Show your work please help me please

The probability that the newspaper Mario buys has reviewed both performances, given that he buys the first newspaper, is 0.72 or 72%.

a) Let's assume the probability of the first newspaper reviewing his performance is 2/7 and the probability of the second newspaper reviewing his performance is 5/7. The probability of both papers reviewing his performance is (2/7) * (5/7) = 10/49.

b)Since Clarissa buys both newspapers, there are two scenarios: either the first newspaper reviews her performance and the second one doesn't, or the second newspaper reviews her performance and the first one doesn't. The probability of only one paper reviewing her performance is 2 * (3/7) * (4/7) = 24/49.

c) If Mario buys one newspaper at random, there is a 2/7 chance that he buys the first newspaper and a 5/7 chance that he buys the second newspaper. Since each newspaper has reviewed one performance, the probability that the newspaper he buys has reviewed both performances is (2/7) * (5/7) = 10/49.

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En un número decimal, la parte que está antes del punto decimal se conoce como la parte entera del número, mientras que la parte que está después del punto decimal se conoce como la parte decimal.

La parte entera representa la cantidad completa de unidades o números enteros que se encuentran en el número. Por ejemplo, en el número decimal 12.345, la parte entera es 12, lo que significa que hay 12 unidades completas.

La parte decimal representa una fracción o porción del número que es menor que uno. En el ejemplo anterior, la parte decimal es .345, lo que significa que es una fracción de una unidad. La parte decimal se puede expresar en forma de fracción o como un número decimal.

La relación entre la parte entera y la parte decimal es que, juntas, representan el número decimal completo. Por ejemplo, el número decimal 12.345 se compone de la parte entera 12 y la parte decimal .345. La parte entera siempre se coloca a la izquierda del punto decimal, mientras que la parte decimal se coloca a la derecha del punto decimal.

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The trigonometric ratio include the following:

In order to determine each of the trigonometric ratios, we would apply each of the trigonometric ratios because the given side lengths represent the adjacent side, opposite side and hypotenuse of a right-angled triangle.

cos(θ) = Adj/Hyp, sin(θ) = Opp/Hyp, tan(θ) = Opp/Adj

Where:

sin(θ) = Opp/Hyp

sin B = b/c

sin(θ) = Opp/Hyp

sin A = a/c

tan(θ) = Opp/Adj

tan A = a/b

cos(θ) = Adj/Hyp

cos B = a/c

cos(θ) = Adj/Hyp

cos A = b/c

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Complete Question:

Use the right triangle to determine the each of the trigonometric ratio.

The universal class is considered a proper class because it contains all sets but cannot itself be treated as a set due to the limitations imposed by set theory. Recognizing it as a proper class ensures the consistency and avoids the paradoxes that would arise if it were treated as a set.

In set theory, classes are collections of objects that share a common property or satisfy a specific condition. A proper class is a class that is not a set, meaning it cannot be treated as an object within the theory of sets. One example of a proper class is the universal class, denoted as "V" in Von Neumann–Bernays–Gödel set theory (NBG) or "U" in Morse–Kelley set theory (MK).

The universal class is defined as the class that contains all sets. It encompasses every possible set that can be constructed within the theory of sets. It is important to note that the universal class itself is not considered a set because it would lead to logical inconsistencies and paradoxes, such as Russell's paradox.

To understand why the universal class is considered a proper class, we need to delve into the foundations of set theory. In most set theories, including Zermelo-Fraenkel set theory (ZF), which is the most commonly used, there is a principle called the limitation of size. This principle ensures that only sets of a certain size can be considered objects within the theory.

The limitation of size prevents the universal class from being considered a set because it would violate the size restrictions. If the universal class were a set, it would contain itself as an element, leading to the well-known Russell's paradox. Russell's paradox arises from the assumption that there exists a set of all sets that do not contain themselves as elements. If the universal class were a set, it would both contain and not contain itself, creating a contradiction.

To avoid such paradoxes and maintain consistency within set theory, the universal class is treated as a proper class. This means that it is not regarded as an object that can be manipulated or operated upon like a set. Instead, it serves as a collection that encompasses all sets, acting as a foundational concept within set theory.

In summary, the universal class is considered a proper class because it contains all sets but cannot itself be treated as a set due to the limitations imposed by set theory. Recognizing it as a proper class ensures the consistency and avoids the paradoxes that would arise if it were treated as a set.

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Step-by-step explanation:

1/2×13×12= 78

12×14= 132

13×14= 182

total surface area

= (78×2)+(132×2)+182

= 602 ft^2

The value of cos (A+B) in simplest form is -136/305.

