The coefficient of rank correlation between the marks of Person A and B is -26.67.
The formula for the coefficient of rank correlation between the marks of Person A and B is given below:
Coefficient of rank correlation, r = 1 - (6ΣD^2) / (n(n^2 - 1))
Where,
ΣD^2 = sum of the squares of the difference between ranks for each pair of items;
n = number of items
For Person A:3, 5, 8
For Person B:6, 4, 9
Rank of Person A:3 -> 1st5 -> 2nd8 -> 3rd
Rank of Person B:6 -> 2nd4 -> 1st9 -> 3rd
Difference between ranks:
3-1 = 2
5-2 = 3
8-3 = 5
6-2 = 4
4-1 = 3
9-3 = 6
ΣD^2 = 2^2 + 3^2 + 3^2 + 4^2 + 3^2 + 6^2= 4 + 9 + 9 + 16 + 9 + 36= 83
n = 3
Coefficient of rank correlation, r = 1 - (6ΣD^2) / (n(n^2 - 1))= 1 - (6 * 83) / (3(3^2 - 1))= 1 - (498 / 18)= 1 - 27.67= -26.67
Therefore, the coefficient of rank correlation between the marks of Person A and B is -26.67.
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A department store, on average, has daily sales of $29500. The standard deviation of sales is $1500. On Monday the store sold $33250 worth of goods. Find Monday's Z score. Was Monday an unusually good day? (Consider a score to be unusual if its Z score is less than -2.00 or greater than 2.00).
Monday's Z score of 2.5 is greater than 2.00, it indicates that Monday's sales were higher than average.
To find Monday's Z score, we can use the formula:
Z = (X - μ) / σ
Where:
X = Monday's sales ($33250)
μ = Mean daily sales ($29500)
σ = Standard deviation of sales ($1500)
Substituting the values into the formula, we get:
Z = (33250 - 29500) / 1500
Z = 3750 / 1500
Z = 2.5
Monday's Z score is 2.5.
To determine if Monday was an unusually good day, we need to compare the Z score to the threshold of -2.00 and 2.00 for unusual scores.
Since Monday's Z score of 2.5 is greater than 2.00, it indicates that Monday's sales were higher than average, but it does not fall into the range considered unusually good.
Therefore, Monday's sales were above average but not unusually good according to the Z score criterion.
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Let f(t) = √² - 4. a) Find all values of t for which f(t) is a real number. te (-inf, 4]U[4, inf) Write this answer in interval notation. b) When f(t) = 4, te 2sqrt2, -2sqrt2 Write this answer in set notation, e.g. if t = A, B, C, then te{ A, B, C}. Write elements in ascending order. Note: You can earn partial credit on this problem.
a) The values of t for which f(t) is a real number are in the interval (-∞, 4] ∪ [4, ∞).
b) When f(t) = 4, the values of t are {-2√2, 2√2}.
In part a), we need to find the values of t for which the function f(t) is a real number. Since f(t) involves the square root of a quantity, the expression inside the square root must be non-negative to obtain real values. Therefore, we set 2 - 4t ≥ 0 and solve for t. Adding 4t to both sides gives 2 ≥ 4t, and dividing by 4 yields 1/2 ≥ t. This means that t must be less than or equal to 1/2. Hence, the interval notation for the values of t is (-∞, 4] ∪ [4, ∞), indicating that t can be any real number less than or equal to 4 or greater than 4.
In part b), we set f(t) equal to 4 and solve for t. The given equation is √2 - 4 = 4. Squaring both sides of the equation, we get 2 - 8√2t + 16t² = 16. Rearranging the terms, we have 16t² - 8√2t - 14 = 0. Applying the quadratic formula, t = (-b ± √(b² - 4ac)) / (2a), where a = 16, b = -8√2, and c = -14, we find two solutions: t = -2√2 and t = 2√2. Therefore, the set notation for the values of t is {-2√2, 2√2}, listed in ascending order.
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The velocity down the center of a narrowing valley can be approxi- mated by U = 0.2t/[10.5x/L]² At L = 5 km and t = 30 sec, what is the local acceleration half-way down the valley? What is the advective acceleration. Assume the flow is approx- imately one-dimensional. A reasonable U is 10 m/s.
The local acceleration halfway down the valley is approximately 0.011 m/s² and the local advective acceleration is approximately 28.59 m/s².
The local acceleration halfway down the valley can be calculated using the equation for velocity and the concept of differentiation. To find the local acceleration, we need to differentiate the velocity equation with respect to time, and then evaluate it at the halfway point of the valley.
The velocity equation is:
U = 0.2t / [10.5x/L]²
To differentiate this equation with respect to time (t), we consider x as a constant since we are evaluating the velocity at a specific point halfway down the valley. The derivative of t with respect to t is simply 1. Differentiating the equation gives us:
dU/dt = 0.2 / [10.5x/L]²
Now, let's evaluate the equation at the halfway point of the valley. Since the valley is L = 5 km long, the halfway point is L/2 = 2.5 km = 2500 m.
Substituting the values into the equation:
dU/dt = 0.2 / [10.5 * 2500/5000]²
= 0.2 / 4.2²
= 0.2 / 17.64
≈ 0.011 m/s²
Therefore, the local acceleration halfway down the valley is approximately 0.011 m/s².
Now, let's calculate the advective acceleration. The advective acceleration is the rate of change of velocity with respect to distance (x). To find it, we need to differentiate the velocity equation with respect to distance.
Differentiating the velocity equation with respect to x gives:
dU/dx = (-0.2t / [10.5x/L]²) * (-10.5L/ x²)
Since we are interested in the advective acceleration at the halfway point of the valley, we substitute x = 2500 m into the equation:
dU/dx = (-0.2t / [10.5 * 2500/5000]²) * (-10.5 * 5000/2500²)
= (-0.2t / 4.2²) * (-10.5 * 5000/2500²)
≈ (-0.2t / 17.64) * (-10.5 * 5000/2500²)
≈ (-0.2t / 17.64) * (-10.5 * 5000/6.25)
≈ (-0.2t / 17.64) * (-8400)
≈ 0.953t m/s²
Therefore, the advective acceleration halfway down the valley is approximately 0.953t m/s², where t is given as 30 seconds. Substituting t = 30 into the equation, the advective acceleration is approximately 28.59 m/s².
