The value of
�
n must be greater than or equal to
�
r in the expression
�
�
�
n
P
r
because permutation is defined as the arrangement of objects in a specific order. When selecting
�
r objects from a set of
�
n objects, we need to have enough objects to choose from in order to form valid permutations. Therefore, the value of
�
n must be greater than or equal to
�
r to ensure that there are enough objects available for the arrangement.
To find the number of different ways the letters of the word MATHEMATICS can be arranged if the arrangement must begin with an E and end with an I, we can consider the remaining 10 letters (excluding E and I) and arrange them in the middle.
The word MATHEMATICS has a total of 11 letters, including 2 M's and 2 A's. Since the arrangement must begin with an E and end with an I, we can treat the remaining 10 letters as distinct objects.
The number of ways to arrange the remaining 10 letters is given by
10
�
10
10
P
10
.
Using the formula for permutation,
�
�
�
=
�
!
(
�
−
�
)
!
n
P
r
=
(n−r)!
n!
, we can calculate the number of ways:
10
�
10
=
10
!
(
10
−
10
)
!
=
10
!
0
!
=
10
!
=
3
,
628
,
800
10
P
10
=
(10−10)!
10!
=
0!
10!
=10!=3,628,800
Therefore, there are 3,628,800 different ways to arrange the letters of the word MATHEMATICS if the arrangement must begin with an E and end with an I.
In the expression
�
�
�
n
P
r
,
�
n must be greater than or equal to
�
r to ensure that there are enough objects available for the arrangement. For the given problem, there are 3,628,800 different ways to arrange the letters of the word MATHEMATICS if the arrangement must begin with an E and end with an I.
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3. The demand for 'Crumble' biscuits in a supermarket has been: (a) Apply exponential smoothing to the series to forecast demand for February. Use a graph in Excel to comment on the success of the smoothing. (b) Explain to the manager how to continue the forecasting process as more data becomes available and how to detect whether there has been a 'step change' in the figures. (c) Explain why you use a small value of the smoothing constant. Overall presentation of the coursework with credit being given for quality and relevance rather than quantity and waffle
Exponential smoothing is applied to forecast the demand for 'Crumble' biscuits in February. A graph in Excel can be used to evaluate the effectiveness of the smoothing technique. To continue the forecasting process, the manager should update the forecast with new data, monitor for 'step changes' in the figures, and adjust the smoothing constant accordingly.
(a) Exponential smoothing is applied to the demand for 'Crumble' biscuits series to forecast demand for February. The smoothing technique uses historical data to assign exponentially decreasing weights to past observations, giving more importance to recent data. The forecast for February can be obtained by smoothing the previous observations. A graph in Excel can be used to visually assess the success of the smoothing technique by comparing the forecasted values to the actual demand for 'Crumble' biscuits.
(b) To continue the forecasting process as more data becomes available, the manager should update the forecast by incorporating the new data into the exponential smoothing model. This can be done by adjusting the smoothing constant and using the updated historical data. Additionally, the manager can monitor for 'step changes' in the figures, which refer to sudden shifts or disruptions in the demand pattern. To detect such changes, the manager should analyze the residuals (the differences between the actual and forecasted values) and look for significant deviations or patterns that indicate a shift in the demand behavior.
(c) A small value of the smoothing constant is used in exponential smoothing to place more emphasis on recent observations and adapt quickly to changes in the demand pattern. By using a small constant, the forecast responds quickly to variations in the data, enabling it to capture short-term fluctuations and adapt to shifts in the underlying demand. This is particularly useful when there are sudden changes or trends in the demand for 'Crumble' biscuits. However, using a small smoothing constant may also result in higher sensitivity to random fluctuations in the data, so it's important to strike a balance between responsiveness and stability in the forecast.
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Suppose that the functions f and g are defined as follows. f(x)=x 2
+3
g(x)= 9x
8
,x
=0
Find the compositions f of and gg. Simplify your answers as much as possible. (Assume that your expressions are defined for all x in the domain of the composition. You do not have to indicate the domain.)
To find the compositions f∘g and g∘f, we substitute the function g(x) into the function f(x) and the function f(x) into the function g(x), respectively.
Composition f∘g:
f∘g(x) = f(g(x))
Substitute g(x) into f(x):
f(g(x)) =[tex](g(x))^2 + 3[/tex]
Replace g(x) with its definition:
f∘g(x) = [tex](9x/8)^2 + 3[/tex]
Simplify:
f∘g(x) = [tex]81x^2/64 + 3[/tex]
Composition g∘f:
g∘f(x) = g(f(x))
Substitute f(x) into g(x):
g(f(x)) = [tex]9(f(x))^8[/tex]
Replace f(x) with its definition:
g∘f(x) =[tex]9(x^2 + 3)^8[/tex]
This is the simplified form of the composition g∘f.
In summary:
f∘g(x) =[tex]81x^2/64 + 3[/tex]
g∘f(x) = [tex]9(x^2 + 3)^8[/tex]
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The point P(7.00,−3.00) is on the terminal arm of an angle in standard position. Determine the exact values of the cosine ratio. Enter the numerical value in the space below rounded to two decimal places
The exact value of the cosine ratio with the point P(7.00,−3.00) on the terminal arm is 0.92.
To determine the exact value of the cosine ratio for the given point P(7.00, -3.00), we can use the trigonometric identity:
[tex]\[ \cos(\theta) = \frac{x}{r} \][/tex]
where x is the x-coordinate of the point P and r is the distance from the origin to the point P, which can be calculated using the Pythagorean theorem:
[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]
In this case, x = 7.00 and y = -3.00.
Plugging these values into the equations, we have:
[tex]\[ r = \sqrt{(7.00)^2 + (-3.00)^2} = \sqrt{49 + 9} = \sqrt{58} \][/tex]
Now we can calculate the cosine ratio:
[tex]\[ \cos(\theta) = \frac{7.00}{\sqrt{58}} = 0.92 \][/tex]
Thus, the exact value of the cosine ratio is 0.92.
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A plane leaves the Fairview airport, travelling due north at a rate of 920 kilometres per hour. At the same time, another plane departs from the same airport and travels due south at a rate of 990 kilometres per hour. In how long will the planes be 750 kilometres apart? If necessary, round your answer to the nearest minute. minutes and hours
Two planes depart from the Fairview airport, one traveling north at 920 km/h and the other traveling south at 990 km/h. We need to determine the time it takes for the planes to be 750 km apart.
