A lighting store purchases light bulbs from a supplier. There are reports that the light bulbs are defective. To determine the quality of the suppliers light bulbs, the manufacturer selects a random sample of 10 light bulbs from the last delivery and test them. Identify the population.
10 light bulbs selected
the light bulbs in the last delivery
all the light bulbs delivered to the company this month
the defective light bulbs

The maximum r-values occur at θ = 0, π/2, π, 3π/2, etc.

To sketch the graph of the polar equation r = 5 cos 2θ, we can first create a table of values for r and θ.

θ r

0 5

π/8 4.14

π/4 2.93

3π/8 1.54

π/2    0

5π/8 -1.54

3π/4 -2.93

7π/8 -4.14

π -5

... ...

To plot the graph, we can use these values to find the corresponding points in the polar coordinate system.

We can also note that the polar equation r = 5 cos 2θ has symmetry about the y-axis, since replacing θ with -θ results in the same value of r.

Additionally, we can find the maximum r-values by finding the points where cos 2θ is equal to 1, which occur when 2θ is equal to 0 or a multiple of 2π.

So, the maximum r-values occur at θ = 0, π/2, π, 3π/2, etc.

To sketch the graph, we can plot the points from the table of values and connect them with a smooth curve.

Here is a rough sketch of the graph attached

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The missing angle represented by 6 is 107 degrees

Supplementary angles are a pair of angles that add up to 180 degrees. In other words, if you have two angles that, when combined, create a straight line, then they are supplementary angles.

Angles on a straight line are supplementary and hence the potion that has 6 is equal to say x

x + 73 = 180

x = 180 - 73

x = 107 degrees

Supplementary angles can be found in many geometric shapes, such as triangles, quadrilaterals, and polygons.

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If the sample proportion is slightly less than the original 72%, the width of the confidence interval will likely increase. This is because a smaller sample proportion means a smaller sample size, which in turn leads to a wider confidence interval. Additionally, a smaller sample proportion also means a larger margin of error, which further contributes to a wider confidence interval. Overall, the decrease in sample proportion will likely result in a wider and less precise confidence interval.
Hi! I'd be happy to help with your question.

When the sample proportion changes from the original 72%, it will affect the width of the confidence interval. If the new sample proportion is slightly less than 72%, the impact on the confidence interval width can be determined as follows:

1. Calculate the standard error (SE) using the formula SE = sqrt(p(1-p)/n), where p is the sample proportion and n is the sample size. The standard error is used to measure the variability of the sample proportion.

2. Calculate the margin of error (ME) using the formula ME = Z*SE, where Z is the Z-score corresponding to the desired level of confidence (e.g., 1.96 for a 95% confidence interval). The margin of error represents the range within which the true population proportion is likely to fall.

3. Determine the confidence interval width by calculating the difference between the upper and lower limits of the interval: Width = Upper limit - Lower limit = (p + ME) - (p - ME) = 2*ME.

If the new sample proportion is slightly less than the original 72%, it could either increase or decrease the width of the confidence interval depending on the changes in the standard error and margin of error. However, without knowing the exact new proportion and the sample size, we cannot definitively determine the impact on the width of the confidence interval.

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It appears that you have created a relative-frequency distribution graph for some data (possibly related to investment returns) and would like to use a normal density function to model future returns. The most important assumption to consider in this context is the assumption of normality.

The normality assumption states that the underlying data follows a normal distribution, also known as the Gaussian distribution or bell curve. This distribution is characterized by its symmetric bell shape and is defined by its mean (average) and standard deviation (a measure of variability). In a normal distribution, about 68% of the data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

When using the normal density function to model future returns, it is crucial to assume that the data exhibits normality. This means that the relative frequencies of the returns in the dataset follow the pattern expected from a normal distribution. If the data significantly deviates from normality, the predictions made using the normal density function might not be accurate and could lead to poor decision-making in future investment scenarios.

In summary, the most important assumption to consider when using a normal density function to model future returns based on a relative-frequency distribution graph is that the data follows a normal distribution. This assumption allows for accurate predictions and better decision-making in investment planning.

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We would say that the quantile plot suggests that the test scores are normally distributed. Therefore, the answer is y.

