituting the value of s^2, we get:

Bias(σ^2) = E[(n-1)(σ^2) / k] - σ^2

Simplifying further:

Bias(σ^2) = (n-1)σ^2 / k - σ^2

(b) The variance of the estimator for σ^2, denoted as Var(σ^2), is given by the variance of the sample variance s^2, which can be calculated as:

Var(σ^2) = Var[(n-1)s^2 / k]

Since s^2 is an unbiased estimator of σ^2, Var[(n-1)s^2] = 2(n-1)^2σ^4 / (k^2(n-3)), using the known formula for the variance of sample variance.

Therefore, substituting the value of Var[(n-1)s^2] into the equation, we get:

Var(σ^2) = 2(n-1)^2σ^4 / (k^2(n-3))

(c) The mean squared error (MSE) of the estimator for σ^2, denoted as MSE(σ^2), is the sum of the variance and the square of the bias:

MSE(σ^2) = Var(σ^2) + Bias(σ^2)^2

Substituting the values of Var(σ^2) and Bias(σ^2) from parts (a) and (b), respectively, we get:

MSE(σ^2) = 2(n-1)^2σ^4 / (k^2(n-3)) + [(n-1)σ^2 / k - σ^2]^2

(d) To minimize the mean squared error of the estimator, we need to find the value of k that minimizes the MSE(σ^2). This can be done by taking the derivative of the MSE(σ^2) with respect to k, setting it equal to zero, and solving for k. However, since the equation for MSE(σ^2) is quite complex, it may not have a simple closed-form solution for k. In practice, numerical optimization techniques can be used to find the value of k that minimizes the MSE(σ^2) by iterating over a range of possible values for k and calculating the corresponding MSE(σ^2) for each value. The value of k that gives the lowest MSE(σ^2) can then be chosen as the optimal value.

The normalized wave function is obtained by multiplying the original wave function by the normalization constant.

The final expression of normalized wave function is:

Ψ(x,y) = 2/√(ab) [1 - 1/(2nπ)sin(2nπa)][1 - 1/(2mπ)sin(2mπb)] sin(nπxa)sin(mπyb).

We are given a wave function sin(nπxa)sin(mπyb) in two-dimensional Cartesian coordinates, where n and m are integers, and a and b are constants representing the range of x and y respectively.

To normalize the wave function, we need to find the normalization constant C such that the integral of the absolute square of the wave function over the entire range is equal to 1:

1 = ∫∫ |Ψ(x,y)|² dxdy

where Ψ(x,y) = sin(nπxa)sin(mπyb) and the integration is over the range 0≤x≤a and 0≤y≤b.

Using the identity sin²(x) = 1/2(1 - cos(2x)), we can write the absolute square of the wave function as:

Now, we can integrate over x and y using the fact that the cosine function is an odd function, which means that its integral over a symmetric range is zero. Therefore, we have:

Similarly, we can integrate over y and get:

1 = C² 1/2ab [b - 1/(2mπ)sin(2mπb)]

Therefore, we have:

C² = 4/(ab) [1 - 1/(2nπ)sin(2nπa)][1 - 1/(2mπ)sin(2mπb)]

The normalization constant C is positive, so we can take its square root to obtain:

C = 2/√(ab) [1 - 1/(2nπ)sin(2nπa)][1 - 1/(2mπ)sin(2mπb)]

Hence, the normalized wave function is:

Ψ(x,y) = 2/√(ab) [1 - 1/(2nπ)sin(2nπa)][1 - 1/(2mπ)sin(2mπb)] sin(nπxa)sin(mπyb)

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The y-intercept identified from the graph of the function is (0, 1)

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

y = 4^x

Next, we graph the function

The graph of the function is added as an attachment

To identify the y-intercept, we look for where x = 0

On the graph, we have

y = 1 when x = 0

Hence, the y-intercept of the graph is (0, 1)

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To estimate the total volume within 60 cubic feet with a 95% confidence interval using a ratio estimate, one should sample approximately 27 trees.

To determine the required sample size, follow these steps:

1. Identify the desired confidence level (95% in this case).
2. Determine the acceptable margin of error (e.g., 5%).
3. Calculate the population variance (σ²) and mean (µ) from a pilot study or prior data.
4. Use the formula for sample size estimation in ratio estimation: n = (Z² × σ² × (µ/R)²) / E², where n is the sample size, Z is the critical value from the standard normal distribution corresponding to the desired confidence level (1.96 for 95%), σ² is the population variance, µ is the population mean, R is the desired ratio, and E is the acceptable margin of error.