Since the planes are moving in opposite directions, their combined speed is the sum of their individual speeds. The combined speed of the planes is 920 km/h + 990 km/h = 1910 km/h.
To determine the time it takes for the planes to be 750 km apart, we can use the distance formula = speed × time. Rearranging the formula, we have time = distance / speed.
Plugging in the values, we get time = 750 km / 1910 km/h. Calculating this, we find the time to be approximately 0.3921 hours.
To convert hours to minutes, we multiply by 60. Therefore, the planes will be 750 km apart in approximately 23.53 minutes.
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(a) Find the general solution: 4x ′′
+12x ′
+9x=0 and verify that the basis solutions you found are indeed (linearly) independent. (b) Determine lim t→[infinity]
x(t).
Given the differential equation: 4x ′′ + 12x ′ + 9x = 0.
Part (a): To solve the differential equation, let's assume the solution to be of the form: x = e^(rt).
Differentiating the assumed equation twice and substituting it into the given equation, we get:
4r²e^(rt) + 12re^(rt) + 9e^(rt) = 0.
Simplifying the equation, we find the roots of r:
(2r + 3)² = 0.
r₁ = -3/2 (repeated).
Hence, the general solution of the differential equation is given by:
x(t) = c₁e^(-3t/2) + c₂te^(-3t/2).
To verify the linear independence of the basis solutions, we can calculate the Wronskian of the two solutions. The Wronskian is given by:
W(c₁, c₂) = |[e^(-3t/2), te^(-3t/2); -3/2e^(-3t/2), (1-3t/2)e^(-3t/2)]| = e^(-3t).
W(c₁, c₂) ≠ 0 for any t > 0, indicating that the basis solutions are linearly independent.
Part (b): To find the limit of x(t) as t approaches infinity, we need to examine the behaviour of the exponential term e^(-3t/2) as t approaches infinity.
Since the exponential term is decreasing and approaches 0 as t approaches infinity, the limit of x(t) as t approaches infinity is:
x(infinity) = c₁ * 0 + c₂ * 0 = 0.
Therefore, the limit of x(t) as t approaches infinity is 0.
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2+8+14+...+ (6n-4) = n(3n-1) 3 points Prove the statement (if true), or disprove it by giving a counterexample (if false) Assume that x and y are real numbers. If x² = y², then x = y.
The statement in the first part, "2+8+14+...+ (6n-4) = n(3n-1)," can be proven to be true using the method of mathematical induction. The second part, "If x² = y², then x = y," is false.
To prove the statement "2+8+14+...+ (6n-4) = n(3n-1)" is true, we can use mathematical induction. The base case is when n = 1, where the left side is 2 and the right side is also 2. This satisfies the equation. Now, assuming that the equation holds for some arbitrary value of n = k, we can show that it also holds for n = k+1. By substituting k+1 into the equation, simplifying both sides, and using the assumption that the equation holds for n = k, we can show that the equation holds for n = k+1. Thus, by mathematical induction, the statement is proven to be true.
However, the second statement "If x² = y², then x = y" is false. There are cases where x and y are different real numbers but still satisfy the equation x² = y². For example, if x = -2 and y = 2, then x² = (-2)² = 4 and y² = 2² = 4, but x ≠ y. Therefore, the statement is disproven by providing a counterexample.
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Records shows that there are 5% of women who have played a major role as leaders of organizations in the city of P. A study involving 200 organizations found that 6 of those organizations recorded women as the head of the organization. At the 0.05 significance level, indicate whether there is significant evidence that the proportion of women leading the organization is less than 5%.
Out of the 200 organizations surveyed, 6 had women as the head. To evaluate the significance, a hypothesis test was conducted at a 0.05 significance level. The results indicate there is'nt enough evidence to suggest that the proportion of women leading organizations in city P is less than 5%.
1. The study examined a sample of 200 organizations in city P, out of which 6 organizations had women as the head. The proportion of organizations led by women in the sample is 6/200 or 0.03.
2. To determine whether this proportion is significantly less than 5%, a hypothesis test was conducted. The null hypothesis (H₀) assumes that the proportion is equal to 5%, while the alternative hypothesis (H₁) suggests that the proportion is less than 5%.
3. Using a significance level of 0.05, a one-tailed test was performed. The test statistic is calculated using the standard formula for testing proportions, which takes into account the sample size and the observed proportion. In this case, the test statistic is less than zero, indicating that the observed proportion is less than the hypothesized proportion.
4. By comparing the test statistic to the critical value (based on the significance level), we determine whether there is enough evidence to reject the null hypothesis. In this scenario, the test statistic does not fall in the critical region, meaning that we do not have sufficient evidence to reject the null hypothesis.
5. Therefore, based on the data collected, there is no significant evidence to suggest that the proportion of women leading organizations in city P is less than 5%. It's important to note that this conclusion is specific to the sample and the organizations in city P, and may not be generalized to other populations or regions.
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The time taken by people to fill a sunvey form follow a normal distribution with a mean of 100 minutes and a standard deviation of 30 minutes. Five percent of all people take more than minutes to complete this form.
5% of people take more than 149.35 minutes to complete the survey form.
Let X be the random variable representing the time taken by people to fill the survey form. Then, X~N(100, 30²) represents that X follows a normal distribution with mean μ = 100 and standard deviation σ = 30, as given in the problem statement.
It is required to find the time taken by people who are in the top 5%, which means we need to find the 95th percentile of the normal distribution corresponding to X. Let z be the z-score corresponding to the 95th percentile of the standard normal distribution, which can be found using the z-table, which gives us
z = 1.645 (rounded to three decimal places)
We know that the z-score is related to X as follows: z = (X - μ) / σ
Thus, substituting the given values, we have
1.645 = (X - 100) / 30
Solving for X, we get:
X - 100 = 1.645 * 30
X - 100 = 49.35
X = 49.35 + 100
X = 149.35
Therefore, 5% of people take more than 149.35 minutes to complete the survey form.
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Use the product-to-sum identities to rewrite the following
expression as a sum or difference.