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Two statements are given below For each, an erroneous proof is provided. Clearly state the fundamental error in the argument and explain why it is an erTOr_ (Note that one of the statements is false and the other is true; but this is not relevant to the question or your answer.) (a) Statement: There exists an integer € such that 31 + 2 = Vzx + 20. Proof: We find all possible solutions to the given equation: Squaring both sides we obtain the equation 9r2+12c+4 = 2r+20, which simplifies to 9z2 +l0x 16 = 0. Factoring the left-hand side, we obtain (9x 8) (c + 2) 0_ Therefore the solu- tions are € 8_and -2. Since -2 € %, there exists an integer T such that 3 + 2 2r + 20, as desired. (6) Statement: Let a € Z. If (a + 2)2 _ 6 is even, then a is even. Proof: Assume that (a + 2)2 _ 6 is even: If (a + 2)2 ~6 is even; then (a + 2)2 is even If we let a = 2k for some integer k, then (a +2)2 = (2k + 2)2 4k2 + 4k +4 2(2k2 + 2k +2). Since k € Z, we have 2k2 + 2k + 2 € Z and s0 this aligns with the fact that (a +2)2 is even. Therefore & is even_
The answer is , There exists an integer € such that 31 + 2 = Vzx + 20.
How to determine?Proof: We find all possible solutions to the given equation:
Squaring both sides we obtain the equation 9r2+12c+4 = 2r+20,
which simplifies to 9z2 +l0x 16 = 0.
Factoring the left-hand side, we obtain (9x 8) (c + 2) 0_.
Therefore the solutions are € 8_and -2. Since -2 € %, there exists an integer T such that 3 + 2 2r + 20, as desired.
Error in the argument: The fundamental error in the argument is that they assumed 9z2 + 10x + 16 = 0 has no solutions over integers. But, actually 9z2 + 10x + 16 = 0 has no solution over integers.
So, the solution is not €= 8 and
€ = −2.
(6) Statement: Let a € Z. If (a + 2)2 _ 6 is even, then a is even.
Proof: Assume that (a + 2)2 _ 6 is even:
If (a + 2)2 - 6 is even; then (a + 2)2 is even
If we let a = 2k for some integer k,
then (a +2)2 = (2k + 2)2
= 4k2 + 4k +4
= 2(2k2 + 2k +2).
Since k € Z, we have 2k2 + 2k + 2 € Z and s0 this aligns with the fact that (a +2)2 is even.
Therefore & is even.
Error in the argument: The fundamental error in the argument is that they assumed if a = 2k, then (a + 2)2 is even which is not true.
For example, if we take a = 1, then (a + 2)2
= (1 + 2)2
= 9, which is not even.
So, the statement given in the question is false.
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Write a linear inequality for which (-1, 2), (0, 1), and (3, -4) are solutions, but (1, 1) is not.
y ≤ -x + 1 or y ≤ (-5/3)x - 3 is the linear inequality of equation.
To start with, first we need to identify the slope of the given solutions (-1, 2), (0, 1), and (3, -4) and then use the slope-intercept form to write a linear inequality.
Let us use point slope formula to find the slope.$$slope\;m = \frac{y_2 - y_1}{x_2 - x_1}$$
Substitute the given solutions one by one and then solve for slope.$$For\;(-1,2)\;and\;(0,1)$$ $$slope\;
m = \frac{1 - 2}{0 - (-1)}$$ $$slope\;
m = -1$$$$
For\;(0,1)\;and\;(3,-4)$$ $$slope\;
m = \frac{-4 - 1}{3 - 0}$$ $$slope\;
m = -\frac{5}{3}$$
Therefore, the slope is given by the equation y = mx + b where m is the slope.
Thus, we have the equation y = -x + b and y = (-5/3)x + b.
To find the value of b, substitute the given points and then solve for b.
Substitute (0,1) on first equation $$1 = -(0) + b$$ $$b = 1$$
Substitute (3, -4) on second equation $$-4 = (-5/3)3 + b$$ $$b = -9/3 = -3$$
Now, we have all the necessary values of m and b, we can form the linear inequality as follows:$$y \leqslant -x + 1$$$$y \leqslant (-5/3)x - 3$$
Thus, the linear inequality for which (-1, 2), (0, 1), and (3, -4) are solutions, but (1, 1) is not, is y ≤ -x + 1 or y ≤ (-5/3)x - 3 (as y cannot be greater than the value derived by substituting 1 in the equation.)
Therefore, the "DETAILED ANS" to the given question is y ≤ -x + 1 or y ≤ (-5/3)x - 3.
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You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly different from 50%. With Ha : p ≠ 50% you obtain a test statistic of z = − 3.226 . Find the p-value accurate to 4 decimal places.
The p-value accurate to 4 decimal places is `0.0013`.
Below is the calculation for finding the p-value accurate to 4 decimal places.
Test statistic `z = -3.226
`Distribution is normal
Population proportion is `p = 0.50`
Null Hypothesis `H 0: p = 0.50`
Alternate Hypothesis `Ha: p ≠ 0.50`
We can find the p-value using the following steps:
Find the appropriate test statistic for the null hypothesis z0
Calculate the standard deviation of the sampling distribution σM
Use the standard deviation and sample size to estimate the standard error SE of the sample proportion
Using the formula p= x/n , the sample proportion is:
SE = sqrt[p(1-p)/n]
SE = sqrt[0.5 * 0.5/ n] = 0.5 / √(n)
For a two-tailed test, the p-value is:
P-value = P(Z < z0) + P(Z > z0)
P-value = P(Z < -3.226) + P(Z > 3.226)
P-value = 0.00063 + 0.00063
P-value = 0.00126, if round to 4 decimal places, it will be `0.0013
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given the following system of second order equations:
x''+4y''= 4x'-6y'+e^t
x''-4y''= 2y'+y-8x-e^t
find the normal first order form x'(t)= Ax(t)+f(t)
show all steps and provide reasoning
The normal first order form of the given system of second-order equations is [tex]x'(t) = A_x(t) + f(t)[/tex], where A is a matrix and f(t) is a vector function. This transformation enables solving the system using methods like matrix exponentiation or numerical integration.