The standard formula for the cosine of the sum of two angles is:

cos(A+B) = cos A cos B - sin A sin B

We are given that cos A = 11/61 and sin B = 3/5. We can use this information to find the values of sin A and cos B by using the Pythagorean identity:

sin^2 A +[tex]cos^2[/tex]A = 1  =>  sin A = [tex]\sqrt(1 - cos^2[/tex] A)

cos^2 B + [tex]sin^2[/tex] B = 1 => cos B = [tex]\sqrt(1 - sin^2 B)[/tex]

Substituting the given values, we get:

sin A =[tex]\sqrt(1 - (11/61)^2) ~~ 60/61[/tex]

cos B = [tex]\sqrt(1 - (3/5)^2) = 4/5[/tex]

Now we can use these values to find the cosine of the sum of the angles:

cos(A+B) = cos A cos B - sin A sin B

= (11/61)(4/5) - (60/61)(3/5)

= (44/305) - (36/61)

= (44 - 180)/305

= -136/305

Note that since A and B are positive acute angles, A+B is also an acute angle, which means that its cosine is negative.

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The measures of the angles are ∠PQR = 96° and ∠PSR = 84°

Given is a cyclic quadrilateral, with angles ∠PQR = (7x - 2)° and ∠PSR = (5x+14)°.

We need to find the measure of the angles PQR and PSR,

So,

We know that the cyclic quadrilaterals have their opposite angles supplementary,

So,

∠PQR + ∠PSR = 180°

7x - 2 + 5x + 14 = 180°

12x + 12 = 180°

12x = 168°

x = 14°

Put the value of x in the angles, we get,

∠PQR = (7x - 2)°

∠PQR = 7 × 14 - 2

∠PQR = 96°

And,

∠PSR = 5 × 14 - 2

∠PSR = 84°

Hence the measures of the angles are ∠PQR = 96° and ∠PSR = 84°

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The solution of the equation 2|2x - 1| - 8 = 18 is x = 7 or x = -6.

The solution of the equation 2|2x - 1| - 8 = 18 can be obtained as follows:

Step 1: Add 8 to both sides of the equation2|2x - 1| = 26

Step 2: Divide both sides of the equation by 2|2x - 1| / 2 = 26 / 2|2x - 1| = 13

Step 3: The absolute value of a number is always positive, so we can divide the equation into two separate equations.|2x - 1| = 13 or -|2x - 1| = 13

Step 4: Solve for x in each equation.

Solution for the first equation:|2x - 1| = 132x - 1 = 13 or 2x - 1 = -13 2x = 14 or 2x = -12 x = 7 or x = -6

Solution for the second equation:-|2x - 1| = 13|2x - 1| = -13

There is no solution to this equation because the absolute value of a number cannot be negative.

Therefore, the solution of the equation 2|2x - 1| - 8 = 18 is x = 7 or x = -6.

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The equation for the water level, L, after d days can be written as:

L = 340 - 0.5d

To find in how many days the water level will be 260m, we can substitute L = 260 into the equation and solve for d:

260 = 340 - 0.5d

0.5d = 340 - 260

0.5d = 80

d = 80 / 0.5

d = 160

Therefore, the water level will be 260m after 160 days.

Answer:

Minimum value = -6

Step-by-step explanation:

The minimum value is the y-coordinate of the minimum on the parabola.

Currently 3/4x^2 + 6x + 6 is in standard form, whose general equation is:

ax^2 + bx + c

Step 1:  We can first find the x-coordinate of the maximum using the following formula:

-b / 2a

From the function, we see that 6 is b and 3/4 is a.  Now, we plug these values into the formula and simplify:

-6 / (2 * 3/4)

-6 / (6/4)

-6 / (3/2)

-6 * 2/3

-12/3

-4

Step 2:  Now we can plug in -4 for x in the function to find the minimum value:

y = 3/4(-4)^2 + 6(-4) + 6

y = 3/4(16) -24 + 6

y = 48/4 - 24 + 6

y = 12 - 24 + 6

y = -12 + 6

y = -6

Thus, the minimum value of the function y = 3/4x^2 + 6x + 6 is -6

Answer:

-3125

Step-by-step explanation:

To find the value of a₆, we can use the given recursive formula and substitute the values for a₁ and the preceding terms until we reach the desired term.

