suppose that a and b vary inversely and that b = 5/3 when a=9. Write a function that models the inverse variation

Answer: A. 6n-11

Step-by-step explanation:

First, ignore the parenthesis because it is addition and subtraction so they are commutative. 10n-4n = 6n and -8-3 is the same as -8+-3 which is -11. Combining the answer gives 6n-11.

The series Σj=1∞j³-2² is converges.

To find out if the series converges or not, we will use the p-series test.

The p-series test states that if Σj=1∞1/p is less than or equal to 1, then the series Σj=1∞1/jp converges.

If Σj=1∞1/p is greater than 1, then the series Σj=1∞1/jp diverges. If Σj=1∞1/p equals 1, then the test is inconclusive.

Let's apply the p-series test to the given series. p = 3 - 2².

Therefore, 1/p = 1/(3 - 2²). Σj=1∞1/p = Σj=1∞3/[(3 - 2²) × j³].

Using the limit comparison test, we compare the given series with the p-series of the form Σj=1∞1/j³.

Let's take the limit of the ratio of the terms of the two series as j approaches infinity. lim(j→∞)(3/[(3 - 2²) × j³])/(1/j³) = lim(j→∞)3(3²)/(3 - 2²) = 9/5.

Since the limit is a finite positive number, the given series converges by the limit comparison test. Therefore, the series Σj=1∞j³-2² converges.

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Rounding to the nearest thousandth, the solution to the equation ln 2 + ln x = 1 is x ≈ 1.359.

To simplify and solve the equation ln 2 + ln x = 1, we can use the properties of logarithms. First, we can apply the property of logarithmic addition, which states that:

ln(a) + ln(b) = ln(ab)

Using this property, we can rewrite the equation as:

ln(2x) = 1

Next, we can exponentiate both sides of the equation using the property that [tex]e^(ln(x)) = x.[/tex]

Therefore, [tex]e^(ln(2x)) = e^1[/tex], which simplifies to 2x = e.

To solve for x, we divide both sides of the equation by 2:

x = e/2

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The vertices of triangle C'D'E, after reflection are determined as: B. C'(3, 0), D'(7, 1), E'(2, 4)

When a triangle is reflected over the y-axis, the x-coordinates of its vertices are negated while the y-coordinates remain the same.

Given the vertices of triangle CDE as:

C(-3, 0)

D(-7, 1)

E(-2, 4)

To find the vertices of triangle C'D'E, we negate the x-coordinates of each vertex:

C' = (3, 0)

D' = (7, 1)

E' = (2, 4)

Therefore, the vertices of triangle C'D'E are:

B. C'(3, 0), D'(7, 1), E'(2, 4)

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The mean, median, mode and range of the given set of numbers would be 2.821mm, 2.835mm, 2.85mm and 0.12mm respectively.

Given set of numbers is as follows:

{2.81mm, 2.90mm, 2.78mm, 2.85mm, 2.82mm, 2.85mm, 2.81mm, 2.85mm}

To find the mean, median, mode and range of the given set of numbers, we have;

Mean:

To find the mean of the given set of numbers, we add all the numbers and divide by the total number of numbers. Here, we have;2.81+2.90+2.78+2.85+2.82+2.85+2.81+2.85=22.57mm

Now, the total numbers of the given set are 8.

Hence;

Mean=22.57/8= 2.82125mm ≈ 2.821mm

Median:

The median is the middle number when all the numbers are arranged in ascending or descending order. Here, the given set of numbers in ascending order is as follows;

{2.78mm, 2.81mm, 2.81mm, 2.82mm, 2.85mm, 2.85mm, 2.85mm, 2.90mm}

Here, the middle numbers are 2.82mm and 2.85mm.

Hence, the median=(2.82+2.85)/2= 2.835mm

Mode:

The mode is the most frequently occurring number. Here, the number 2.85mm occurs most frequently.