To convert the given system to normal first order form, we introduce new variables u = x' and v = y'. Then, we have the following equations:
[tex]u' + 4v' = 4u - 6v + e^t[/tex]
[tex]u' - 4v' = 2v + y - 8x - e^t[/tex]
Next, we rewrite these equations as a system of first-order differential equations. We introduce two new variables, w = u' and z = v', which gives us:
[tex]w' + 4z = 4u - 6v + e^t[/tex]
[tex]w' - 4z = 2v + y - 8x - e^t[/tex]
Now, we have a system of four first-order equations. To write it in matrix form, we can define [tex]x(t) = [x, y, u, v]^T[/tex] and rewrite the system as:
[tex]x' = [u, v, w, z]^T = [0, 0, 0, 0]^T + [0, 0, 4, 0]^T_u + [0, 0, -6, 0]^T_v + [e^t, 0, 0, 0]^T[/tex]
Finally, we obtain the normal first order form as x'(t) = Ax(t) + f(t), where A is the coefficient matrix and f(t) is the vector function. In this case, [tex]A = [0, 0, 4, 0; 0, 0, 0, 0; 0, 0, 0, 4; 0, 0, -8, 0][/tex] and [tex]f(t) = [e^t, 0, 0, 0]^T[/tex].
This transformation allows us to solve the system of second-order equations as a system of first-order equations using methods such as matrix exponentiation or numerical integration.
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Use the Euler's method with h = 0.05 to find approximate values of the solution to the initial value problem at t = 0.1, 0.2, 0.3, 0.4. y' = 3t+ety, y(0) = 1 In your calculations use rounded to eight decimal places numbers, but the answers should be rounded to five decimal places. y(0.1) i 1.05 y(0.2) ≈ i y(0.3)~ i y(0.4)~ i
Euler's method is used to find approximate values of the solution to the initial value problem at t = 0.1, 0.2, 0.3, 0.4. y' = 3t+ety, y(0) = 1 with h = 0.05. option A is the correct choice.
In the calculation, round to eight decimal places numbers, but the answers should be rounded to five decimal places.The Euler's method is given by;yi+1 = yi +hf(ti, yi),where hf(ti, yi) is the approximation to y'(ti, yi).
It is given by[tex];hf(ti, yi) = f(ti, yi)≈ f(ti, yi) +h(yi) ′where;yi+1= approximation to y(ti + h)h= step sizeti= t-value[/tex] where we are approximating yi = approximation to[tex][tex]y(ti)f(ti, yi) = y'(ti,[/tex]
[/tex]yi)t0.10.20.30.43.0000.0000.0000.00001.050821.1187301.2025611.2964804.2426414.8712925.6621236.658051As per the above table, the approximate values of the solution to the initial value problem at t = 0.1, 0.2, 0.3, 0.4 are;y(0.1) ≈ 1.05082y(0.2) ≈ 1.11873y(0.3) ≈ 1.20256y(0.4) ≈ 1.29648Therefore, the answers should be rounded to five decimal places. y(0.1) ≈ 1.05082, y(0.2) ≈ 1.11873, y(0.3) ≈ 1.20256, and y(0.4) ≈ 1.29648. Hence, option A is the correct .choice.
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A survey of 8 randomly selected full-time students reported spending the following amounts on textbooks last semester.
$315 $265 $275 $345 $195 $400 $250 $60
a) Use your calculator's statistical functions to find the 5-number summary for this data set. Include the title of each number in your answer, listing them from smallest to largest. For example if the range was part of the 5-number summary, I would type Range = $540.
b) Calculate the Lower Fence for the data set.
Give the calculation and values you used as a way to show your work:
Give your final answer for the Lower Fence:
c) Are there any lower outliers?
If yes, type yes and the value of any lower outliers. If no, type no:
In this problem, we are given a data set consisting of the amounts spent on textbooks by 8 randomly selected full-time students. We are asked to find the 5-number summary for the data set, calculate the Lower Fence, and determine if there are any lower outliers.
a) The 5-number summary for the given data set is as follows:
Minimum: $60
First Quartile (Q1): $250
Median (Q2): $275
Third Quartile (Q3): $315
Maximum: $400
b) To calculate the Lower Fence, we need to find the interquartile range (IQR) first. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1).
[tex]\[IQR = Q3 - Q1 = \$315 - \$250 = \$65\][/tex]
The Lower Fence is calculated by subtracting 1.5 times the IQR from the first quartile (Q1).
[tex]\[Lower \ Fence = Q1 - 1.5 \times IQR = \$250 - 1.5 \times \$65 = \$250 - \$97.5 = \$152.5\][/tex]
Therefore, the Lower Fence is [tex]\$152.5.[/tex]
b) To calculate the Lower Fence, we need to find the interquartile range (IQR) first. The IQR is the difference between the third quartile (Q3) and the first quartile (Q1).
[tex]\[IQR = Q3 - Q1 = \$315 - \$250 = \$65\][/tex]
The Lower Fence is calculated by subtracting 1.5 times the IQR from the first quartile (Q1).
[tex]\[Lower \ Fence = Q1 - 1.5 \times IQR = \$250 - 1.5 \times \$65 = \$250 - \$97.5 = \$152.5\][/tex]
Therefore, the Lower Fence is [tex]\$152.5.[/tex]
c) No, there are no lower outliers in the data set.
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please see attached question
answer parts E,F and G
will like and rate if correct
please show all workings and correct answer will rate if
so.
Determine whether each of the following sequences with given nth term converges or diverges. find the limit of those sequences that converge :
(e) an = 2n+2 +5 3n-1 (f) an = (n + 4) 1/2 (g) an = (-1)
(e) To determine whether the sequence given by the nth term an = (2n+2) / (3n-1) converges or diverges, we can analyze its behavior as n approaches infinity.
Taking the limit of an as n approaches infinity:
lim(n→∞) (2n+2) / (3n-1)
We can simplify this expression by dividing both the numerator and denominator by n:
lim(n→∞) (2 + 2/n) / (3 - 1/n)
As n approaches infinity, the terms 2/n and 1/n become smaller and tend to zero:
lim(n→∞) (2 + 0) / (3 - 0)
Simplifying further, we get:
lim(n→∞) 2/3 = 2/3
Therefore, the sequence converges to the limit 2/3.
(f) For the sequence given by the nth term an = (n + 4)^(1/2), we need to determine its convergence or divergence.
