(a) If y=-x² + 4x + 5
(i) Find the z and y intercepts.
(ii) Find the axis of symmetry and the maximum value of the parabola
(iii) Sketch the parabola showing and labelling the r and y intercepts and its vertex (turning point).

The required solution of the series is: y = 2 - x - (2/3)x² + (2/9)x³ - (8/45)x⁴ + (2/1575)x⁵ + ...

The given differential equation is y″ - (x / (x + 1)) y′ + y / (x + 1) = 0 and initial conditions y(0) = 2 and y′(0) = -1.

Using the power series method, we assume that the solution of the differential equation can be written in the form of power series as:

y = ∑(n = 0)^(∞) aₙxⁿ

Differentiating y once and twice, we get

y′ = ∑(n = 1)^(∞) naₙx^(n - 1) and

y″ = ∑(n = 2)^(∞) n(n - 1)aₙx^(n - 2)

Substitute y, y′, and y″ in the differential equation and simplify the equation:

∑(n = 2)^(∞) n(n - 1)aₙx^(n - 2) - ∑(n = 1)^(∞) [(n / (x + 1))aₙ + aₙ₋₁]x^(n - 1) + ∑(n = 0)^(∞) aₙx^(n - 1) / (x + 1) = 0

Rearranging the terms, we get

aₙ(n + 1)(n + 2) - aₙ(x / (x + 1)) - aₙ₋₁

= 0aₙ(x / (x + 1))

= aₙ(n + 1)(n + 2) - aₙ₋₁a₀ = 2 and

a₁ = -1

Let's find some of the coefficients:

a₂ = - 2a₀ / 3,

a₃ = 2a₀ / 9 - 5a₁ / 18,

a₄ = - 8a₀ / 45 + 2a₁ / 15 + 49a₂ / 360,

a₅ = 2a₀ / 1575 - a₁ / 175 - 59a₂ / 525 + 469a₃ / 4725 + 4307a₄ / 141750...

The solution of the differential equation that satisfies the initial conditions is:

y = 2 - x - (2/3)x² + (2/9)x³ - (8/45)x⁴ + (2/1575)x⁵ + ...

Therefore, the required solution is: y = 2 - x - (2/3)x² + (2/9)x³ - (8/45)x⁴ + (2/1575)x⁵ + ...

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(a) To find the derivative of the function f(x) = -9x + 6, we differentiate term by term. The derivative of -9x is -9, and the derivative of 6 is 0. Therefore, f'(x) = -9.

(b) To find the critical points, we set f'(x) equal to zero and solve for x:

-9 = 0. Since there is no solution to this equation, there are no critical points. (c) Since there are no critical points, we cannot classify any extrema. (d) However, in this case, we can still evaluate the second derivative at x = -3 to determine if it is a maximum, minimum, or saddle point. Taking the derivative of f'(x) = -9 with respect to x gives us f"(x) = 0, which is a constant value.

(e) Similarly, we can evaluate the second derivative at x = +3 to determine the nature of the extremum. Evaluating f"(x) at x = +3 gives us f"(x) = 0, which is also a constant value.

Since the second derivative is zero at both x = -3 and x = +3, we cannot determine the nature of the extrema using the second derivative test. In this case, further analysis is needed to determine if these points are maximum, minimum, or saddle points. In summary, the function f(x) = -9x + 6 has no critical points, and therefore no extrema can be classified. The second derivative is zero at x = -3 and x = +3, which means we need additional information or methods to determine the nature of the extrema at these points.

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The exact value of sin 285° using half angle identity is given as ±√[2 + √[3]]/2.

Half angle identities refer to the trigonometric identities which represent trigonometric functions in terms of half of the angle of the given function.

Trigonometric functions sine, cosine and tangent can be represented using half angle identities as follows:

sin(θ/2) = ±√[1 − cos(θ)]/2cos(θ/2)

= ±√[1 + cos(θ)]/2tan(θ/2)

= ±√[1 − cos(θ)]/[1 + cos(θ)]

Given, we have to find the exact value of sin 285° using half angle identity.

Let us write the given angle 285° in terms of a smaller angle using the reference angle theorem as follows:

285° = 360° - 75°

We know that sin(θ) = sin(θ - 2π)

Therefore, sin(285°) = sin(285° - 2π)

Now, substituting the value of sin(θ) in half angle identity of sine:

sin(θ/2) = ±√[1 − cos(θ)]/2sin(285°/2)

= ±√[1 - cos(570°)]/2

= ±√[1 - cos(210°)]/2

Here, we need to find the value of cos(210°).cos(210°)

= cos(360° - 150°)

= cos(150°)

= -√[3]/2

By substituting the value of cos(210°) in half angle identity of sine, we get:

sin(285°/2)

= ±√[1 - (-√[3]/2)]/2

= ±√[2 + √[3]]/2

Thus, the exact value of sin 285° using half angle identity is given as ±√[2 + √[3]]/2.

