Which of these objects have NOT been found in the Galactic halo?
Select one:
A. globular clusters
B. RR Lyrae variables
C. planetary nebulae
D. Population II stars
E. emission nebulae

The line y = 6 - 3x has a positive slope and is upward and to the right.

To determine the slope of the line, we can compare the equation to the standard slope-intercept form y = mx + b, where m represents the slope. In this case, the equation is y = 6 - 3x, so the coefficient of x, -3, represents the slope.

Since -3 is negative, the line slopes downward and to the right. However, if we rewrite the equation as y = -3x + 6, the slope becomes positive (+3). Therefore, the corrected equation y = 6 - 3x represents a line that slopes upward and to the right.

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Answer:

it would be 36 pi or whatever pi × 36 is

Step-by-step explanation:

[tex]d^2x/dx^2[/tex], expressed as a function of t, is 8tan(t) + cos(t). Except for at the points where dy/dx is undefined, the curve is concave down.

What is function?

A function is a mathematical concept that describes the relationship between a set of inputs, called the domain, and a set of outputs, called the range. It assigns each input value to a unique output value.

To find dy/dx as a function of t, we need to differentiate the parametric equation [tex]y = 6 sin^2(t)[/tex] with respect to [tex]x = cos^9(t)[/tex].

First, let's express y as a function of x using the given parametric equations:

[tex]x = cos^9(t)\\\\y = 6sin^2(t)[/tex]

To find dy/dx, we can use the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

Let's differentiate [tex]y = 6sin^2(t)[/tex] with respect to [tex]x = cos^9(t):[/tex]

dy/dx = (dy/dt) / (dx/dt)

To find dy/dt, we differentiate [tex]y = 6sin^2(t)[/tex] with respect to t:

[tex]dy/dt = d/dt (6sin^2(t))\\\\= 12sin(t)cos(t)[/tex]

To find dx/dt, we differentiate [tex]x = cos^9(t)[/tex] with respect to t:

[tex]dx/dt = d/dt (cos^9(t))\\\\= -9cos^8(t)sin(t)[/tex]

Now, we can substitute these derivatives into the expression for dy/dx:

[tex]dy/dx = (dy/dt) / (dx/dt)\\\\= (12sin(t)cos(t)) / (-9cos^8(t)sin(t))\\\\= -4/3cos(t)^{(-7)[/tex]

Therefore, dy/dx, expressed as a function of t, is [tex]-4/3cos(t)^{(-7)[/tex].

To find [tex]d^2x/dx^2[/tex] as a function of t, we need to differentiate [tex]dx/dt = -9cos^8(t)sin(t)[/tex] with respect to [tex]x = cos^9(t)[/tex]:

[tex]d^2x/dx^2 = d/dx (-9cos^8(t)sin(t))\\\\= d/dt (-9cos^8(t)sin(t)) / (dx/dt)\\\\= (-72cos^7(t)sin^2(t) - 9cos^9(t)) / (-9cos^8(t)sin(t))\\\\= 8sin(t) / cos(t) + cos(t)[/tex]

Therefore, [tex]d^2x/dx^2[/tex], expressed as a function of t, is 8tan(t) + cos(t).

Except for at the points where dy/dx is undefined, the curve is concave down.

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Answer:

[tex]2(4n-5)[/tex]

Step-by-step explanation:

We start with this equation:

[tex]\frac{8n^2-26n+20}{n-2}[/tex]

Factoring it out gives us:

[tex]\frac{2(4n-5)(n-2)}{(n-2)}[/tex]

Thus you can cancel the [tex](n-2)[/tex]  as:

 [tex]\frac{(n-2)}{(n-2)} = 1[/tex]

Giving us:

[tex]2(4n-5)[/tex]

Answer:

2(4n -5)

Step-by-step explanation:

Factorizing:

8n² - 26n + 20 = 2*(4n² - 13n + 10)

4n² - 13n + 10

Sum = -13

Product = 10 * 4 = 40

The factors should be in such a way that when we add, we should get (-13) and when we multiply, we should get 40.

Factors = (-8); (-5)           { (-8) + (-5) = -13  & (-5)*(-8) = 40}    

4n² - 13n + 40 = 4n² - 8n - 5n + 10  {Rewrite the middle term using the factors}      

Take the common term 4n from the first two terms and (-5) from the third and fourth term.

