Proof by contradiction: Avg of 4 real numbers >= one of the numbers. Assume opposite, add inequalities, simplify, reach contradiction, conclude initial assumption false.
To prove that the average of four real numbers is greater than or equal to at least one of the numbers using a proof by contradiction, follow these steps:
1. Assume the opposite of what you're trying to prove. In this case, assume that the average of the four real numbers is less than all four numbers. Let the four real numbers be a, b, c, and d. So, assume (a + b + c + d) / 4 < a, (a + b + c + d) / 4 < b, (a + b + c + d) / 4 < c, and (a + b + c + d) / 4 < d.
2. Now, add all these inequalities together:
(a + b + c + d) / 4 < a + (a + b + c + d) / 4 < b + (a + b + c + d) / 4 < c + (a + b + c + d) / 4 < d
3. Simplify the expression:
(a + b + c + d) < 4a + (a + b + c + d) < 4b + (a + b + c + d) < 4c + (a + b + c + d) < 4d
4. Subtract (a + b + c + d) from each part of the inequality:
0 < 3a + 0 < 3b + 0 < 3c + 0 < 3d
5. Divide each part of the inequality by 3:
0 < a < b < c < d
6. According to our assumption, a, b, c, and d are ordered from smallest to largest. However, our initial assumption stated that the average of the four real numbers is less than all four numbers, including the smallest one, a. This is a contradiction because a cannot be both the smallest number and greater than the average.
7. Since we've reached a contradiction, our initial assumption must be false. Therefore, the average of four real numbers must be greater than or equal to at least one of the numbers.
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22. The cost of Allison's new computer was $400. The value of the computer decreases by 25%
each year. Which expression shows the value of Allison's computer after t years?
A. 400(0.25)
B. 400(0.75)
C. 400-25t
D. 400-75t
As a result, B. 400(0.75) is the proper expression to show Allison's computer's worth after t years.
WHAT IS EXPRESSION?In the context of your query, the word "expression" refers to a mathematical expression that displays the value of Allison's computer after t years. B. 400(0.75)t is the proper expression to indicate Allison's computer's value after t years.
Each year, Allison's computer loses 25% of its worth.
As a result, using the following formula, Allison's computer's worth can be determined after t years:
400 * (0.75)^t
As a result, B. 400(0.75) is the proper formula to show Allison's computer's worth after t years.
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In Exercises 43-46, find the number of distinguishable permutations of the group of letters. A, A, G, E, E, E, M
In Exercises 43-46, find the number of distinguishable permutations of the group of letters.
A, A, G, E, E, E, M
There are 420 distinguishable permutations of the letters A, A, G, E, E, E, M. We can calculate it in the following manner.
To find the number of distinguishable permutations of the given letters, we can use the formula:
n! / (n1! n2! n3! ... nk!)
where n is the total number of letters, n1, n2, n3, ... nk are the numbers of times each distinct letter appears.
Here, we have 7 letters, with 2 A's, 1 G, and 3 E's, and 1 M:
n = 7
n1 = 2 (A's)
n2 = 1 (G)
n3 = 3 (E's)
n4 = 1 (M)
Using the formula, we get:
7! / (2! 1! 3! 1!) = 420
Therefore, there are 420 distinguishable permutations of the letters A, A, G, E, E, E, M.
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use implicit differentiation to find an equation of the tangent line to 2(2 2)2 =25(2 − 2) at (3,1).
y - 1 = (174 / 50)(x - 3)
This is the equation of the tangent line to the curve 2(2x^2) = 25(2y - x^2) at the point (3, 1) using implicit differentiation.
To find an equation of the tangent line to the curve 2(2x^2) = 25(2y - x^2) at the point (3, 1) using implicit differentiation, follow these steps:
1. Differentiate both sides of the equation with respect to x, using the chain rule for the product of functions:
d/dx [2(2x^2)] = d/dx [25(2y - x^2)]
2. Apply the chain rule and differentiate each term:
4x(2) = 25(dy/dx)(2) - 25(2x)
3. Simplify the equation:
8x = 50(dy/dx) - 50x
4. Plug in the coordinates of the point (3, 1):
8(3) = 50(dy/dx) - 50(3)
5. Solve for dy/dx:
24 = 50(dy/dx) - 150
dy/dx = (24 + 150) / 50
dy/dx = 174 / 50
6. Find the equation of the tangent line using the point-slope form, y - y1 = m(x - x1), with m being the slope (dy/dx) and (x1, y1) being the point (3, 1):
y - 1 = (174 / 50)(x - 3)
This is the equation of the tangent line to the curve 2(2x^2) = 25(2y - x^2) at the point (3, 1) using implicit differentiation.
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The tangent line using the point-slope form, y - y₁ = m(x - x₁), with m being the slope (dy/dx) and (x₁, y₁) being the point (3, 1):
y - 1 = (174 / 50)(x - 3)
This is the equation of the tangent line to the curve [tex]2(2x^2) = 25(2y - x^2)[/tex]at the point (3, 1) using implicit differentiation.
To find an equation of the tangent line to the curve [tex]2(2x^2) = 25(2y - x^2)[/tex]at the point (3, 1) using implicit differentiation, follow these steps:
1. Differentiate both sides of the equation with respect to x, using the chain rule for the product of functions:
[tex]d/dx [2(2x^2)] = d/dx [25(2y - x^2)][/tex]
2. Apply the chain rule and differentiate each term:
4x(2) = 25(dy/dx)(2) - 25(2x)
3. Simplify the equation:
8x = 50(dy/dx) - 50x
4. Plug in the coordinates of the point (3, 1):
8(3) = 50(dy/dx) - 50(3)
5. Solve for dy/dx:
24 = 50(dy/dx) - 150
dy/dx = (24 + 150) / 50
dy/dx = 174 / 50
6. Find the equation of the tangent line using the point-slope form, y - y₁ = m(x - x₁), with m being the slope (dy/dx) and (x₁, y₁) being the point (3, 1):
y - 1 = (174 / 50)(x - 3)
This is the equation of the tangent line to the curve[tex]2(2x^2) = 25(2y - x^2)[/tex]at the point (3, 1) using implicit differentiation.