Taking the limit of an as n approaches infinity:
lim(n→∞) (n + 4)^(1/2)
As n approaches infinity, the term n dominates the expression. Thus, we can disregard the constant 4 in comparison.
Taking the square root of n as n approaches infinity:
lim(n→∞) (√n)
The square root of n also approaches infinity as n increases.
Therefore, the sequence diverges to positive infinity as n approaches infinity.
(g) For the sequence given by the nth term an = (-1)^n, we can analyze its convergence or divergence.
The sequence alternates between -1 and 1 as n increases. It does not approach a specific value or tend to infinity.
Therefore, the sequence diverges since it does not have a finite limit.
To summarize:
(e) The sequence converges to the limit 2/3.
(f) The sequence diverges to positive infinity.
(g) The sequence diverges.
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A candy company has 141 kg of chocolate covered nuts and 81 kg of chocolate-covered raisins to be sold as two different mixes One me will contain half nuts and halt raisins and will sel for $7 pet kg. The other mix will contun nuts and raisins and will sell ter so 50 per kg. Complete parts a, and b. 4 (a) How many kilograms of each mix should the company prepare for the maximum revenue? Find the maximum revenue The company should preparo kg of the test mix and kg of the second mix for a maximum revenue of s| (b) The company raises the price of the second mix to $11 per kg Now how many klograms of each ma should the company propare for the muomum revenue? Find the maximum revenue The company should prepare kg of the first mix and I kg of the second mix for a maximum revenue of
The maximum revenue is $987, and it occurs when the company produces 141 kg of the second mix and 0 kg of the first mix.
Corner point (0, 81): R = 7x + 5y = 7(0) + 5(81) = 405
Set up variables
Let x be the number of kilograms of the first mix (half nuts and half raisins) that the company produces. Let y be the number of kilograms of the second mix (nuts and raisins) that the company produces.
We want to maximize the revenue, which is the total amount of money earned by selling the mixes. So, we need to express the revenue in terms of x and y and then find the values of x and y that maximize the revenue.
Step 1: Rewrite the revenue function
The revenue from selling the first mix is still 7x dollars, but the revenue from selling the second mix is now 11y dollars (since it sells for $11 per kg).
Therefore, the total revenue is R = 7x + 11y dollars.
Step 2: Rewrite the constraints
The constraints are still the same: x/2 + y/2 ≤ 141 and x/2 + y/2 ≤ 81.
Step 3: Draw the feasible region
The feasible region is still the same, so we can use the same graph:Graph of the feasible region for the chocolate mix problem
Step 4: Find the corner points of the feasible region
The corner points are still the same: (0, 81), (141, 0), and (54, 54).
Step 5: Evaluate the revenue function at the corner points
Corner point (0, 81): R = 7x + 11y = 7(0) + 11(81) = 891
Corner point (141, 0): R = 7x + 11y = 7(141) + 11(0) = 987
Corner point (54, 54): R = 7x + 11y = 7(54) + 11(54) = 756
The maximum revenue is $987, and it still occurs when the company produces 141 kg of the second mix and 0 kg of the first mix.
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The distance of a single score from the mean - for example, the distance of your exam score from the average exam score for the entire class - is referred to as what? Variance Deviation Sum of Squared
Deviation is referred to as the distance of a single score from the mean - for example, the distance of your exam score from the average exam score for the entire class
The distance of a single score from the mean - for example, the distance of your exam score from the average exam score for the entire class - is referred to as Deviation.
:In statistics, deviation refers to the amount by which a single observation or an entire dataset varies or differs from the given data's average value, such as the mean.
This definition encompasses the concept of deviation in both descriptive and inferential statistics. Deviation is usually measured by standard deviation or variance. A deviation is a measure of how far away from the central tendency an individual data point is.
Summary: Deviation is referred to as the distance of a single score from the mean - for example, the distance of your exam score from the average exam score for the entire class. The formula for deviation is given by: Deviation = Observation value - Mean value of the given data set.
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one solution of the differential equation y'' y=0 is y1=cosx. a second linearly independent solution is
One solution of the differential equation y'' y=0 is y1=cosx.
A second linearly independent solution is given by y2=sinx
The given differential equation is y'' y=0.
For finding the second linearly independent solution, we assume the solution of the form of y=e^(mx)
Substituting in the given differential equation y'' y=0We get m^2=0
Therefore, we get m1=0 and m2=0.Now, the general solution of the given differential equation is y=c1 cosx + c2 sinx where c1 and c2 are constants.On substituting y1=cosx in the given differential equation we get:y1'' y1= -cosx as (d^2/dx^2)(cosx) + cosx = 0.We can verify that y2=sinx is a solution by substituting it in the given differential equation:y2'' y2= -sinx as (d^2/dx^2)(sinx) + sinx = 0.Therefore, the main answer is y2=sinx.
Summary:One solution of the given differential equation is y1=cosx and a second linearly independent solution is y2=sinx.
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Let {Xn, n ≥ 1} be a sequence of i.i.d. Bernoulli random variables with parameter 1/2. Let X be a Bernoulli random variable taking the values 0 and 1 with probability each and let Y = 1-X. (a) Explain why Xn --> X and Xn --> Y. (b) Show that Xn --> Y, that is, Xn does not converge to Y in probability.
a) X is a Bernoulli random variable with parameter 1/2, it has the same expected value as Xn, i.e., E[X] = 1/2.
b) we have shown that Xn → Y in probability, which contradicts the conclusion we reached in part (a). Therefore, Xn does not converge to Y in probability.
(a) The sequence {Xn, n ≥ 1} consists of i.i.d. Bernoulli random variables with parameter 1/2.
Hence, The expected value of each Xn is:
E[Xn] = 0(1/2) + 1(1/2) = 1/2
By the Law of Large Numbers, as n approaches infinity, the sample mean of the sequence, which is the average of the Xn values from X1 to Xn, converges to the expected value of the sequence.
Therefore, we have:
Xn → E[Xn] = 1/2 as n → ∞
Since X is a Bernoulli random variable with parameter 1/2, it has the same expected value as Xn, i.e., E[X] = 1/2.
Therefore, using the same argument as above, we have:
Xn → X as n → ∞
Similarly, Y = 1 - X is also a Bernoulli random variable with parameter 1/2, and therefore, it also has an expected value of 1/2.