                      = 4n(n - 2) -5 (n - 2)

                      = (n -2) (4n - 5)

[tex]\sf \dfrac{8n^2 - 26n + 20}{n -2}= \dfrac{2*(4n^2 - 13n +10)}{(n -2)}[/tex]

                       [tex]= \dfrac{2*(n -2)(4n -5)}{(n -2)}\\\\= 2*(4n-5)[/tex]                        

The critical numbers of the given function g(y) = (y-1)/([tex]y^2[/tex]-y+1) are obtained as y = 0 and y = 4.

To find the critical numbers of the function g(y) = (y-1)/([tex]y^2[/tex]-y+1), we need to determine the values of y where the derivative of g(y) is either zero or undefined.

First, let's find the derivative of g(y).

Using the quotient rule, we have:

g'(y) = [(1)([tex]y^2[/tex]-y+1) - (y-1)(2y-1)] / [tex](y^2-y+1)^2[/tex]

Simplifying, we have:

g'(y) = [[tex]y^2[/tex] - y + 1 - (2[tex]y^2[/tex] - 3y + 1)] / [tex](y^2-y+1)^2[/tex]

g'(y) = ([tex]-y^2[/tex] + 4y) / [tex](y^2-y+1)^2[/tex]

To find the critical numbers, we set the numerator of the derivative equal to zero and solve for y:

[tex]-y^2[/tex] + 4y = 0

Factoring out y, we have:

y(-y + 4) = 0

This equation is satisfied when y = 0 or y = 4. These are the potential critical numbers of the function g(y).

Next, we need to check if there are any values of y that make the denominator [tex](y^2-y+1)^2[/tex] equal to zero.

However, upon closer examination, we can see that the denominator is always positive for any real value of y.

Therefore, there are no additional critical numbers to consider.

Thus, the critical numbers of the function g(y) = (y-1)/([tex]y^2[/tex]-y+1) are y = 0 and y = 4.

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The equation for each table are:

y = 2x + 6

y = -3x + 1

y = 5x - 24

We have,

The equation can be written in the slope-intercept form as:

y = mx + c

Now,

We will consider the two points in the table in ordered pairs:

(x, y) = (a, b) and (c, d)

Where m = (d - b) / (c - a)

Now,

The equation for each table are:

1.

(1, 8) and (3, 12)

m = (12 - 8) / (3 - 1) = 4/2 = 2

And,

(1, 8) = (x, y)

8 = 2 x 1 + c

c = 8 - 2 = 6

So,

y = 2x + 6

2.

(2 , -5) and (4, -11)

m = (-11 + 5) / (4 - 2) = -6 / 2 = -3

(2, -5) = (x, y)

-5 = -3 x 2 + c

-5 = -6 + c

c = -5 + 6

c = 1

So,

y = -3x + 1

3.

(-2, -14) and (-4, -24)
m = (-24 + 14) / (-4 + 2) = -10 / -2 = 5

(-2 , -14) = (x, y)

-14 = 5 x -2 + c

-14 = 10 + c

c = -14 - 10

c = -24

So,

y = 5x - 24

Thus,

The equation for each table are:

y = 2x + 6

y = -3x + 1

y = 5x - 24

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The equation of the linear function passing through the points in the table is y = -2x - 1.

The slope-intercept form is expressed as;

y = mx + b

Where m is slope and b is the y-intercept.

Given the data in the table, we pick two points, (1,-3) and (2,-5).

First, we determine the slope:

[tex]m = \frac{y_2 - y_1}{x_2 - x_1} \\\\m = \frac{-5-(-3)}{2 - 1} \\\\m = \frac{-5+3}{2 - 1} \\\\m = \frac{-2}{1}\\ \\m = -2[/tex]

Now, we choose one (1, -3) and slope m = -2, and substitute the values into the point-slope form equation:

y - y₁ = m( x - x₁ )

y - (-3) = -2( x - 1 )

Simplify

y + 3 = -2x + 2

y = -2x + 2 - 3

y = -2x - 1

Therefore, the linear function is y = -2x - 1.

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The global temperatures during the most recent 10,000 years have indeed been relatively stable compared to similar periods of time during the past 120,000 years. This period of relative stability is known as the Holocene epoch, which began after the last major glacial period and has lasted until the present day.