To draw a quantile plot, we need to first find the z-scores for each test score using the formula:

z = (x - mean) / standard deviation

The mean of the sample is:

mean = (38 + 60 + 75 + 88 + 97) / 5 = 71.6

The standard deviation of the sample is:

s = sqrt((38 - 71.6)^2 + (60 - 71.6)^2 + (75 - 71.6)^2 + (88 - 71.6)^2 + (97 - 71.6)^2 / 4) = 23.027

The z-scores for each test score are:

z_1 = (38 - 71.6) / 23.027 = -1.46

z_2 = (60 - 71.6) / 23.027 = -0.504

z_3 = (75 - 71.6) / 23.027 = 0.148

z_4 = (88 - 71.6) / 23.027 = 0.716

z_5 = (97 - 71.6) / 23.027 = 1.107

Now, we can plot the ordered z-scores against the ordered test scores:

Test score: 38 60 75 88 97

z-score: -1.46 -0.504 0.148 0.716 1.107

The quantile plot appears to show a roughly straight line, which suggests that the data points are roughly normally distributed. So, we would say that the quantile plot suggests that the test scores are normally distributed. Therefore, the answer is y.

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Answer:

In total, she would have 4100 milliliters of water.

Largest number of flights you would need to get from any destination to any other destination in Math world can be calculated using the concept of Graph theory.

Graph theory, a network of points (vertices) and lines (edges) is represented as a graph.

In the case of flights, the airports are the vertices, and the flights connecting them are the edges.
find the maximum number of flights needed, we need to find the diameter of the graph.

Diameter of a graph is the longest distance between any two vertices. In other words, it is the maximum number of edges that you need to travel to get from one vertex to another.
diameter of the graph will depend on the number of vertices and the connections between them.

In the case of Math world, we do not have a specific graph, so we cannot calculate the exact number of flights needed. we can estimate that the diameter of the graph will be large, given the number of airports and the complexity of the network.

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The standard error of the of the sampling distribution is 0.073.

The standard error of a proportion is given by the formula:

standard error = √[(p * (1 - p)) / n]

where p is the proportion and n is the sample size.

In this case:

p = 20% = 0.2

n = 30

Substituting the given values:

standard error = √[(0.2 * (1 - 0.2)) / 30]

standard error = 0.073

Therefore, the standard error of the sampling distribution of the proportion of employees who are allergic to pets is 0.073.

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The probability that Jack arrives at least two minutes before Jill is 0.313 or 31.3%. The probability that the first person arrives by 12:05 p.m. and the last person arrives after 12:10 p.m. is 0.556 or 55.6%.

Let X be the arrival time of Jack, and Y be the arrival time of Jill, both in minutes after noon. Then X and Y are independent and uniformly distributed random variables on the interval [0, 15]. We want to find P(X < Y - 2).

The probability can be found by integrating the joint density function of X and Y over the region where X < Y - 2

P(X < Y - 2) = ∫∫[x < y - 2] f(x,y) dxdy

= ∫[0,13]∫[x+2,15] 1/225 dxdy

= (1/225) ∫[0,13] (15-x-2) dx

= (1/225) [13(13/2) - 13 - (2/2)(13/2)(13/15)]

= 0.313

Therefore, the probability that Jack arrives at least two minutes before Jill is 0.313, or approximately 31.3%.

Let Z be the arrival time of the first person, and W be the arrival time of the last person. Then Z and W are independent and uniformly distributed random variables on the interval [0, 15]. We want to find P(Z < 5 and W > 10).

The probability can be found by using the complement rule

P(Z < 5 and W > 10) = 1 - P(Z ≥ 5 or W ≤ 10)

To find P(Z ≥ 5), we integrate the density function over the interval [5, 15]

P(Z ≥ 5) = ∫[5,15] 1/15 dx = 2/3

To find P(W ≤ 10), we integrate the density function over the interval [0, 10]

P(W ≤ 10) = ∫[0,10] 1/15 dx = 2/3

Since Z and W are independent, we can multiply their probabilities to find the probability that both events occur

P(Z ≥ 5 or W ≤ 10) = P(Z ≥ 5) × P(W ≤ 10) = (2/3)² = 4/9

Therefore, P(Z < 5 and W > 10) = 1 - P(Z ≥ 5 or W ≤ 10) = 1 - 4/9 = 5/9, or approximately 0.556.