Given:

a₁ = 1

aₙ = -5aₙ₋₁

Let's calculate the values step by step:

a₂ = -5a₁ = -5(1) = -5

a₃ = -5a₂ = -5(-5) = 25

a₄ = -5a₃ = -5(25) = -125

a₅ = -5a₄ = -5(-125) = 625

a₆ = -5a₅ = -5(625) = -3125

Therefore, the value of a₆ is -3125.

Hope this helps!

From the given system of linear equations there are no solutions.

The given system of equations are 5x+y=3 and 10x+2y=-6.

Here, 5x+y=3 -----(i) and 10x+2y=-6 ⇒ 5x+y=-3 --------(ii)

Subtract equation (i) from equation (ii), we get

5x+y-(5x+y)=-3-3

5x+y-5x-y=-6

0≠ -6

There are no solutions

Hence, from the given system of linear equations there are no solutions.

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Answer:

Step-by-step explanation:

After 9 months, both Plant A and Plant B will be 172 cm tall.

To find the number of months it will take for Plant A and Plant B to be the same height, we need to set up an equation. Let's use the variable "t" to represent the number of months.

The height of Plant A after t months can be represented as: 64 + 12t

The height of Plant B after t months can be represented as: 28 + 16t

To find the number of months when both plants will be the same height, we set the two expressions equal to each other:

64 + 12t = 28 + 16t

Simplifying the equation:

12t - 16t = 28 - 64

-4t = -36

Dividing both sides of the equation by -4:

t = -36 / -4

t = 9

Therefore, after 9 months, Plant A and Plant B will be the same height. To find the height they will reach at that time, we substitute t = 9 into either equation. Let's use the equation for Plant A:

Height of Plant A after 9 months = 64 + 12 * 9

= 64 + 108

= 172 cm

Therefore, after 9 months, both Plant A and Plant B will be 172 cm tall.

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Let's call the amount Susan invested in the fund that paid a 12% profit "x" and the amount she invested in the stock that suffered a 4% loss "y".

We know that Susan's overall net profit was $2,160, so the total amount of profit she earned from both investments was $2,160.

The profit from the investment in the fund that paid a 12% profit was 0.12x (since the profit rate was 12%). The loss from the investment in the stock that suffered a 4% loss was -0.04y (since the profit rate was negative 4%).

We can set up two equations based on this information:

The total amount invested was $22,000:

x + y = 22,000

The total profit was $2,160:

0.12x - 0.04y = 2,160

We can use the first equation to solve for one of the variables in terms of the other:

x = 22,000 - y

Now we can substitute this expression for "x" into the second equation:

0.12(22,000 - y) - 0.04y = 2,160

Simplifying:

2,640 - 0.12y - 0.04y = 2,160

Combining like terms:

2,640 - 0.16y = 2,160Subtracting 2,640 from both sides:

-0.16y = -480

Dividing both sides by -0.16:

y = 3,000

So Susan invested $3,000 in the stock that suffered a 4% loss.

To find the amount she invested in the fund that paid a 12% profit, we can substitute this value for "y" in one of the equations we set up earlier:

x + y = 22,000

x + 3,000 = 22,000

x = 19,000

So Susan invested $19,000 in the fund that paid a 12% profit.

Therefore, the amount invested at 12% was $19,000 and the amount invested in stock was $3,000.

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The amount invested in the in account that had an 12% profit is $19,000

The amount invested in the in account that had a 4% loss is $3,000.

[tex]\sf a + b = 22,000[/tex] equation 1

[tex]\sf 0.12a - 0.04b = 2160[/tex] equation 2

Where:

Multiply equation 1 by 0.12

[tex]\sf 0.12a + 0.12b = 2640[/tex] equation 3

Subtract equation 2 from equation 3

[tex]\sf 0.16b = 480[/tex]

Divide both sides of the equation by 0.16

[tex]\sf b = 3000[/tex]

Subtract 3,000 from 22,000

[tex]\sf a = 22,000 - 3,000 = 19,000[/tex]

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If jill has graded 45% of her assessments and she has 27 assessments graded then total assessments are 60

The total number of assessments Jill needs to grade as "x".

According to the information provided, Jill has already graded 45% of her assessments, which is equivalent to 0.45 when expressed as a decimal.

So, we can set up the proportion:

(graded assessments) / (total assessments) = (graded percentage) / 100

Substituting the known values:

27 / x = 45 / 100

To solve for x, we can cross-multiply and solve the resulting equation:

100 × 27 = 45×x

2700 = 45x

Divide both sides of the equation by 45:

2700 / 45 = x

60 = x

Therefore, Jill needs to grade a total of 60 assessments.