Hence, the mode is 2.85mm

Range:The range of the given set of numbers is the difference between the highest and lowest number in the set. Here, the highest number is 2.90mm and the lowest number is 2.78mm. Hence, the range= 2.90-2.78=0.12mm

Therefore, the mean, median, mode and range of the given set of numbers are as follows:

Mean= 2.821mm

Median= 2.835mm

Mode= 2.85mm

Range= 0.12mm

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We can modify the 'Example3.m' function such that it prints a warning if the entered marks in any subject are less than30% as follows:

2.  Function x = Subject (English, Math, Chemistry)

English = input ('English mark')

Math = input ('Math mark')

Chemistry = input ('Chemistry mark')

if subject < 30 (Warning: Mark is less than 30%. Cannot proceed)

end output;

3. Function x = Example 3

English = input ('English mark')

Maths = input ('Math mark')

Chemistry = input ('Chemistry mark')

x = (English+Maths+Chemistry)/3;

end

To modify the function, we have to input the value as shown above. The next thing to do will be to enter a condition such that if marks represented by y in the above function are less than 30, then the code will be terminated.

Also, the function for average marks can be gotten by inputting the marks and then dividing by the total number.

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Complete Question:

2. Modify 'Example3.m' function such that it prints a warning if the entered marks in any subject are less than \( 30 \% \).

3: Calculate average marks

To modify the 'Example3.m' function to print a warning if the entered marks in any subject are less than 30%, you can add a conditional statement within the code. Here's an example of how you can implement this:

function averageMarks = Example3(marks)

   % Check if any subject marks are less than 30%

   if any(marks < 0.3)

       warning('Some subject marks are less than 30%.');

   end

   % Calculate the average marks

   averageMarks = mean(marks);

end

In this modified version, the `if` statement checks if any marks in the `marks` array are less than 0.3 (30%). If this condition is true, it prints a warning message using the `warning` function. Otherwise, it proceeds to calculate the average marks as before.

Make sure to replace the original 'Example3.m' function code with this modified version in order to incorporate the warning functionality.

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Answer:

What are you trying to find???

Step-by-step explanation:

If it is median, then it is the line in the middle of the box, which is on 19.

The function f(x) = x/(x-1) is neither surjective nor injective.

To determine whether the function f(x) = x/(x-1) is surjective, injective, or neither, let's analyze each property separately:

1. Surjective (Onto):

A function is surjective (onto) if every element in the codomain has at least one preimage in the domain. In other words, for every y in the codomain, there exists an x in the domain such that f(x) = y.

Let's consider the function f(x) = x/(x-1):

For f(x) to be surjective, every real number y in the codomain (R) should have a preimage x such that f(x) = y. However, there is an exception in this case. The function has a vertical asymptote at x = 1 since f(1) is undefined (division by zero). As a result, the function cannot attain the value y = 1.

Therefore, the function f(x) = x/(x-1) is not surjective (onto).

2. Injective (One-to-One):

A function is injective (one-to-one) if distinct elements in the domain map to distinct elements in the codomain. In other words, for any two different values x1 and x2 in the domain, f(x1) will not be equal to f(x2).

Let's consider the function f(x) = x/(x-1):

Suppose we have two distinct values x1 and x2 in the domain such that x1 ≠ x2. We need to determine if f(x1) = f(x2) or f(x1) ≠ f(x2).

If f(x1) = f(x2), then we have:

x1/(x1-1) = x2/(x2-1)

Cross-multiplying:

x1(x2-1) = x2(x1-1)

Expanding and simplifying:

x1x2 - x1 = x2x1 - x2

x1x2 - x1 = x1x2 - x2

x1 = x2

This shows that if x1 ≠ x2, then f(x1) ≠ f(x2). Therefore, the function f(x) = x/(x-1) is injective (one-to-one).

In summary:

- The function f(x) = x/(x-1) is not surjective (onto) because it cannot attain the value y = 1 due to the vertical asymptote at x = 1.

- The function f(x) = x/(x-1) is injective (one-to-one) as distinct values in the domain map to distinct values in the codomain, except for the undefined point at x = 1.

Thus, the function f(x) = x/(x-1) is neither surjective nor injective.

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Answer:

the answer is -109

Step-by-step explanation:

To factorize 4x2 + 9x - 13 completely, we will make use of splitting the middle term method. Let's start by multiplying the coefficient of the x2 term and the constant

term 4(-13) = -52. Our aim is to find two

numbers that multiply to give -52 and add up to 9. The numbers are +13 and

-4Therefore, 4x2 + 13x - 4x - 13 = ONow,

group the first two terms together and the last two terms together and factorize them out4x(x + 13/4) - 1(× + 13/4) = 0(x + 13/4)(4x - 1)

= OTherefore, the fully factorised form of 4x2 + 9x - 13 is (x + 13/4)(4x - 1).

If det(A)= -1, then A is a singular matrix is true.