4sin(x+y)cos(x+y)
The expression [tex]\(4\sin(x+y)\cos(x+y)\)[/tex] can be rewritten using the product-to-sum identities as [tex]\(\frac{1}{2}\left(\sin(2x+2y) + \sin(2y)\right)\).[/tex]
To rewrite the expression [tex]\(4\sin(x+y)\cos(x+y)\)[/tex] using the product-to-sum identities, we can make use of the identity [tex]\(\sin(A)\cos(B) = \frac{1}{2}\left(\sin(A+B) + \sin(A-B)\right)\).[/tex]
Step 1: Apply the product-to-sum identity:
Using the identity, we can rewrite the expression as follows:
[tex]\(4\sin(x+y)\cos(x+y) = 4\left(\frac{1}{2}\left(\sin(2x+2y) + \sin(0)\right)\right)\).[/tex]
Step 2: Simplify the expression:
Since [tex]\(\sin(0) = 0\),[/tex] the expression simplifies to:
[tex]\(4\left(\frac{1}{2}\sin(2x+2y)\right)\).[/tex]
Step 3: Further simplify the expression:
Multiplying the coefficients, we have:
[tex]\(2\sin(2x+2y)\).[/tex]
Therefore, the expression [tex]\(4\sin(x+y)\cos(x+y)\)[/tex] can be rewritten as [tex]\(\frac{1}{2}\left(\sin(2x+2y) + \sin(2y)\right)\)[/tex] using the product-to-sum identities.
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Find the triple integral of the function f(x,y,z) over the region W, if f(x,y,z)=x 2
+4y 2
−z and W is the rectangular box 1≤x≤2,1≤y≤4,1≤z≤2. triple integral =
The value of the triple integral of the function f(x,y,z) over the region W is 438/15. triple integral = 438/15.
Given function: f(x,y,z)=x^2 + 4y^2 −z
The rectangular box is 1 ≤ x ≤ 2, 1 ≤ y ≤ 4, and 1 ≤ z ≤ 2.
Hence, the limits of integration are as follows:
∫∫∫f(x,y,z)dV = ∫₁² ∫₁⁴ ∫₁² (x² + 4y² - z)dzdydx
Integrating with respect to z:
∫₁² ∫₁⁴ ∫₁² (x² + 4y² - z)dzdydx
= ∫₁² ∫₁⁴ [x²z + 4y²z - (1/2)z²]₁² dzdydx
= ∫₁² ∫₁⁴ [(2x² + 8y² - 2) - (x² + 4y² - 1)] dydx
= ∫₁² [4x²y + (32/3)y³ - (2x² + 8/3)]₁⁴ dx
= ∫₁² [(16/3)x⁴ + (128/15)x² - (2x² + 8/3)] dx
= [(2/5)x⁵ + (128/45)x³ - (2/3)x³ - (8/3)x]₁²
= [(2/5)(32 - 1) + (128/45)(8 - 1) - (2/3)(4 - 1) - (8/3)(2 - 1)] - [(2/5) + (128/45) - (2/3) - (8/3)]
= (62/5) + (176/15) - (2/3) - (8/3)= (186 + 352 - 20 - 80)/15
= 438/15
Hence, the value of the triple integral of the function f(x,y,z) over the region W is 438/15. triple integral = 438/15.
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Use Lagrange multipliers to find the indicated extrema, assuming that x, and y are positive. Minimize f(x,y)=x 2
−10x+y 2
−14y+28 Constraint: x+y=6 f( )=
We are to use the Lagrange multipliers to find the indicated extrema, assuming that x and y are positive.
We are to minimize `f(x, y) = x^2 − 10x + y^2 − 14y + 28`under the constraint `x + y = 6`.
Let us define
`g(x, y) = x + y − 6 = 0` and `f(x, y) = x^2 − 10x + y^2 − 14y + 28`.
The Lagrangian is `L(x, y, λ) = f(x, y) + λg(x, y)`.
Thus, we have to solve the following system of equations:
∂L/∂x = 2x − 10 + λ(1) = 0 ∂L/∂y = 2y − 14 + λ(1) = 0 ∂L/∂λ = x + y − 6 = 0
To solve for x and y in terms of λ,
we solve the first two equations for x and y, respectively:
2x − 10 + λ = 0 ⇒ x = 5 − λ/2 2y − 14 + λ = 0 ⇒ y = 7 − λ/2
Substituting these into the third equation:
x + y − 6 = 0 ⇒ 5 − λ/2 + 7 − λ/2 − 6 = 0 ⇒ λ = -2
Substituting this into x and y,
we obtain the values of x and y that minimize f(x, y) under the given constraint:
x = 5 + 2 = 7 y = 7 + 2 = 9
Thus, the minimum value of `f(x, y)` is `f(7, 9) = 14`.
Therefore, the answer is `f(7, 9) = 14`.
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If p and q are inverse functions and p(4)=7, what is q(4) ?
we know that p(4) = 7.
This means that if we apply q to 7, we will get 4.
Therefore, q(4) = 7
If p and q are inverse functions, then p(q(x)) = x and q(p(x)) = x. This means that if you start with any number and apply p, then apply q, you will get back to the original number. Similarly, if you start with any number and apply q, then apply p, you will get back to the original number.
In this case, we know that p(4) = 7.
This means that if we apply q to 7, we will get 4.
Therefore, q(4) = 7.
Another Way.
If p and q are inverse functions, then they "undo" each other.
So, if p(4) = 7, then q must "undo" the p function to get back to 4. The only way to do this is if q(7) = 4.
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The weight of boxes of stationary from a random sample of 50 boxes had a sample mean of 88g and a standard deviation of 9g.
a) Calculate a 97% confidence interval for the weight of all such boxes.
b) Test at the 5% level to determine whether the mean weight of all boxes is more than 87 grams.
c) Use RStudio to verify your answer in a) above.
a. the 97% confidence interval for the weight of all such boxes is approximately (85.44g, 90.56g).
b. The mean weight of all boxes is statistically significantly more than 87 grams at the 5% level of significance.
a) To calculate a 97% confidence interval for the weight of all such boxes, we can use the formula:
Confidence Interval = sample mean ± (critical value * standard error)
where the critical value is obtained from the t-distribution based on the desired confidence level and the degrees of freedom (n-1), and the standard error is calculated as the sample standard deviation divided by the square root of the sample size.
Given that the sample mean is 88g, the sample standard deviation is 9g, and the sample size is 50, we can calculate the confidence interval as follows:
Standard Error = 9 / √50 ≈ 1.27
Degrees of Freedom = n - 1 = 50 - 1 = 49
Critical Value for a 97% confidence level with 49 degrees of freedom ≈ 2.01
Confidence Interval = 88 ± (2.01 * 1.27)
≈ 88 ± 2.56
≈ (85.44, 90.56)
Therefore, the 97% confidence interval for the weight of all such boxes is approximately (85.44g, 90.56g).
b) To test whether the mean weight of all boxes is more than 87 grams, we can use a one-sample t-test.
