SOLVE FOR 20 pls hurry

1) If the couple must sit together, there are 240 ways that they can be seated.

2) If the couple must not be together, there are 720 ways that they can be seated.

1) If the couple must sit together, we can think of them as one unit. So, there are 5 units to be seated around the table. We can choose any one of these units to be the starting point, and then arrange the other units clockwise around the table.

The couple can be arranged in two different ways within their unit. So, the total number of ways they can be seated is: 5! × 2 = 240.

2) If the couple must not sit together, we can think of them as separate units. So, there are 6 units to be seated around the table. We can choose any one of these units to be the starting point, and then arrange the other units clockwise around the table.

The couple cannot be seated next to each other, so there are 3 possible units where they cannot be seated next to each other. Once we have chosen one of these units for the starting point, there are 2 ways to arrange the couple within their unit. Therefore, the total number of ways they can be seated is: 5! × 2 × 3 = 720.

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All three parts of the given problem regarding trails, paths and walks have been solved below.

A path is defined as a way in which we can traverse the graph such that neither vertices nor the edges are repeated. In the given graph we can see that for going from a to c there are 4 ways that satisfy the condition of a path. A Trail is an open walk in which no edge is repeated but the vertex can be repeated. So, in the given graph we can see that there are 3!(4) ways in which we can go from a to b and then c and 4 paths in which we can go directly from a to c.

The walk is a sequence of vertices and edges of a graph that can be closed if we come to the same point after traversing the graph or open if we go from one point to another. In the given graph since we go from a point to c point the walk is open but the length of edges of the walk is not defined so if we take the no. of edge length to be 4

there can be 44 ways

so,

number of paths from a to c = 4

number of trails from a to c = 4+3!(4)= 28

number of paths from a to c = 4+3!(4)+2!(6)+1!(4)= 44

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Based on the properties of similar triangles, the two true statements are:

In Mathematics, two (2) triangles are said to be similar when the ratio of their corresponding side lengths are equal and their corresponding angles are congruent.

Based on the properties of similar triangles, we have the following points:

In this scenario, we can can logically deduce that the two true statements are:

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Answer: B, D

Step-by-step explanation: AXC-CXB, ACB-AXC

The flux of the vector-field F = 1i + 1j + 2k across the surface S is 2. We find out the flux of the vector-field using Green's Theorem.

Flux form of Green's Theorem for the given vector-field

φ = ∫ F.n ds

= ∫∫ F. divG.dA

Here G is equivalent to the part of the plane = 2x+4y+z = 2.

and given F = 1i + 1j + 2k

divG = div(2x+4y+z = 2) = 2i + 4j + k

Flux = ∫(1i + 1j + 2k) (2i + 4j + k) dA

φ = ∫ (2 + 4 + 2)dA

= 8∫dA

A = 1/2 XY (on the given x-y plane)

2x+4y =2

at x = 0, y = 1/2

y = 0, x = 1

1/2 (1*1/2) = 1/4

Therefore flux = 8*1/4 = 2

φ = 2.

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The angle subtended by arc BC on center of circle is of 2 times 65 = 130

What is Area of circle ?

Area of circle can be defined as the product of pi and the square of radius of a circle.

Using the property that angle subtended by an arc on center is double than on circumference, you can find measure of arc BC.

The angle subtended by arc BC on center of circle is of 2 times 65 = 130

Given,

The angle subtended by arc BC on circumference is 65 degrees.

We have to find  Length of arc BC

Relation between angle at center and angle at circumference by an arc:

"Angle subtended by an arc of a circle on its center is double than angle subtended by it on circumference"

Therefore , angle subtended by arc BC on center of circle is of 2 times 65 = 130 .

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The expression x + 80 represents the number of coins Arthur has now, since he x pennies and an additional 6 dimes and 4 nickels.

We shall convert the pennies, dimes and nickels into same unit before adding to get the expression for the number of coins Arthur will have in total.

A penny = 1 cent, so

x pennies = x cents

A dime = 10 cents so;

6 dimes = 6 × 10cents

6 dimes = 60 cents

A nickel = 5 cents so;

4 nickels = 4 × 5 cents

4 nickels = 20 cents

Arthur's coins in total = (x + 60 + 40) cents

Arthur's coins in total = (x + 80) cents.

Therefore, the expression expression x + 80 represents the number of coins Arthur will have in total, wether in pennies, dimes or nickels.