Singular matrices are matrices whose determinant is zero. A non-singular matrix is one whose determinant is non-zero or whose inverse exists. A matrix is invertible if and only if its determinant is not zero. A square matrix whose determinant is equal to zero is known as a singular matrix. It is not possible to obtain its inverse since it does not exist because det(A) = 0 and the matrix has infinite solutions. The determinant of a matrix A can be represented by det(A) or |A|. det(A) is defined as follows:

If det(A)= -1, then A is a singular matrix.

Hence, the statement det(A)= -1, then A is a singular matrix is true.

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(a) p: "There is one and only one real solution to the equation [tex]x^2[/tex]."

(b) p -> q: "If it is sunny, then I will go for a walk."

(c) r: "Either I will go shopping or I will stay at home."

(d) "If it is sunny, then I will go for a walk."

(e) "I will go shopping or I will stay at home."

(f) p(a): "A is a prime number."

(a) Let p be the proposition "There is one and only one real solution to the equation [tex]x^2[/tex]."

Propositional logic representation: p

(b) q: "If it is sunny, then I will go for a walk."

Propositional logic representation: p -> q

(c) r: "Either I will go shopping or I will stay at home."

Propositional logic representation: r

(d) "If it is sunny, then I will go for a walk."

English representation: If it is sunny, I will go for a walk.

(e) "I will go shopping or I will stay at home."

English representation: I will either go shopping or stay at home.

(f) p(a): "A is a prime number."

Propositional logic representation: p(a)

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a) Vector (0,2,1) can be expressed as a linear combination of u and v.

b) Vector (2,1,8) cannot be expressed as a linear combination of u and v.

c) Vector (0,0,0) can be expressed as a linear combination of u and v.

To determine if a vector can be expressed as a linear combination of u and v, we need to check if there exist scalars such that the equation a*u + b*v = vector holds true.

a) For vector (0,2,1):

We can solve the equation a*(0,1,2) + b*(-1,2,1) = (0,2,1) for scalars a and b. By setting up the system of equations and solving, we find that a = 1 and b = 2 satisfy the equation. Therefore, vector (0,2,1) can be expressed as a linear combination of u and v.

b) For vector (2,1,8):

We set up the equation a*(0,1,2) + b*(-1,2,1) = (2,1,8) and try to solve for a and b. However, upon solving the system of equations, we find that there are no scalars a and b that satisfy the equation. Therefore, vector (2,1,8) cannot be expressed as a linear combination of u and v.

c) For vector (0,0,0):

We set up the equation a*(0,1,2) + b*(-1,2,1) = (0,0,0) and solve for a and b. In this case, we can observe that setting a = 0 and b = 0 satisfies the equation. Hence, vector (0,0,0) can be expressed as a linear combination of u and v.

In summary, vector (0,2,1) and vector (0,0,0) can be expressed as linear combinations of u and v, while vector (2,1,8) cannot.

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The constant value under the radical sign is 2.

We are given the radical expression

√50x⁵ y³ z

which we have to simplify it as much as possible. The constant value under the radical sign can also be found in the simplified expression. We know that

[tex]$\sqrt{a^2b}=\left|a\right|\sqrt{b}$[/tex] for all a and b ≥ 0.

Firstly, we factorize 50x⁵ as:

[tex]$$50x^5=2\cdot 5^2\cdot x^5x^{2}[/tex]

       [tex]= 2\cdot 5^2\cdot (x^2)^2\cdot x$$[/tex]

So,

[tex]$$\sqrt{50x^5y^3z}=\sqrt{2\cdot 5^2\cdot (x^2)^2\cdt x\cdot y^2\cdot y\cdot z}$$[/tex]

Next, using the properties of radicals, we can split the expression as follows:

[tex]$$\sqrt{2}\cdot 5\cdot (x^2)\cdot \sqrt{xyz}$$[/tex]

Now, we have to check if there are any other perfect square factors inside the radical sign. We know that:

[tex]$x^2 = x\cdot x$[/tex]

hence,

[tex]$$\sqrt{2}\cdot 5\cdot x\cdot x\cdot \sqrt{yz}=\sqrt{2}\cdot 5x^2\cdot \sqrt{yz}$$[/tex]

Therefore, the radical expression [tex]$\sqrt{50x^5y^3z}$[/tex] is simplified as [tex]$\sqrt{2}\cdot 5x^2\cdot \sqrt{yz}$[/tex].

