please help and hurry

The midpoint of the line segment joining the points corresponding to the complex numbers 6 + i and 2 + 9i in the complex plane is 4 + 5i.

The midpoint of a line segment can be found by taking the average of the x-coordinates and the average of the y-coordinates of the two endpoints. In the complex plane, the x-coordinate corresponds to the real part of the complex number and the y-coordinate corresponds to the imaginary part.

So, for the complex numbers 6 + i and 2 + 9i, we can find the midpoint by averaging the real parts (6 and 2) and the imaginary parts (1 and 9):
Midpoint = ((6+2)/2, (1+9)/2) = (4, 5)
In complex number form, this would be 4 + 5i.
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Answer:

3

Step-by-step explanation:

Hice una ecuación: 2x+4=10. Luego, reste 4 de cada lado para obtener: 2x = 6, lo que resulta en: x = 3.

The integral is a bit complex. Therefore, the final answer for the integral will be the sum of the above two integrals. ∫S f(x, y, z) ds = ∫0³ (1 + V)i + (2t)Vj - 4t³k √(1 + 4t²V² + 4t⁶) dt + ∫0³ (27 + 81V - t⁴) √(1 + 4t²V² + 4t⁶) dt.

We are given the function f(x, y, z) = x + Vy - z².

We need to integrate this over the path given by C1 and C2 from (0,0,0) to (3,9,3).

The path is given by C1: r(t) = ti + t²j,

where 0 ≤ t ≤ 3 and C2: r(t) = 3i + 9j + tk,

where 0 ≤ t ≤ 3.Substituting these values in the function, we get:f(r(t)) = r(t)i + Vr(t)j - z²

= ti + t²j + V(ti + t²)k - (tk)²

= ti + t²j + Vti + Vt² - t²k²

= ti + t²j + Vti + Vt² - t⁴

Taking the derivative of the above function, we get:

∂f/∂t = i + 2tj + V(i + 2tk) - 4t³k

= (1 + V)i + (2t)Vj - 4t³k

The magnitude of dr/dt is given by:

|dr/dt| = √[∂x/∂t² + ∂y/∂t² + ∂z/∂t²]²

= √[1² + 4t²V² + 4t⁶]

We need to find ∫S f(x, y, z) ds over the path C1 and C2,

which is given by:

∫S f(x, y, z) ds

= ∫C1 f(r(t)) |dr/dt| dt + ∫C2 f(r(t)) |dr/dt| dt

Substituting the values in the above equation, we get:

∫S f(x, y, z) ds = ∫0³ (1 + V)i + (2t)Vj - 4t³k √(1 + 4t²V² + 4t⁶) dt + ∫0³ (27 + 81V - t⁴) √(1 + 4t²V² + 4t⁶) dt

The integral is a bit complex. Therefore, this cannot be solved here. The final answer for the integral will be the sum of the above two integrals.

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Answer:

8 cups

Step-by-step explanation:

\(6\frac{5}{8} + \frac{11}{8}\\\\=\frac{53}{8} + \frac{11}{8}\\\\=\frac{64}{8}\\\\=8\)

Answer:

15m long

Step-by-step explanation:

The expression that is equivalent to 1/3y is 1/6y + 1/6(y + 12) - 1/6 * 12

The expression is given as:

1/3y

Express 1/3y as 1/6y + 1/6y

So, we have:

1/6y + 1/6y

Factor out y

1/6y + 1/6(y)

Add 0 to the above equation

1/6y + 1/6(y + 0)

Express 0 as 12 - 12

1/6y + 1/6(y + 12 - 12)

Expand

1/6y + 1/6(y + 12) - 1/6 * 12

Evaluate the product

1/6y + 1/6(y + 12) - 2

Hence, the expression that is equivalent to 1/3y is 1/6y + 1/6(y + 12) - 1/6 * 12

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The expression that is equivalent to 1/3y is 1/6y + 1/6(y + 12) - 1/6 * 12

The expression is given as:

1/3y

Express 1/3y as 1/6y + 1/6y

So, we have:

1/6y + 1/6y

Factor out y

1/6y + 1/6(y)

Add 0 to the above equation

1/6y + 1/6(y + 0)

Express 0 as 12 - 12

1/6y + 1/6(y + 12 - 12)

Expand

1/6y + 1/6(y + 12) - 1/6 * 12

Evaluate the product

1/6y + 1/6(y + 12) - 2

Therefore, the expression that is equivalent to 1/3y is 1/6y + 1/6(y + 12) - 1/6 * 12

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Answer:

3x + 21

Step-by-step explanation:

3(x + 7)           Distribute

3x + 21

Answer:

B I think is the answer

Step-by-step explanation:

46

Answer:

Step-by-step explanation:

To solve the given Integer Programming (IP) problem using the Branch-and-Bound algorithm, we follow these steps:

Initialization: Start with an initial relaxation by solving the linear programming (LP) relaxation of the problem. Ignore the integrality constraints and find the optimal solution.