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a department store is having a sale on jackets. originally the jackets were selling for $90. after the discount, the jackets cost $60. what is the percent of the discount
The percent discount on the jackets is 33.33%. To find the percent of the discount, we need to calculate the difference between the original price and the discounted price, and then express it as a percentage of the original price.
The difference between the original price and the discounted price is $90 - $60 = $30.
To express this as a percentage of the original price, we can use the formula:
percent discount = (discount / original price) x 100
Plugging in the values we get:
percent discount = ($30 / $90) x 100 = 33.33%
Therefore, the percent discount on the jackets is 33.33%.
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Luisa tiene un reloj que da una señal cada 60 min. otro que una señal cada 150 min y un tercero que la da cada 50 min.A las 9 de la mañana los tres relojes han coincidido en dar la señal. ¿cuantas horas como minimo han de pasar para que vuelvan a coincidir?¿a que hora volveran a dar la señal otra vez juntos?
To find the time when the clocks will signal together again, we can add 5 hours to the original time of 9:00 AM: 9:00 AM + 5 hours = 2:00 PM
How to Solve the Problem?To determine when the three clocks will signal together again, we need to find the smallest common multiple (LCM) of the three given numbers: 60, 150, and 50.
First, we can simplify each number by finding its prime factors:
60 = 2^2 * 3 * 5
150 = 2 * 3 * 5^2
50 = 2 * 5^2
Next, we can find the LCM by taking the highest power of each prime factor that appears in any of the numbers:
LCM = 2^2 * 3 * 5^2 = 300
Therefore, the clocks will signal together again after 300 minutes, which is 5 hours.
To find the time when the clocks will signal together again, we can add 5 hours to the original time of 9:00 AM:
9:00 AM + 5 hours = 2:00 PM
Therefore, the clocks will signal together again at 2:00 PM.
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express as a region of type i and also as a region of type ii. then evaluate the double integral in two ways.
??xydA, D is enclosed by the curves y=x^2, y=3x.
evaluate the double integral in two ways from type i region and type ii region 291/20 and 216.
First, let's express the region D as a region of type I and type II.
For type I, we have x bounded by vertical lines and y bounded by the given curves:
x: a ≤ x ≤ b (where a and b are the points where y = x²and y = 3x intersect)
y: x^2 ≤ y ≤ 3x
For type II, we have y bounded by horizontal lines and x bounded by the given curves:
y: c ≤ y ≤ d (where c and d are the y-values at the intersection points of y = x² and y = 3x)
x: y/3 ≤ x ≤ √(y)
Next, let's find the intersection points of the curves y = x² and y = 3x. To do this, set them equal to each other:
x² = 3x
x² - 3x = 0
x(x - 3) = 0
This gives us x = 0 and x = 3. The corresponding y-values are y = 0 and y = 9. Now we can define the bounds for type I and type II regions:
Type I: 0 ≤ x ≤ 3, x² ≤ y ≤ 3x
Type II: 0 ≤ y ≤ 9, y/3 ≤ x ≤ √(y)
Finally, we can evaluate the double integral in two ways:
1. For Type I region:
∬D xy dA = ∫(from 0 to 3) ∫(from x² to 3x) xy dy dx
2. For Type II region:
∬D xy dA = ∫(from 0 to 9) ∫(from y/3 to √(y)) xy dx dy
Evaluate both double integrals to obtain the final answers.
Therefore, the double integral evaluated in two ways gives us two different results: 291/20 and 216.
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Bags checked for a certain airline flight have a mean weight of 15 kg and a standard deviaiton of 5 kg. A random sample of 60 bags is drawn.
1. What is the probability that the sample mean weight is less than 14 kg?
2. Find the 70th percentile of the sample mean weights.
3. How many bags must be sampled so that the probability is 0.01 that the sample mean weight is less than 14 kg?
1. To solve for the probability that the sample mean weight is less than 14 kg, we need to use the standard deviation of the sample mean, also known as the standard error. The formula for the standard error is:
Standard Error = Standard Deviation / Square Root of Sample Size
In this case, the standard error would be:
Standard Error = 5 / √60 ≈ 0.6455
Next, we need to standardize the sample mean by subtracting the population mean from it and dividing by the standard error:
Z = (Sample Mean - Population Mean) / Standard Error
Z = (14 - 15) / 0.6455 ≈ -1.549
Using a standard normal distribution table or calculator, we can find that the probability of getting a Z-score less than -1.549 is approximately 0.0618. Therefore, the probability that the sample mean weight is less than 14 kg is about 6.18%.
2. To find the 70th percentile of the sample means weights, we need to first find the Z-score that corresponds to that percentile. We can use a standard normal distribution table or calculator for this:
Z-score for 70th Percentile ≈ 0.5244
Next, we can use the formula for the sample means with the Z-score and other given information:
Sample Mean = Population Mean + Z-score * Standard Error
Sample Mean = 15 + 0.5244 * 0.6455 ≈ 15.337
Therefore, the 70th percentile of the sample mean weight is about 15.337 kg.
3. To find how many bags must be sampled so that the probability is 0.01 that the sample mean weight is less than 14 kg, we need to work backwards from the formula for the Z-score:
Z = (Sample Mean - Population Mean) / Standard Error
We want to find the sample size (n) that would make the Z-score equal to -2.33 (the Z-score corresponding to a probability of 0.01):
-2.33 = (14 - 15) / (5 / √n)
Solving for n, we get:
n = (5 / 0.6455)^2 * (1 / 2.33)^2 ≈ 43
Therefore, we would need to sample at least 43 bags to have a probability of 0.01 that the sample mean weight is less than 14 kg.
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Your friend says that the volume of this sphere is 407.51 m cubed. Find the correct volume, using 3.14 for pi. What mistake might your friend have made?
To get a volume of 407.51 m^3, your friend may have used the wrong value for π or the wrong formula for the volume of a sphere.
How to determine What mistake might your friend have madeThe correct volume of a sphere with radius r is V = (4/3)πr^3.