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To find the battery life of the latest model 10 months from now, we need to calculate the cumulative increase in battery life over the 10-month period.

The battery life of each model increases by 4% compared to the previous model. Therefore, the battery life of the second model is [tex]\displaystyle 100\% + \dfrac{4}{100} = 104\%[/tex] of the first model's battery life. Similarly, the battery life of the third model is [tex]\displaystyle 104\% + \dfrac{4}{100} = 108.16\%[/tex] of the second model's battery life, and so on.

Using this pattern, the battery life of the latest model 10 months from now can be calculated as follows:

[tex]\displaystyle 632.0 \, \text{minutes} \times \left(1 + \dfrac{4}{100}\right)^{10}[/tex]

Simplifying this expression, we get:

[tex]\displaystyle 632.0 \times \left(1.04\right)^{10}[/tex]

Calculating this expression, we find that the latest model's battery life 10 months from now is approximately 1057.1 minutes.

Therefore, the correct answer is A) 1057.1.

[tex]\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}[/tex]

♥️ [tex]\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}[/tex]

The set of solutions to the homogeneous system forms a linear subspace of R³, since it can be expressed as a linear combination of vectors with a parameter t.

To solve the homogeneous system of linear equations:

3x₁ - x₂ + x₃ = 0

-x₁ + 7x₂ - 2x₃ = 0

2x₁ + 6x₂ - x₃ = 0

We can rewrite the system in matrix form as AX = 0, where A is the coefficient matrix and X is the vector of variables:

A = [[3, -1, 1], [-1, 7, -2], [2, 6, -1]]

X = [x₁, x₂, x₃]

To find the solutions, we need to find the null space of the matrix A, which corresponds to the vectors X that satisfy AX = 0.

By performing Gaussian elimination on the augmented matrix [A|0] and row reducing it to reduced row-echelon form, we obtain:

[[1, 0, -1/3, 0], [0, 1, 1/3, 0], [0, 0, 0, 0]]

This shows that the system has infinitely many solutions and can be parameterized by setting x₃ = t, where t is a parameter. The solutions can then be expressed as:

x₁ = t/3

x₂ = -t/3

x₃ = t

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Alpha can be interpreted as false positives or the significance level in hypothesis testing.

Alpha refers to the significance level in hypothesis testing, which is the predetermined threshold used to determine whether to reject the null hypothesis.

It represents the probability of rejecting the null hypothesis when it is actually true, leading to a Type I error.

A false positive occurs when the test incorrectly concludes that there is a significant effect or relationship, even though it does not exist in reality.

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To solve the equation -1 = 6 sin(2t), we can apply our knowledge of solving cosine equations to solve it. The reason is that the sine function is closely related to the cosine function.

We can use a trigonometric identity to convert the sine equation into a cosine equation.

The trigonometric identity we can use is sin²θ + cos²θ = 1. By rearranging this identity, we get cos²θ = 1 - sin²θ. We can substitute this expression into our equation to obtain a cosine equation.

-1 = 6 sin(2t)

-1 = 6 * √(1 - cos²(2t))  [Using the identity cos²θ = 1 - sin²θ]

-1 = 6 * √(1 - cos²(2t))

Now we have a cosine equation that we can solve. Let's denote cos(2t) as x:

-1 = 6 * √(1 - x²)

Squaring both sides of the equation to eliminate the square root:

1 = 36(1 - x²)

36x² = 36 - 1

36x² = 35

x² = 35/36

Taking the square root of both sides:

x = ±√(35/36)

Now that we have the value of x, we can find the values of 2t by taking the inverse cosine:

cos(2t) = ±√(35/36)

2t = ±cos⁻¹(√(35/36))

t = ±(1/2)cos⁻¹(√(35/36))

So, we have solved the equation -1 = 6 sin(2t) by converting it into a cosine equation. This demonstrates how we can apply our knowledge of solving cosine equations to solve sine equations by using trigonometric identities and the relationship between the sine and cosine functions.

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The zeros of the function in the given graph are x = 0 and x = 5

The zeros of a function on a graph, also known as the x-intercepts or roots, are the points where the graph intersects the x-axis. Mathematically, the zeros of a function f(x) are the values of x for which f(x) equals zero.

In other words, if you plot the graph of a function on a coordinate plane, the zeros of the function are the x-values at which the corresponding y-values are equal to zero. These points represent the locations where the function crosses or touches the x-axis.