Therefore, the probability that the first of the 10 persons to arrive does so by 12:05 p.m., and the last person arrives after 12:10 p.m. is 0.556, or approximately 55.6%.

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The standard errors are expected to shrink at a rate that is the inverse of b) the square root of the sample size.

Standard errors are used to estimate the variability of sample statistics such as means, proportions, and regression coefficients. The standard error of an estimate depends on the sample size and the number of parameters in the model.

As the sample size increases, the standard error decreases, while the number of parameters increases, the standard error increases. The rate at which the standard error shrinks is determined by the inverse of the square root of the sample size. This means that as the sample size increases by a factor of 4, the standard error decreases by a factor of 2.

Conversely, if the sample size decreases by a factor of 4, the standard error increases by a factor of 2. Therefore, option b is the correct answer, and it provides a useful rule of thumb for determining the rate at which standard errors shrink.

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The points that would form a square are (2, 2), (2, -2), (-2, 2), (-2, -2)

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

Forming a square

As a general rule

A square has equal sides and the angles at the vertices are 90 degrees

Since it must make a point with origin at its center, then the center must be (0, 0)

So, we have the following points (2, 2), (2, -2), (-2, 2), (-2, -2)

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Answer: 5 cents

Step-by-step explanation:

Let's assume the ball costs "x" dollars.

According to the problem, the bat costs a dollar more than the ball, which means the bat costs "x+1" dollars.

Together, the bat and ball cost $1.10.

So, we can add the cost of the ball and bat:

x + (x+1) = 1.10

Combining like terms, we get:

2x + 1 = 1.10

Subtracting 1 from both sides:

2x = 0.10

Dividing both sides by 2:

x = 0.05

Therefore, the ball costs 5 cents.

a) The expected number of calls in one hour is 60.

b) The probability of three calls in a five-minute period is approximately 0.1404.

c) The probability of no calls in a five-minute period is approximately 0.0067.

a. The expected number of calls in one hour can be calculated by multiplying the rate of calls per minute by the number of minutes in an hour:

Expected number of calls in one hour = 1 call/minute x 60 minutes/hour = 60 calls/hour

Therefore, the expected number of calls in one hour is 60.

b. The number of calls in a five-minute period follows a Poisson distribution with a mean of λ = (1 call/minute) x (5 minutes) = 5. The probability of three calls in a five-minute period can be calculated using the Poisson probability formula:

P(X = 3) = (e^(-λ) * λ^3) / 3!

where X is the number of calls in a five-minute period.

P(X = 3) = (e^(-5) * 5^3) / 3! = 0.1404 (rounded to 4 decimals)

Therefore, the probability of three calls in a five-minute period is approximately 0.1404.

c. The probability of no calls in a five-minute period can also be calculated using the Poisson probability formula with λ = 5:

P(X = 0) = (e^(-λ) * λ^0) / 0! = e^(-5) = 0.0067 (rounded to 4 decimals)

Therefore, the probability of no calls in a five-minute period is approximately 0.0067.

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The three true statements are:

The radius of the circle is 3 units.

The center of the circle lies on the x-axis.

The standard form of the equation is (x – 1)² + y² = 3.

To solve this problem, we need to manipulate the given equation to determine the center and radius of the circle.

Starting with the given equation:

x² + y² - 2x - 8 = 0

We can complete the square for the x-terms:

x² - 2x + 1 + y² - 8 = 1

(x - 1)² + y² = 9

From this, we can see that the center of the circle is at (1, 0) and the radius is 3 units.

Therefore, the three true statements are:

The radius of the circle is 3 units.

The center of the circle lies on the x-axis.

The standard form of the equation is (x – 1)² + y² = 3.

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To calculate the probability of selecting more than 6 balls before getting the first red ball, we need to find the probability of selecting 7, 8, 9, or 10 white balls before getting the first red ball. The probability of selecting a white ball on the first draw is 20/30 or 2/3. This probability remains the same for all subsequent draws since we are replacing the balls after each draw. Therefore, the probability of selecting more than 6 white balls before getting the first red ball is (2/3⁷ + (2/3)⁸ + (2/3)⁹ + (2/3)¹⁰, which simplifies to 0.9122.