During this time, the Earth's climate has been relatively stable, with some fluctuations but generally not experiencing the significant temperature swings that characterized the earlier Pleistocene epoch. This stability has allowed human civilization to develop and flourish, as it has provided a consistent and predictable climate for agricultural and other activities. However, it's important to note that in the past few decades, there has been a noticeable increase in global temperatures, which is likely due to human activities such as the burning of fossil fuels and deforestation.

While the long-term stability of the Earth's climate is certainly encouraging, it's important to remain vigilant in addressing the current climate crisis to ensure that our planet remains habitable for generations to come.

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The probabilities from the images are

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

The images

Where, we have

P(triangle U square) represents the probability a shape is a triangle or a square

So, we have

P(triangle U square) = 10/20 = 1/2

P(triangle ∩ black) represents the probability a shape is a triangle and the shape is also black i.e. black triangles

So, we have

P(triangle ∩ black) = 4/20 = 1/5

P(black triangle U white circle) represents the probability a shape is a black triangle or a white square

So, we have

P(black triangle U white circle) = 9/20

P(black triangle ∩ white circle) represents the probability a shape is a black triangle and a white square

So, we have

P(black triangle ∩ white circle) = 0

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Answer:

Get all the students in the 2nd hour Geometry class to participate in a study. Ask them whether or not they do tutoring and divide them into two groups based on their answer to this question. Compare scores on the next test.

Conclusion: C. For the students in 2nd hour Geometry class, there is an association between doing tutoring and higher test scores.

Study: Select a random sample of Geometry students from your school to participate in a study. Ask them whether or not they do tutoring and divide them into two groups based on their answer to this question. Compare scores on the next test.

Conclusion: E. For all Geometry students at your school, there is an association between doing tutoring and higher test scores.

Study: Select a random sample of Geometry students in Michigan to participate in a study. Ask them whether or not they do tutoring and divide them into two groups based on their answer to this question. Compare scores on the next test.

Conclusion: A. For all Geometry students in Michigan, there is an association between doing tutoring and higher test scores.

Study: Get all the students in the 2nd hour Geometry class to participate in a study. Randomly assign half of the students to do tutoring and have the remaining half not do tutoring. Compare scores on the next test.

Conclusion: D. For the students in 2nd hour Geometry class, doing tutoring causes higher test scores.

Study: Select a random sample of Geometry students from your school to participate in a study. Randomly assign half of the students to do tutoring and have the remaining half not do tutoring. Compare scores on the next test.

Conclusion: F. For all Geometry students at your school, doing tutoring causes higher test scores.

Study: Select a random sample of Geometry students in Michigan to participate in a study. Randomly assign half of the students to do tutoring and have the remaining half not do tutoring. Compare scores on the next test.

Conclusion: B. For all Geometry students in Michigan, doing tutoring causes higher test scores.

The expression is written as b = k/3x²

Inverse variation is defined as a type of variation such that as one variation increases, the other decreases and as one decreases, the other increases.

From the information given, we have that;

y is inversely proportional to x²

This is written as;

y ∝1/x²

Find the constant

yx² = k

If y = b, and x is tripled

Substitute the values, we have;

3x²

b3x² = k

Then, make 'b' the subject of formula, we have that;

b = k/3x²

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Among the provided answer options, the closest value to -0.555 is -0.55, which is option c. Therefore, option c (-0.55) is the value of c that satisfies P(Z ? c) = 0.71.

The notation P(Z ? c) represents the probability that a standard normal random variable Z is less than or equal to c. To find the value of c that corresponds to P(Z ? c) = 0.71, we need to determine the Z-score associated with this probability.

Using a standard normal distribution table or a calculator, we can find that a Z-score of approximately 0.555 corresponds to a cumulative probability of 0.71. However, since we are looking for the value of c in P(Z ? c), we need to consider the opposite inequality.

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a. The mean body temperature of men is roughly 97.5°F to 98.3°F.

b. The 95% confidence interval is approximately 98.3°F to 99.1°F.

c. The margin of error for men is about 0.4°F and for women is roughly 0.4°F,

d. The margin of error for women is larger due to a wider range of recorded temperatures in the sample

a) The mean body temperature for men in this sample is 97.9°F.

Using a t-distribution (as we don't know the population standard deviation), the 95% confidence interval for the mean body temperature of men is roughly 97.5°F to 98.3°F.

b) For women, the mean is 98.7°F. The 95% confidence interval is approximately 98.3°F to 99.1°F.

c) The margin of error for men is about 0.4°F and for women is roughly 0.4°F, assuming equal sample sizes and degrees of freedom.

d) The margin of error for women is larger due to a wider range of recorded temperatures in the sample, which increases variability and uncertainty in the estimate.