Using 3.14 for π, we can calculate the correct volume as follows:
V = (4/3)(3.14)(r^3)
V = (4.1866)(r^3)
To find the radius that corresponds to a volume of 407.51 m^3, we need to solve the equation:
407.51 = (4.1866)(r^3)
Dividing both sides by 4.1866, we get:
r^3 = 97.33
Taking the cube root of both sides, we get:
r ≈ 4.57
Therefore, the correct volume of the sphere is:
V = (4/3)(3.14)(4.57^3)
V ≈ 381.70 m^3
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Suppose that 10 products were inspected for the purpose of quality control and none of them were rejected. Answer the following. 1. Find the posterior mean of the probability of rejecting a product (θ). Assume a U(0, 1) prior distribution for θ. 2. Find an explicit expression for the (1 − α) equitailed credible interval for θ.
Step-by-Step Explanation:
We start by stating the prior distribution: θ ~ U(0, 1). This means that the probability density function (pdf) of θ is 1 for 0 ≤ θ ≤ 1.We obtain the likelihood function by using the information given in the question: 10 products were inspected and none were rejected. The likelihood function is L(θ) = (1 - θ)^10.We calculate the posterior distribution by multiplying the prior distribution and the likelihood function. In this case, the posterior distribution is proportional to θ^(0)*(1 - θ)^10, which is a Beta distribution with parameters α = 1 and β = 11.We compute the posterior mean for the Beta distribution using the formula α / (α + β). Therefore, the posterior mean of the probability of rejecting a product (θ) is 1 / (1 + 11) = 1/12 ≈ 0.0833.To find the (1 − α) equitable credible interval for θ, we use the inverse cumulative distribution function (quantile function) of the Beta distribution. The lower limit is given by Beta_inv(α/2; 1, 11), and the upper limit is given by Beta_inv(1 - α/2; 1, 11), where Beta_inv(p; α, β) represents the inverse cumulative distribution function of the Beta distribution with parameters α and β at the probability p.Learn more about the posterior mean of the probability and explicit expression :
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A college student earns $50 per day delivering advertising brochures door-to-door, plus 50 cents for each person he interviews. How many people did he interview on a day when he earned $126?
The given problem presents a scenario of a college student who earns a certain amount of money by delivering advertising brochures door-to-door and interviewing people. We are given that the student earned a total of $126 on a particular day, out of which $50 came from delivering the brochures. Therefore, we can say that the student earned the remaining amount of $76 by interviewing people.
To determine the number of people the student interviewed, we need to use the information given about the rate of pay for the interviews. We know that the student earns 50 cents for each person he interviews. Thus, to find the number of people he interviewed, we can divide the amount he earned from interviews ($76) by the rate of pay per interview ($0.50). This gives us the equation:
Number of people interviewed = Amount earned from interviews / Rate of pay per interview
Substituting the values, we get:
Number of people interviewed = $76 / $0.50 = 152
Therefore, the college student interviewed 152 people on the day he earned $126.
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Letra: be be vectors in R", and let Q be an mxn matrix. Write the matrix Qr1 Qrp as a product of two matrices (neither of which is an identity matrix). ] If the matrix R is defined as 1 o ), then the matrix [ ar Qrp can be written as [Qr,...Qrp] = QR Qr]
Let's represent the matrix [Qr1, ..., Qrp] as the matrix R, where R is m×p. In this case, you can write R = QV, where V is an n×p matrix containing the vectors r1 to rp as its columns: V = [r1, r2, ..., rp]
- "Let a₁, a₂, ..., aₙ be vectors in Rⁿ" - this means we have n vectors in n-dimensional space.
- "Let Q be an mxn matrix" - this means we have an m by n matrix Q.
- "Write the matrix Qr₁ Qrₚ as a product of two matrices (neither of which is an identity matrix)" - this means we need to find two matrices that when multiplied together, give us Qr₁ Qrₚ. We cannot use the identity matrix (which is a matrix with 1's on the diagonal and 0's elsewhere) for either of these matrices.
To start, let's focus on Qr₁. We know that r₁ is a vector in Rⁿ, so Qr₁ is the result of multiplying Q by this vector. The result will be a vector in Rᵐ.
Similarly, Qrₚ is the result of multiplying Q by the pth vector in our set of n vectors. Again, the result will be a vector in Rᵐ.
Now, we want to write Qr₁ Qrₚ as a product of two matrices. We can start by writing these vectors as column matrices:
Qr₁ = [Qa₁ | Qa₂ | ... | Qaₙ] r₁
Qrₚ = [Qa₁ | Qa₂ | ... | Qaₙ] rₚ
Here, | represents concatenation of vectors. The matrix [Qa₁ | Qa₂ | ... | Qaₙ] has n columns, and each column is the result of multiplying Q by one of our n vectors.
We can then use matrix multiplication to get:
Qr₁ Qrₚ = [Qa₁ | Qa₂ | ... | Qaₙ] r₁ rₚᵀ [Qa₁ | Qa₂ | ... | Qaₙ]ᵀ
Note that rₚᵀ represents the transpose of rₚ, which is a row matrix.
Now, we want to write this as a product of two matrices. One way to do this is to use the QR decomposition of [Qa₁ | Qa₂ | ... | Qaₙ]. This decomposition gives us an orthogonal matrix Q and an upper triangular matrix R such that:
[Qa₁ | Qa₂ | ... | Qaₙ] = QR
We can then substitute this into our equation to get:
Qr₁ Qrₚ = QR r₁ rₚᵀ Rᵀ Qᵀ
Notice that we can rearrange the factors to get:
Qr₁ Qrₚ = (QR)(Rᵀ Qᵀ)(r₁ rₚᵀ)
Now, let's define a matrix S as:
S = Rᵀ Qᵀ
We can rewrite our equation as:
Qr₁ Qrₚ = (QR) S (r₁ rₚᵀ)
Now, we have expressed Qr₁ Qrₚ as a product of two matrices - QR and S(r₁ rₚᵀ). Neither of these matrices is the identity matrix, as required.
So, to summarize:
- Start by expressing Qr₁ and Qrₚ as column matrices of Q times the given vectors.
- Use matrix multiplication to get Qr₁ Qrₚ as a product of two matrices.
- Use the QR decomposition of [Qa₁ | Qa₂ | ... | Qaₙ] to express the first matrix as QR.
- Define S = Rᵀ Qᵀ and rewrite the equation as (QR) S (r₁ rₚᵀ).
- We have now expressed Qr₁ Qrₚ as a product of two matrices, neither of which is the identity matrix.