Finding the zeros of a function is important because it helps determine the points where the function changes signs or crosses the x-axis, which can provide valuable information about the behavior and properties of the function.

The zeros of the function of this graph is at point x = 0 and x = 5

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The values of FAB, FBC, and FBE can be expressed in terms of Fbx and Fby as follows:

Given equations are:

To solve the given system of equations such that Fab, Fbc, and Fbe are in terms of only Fbx and Fby, we need to substitute the values of FBC and FBE in terms of Fbx and Fby in equation (1).

Substituting the value of FBC from equation (2) into equation (1), we get:

FAB = FBX - (4/5)((125/68)(196/75FBy - 32/25FBX + 138/125FBE) + (125/432)(189/50FBX - 74/125((125/68)(196/75FBy - 32/25FBX + 138/125FBE)) - 5/2FAB))

Simplifying the above equation, we get:

FAB = (35/54)FBX - (196/375)FBy - (69/200)FBE

Therefore, FAB is in terms of Fbx, Fby, and Fbe.

We can also substitute the values of FAB and FBE in terms of Fbx and Fby in equation (2). Substituting the values of FAB and FBE in equation (2), we get:

FBC = (125/68)(196/75FBy - 32/25FBX + 138/125((125/432)(189/50FBX - 74/125((125/68)(196/75FBy - 32/25FBX + 138/125FBE)) - 5/2((35/54)FBX - (196/375)FBy - (69/200)FBE)))

Simplifying the above equation, we get:

FBC = (5/68)FBX + (49/300)FBy - (1/27)FBE

Therefore, FBC is in terms of Fbx, Fby, and Fbe.

Similarly, substituting the values of FAB and FBC in terms of Fbx and Fby in equation (3), we get:

FBE = (25/432)FBX - (49/300)FBy + (1/27)((125/68)(196/75FBy - 32/25FBX + 138/125((35/54)FBX - (196/375)FBy - (69/200)FBE)))

Simplifying the above equation, we get:

FBE = (25/432)FBX - (49/300)FBy + (7/108)FBE

Therefore, FBE is in terms of Fbx and Fby.

Hence, we have obtained the values of FAB, FBC, and FBE in terms of only Fbx and Fby.

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The given system of differential equations is:

Jx' = Ax + By

y' = Cx + Dy

To find the solution to the given system of differential equations, let's first rewrite the system in matrix form:

Jx' = A*x + B*y

y' = C*x + D*y

where

J = [-67 -210]

A = [21 66]

B = [0]

C = [0]

D = [1]

Now, let's solve the system using the initial conditions. We'll differentiate both sides of the second equation with respect to t:

y' = C*x + D*y

y'' = C*x' + D*y'

Substituting the values of C, x', and y' from the first equation, we have:

y'' = 0*x + 1*y' = y'

Now, we have a second-order ordinary differential equation for y(t):

y'' - y' = 0

This is a homogeneous linear differential equation with constant coefficients. The characteristic equation is:

r^2 - r = 0

Factoring the equation, we have:

r(r - 1) = 0

So, the solutions for r are r = 0 and r = 1.

Therefore, the general solution for y(t) is:

y(t) = c1*e^0 + c2*e^t

y(t) = c1 + c2*e^t

Now, let's solve for c1 and c2 using the initial conditions:

At t = 0, y(0) = -5:

-5 = c1 + c2*e^0

-5 = c1 + c2

At t = 0, y'(0) = 17:

17 = c2*e^0

17 = c2

From the second equation, we find that c2 = 17. Substituting this into the first equation, we get:

-5 = c1 + 17

c1 = -22

So, the particular solution for y(t) is:

y(t) = -22 + 17*e^t

Now, let's solve for x(t) using the first equation:

Jx' = A*x + B*y

Substituting the values of J, A, B, and y(t), we have:

[-67 -210] * x' = [21 66] * x + [0] * (-22 + 17*e^t)

[-67 -210] * x' = [21 66] * x - [0]

[-67 -210] * x' = [21 66] * x

Now, let's solve this system of linear equations for x(t). However, we can see that the second equation is a multiple of the first equation, so it doesn't provide any new information. Therefore, we can ignore the second equation.