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Answer:

the cost to rent one water tank is $226.08.

Step-by-step explanation:

Given:

Radius (r) = 3 feet

Height (h) = 2 feet

Cost per cubic foot = $4

The volume of a cylinder is given by the formula:

Volume = π * r^2 * h

Substituting the given values:

Volume = 3.14 * (3 feet)^2 * 2 feet

Volume = 3.14 * 9 square feet * 2 feet

Volume = 56.52 cubic feet

The cost to rent one water tank is calculated by multiplying the volume by the cost per cubic foot:

Cost = Volume * Cost per cubic foot

Cost = 56.52 cubic feet * $4

Cost = $226.08

Step-by-step explanation:

To find the cost to rent one water tank, we need to calculate its volume and then multiply it by the rental cost per cubic foot.

The formula for the volume of a cylinder is given by:

V = πr²h

where V is the volume, π is a mathematical constant approximately equal to 3.14, r is the radius of the cylinder, and h is the height of the cylinder.

In this case, the radius (r) is 3 feet, and the height (h) is 2 feet. Plugging these values into the formula, we get:

V = 3.14 * (3^2) * 2

V = 3.14 * 9 * 2

V = 56.52 cubic feet

Now, we can multiply the volume by the rental cost per cubic foot:

Cost = V * $4

Cost = 56.52 * $4

Cost = $226.08

Therefore, the cost to rent one water tank is $226.08.

Answer:

1048 student tickets were sold and 1452 adult tickets were sold

Step-by-step explanation:

We can use a system of equations to find the quantity of both the adult and student tickets.  We can allow A to represent the quantity of adult tickets and S to represent the quantity of student tickets.

First equation:  (For the context of my explanation, revenue is defined as the product of price of an item and the quantity) We know that the sum of the revenues earned from the adult tickets equals the total revenue as

(price of adult tickets * quantity of adult tickets) + (price of student tickets * quantity of student tickets) = total revenue

Since we know that the adult tickets cost $3, the student tickets cost $2, and the total revenue earned was $6452, our first equation is:

3A + 2S = 6452

Second Equation:  We further know that the sum of the quantities of adult and student tickets equals the total amount of tickets sold as

quantity of adult tickets + quantity of student tickets = total amount of tickets sold

Since we know that there were 2500 fans in the gym, our second equation is:

A + S = 2500

Method to solve:  We can isolate A in the second equation by subtracting S from both sides.  This will allow us to substitute it in the first equation to first solve for S , the quantity of student tickets sold:

Step 1:  Isolating S in second equation:

(A + S = 2500) - S

A = -S + 2500

Step 2:  Plugging in (substituting) A = -S + 2500 for A in 3A + 2S = 6452:

3(-S + 2500) + 2S = 6452

-3S + 7500 + 2S = 6452

-S + 7500 = 6452

-S = -1048

S = 1048

Now that we know the quantity of student tickets sold was 1048, we can plug in 1048 for S in any of the two equations in our system to solve for A, the quantity of adult tickets sold.  Let's use the first equation:

Step 3:  Plugging in 1048 for S in 3A + 2S = 6452

3A + 2(1048) = 6452

3A + 2096 = 6452

3A = 4356

A = 1452

Thus, the quantity of adult tickets sold was 1452.

Optional Step 4:  We can check that we've found the correct answers by plugging in 1048 for S and 1452 for A in both equations in our system and checking that we get 6452 for the first equation and 2500 for the second equation:

Plugging in 1048 for S and 1452 for A in 3A + 2S = 6452 (i.e., the first equation in our system):

3(1452) + 2(1048) = 6452

4356 + 2096 = 6452

6452

Plugging in 1048 for S and 1452 for a in A + S = 2500 (i.e., the second equation in our system):

1048 + 1452 = 2500

2500 = 2500

A. Matrix AB is [tex]\left[\begin{array}{cc}24&37\\6&25\end{array}\right][/tex]

B. Matrix BA is [tex]\left[\begin{array}{cc}10&-12\\-1&39\end{array}\right][/tex]

C. No, in general, AB does not always equal BA.

A. To find AB we say [tex]\left[\begin{array}{cc}-3&5\\1&3\end{array}\right][/tex] × [tex]\left[\begin{array}{cc}-3&1\\3&8\end{array}\right][/tex]

which becomes [tex]\left[\begin{array}{cc}-3*-3 + 5*3& -3*3 + 5*8\\1*3 +3*3&1*1 + 8*3\end{array}\right][/tex] ⇒   [tex]\left[\begin{array}{cc}24&37\\6&25\end{array}\right][/tex]