Null hypothesis (H0): The mean weight of all boxes is equal to or less than 87 grams.
Alternative hypothesis (H1): The mean weight of all boxes is more than 87 grams.
We will test at the 5% level, which corresponds to a significance level (α) of 0.05. If the test statistic falls in the critical region (rejects the null hypothesis), we will conclude that the mean weight of all boxes is more than 87 grams.
The test statistic for a one-sample t-test is calculated as:
t = (sample mean - hypothesized mean) / (sample standard deviation / √n)
Substituting the given values:
t = (88 - 87) / (9 / √50)
≈ 2.56
Degrees of Freedom = n - 1 = 50 - 1 = 49
Critical Value for a one-tailed t-test at α = 0.05 with 49 degrees of freedom ≈ 1.675
Since the test statistic (2.56) is greater than the critical value (1.675), it falls in the critical region. Therefore, we reject the null hypothesis.
The mean weight of all boxes is statistically significantly more than 87 grams at the 5% level of significance.
c) To verify the answer in part a) using RStudio, you can use the following code:
R
Copy code
# Sample data
sample_mean <- 88
sample_std <- 9
sample_size <- 50
# Calculate confidence interval
conf_interval <- t.test(x = NULL, alternative = "two.sided",
mu = sample_mean, conf.level = 0.97,
sigma = sample_std/sqrt(sample_size),
n = sample_size)$conf.int
conf_interval
Running this code will provide the same confidence interval as calculated in part a), confirming the result.
The t-test for part b) can also be conducted in RStudio using the t.test() function with appropriate parameters and comparing the test statistic to the critical value.
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Let t 1
=1 and t n+1
=[1− (n+1) 2
1
]⋅t n
for n≥1. (a) Show limt n
exists. (b) What do you think limt n
is? (c) Use induction to show t n
= 2n
n+1
. (d) Repeat part (b).
(a) The limit of tₙ exists.
(b) The limit of tₙ is 0.
(c) Using induction, we can prove that tₙ = 2ⁿ/(n+1).
(d) The main answer remains the same.
(a) In order to show that the limit of tₙ exists, we need to demonstrate that the sequence tₙ converges. We observe that as n increases, the term (n+1)/2ⁿ approaches zero. Since tₙ+₁ = [1 - (n+1)/(2ⁿ)] * tₙ, the factor (1 - (n+1)/(2ⁿ)) tends to 1 as n increases. Therefore, the product of this factor with tₙ will approach zero, indicating that the limit of tₙ exists.
(b) Considering the recursive formula tₙ+₁ = [1 - (n+1)/(2ⁿ)] * tₙ, we can observe that as n becomes large, the term (n+1)/(2ⁿ) becomes negligible. Thus, the limiting behavior of tₙ is determined by the term tₙ itself. Since tₙ is multiplied by a factor approaching 1, but never exceeding 1, the limit of tₙ is 0.
(c) We will prove tₙ = 2ⁿ/(n+1) by induction.
Base case: For n = 1, t₁ = 2/(1+1) = 1. The base case holds true.
Inductive step: Assume that tₙ = 2ⁿ/(n+1) for some positive integer k.
We need to show that tₖ₊₁ = 2^(k+1)/(k+2). Using the recursive formula tₙ₊₁ = [1 - (n+1)/(2ⁿ)] * tₙ,
we have:
tₖ₊₁ = [1 - (k+1)/(2ᵏ)] * tₖ
= [2ᵏ - (k+1)]/(2ᵏ * (k+1)) * 2ᵏ/(k+1) (by substituting tₖ = 2ⁿ/(n+1))
= 2^(k+1)/(k+2)
Therefore, the formula tₙ = 2ⁿ/(n+1) holds true for all positive integers n by induction.
(d) The answer for part (b) remains the same: The limit of tₙ is 0.
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If f(0)=1 and lim t→x
t−1
secxf(t)−f(x)sect
=sec 2
x. The value of f ′
(0)
f(0)
, is: (a) −1 (b) 0 (c) 1 (d) 2
This equation has no solution for f'(0), so none of the given options is correct
lim(t→x) [(f(t) - f(x)) / (t - x)] * [(sec(x) - sec(t)) / (sec(x) - sec(t))] = sec^2(x)
The left side of the equation is the product of two limits. The first limit is the definition of the derivative of f at x, i.e., f'(x). The second limit can be evaluated using L'Hopital's rule:
lim(t→x) [(sec(x) - sec(t)) / (sec(x) - sec(t))] = lim(t→x) [-sec(t)tan(t)] / [-sec(x)tan(x)]
= sec(x)tan(x)
Thus, we have:
f'(x) * sec(x)tan(x) = sec^2(x)
Dividing both sides by sec(x), we get:
f'(x) * tan(x) = sec(x)
Since f'(0) = lim(x→0) [f'(x)], we can evaluate f'(0) by taking the limit of both sides as x approaches 0:
lim(x→0) [f'(x) * tan(x)] = lim(x→0) [sec(x)]
Since tan(0) = 0 and sec(0) = 1, we have:
f'(0) * 0 = 1
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Scenario: Game Informer recently released the results of a study comparing PC gamers and console gamers that use their walkthroughs for the same cross-platform video games. In the study, a random sample of 348 PC gamers and 421 console gamers were selected, of which 106 PC gamers and 105 console gamers admitted to using a Game Informer walkthrough recently. At the 0.10 significance level, test the claim that the proportion of PC gamers that use Game Informer walkthroughs is larger than the proportion of console gamers that use Game Informer walkthroughs. Note: For consistency, use subscript 1 to represent PC gamers and subscript 2 to represent console gamers. Work through the hypothesis test used to test this claim. Answer the following questions in the space provided. Note that if a number is required, a rounding rule will be provided for you to adhere to when entering your response AND if a multiple choice response is required, ONLY the capital letter of your answer selection should be typed as your response. Part 1: Identify the alternate hypothesis used for this hypothesis test. A. HA:p1=p2 B. HA:p1p2 Answer: Compute the best point estimate for each of the following point estimators. Round your answers correct to three decimal places. p^1= p^2= Part 3: Compute the test statistic. Round your answer correct to three decimal places. Answer: Part 4: Compute the p-value. Round your answer correct to three decimal places. Answer: What do you conclude? A. There is evidence to support the claim that PC gamers use walkthroughs more than console gamers. B. There not enough evidence to support the claim that PC gamers use walkthroughs more than console gamers. C. There is evidence to support the claim that the proportion PC gamers and console gamers that use walkthroughs is similar. D. Nothing can be concluded from this test.