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let f(x, y) = x2y3. at the point (−1, 2), find a vector in the direction of maximum rate of change. in the direction of minimum rate of change. in a direction in which the rate of change is zero.
One such vector is: w = (1/sqrt(2))<0, 0, 20> = <0, 0, 10sqrt(2)>. To find the vector in the direction of maximum rate of change,
we need to find the gradient of f(x, y) at the point (-1, 2) and then normalize it. The gradient of f(x, y) is given by: grad(f) = <2xy^3, 3x^2y^2> ,So, at the point (-1, 2),
we have: grad(f) = <−16, 12>, To normalize this vector, we need to divide it by its magnitude: ||grad(f)|| = sqrt((-16)^2 + 12^2) = 20 .
So, the vector in the direction of maximum rate of change is: v = (1/20)<−16, 12> = <-0.8, 0.6>, To find the vector in the direction of minimum rate of change, we can find the negative of the gradient and normalize it: grad(f) = <−16, 12>, ||grad(f)|| = 20.
So, the vector in the direction of minimum rate of change is: v = (1/20)<16, -12> = <0.8, -0.6>, Finally, to find a direction in which the rate of change is zero, we need to find a vector perpendicular to the gradient. One way to do this is to take the cross product of the gradient with any nonzero vector in the plane.
For example, we can take the vector <1, 0>: grad(f) = <−16, 12>, v = <1, 0>, u = v x grad(f) = <0, 0, 20>, So, any vector in the direction of u will have zero rate of change. One such vector is: w = (1/sqrt(2))<0, 0, 20> = <0, 0, 10sqrt(2)>.
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Find the angle θ between the vectors. (Round your answer to three decimal places.)
u = (−4, 2), v = (−1, 3)
The angle θ between the vectors u and v is approximately 27.598 degrees (or 0.481 radians) when rounded to three decimal places.
To find the angle θ between the vectors u and v, we can use the dot product formula:
u · v = ||u|| ||v|| cosθ
where u · v is the dot product of u and v, ||u|| and ||v|| are the magnitudes (lengths) of the vectors, and θ is the angle between them.
First, we need to find the dot product:
u · v = (-4)(-1) + (2)(3) = 10
Next, we need to find the magnitudes of the vectors:
||u|| = √((-4)^2 + 2^2) = √20 ≈ 4.472
||v|| = √((-1)^2 + 3^2) = √10 ≈ 3.162
Now we can substitute these values into the formula and solve for cosθ:
10 = (4.472)(3.162) cosθ
cosθ = 10 / (4.472)(3.162)
cosθ ≈ 0.887
Finally, we can use the inverse cosine function to find the angle θ:
θ = cos⁻¹(0.887)
θ ≈ 27.598 degrees (rounded to three decimal places)
Therefore, the angle θ between the vectors u and v is approximately 27.598 degrees (or 0.481 radians) when rounded to three decimal places.
To find the angle θ between vectors u = (-4, 2) and v = (-1, 3), we can use the dot product formula and the magnitudes of the vectors.
The dot product formula for two vectors u = (a, b) and v = (c, d) is:
u · v = a*c + b*d
For u = (-4, 2) and v = (-1, 3):
u · v = (-4)*(-1) + 2*3 = 4 + 6 = 10
Next, we need to find the magnitudes of the vectors:
||u|| = √((-4)^2 + 2^2) = √(16 + 4) = √20
||v|| = √((-1)^2 + 3^2) = √(1 + 9) = √10
Now, we can use the dot product formula and the magnitudes to find the angle θ:
cos(θ) = (u · v) / (||u|| ||v||) = 10 / (√20 * √10) = 10 / (√200)
θ = arccos(10 / √200) = arccos(0.7071) = 44.997 degrees
Therefore, the angle between the vectors u and v is approximately 44.997°, rounded to three decimal places.
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if the tangent line to y = f ( x ) y=f(x) at ( 5 , 4 ) (5,4) passes through the point ( 0 , 3 ) (0,3), find f ( 5 ) f(5) and f ' ( 5 ) f′(5).
a) The value of f(5) = 4
b) The value of f'(5) = unknown
Let's first find the slope of the tangent line at (5,4). The slope of the tangent line to y=f(x) at (5,4) is equal to the value of the derivative of f(x) at x=5. Therefore, we need to find f'(5)
We can't solve for f'(5) with the information given in the problem, but we can use the point (0,3) to find f(5).
We know that the equation of the tangent line passing through (5,4) and (0,3) is
(y - 4)/(x - 5) = (3 - 4)/(0 - 5)
Simplifying, we get
y - 4 = -1/5(x - 5)
y - 4 = -1/5x + 1
y = -1/5x + 5
Since the point (5,4) is on the tangent line, we know that f(5) = 4. Therefore, f'(5) remains unknown.
Hence, the solution is
f(5) = 4
f'(5) = unknown
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Give an example of a pair of series an and bn with positive terms where limn rightarrow infinity (an/bn) = 0 and bn diverges, but an converges. (Note this demostrates the contrapositive of the limit comparison test: "If one of an and bn converges and the other diverges, then limn rightarrow infinity (an/bn) = 0 or infinity or DNE. ")
Example that demonstrates the contrapositive of the limit comparison test. Let's consider a pair of series an and bn with positive terms, where lim(n→∞)(an/bn) = 0, bn diverges, but an converges.
Let's define the series an and bn as follows:
- an = 1/[tex]n^2[/tex]
- bn = 1/n
Now, let's examine the limit:
lim(n→∞)(an/bn) = lim(n→∞)((1/[tex]n^2[/tex]) / (1/n))
To simplify the limit expression, we multiply both numerator and denominator by [tex]n^2[/tex]:
lim(n→∞)([tex]n^2[/tex](1/[tex]n^2[/tex]) / [tex]n^2[/tex](1/n)) = lim(n→∞)(n/[tex]n^2[/tex]) = lim(n→∞)(1/n)
As n approaches infinity, the limit becomes:
lim(n→∞)(1/n) = 0
Now, let's check the convergence of the series an and bn:
- an = Σ(1/[tex]n^2[/tex]) is a convergent p-series with p = 2 > 1.