B. To find Matrix BA  [tex]\left[\begin{array}{cc}-3&1\\3&8\end{array}\right][/tex] ×  [tex]\left[\begin{array}{cc}-3&5\\1&3\end{array}\right][/tex]

Which becomes [tex]\left[\begin{array}{cc}-3*3 + 1*1 &-3*5 + 1*3\\3*-3 + 8*1&3*5 + 8*3\end{array}\right][/tex] ⇒  [tex]\left[\begin{array}{cc}10&-12\\-1&39\end{array}\right][/tex]

C. For matrices A and B such that AB and BA both exist, AB will equal BA if and only if the matrices commute. A matrix commutes with another matrix if the order in which they are multiplied does not matter.

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Answer: The values of the six trigonometric functions for the angle formed by the positive x-axis and the line segment from (0,0) to (4,-3) are:

sinθ = -3/5

cosθ = 4/5

tanθ = -3/4

cscθ = -5/3

secθ = 5/4

cotθ = -4/3

Step-by-step explanation: To find the values of the six trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent) of an angle formed by the positive x-axis and a line segment, we need to determine the lengths of the sides of the right triangle formed.

In this case, the line segment goes from (0,0) to (4,-3). The horizontal length (adjacent side) is 4 units, and the vertical length (opposite side) is -3 units (since it goes downward).

Now, we can calculate the trigonometric functions:

Sine (sinθ) = opposite/hypotenuse = -3/5

Cosine (cosθ) = adjacent/hypotenuse = 4/5

Tangent (tanθ) = opposite/adjacent = -3/4

Cosecant (cscθ) = 1/sinθ = -5/3

Secant (secθ) = 1/cosθ = 5/4

Cotangent (cotθ) = 1/tanθ = -4/3

Therefore, the values of the six trigonometric functions for the angle formed by the positive x-axis and the line segment from (0,0) to (4,-3) are:

sinθ = -3/5

cosθ = 4/5

tanθ = -3/4

cscθ = -5/3

secθ = 5/4

cotθ = -4/3

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Answer:

11.8 cm²

Step-by-step explanation:

Cosine rule: a ² = b ² + c ²  - 2bc COS A

COS A = (b ²  +  c ²   -  a ²) / (2bc)

Area of triangle = ½ ab sin C

name the angle that joins the '3' side to either of the '8' sides. label that angle as A.

now, Cos A = (8² + 3² - 8²) / [(2)(3)(8)]

= 3/16.

A = inverse Cos (3/16)

= 79.193°.

Area of triangle = 1/2 ab Sin C (we'll just use letter A instead of C)

= 1/2 (3)(8) sin (3/16)

= 11.8 cm².

another way you can find it is by this simpler method:

think of one side of 8, the side of 3 and the height that you do not know by using Pythagoras' Theorem for right-angled triangle. (a² + b² = c²). go halfway up the '3' side, then cut across straight to where the '8' sides meet. we now have right-angled triangle. let's call the height you don't know h.

we have 1.5² + h² = 8²

h² = 8² - 1.5²

h² = 61.75

h = √61.75

so area = 1/2 X 1.5 X √61.75

= 11.8 cm²

The expression [2 x (5 + 1)] ÷ 2 evaluates to 6 when applying the appropriate order of operations

When evaluating the expression [2 x (5 + 1)] ÷ 2, the true statement is that the expression simplifies to 6. The order of operations, also known as the PEMDAS rule, guides us in performing the calculations correctly.

According to the PEMDAS rule, we start by evaluating the expression inside the parentheses, which is (5 + 1), resulting in 6. Next, we multiply 2 by the result of the parentheses, giving us 12. Finally, we divide 12 by 2, which yields the final answer of 6.

This process ensures that we follow a consistent and agreed-upon method for evaluating mathematical expressions. By adhering to the PEMDAS rule, we maintain clarity and precision in our calculations, allowing us to obtain the correct result.

In conclusion, the expression [2 x (5 + 1)] ÷ 2 evaluates to 6 when applying the appropriate order of operations.

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The exponential function that could be represented by the graph is given as follows:

[tex]y = 10^x[/tex]

An exponential function has the definition presented as follows:

[tex]y = ab^x[/tex]

In which the parameters are given as follows:

The graph crosses the y-axis at y = 1, hence the parameter a is given as follows:

a = 1.