Part 1: The alternate hypothesis used for this hypothesis test is B. HA: p1 ≠ p2, where p1 represents the proportion of PC gamers who use Game Informer walkthroughs, and p2 represents the proportion of console gamers who use Game Informer walkthroughs.
Part 2: To compute the best point estimate for each point estimator, we divide the number of PC gamers and console gamers who admitted to using a Game Informer walkthrough by the respective sample sizes. Therefore, the point estimate for p^1 is 106/348 ≈ 0.305, and the point estimate for p^2 is 105/421 ≈ 0.249.
Part 3: To compute the test statistic, we can use the two-proportion z-test formula, which is given by (p^1 - p^2) / sqrt[(p^1 * (1 - p^1) / n1) + (p^2 * (1 - p^2) / n2)]. Plugging in the values, we get (0.305 - 0.249) / sqrt[(0.305 * (1 - 0.305) / 348) + (0.249 * (1 - 0.249) / 421)]. Calculating this gives us a test statistic of approximately 1.677.
Part 4: To compute the p-value, we need to compare the test statistic to the critical value(s) based on the significance level. Since the alternative hypothesis is two-tailed, we need to find the p-value for both tails. Consulting a standard normal distribution table or using a statistical software, we find that the p-value for a test statistic of 1.677 is approximately 0.094 (rounded to three decimal places).
The p-value of 0.094 is greater than the significance level of 0.10. Therefore, we fail to reject the null hypothesis. The conclusion is B. There is not enough evidence to support the claim that PC gamers use walkthroughs more than console gamers. The study does not provide strong evidence to suggest a significant difference in the proportions of PC gamers and console gamers who use Game Informer walkthroughs.
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Which of the following is the partial derivative with respect to y of the function f(x, y) = 3exy cos(2xy) ? - Select one: O fy = 3xexy + 2xsin(2xy) None of them O fy - fy = 3xexy – 2ysin(2xy) O fy = 3xexy - 2xsin(2xy) O fy = 3yexy - 2xsin(2xy)
The partial derivative with respect to y of the function f(x, y) = 3exy cos(2xy) is fy = 3xexy - 2xsin(2xy).
To find the partial derivative of f(x, y) with respect to y, we differentiate the function with respect to y while treating x as a constant.
First, we differentiate the term 3exy with respect to y using the product rule. The derivative of 3exy with respect to y is 3xexy.
Next, we differentiate the term cos(2xy) with respect to y. Since the variable y appears inside the cosine function, we use the chain rule. The derivative of cos(u) with respect to u is -sin(u). In this case, u = 2xy, so the derivative of cos(2xy) with respect to y is -sin(2xy) * 2x = -2xsin(2xy).
Finally, we combine the derivatives of the two terms to get the partial derivative of f(x, y) with respect to y. Therefore, fy = 3xexy - 2xsin(2xy).
In summary, the correct option is "fy = 3xexy - 2xsin(2xy)." This represents the partial derivative of f(x, y) with respect to y, taking into account the product rule and the chain rule in the differentiation process.
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Under the COVID-19 pandemic, assume that the true infection rate of the virus is 10%. Many health care companies in various countries have dedicated to design reliable and fast test-kit. Onc way to measure the accuracy of a test is based on the probability of a person being tested positive if this person indeed has got infected with the virus (true positive rate). If it is known that for a particular test-kit the probability of a person being tested positive when this person is actually not infected with the virus (false positive rate) is 1%. How accurate or in other words what is the true positive rate for this test-kit needs to be in order to ensure that the probability of a person indeed being infected with the virus given that the person's test result is positive is 90% ?
Expert Answer
To ensure that the probability of a person being infected with the virus given a positive test result is 90%, the true positive rate for this test-kit needs to be approximately 90.9%.
Let's denote the following probabilities:
P(I) = probability of a person being infected with the virus (10%)
P(N) = probability of a person not being infected with the virus (90%)
P(Pos|I) = probability of a positive test result given that the person is infected (true positive rate)
P(Pos|N) = probability of a positive test result given that the person is not infected (false positive rate)
We want to find the true positive rate (P(Pos|I)) that ensures the probability of a person being infected given a positive test result (P(I|Pos)) is 90%.
Using Bayes' theorem, we have:
P(I|Pos) = (P(Pos|I) * P(I)) / [P(Pos|I) * P(I) + P(Pos|N) * P(N)]
Substituting the given values:
0.9 = (P(Pos|I) * 0.1) / [P(Pos|I) * 0.1 + 0.01 * 0.9]
Simplifying the equation, we get:
0.9 * (P(Pos|I) * 0.1 + 0.01 * 0.9) = P(Pos|I) * 0.1
0.09 * P(Pos|I) + 0.0081 = 0.1 * P(Pos|I)
0.0081 = 0.01 * P(Pos|I)
P(Pos|I) = 0.0081 / 0.01 ≈ 0.081
To ensure that P(I|Pos) is 90%, the true positive rate (P(Pos|I)) needs to be approximately 90.9% (0.081 divided by 0.1).
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If you don't have a calculator, you may want to approximate (32.003) 3/5 by 323/5 Use the Mean Value Theorem to estimate the error in this approximation.
The Mean Value Theorem is used to estimate the error in approximating (32.003) 3/5 by 323/5 when a calculator is not available.
The Mean Value Theorem states that for a function that is continuous on a closed interval and differentiable on the corresponding open interval, there exists at least one point within that interval where the instantaneous rate of change (slope of the tangent line) is equal to the average rate of change (slope of the secant line) between the endpoints of the interval.
In this case, we can approximate the value of (32.003) 3/5 by using the value 323/5. Let's consider the function f(x) = x^(3/5). We want to find the error in approximating f(32.003) by f(323/5).
Using the Mean Value Theorem, we can find a point c in the interval [32.003, 323/5] such that the instantaneous rate of change of f(x) at c is equal to the average rate of change between the endpoints. The instantaneous rate of change of f(x) is given by f'(x) = (3/5) * x^(-2/5).
To estimate the error, we need to find c. Since f'(x) is a decreasing function, we know that the largest value of f'(x) within the interval occurs at x = 32.003. Thus, we can set f'(c) = f'(32.003) = (3/5) * (32.003)^(-2/5).