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Answer:

Step-by-step explanation:

B - 6 million

b. 6 Million

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Based on the given table, the profit-maximizing level of output for company Econislife is 4 units. At this quantity, the company's profit is $40.

To determine the profit-maximizing level of output, we need to analyze the relationship between total revenue (TR) and total costs (TC) for different quantities of output. The profit can be calculated as the difference between total revenue and total costs.

Looking at the table, we can see that as the quantity increases from 1 to 5, both total revenue and total costs increase. However, at a certain point, the increase in total costs outpaces the increase in total revenue, leading to a decrease in profit.

To find the profit-maximizing level of output, we compare the profits at different quantities.

Quantity 1: TR = $28, TC = $40, Profit = TR - TC = $28 - $40 = -$12

Quantity 2: TR = $56, TC = $45, Profit = TR - TC = $56 - $45 = $11

Quantity 3: TR = $84, TC = $55, Profit = TR - TC = $84 - $55 = $29

Quantity 4: TR = $112, TC = $72, Profit = TR - TC = $112 - $72 = $40

Quantity 5: TR = $140, TC = $120, Profit = TR - TC = $140 - $120 = $20

From the above calculations, it is evident that the profit is highest at Quantity 4. Therefore, the profit-maximizing level of output for company Econislife is 4 units. At this quantity, the company's profit is $40.

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To find the dimensions of the largest room that can be built with a fixed perimeter of 600 feet.

We need to divide the perimeter by 2 and use that as the sum of two adjacent sides. Let's call the length of the rectangle "l" and the width "w".
So we have:  2l + 2w = 600
Simplifying:  l + w = 300
We want to maximize the area of the rectangle, which is given by:  A = lw
We can solve for one variable in terms of the other:  l = 300 - w
Substituting into the area equation:
A = (300 - w)w
A = 300w - w^2
To maximize the area, we need to find the value of w that makes the derivative of A with respect to w equal to 0:  dA/dw = 300 - 2w = 0
w = 150
So the width of the rectangle is 150 feet. Substituting back into the perimeter equation: l + 150 = 300
l = 150
So the length of the rectangle is also 150 feet.
Therefore, the largest room that can be built has dimensions 150 ft by 150 ft, and its area is:  A = lw = 150 * 150 = 22,500 ft^2
The dimensions of the largest rectangular room a carpenter can build with a fixed perimeter of 600 feet are 150 ft by 150 ft. The area of this room is 22,500 ft². This is because when the length and width are equal, the area of the rectangle is maximized.

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The probability that the jury makes the correct decision is approximately 0.7063.

To solve this problem, we need to find the probability that the jury makes the correct decision if the defendant is guilty. Let's break down the problem into smaller steps.

We know that the probability of a single juror making the correct decision is 0.80. If the defendant is guilty, then the probability of a juror making the correct decision is still 0.80. Therefore, the probability that a single juror makes the correct decision if the defendant is guilty is 0.80.

We can use the binomial distribution formula to determine the probability of at least 10 out of 12 jurors making the correct decision. The formula is:

P(X ≥ k) = 1 - Σ(i=0 to k-1) [n!/(i!(n-i)!) x \(p^i \times (1-p)^{(n-i)}\) ]

where:

P(X ≥ k) is the probability of at least k successes

n is the total number of trials (in this case, 12 jurors)

p is the probability of success in a single trial (in this case, 0.80)

k is the number of successes we want to find the probability of (in this case, 10)

Plugging in the values, we get:

P(X ≥ 10) = 1 - Σ(i=0 to 9) [12!/(i!(12-i)!) x \(0.80^i \times (1-0.80)^{(12-i)}\)]

Using a calculator or software, we can calculate this to be approximately 0.7063.

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The probability of getting a sum that is a multiple of 5 when throwing two fair dice is 1/7.

At the point when two fair dice are tossed, the complete number of results is 36 (6 potential results for each bite the dust). To track down the likelihood of getting a total that is a different of 5, we want to count the great results.

The products of 5 that can be gotten are 5 and 10. The potential mixes to get an amount of 5 are (1, 4), (2, 3), (3, 2), and (4, 1). The conceivable mix to get an amount of 10 is (5, 5).

In this way, there are 5 good results out of 36 all out results. In this way, the likelihood of getting a total that is a numerous of 5 will be 5/36, or roughly 0.1389, which can be improved as 1/7 in its least complex structure.