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The true statements about the normal distribution are A. Approximately 95% of scores/values will fall between +/- 2 standard deviations from the mean and C. The majority of scores/values will fall within +/- 1 standard deviation of the mean.

In a normal distribution, approximately 95% of the scores/values will fall within two standard deviations (plus or minus) from the mean. This means that the distribution is symmetric, and the majority of values are concentrated around the mean. Therefore, statement A is true.

Regarding statement C, in a normal distribution, the majority of scores/values (around 68%) will fall within one standard deviation (plus or minus) from the mean. This shows that the distribution is relatively tightly clustered around the mean. Hence, statement C is also true.

Statement B is not true for the normal distribution. In a normal distribution, the tails on both sides of the distribution have equal lengths, making it a symmetric bell-shaped curve. Therefore, the right tail is not longer than the left tail.

Statement D is also not true. While the vast majority of scores/values fall within three standard deviations from the mean, it is not accurate to say that 100% of the values will fall within this range. The normal distribution extends infinitely in both directions, so there is a small possibility of extreme values lying beyond three standard deviations from the mean.

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If we can reject H₀ at the 5% level of significance, then we can also reject the same H₀ with the same H₁ at the 10% level of significance.

If we can reject the null hypothesis H₀ at the 5% level of significance, then it implies that the probability of getting a sample mean, as extreme as the one we have observed, under the null hypothesis is less than 5%. Hence, we can reject the null hypothesis at the 5% level of significance.

Similarly, if we consider the 10% level of significance, then it implies that the probability of getting a sample mean as extreme as the one we have observed under the null hypothesis is less than 10%. Hence, if we can reject the null hypothesis at the 5% level of significance, then we can also reject it at the 10% level of significance. Therefore, if we reject H₀ with a given H₁ at a higher level of significance, we will surely reject H₀ at a lower level of significance.

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The graph of the function y = 2x + 1 is added as an attachment

The slope is 2 and the y-intercept is 1

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

y = 2x + 1

The above function is an linear function that has been transformed as follows

Next, we plot the graph using a graphing tool by taking not of the above transformations rules

The graph of the function is added as an attachment

From the graph, we have

Slope = 2

y-intercept = 1

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The answer is - based on the equations, the propositions (1) peg (2) p-q (3) q-r (4) 'r - c. Inconsistent.

Determine if the propositions (1) p^eg (2) p-q (3) q-r (4) r are consistent or inconsistent.

Consistent:

A set of propositions is said to be consistent if all propositions in the set can be true simultaneously.

Inconsistent:

A set of propositions is said to be inconsistent if all propositions in the set cannot be true simultaneously.

(1) p ^ eg

This is inconsistent since if we assume p to be true, then eg becomes false, and if we assume eg to be true, then p becomes false.

Thus they cannot be true at the same time.

(2) p - q.

This is consistent since both propositions can be false at the same time.

(3) q - r

This is consistent since both propositions can be false at the same time.

(4) r.

This is consistent since it is a single proposition.

Therefore, options (b), (d), and (e) can be eliminated.

Hence, the correct option is (c) Inconsistent.

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Therefore, the correct statement is: a) The total sum of squares (SST) is larger than or equal to 342.

The sum of squares regression (SSR) represents the sum of the squared differences between the predicted values and the mean of the dependent variable. It measures the amount of variation in the dependent variable that is explained by the regression model.

If the SSR is 342, it means that the regression model is able to explain 342 units of variation in the dependent variable. Since SSR is a measure of explained variation, it must be true that the total sum of squares (SST) is larger than or equal to 342. SST represents the total variation in the dependent variable.

The other statements (b, c, and d) are not necessarily true based on the given information about SSR. The sign of the slope of the regression line or the magnitude of the error sum of squares cannot be determined solely from the value of SSR.

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The given question involves analyzing the properties of i.i.d. random variables with a specific probability density function (pdf). In part (a), we are asked to find the conditional expectation and variance of X.

(a) To find the conditional expectation and variance of X, we can use the law of total expectation and the law of total variance. The given hint suggests using these laws to calculate the desired quantities.

(b) The task in this part is to show that as the sample size increases to infinity, the probability that the maximum value of the sample equals a specific value approaches 1 - 1/e. This can be achieved by analyzing the properties of the maximum value, considering the behavior of extreme values, and using mathematical techniques such as limit theorems.

(c) In this part, you are asked to design a simulation study to demonstrate the convergence of the maximum value. This involves generating multiple samples of different sizes (e.g., 100, 250, 500, 1000, 2000, 5000) from the given distribution and calculating the probability that the maximum value equals a specific value (ô). By comparing the probabilities obtained from the simulation study with the theoretical result from part (b), you can demonstrate the convergence.