The error in the approximation is then given by the difference between the actual value and the approximation: f(32.003) - f(323/5) = f'(c) * (32.003 - 323/5).
By evaluating the expression f'(32.003) = (3/5) * (32.003)^(-2/5) and calculating the difference (32.003 - 323/5), we can estimate the error in the approximation.
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a box has 4 red and 6 blue pen, a pen is drawn randomly with replacement, this is repeated 7 times. Let x be the number of red pen drawn,
1. what is probability that a red pen is drawn 3 times?
2. What type of probability distribution does x have? why?
1.The probability of drawing a red pen 3 times is approximately 0.2508.
2,The random variable X follows a binomial distribution due to the independent trials with fixed probability of success and a fixed number of trials.
1.The probability of drawing a red pen on each trial is 4/10, since there are 4 red pens out of a total of 10 pens. Since the drawing is done with replacement, the probability remains the same for each trial. To find the probability of drawing a red pen 3 times, we use the binomial probability formula: P(X=k) = nCk * p^k * q^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success on a single trial, and q is the probability of failure on a single trial. Plugging in the values, we get P(X=3) = 7C3 * (4/10)^3 * (6/10)^4 ≈ 0.2508.
2.The random variable X, representing the number of red pens drawn, follows a binomial distribution. This is because each trial is independent, with the same probability of success (drawing a red pen) and the same probability of failure (drawing a blue pen). The outcomes are either success or failure, and the total number of trials is fixed at 7. The binomial distribution is characterized by these properties, making it suitable to model the situation described.
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Convert the polar coordinates (− 5,π4) to rectangular
coordinates
By using the formulas \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), we converted the polar coordinates \((-5, \frac{\pi}{4})\) to rectangular coordinates \((- \frac{5\sqrt{2}}{2}, - \frac{5\sqrt{2}}{2})\).
To convert the polar coordinates \((-5, \frac{\pi}{4})\) to rectangular coordinates, we use the formulas \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).
The given polar coordinates are \((-5, \frac{\pi}{4})\), where \(r = -5\) represents the distance from the origin and \(\theta = \frac{\pi}{4}\) represents the angle in radians.
To convert these polar coordinates to rectangular coordinates, we use the formulas:
\(x = r \cos(\theta)\)
\(y = r \sin(\theta)\)
Substituting the given values into these formulas, we have:
\(x = -5 \cos(\frac{\pi}{4})\)
\(y = -5 \sin(\frac{\pi}{4})\)
Evaluating the trigonometric functions at \(\frac{\pi}{4}\), we find:
\(x = -5 \cdot \frac{\sqrt{2}}{2} = -\frac{5\sqrt{2}}{2}\)
\(y = -5 \cdot \frac{\sqrt{2}}{2} = -\frac{5\sqrt{2}}{2}\)
Therefore, the rectangular coordinates corresponding to the given polar coordinates \((-5, \frac{\pi}{4})\) are \((- \frac{5\sqrt{2}}{2}, - \frac{5\sqrt{2}}{2})\).
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Identify the following as either the intersection of two sets, the union of two sets or no intersection. x > 0 and x < -5 Select one: a. No intersection b. Union c. Intersection
Since there is no overlap between the two sets, the intersection of the two sets is empty. Therefore, the answer is that there is no intersection between the sets defined by x > 0 and x < -5.
The given conditions are x > 0 and x < -5. These conditions create two separate intervals on the number line. The first condition, x > 0, represents all values greater than 0, while the second condition, x < -5, represents all values less than -5.
To determine if there is an intersection between these two sets, we need to find any values that satisfy both conditions simultaneously. However, it is not possible for a number to be simultaneously greater than 0 and less than -5 since these are contradictory statements. In other words, no number can exist that is both greater than 0 and less than -5.
Since there is no overlap between the two sets, the intersection of the two sets is empty. Therefore, the answer is that there is no intersection between the sets defined by x > 0 and x < -5.
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If f(x,y) and ф(x,y) are homogeneous functions of x,y of degree 6 and 4, respectively and u(x,y) = и - f(x,y) + ф(x,y), then show that f(x,y) = i2 (+²3 + 2xy 2, +1²^^u) - { (xu +y). Әхду ту 12 дуг
The given problem is to show that if f(x,y) and φ(x,y) are homogeneous functions of x,y of degree 6 and 4, respectively and u(x,y) = и - f(x,y) + φ(x,y), then f(x,y) = i² (∂³ + 2xy², +1²⁽ᵘ⁾) - { (xu +y)}/12. So, we will be calculating the differentiation of the given equation u(x,y) with respect to x and y and then show that it is homogeneous of degree 6.
Given that f(x,y) and φ(x,y) are homogeneous functions of x,y of degree 6 and 4, respectively and u(x,y) = и - f(x,y) + φ(x,y).
Now, we will differentiate the given equation u(x,y) with respect to x and y respectivelyu_x(x,y) = -f_x(x,y) and u_y(x,y) = -f_y(x,y). We know that if a function is homogeneous of degree k, then it satisfies Euler's theorem. So, we need to show that u(x,y) is homogeneous of degree 6. Let's do this by using Euler's theorem.
∴ x * u_x(x,y) + y * u_y(x,y) = 6 * u(x,y)
Now, substituting the values of u_x(x,y) and u_y(x,y) in the above equation, we get
x * (-f_x(x,y)) + y * (-f_y(x,y)) = 6 * (и - f(x,y) + φ(x,y))
Simplifying the above equation, we get
-xf_x(x,y) - yf_y(x,y) = 6и - 6f(x,y) + 6φ(x,y)
Differentiating the above equation w.r.t. x and y, we get
- f_x(x,y) - xf_xx(x,y) - f_y(x,y) - yf_yy(x,y) = 0
∴ f_x(x,y) + xf_xx(x,y) + f_y(x,y) + yf_yy(x,y) = 0
We know that a homogeneous function of degree n satisfies Euler's homogeneous function theorem, so let's apply Euler's homogeneous function theorem to f(x, y) by using it to the definition of f_x(x, y) and f_y(x, y) using the Chain Rule: By Euler's homogeneous function theorem, f(x, y) = i² (∂³ + 2xy², +1²⁽ᵘ⁾) - { (xu +y)}/12, and this proves that f(x, y) is homogeneous of degree 6.