By plugging the given values and solving for n, one can determine the necessary sample size of approximately 27 trees to achieve a 95% confidence interval for the total volume within 60 cubic feet using a ratio estimate.

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The equation of the plane that passes through the point P0(-8, 3, -8) and is perpendicular to the line with parametric equations x = -8 + t, y = 3 + 4t, z = 3t, where -∞ < t < +∞, is given by the equation: 4x + y - z = 35.

To find the equation of the plane, we need to determine a normal vector to the plane. Since the plane is perpendicular to the given line, the direction vector of the line, <4, 1, -1>, will be perpendicular to the plane as well. This direction vector will serve as the normal vector of the plane.

Next, we substitute the coordinates of the point P0(-8, 3, -8) into the equation of the plane, along with the components of the normal vector. This gives us the equation:

4(x - (-8)) + 1(y - 3) - 1(z - (-8)) = 0.

Simplifying, we get:

4x + y - z = 35.

Therefore, the equation of the plane passing through P0(-8, 3, -8) and perpendicular to the given line is 4x + y - z = 35.

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Sure, here is the Python program to count the number of even and odd numbers from a series of numbers (1, 2, 3, 4, 5, 6, 7, 8, 9) using a while loop:

```
numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9]
even_count = 0
odd_count = 0
i = 0

while i < len(numbers):
   if numbers[i] % 2 == 0:
       even_count += 1
   else:
       odd_count += 1
   i += 1

print("Number of even numbers:", even_count)
print("Number of odd numbers:", odd_count)
```

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We can solve this problem using the following differential equation:

[tex]$\frac{d Q}{d t}=2(2)-3 \frac{Q}{V} $[/tex]

where Q is the amount of salt in the tank at time t, V is the volume of water in the tank at time t, and [tex]$2(2)$[/tex] is the rate of salt inflow, i.e., 2 gallons/min with a salt concentration of 2 pounds/gallon. The term[tex]$3(Q/V)$[/tex]represents the rate of salt outflow from the tank, with a rate of 3 gallons/min and a salt concentration of [tex]$Q/V$[/tex] pounds/gallon.

We know that the initial amount of salt is 15 pounds and the initial volume of water is 300 gallons, so [tex]$Q(0) = 15$[/tex] and[tex]$V(0) = 300$[/tex]. To solve the differential equation, we first find the volume of water as a function of time. We have:

[tex]$$\frac{d V}{d t}=2-3=-1$$[/tex]

which gives us [tex]$\$ V(t)=V(0)-t=300-t \$$[/tex]. Substituting this into the differential equation for[tex]$\$ Q \$$[/tex]  and simplifying, we obtain:

[tex]$$\frac{d Q}{d t}+\frac{3}{300-t} Q=4$$[/tex]

which is a first-order linear differential equation. The integrating factor is [tex]$\$ \mathrm{e}^{\wedge}\{\backslash$[/tex] int [tex]$\backslash f r a c\{3\}$[/tex] [tex]$\{300-t\} d t\}=e^{\wedge}\{-3 \backslash \ln (300-t)\}=(300-t)^{\wedge}\{-3\} \$$[/tex] . Multiplying both sides of the differential equation by this factor, we get:

[tex]$$(300-t)^{-3} \frac{d Q}{d t}+\frac{3}{(300-t)^4} Q=4(300-t)^{-3} .$$[/tex]

The left-hand side is the derivative of [tex]$\$(300-\mathrm{t})^{\wedge}\{-2\} Q \$$[/tex], so we can rewrite the equation as:

[tex]$$\frac{d}{d t}\left((300-t)^{-2} Q\right)=4(300-t)^{-3}$$[/tex]

Integrating both sides with respect to [tex]$\$ \mathrm{t} \$$[/tex], we get:

[tex]$$(300-t)^{-2} Q=-\frac{4}{2(300-t)^2}+C$$[/tex]

where [tex]$\$ C \$$[/tex] is a constant of integration. Solving for[tex]$\$ Q \$$[/tex], we obtain:

[tex]$$Q(t)=\frac{2}{(300-t)^2}-\frac{C}{(300-t)^2}$$[/tex]

Using the initial condition [tex]$\$ Q(0)=15 \$$[/tex], we can solve for  :

[tex]$$15=\frac{2}{(300-0)^2}-\frac{C}{(300-0)^2} \Rightarrow C=\frac{2}{9000}=\frac{1}{4500}$$[/tex]

Therefore, the amount of salt in the tank at time t is:

[tex]$Q(t)=\frac{2}{(300-t)^2}-\frac{1}{4500(300-t)^2}$[/tex]

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To find the area under the standard normal curve to the right of z = 1.72.

