List two multiples of 17

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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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.

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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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 location of C' after the dilation is given as follows:

C. C'(4, -6).

A dilation can be defined as a transformation that multiplies the distance between every point in an object and a fixed point, called the center of dilation, by a constant factor called the scale factor.

The location of C is given as follows:

C(2,-3).

The scale factor is given as follows:

k = 2.

Hence the location of C' after the dilation is given as follows:

C'(4, -6).

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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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(a) (-2, 2)

(i) To find the polar coordinates (r,θ) of the point (-2,2), we can use the formulas:

r = √(x^2 + y^2)

θ = tan^(-1)(y/x)

Substituting the given values, we have:

r = √((-2)^2 + 2^2) = √8

θ = tan^(-1)(2/-2) = - tan^(-1)(1) = -π/4

Therefore, the polar coordinates (r,θ) of the point (-2,2) are (r,θ) = (√8, -π/4).

(ii) It is not possible for the polar radius r to be negative. Therefore, there are no polar coordinates (r,θ) of the point (-2,2) where r<0.

(b) (3,3√3)

(i) To find the polar coordinates (r,θ) of the point (3,3√3), we can use the formulas:

r = √(x^2 + y^2)

θ = tan^(-1)(y/x)

Substituting the given values, we have:

r = √(3^2 + (3√3)^2) = 6

θ = tan^(-1)((3√3)/3) = π/3

Therefore, the polar coordinates (r,θ) of the point (3,3√3) are (r,θ) = (6, π/3).

(ii) It is not possible for the polar radius r to be negative. Therefore, there are no polar coordinates (r,θ) of the point (3,3√3) where r<0.

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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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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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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 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 two lengths which separate the top 9% and bottom 9% of nails are equal to 5.52 cm and 5.44 cm respectively and any nail length which are outside the calculated range can be rejected.

Mean = 5.48 cm

Standard deviation = 0.03cm

Use the z-score formula to find the corresponding z-scores for the top 9% and bottom 9% of the normal distribution,

For the top 9%, we have,

Attached table,

z = InvNorm(1 - 0.09)

 ≈ 1.34

For the bottom 9%, we have,

z = InvNorm(0.09)

  ≈ -1.34

where InvNorm is the inverse normal cumulative distribution function.

Using the z-score formula, we have,

z = (x - μ) / σ

Rearranging the formula, we can solve for x,

x = μ + zσ

Length that separates the top and bottom 9% of nails is,

For the top 9%, we have,

x = 5.48 + 1.34(0.03)

  ≈ 5.52

For the bottom 9%, we have,

x = 5.48 - 1.34(0.03)

   ≈ 5.44

Therefore, the two lengths that separate the top 9% and bottom 9% of nails are 5.52 cm and 5.44 cm, respectively. Any nail length outside this range can be rejected.

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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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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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a. The probability that one of the selected adults wears glasses is 0.323.

b. The expected number of adults not wearing glasses in this selection is 4.

c. The probability that more than three but no more than five of the selected adults will not be wearing glasses is 0.451.

d. The standard deviation of the number of adults not wearing glasses in this selection is 1.385.

(a) To find the probability that one of the selected adults wears glasses, we can use the binomial distribution formula:

P(X = k) = [tex]{}^nC_k[/tex] × [tex]p^k[/tex] × [tex](1-p)^{(n-k)}[/tex]

where n is the number of trials (in this case, 10), k is the number of successes (in this case, 1), p is the probability of success (in this case, 0.6), and (1-p) is the probability of failure (in this case, 0.4).

Substituting these values, we get:

P(X = 1) = [tex]{}^nC_k[/tex] × [tex]0.6^1[/tex] × [tex]0.4^9[/tex]

= 10 × [tex]0.6^1[/tex] × [tex]0.4^9[/tex]

= 0.323

So the probability that one of the selected adults wears glasses is 0.323.

(b) To find the expected number of adults not wearing glasses, we can use the formula:

E(X) = n × (1 - p)

where n is the number of trials (in this case, 10) and p is the probability of success (in this case, 0.6).

Substituting these values, we get:

E(X) = 10 × (1 - 0.6)

= 4

So the expected number of adults not wearing glasses is 4.

(c) To find the probability that more than three but no more than five of the selected adults will not be wearing glasses, we can use the binomial distribution formula again:

P(3 < X < 6) = P(X = 4) + P(X = 5)

Substituting the values using the formula, we get:

P(X = 4) = 10 choose 4 × [tex]0.4^4[/tex] × [tex]0.6^6[/tex]

= 210 × [tex]0.4^4[/tex] × [tex]0.6^6[/tex]

= 0.250

P(X = 5) = 10 choose 5 × [tex]0.4^5[/tex] × [tex]0.6^5[/tex]

= 252 × [tex]0.4^5[/tex] × [tex]0.6^5[/tex]

= 0.201

Therefore,

P(3 < X < 6) = P(X = 4) + P(X = 5)

= 0.250 + 0.201

= 0.451

So the probability that more than three but no more than five of the selected adults will not be wearing glasses is 0.451.

(d) To find the standard deviation of the number of adults not wearing glasses, we can use the formula:

SD(X) = √(n × p × (1 - p))

Substituting the values, we get:

SD(X) = √(10 × 0.6 × 0.4)

= 1.385

So the standard deviation of the number of adults not wearing glasses is 1.385.

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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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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 following represents the contents of arr as a result of executing the code segment {1, 2, 3, 5, 6, 7}. The correct answer is c.

The for loop in the code segment starts at index 3 and goes up to the second-to-last index of the array (arr.length - 1), so the elements being modified are 4, 5, and 6.

For each iteration of the loop, the element at the current index is replaced with the element at the next index. This effectively "shifts" the values of the array to the left by one position, overwriting the original values.

After the loop completes, the last element in the array (7) is not modified because the loop condition is k < arr.length - 1, not k < arr.length. Therefore, the resulting array is {1, 2, 3, 5, 6, 7}.

The correct answer is c.

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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 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 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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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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Answer:

64

Step-by-step explanation:

[tex]\frac{24}{3} = \frac{y}{8} \\\\(24*8)/3 = y\\\\y=64[/tex]

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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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

The 2 last ones are expressions

Step-by-step explanation:

Expression are often used in parathese

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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