When x is increased by one, y is multiplied by 10, as we have that when x = 1, y = 5 and when x = 2, y = 50, hence the parameter b is given as follows:

b = 10.

Hence the function is given as follows:

[tex]y = 10^x[/tex]

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The value of x using the theorem of intersecting secants is 10

From the question, we have the following parameters that can be used in our computation:

intersecting secants

Using the intersecting secants equation, we have

8 * (8 + x) = 6 * (6 + 18)

Evaluate the like terms

So, we have

8 * (8 + x) = 6 * 24

Divide both sides by 8

8 + x = 6 * 3

So, we have

8 + x = 18

Subtract 8 from both sides

x = 10

Hence, the value of x is 10

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n³a⁹= (9/64)r³ this equation shows that 125ncubica(a³) = 9 cubic r, as desired by cone and sphere

The volume of a sphere is given by the formula V = (4/3)πr³, where r is the radius of the sphere.

Since we have n ice spheres of radius a

the total volume of the ice spheres is given by V_ice = n × (4/3)πa³.

The volume of a cone is given by the formula V_cone = (1/3)πr²h

where r is the radius of the base and h is the height of the cone.

The slant height of the cone is also given as r.

Using the Pythagorean theorem, we can find the height of the cone, h, in terms of r:

h =√(r² - a²)

Substituting this value of h into the formula for the volume of the cone, we get:

V_cone = (1/3)πr² √(r² - a²)

V_ice = V_cone

n(4/3)πa³= (1/3)πr²√(r² - a²)

Simplifying this equation, we can cancel out the common factors of (1/3)π:

4na³= r²√(r² - a²)

Now, let's cube both sides of the equation to eliminate the square root:

(4na³)³ = (r²√(r² - a²))³

64n³a⁹ = r⁶ (r² - a²)

n³a⁹= (9/64)r³

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The angle m∠BAC is 54 degrees.

If two chords intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle.

Therefore, let's find the angle BAC.

Hence,

10x + 4 = 1 / 2 (12x + 15 + 9x - 12)

10x + 4 = 1 / 2 (21x + 3)

10x + 4 = 10.5x + 1.5

10x - 10.5x = 1.5 - 4

-0.5x = -2.5

divide both sides by -0.5

x = -2.5 / -0.5

x = 5

Therefore,

m∠BAC = 10(5) + 4

m∠BAC = 50 + 4

m∠BAC = 54 degrees

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Probability of selecting an athlete that stretches = 7/16

Percentage of athlete that do not stretch and got injured = 51.33%

a)

From table,

Athletes that stretch = Athletes that got injured while stretching + Athletes that do not got injured while stretching

Athletes that stretch = 55 + 295

Athletes that stretch = 350

Total number of athletes = stretch + does not stretch

total number of athletes = 350 + 450 = 800

Probability of an event to occur = Number of favourable outcomes / Total number of outcomes.

Probability of selecting an athlete that stretches = 350/ 800

Probability of selecting an athlete that stretches = 7/16

b)

Given

Athlete that do not stretch = 450

Athlete that got injured that do not stretch = 231

Now,

Percentage of athlete that do not stretch and got injured = 231/450 × 100

Percentage of athlete that do not stretch and got injured = 51.33%

Hence from the data given in table the required values can be found out.

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The solution to the system of equations is x = 4 and y = 8. The lines represented by the equations y = 2x and 6x - 2y = 8 Intersect at the point (4, 8).

To analyze the given equations:

Equation 1: y = 2x

Equation 2: 6x - 2y = 8

We can begin by examining Equation 1, which is in slope-intercept form (y = mx + b). In this equation, the coefficient of x is 2, representing the slope of the line. Therefore, the line described by Equation 1 has a slope of 2.

Now, let's move on to Equation 2. It can be rewritten by rearranging the terms:

6x - 2y = 8

-2y = -6x + 8

Dividing by -2 on both sides:

y = 3x - 4

By comparing Equation 2 with the slope-intercept form (y = mx + b), we can see that its slope is 3.

Comparing the slopes of the two equations, we observe that they are not equal. Since the slopes are different, the lines represented by the equations y = 2x and y = 3x - 4 are not parallel.

To determine if they intersect, we can equate the right-hand sides of the two equations:

2x = 3x - 4

By rearranging terms, we get:

x = 4

Substituting x = 4 back into Equation 1:

y = 2(4)

y = 8

Therefore, the solution to the system of equations is x = 4 and y = 8. The lines represented by the equations y = 2x and 6x - 2y = 8 intersect at the point (4, 8).

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