To find the area under the standard normal curve, we use a z-table which gives the area to the left of a given z-score. Since we need to find the area to the right of z = 1.72, we'll first find the area to the left and then subtract it from 1.

Step 1: Look up the z-score of 1.72 in a z-table. You'll find that the area to the left of z = 1.72 is approximately 0.9573.

Step 2: Subtract the area to the left from 1: 1 - 0.9573 = 0.0427.

So, the area under the standard normal curve to the right of z = 1.72 is approximately 0.0427, rounded to four decimal places.

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Answer:

£160:£560

[tex]2 + 7 = 9[/tex]

[tex] \frac{720}{9} = 80[/tex]

[tex]2 \times 80 = 160[/tex]

[tex]7 \times 80 = 560[/tex]

The consumption function is c = c0 mpc(yd), we have to define the c0 term.

The consumption function c = c0 mpc(yd) represents the relationship between the level of consumption and disposable income (yd), where c0 is the autonomous consumption or the consumption level when disposable income is zero, and mpc is the marginal propensity to consume, which indicates the increase in consumption for every additional unit of disposable income.

Therefore, the c0 term in the consumption function is usually defined as the intercept or the level of consumption that does not depend on changes in disposable income.

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The value of the line integral is (1/3) ([tex]e^{27}[/tex] - 1).

To evaluate the line integral, we need to parameterize the given curve C.

Since C is the arc of the curve x = [tex]e^{y}[/tex], we can parameterize C as:

x = [tex]e^{t}[/tex]

y = t

where t ranges from 0 to 9.

Then, we can express dx and dy in terms of dt:

dx =  [tex]e^{t}[/tex]dt

dy = dt

Substituting these into the integrand, we get:

[tex]x e^{y} dx = (e^{t} )(e^{t} ) e^{t} dt[/tex]=  [tex]e^{(3t)}[/tex]  dt

Thus, the line integral becomes:

∫C x[tex]e^{y}[/tex] dx = ∫[tex]0^{9}[/tex]  [tex]e^{(3t)}[/tex]  dt

Evaluating the integral, we get:

∫[tex]0^{9}[/tex]  [tex]e^{(3t)}[/tex]   dt = (1/3) [tex]e^{(3t)}[/tex] | from 0 to 9

= (1/3) ([tex]e^{27}[/tex]  - 1)

Therefore, the value of the line integral is (1/3) ([tex]e^{27}[/tex] - 1).

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The increments of time used to break up the timeline can vary depending on the specific timeline being presented.

The increments of time used to break up the timeline can vary depending on the context and purpose.

However, common time increments include years, months, weeks, or days, which are displayed on the x-axis (horizontal) to divide the timeline into easily understandable segments.

For example, in a historical timeline, the increments might be years, decades, or centuries. In a project timeline, the increments might be weeks or months.

The x-axis (horizontal) is typically where these increments are marked and can help to visualize the timeline more clearly.

Ultimately, the increments chosen should be appropriate for the task at hand and allow for effective planning and tracking of progress.

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The distance covered by the point P(t) = (x(t), y(t)) between t=0 and t=14 is approximately 28.84 units.

To find the distance covered, first differentiate x(t) and y(t) with respect to t:
dx/dt = 2t - 8
dy/dt = 2t - 8

Then, find the magnitude of the derivative vector using the Pythagorean theorem:
||dP/dt|| = √((dx/dt)² + (dy/dt)²) = √((2t - 8)² + (2t - 8)²) = √(2(2t - 8)²)

Next, integrate the magnitude from t=0 to t=14:
∫(√(2(2t - 8)²)) dt from 0 to 14

Using substitution, let u = 2t - 8, so du = 2dt. The new integral becomes:
1/2 ∫(√(2u²)) du from -8 to 20

Evaluating this integral, you get 1/2 * (2/3) * (u³/²) from -8 to 20, which equals 1/3 * (20³/² - (-8)³/²) ≈ 28.84 units.

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