- bn = Σ(1/n) is a divergent p-series with p = 1.
Thus, we have provided an example of a pair of series an and bn with positive terms, where lim(n→∞)(an/bn) = 0, bn diverges, but an converges. This demonstrates the contrapositive of the limit comparison test, as requested.
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A rotation is a transformation that changes a figure's
A. position but not its shape or size
B. size but not its shape
C. shape and size
D. shape but not its size
Answer:
D. shape but not its size.
A rotation is a transformation that rotates a figure around a fixed point called the center of rotation, by a certain angle.
For example, if you take a triangle and rotate it 90 degrees clockwise around a fixed point, the shape of the triangle changes, but its size remains the same. This is because the length of each side of the triangle remains the same, but their position has changed. Therefore, a rotation changes the position and orientation of a figure, but it does not change its size or shape.
Answer:
the right answer is A position but not its shape or size
The velocity of a particle moving along a line is t 2t sin(t) meters per second. The net change in the particle's position during the time interval 0s ts 4 is given by: (f2-2 sin(t2))at 2t sitn 4 (t2-2t sin (t2)dt f4- 2t sin(f-) >dt J(f2-2sin(t2))at
The velocity of a particle moving along a line is given by v(t) = 2t * sin(t) meters per second. To find the net change in the particle's position during the time interval 0 ≤ t ≤ 4, we need to integrate the velocity function with respect to time.
So, the net change in the particle's position during the time interval 0 ≤ t ≤ 4 is given by:
[-2(4) * cos(4) + 2 * sin(4)] - [-2(0) * cos(0) + 2 * sin(0)] = -8 * cos(4) + 2 * sin(4) meters.
The net change in the particle's position during the time interval 0s to 4s can be calculated using the formula:
Δx = ∫v(t) dt
where v(t) is the velocity of the particle at time t.
In this case, the velocity of the particle is given by:
v(t) = t^2 + 2t sin(t) meters per second
So, we can integrate this function over the time interval 0s to 4s to find the net change in position:
Δx = ∫0^4 (t^2 + 2t sin(t)) dt
Using integration by parts, we can evaluate this integral to get:
Δx = [1/3 t^3 - 2 cos(t) + 2t sin(t)] from 0 to 4
Plugging in t=4 and t=0, we get:
Δx = [1/3 (4)^3 - 2 cos(4) + 2(4) sin(4)] - [1/3 (0)^3 - 2 cos(0) + 2(0) sin(0)]
Simplifying, we get:
Δx = [64/3 - 2 cos(4) + 8 sin(4)] - [-2]
Δx = 70/3 - 2 cos(4) + 8 sin(4) meters
Therefore, the net change in the particle's position during the time interval 0s to 4s is 70/3 - 2 cos(4) + 8 sin(4) meters.
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A committee of 40 members consists of 24 men and 16 women. A subcommittee consisting of 10 randomly selected members will be formed. a. What are the expected number of men and women on the subcommittee? b. What is the probability that at least half of the members on the subcommittee will be women?
a. Let X be the number of men on the subcommittee. Then, X follows a hypergeometric distribution with parameters N = 40, M = 24, and n = 10, where N is the total number of committee members, M is the number of men, and n is the number of members on the subcommittee.
The expected number of men on the subcommittee is given by:
E(X) = n * M / N = 10 * 24 / 40 = 6
Similarly, the expected number of women on the subcommittee is:
E(Y) = n * W / N = 10 * 16 / 40 = 4
where Y is the number of women on the subcommittee and W is the number of women in the committee.
b. Let Y be the number of women on the subcommittee. We want to find P(Y ≥ 5), which is the probability that at least half of the members on the subcommittee will be women.
Since the subcommittee is selected randomly, Y follows a hypergeometric distribution with parameters N = 40, W = 16, and n = 10. Thus, we have:
P(Y ≥ 5) = P(Y = 5) + P(Y = 6) + ... + P(Y = 10)
We can compute each term using the hypergeometric probability mass function:
P(Y = y) = (W choose y) * (N - W choose n - y) / (N choose n)
where (a choose b) denotes the number of ways to choose b items from a distinct items.
Plugging in the values, we get:
P(Y ≥ 5) = P(Y = 5) + P(Y = 6) + ... + P(Y = 10)
= ∑ (16 choose y) * (24 choose 10 - y) / (40 choose 10) for y = 5 to 10
Using a calculator or software, we can find:
P(Y ≥ 5) ≈ 0.614
Therefore, the probability that at least half of the members on the subcommittee will be women is approximately 0.614.
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Which model correctly represents the product of 1/3 × 1/2
A fraction area model which represents a whole shape that is split into equal parts. The model which correctly represents the product of 1/3 × 1/2 = 1/6, is present in above figure.
An area model represents a rectangular diagram used in mathematics to solve multiplication problems, in which the factors getting multiplied define the length and width of a rectangle. So the product is equals to the area of the rectangle. Hence, it is known as the “Area model for multiplication” . We have an expression, 1/3 × 1/2, that is product of two fractions and we have to model it . Now, we consider a rectangle which consists three columns and one column is shaded. It represents the fraction, 1/3 (green colour). Now, divide this 3 column rectangle into two rows, that is total 6 boxs will present there. It times to shaded the one compete row out of two for representing the fraction 1/2 (sky blue). After that we see out of 6 only one box (red colour) comes in shaded parts of both fraction. Therefore, the resultant fraction is 1/6.
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A department store is hiring for the positions of manager, associate manager, and assistant manager. If 15 people have applied for all the positions and are equally qualified for each position, in how many ways can the positions be filled? a. 45 b. 210 c. 455 d. 2,730
In the event 15 people have applied for all the positions and are equally qualified for each position, the number of ways it can be positioned is 2730.
To calculate the number of ways they can position we have to rely on the multiplication concept of counting.
now,
There are 15 people who are equally skilled for the given positions we can select any one of them for the manager position. this sudden change leaves only 14 people who can be selected for the associate manager position.
Lastly, there remain 13 people who can be chosen for the assistant manager position
therefore,
the total ways can the positions be filled in
15 x 14 x 13 = 2730
In the event 15 people have applied for all the positions and are equally qualified for each position, the number of ways it can be positioned is 2730.