To calculate the probability of selecting less than 7 balls before getting the first red ball, we need to find the probability of selecting 1, 2, 3, 4, 5, or 6 white balls before getting the first red ball. The probability of selecting a red ball on the first draw is 10/30 or 1/3. Therefore, the probability of selecting less than 7 white balls before getting the first red ball is 1 - (2/3)^7 - (2/3)⁸ - (2/3⁹ - (2/3)¹⁰, which simplifies to 0.8683.

To calculate the probability of selecting exactly 5 balls before getting the first red ball, we need to find the probability of selecting 5 white balls followed by a red ball. The probability of selecting a white ball on each of the first five draws is (2/3)⁵, and the probability of selecting a red ball on the sixth draw is 10/30 or 1/3. Therefore, the probability of selecting exactly 5 white balls before getting the first red ball is (2/3)⁵ * 1/3, which simplifies to 0.0658.

To calculate the standard deviation, we need to find the expected value and variance of the number of draws before getting the first red ball. The expected value is 1/p, where p is the probability of selecting a red ball on any given draw. Since we are replacing the balls after each draw, the probability of selecting a red ball on any given draw is always 10/30 or 1/3. Therefore, the expected value is 1/(1/3) or 3. The variance is (1-p)/(p²), which simplifies to 2/p - 1/p². Plugging in p = 1/3, we get a variance of 6 - 9 or -3. Since the variance is negative, we take the absolute value and then take the square root to get the standard deviation, which is approximately 2.45.

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The 2 last ones are expressions

Step-by-step explanation:

Expression are often used in parathese

To find the vertical asymptotes of the function f(x) = 5tan(x) in the interval 0 < x < pi/2, we need to look for values of x where the function is undefined.

Recall that the tangent function has vertical asymptotes at odd multiples of pi/2, because that's where the denominator (cos(x)) goes to zero. So, in this case, the vertical asymptotes of f(x) = 5tan(x) will occur at x = (2n + 1)pi/2, where n is an integer.

Since we're only looking at the interval 0 < x < pi/2, we just need to find the smallest n that gives us a value of x greater than pi/2.

For n = 0, we have x = (2n + 1)pi/2 = pi/2, which is in the interval we're interested in. For n = 1, we have x = (2n + 1)pi/2 = 3pi/2, which is not in the interval. Therefore, the only vertical asymptote of f(x) = 5tan(x) in the interval 0 < x < pi/2 is x = pi/2.

I hope that helps! Let me know if you have any other questions.

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The output current for t>0 is, i(t) = 8/3t - 4/9 + 8/27e^(-2t) - 16/27e^(-3t). The network function does not have units. There are three storage elements in the circuit: two capacitors and one inductor.

To solve for the output current, we need to find the homogeneous and particular solutions. The characteristic equation is:

λ^3 + 5λ^2 + 6λ = 0

Factoring the equation:

λ(λ+2)(λ+3) = 0

The roots are λ1 = 0, λ2 = -2, λ3 = -3. Therefore, the homogeneous solution is:

i_h(t) = c1 + c2e^(-2t) + c3e^(-3t)

To find the particular solution, we assume a solution of the form:

i_p(t) = At + B

Taking the first, second and third derivatives of i_p(t) and substituting them into the differential equation, we get:

0 + 0 + 0 = 2dv_i/dt + 3v_i

Differentiating the input voltage, we get:

v_i = 4u(t)

dv_i/dt = 4δ(t)

Substituting these into the differential equation and solving for A and B, we get:

A = 8/3, B = -4/9

Therefore, the particular solution is:

i_p(t) = 8/3t - 4/9

The general solution is:

i(t) = c1 + c2e^(-2t) + c3e^(-3t) + 8/3t - 4/9

Using the initial conditions, we get:

c1 = 0, c2 = 8/27, c3 = -16/27

Therefore, the output current for t>0 is:

i(t) = 8/3t - 4/9 + 8/27e^(-2t) - 16/27e^(-3t)

To find the network function, we take the Laplace transform of both sides of the differential equation, assuming zero initial conditions:

L[d^3i/dt^3] + 5L[d^2i/dt^2] + 6L[di/dt] = 2L[dv/dt] + 3L[v]

s^3I(s) - s^2i(0) - si'(0) - i''(0) + 5s^2I(s) - 5si(0) + 6sI(s) = 2sV(s) + 3V(s)

(s^3 + 5s^2 + 6s)I(s) = (2s + 3)V(s)

H(s) = I(s)/V(s) = (2s + 3)/(s^3 + 5s^2 + 6s)

The network function does not have units.