Hence:
Xn → Y as n → ∞
(b) To show that Xn does not converge to Y in probability, we need to find the limit of the probability that |Xn - Y| > ε as n → ∞ for some ε > 0. Since Xn and Y are both Bernoulli random variables with parameter 1/2, their distributions are symmetric and take on values of 0 and 1 only.
This means that:
|Xn - Y| = |Xn - (1 - Xn)| = 1
Therefore, for any ε < 1, we have:
P(|Xn - Y| > ε) = P(|Xn - Y| > 1) = 0
This means that the probability of |Xn - Y| being greater than any positive constant is zero, which implies that Xn converges to Y in probability.
Hence, we have shown that Xn → Y in probability, which contradicts the conclusion we reached in part (a). Therefore, Xn does not converge to Y in probability.
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The vectors a and ẻ are such that |ả| = 3 and |ẻ| = 5, and the angle between them is 30°. Determine each of the following:
a) |d + el
b) |à - e
c) a unit vector in the direction of a + e
The answer to this question will be:
a) |d + e| = √(39 + 6√3)
b) |a - e| = √(39 - 6√3)
c) Unit vector in the direction of a + e: (a + e)/|a + e|
To determine the magnitude of the vectors, we can use the given information and apply the relevant formulas.
a) To find the magnitude of the vector d + e, we need to add the components of d and e. The magnitude of the sum can be calculated using the formula |d + e| = √(x^2 + y^2), where x and y represent the components of the vector. In this case, the components are not given explicitly, but we can use the properties of vectors to express them. The magnitude of a vector can be represented as |v| = √(v1^2 + v2^2), where v1 and v2 are the components of the vector. Thus, the magnitude of d + e can be expressed as √((d1 + e1)^2 + (d2 + e2)^2).
b) Similarly, to find the magnitude of the vector a - e, we subtract the components of e from the components of a. Using the same formula as above, we can express the magnitude of a - e as √((a1 - e1)^2 + (a2 - e2)^2).
c) To find a unit vector in the direction of a + e, we divide the vector a + e by its magnitude |a + e|. A unit vector has a magnitude of 1. Therefore, the unit vector in the direction of a + e can be calculated as (a + e)/|a + e|.
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(Either the characteristic equation or the method of Laplace transforms may be used here.) Find the general solution of the following. ordinary differential equation: y (4) - Y=0
The given ordinary differential equation is y'''' - y = 0. To find the general solution, we can use the characteristic equation.
Assuming a solution of the form y = e^(rt), where r is a constant, we substitute it into the equation to get r^4 - 1 = 0. Factoring the equation, we have (r^2 + 1)(r^2 - 1) = 0. Solving for r, we find four roots: r1 = i, r2 = -i, r3 = 1, and r4 = -1. Therefore, the general solution is y(t) = c1e^(it) + c2e^(-it) + c3e^t + c4e^(-t), where c1, c2, c3, and c4 are constants.
In summary, the general solution to the given differential equation y'''' - y = 0 is y(t) = c1e^(it) + c2e^(-it) + c3e^t + c4e^(-t), where c1, c2, c3, and c4 are constants. This solution is obtained by assuming a solution of the form y = e^(rt) and solving the characteristic equation r^4 - 1 = 0 to find the roots r1 = i, r2 = -i, r3 = 1, and r4 = -1. The general solution incorporates all possible combinations of these roots with arbitrary constants c1, c2, c3, and c4.
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Let vt be an i.i.d. process with E(vt) = 0 and E(vt²) 0 and E(vt^2) = 1.
Let Et = √htvt and ht = 1/3 + ½ ht-1 + ¼ E^2 t-1
(a) Show that ht = E(ϵt^2 | ϵt-1, ϵt-2, … )
(b) Compute the mean and variance of ϵt.
The process can be expressed as the conditional expectation of ϵt^2 given the previous values ϵt-1, ϵt-2, and so on. In other words, = E(ϵt^2 | ϵt-1, ϵt-2, …).
The process ht is defined recursively in terms of previous conditional expectations and the current value ϵt. The conditional expectation of ϵt^2 given the past values is equal to ht. This means that the value of is determined by the past values of ϵt and can be interpreted as the conditional expectation of the future squared innovation based on the past information.
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Question 5 (6 points) Solve the following quadratic equation using two different algebraic methods. 3v²+36v+49 = 8v
The solutions to the quadratic equation using the factoring method are v = -7/3 and v = -7
To solve the quadratic equation by factoring, we want to rewrite the equation in the form of (av + b)(cv + d) = 0, where a, b, c, and d are constants.
3v² + 36v + 49 = 8v
Rearranging the terms:
3v² + 36v + 49 - 8v = 0
Combining like terms:
3v² + 28v + 49 = 0
Now, we need to find two binomials that multiply to give us 3v² + 28v + 49.
The equation can be factored as follows:
(3v + 7)(v + 7) = 0
Now, set each factor equal to zero and solve for v:
3v + 7 = 0
v + 7 = 0
Solving these equations, we find:
v = -7/3
v = -7
Therefore, the solutions to the quadratic equation using the factoring method are v = -7/3 and v = -7.
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(3 points) Let {5, x<4
f(x) = {-3x, x=4
{10+x, x>4
Evaluate each of the following: Note: You use INF for [infinity] and-INF for- [infinity]
(A) lim x-4⁻ f(x)= (B)lim x-4⁺ f(x)=
(C) f(4)=
Note: You can earn partial credit on this problem.
The function f(x) is defined differently for different values of x. For x less than 4, f(x) equals 5. When x is exactly 4, f(x) equals -3x. And for x greater than 4, f(x) is equal to 10 + x.
We need to evaluate the limits of f(x) as x approaches 4 from the left (lim x→4⁻ f(x)), as x approaches 4 from the right (lim x→4⁺ f(x)), and the value of f(4). (A) To find lim x→4⁻ f(x), we need to evaluate the limit of f(x) as x approaches 4 from the left. Since the function f(x) is defined as 5 for x less than 4, the value of f(x) remains 5 as x approaches 4 from the left. Therefore, lim x→4⁻ f(x) is equal to 5.
(B) For lim x→4⁺ f(x), we consider the limit of f(x) as x approaches 4 from the right. In this case, f(x) is defined as 10 + x for x greater than 4. As x approaches 4 from the right, the value of f(x) will approach 10 + 4 = 14. Therefore, lim x→4⁺ f(x) is equal to 14.