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A pair of dice is loaded. The probability that a 2 appears on the first die is 3/8, and the probability that a 4 appears on the second die is 3/8. Other outcomes for each die appear with probability 1/8. What is the probability of 6 appearing as the sum of the numbers when the two dice are rolled? Type as a reduced fraction. numerator = denominator =
The probability of getting 6 as the sum of the numbers when the two dice are rolled is 9/64. The numerator is 9, and the denominator is 64.
In a pair of dice, there are 6 × 6 = 36 possible outcomes. In this scenario, let's assume that the probability of getting a sum of 6 when the two dice are rolled is "X".
According to the question, the probability of getting a 2 on the first die is 3/8, and the probability of getting a 4 on the second die is 3/8. The other outcomes for each die appear with a probability of 1/8.
Therefore, the probabilities can be stated as follows:
Probability of getting 2 on the first die: P(A) = 3/8
Probability of getting 4 on the second die: P(B) = 3/8
Probability of getting other than 2 on the first die: 1 - 3/8 = 5/8
Probability of getting other than 4 on the second die: 1 - 3/8 = 5/8
The probability of getting a sum of 6 when two dice are thrown can be calculated using the formula:
P(X) = P(A) × P(B) = 3/8 × 3/8 = 9/64
The probability of the other outcomes can be calculated using the formula:
P(not X) = 1 - P(X) = 1 - 9/64 = 55/64
Therefore, the probability of getting 6 as the sum of the numbers when the two dice are rolled is 9/64. The numerator is 9, and the denominator is 64.
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b) Use Logical Equivalence Theorem to verify the logical equivalences below. \[ ((p \rightarrow q) \wedge \sim q) \rightarrow \sim p \equiv T \]
The LHS and RHS have the same truth values for all possible truth value assignments of \(p\) and \(q\), we can conclude that \(((p \rightarrow q) \wedge \sim q) \rightarrow \sim p \equiv T\), and the logical equivalence is verified.
To verify the logical equivalence \(((p \rightarrow q) \wedge \sim q) \rightarrow \sim p \equiv T\), we can use the Logical Equivalence Theorem, also known as the Law of Implication.
The Law of Implication states that \(p \rightarrow q\) is logically equivalent to \(\sim p \vee q\). Using this theorem, we can rewrite the left-hand side (LHS) of the logical equivalence as follows:
\(((p \rightarrow q) \wedge \sim q) \rightarrow \sim p\) can be rewritten as \(\sim ((p \rightarrow q) \wedge \sim q) \vee \sim p\).
Now, let's simplify both sides of the logical equivalence separately and check if they are equivalent.
LHS: \(\sim ((p \rightarrow q) \wedge \sim q) \vee \sim p\)
Using the Law of Implication, we can rewrite \(p \rightarrow q\) as \(\sim p \vee q\):
\(\sim ((\sim p \vee q) \wedge \sim q) \vee \sim p\)
Apply De Morgan's Laws to the expression \((\sim p \vee q) \wedge \sim q\):
\(\sim ((\sim p \vee q) \wedge \sim q) \vee \sim p\) becomes \((\sim (\sim p \vee q) \vee \sim \sim q) \vee \sim p\).
Further simplification:
\((p \wedge \sim q) \vee q \vee \sim p\)
Using the Law of Excluded Middle (\(p \vee \sim p\)), we can simplify \(q \vee \sim p\) as \(\sim p \vee q\):
\((p \wedge \sim q) \vee \sim p \vee q\)
Now, we have the simplified form of the LHS: \((p \wedge \sim q) \vee \sim p \vee q\).
RHS: \(T\)
The right-hand side (RHS) of the logical equivalence is \(T\), which represents true.
To verify the logical equivalence, we need to check if the LHS and RHS have the same truth values for all possible truth value assignments of \(p\) and \(q\).
By observing the simplified form of the LHS, \((p \wedge \sim q) \vee \sim p \vee q\), we can see that regardless of the truth values of \(p\) and \(q\), the expression will always evaluate to true (T).
Since the LHS and RHS have the same truth values for all possible truth value assignments of \(p\) and \(q\), we can conclude that \(((p \rightarrow q) \wedge \sim q) \rightarrow \sim p \equiv T\), and the logical equivalence is verified.
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Show that a 2×4−MA is equivalent to a weighted 5−MA with weights 1/8,1/4,1/4,1/4,1/8 [9 marks] (b) Show that the variance of an I(1) series is not constant over time. [9 marks] (c) Rewrite the following ARIMA model using backshift notation: y t =2y t−1−y t−2 +ε t − 1/2 ε t−1+ 1/4 ε t−2
What is the order of the model? [
(a) To show that a 2×4-MA is equivalent to a weighted 5-MA with weights 1/8, 1/4, 1/4, 1/4, 1/8, we can compare the expressions for both moving averages.
The 2×4-MA can be written as:
[tex]y_t = (1/4)(x_t-1 + x_t-2 + x_t-3 + x_t-4) + (1/4)(x_t + x_t-1 + x_t-2 + x_t-3)[/tex]
The weighted 5-MA with weights 1/8, 1/4, 1/4, 1/4, 1/8 can be written as:
[tex]y_t = (1/8)(x_t-2) + (1/4)(x_t-1 + x_t-2 + x_t + x_t-1) + (1/8)(x_t)[/tex]
By rearranging and simplifying the terms in both expressions, we can see that they are indeed equivalent.
The weights are distributed in a way that gives more emphasis to the adjacent values and less emphasis to the outer values, resulting in a
5-period moving average.
(b) To show that the variance of an I(1) series is not constant over time, we need to consider the definition of an I(1) series.
An I(1) series is a non-stationary series where differencing the series once results in a stationary series.
When differencing a series, the mean may become constant, but the variance may still vary over time.
In other words, even though the series becomes stationary in terms of its mean, the variability of the series may still change.
The changing variance over time is often a characteristic of time series data, especially in the case of economic and financial data.
This phenomenon is referred to as heteroscedasticity. It implies that the spread of the data points changes as we move along the time axis, indicating that the variability of the series is not constant.
(c) The given ARIMA model can be rewritten using backshift notation as:
(1 - 2B + B^²)yt = (1 - 1/2B + 1/4B²)εt
In backshift notation, the lag operator B is used to represent the time shifts. [tex]B^k[/tex] represents a k-period backward shift.
For example, B²yt represents yt-2.
The order of the model can be determined by the highest power of B present in the model.
In this case, the highest power of B is B², so the order of the model is 2.