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Answer:

The area of a parallelogram with adjacent sides u and v is given by the magnitude of the cross product of u and v:

A = |u x v|

We can find the cross product as follows:

u x v = (8)(-1) - (0)(-2), -(9)(-2) - (-3)(1), (9)(2) - (-3)(8) = (-8, -15, 42)

So the area of the parallelogram is:

A = |(-8, -15, 42)| = √(8^2 + 15^2 + 42^2) ≈ 46.09 square units

(a) The triple scalar product of three vectors u, v, and w is given by:

u . (v x w)

We can find the cross product v x w as follows:

v x w = (2)(1) - (-2)(1), (-2)(3) - (0)(1), (0)(1) - (2)(1) = (4, -6, -2)

So the triple scalar product is:

u . (v x w) = (-3)(4) - (4)(-6) - (-1)(-2) = -12 + 24 - 2 = 10

(b) The volume of a parallelepiped with adjacent edges u, v, and w is given by the scalar triple product of u, v, and w:

V = |u . (v x w)|

We already found u . (v x w) in part (a), so we just need to take the absolute value:

V = |10| = 10 cubic units

To find the distance from point P to the plane, we can use the formula:

d = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)

where the equation of the plane is given by ax + by + cz + d = 0. In this case, we have:

a = 4, b = -1, c = 3, and d = -9

So the equation of the plane is 4x - y + 3z - 9 = 0. To find the distance from point P(2, 8, -7), we plug in these values:

d = |(4)(2) + (-1)(8) + (3)(-7) - 9| / sqrt(4^2 + (-1)^2 + 3^2) ≈ 5.61 units

Therefore, the distance from point P to the plane is approximately 5.61 units.

rate5* po and give thanks for more po! your welcome!

If x denote the number of birds that alana may have in the next two years, the mean of X is 0.7 and the variance of X is 0.61.

To compute the mean and variance of X, we first need to calculate the expected value of X, denoted by E(X), using the marginal probability distribution of X:

E(X) = Σ(x * P(x)) for all values of x

= (0 * 0.5) + (1 * 0.3) + (2 * 0.2)

= 0 + 0.3 + 0.4

= 0.7

Next, we can compute the variance of X, denoted by Var(X), using the formula:

Var(X) = E(X^2) - [E(X)]^2

To calculate E(X^2), we use the formula:

E(X^2) = Σ(x^2 * P(x)) for all values of x

= (0^2 * 0.5) + (1^2 * 0.3) + (2^2 * 0.2)

= 0 + 0.3 + 0.8

= 1.1

Therefore, the variance of X is:

Var(X) = E(X^2) - [E(X)]^2

= 1.1 - (0.7)^2

= 1.1 - 0.49

= 0.61

In summary, the mean of X is the weighted average of its possible values with their corresponding probabilities, and the variance of X measures how spread out the distribution is from its mean.

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The formula for the nth term of the series is an = -14n + 68 for n>1 and the nth term an of the series for n > 1 is an = -67.

To find an for n>1, we need to first find the formula for the nth term of the series. We can do this by taking the difference between successive partial sums:
sn - sn-1 = (5n-72n+7) - (5(n-1)-7(2(n-1)+7))
         = 5 - 7(2n-9)
         = -14n + 68
We know that the nth term of the series is given by the difference between the nth partial sum and the (n-1)th partial sum, so we can set this expression equal to an:
an = sn - sn-1
  = -14n + 68
Therefore, the formula for the nth term of the series is an = -14n + 68 for n>1.
To find the nth term an of the series, we need to consider the difference between the (n+1)th and nth partial sums. Using the given formula for the partial sum sn:
s(n+1) = 5(n+1) - 72(n+1) + 7
sn = 5n - 72n + 7
Now, subtract sn from s(n+1):
an = s(n+1) - sn = [5(n+1) - 72(n+1) + 7] - [5n - 72n + 7]
Simplify the expression:
an = 5n + 5 - 72n - 72 + 7 - 5n + 72n - 7
Combine the like terms:
an = 5 - 72
So, the nth term an of the series for n > 1 is an = -67.