Branching: Identify a fractional variable in the LP relaxation solution. Let's say x₁ is fractional with a value of 2.5. Create two subproblems by branching on x₁: one with the constraint x₁ ≤ 2 and the other with the constraint x₁ ≥ 3.

Solve Subproblems: Solve the LP relaxation of each subproblem obtained in the previous step.

Subproblem 1: x₁ ≤ 2

Minimize z = -20x₁ + 10x₂ + 25x₃ - 20x₄

subject to:

x₁ + x₂ + x₃ + 2x₄ ≤ 12

3x₁ + x₂ + 2x₃ + 2x₄ ≥ 10

x₁ ≤ 2

Subproblem 2: x₁ ≥ 3

Minimize z = -20x₁ + 10x₂ + 25x₃ - 20x₄

subject to:

x₁ + x₂ + x₃ + 2x₄ ≤ 12

3x₁ + x₂ + 2x₃ + 2x₄ ≥ 10

x₁ ≥ 3

Solve both subproblems using any LP solver to obtain their optimal solutions.

Branching and Pruning: Check the solutions obtained from the subproblems.

If the optimal solution of a subproblem is infeasible or worse than the current best integer solution, prune the subproblem.

If the optimal solution of a subproblem is integer and better than the current best integer solution, update the best integer solution.

Repeat Steps 2-4 for each remaining subproblem until all subproblems have been solved or pruned.

Termination: Terminate the algorithm when all subproblems have been solved or pruned. The best integer solution found throughout the algorithm is the optimal solution to the IP problem.

Note: The LP relaxation solutions may not necessarily be integer, but the branching and pruning steps ensure that we explore only the relevant subproblems in the search space.

Please note that the above steps outline the general approach of the Branch-and-Bound algorithm. The actual implementation and solution may require additional details and specific calculations based on the given problem.

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Answer:

A. 24 units

Explanation:

The perimeter of the triangle RST is equal to the sum of their sides.

We know that RS = 8 units and ST = 6 units.

Then, by the Pythagorean theorem, we can calculate the length of side RT as follows

Then, the perimeter is equal to

Perimeter = RS + ST + RT

Perimeter = 8 + 6 + 10

Perimeter = 24

So, the answer is A. 24 units

1) Arranging the expressions by growth rate from slowest to fastest:

log3(n), log2(n), n^(2/3), 20n, 4n^2, 3n, n! Stirling's approximation is used to estimate the growth rate of n!. According to Stirling's approximation, n! ≈ (√(2πn)) * ((n/e)^n). 2) Estimating the number of inputs that could be processed in the given cases: (a) For the algorithm with time complexity T(n) = 3 * 2^n: On the new machine that is 64 times as fast, we could process 6 more inputs in the same time. (b) For the algorithm with time complexity T(n) = n^2: On the new machine that is 64 times as fast, we could process 4096 times more inputs in the same time. (c) For the algorithm with time complexity T(n) = 8n: On the new machine that is 64 times as fast, we could process 512 times more inputs in the same time.

1) Arranging the expressions by growth rate from slowest to fastest:

log 3​

n, log 2​

n, n 2/3, 4n^2, 20n, 3n, n!

Stirling's approximation is used to estimate the growth rate of n!. According to Stirling's approximation, n! ≈ (√(2πn))(n/e)^n.