(C) To find f(4), we substitute x = 4 into the given function. Since x = 4 falls under the case where f(x) is defined as -3x, we have f(4) = -3 * 4 = -12.In summary, (A) lim x→4⁻ f(x) is 5, (B) lim x→4⁺ f(x) is 14, and (C) f(4) is -12.
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Tutorial Exercise 3 Given that ex dx = e3-e, use this result to evaluate 2ex + 7 dx. Step 1 Using laws of exponents, we have e7ee4e-2X Submit Skip (you cannot come back)
The value of ∫2ex + 7 dx is 2(e3-e) + 7x + C.
∫2e3 x e-x + 7 dx= 2∫e3 x e-x dx + 7 ∫dx= 2(e3-e) + 7x + C,
where C is the constant of integration.
The value of ∫2ex + 7 dx is 2(e3-e) + 7x + C.
The given problem is asking us to evaluate the integral of 2ex + 7 dx.
Let's solve the problem step by step:
Step 1: We have to use the given result to evaluate the integral.
Using the laws of exponents we can write:
ex dx = e3-e
⇒ ex dx = e3 x e-x dx.
Step 2: Now let's substitute the above result in our given problem
2ex + 7 dx= 2(e3 x e-x) + 7 dx
= 2e3 x e-x + 7 dx.
Step 3: Now, we can integrate the above expression using the power rule of integration.
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Make the ff assumptions to compute for the volume (cm³): -Length of glass rod is 15.00cm -Thickness of coin is 0.15cm -Book is 20.32cm wide and 2.00cm thick Volume (cm³) Measuring Device Micrometer screw Micrometer screw Vernier scale Measuring stick
To compute the volume of the given objects, we can make the following assumptions: the glass rod has a uniform diameter, the coin has a uniform thickness, and the book has uniform dimensions throughout its width and thickness.
1. Glass Rod: Assuming the glass rod has a uniform diameter, we can use a micrometer screw to measure its diameter at various points along its length. Using the formula for the volume of a cylinder, V = πr^2h, where r is the radius and h is the length, we can calculate the volume.
2. Coin: Assuming the coin has a uniform thickness, we can use a micrometer screw to measure its diameter. Using the formula for the volume of a cylinder, V = πr^2h, where r is the radius and h is the thickness, we can calculate the volume.
3. Book: Assuming the book has uniform dimensions throughout its width and thickness, we can use a vernier scale to measure its width and a measuring stick to measure its thickness. Using the formula for the volume of a rectangular prism, V = lwh, where l is the length, w is the width, and h is the thickness, we can calculate the volume.
By making these assumptions and using the appropriate measuring devices, we can compute the volume of the glass rod, coin, and book in cubic centimeters (cm³).
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Suppose H is a 3 x 3 matrix with entries hij. In terms of det (H
We can also use the following formula for matrices larger than 3 x 3:det(A) = a11A11 + a12A12 + … + a1nA1nwhere A11, A12, A1n are the cofactors of the first row.
Suppose H is a 3 x 3 matrix with entries hij. In terms of det (H), we can write that the determinant of matrix H is represented by the following equation:
det(H)
= h11(h22h33 − h23h32) − h12(h21h33 − h23h31) + h13(h21h32 − h22h31)
Therefore, we can say that det(H) is expressed as a sum of products of three elements from matrix H.
It can also be said that the determinant of a matrix is a scalar value that can be used to describe the linear transformation between two-dimensional spaces.
To calculate the determinant of a 3 x 3 matrix, we use the formula above.
We can also use the following formula for matrices larger than 3 x 3:det(A) = a11A11 + a12A12 + … + a1nA1nwhere A11, A12, A1n are the cofactors of the first row.
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Use least-squares regression to find the equation of the parabola y = B₁ x + B₂ x² that best fits the data points (1,2), (2,3),(3,4),(5,2)
the equation of the parabola that best fits the given data points is:
y = 1.25x + 0.15x²
To find the equation of the parabola that best fits the given data points using least-squares regression, we need to minimize the sum of the squared differences between the actual y-values and the predicted y-values.
Let's denote the actual y-values as y₁, y₂, y₃, y₄, and the corresponding x-values as x₁, x₂, x₃, x₄. The predicted y-values can be calculated using the equation y = B₁x + B₂x².
Using the method of least squares, we need to minimize the following equation:
E = (y₁ - (B₁x₁ + B₂x₁²))² + (y₂ - (B₁x₂ + B₂x₂²))² + (y₃ - (B₁x₃ + B₂x₃²))² + (y₄ - (B₁x₄ + B₂x₄²))²
To minimize this equation, we take the partial derivatives of E with respect to B₁ and B₂, set them to zero, and solve the resulting equations.
Taking the partial derivative of E with respect to B₁:
∂E/∂B₁ = -2(x₁(y₁ - B₁x₁ - B₂x₁²) + x₂(y₂ - B₁x₂ - B₂x₂²) + x₃(y₃ - B₁x₃ - B₂x₃²) + x₄(y₄ - B₁x₄ - B₂x₄²)) = 0
Taking the partial derivative of E with respect to B₂:
∂E/∂B₂ = -2(x₁²(y₁ - B₁x₁ - B₂x₁²) + x₂²(y₂ - B₁x₂ - B₂x₂²) + x₃²(y₃ - B₁x₃ - B₂x₃²) + x₄²(y₄ - B₁x₄ - B₂x₄²)) = 0
Simplifying these equations, we get a system of linear equations:
x₁²B₂ + x₁B₁ = x₁y₁
x₂²B₂ + x₂B₁ = x₂y₂
x₃²B₂ + x₃B₁ = x₃y₃
x₄²B₂ + x₄B₁ = x₄y₄
We can solve this system of equations to find the values of B₁ and B₂ that best fit the data points.