By following the given instructions and applying the relevant statistical concepts and techniques, you will be able to answer each part of the question and provide a thorough analysis.

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The study involving a woman's ability to identify the pouring order of tea and milk is an experiment with the explanatory variable being the order of pouring and the response variable being the correct identification; the parameter is the probability of correct identification, and the statistic is the observed proportion; the null hypothesis assumes guessing, and the alternative hypothesis suggests better than chance performance; without calculating the p-value, no conclusion can be drawn about the woman's ability.

This is an Experiment because the woman was presented with cups and asked to identify which had been poured first. The researcher controlled the cups' contents and the order in which they were presented. The parameter is the probability (p) of correctly identifying the pouring order of tea and milk.

The statistic is the observed proportion (p-hat) of cups correctly identified as having tea poured first. Null hypothesis (H0): The woman's ability to identify the pouring order is based on guessing alone (p = 0.5). Alternative hypothesis (Ha): The woman's ability to identify the pouring order is better than chance (p > 0.5).

To approximate the p-value, we need more information such as the sample size or the number of successful identifications. Without this information, it is not possible to calculate the p-value or determine statistical significance.

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To find the sum of squares (SS) for the given equation, we need to calculate the sum of squares of individual terms. The options provided are decimal values, and we need to determine which one is the closest.

The given equation is SS bet = (20²/5) + (45²/5) + (35²/5) + (100²/15). To calculate the SS, we need to square each term and then sum them up. Let's perform the calculations:

SS bet = (20²/5) + (45²/5) + (35²/5) + (100²/15)

= (400/5) + (2025/5) + (1225/5) + (10000/15)

= 80 + 405 + 245 + 666.67

= 1396.67

Now we compare this value with the options provided. Among the options, the closest approximation to the calculated SS value of 1396.67 is option D: 61.40.

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A mathematical entity known as a vector denotes both magnitude and direction. It is frequently used to express things like distance, speed, force, and acceleration.  Option c is the correct answer.

A vector can be represented visually by an arrow or a directed line segment.

We can examine if there are scalars A and B such that Z = A * U + B * V to see if the vector Z = [4, -12, 9] belongs to the span of the vectors U = [-12, 27, 4] and V = [-4, -3, 9].

Putting the equation together, we have:

A* [-12, 27, 4] + B* [-4, -3, 9] = Z = A * U + B * V [4, -12, 9]

When the right side of the equation is expanded, we obtain:

[4, -12, 9] is equivalent to [-12A - 4B, 27A - 3B, 4A + 9B]

At this point, we may compare the appropriate elements on both sides:

4A + 9B = 9 -12A - 4B = 4 27A - 3B = -12

To determine the values of A and B, we can solve this system of equations. By condensing the equations, we obtain:

27A - 3B = -12 --> -

12A - 4B = 4 --> 

3A + B = -1 9A - B 

= -4 4A + 9B 

= 9

A = -1 and B = 4 are the results of solving this system of equations.

Z, therefore, equals -1 * U plus 4 * V.

The result of substituting the values of U and V is:

Z = -1 * [-12, 27, 4] + 4 * [-4, -3, 9]

Z = [12, -27, -4] + [-16, -12, 36]

Z = [-4, -39, 32]

Thus, Z = [-4, -39, 32].

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Therefore, the specific value for the test statistic and p-value cannot be determined without knowing the degrees of freedom, which depends on the sample size (n).

The test statistic for this sample can be calculated using the formula:

[tex]t = (d - μd) / (sd / √(n))[/tex]

Substituting the given values:

d = -26 (average difference)

μd = 0 (null hypothesis mean)

sd = 33.4 (standard deviation of differences)

n = 8 (sample size)

Plugging in these values, the test statistic is:

[tex]t = (-26 - 0) / (33.4 / √(8))[/tex]

The p-value for this sample can be obtained by comparing the test statistic to the t-distribution with (n - 1) degrees of freedom and determining the probability of obtaining a more extreme value.

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To find the Fourier Series expansion of the given function f(x), we need to determine the coefficients of the series. The Fourier Series representation of f(x) is given by:

f(x) = a₀/2 + Σ(aₙcos(nπx/2) + bₙsin(nπx/2))

To find the coefficients a₀, aₙ, and bₙ, we can use the formulas:

a₀ = (1/2)∫[0,2] f(x) dx

aₙ = ∫[0,2] f(x)cos(nπx/2) dx

bₙ = ∫[0,2] f(x)sin(nπx/2) dx

Let's calculate these coefficients step by step.

1. Calculation of a₀:

a₀ = (1/2)∫[0,2] f(x) dx

Since f(x) is defined differently for different intervals, we need to split the integral into two parts:

a₀ = (1/2)∫[0,1] x dx + (1/2)∫[1,2] 1 dx

  = (1/2) * [(1/2)x²]₀¹ + (1/2) * [x]₁²

  = (1/2) * [(1/2) - 0] + (1/2) * [2 - 1]

  = (1/2) * (1/2) + (1/2) * 1

  = 1/4 + 1/2

  = 3/4

So, a₀ = 3/4.