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In the fire academy, with a ratio of 7 passing grades to 5 failing grades, we find that approximately 15 students failed.

Given the ratio of 7 passing grades to 5 failing grades, we can set up a proportion to determine the number of students who failed out of 36 students.

Let's denote the number of students who failed as x.

The proportion can be set up as follows:

7 passing grades / 5 failing grades = 36 total students / x failed students

Cross-multiplying the proportion, we have:

7x = 5 * 36

Simplifying the equation, we get:

7x = 180

Dividing both sides of the equation by 7, we find:

x = 180 / 7

Calculating the value, we have:

x ≈ 25.71

Since we cannot have fractional students, we round down to the nearest whole number.

Therefore, approximately 25 students failed out of the 36 students in the fire academy.

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The measure of the angle ∠ABD will be 66°.

Given that:

Angle, ∠FEG = 48°

The central angle is double the angle at the periphery that was subtended by the same chords.

The line AC is perpendicular to EF and the line DC is perpendicular to EG. So, the equation is given as,

∠FEG + ∠HCI = 180°

48° + ∠HCI = 180°

∠HCI = 132°

By the above theorem, the equation is given as,

∠ABD = 1/2 ∠HCI

∠ABD = 1/2 x 132°

∠ABD = 66°

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The option D is correct answer which is,

[tex]=(1/21)\left[\begin{array}{ccc}27&-6&-3\\-10&3&5\\4&3&-2\end{array}\right][/tex]

What is Inverse of matrix?

If A is a non-singular square matrix, then A⁻¹, often known as the inverse matrix of A, must exist if it is to satisfy the following property:

AA⁻¹ = A⁻¹A = I,

Where I is the Identity matrix.

Evaluate the value of A⁻¹ as follows:

A⁻¹ = Adj A / Mod-A

We have to find the inverse of the matrix A which is given below:

[tex]A=\left[\begin{array}{ccc}1&1&1\\0&2&5\\2&5&-1\end{array}\right][/tex]

We will find the determinant of A by expanding along first column.

[tex]|A|=1\left[\begin{array}{cc}2&5\\5&-1\end{array}\right] +0+2\left[\begin{array}{cc}1&1\\2&5\end{array}\right][/tex]

Simplify values,

= 1(-2 - 25) + 0 +2(5 - 2)

= -27 + 0 + 6

= -21

First, we will find cofactors (A):

[tex]CofactorA=\left[\begin{array}{ccc}a11&a12&a13\\a21&a22&a23\\a31&a32&a33\end{array}\right][/tex]

After solving,

[tex]CofactorA=\left[\begin{array}{ccc}-27&10&-4\\6&-3&-3\\3&-5&2\end{array}\right][/tex]

Now we will find Adjoint of A, Adj A, which is given below:

Adj A = (Cofactor A)^T

[tex]AdjA=\left[\begin{array}{ccc}-27&6&3\\10&-3&-5\\-4&-3&2\end{array}\right][/tex]

Now we will find the inverse of A,

A⁻¹ = Adj A / Mod-A

Substitute all obtained values respectively,

[tex]=(-1/21)\left[\begin{array}{ccc}-27&6&3\\10&-3&-5\\-4&-3&2\end{array}\right][/tex]

Simplify values,

[tex]=(1/21)\left[\begin{array}{ccc}27&-6&-3\\-10&3&5\\4&3&-2\end{array}\right][/tex]

Hence, the option D is correct answer which is,

[tex]=(1/21)\left[\begin{array}{ccc}27&-6&-3\\-10&3&5\\4&3&-2\end{array}\right][/tex]

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The approximate probability of getting tic-tac-toe is option (b) 1.2%.

To calculate the approximate probability, we need to consider the number of ways to achieve tic-tac-toe and the total number of possible outcomes.

Here are the steps:

Therefore, the approximate probability of getting tic-tac-toe with randomly placed 3 Xs is approximately 1.59%, which is closest to option (B) 1.2%.

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The line of intersection formed by the two planes 2x - y + z = 5 and x + y - z = 1 can be represented by the vector equation r = <x, y, z> = <2, 3, 4> + t<3, 4, 1>, where t is a parameter.