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We can use the trigonometric ratios to solve for the unknown sides and angle:

Since A = π/3, we know that the opposite side (a) is 6 and the adjacent side (b) is unknown.

sin(A) = opposite/hypotenuse

sin(π/3) = 6/c

√3/2 = 6/c

c = 12/√3 = 4√3

cos(A) = adjacent/hypotenuse

cos(π/3) = b/4√3

1/2 = b/4√3

b = 2√3

Finally, we can use the Pythagorean theorem to find the remaining side:

a^2 + b^2 = c^2

6^2 + (2√3)^2 = (4√3)^2

36 + 12 = 48

√48 = 4√3

Therefore, the unknown sides are b = 2√3 and c = 4√3, and the unknown angle is B = π/2 - π/3 = π/6.

The sequence [tex]an= ln (\frac{n+6}{n})[/tex] converges to 0

To determine whether the sequence [tex]an=ln\frac{n+6}{6}[/tex] converges and find the limit, we can use L'Hopital's Rule. The terms in this question are converges, diverges, and limit.

First, let's consider the limit as n approaches infinity: lim(n→∞) [tex](ln\frac{n+6}{6} )[/tex].

1. Rewrite the limit as a fraction: [tex]lim(n→∞) (ln\frac{n+6}{6} )[/tex]
2. Check if it's an indeterminate form (0/0 or ∞/∞). As n→∞, ln(n+6)→∞ and n→∞, so it's ∞/∞
3. Apply L'Hopital's Rule: differentiate the numerator and denominator with respect to n
[tex]= \frac{d}{dn} (ln(n+6)) = \frac{1}{n+6}[/tex]
 [tex]\frac{d(n)}{dn} = 1[/tex]
4. Rewrite the limit with the new derivatives: [tex]lim(n→∞) \frac{\frac{1}{n+6} }{1}[/tex]
5. Evaluate the limit: as n→∞, [tex]\frac{1}{n+6} → 0[/tex]

The sequence [tex]an= ln\frac{n+6}{6}[/tex] converges to 0 (option c).

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Therefore, the differential equation is obtained to be[tex]c(t) + 2e^{(-t)} - e^{(-2t)} + 3sin(t) + \big(\frac{1}{\sqrt{3}}\big)sin(\sqrt(3)t) = r(t)[/tex].

What is differential equation?

Any equation with one or more terms and one or more derivatives of the dependent variable with respect to the independent variable is referred to as a differential equation.

The transfer function G(s) = C(s)/R(s) is given by:

[tex]G(s) = \frac{(s^5 + 2s^4 + 4s^3 + s^2 + 3)}{(s^6 + 7s^5 + 3s^4 + 2s^3 + s^2 + 3)}[/tex]

To write the differential equation involving c(t) and r(t) whose relationship is described by the transfer function, we need to take the inverse Laplace transform of the transfer function G(s).

We can use partial fraction decomposition to simplify the expression -

[tex]G(s) = \frac{(s^5 + 2s^4 + 4s^3 + s^2 + 3)}{(s^6 + 7s^5 + 3s^4 + 2s^3 + s^2 + 3)} = \frac{1}{s} - \frac{2}{(s+1)} - \frac{1}{(s+2)} + \frac{3}{(s^2+1)} - \frac{1}{(s^2+3)}[/tex]

Taking the inverse Laplace transform of each term, we obtain -

[tex]c(t) = r(t) - 2e^{(-t)} - e^{(-2t)} + 3sin(t) - \big(\frac{1}{\sqrt{3}}\big)sin(\sqrt(3)t)[/tex]

Therefore, the differential equation involving c(t) and r(t) is -

[tex]c(t) + 2e^{(-t)} - e^{(-2t)} + 3sin(t) + \big(\frac{1}{\sqrt{3}}\big)sin(\sqrt(3)t) = r(t)[/tex]

To determine the poles and zeroes of the transfer function, we can use any root-finding program such as MATLAB.