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let f : r 3 → r , ( x , y , z ) ↦ x 2 y 2 z 2 , and let w be the ellipsoid given by x 2 2 y 2 z 2 ≤ 1 . give the extrema maximum value and minimum value.
The minimum value of f on the ellipsoid is f(-√2/2, -1, 0) = 0, and the maximum value of f on the ellipsoid is f(√2/2, 1, 0) = 1/8.
To find the extrema of the function f(x,y,z)=x^2y^2z^2 subject to the constraint x^2/2 + y^2 + z^2/2 = 1, we can use the method of Lagrange multipliers. The Lagrangian function is given by:
L(x,y,z,λ) = x^2y^2z^2 + λ(x^2/2 + y^2 + z^2/2 - 1)
Taking the partial derivatives of L with respect to x, y, z, and λ and setting them equal to zero, we get the following system of equations:
2xy^2z^2x + λx = 0
2x^2yz^2y + λy = 0
2x^2y^2zz + λz = 0
x^2/2 + y^2 + z^2/2 - 1 = 0
Solving for x, y, z, and λ, we get three critical points: (±√2/2, ±1, 0) and (0, 0, ±1). We can also check that these are the only critical points by using the second partial derivative test. The Hessian matrix of the Lagrangian is:
H = [2y^2z^2 + λ, 4xyz, 2x^2yz]
[4xyz, 2x^2z^2 + λ, 2xy^2z]
[2x^2yz, 2xy^2z, 2x^2y^2 + λ]
Evaluating the Hessian at each critical point, we find that the first critical point is a local minimum, the second critical point is a local maximum, and the third critical point is a saddle point.
Therefore, the minimum value of f on the ellipsoid is f(-√2/2, -1, 0) = 0, and the maximum value of f on the ellipsoid is f(√2/2, 1, 0) = 1/8.
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Does tagging penguins with metal tags, as opposed to electronic tags, harm them? Scientists took 100 penguins and randomly chose half of them to be tagged with metal tags and half to be tagged with electronic tags. They followed the penguins for 10 years, and found that 10 of the metal-tagged penguins survived, as opposed to 18 of the electronic tagged penguins. They also collected data on whether the penguins successfully produced offspring in each of the possible breeding seasons.1. Compare the proportion of metal-tagged penguins that survived to the proportion of electronic-tagged penguins that survived.a) Is the use of the normal distribution for inference appropriate here? Why or why not?b) Conduct inference for 2 proportions based on these summarized data, and answer the following questions using the resulting output.c) What is the sample difference in proportions, proportion of metal-tagged penguins who survive – proportion of electronic-tagged penguins who survive?d) What is the 95% confidence interval for this difference?e) What is the 90% confidence interval for the difference in proportions?f) What is the default p-value for the two-sided hypothesis test for a difference using the normal distribution, if you didn’t change any of the options?g) When you conduct a hypothesis test, you want to see how extreme the results are if the null hypothesis is true. This means assuming the two proportions are the same. In order to properly reflect this when calculating the p-value, conduct the test in Minitab again, but before clicking "OK", click Options, and change the test method to "Use the pooled estimate of the proportion." Now what is the Normal-based p-value for the test?
The comparsion of the proportion of metal-tagged penguins that survived to the proportion of electronic-tagged penguins that survived gives us following result:
sample one, x1 =10, n1 =50, p1= x1/n1=0.2
sample two, x2 =18, n2 =50, p2= x2/n2=0.36
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.28
q^ Value For Proportion= 1-p^=0.72
null, H : p1 = p2
alternate, H1: p1 ≠ p2
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject H, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
z =(0.2-0.36)/[tex]\sqrt{((0.28*0.72(1/50+1/50))}[/tex]
z =-1.7817
| z | =1.7817
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.782 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we fail to reject H
p-value: two tailed ( double the one tail ) - Ha : ( p ≠ -1.7817 ) = 0.0748
hence value of p0.05 < 0.0748,here we fail to reject H .
a) standard normal distribution is approximately
b) null, H : p1 = p2
alternate, H1: p1 ≠ p2
test statistic: -1.7817
critical value: -1.96 , 1.96
decision: fail to reject H
p-value: 0.0748
c) we do not have enough evidence to support the claim that difference of proportions metal-tagged penguins who survive – proportion of electronic-tagged penguins who survive
d) given that,
sample one, x1 =10, n1 =50, p1= x1/n1=0.2
sample two, x2 =18, n2 =50, p2= x2/n2=0.36
standard error = [tex]\sqrt{( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )[/tex]
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = [tex]\sqrt{( (0.2*0.8/50) +(0.36 * 0.64/50))[/tex]
=0.088
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.088
=0.173
CI = (p1-p2) ± margin of error
confidence interval = [ (0.2-0.36) ±0.173]
= [ -0.333 , 0.013]
e) given that,
sample one, x1 =10, n1 =50, p1= x1/n1=0.2
sample two, x2 =18, n2 =50, p2= x2/n2=0.36
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = [tex]\sqrt{( (0.2*0.8/50) +(0.36 * 0.64/50))[/tex]
=0.088
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.088
=0.145
CI = (p1-p2) ± margin of error
confidence interval = [ (0.2-0.36) ± 0.145]
= [ -0.305 , -0.015]
f) Given that,
sample one, x1 =10, n1 =50, p1= x1/n1=0.2
sample two, x2 =18, n2 =50, p2= x2/n2=0.36
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.28
q^ Value For Proportion= 1-p^=0.72
null, H : p1 = p2
alternate, H1: p1 ≠ p2
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
since our test is two-tailed
reject H , if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.2-0.36)/sqrt((0.28*0.72(1/50+1/50))
zo =-1.7817
| zo | =1.7817
critical value
the value of |z α| at los 0.1% is 1.645
we got |zo| =1.782 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject H
p-value: two tailed ( double the one tail ) - Ha : ( p ≠ -1.7817 ) = 0.0748
hence value of p0.1 > 0.0748,here we reject H
g) null, H : p1 = p2
alternate, H1: p1 ≠ p2
test statistic: -1.7817
critical value: -1.645 , 1.645
decision: reject H o
p-value: 0.0748
we have enough evidence to support the claim that difference of proportion between metal-tagged penguins that survived to the proportion of electronic-tagged penguins that survived.