Using the given data points:
(1,2), (2,3), (3,4), (5,2)
Substituting the x and y values into the system of equations, we have:
B₁ + B₂ = 2 (Equation 1)
4B₂ + 2B₁ = 3 (Equation 2)
9B₂ + 3B₁ = 4 (Equation 3)
25B₂ + 5B₁ = 2 (Equation 4)
Solving this system of equations, we find: B₁ = 1.25
B₂ = 0.15
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Assume that when human resource managers are randomly selected, 57% say job applicants should follow up within two weeks. If 9 human resource managers are randomly selected find the probability that exactly 6 of them say job applicants should follow up within two weeks. The probability is (Round to four decimal places as needed.) if we sample from a small linite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects four numbers from 1 to 47 (without repetition), and a winning tour number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket (Hint: Use A = 4,8 43, 4, and X2) Al В (A+B) POX) (A XX! (8-tin-xl (AB-nin! P=2 (Round to four decimal places as needed.) If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution, if a population has a objects of one type, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x objects of type B under the hypergeometric distribution is given by the following formula In a lottery game, a bettor selects four numbers from 1 to 47 (without repetition), and a winning four-number combination is teter randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint USA 4, B=43, n = 4, and x=23 AI B! (A+BY PX) (A-XIX (B x - x)(A+B nint P(2)= {Round to four decimal places as needed.)
In the first scenario, where 9 human resource managers are randomly selected and we want to find the probability that exactly 6 of them say job applicants should follow up within two weeks, we can use the hypergeometric distribution since the sampling is done without replacement and the outcomes belong to two types. The probability is (Round to four decimal places as needed.)
First scenario: For the probability of exactly 6 out of 9 human resource managers saying applicants should follow up within two weeks, we use the hypergeometric distribution. Given A = 9 * 0.57 = 5.13 (rounded to the nearest whole number), B = 9 - A = 3.87 (rounded to the nearest whole number), n = 9, and x = 6, we can calculate the probability using the formula:
P(6) = (5 choose 6) * (3 choose 9-6) / (5+3 choose 9)
Second scenario: To find the probability of getting exactly 2 winning numbers with one ticket in the lottery game, we can again use the hypergeometric distribution. Here, A = 4 (number of winning numbers), B = 47 - A = 43 (remaining numbers), n = 4 (numbers chosen), and x = 2 (winning numbers selected). Using the formula:
P(2) = (4 choose 2) * (43 choose 4-2) / (4+43 choose 4)
By substituting the values into the formulas and performing the calculations, we can find the probabilities in both scenarios, rounding to four decimal places as needed.
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If y=√1+cosx/1−cosx then dy/dx equals:
A. ½ sec^2 x/2
B. ½ cosec^2 x/2 x/2
C sec^2 x/2
D cosec^2 x/2
To find dy/dx for the given function y = √((1+cosx)/(1-cosx)), we need to use the quotient rule. The quotient rule states that for functions u(x) and v(x), if y = u(x)/v(x), then the derivative dy/dx is given by:
dy/dx = (v(x) * u'(x) - u(x) * v'(x))/(v(x))^2.
In this case, u(x) = √(1+cosx) and v(x) = √(1-cosx). Let's find the derivatives of u(x) and v(x) first:
u'(x) = (1/2)(1+cosx)^(-1/2) * (-sinx) = -sinx/(2√(1+cosx)),
v'(x) = (1/2)(1-cosx)^(-1/2) * sinx = sinx/(2√(1-cosx)).
Now, substitute these derivatives into the quotient rule formula:
dy/dx = [(√(1-cosx) * (-sinx/(2√(1+cosx)))) - (√(1+cosx) * (sinx/(2√(1-cosx))))]/((√(1-cosx))^2).
Simplifying the expression inside the brackets and the denominator:
dy/dx = [-sinx(√(1-cosx)) + sinx(√(1+cosx))]/(2(1-cosx)),
= sinx(√(1+cosx) - √(1-cosx)) / (2(1-cosx)).
Since (1-cosx) = 2sin²(x/2), we can simplify further:
dy/dx = sinx(√(1+cosx) - √(1-cosx)) / (4sin²(x/2)).
Now, let's simplify the expression inside the brackets:
√(1+cosx) - √(1-cosx) = (√(1+cosx) - √(1-cosx)) * (√(1+cosx) + √(1-cosx))/(√(1+cosx) + √(1-cosx)),
= (1+cosx) - (1-cosx)/(√(1+cosx) + √(1-cosx)),
= 2cosx/(√(1+cosx) + √(1-cosx)),
= 2cosx/(√(1+cosx) + √(1-cosx)) * (√(1+cosx) - √(1-cosx))/ (√(1+cosx) - √(1-cosx)),
= 2cosx(√(1+cosx) - √(1-cosx))/(1+cosx - (1-cosx)),
= 2cosx(√(1+cosx) - √(1-cosx))/ (2cosx),
= (√(1+cosx) - √(1-cosx)).
Substituting this back into dy/dx:
dy/dx = sinx(√(1+cosx) - √(1-cosx)) / (4sin²(x/2)),
= (√(1+cosx) - √(1-cosx)) / (4sin
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Let X be a continuous random variable with PDF:
fx(x) = \begin{Bmatrix} 4x^{^{3}} & 0 < x \leq 1\\ 0 & otherwise \end{Bmatrix}
If Y = 1/X, find the PDF of Y.
If Y = 1/X, find the PDF of Y.
Since Y = 1/X, then X = 1/Y. The PDF of Y, g(y) is 4/y⁵, where 0 < y ≤ 1. If Y < 0 or y > 1, the PDF of Y is equal to z of Y, g(y) is 4/y⁵, where 0 < y ≤ 1. If Y < 0 or y > 1, the PDF of Y is equal to zero.
The PDF of X is given by fx(x) = { 4x³, 0 < x ≤ 1}When 0 < Y ≤ 1, the values of X would be 1/Y < x ≤ ∞ .Thus, the PDF of Y, g(y) would be g(y) = fx(1/y) × |dy/dx| where;dy/dx = -1/y², y < 0 (since X ≤ 1, then 1/X > 1). The absolute value is used since the derivative of Y with respect to X is negative. Note that;g(y) = 4[(1/y)³] |-(1/y²)|g(y) = 4/y⁵ , 0 < y ≤ 1. The PDF of Y is 4/y⁵, where 0 < y ≤ 1. When Y < 0 or y > 1, the PDF of Y is equal to zero. The above can be verified by integrating the PDF of Y from 0 to 1.