2) Estimating the number of inputs that could be processed in the given cases:

(a) For the algorithm with time complexity T(n) = 3 * 2^n:

On the new machine that is 64 times as fast, the time taken for n inputs would be t/64 seconds. To find the number of inputs that can be processed in t seconds on the new machine, we need to solve the equation:

t/64 = 3 * 2^n

Simplifying the equation:

2^n = (t/64)/3

2^n = t/192

n = log2(t/192)

(b) For the algorithm with time complexity T(n) = n^2:

On the new machine that is 64 times as fast, the time taken for n inputs would be t/64 seconds. To find the number of inputs that can be processed in t seconds on the new machine, we need to solve the equation:

(t/64) = n^2

n^2 = t/64

n = sqrt(t/64)

(c) For the algorithm with time complexity T(n) = 8n:

On the new machine that is 64 times as fast, the time taken for n inputs would be t/64 seconds. To find the number of inputs that can be processed in t seconds on the new machine, we need to solve the equation:

(t/64) = 8n

n = (t/64)/8

n = t/512

Note: In all cases, the estimates assume that the time complexity remains the same on the new machine.

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The  particular solution of the differential equation that satisfies the initial condition  \(\frac{du}{dv} = uvsin v^{2} u(0)= e^{4}\)  is  \(u(v)=sin v^{2} + e^{4}-\frac{1}{2}\)

To find the particular solution of the differential equation that satisfies the initial condition                                                              

 \(\frac{du}{dv} = uv sin v^{2} u(0) = e^{4}\)

we must follow the steps below: Initial Equation \(\frac{du}{dv} = uv sin\)

v²Separating variables gives us: \(u du = v sinv^{2} dv\)

a) Integrating both sides, we have:

\(\int\limits {u} \, du = \int\limits {v sin v^{2} } \, dv(\frac{u^{2} }{2} )\)

\(= \frac{-1}{2} cos v^{2} +C1 u^{2}\)

\(= -cosv^{2} + C2\)

b) Differentiate both sides to get:

\(2u \frac{du}{dv} = 2v^{2} cosv^{2}\)

c) We replace du/dv in the equation above with the equation given in the initial condition to obtain:                                                      

  = \(2u (uv) sinv^{2}\)

= \(2v^{2} cosv²u\)

= \(2v cosv^{2} dv + C3\)

d) Integrating both sides gives:

\(\int\limits {u} \, dv = \int\limits {2v cos v^{2} } \, dv + \frac{C3}{2v} (v)\)

\(= sinv^{2} + C4 + \frac{C3}{2}\)

e) We substitute the initial condition to obtain the value of

\(C4.e^{2}= sin 0^{2}+ C4+ \frac{C3}{2C4}\)

\(= e^{4} - \frac{1}{2}\)

The particular solution is obtained by replacing the value of C4 and is as follows: \(u(v) = sin v^{2} + e^{2} - \frac{1}{2}\)

Answer: \(u(v) = sin v^{2} + e^{2} - \frac{1}{2}\)

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square

its the middle bit that looks darker than the rest.

The use of regression modeling in retail analytics can help businesses make data-driven decisions that ultimately lead to increased profits and growth.

Based on the information given, it seems that the large sports supplier is interested in predicting the annual revenue of a particular store based on various factors, such as population, promotion expenditure, and distance from the city center. This is a common approach in retail analytics, where regression models are often used to predict sales or revenue based on different variables.

By constructing a regression model, the sports supplier can gain valuable insights into which factors are most strongly associated with revenue, and how they can optimize their operations to increase sales. For example, they may find that stores located closer to the city center tend to have higher revenue, or that increased promotion expenditure leads to a greater increase in revenue in smaller towns.

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Answer:

Rajiv bought 23,382 balls.

Step-by-step explanation:

Given that Rajiv bought 1325 packs of red balls, 1238 packs of yellow balls and 1334 packs of green balls, and there were 6 balls in each pack, to determine how many balls did Rajiv buy the following calculation must be performed:

(1,325 + 1,238 + 1,334) x 6 = X

3.897 x 6 = X

23.382 = X

Therefore, Rajiv bought 23,382 balls.

A number is divisible by 6, if the last digit is even and the sum of the digits is divisible by 3

Hope this will help...

Answer:

\( {x}^{2} - 15x + 54 \\ (x - 9)(x -6 )\)

____o___o___

\(3 {x}^{2} + 22x + 7 \\ \\ ( 3x+1)( x+7 )\)

I hope I helped you^_^

Answer:

8

Step-by-step explanation:

Combine the X's to become 1x and then add one to both sides. Also there is this site called . Its a calc. U can use that instead of brainly for the answers. Saves u points.