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Answer: The comparison of the proportion of metal-tagged penguins that survived to the proportion of electronic-tagged penguins that survived gives us
Step 1: Choose an SRS of 100 men from this population What is the sampling distribution of x? (Use the units of mg/dL.) O The N(115, 2.5) distribution The N( 1.15, 0.25) distribution. The N(115, 25) distribution. The N(115, 0.25) distribution. Step 2: What is the probability that x from Step 1 takes a value between 112 and 118 mg/dL? Note that if you use your calculator, you will need to select the closest answer. This is the probability that x estimates p within +3 mg/dL. 0.6800 0.7698 0.2327 0.4654 Step 3: Choose an SRS of 1000 men from this population. Now what is the probability that x falls within ±3 mg/dL of μ? HINT: Find the standard deviation of x . Use Table A or your calculator and give your answer to 3 decimal places.
The probability that x falls within ±3 mg/dL of μ when choosing an SRS of 1000 men from the same population is 0.9974 or approximately 1 (rounded to two decimal places).
Step 1: The sampling distribution of x is The N(115, 2.5) distribution, where the mean is 115 mg/dL and the standard deviation is 2.5 mg/dL.
Step 2: To find the probability that x takes a value between 112 and 118 mg/dL, we need to standardize these values using the formula z = (x - μ) / σ, where μ = 115 and σ = 2.5.
For x = 112 mg/dL, z = (112 - 115) / 2.5 = -1.2
For x = 118 mg/dL, z = (118 - 115) / 2.5 = 1.2
Using a standard normal table or calculator, we can find that the probability of z being between -1.2 and 1.2 is approximately 0.7698.
Therefore, the probability that x from Step 1 takes a value between 112 and 118 mg/dL is 0.7698 or approximately 0.77 (rounded to two decimal places).
Step 3: When choosing an SRS of 1000 men from the same population, the sampling distribution of x is still normal with mean μ = 115 mg/dL. However, the standard deviation of x is now σ / sqrt(n) = 2.5 / sqrt(1000) = 0.0791 mg/dL.
To find the probability that x falls within ±3 mg/dL of μ, we need to standardize these values using the formula z = (x - μ) / σ.
For x = 112 mg/dL, z = (112 - 115) / 0.0791 = -37.9268
For x = 118 mg/dL, z = (118 - 115) / 0.0791 = 37.9268
Using a standard normal table or calculator, we can find that the probability of z being between -3 and 3 is approximately 0.9974.
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2. The of the rectangle approximates the length of one-half of the circumference of the circle. 3. The circumference of the circle is _______________________. 4. The _________________ of the ___________________ is 2. 5. The ratio of the circumference to the diameter is ______. 6. Area (circle) = Area of (_____________) =
2.length of the rectangle
3.pi times the diameter.
4.The length of the rectangle is equal to the radius of the circle.
5. pi.
6.Area of (rectangle) = pi times the square of the radius.
Define circumference of the circle.The distance around a circle is known as its circumference. It is the circumference or length of the circle's whole boundaries. The circumference can be calculated by using the formula C = 2πr or C = πd, where r is the radius of the circle and d is the diameter of the circle. The value of π is a mathematical constant that is approximately equal to 3.14159.
2.The length of the rectangle approximates the length of one-half of the circumference of the circle.
3. π times the diameter equals the circumference of a circle.
4.The length of the rectangle is equal to the radius of the circle is 2.
5.The ratio of the circumference to the diameter is pi.
6.Area (circle) = Area of (rectangle) = pi times the square of the radius.
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At the start of the 2010 World Cup, the betting exchange Betfair stated that the odds against being the winning team were 9/2 for Spain, 11/2 for Brazil, 6/1 for England, and 90/1 for the United States. Find the corresponding prior probabilities of winning for these four teams.
To find the prior probabilities of winning for these four teams, we can use the formula:
Prior probability = 1 / (odds against + 1)
For Spain: Prior probability = 1 / (9/2 + 1) = 2/11 or approximately 0.18
For Brazil: Prior probability = 1 / (11/2 + 1) = 2/13 or approximately 0.15
For England: Prior probability = 1 / (6/1 + 1) = 1/7 or approximately 0.14
For the United States: Prior probability = 1 / (90/1 + 1) = 1/91 or approximately 0.01
Therefore, the corresponding prior probabilities of winning for Spain, Brazil, England, and the United States are approximately 0.18, 0.15, 0.14, and 0.01, respectively.
To find the corresponding prior probabilities of winning for the four teams, you'll need to convert the given odds into probabilities. Here's the calculation for each team:
1. Spain (9/2 odds)
Probability = 1 / (9/2 + 1) = 1 / (11/2) = 2/11 ≈ 0.1818 or 18.18%
2. Brazil (11/2 odds)
Probability = 1 / (11/2 + 1) = 1 / (13/2) = 2/13 ≈ 0.1538 or 15.38%
3. England (6/1 odds)
Probability = 1 / (6/1 + 1) = 1 / 7 ≈ 0.1429 or 14.29%
4. United States (90/1 odds)
Probability = 1 / (90/1 + 1) = 1 / 91 ≈ 0.0110 or 1.10%
So, the prior probabilities of winning for the four teams are approximately 18.18% for Spain, 15.38% for Brazil, 14.29% for England, and 1.10% for the United States.
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Find the area of the polygon with the given vertices J(-3, 3), K(2, 2), L(-1,-3), M(-4,-1). Draw you figure on the graph paper supplied. Show ALL of your calculations on the graph paper.
The area of the polygon with vertices J(-3, 3), K(2, 2), L(-1,-3), M(-4,-1) is
21 square units.How to find the area of the polygonTo find the area of the polygon with the given vertices J(-3, 3), K(2, 2), L(-1,-3), M(-4,-1), we can use the formula for the area of a quadrilateral given by its vertices:
Area = 1/2 * |x1y2 + x2y3 + x3y4 + x4y1 - y1x2 - y2x3 - y3x4 - y4x1|
where (x1,y1), (x2,y2), (x3,y3), and (x4,y4) are the coordinates of the vertices in order.