Simplifying the first equation, we have:

-67 * x' - 210 * x' = 21 * x

Combining like terms:

-277 * x' = 21 * x

Dividing both sides by -277:

x' = -21/277 * x

Integrating both sides with respect to t:

∫(1/x) dx = ∫(-21/277) dt

ln|x| = (-21/277) * t + C

Taking the exponential of both sides:

|x| = e^((-21/277) * t + C)

Since x can be positive or negative, we have two cases:

Case 1: x > 0

x = e^((-21/277) * t + C)

Case 2: x < 0

x = -e^((-21/277) * t + C)

Therefore, the solution to the

given system of differential equations is:

x(t) = C1 * e^((-21/277) * t) for x > 0

x(t) = -C2 * e^((-21/277) * t) for x < 0

y(t) = -22 + 17 * e^t

where C1 and C2 are constants determined by additional initial conditions or boundary conditions.

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The quadratic function has only one x-intercept if m = 3.

A quadratic function of the form:

y = ax² + bx + c

Has one solution only if the discriminant D = b² -4ac is equal to zero.

Here the quadratic function is:

y = x² + 10x + 28 - m

The discriminant is:

(10)² -4*1*(28 - m)

And that must be zero, so we can solve the equation:

(10)² -4*1*(28 - m) = 0

100 - 4*(28 - m) =0

100 = 4*(28 - m)

100/4 = 28 - m

25 = 28 - m

m = 28 - 25 = 3

m = 3

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a)  The estimated interquartile range for the masses of the emperor penguins is 30 kg - 25 kg = 5 kg.

b) The median mass of the king penguins would be M kg, with Q1 being M - 3.5 kg and Q3 being M + 3.5 kg.

c) Without the specific value of M, we cannot make a direct comparison between the median masses of the two species. By comparing interquartile range  values, we can infer that the masses of the king penguins have a larger spread or variability within the interquartile range compared to the emperor penguins.

a) To estimate the interquartile range for the masses of the emperor penguins, we can use the cumulative frequency table provided. The median mass is given as 23 kg, which means that 50% of the emperor penguins have a mass of 23 kg or less. Since the cumulative frequency at this point is 20, we can infer that there are 20 emperor penguins with a mass of 23 kg or less.

b) Given that the interquartile range for the masses of the king penguins is 7 kg, we can apply a similar approach to estimate the median mass of the king penguins. Since the interquartile range represents the range between Q1 and Q3, which covers 50% of the data, the median will lie halfway between these quartiles.

c) To make comparisons between the masses of the emperor and king penguins, we can consider the following two aspects:

Overall, based on the available information, it is challenging to make specific comparisons between the masses of the two penguin species without knowing the exact values for the median mass of the

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The second equation after eliminating the variable x is 0.5y + 3.5z = 11.5.

To solve the system of linear equations using Gaussian elimination, we'll perform row operations to eliminate variables one by one. Let's start with the given system of equations:

Eliminate x from equations 2 and 3:

To eliminate x, we'll multiply equation 1 by -1.5 and add it to equation 2. We'll also multiply equation 1 by -2 and add it to equation 3.

New equation 3: (4x - 2y + 3z) - 2 * (2x + y - z) = -9 - 2 * 1

Simplifying the equation 3: 4x - 2y + 3z - 4x - 2y + 2z = -9 - 2

Simplifying further: -0.5y - 3.5z = -11.5

So, the second equation after eliminating the variable x is 0.5y + 3.5z = 11.5.

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The number of students who offer three subjects is 11.  

Given that, In a class of 100 students,35 students offer History (H),43 students offer Geography (G) and50 students offer Economics (E).

14 students offer History and Geography,13 students offer Geography and Economics,11 students offer History and Economics.

Let X be the number of students who offer three subjects (H, G, E).Then the number of students who offer only two subjects = (14 + 13 + 11) - 2X= 38 - 2X

Now, the number of students who offer only one subject

= H - (14 + 11 - X) + G - (14 + 13 - X) + E - (13 + 11 - X)

= (35 - X) + (43 - X) + (50 - X) - 2(14 + 13 + 11 - 3X)

= 128 - 6X

The number of students who offer none of the subjects

= 100 - X - (38 - 2X) - (128 - 6X)

= - 66 + 9X

From the given problem, it is given that the number of students who offer none of the subjects is four times the number of those who offer three subjects.

So, -66 + 9X = 4XX = 11

Hence, 11 students offer three subjects.

Therefore, the number of students who offer three subjects is 11.

In conclusion, the number of students who offer three subjects is 11.