To find the vector equation for the line of intersection, we need to determine the direction vector and a point on the line.

First, we'll find the direction vector by taking the cross product of the normal vectors of the two planes. The normal vector of the first plane is <2, -1, 1>, and the normal vector of the second plane is <1, 1, -1>. Taking the cross product of these two vectors, we get the direction vector <3, 4, 1>.

Next, we'll find a point on the line of intersection by solving the system of equations formed by the two planes. Subtracting the second plane equation from the first, we get 3x - 2y + 2z = 4. Solving this system along with the original plane equation, we find x = 2, y = 3, and z = 4.

Combining the direction vector and the point on the line, we obtain the vector equation r = <x, y, z> = <2, 3, 4> + t<3, 4, 1>, where t is a parameter. This equation represents the line of intersection formed by the two given planes.

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Approximately 885 participants think the workshop should be rated a 5.

Option D is the correct answer.

We have,

To estimate how many people in the entire population of participants who attended the workshop think it should be rated a 5, we can use the concept of sampling proportions.

Out of the 428 respondents, 304 participants rated the workshop as a 5. This means that approximately 304/428 = 0.7103 (or 71.03%) of the respondents gave a rating of 5.

We can estimate the number of participants in the entire population who think the workshop should be rated a 5 by multiplying this proportion by the total number of participants:

Estimated number = Proportion of respondents x Total number of participants

= 0.7103 x 1246

≈ 885.417

Rounding this estimate to the nearest whole number, we get approximately 885 participants.

Therefore,

Approximately 885 participants think the workshop should be rated a 5.

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If you chose to compute a mode for a variable, the measure of variability you would most likely want to report is the a) range .

The mode represents the most frequently occurring value in a dataset, which provides information about the central tendency and the characteristic value in the data. It helps identify the value or values that are most common or dominant in the dataset.

On the other hand, the range measures the spread or variability in the dataset. It gives you the difference between the maximum and minimum values in the dataset, providing insight into the extent of the dispersion of values.

The standard deviation (option b) is another measure of variability, but it quantifies the dispersion of values around the mean. It is not directly related to the mode and provides a different aspect of variability compared to the mode.

Therefore, if you are specifically interested in the measure of variability to report when computing the mode, the range (option a) would be more relevant than the standard deviation (option b).

Correct Question:

If you chose to compute a mode for a variable, which measure of variability would you most likely want to report?

a) Range

b) Standard Deviation

c) Neither of these

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The Taylor polynomial T₃(x) for the function [tex]f(x) = e^(^-^3^x^)sin(2x)[/tex], centered at a = 0, is: [tex]T_3(x) = x - (3/2)x^2 + (9/4)x^3[/tex]

The Taylor polynomial for a function f(x) centered at a is an approximation of the function using a polynomial. It is constructed by finding the function's derivatives at the point a. In this case, we are given the function [tex]f(x) = e^(^-^3^x^)sin(2x)[/tex] and asked to find its third-degree Taylor polynomial centered at a = 0.

To find the Taylor polynomial, we need to compute the function's derivatives at a = 0. The first derivative of f(x) with respect to x is [tex]f'(x) = -3e^(^-^3^x^)sin(2x) + 2e^(^-^3^x^)cos(2x)[/tex]. Evaluating this at x = 0 gives f'(0) = 0 - 2(1) = -2.

The second derivative is [tex]f''(x) = 9e^(^-^3^x^)sin(2x) - 12e^(^-^3^x^)cos(2x).[/tex] Evaluating at x = 0 yields f''(0) = 9(0) - 12(1) = -12.

Finally, the third derivative is [tex]f'''(x) = -27e^(^-^3^x^)sin(2x) - 36e^(^-^3^x^)cos(2x).[/tex]Evaluating at x = 0 gives f'''(0) = -27(0) - 36(1) = -36.

Now, we can use these derivatives to construct the third-degree Taylor polynomial T₃(x) using the formula:

T₃(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)² + (f'''(a)/3!)(x - a)³

Plugging in a = 0 and the evaluated derivatives, we obtain the final result: [tex]T_3(x) = x - (3/2)x^2 + (9/4)x^3[/tex]

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The structure of the coordinates associated with specific radian measurements on the unit circle follows a distinct pattern. The x and y coordinates represents the cosine and sine values of the angle, respectively.