In rundown, the likelihood of getting a total that is a numerous of 5 while tossing two fair dice is 1/7.

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The rate of change of volume is 418.6  inch³/s .

Rate of increase of radius with time is ( dr/dt ) = 5

rate of decrease of height with time is (dh/dt ) = -3

volume of a cone is given by (V) =  \(\frac{1}{3} \pi r^{2}h\)

Now the rate of change of volume  with respect to time is given by dV/dt

∴ dV/dt = d( \(\frac{1}{3} \pi r^{2}h\)) / dt

dV/dt =  \(\frac{1}{3} \pi (2rh\frac{dr}{dt} + r^{2}\frac{dh}{dt} )\)

value of radius and height when volume is changing are :

r = 20 inches

h = 40 inches

dV/dt = \(\frac{1}{3} \pi (\) 2×20×40 ×5 - \(20^{2}\)×3 )

dV/dt = \(\frac{1}{3} \pi\)(1600 - 1200)

dV/dt = (400/3) 3.14

dV/dt = 418.66 inch³/s

So the rate of change of the volume is 418.66 inch³/s.

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The investment problem is modeled by an initial value problem (IVP) where the rate of change of the investment is determined by the initial investment, monthly deposits, and the interest rate.

a) The investment problem can be modeled by an initial value problem where the rate of change of the investment, y(t), is given by the initial investment, monthly deposits, and the interest rate. The IVP can be written as:

dy/dt = 0.09y + 200, y(0) = 500.

b) To solve the IVP, we can use an integrating factor to rewrite the equation in the form dy/dt + P(t)y = Q(t), where P(t) = 0.09 and Q(t) = 200. Solving this linear first-order differential equation, we obtain the solution for y(t).

c) To find the value of the investment after 2 years, we substitute t = 2 into the obtained solution for y(t) and calculate the corresponding value.

d) After 2 years, the monthly deposit increases to $300. To find the value of the investment a year later, we substitute t = 3 into the solution and calculate the value accordingly.

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Step-by-step explanation:

First one

5,5√5,25

Second one

-3,12,-48

Answer:

4:5

Step-by-step explanation:

5=women

4=men

(5+4=9)

4:5

Answer:

5:4

Step-by-step explanation:

4/9 are men

so, 5/9(women) : 4:9(men)

times 9 both side

5(women):4(men)

Isolate the variable by dividing each side by factors that don't contain the variable.

x = 20

Step-by-step explanation:

0.5x+8=-7.2

0.5x=-7.2-8

0.5x=-15.2

×=-15.2÷0.5

x=30.4

6/50 fraction of all people in the company are women under age of 30 In a company where the ratio of men to women i 3:2.

In a company the ratio of men to women  3:2

30% of the women are under the age of 30.

suppose there are 500 people in the company

then 300 are men and 200 are women

30% of 200 = 200*30/100= 60 women are under 30

60/500= 6/50 women are under 30

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Answer:

x = 10/7 = 1.429

Step-by-step explanation:

Answer:

X = 1.030714

Step-by-step explanation:

Let's start by substituting the right-hand side of the equation into the left-hand side:

What does the math equation mean?

Two expressions are combined by the equal sign to form a mathematical statement known as an equation. For instance, a formula might be 3x - 5 = 16. After solving this equation, we learn that the value of the variable x is 7.

4y - 3 = 4(y + 1)

4y - 3 = 4y + 4

Now, we can isolate y by subtracting 4y from both sides:

0 = y + 4

-4 = y

So, there is a solution to the equation: y = -4. This means that Dawson is incorrect in saying that the equation has no solution.

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Its B10 Since 17 + 10 = 27.

Hope this helps <33!!!!

Answer:

The answer is B/10

Step-by-step explanation:

if you subtract it you will get 10 as your answer.

Answer:

16

16. 16-4 = 12.

1) The event is considered as rare.

2) It is called a rare event event because the probability is less than 5% and is well below threshold.

1) The probability of winning the grand prize is 0.005, which means there is a 0.5% chance of winning the prize.

Since this probability is low, Michele winning the grand prize is considered a rare event.

The term "rare event" is often used to describe an event with a low probability of occurrence.

A rare event typically has a probability of less than 5%, and in this case, the probability is well below that threshold. A probability of less than 0.05 is generally considered a rare event.

The fact that Michele won the grand prize means that she defied the odds, which makes the event rare.