2. Calculation of aₙ:

aₙ = ∫[0,2] f(x)cos(nπx/2) dx

Again, we need to split the integral into two parts:

For the interval [0,1]:

aₙ₁ = ∫[0,1] xcos(nπx/2) dx

Integrating by parts, we have:

aₙ₁ = [x(2/nπ)sin(nπx/2)]₀¹ - ∫[0,1] (2/nπ)sin(nπx/2) dx

    = [(2/nπ)sin(nπ/2) - 0] - (2/nπ)∫[0,1] sin(nπx/2) dx

    = (2/nπ)sin(nπ/2) - (2/nπ)(-2/π)cos(nπx/2)]₀¹

    = (2/nπ)sin(nπ/2) + (4/n²π²)cos(nπ/2) - (2/n²π²)cos(nπ)

    = (2/nπ)sin(nπ/2) + (4/n²π²)cos(nπ/2) - (2/n²π²)(-1)^n

For the interval [1,2]:

aₙ₂ = ∫[1,2] 1cos(nπx/2) dx

    = ∫[1,2] cos(nπx/2) dx

    = [(2/nπ)sin(nπx/2)]₁²

    = (2/nπ)(sin(nπ) - sin(nπ/2))

    = (2/nπ)(0 - 1)

    = -2/nπ

Therefore, aₙ = aₙ₁ + aₙ₂

    = (2/nπ)sin(nπ/2)

+ (4/n²π²)cos(nπ/2) - (2/n²π²)(-1)^n - 2/nπ

3. Calculation of bₙ:

bₙ = ∫[0,2] f(x)sin(nπx/2) dx

For the interval [0,1]:

bₙ₁ = ∫[0,1] xsin(nπx/2) dx

Using integration by parts, we have:

bₙ₁ = [-x(2/nπ)cos(nπx/2)]₀¹ + ∫[0,1] (2/nπ)cos(nπx/2) dx

    = [-x(2/nπ)cos(nπ/2) + 0] + (2/nπ)∫[0,1] cos(nπx/2) dx

    = -(2/nπ)cos(nπ/2) + (2/nπ)(2/π)sin(nπx/2)]₀¹

    = -(2/nπ)cos(nπ/2) + (4/n²π²)sin(nπ/2)

For the interval [1,2]:

bₙ₂ = ∫[1,2] sin(nπx/2) dx

    = [-2/(nπ)cos(nπx/2)]₁²

    = -(2/nπ)(cos(nπ) - cos(nπ/2))

    = 0

Therefore, bₙ = bₙ₁ + bₙ₂

    = -(2/nπ)cos(nπ/2) + (4/n²π²)sin(nπ/2)

Now we have obtained the coefficients of the Fourier Series expansion for the given function f(x). We can plot the points and draw the graph.

Using the provided data:

Dogs Stride length (meters): 1.5, 1.7, 2.0, 2.4, 2.7, 3.0, 3.2, 3.5, 2, 3.5

Speed (meters per second): 3.7, 4.4, 4.8, 7.1, 7.7, 9.1, 8.8, 9.9

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The three inequalities for the sum of whole numbers are: x ≥ 0, y ≥ 0, x + y > 20.

The sum of two whole numbers is greater than 20.

The three inequalities for the statement above are given by x+y > 20 where x and y are whole numbers.

Whole numbers are positive integers that do not have any fractional or decimal parts.

In other words, whole numbers are numbers like 0, 1, 2, 3, 4, and so on, which are not fractions or decimals.

The inequalities for the above statement are: x ≥ 0, y ≥ 0, and x + y > 20.

Therefore, the correct option is:x ≥ 0, y ≥ 0, x + y > 20.

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The orthogonal decomposition of vector v with respect to vectors w1 and w2 is v = [5, -3, 4] = [4.5, -2, 4.5] + [0.5, -1, -0.5].

To find the orthogonal decomposition of vector v with respect to vector w, we need to find the projection of v onto the subspace spanned by w and subtract it from v.

Given:

v = [5, -3, 4]

w1 = [1, 2, 1]

w2 = [1, -1, 1]

First, we need to find the projection of v onto the subspace spanned by w. To do this, we calculate the projection vector p:

p = ((v · w1) / (w1 · w1)) * w1 + ((v · w2) / (w2 · w2)) * w2

where · represents the dot product.