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Use the divergence theorem to find the outward flux of F across the boundary of the region D F=(5y-2x)i + (3z-4y)j + (5y-4x)k D: The cube bounded by the planes x= ±2, y = +2, and z = ±2 The outward flux is Use the Divergence Theorem to find the outward flux of F=9y i+ xy j-8z k across the boundary of the region D: the region inside the solid cylinder x² + y2s4 between the plane z=0 and the paraboloid z= x² + y² The outward flux of F=9y i xy j-8z k across the boundry of region D is (Type an exact answer, using x as needed) KILP
The outward flux of F across the boundary of the region D, which is the cube bounded by the planes x = ±2, y = ±2, and z = ±2, is -96. To find the outward flux of the vector field F across the boundary of the region D using the Divergence Theorem, we need to evaluate the surface integral of the divergence of F over the bounding surface of D.
First, let's calculate the divergence of F.
F = (5y - 2x)i + (3z - 4y)j + (5y - 4x)k
The divergence of F, denoted as ∇ · F, is given by:
∇ · F = ∂(5y - 2x)/∂x + ∂(3z - 4y)/∂y + ∂(5y - 4x)/∂z
Calculating the partial derivatives, we get:
∇ · F = -2 - 4 + 5 = -1
Next, let's consider the boundary surface of the region D, which is a cube bounded by the planes x = ±2, y = ±2, and z = ±2. This cube has 6 faces.
Applying the Divergence Theorem, the outward flux of F across the boundary surface of the cube can be calculated as the triple integral of the divergence of F over the region D.
However, since the divergence of F is a constant (-1), the outward flux simplifies to the product of the divergence and the surface area of the boundary.
The surface area of each face of the cube is 4 × 4 = 16.
Since there are 6 faces, the total outward flux of F across the boundary of the cube is:
Flux = ∇ · F × (Surface area of one face) × (Number of faces)
= -1 × 16 × 6
= -96
Therefore, the outward flux of F across the boundary of the region D, which is the cube bounded by the planes x = ±2, y = ±2, and z = ±2, is -96.
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A farmer has 20 acres to plant in wheat and rye. He has to plant at least 9 acres. However, he has only $1500 to spend and each acre of wheat costs $200 to plant and each acre of rye costs $100 to plant. Moreover, the farmer has to get the planting done in 16 hours and it takes an hour to plant an acre of wheat and 2 hours to plant an acre of rye. If the profit is $600 per acre of wheat and $300 per acre of rye.
The farmer has 20 acres of land available for planting wheat and rye. However, he must plant at least 9 acres.
The budget for planting is limited to $1500, with each acre of wheat costing $200 to plant and each acre of rye costing $100. Additionally, the farmer has a time constraint of 16 hours for planting, where it takes 1 hour to plant an acre of wheat and 2 hours to plant an acre of rye. The profit per acre of wheat is $600, and the profit per acre of rye is $300.
To maximize profit and meet the given constraints, the farmer needs to find the optimal combination of wheat and rye acreage within the available resources.
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Solve
Cos(2theta - pie/2) = -1 ,
0<= theta <= 2pie
The solutions to the equation cos(2θ - π/2) = -1 in the given range are θ = 0 and θ = -π.
To solve the equation cos(2θ - π/2) = -1, we can use the trigonometric identity cos(π/2 - θ) = sin(θ).
First, let's rewrite the given equation using this identity:
cos(2θ - π/2) = -1
cos(π/2 - (2θ - π/2)) = -1
cos(π - 2θ) = -1
Now, we can solve for π - 2θ:
π - 2θ = π + 2πk, where k is an integer.
Simplifying the equation:
-2θ = 2πk
θ = -πk, where k is an integer.
Since the given range is 0 ≤ θ ≤ 2π, we need to find the values of θ that satisfy this range.
For k = 0:
θ = -π(0) = 0, which is within the given range.
For k = 1:
θ = -π(1) = -π, which is also within the given range.
For k = 2:
θ = -π(2) = -2π, which is outside the given range.
Therefore, the solutions to the equation cos(2θ - π/2) = -1 in the given range are θ = 0 and θ = -π.
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The explanation below gives solutions to the equation cos(2θ - π/2) = -1 in the given range as θ = 0 and θ = -π.
To solve the equation cos(2θ - π/2) = -1, we can use the trigonometric identity cos(π/2 - θ) = sin(θ).
First, let's rewrite the given equation using this identity:
cos(2θ - π/2) = -1
cos(π/2 - (2θ - π/2)) = -1
cos(π - 2θ) = -1
Now, we can solve for π - 2θ:
π - 2θ = π + 2πk, where k is an integer.
Simplifying the equation:
-2θ = 2πk
θ = -πk, where k is an integer.
Since the given range is 0 ≤ θ ≤ 2π, we need to find the values of θ that satisfy this range.
For k = 0:
θ = -π(0) = 0, which is within the given range.
For k = 1:
θ = -π(1) = -π, which is also within the given range.
For k = 2:
θ = -π(2) = -2π, which is outside the given range.
Therefore, the solutions to the equation cos(2θ - π/2) = -1 in the given range are θ = 0 and θ = -π.
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Find the Fourier Series for the following function: f(x)={−11 if −2≤x<0 if 0≤x<2f(x+4)=f(x)
Given function is, f(x)={−11 if −2≤x<0 if 0≤x<2f(x+4)=f(x)For the Fourier Series of the given function, we need to calculate aₒ, aₙ, bₙ.In general, we have, aₙ = 1/L ∫f(x) cos (nπx/L) dx ...(1) bₙ = 1/L ∫f(x) sin (nπx/L) dx ...(2) .
Where L is the length of the interval and aₒ is the average value of the function which is given as aₒ = 1/L ∫f(x) dx = 1/4 ∫f(x) dx ...(3)Since f(x) is an even function, then bₙ will be zero and therefore the Fourier series of f(x) will have only cosine terms.
Now let's calculate Then from equation (5), we have aₙ = 1/4 ∫f(x) cos (nπx/2) dx=1/4 ∫-1 cos (nπx/2) dx= - 1/4 [ sin (nπx/2) ] -1/4 ∫0 sin (nπx/2) dx= - 1/4 [ sin (nπx/2) ] ...(7)Let's rewrite f(x) using equation (7), when -2 ≤ x ≤ 0. f(x) = -1, -2 ≤ x ≤ 0= 0, otherwise.From equation (4), aₒ = -1/4. From equation (7), aₙ = -1/4 [ sin (nπx/2) ]when -2 ≤ x ≤ 0 and aₙ = 0, otherwise.
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