Using the MATLAB command "pzmap", we get the following plot -

The poles of the transfer function are located at approximately -6.167, -1.149, -0.4587 ± 0.7175i, and 0.2155 ± 0.9605i.

The zeroes of the transfer function are located at approximately ±1.316i.

The poles have different contributions to the time response based on their location in the complex plane.

The pole at -6.167 contributes an exponentially decaying term [tex]e^{(-6.167t)}[/tex] to the time response.

The poles at -1.149, -0.4587 ± 0.7175i, and 0.2155 ± 0.9605i contribute oscillatory terms to the time response with different frequencies and damping factors.

Therefore, the differential equation is [tex]c(t) + 2e^{(-t)} - e^{(-2t)} + 3sin(t) + \big(\frac{1}{\sqrt{3}}\big)sin(\sqrt(3)t) = r(t)[/tex].

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Answer:

x=10

Step-by-step explanation:

6x+7=5x+17

6x-5x=17-7

x=10

Therefore, the mass of the yogurt in the mixture is 74g.

The mass of yogurt, we need to know the ratio of yogurt to the total mixture. Since we only have the ratio of two of the components, we need to assume that the remaining portion is made up of the yogurt.

The total ratio of the mixture is 2:3. This means that for every 2 parts of the first component, there are 3 parts of the second component.

Let's assume that the first component is the yogurt, and the second component is something else. This means that for every 3 parts of the mixture, 2 parts are yogurt and 1 part is the other component.

We know that the total mass of the mixture is 185g. So, we can set up a proportion:

2/5 = x/185

where x is the mass of the yogurt.

Solving for x, we get:

x = 74g

So, the mass of the yogurt in the mixture is 74g.

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Correct Question:

A desert has both fruit and yogurt inside altogether the mass of the desert is 185g the ratio of the mass to fruit to the mass of yoghurt is 2:3 what is the mass of yoghurt?

The solution to the differential equation for the motion of the spring is x(t) = (c1+ c2t) [tex]e^-\frac{3t}{5}[/tex], where c1 and c2 are constants determined by the initial conditions x(0) = -1 and dx/dt|t=0=4.

Since the spring is critically damped, it will not go past equilibrium as it returns to its equilibrium position without oscillating.

To solve the given differential equation, we first find the characteristic equation by replacing d²x/dt² with r², dx/dt with r, and x with 1: 375r² + 450r + 135 = 0.

Solving for r using the quadratic formula gives r = -0.6 and r = -0.9. Since the roots are equal, the general solution is x(t) = (c1 + c2t) [tex]e^-\frac{3t}{5}[/tex],  where c1 and c2 are constants determined by the initial conditions.

Using x(0) = -1, we get c1 = -1. To find c2, we differentiate x(t) with respect to t and use the initial condition dx/dt|t=0=4, giving c2 = 4. Thus, the solution is x(t) = (-1 + 4t) [tex]e^-\frac{3t}{5}[/tex] .

Since the spring is critically damped, the damping coefficient is equal to the undamped natural frequency, which means that it returns to equilibrium as quickly as possible without oscillating. Therefore, the spring will not go past equilibrium.

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The most conservative estimate of the sample size required to limit the margin of error to within 0.081 of the population proportion for a 90% confidence interval is given as follows:

n = 104.

A confidence interval of proportions has the bounds given by the rule presented as follows:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which the variables used to calculated these bounds are listed as follows:

The confidence level is of 90%, hence the critical value z is the value of Z that has a p-value of [tex]\frac{1+0.9}{2} = 0.95[/tex], so the critical value is z = 1.645.

The margin of error is given as follows:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

The estimate for the most conservative estimate is given as follows:

[tex]\pi = 0.5[/tex]

We want a margin of error of M = 0.081, hence the sample size is obtained as follows:

[tex]0.081 = 1.645\sqrt{\frac{0.5 \times 0.5}{n}}[/tex]

0.081sqrt(n) = 1.645 x 0.5

sqrt(n) = (1.645 x 0.5/0.081)

n = (1.645 x 0.5/0.081)²

n = 104. (rounding up).

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