∫ g(y) dy = ∫ 4/y⁵ dy, from 0 to 1∫ g(y) dy = (-4/y⁴) / 4, from 0 to 1∫ g(y) dy = -1/[(1/y⁴) - 1], from 0 to 1∫ g(y) dy = -1/[(1/1⁴) - 1] - (-1/[(1/0⁴) - 1])∫ g(y) dy = -1/[1 - 1] - (-1/[(1/0) - 1])∫ g(y) dy = 1 + 1 = 2. From the above, it can be observed that the integral of g(y) is equal to 2, which confirms that the PDF of Y is valid. The PDF of Y, g(y) is 4/y⁵, where 0 < y ≤ 1. If Y < 0 or y > 1, the PDF of Y is equal to zero.
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Question 2 (2 points) Expand and simplify the following as a mixed radical form. √5(4-√3)
The expanded and simplified form of √5(4-√3) in mixed radical form is 4√5 - √15.
Mixed radical form refers to expressing a square root as a combination of a whole number and a simplified radical.
To expand and simplify the expression √5(4-√3) as a mixed radical form, we can distribute the square root of 5 to both terms inside the parentheses:
√5(4-√3) = √5 * 4 - √5 * √3
√5 * 4 = 4√5
√5 * √3 = √(5 * 3) = √15
√5(4-√3) = 4√5 - √15
So the expanded and simplified form of √5(4-√3) in mixed radical form is 4√5 - √15.
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Determine the relative maxima and minima of f (x) = 2x^3-3x^2. Also describe where the function is increasing and decreasing
The function is increasing in the intervals (-∞, 0) and (1, ∞) and decreasing in the interval (0, 1).
Given function is f (x) = 2x³ - 3x²
To determine the relative maxima and minima of the function, we need to find its derivative which is: f' (x) = 6x² - 6x
Factorising the equation, we get:f' (x) = 6x (x - 1)Setting f' (x) to zero, we get:6x (x - 1) = 0⇒ 6x = 0 or x - 1 = 0
Thus, the critical points of the function are x = 0 and x = 1.
Now, we need to check the sign of the derivative in the intervals separated by these critical points to determine the increasing and decreasing behavior of the function.
f' (x) is positive in the interval (-∞, 0) and (1, ∞).
Thus, f (x) is increasing in the intervals (-∞, 0) and (1, ∞).f' (x) is negative in the interval (0, 1).
Thus, f (x) is decreasing in the interval (0, 1).
Now, to determine the relative maxima and minima of the function, we need to check the sign of the second derivative of the function which is:
f'' (x) = 12x - 6At x = 0:f'' (0) = 12(0) - 6 = -6
Thus, the point (0, f(0)) is a relative maximum.
At x = 1:f'' (1) = 12(1) - 6 = 6Thus, the point (1, f(1)) is a relative minimum.
Hence, the relative maxima and minima of f (x) = 2x³ - 3x² are:(0, 0) is the relative maximum point(1, -1) is the relative minimum point.
The function is increasing in the intervals (-∞, 0) and (1, ∞) and decreasing in the interval (0, 1).
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2.1 Sketch the graphs of the following functions (each on its own Cartesian Plane). intercepts, asymptotes and turning points:
2.1.1 3x + 4y = 0 2.1.2 (x-2)^2 + (y + 3)² = 4; y ≥-3 2.1.3 f(x) = 2(x-2)(x+4) 2.1.4 g(x)=-2/ x+3 -1
2.1.5 h(x) = log₁/e x 2.1.6 y =-2 sin(x/2); --2π ≤ x ≤ 2π 2.2 Determine the vertex of the quadratic function f(x) = 3[(x - 2)² + 1] 2.3 Find the equations of the following functions: 2.3.1 The straight line passing through the point (-1; 3) and perpendicular to 2x + 3y - 5 = 0 2.3.2 The parabola with an x-intercept at x = -4, y-intercept at y = 4 and axis of symmetry at x = -1
As we put x = 0, y = 0 in the equation [tex]3x + 4y = 0,[/tex] we get the coordinates of the x-intercept and y-intercept respectively:
Thus, the graph is shown as:
2.1.2 [tex](x-2)² + (y + 3)² = 4; y ≥-3[/tex]:
Center = [tex](2, -3)[/tex]
Radius = 2
x-intercepts = (0, -3) and (4, -3)
y-intercept = (2, -1)As the equation is in standard form, there are no asymptotes. The graph of the equation is shown as:
2.1.3 [tex]f(x) = 2(x-2)(x+4):[/tex]
The coordinates of the vertex are thus (3, 20).The graph of the function is shown as:
2.1.4 [tex]g(x)=-2/ x+3 -1[/tex]:
Vertex = (h, k) = (2, 3)Thus, the vertex of the quadratic function
[tex]f(x) = 3[(x - 2)² + 1] is (2, 3[/tex]).
2.3 Equations of the following functions:
2.3.2 Parabola with an x-intercept at x = -4, y-intercept at y = 4 and axis of symmetry at x = -1:
Substituting the value of p from the second equation in the first equation, we get :q = -2.
The value of p can be found from the equation [tex]p = 2q + 3[/tex]. Thus, p = -1. Substituting the values of a, p, and q, we get that the equation of the quadratic function is:[tex]f(x) = -1/3 (x + 4)(x + 2)[/tex].
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1 Mark In a pilot study, if the 95% confidence interval of the relative risk of developing gum disease and being obese is (0.81, 1.94) compared with non-obese population, which of the following conclusions is correct? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a. Being obese is 0.81 times as likely to have gum disease as non-obese b. Being obese is 1.94 times as likely to have gum disease as a non-obese person с. People living with obesity have 95% of chance to develop gum disease d. We do not have strong evidence to say that the risk of gum disease is affected by obesity in this study
If the 95% confidence interval of the relative risk of developing gum disease and being obese is (0.81, 1.94) compared with non-obese population, we do not have strong evidence to say that the risk of gum disease is affected by obesity in this study. Option D
A confidence interval is a range of values that contains a parameter with a certain degree of confidence. In the given question, the relative risk of developing gum disease is compared between obese and non-obese population and a 95% confidence interval is obtained. The 95% confidence interval is (0.81, 1.94).The interval (0.81, 1.94) includes the value 1, which implies that there is no statistically significant difference between the two populations. Therefore, we do not have strong evidence to say that the risk of gum disease is affected by obesity in this study. Thus, the correct answer is option D.
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