Calculating the dot products:

v · w1 = 51 + (-3)2 + 41 = 5 - 6 + 4 = 3

w1 · w1 = 11 + 22 + 11 = 1 + 4 + 1 = 6

v · w2 = 51 + (-3)(-1) + 41 = 5 + 3 + 4 = 12

w2 · w2 = 11 + (-1)(-1) + 11 = 1 + 1 + 1 = 3

Now, we can calculate the projection vector p:

p = (3/6) * [1, 2, 1] + (12/3) * [1, -1, 1]

= [1/2, 1, 1/2] + [4, -4, 4]

= [4.5, -2, 4.5]

Finally, we can find the orthogonal decomposition of v:

v = p + v_perp

where v_perp is the component of v orthogonal to the subspace spanned by w. To find v_perp, we subtract p from v:

v_perp = v - p

= [5, -3, 4] - [4.5, -2, 4.5]

= [0.5, -1, -0.5]

Therefore, the orthogonal decomposition of v with respect to w is:

v = [4.5, -2, 4.5] + [0.5, -1, -0.5]

= [5, -3, 4]

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We have data on exam grades divided by gender. The table provides information on the number of observations, standard deviations, and means for male and female students.

(a) To test if the standard deviation of exam grades for male students is larger than that of female students, we can use an F-test. The F-test compares the ratio of the variances between the two groups. In this case, we compare the variance of grades for males to the variance of grades for females. If the calculated F-statistic is greater than the critical F-value at a 1% significance level, there is evidence that the standard deviation of grades for male students is larger.

(b) To assess if there is a statistically significant difference in the average grades between male and female students, we can use a two-sample t-test. This test compares the means of two independent groups. We compare the mean grades for males to the mean grades for females. If the calculated t-statistic is greater than the critical t-value at a 1% significance level, we conclude that there is a statistically significant difference in average grades between the two genders.

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The solution to the differential equation is: ln(y) - x = e-x dx - 1/2.

(i) (x – 4) y4dx – 23 (y2 – 3) dy = 0The differential equation (i) can be solved using the method of separable variables.

To do this, first we rearrange the terms to obtain it in the following form: dy/(y^2 - 3) = (x - 4)dx/23y4.

The integral form of the equation is thus: ∫dy/(y^2 - 3) = ∫(x - 4)/23y4dx.

Note that we need to integrate both sides with respect to their variables.

Hence we proceed to obtain the solutions by integration as follows:

∫dy/(y^2 - 3) = ∫(x - 4)/23y4dx= (1/2√3) ln(|(y-√3)/(y+√3)|) = (1/345)y-3 + C.

where C is the constant of integration that we have to find.

To get the constant of integration C, we use the initial condition where y(0) = 2.

Substituting y(0) = 2 into the equation (1/2√3) ln(|(y-√3)/(y+√3)|) = (1/345)y-3 + C, we obtain: C = (1/2√3) ln(|(2-√3)/(2+√3)|) - (1/345)(2)-3= - 0.0837.

Hence the solution to the differential equation is:(1/2√3) ln(|(y-√3)/(y+√3)|) = (1/345)y-3 - 0.0837(ii) e-y (1+ = 1, y(0) = 1.

The differential equation (ii) can be solved using the method of separable variables.

To do this, we first arrange the terms to obtain it in the following form: (1/y) dy - 1 = -x dx.e-x dx = ∫1/(y) dy - ∫1 dx = ln(y) - x + C. where C is the constant of integration that we have to find.

To obtain C, we use the initial condition where y(0) = 1.e-x dx = ln(1) - 0 + C= C.

Hence the solution to the differential equation is: ln(y) - x = e-x dx + C. Substituting y = 1 when x = 0, we have: ln(1) - 0 = e-0(1/2) + C.C = - 1/2 Therefore the solution to the differential equation is: ln(y) - x = e-x dx - 1/2.

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The probability that both cards are Kings is 1/221. Option (B) is the correct answer.

Solution: Given: We have two cards that are dealt successively (without replacement) from a shuffled deck of 52 playing cards. We need to find the probability that both cards are Kings. There are 52 cards in a deck of cards. There are four kings in a deck of cards.

Therefore, Probability of getting a king card = 4/52

After selecting one king card, the number of cards remaining in the deck is 51.

Therefore, Probability of getting second king card = 3/51

Required probability of getting both kings is the product of both probabilities.

P(both king cards) = P(first king card) × P(second king card)

= 4/52 × 3/51

= 1/221

Therefore, the probability that both cards are Kings is 1/221.Option (B) is the correct answer.