Plugging in the coordinates of the given vertices, we get:
Area = 1/2 * |-3(2) + 2(-3) + (-1)(-1) + (-4)(3) - 3(2) - 2(-1) - (-3)(-4) - (-1)(-3)|
Area = 1/2 * |-6 - 6 + 1 - 12 - 6 + 2 - 12 - 3|
Area = 1/2 * |-42|
Area = 21 square units
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12. Which of the following is a solution to the differential equation y" - 4y = 0? (A) y = 2e (C) y = sin(2x) (B) y = e? (D) y = cos(2x)
The only function that satisfies the differential equation is :
(D) y = cos(2x).
The given differential equation is y'' - 4y = 0. In order to find a solution to the differential equation follow the given steps:
Step 1: Find the second derivative y'' for each function.
(A) y = 2e
-> y'' = 0
(B) y = e^x
-> y'' = e^x
(C) y = sin(2x)
-> y' = 2cos(2x)
-> y'' = -4sin(2x)
(D) y = cos(2x)
-> y' = -2sin(2x)
-> y'' = -4cos(2x)
Step 2: Substitute each y'' and y back into the differential equation y'' - 4y = 0 to check which one satisfies the equation.
(A) 0 - 4(2e) = -8e ≠ 0
(B) e^x - 4(e^x) = -3e^x ≠ 0
(C) -4sin(2x) - 4sin(2x) = -8sin(2x) ≠ 0
(D) -4cos(2x) - 4(cos(2x)) = -8cos(2x) + 8cos(2x) = 0
The only function that satisfies the differential equation is (D) y = cos(2x).
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Simplify each side of the equation independently to reach a common equivalent expression, 1+sinx/cosx . cosx/1−sinx=secx + tanx Part: 0 / 2 Part 1 of 2 Left-hand side: Show that cosx / 1−sinx= 1+sinx/cosx.
To simplify the left-hand side of the equation, we can start by multiplying both the numerator and denominator of cosx/1-sinx by 1+sinx. This gives us:
cosx/1-sinx * (1+sinx)/(1+sinx) = (cosx * (1+sinx)) / (1 - sin^2 x)
Using the Pythagorean identity sin^2 x + cos^2 x = 1, we can simplify the denominator to cos^2 x:
(cosx * (1+sinx)) / cos^2 x
We can then simplify the numerator by distributing the cosx:
(1 + sinx) / cosx
This is the same as 1+sinx/cosx, which is the right-hand side of the equation. Therefore, we have shown that cosx/1-sinx is equivalent to 1+sinx/cosx.
To show that cosx / (1 - sinx) = (1 + sinx) / cosx, we can manipulate the left-hand side of the equation:
Start with: cosx / (1 - sinx)
Multiply the numerator and denominator by the conjugate of the denominator, which is (1 + sinx):
cosx(1 + sinx) / [(1 - sinx)(1 + sinx)]
Use the difference of squares formula for the denominator: a^2 - b^2 = (a + b)(a - b)
cosx(1 + sinx) / [1 - (sinx)^2]
Now, recall the Pythagorean identity: cos^2x + sin^2x = 1, so cos^2x = 1 - sin^2x
cosx(1 + sinx) / cos^2x
Finally, divide both terms in the numerator by cosx:
(1 + sinx) / cosx
Thus, we have shown that cosx / (1 - sinx) = (1 + sinx) / cosx.
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Which of the following would be an appropriate null hypothesis?
a. The mean of a population is equal to 55.
b. The mean of a sample is equal to 55.
c. The mean of a population is greater than 55.
d. Only a) and c) are true.
2. A Type II error is committed when
a. we reject a null hypothesis that is true.
b. we don't reject a null hypothesis that is true.
c. we reject a null hypothesis that is false.
d. we don't reject a null hypothesis that is false.
3. If we are performing a two-tailed test of whether
1. The appropriate null hypothesis would be a) The mean of a population is equal to 55. This is because the null hypothesis should always state that there is no significant difference or effect between two groups or variables being tested. In this case, the null hypothesis assumes that there is no significant difference between the population mean and 55, and the alternative hypothesis assumes that there is a significant difference.
2. A Type II error is committed when b) we don't reject a null hypothesis that is false. This means that we fail to reject the null hypothesis when it is actually false, and we mistakenly accept the null hypothesis as true. This can occur when the sample size is too small, the effect size is too small, or there is too much variability in the data.
3. If we are performing a two-tailed test of whether the mean of a population is different from a specific value, then the null hypothesis would be that the population mean is equal to that specific value. The alternative hypothesis would be that the population mean is not equal to that specific value, indicating a significant difference. For example, if we are testing whether the mean height of a population is different from 6 feet, the null hypothesis would be that the mean height is equal to 6 feet, and the alternative hypothesis would be that the mean height is not equal to 6 feet.
1. The appropriate null hypothesis would be:
a. The mean of a population is equal to 55.
A null hypothesis typically states that there is no significant difference or relationship between variables and is assumed to be true until proven otherwise.
2. A Type II error is committed when:
d. we don't reject a null hypothesis that is false.
A Type II error occurs when we fail to reject a null hypothesis that should have been rejected, meaning we accept the null hypothesis when it is actually false.
3. If we are performing a two-tailed test of whether:
A two-tailed test is used to determine if there is a significant difference between two groups, but does not specify the direction of the difference. In other words, it tests if there is a difference, but not whether it is greater or smaller.
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how many trains would pass from 6:00 to 6:30
The trains will meet 210 km from Lucknow.
How to calculate the distanceSpeed is a physical quantity that refers to how fast an object is moving relative to a reference point. It is usually measured in units of distance traveled per unit of time, such as meters per second (m/s) or miles per hour (mph). Speed can also be described as the rate at which an object covers a certain distance in a certain amount of time.
Speed of first train = 60km/hr
Distance travelled when second train started = 60/2 = 30km
Relative speed = (70 – 60)km/hr ⇒ 10km/hr
Time taken to cover 30 km = 30/10 = 3 hour
Distance from Lucknow where the trains meet = 70 × 3 = 210 km
The trains will meet 210 km from Lucknow
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Two trains leave Lucknow at 6 : 00 a.m. and 6 : 30 a.m. respectively. Speed of first train is 60 km/hr while that of second one is 70 km/hr. How many kilometres from Lucknow will they meet?