Answer: 8

Step-by-step explanation:

the answer should be 175

1. Definition of a correspondence: A correspondence is a mathematical concept that defines a relation between two sets, where each element in the first set is associated with one or more elements in the second set. It can be thought of as a rule that assigns elements from one set to elements in another set based on certain criteria or conditions.

2. Definition of a fixed point of a correspondence: In the context of a correspondence, a fixed point is an element in the first set that is associated with itself in the second set. In other words, it is an element that remains unchanged when the correspondence is applied to it.

3. Set of (mixed strategy) best replies in normal form games: In a normal form game, the set of (mixed strategy) best replies for a given player i is the collection of strategies that maximize the player's expected payoff given the strategies chosen by the other players. It represents the optimal response for player i in a game where all players are using mixed strategies.

Best reply correspondence: The "best reply correspondence," denoted by B in class, is a correspondence that assigns to each mixed strategy profile the set of best replies for each player. It maps a mixed strategy profile to the set of best responses for each player.

4. Nash equilibrium and fixed point of best reply correspondence: A mixed strategy profile α∗ is a Nash equilibrium if and only if it is a fixed point of the best reply correspondence. This means that when each player chooses their best response strategy given the strategies chosen by the other players, no player has an incentive to unilaterally change their strategy. The mixed strategy profile remains stable and no player can improve their payoff by deviating from it.

5. Brower's fixed point theorem: Brower's fixed point theorem states that any continuous function from a closed and bounded convex subset of a Euclidean space to itself has at least one fixed point. In other words, if a function satisfies these conditions, there will always be at least one point in the set that remains unchanged when the function is applied to it.

6. Proving Brower's theorem using Kakutani's fixed point theorem: Kakutani's fixed point theorem is a more general version of Brower's fixed point theorem. By using Kakutani's theorem, we can prove Brower's theorem as a corollary.

Kakutani's theorem states that any correspondence from a non-empty, compact, and convex subset of a Euclidean space to itself has at least one fixed point. Since a continuous function can be seen as a special case of a correspondence, Kakutani's theorem can be applied to prove Brower's theorem.

7. Conditions for Kakutani's fixed point theorem: Kakutani's fixed point theorem requires several conditions to hold in order to guarantee the existence of a fixed point. These conditions include non-emptiness, compactness, convexity, and upper semi-continuity of the correspondence.

If any of these conditions are not satisfied, the conclusion of Kakutani's theorem does not hold, and there may not be a fixed point.

8. Example without a fixed point: An example without a fixed point can be a correspondence that does not satisfy the condition of closedness in the relative topology of S², where S is the set where the correspondence is defined. This means that there is a correspondence that maps elements in S to other elements in S, but there is no element in S that remains unchanged when the correspondence is applied.

This is a valid counterexample because it shows that even if all other conditions of Kakutani's theorem are satisfied, the lack of closedness in the relative topology can prevent the existence of a fixed point.

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For 5 chairs, the machine needs 25 minutes.

A percentage in mathematics is an equation that connects two or more ratios and declares that they are equal. Two values are compared in a ratio, which is frequently stated as a fraction or the division of two integers. A percentage, which usually takes the form: expresses the equivalence of two or more ratios.

a/b = c/d

Given:

Rate of production = 2 chairs every 10 minutes

Time taken to make 5 chairs

2 chairs / 10 minutes = 5 chairs / x minutes

Cross-multiplying,

2 * x = 10 * 5

2x = 50

Dividing both sides by 2:

x = 50 / 2

x = 25

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Complete Question: a machine factory makes chairs at a rate 2 chair every 10 minutes how much time does machine take to make five chairs?

Answer:

Not proportional

Step-by-step explanation:

1/5 = 0.2

2/8 = 0.25

6/18 = 0.33

etc...

9514 1404 393

Answer:

Step-by-step explanation:

The drawing is a bit unfortunate, as it does not show the angles true to size.

__

When the altitude (EG) of a triangle (∆DFG) is also a median, then that triangle is isosceles. Both halves are congruent right triangles.

  ∆GED ≅ ∆GEF

Since these are right triangles, the acute angle not given (∠EGF) is the complement of the one that is given.

  m∠EGF = 90° -m∠F = 90° -40°

  m∠EGF = 40°

__

Of course, angle D in ∆GED is identical to its counterpart in ∆GEF, angle F, so ...

  m∠D = 40°

__

Because the lines are parallel, angle ABE corresponds (and is congruent) to angle DEG, which is marked as a right angle.

  m∠ABE = 90°

Answer:

12

Step-by-step explanation:

12 × 3= 36

36 + 4= 40

so the answer is 12

Answer:

Not true with all numbers!