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We are required to solve the given initial-value problem using Laplace transform

where;$$y'' + y = f(t),\ y(0) = 1,\ y'(0) = 0,$$and$$f(t) =\begin{cases}1,&0\leq t<\frac{\pi}{2}\\ \sin(t),&t\geq\frac{\pi}{2} \end{cases}$$Given, $$y(t) =\left(\right)+\left(\right)u(t-\frac{\pi}{2})$$

Taking Laplace Transform of the given equation,$$\mathcal{L}\left[y''+y\right]=\mathcal{L}\left[f(t)\right]$$$$\mathcal{L}\left[y''\right]+\mathcal{L}\left[y\right]=\mathcal{L}\left[f(t)\right]$$$$s^2Y(s)-sy(0)-y'(0)+Y(s)=\frac{1}{s}+\mathcal{L}\left[\sin(t)\right]u\left(t-\frac{\pi}{2}\right)$$$$s^2Y(s)+Y(s)=\frac{1}{s}+\frac{\exp\left(-\frac{\pi s}{2}\right)}{s^2+1}$$$$\left(s^2+1\right)Y(s)=\frac{1}{s}+\frac{\exp\left(-\frac{\pi s}{2}\right)}{s^2+1}$$$$Y(s)=\frac{1}{s\left(s^2+1\right)}+\frac{\exp\left(-\frac{\pi s}{2}\right)}{\left(s^2+1\right)^2}$$

We know that the inverse Laplace transform

of$$\mathcal{L}^{-1}\left[\frac{1}{s\left(s^2+a^2\right)}\right]=\frac{1}{a}\cos(at)$$

Hence,

$$y(t)=\frac{1}{1}\cos(t)+\frac{1}{2}\exp\left(-\frac{\pi}{2}\right)t\sin(t)$$$$y(t)=\cos(t)+\frac{1}{2}t\sin(t)\exp\left(-\frac{\pi}{2}\right)$$

[tex]Therefore, $$y(t)=\cos(t)+\frac{1}{2}t\sin(t)\exp\left(-\frac{\pi}{2}\right)$$This is the required solution.[/tex]

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a) Odds: 1 in (50 choose 4).

b) Odds: (1/50) * (1/49) * (1/48) * (1/47).

a) To calculate the odds in favor of ending up with 1 Cherry Passion Fruit, 1 Mandarin Orange Mango, 1 Strawberry Banana, and 1 Pineapple Pear when grabbing 4 Jelly Bellies, we need to consider the number of favorable outcomes and the total number of possible outcomes.

Since there is only one Cherry Passion Fruit, one Mandarin Orange Mango, one Strawberry Banana, and one Pineapple Pear in the bag, the number of favorable outcomes is 1. The total number of possible outcomes can be calculated by the combination formula, which is C(50, 4) = 50! / (4! * (50-4)!). This simplifies to 50! / (4! * 46!).

Therefore, the odds in favor can be calculated as: Odds in favor = Number of favorable outcomes / Total number of possible outcomes = 1 / (50! / (4! * 46!)).

b) To calculate the odds in favor of eating a Mixed Berry, then a Pineapple Pear, then a Mandarin Orange Mango, and finally a Cherry Passion Fruit when selecting Jelly Bellies one at a time, we need to consider the number of favorable outcomes and the total number of possible outcomes.

Since the Jelly Bellies are selected one at a time, the probability of getting a Mixed Berry first is 1/50. After selecting the Mixed Berry, there are now 49 Jelly Bellies left, so the probability of getting a Pineapple Pear next is 1/49. Similarly, the probability of getting a Mandarin Orange Mango next is 1/48, and the probability of getting a Cherry Passion Fruit last is 1/47.

To calculate the odds in favor, we multiply the individual probabilities: Odds in favor = (1/50) * (1/49) * (1/48) * (1/47).

Please note that these calculations assume that each Jelly Belly is equally likely to be selected and that the Jelly Bellies are selected without replacement.

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The solution of the given system of equations is:x1= 1x2 =-2x3 = -2/5x4 = 1.

The system of linear equations given is:

X12x2-x3 + x4 = − 13x1+5x2-4x3 − x4 = −46x1+5x27x3 − 2 x4 = −15x1+5x2 −6x3 − x4 =-4

The system can be written in the augmented matrix form as: [1 2 -1 1 -1][3 5 -4 -1 -4][6 5 2 -7 -1][5 5 -6 -1 -4]

To solve the system of equations by modifying it to REF and to RREF using equivalent elementary operations, we need to perform the following operations: Interchange two rows Add or subtract a multiple of one row to another row Multiply a row by a nonzero scalar

These operations should be used to obtain the row-echelon form (REF) and then reduced row-echelon form (RREF) of the augmented matrix. Row Echelon Form To obtain the REF of the matrix, we will use elementary operations to eliminate the first nonzero element of every row below the leading coefficient of the previous row.

The REF of the given matrix is: [1 2 -1 1 -1][0 -1 1 -4 1][0 0 10 -17 5][0 0 0 -9 -9]

Reduced Row Echelon Form

To obtain the RREF of the matrix, we will further use elementary operations to eliminate all elements below the leading coefficients of the previous rows.

The RREF of the given matrix is: [1 0 0 0 -1][0 1 0 0 -2][0 0 1 0 -2/5][0 0 0 1 1]

Therefore, the solution of the given system of equations is:x1= 1x2 =-2x3 = -2/5x4 = 1.