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It should be 33.49 sorry if I’m a little off!

cosh z = cos(iz) is true for all complex numbers z. The solutions to cosh z = 0 are z = (2n + 1)πi/2, where n is an integer.

To show that cosh z = cos(iz) is true for all complex numbers z, we can start by expressing the definitions of cosh z and cos(iz) in terms of exponentials. The hyperbolic cosine function is defined as cosh z = (e^z + e^(-z))/2, and the cosine function of the imaginary part of z is cos(iz) = (e^(iz) + e^(-iz))/2.

By substituting iz for z in the definition of cosh z, we get cosh(iz) = (e^(iz) + e^(-iz))/2. Using Euler's formula e^(ix) = cos(x) + isin(x), we can rewrite this expression as cosh(iz) = cos(z)/2 + i(sin(z)/2).

Now, let's express cos(iz) using Euler's formula as cos(iz) = cos(-z)/2 + i(sin(-z)/2) = cos(z)/2 - i(sin(z)/2).

We can observe that cosh(iz) and cos(iz) have the same real part (cos(z)/2) and differ only in the sign of the imaginary part. Therefore, cosh z = cos(iz) holds true for all complex numbers z.

To solve cosh z = 0, we set cosh z equal to zero and solve for z. The equation cosh z = 0 implies that (e^z + e^(-z))/2 = 0. Multiplying both sides by 2 and rearranging, we have e^z + e^(-z) = 0.

Let's substitute e^z with a new variable, say w. The equation becomes w + 1/w = 0, which is a quadratic equation. Multiplying through by w, we get w^2 + 1 = 0. Solving for w, we find w = ±i.

Substituting e^z back in for w, we have e^z = ±i. Taking the natural logarithm of both sides, we get z = ln(±i). Using the properties of the complex logarithm, we have ln(±i) = ln(e^((2n + 1)πi/2)) = (2n + 1)πi/2, where n is an integer.

Therefore, the solutions to cosh z = 0 are z = (2n + 1)πi/2, where n is an integer.

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We know that the diagonals of a square bisect each other which results in the formation of 4 - 90° angles . Which means ZX = 2 × WT . Then ,

= ZX = 2 × 6

= ZX = 12 ( ZX is a diagonal )

Side XY =

XY = √2 d/2

XY = √2 . 12/2

XY = 8.48528

m∠WTZ = 90° ( as the diagonals of a square intersect to form a complete angle ( 360° ) which is divided into four equal parts , each part being 90° )

m∠WTZ = 90°

m∠WYX = 45° ( WYX is half of one of the interior angles in this square which will mean it will be equal to 45° )

Therefore , m∠WYX = 45°

ZX = 12

XY = 8.48528

m∠WTZ = 90°

m∠WYX = 45°

Answer :-

step by step explanation:-

Let the time be x

principal = rs 8000

Amount = rs 13824

rate = 20% p.a

A = P(1+r/100)^n

13824 = 8000 ( 1 + 20/100)^n

=> 13824/8000 = (120/100)^n

=> 13824/8000 = (24/20)^n

=> (24/20)³ = (24/20)^n

=> 3 = n

=> n = 3 years

The statement ∀x(∃y(y^2 = x) → x > 0) is true, as every real number x with a corresponding y such that y^2 = x is greater than 0. The statement ∃x, y, z(x + y = xz) is false, as it does not hold for all real numbers.

∀x(∃y(y^2 = x) → x > 0):

This statement asserts that for every real number x, if there exists a real number y such that y^2 = x, then x must be greater than 0. We can prove this statement to be true. Suppose there exists a real number x such that (∃y)(y^2 = x). In this case, we can take the square root of x, which will yield a real number y such that y^2 = x. Since the square of any real number is non-negative, y^2 = x implies x > 0. Therefore, the statement is true.

∃x, y, z(x + y = xz):

This statement claims the existence of real numbers x, y, and z such that x + y = xz. We can disprove this statement by showing that it is not true for all real numbers. Suppose we choose x = 0, y = 1, and z = 2. Plugging these values into the equation, we get 0 + 1 = 0 * 2, which simplifies to 1 = 0. Since this equation is false, we have found a counterexample that disproves the statement. Hence, the statement is false.

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Answer:

n=15          j=12

Step-by-step explanation:

The scale factor is 1.5

Match up the triangles, and you can see that 12/8=1.5

10*1.5=j, or 15

18/1.5=12