Step-by-step explanation:

see.Ex.3x3=9+4=13 not 40

The 95% confidence interval for the given population proportion is between 0.1716 to 0.2884.

The confidence interval for a population proportion is calculated by the formula,

\(C.I = \bar{p}\pm z_{\alpha/2}\sqrt{\frac{\bar{p}(1-\bar{p})}{n} }\)

Where \(\bar{p}\) is the sample proportion and α is the level of significance.

It is given that,

The statistics reported on CNBC projects, a surprising number of motor vehicles are not covered by insurance. The sample results are

The sample size n = 200;

The number of successes = 46

So, the sample proportion \(\bar{p}\) = 46/200 = 0.23

For a 95% confidence interval, the level of significance is

α = 1 - 95/100 = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

Then, the z-score for the value 0.025 is

\(z_{\alpha/2}\) = 1.96 (from the table)

Thus, the confidence interval is

\(C.I = \bar{p}\pm z_{\alpha/2}\sqrt{\frac{\bar{p}(1-\bar{p})}{n} }\)

⇒ C.I = 0.23 ± (1.96) × \(\sqrt{\frac{0.23(1-0.23)}{200} }\)

⇒ C.I = 0.23 ± 1.96 × 0.0298

⇒ C.I = 0.23 ± 0.0584

So, the upper limit is 0.23 + 0.0584 = 0.2884 and the lower limit is 0.23 - 0.0584 = 0.1716.

Therefore, the 95% confidence interval for the given population proportion lies between 0.1716 to 0.2884.

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Answer:

not exactly sure what is being asked butttt i think it's 90

Step-by-step explanation:

Answer:

(15÷8+2÷5)+(-7÷8)

=(1.875+0.4)- 0.875

= 2.275 - 0.875

=== 1.4

Step-by-step explanation:

mark me as brainlist

Answer: ​7/5, or 1.4, 1.4 in fraction is 7/5

​​  

Step-by-step explanation:

Probability is unnecessary to predict a fixed event.

Using translation concepts, it is found that the option that describes the graph of \(y = -\frac{1}{4}\arccos{(x - 3)} + 1\) compared to the graph of \(y = \arccos{x}\) is:

The parent function is:

\(y = \arccos{x}\)

In this problem:

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The mean (μ) of a hypergeometric distribution is given by: 8.181

The standard deviation (σ) of a hypergeometric distribution is given by:

1.644

Here, we have,

To find the mean value and standard deviation of the number of projects not among the first 15 that are from the second section, we need to calculate the probabilities for different numbers of projects from the second section.

Let's denote:

N1: Number of students in the first section (25)

N2: Number of students in the second section (30)

N: Total number of projects graded (15)

To calculate the probability of exactly 10 projects being from the second section, we can use the hypergeometric distribution.

The formula for the hypergeometric distribution is:

P(X = k) = (C(N2, k) * C(N1, N - k)) / C(N1 + N2, N)

Where:

X is the random variable representing the number of projects from the second section among the first 15 graded projects.

C(a, b) is the binomial coefficient, also known as "a choose b."

Using this formula, we can calculate the probability for X = 10:

P(X = 10) = (C(30, 10) * C(25, 15 - 10)) / C(55, 15)

Next, we can calculate the mean and standard deviation.

The mean (μ) of a hypergeometric distribution is given by:

μ = N * (N2 / (N1 + N2))

  = 15 * (30 / (25 + 30) )

  = 8.181

The standard deviation (σ) of a hypergeometric distribution is given by:

σ = √(N * (N1 / (N1 + N2)) * (N2 / (N1 + N2)) * ((N1 + N2 - N) / (N1 + N2)) )

  = √(15 * (25 / (25 + 30)) * (30 / (25 + 30)) * (25+30 - 15 )/(25+30)) )

  = 1.644

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complete question:

An Instructor Who Taught Two Sections Of Engineering Statistics Last Term, The First With 25 Students And The Second With 30, Decided To Assign A Term Project. After All Projects Had Been Turned In, The Instructor Randomly Ordered Them Before Grading. Consider The First 15 Graded Projects. (A) What Is The Probability That Exactly 10 Of These Are From The

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects.

what are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?