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Answer:

The probability that at least two motorbikes out of the ten have defective lights is 0.1445.

Step-by-step explanation:

According to the survey, the probability of a motorbike having defective lights is 7 %. which can be expressed as 0.07.

The probability that at least two bikes have defective lights is the probability can be from two, three, four, ... up to ten defective bikes. the sum of these probabilities is the probability of at least two defective bikes.

P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + ... + P(X = 10)

By using the binomial probability formula we can calculate P(X = k):

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

Where :

calculation:

P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) + ... + P(X = 10)

P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)

P(X ≥ 2) = 1 - C(10, 0) * p^0 * (1 - p)^(10 - 0) - C(10, 1) * p^1 * (1 - p)^(10 - 1)

P(X ≥ 2) = 1 - (1 - p)^10 - 10 * p * (1 - p)^9

P(X ≥ 2) = 1 - (1 - 0.07)^10 - 10 * 0.07 * (1 - 0.07)^9

P(X ≥ 2) = 0.1445

Therefore the probability that at least two motorbikes out of the ten have defective lights is 0.1455.

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The series ∑ₖ₌₂ (4k/k²)ᵏ diverges because the root test shows that the limit of the nth root is 4, greater than 1.

To determine whether the series converges or diverges, we apply the root test. Taking the nth root of the terms, we get 4(k/n)^(-1/n).

As n approaches infinity, (k/n) approaches a constant value. Since the exponent -1/n tends to 0, the limit of the nth root simplifies to 4.

According to the root test, if the limit of the nth root is less than 1, the series converges; if it is greater than 1, the series diverges.

In this case, the limit is 4, which is greater than 1. Thus, the series diverges.


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The product of two functions, f(x) and g(x), where f(x) tends to infinity as x tends to 1 and g(x) is always greater than 1/2022, will also tend to infinity as x tends to 1.

To prove that f(x)g(x) tends to infinity as x tends to 1, we need to show that the product of f(x) and g(x) becomes arbitrarily large for values of x close to 1.

Given that f(x) tends to infinity as x tends to 1, we can say that for any M > 0, there exists a number δ > 0 such that if 0 < |x - 1| < δ, then f(x) > M. This means that we can find a value of f(x) as large as we want by choosing an appropriate value of M.

Now, we are given that g(x) > 1/2022 for every x. This implies that g(x) is always greater than a positive constant value, namely 1/2022. Let's call this constant value C = 1/2022.

Considering the product f(x)g(x), we can see that if we choose a value of x close to 1, the value of f(x) tends to infinity, and g(x) is always greater than C = 1/2022. Therefore, the product f(x)g(x) will also tend to infinity.

To illustrate this further, let's suppose we choose an arbitrary large number N. We can find a corresponding value of M such that for f(x) > M, the product f(x)g(x) will be greater than N. This is because g(x) is always greater than C = 1/2022.

In conclusion, since f(x) tends to infinity as x tends to 1 and g(x) is always greater than 1/2022, the product f(x)g(x) will also tend to infinity as x tends to 1. The constant factor of 1/2022 does not affect the tendency of f(x)g(x) to approach infinity.

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The stem and leaf plot for the data is plotted below. With 51 being a potential outlier as it is significantly lower than other values in the data.

Given the data :

The stem and leaf plot for the given data is illustrated below :

5 | 1

7 | 6 7 8 9

8 | 1 2 4 6

9 | 9

Outliers are values which shows significant deviation from other values within a set of data.

From the data, the value 51 seem to be a potential outlier value as it differs significantly when compared to other values in the data.

Therefore, there is a potential outlier which is 51 because it differs significantly from other values in distribution.

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The test scores of IBM students are normally distributed with a mean of 950 and a standard deviation of 200. Using this information, we can answer the following questions: a) the percentage of students with scores higher than 1390, b) the percentage of students with scores between 1100 and 1200, c) the minimum and maximum values of the middle 87.4% of scores, and d) the number of students who took the exam if there were 165 students who scored above 1432.

a) To find the percentage of students with scores higher than 1390, we need to calculate the area under the normal distribution curve to the right of the score 1390. Using a standard normal distribution table or a graphing tool, we can find the corresponding z-score for 1390. Once we have the z-score, we can determine the proportion or percentage of the distribution to the right of that z-score, which represents the percentage of students with scores higher than 1390.

b) To find the percentage of students with scores between 1100 and 1200, we need to calculate the area under the normal distribution curve between these two scores. Similar to the previous question, we can convert the scores to their corresponding z-scores and find the area between the two z-scores using a standard normal distribution table or a graphing tool.

c) To find the minimum and maximum values of the middle 87.4% of the scores, we need to locate the z-scores that correspond to the 6.3% area on each tail of the distribution. By finding these z-scores and converting them back to the original scores using the mean and standard deviation, we can determine the minimum and maximum values of the middle 87.4% of the scores.

d) To determine the number of students who took the exam based on the information about the number of students who scored above 1432, we need to calculate the area under the normal distribution curve to the right of the score 1432.