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Answer:

bc = 8

Step-by-step explanation:

We are given that,

ab = 20, (i)

ac = 12, (ii)

and,

c is between a and b,

we have to find bc,

Since c is between ab, so,

ab = ac + bc

which gives,

bc = ab - ac

bc = 20 - 12

bc = 8

Answer:

Hexagon

Step-by-step explanation:

Since the hexagon has more sides it should cover more space

Given the graph of a function y and three transformations as follows:

1. Reflect the graph of y about the x-axis2. Shift the graph of y 5 units up 3.

Shift the graph of y 6 units to the left to find the final function after the above transformations are applied to the graph of y, we use the following transformation rules:1. Reflect the part about the x-axis: Multiply the process by -12. Shift the function up or down: Add or subtract the shift amount to function 3. Shift the position left or right: Replace x with (x ± h) where h is the shift amount.

Here, the given function is y. So we have y = f(x)After reflecting the position about the x-axis, we have:y = -f(x)After shifting the reflected function 5 units up, we have:[tex]y = -f(x) + 5[/tex] After shifting the above part 6 units to the left, we have[tex]:y = -f(x + 6) + 5[/tex]

Thus, the function that is finally graphed after the above transformations are applied to the graph of y in the given order is[tex]y = -f(x + 6) + 5[/tex] where f(x) is the original function.

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The possible values for the divisor d are 1 and 37.

Let's denote the original number as x and the divisor as d.

According to the given information:

x divided by d leaves a remainder of 24. We can express this as x ≡ 24 (mod d).

2x divided by d leaves a remainder of 11. This can be expressed as 2x ≡ 11 (mod d).

We can rewrite these congruence equations as:

x ≡ 24 (mod d) -- Equation 1

2x ≡ 11 (mod d) -- Equation 2

To find the divisor, we need to find a value of d that satisfies both equations simultaneously.

Let's solve these congruence equations:

From Equation 1, we can write:

x = 24 + kd -- Equation 3, where k is an integer

Substituting Equation 3 into Equation 2:

2(24 + kd) ≡ 11 (mod d)

48 + 2kd ≡ 11 (mod d)

48 ≡ 11 (mod d)

48 - 11 ≡ 0 (mod d)

37 ≡ 0 (mod d)

This implies that d divides 37 without any remainder. The divisors of 37 are 1 and 37.

Therefore, the possible values for the divisor d are 1 and 37.

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Contrary to the initial claim, the calculated area is zero, not equal to 3/8π square units. It is possible that an error was made in the formulation or the intended astroid equation.

To show that the area enclosed by the astroid defined by the parametric equations x = cos^3(t) and y = sin^5(t) is equal to 3/8π square units, we can use the formula for finding the area of a plane curve given by parametric equations.

The formula for finding the area A enclosed by the curve described by parametric equations x = f(t) and y = g(t) over an interval [a, b] is:

A = ∫[a,b] |(f(t) * g'(t))| dt

In this case, we have x = cos^3(t) and y = sin^5(t). To find the area enclosed by the astroid, we need to determine the interval [a, b] over which we want to calculate the area.

Since the astroid completes one full loop as t varies from 0 to 2π, we can choose the interval [0, 2π] to calculate the area.

Now, we can calculate the area by evaluating the integral:

A = ∫[0,2π] |(cos^3(t) * (5sin^4(t)cos(t)))| dt

Simplifying the integrand:

A = ∫[0,2π] |(5cos^4(t)sin^4(t)cos(t))| dt

Using the fact that sin^2(t) = 1 - cos^2(t), we can rewrite the integrand as:

A = ∫[0,2π] |(5cos^4(t)(1-cos^2(t))cos(t))| dt

Expanding and simplifying further:

A = ∫[0,2π] |(5cos^5(t) - 5cos^7(t))| dt

Now, we can integrate term by term:

A = ∫[0,2π] (5cos^5(t) - 5cos^7(t)) dt

Evaluating the integral over the interval [0,2π], we obtain:

A = [(-cos^6(t)/6) + (cos^8(t)/8)]|[0,2π]

Plugging in the upper and lower limits:

A = [(-cos^6(2π)/6) + (cos^8(2π)/8)] - [(-cos^6(0)/6) + (cos^8(0)/8)]

Simplifying:

A = (1/6 - 1/8) - (1/6 - 1/8)

A = 1/8 - 1/8

A = 0

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