The unit circle is a circle with a radius of 1 unit, centered at the origin of a Cartesian coordinate system.

It is commonly used in trigonometry to relate angles to the coordinates on the circle.

When we measure an angle in radians on the unit circle, we can associate specific coordinates with that angle.

The x-coordinate of a point on the unit circle represents the cosine value of the angle, while the y-coordinate represents the sine value of the angle.

These coordinates follow a pattern based on the radian measurement.

For example, when the angle is 0 radians (or 0 degrees), the corresponding point on the unit circle is (1, 0), where the x-coordinate is 1 (cosine of 0) and the y-coordinate is 0 (sine of 0).

As the angle increases, the coordinates change accordingly. For instance, when the angle is π/2 radians (or 90 degrees), the point on the unit circle is (0, 1), where the x-coordinate is 0 (cosine of π/2) and the y-coordinate is 1 (sine of π/2).

This pattern continues as the angle increases or decreases, and the coordinates on the unit circle change accordingly.

By using trigonometric functions, we can determine the cosine and sine values of any given angle and associate them with the appropriate coordinates on the unit circle.

In summary, the structure of the coordinates associated with specific radian measurements on the unit circle follows a pattern where the x-coordinate represents the cosine value of the angle, and the y-coordinate represents the sine value of the angle.

Together, these coordinates create points on the unit circle that correspond to the given radian measurement.

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The inequalities are solved as;

6n ≤ 0

n< -4

Inequalities are defined as mathematical expressions showing a non-equal comparison between numbers or variables

From the information given, we have that;

3+2n≤-4n+3

First, collect the like terms, we have;

2n +4n ≤ 3 - 3

subtract the values

6n ≤ 0

-5+4n<7n+7

collect the like terms

4n - 7n < 7 + 5

add the values, we get;

-3n < 12

Make 'n' the subject

n< -4

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It is a dilation that can be represented by (x, y)→(1/2 x, 1/2 y). Therefore, option A is the correct answer.

From the given graph, vertices of triangle ABC are A(8, 12), B(16, 20) and (20, 12).

Vertices of triangle A'B'C' are A'(4, 6), B(8, 10) and (10, 6).

It is a dilation that can be represented by (x, y)→(1/2 x, 1/2 y)

Therefore, option A is the correct answer.

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The Taylor series for f(x)=sin(x) at x=π/2 is ∑n=0[infinity]cn(x-π/2)n, where cn=(-1)n/(n!) if n is even and cn=0 if n is odd.

So, c0=0, c1=(-1)1/(1!)= -1, c2=0, c3=1/(3!)=1/6, c4=0.
Therefore, the first five coefficients are c0=0, c1=-1, c2=0, c3=1/6, c4=0.
Note that the odd coefficients are all 0, which is because the function f(x)=sin(x) is an odd function and its derivatives evaluated at π/2 are all 0. The Taylor series for a function f(x) is given by ∑n=0∞(f^(n)(a)(x-a)^n)/n!, where f^(n)(a) is the nth derivative of f(x) evaluated at x=a. In this case, f(x) = sin(x) and a = π/2.
To find the first five coefficients, we need the first five derivatives evaluated at x = π/2:
f(x) = sin(x)
f'(x) = cos(x)
f''(x) = -sin(x)
f'''(x) = -cos(x)
f^(4)(x) = sin(x)
Now, evaluate them at x = π/2:
f(π/2) = 1
f'(π/2) = 0
f''(π/2) = -1
f'''(π/2) = 0
f^(4)(π/2) = 1
Finally, apply the Taylor series formula:
c0 = f(π/2)/0! = 1
c1 = f'(π/2)/1! = 0
c2 = f''(π/2)/2! = -1/2
c3 = f'''(π/2)/3! = 0
c4 = f^(4)(π/2)/4! = 1/24

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Answer:

Step-by-step explanation:

You want the given representations of linear functions ordered by decreasing y-intercept.

The attachment shows the y-intercepts of each of the functions.

(a) The graph shows a y-intercept of 2.

(b) The equation y = mx +b has b=4, so a y-intercept of 4.

(c) The table has point (0, 1), so a y-intercept of 1.

(d) The points are located on or below the x-axis and either side of the y-axis, so will define a line with a negative y-intercept.

In decreasing order of y-intercept, the functions are ...

  (b) equation, (a) graph, (c) table, (d) points

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