By using the same method as in question a), we can find the corresponding z-score for 1432 and determine the proportion or percentage of the distribution to the right of that z-score. We can then calculate the number of students by multiplying this proportion by the total number of students.

By utilizing the properties of the normal distribution and performing the necessary calculations using z-scores and area calculations, we can answer the given questions and provide a graphical representation of the distribution to aid in understanding the solutions.

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The true statements are:

- A and B are independent, then A and B are also independent.

- If the probability of event A is influenced by the occurrence of event B, then the two events are dependent.

- If the event A equals the event ∅, then the probability of the complement of A is 1.

A. "A and B are independent, then A and B are also independent."

This statement is true.

If A and B are independent events, it means that the occurrence of A does not affect the probability of B, and vice versa. In this case, if A and B are independent, then A and B are also independent.

B. "Event A and its complement [tex]A^c[/tex] are mutually exclusive events."

This statement is false.

Mutually exclusive events are events that cannot occur simultaneously.

C. "A and [tex]A^c[/tex] are independent events."

This statement is false. A and [tex]A^c[/tex] are complements of each other, meaning if one event occurs, the other cannot occur. Therefore, they are dependent events.

D. "Event A equals the event ∅, then the probability of the complement of A is 1."

This statement is true.

If A is an empty set (∅), it means that A does not occur. The complement of A, denoted as [tex]A^c[/tex], represents the event that A does not occur.

E. "If the probability of event A is influenced by the occurrence of event B, then the two events are dependent."

This statement is true. If the probability of event A is influenced by the occurrence of event B, it suggests that the two events are not independent.

The occurrence of event B affects the likelihood of event A, indicating a dependency between the two events.

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The question attached here is incomplete, the complete question is:

Which of the following statements are TRUE?

There may be more than one correct answer, please select that are

A and B are independent, then [tex]A^c[/tex] and B are also independent

Event A and its complement [tex]A ^ c[/tex] are mutually exclusive event.

A and [tex]A^c[/tex] 1 independent event

If event A equals event B, then the probability of their intersection is 1.

Given that the graph is z = V-103- 2eIm(-)V_I), S is below the real axis. Therefore, the correct option is (I).

We are to determine what is true about the graph S which is non-empty. The choices to choose from are:(I) S is below the real axis(II) S is a circle. Let's re-arrange the given expression;

z = V-103- 2eIm(-)V_I)...... Equation (1)Let V = a + ib Where a is the real part of V, and b is the imaginary part of V, then substituting in Equation (1) yields z = sqrt(a² + b²) - 103 - 2e^(-b)cos(a) + i2e^(-b)sin(a)...... Equation (2)Equation (2) is in the form z = f(a, b), which is a function of two variables.

Therefore, the graph S is a surface in the three-dimensional coordinate system of a, b, and z. In general, for any function f(x, y) of two variables x and y, there are several ways to represent the graph of f. For instance, we can use a contour plot or a three-dimensional surface plot.

However, it is not easy to determine the exact shape of the surface S from Equation (2) without plotting it. However, there is one thing we can tell about the graph of Equation (2) based on the given expression for z. Since z is the difference between the magnitude of V and a constant (103 - 2e^(-b)cos(a)), we can see that z is always non-negative. That is, z >= 0. Geometrically, this means that the graph S lies above or on the real axis of the three-dimensional coordinate system of a, b, and z. Therefore, the correct option is (I) only: S is below the real axis. Option (II) is not true in general, since the graph S can have various shapes, not just circles.

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It will take 13.5 years . The appropriate formula needed to solve the application problem of determining the length of time needed for the population of 100 frogs to triple with an annual growth rate parameter of 8%, compounding continuously is A = Pe^rt.

Step by step answer:

Given, P = 100 (initial population) The annual growth rate parameter is 8%, compounding continuously. So, r = 0.08 (annual growth rate)We need to determine the time needed for the population to triple. Let's say t years. So, we have to find out when the population (A) becomes three times the initial population (P).i.e. A = 3P

Substitute the given values in the formula: A = Pe^(rt)3P = 100e^(0.08t)

Divide both sides by 100:3 = e^(0.08t)

Take the natural logarithm of both sides: ln3 = ln(e^(0.08t))

Use the property of logarithms that ln(e^(x)) = x:ln3

= 0.08t

Divide both sides by 0.08:t = ln3/0.08t

= 13.5 years

Therefore, it will take 13.5 years for the population of 100 frogs to triple with an annual growth rate parameter of 8%, compounding continuously.

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