what is the y-intercept of this radical function f(x)=^3√x+1-3

A. y=-2
B. y=1
C. y=8
D. y=26

​

The transformation after the dilation is a reflection over the y-axis. The correct option is D.

A dilation about the origin will not change the orientation of the shape. However, a reflection over the y-axis will flip the shape over the y-axis, which is what we see in the transformation from A to A'.

The other options are incorrect because they would change the orientation of the shape in addition to moving it. A rotation 90° counterclockwise would rotate the shape 90° counterclockwise, a translation down 9 units would move the shape down 9 units, and a reflection over the x-axis would flip the shape over the x-axis.

A is the original shape, and A' is the transformed shape. The transformation from A to A' is a reflection over the y-axis.

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The additional time it will take for the liquid's temperature to decrease from 64°F to 32°F is 2 minutes, considering a rate of temperature decrease of 16°F per minute.

To find the additional time it will take for the liquid's temperature to decrease from 64°F to 32°F, we can consider the rate at which the temperature is decreasing.

From the given information, we know that the liquid's temperature decreases from 160°F to 64°F in 6 minutes. This means that the temperature decreases by 160°F - 64°F = 96°F in 6 minutes.

Therefore, the rate of temperature decrease is 96°F / 6 minutes = 16°F per minute.

To find the additional time it will take for the temperature to decrease from 64°F to 32°F, we can calculate the difference in temperature and divide it by the rate of temperature decrease

Temperature difference = 64°F - 32°F = 32°F

Additional time = Temperature difference / Rate of temperature decrease

Additional time = 32°F / 16°F per minute = 2 minutes

Therefore, it will take an additional 2 minutes for the temperature to decrease from 64°F to 32°F.

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The correct statement regarding left-skewed distributions is option b) The distance from the smallest observation to Q1 is larger than the distance from Q3 to the largest observation.

In left-skewed distributions, also known as negatively skewed distributions, the tail of the distribution is elongated to the left, meaning that the left side of the distribution has a longer or heavier tail compared to the right side. This results in a majority of the data concentrated towards higher values on the right side and a few extreme values on the left.

Option b) states that the distance from the smallest observation to Q1 is larger than the distance from Q3 to the largest observation. This statement is correct because in left-skewed distributions, Q1 (the first quartile) represents the 25th percentile of the data and is closer to the lower values. The smallest observation is even further to the left, resulting in a larger distance between the smallest observation and Q1. On the other hand, Q3 (the third quartile) is closer to the higher values, and the largest observation is even further to the right. Therefore, the distance from the smallest observation to Q1 is larger than the distance from Q3 to the largest observation.

Options a), c), and d) do not accurately describe the characteristics of left-skewed distributions and are therefore incorrect.

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a) The sοlutiοn tο the differential equatiοn y' - 2y = u(t) with initial cοnditiοn y(0) = 0 is y(t) = -1/2 - 1/2[tex]\rm e^{(2t)[/tex].

b)  1 + 1/2[tex]\rm e^{(2t)[/tex].

c)  (A/10)sin(2t) + (1 - A/10)[tex]\rm e^{(2t)[/tex].

d)  (8/10)(1 - [tex]\rm e^{(-10t))[/tex].

e)  (1/60)(1 - [tex]\rm e^{(-8t))[/tex].

f)  (1/13)(1 - cοs(2t)).

a) Tο sοlve the differential equatiοn y' - 2y = u(t) with initial cοnditiοn y(0) = 0 using Laplace transfοrms, we'll apply the Laplace transfοrm tο bοth sides οf the equatiοn.

Taking the Laplace transfοrm οf the differential equatiοn, we have:

sY(s) - y(0) - 2Y(s) = 1/s,

where Y(s) represents the Laplace transfοrm οf y(t).

Substituting the initial cοnditiοn y(0) = 0, we get:

sY(s) - 0 - 2Y(s) = 1/s.

Nοw, we can sοlve fοr Y(s):

(s - 2)Y(s) = 1/s,

Y(s) = 1/[s(s - 2)].

Next, we need tο find the inverse Laplace transfοrm οf Y(s). We can use partial fractiοn decοmpοsitiοn tο simplify the expressiοn:

Y(s) = A/s + B/(s - 2),

where A and B are cοnstants tο be determined.

Multiplying bοth sides by s(s - 2), we have:

1 = A(s - 2) + Bs.

Setting s = 0, we get:

1 = -2A,

A = -1/2.

Setting s = 2, we get:

1 = -2B,

B = -1/2.

Sο, the partial fractiοn decοmpοsitiοn becοmes:

Y(s) = -1/2s - 1/2(s - 2),

Nοw, we can take the inverse Laplace transfοrm:

y(t) = -1/2 - 1/2[tex]\rm e^{(2t)[/tex].

Therefοre, the sοlutiοn tο the differential equatiοn y' - 2y = u(t) with initial cοnditiοn y(0) = 0 is y(t) = -1/2 - 1/2[tex]\rm e^{(2t)[/tex].

The rest of them can be calculated in the same way,

b) The sοlutiοn tο the differential equatiοn y' - 2y = u(t) with initial cοnditiοn y(0) = 1 is y(t) = 1 + 1/2[tex]\rm e^{(2t)[/tex].

c) The sοlutiοn tο the differential equatiοn y' + 10y - Asin(2t)u(t) with initial cοnditiοn y(0) = 1 is y(t) = (A/10)sin(2t) + (1 - A/10)[tex]\rm e^{(2t)[/tex].

d) The sοlutiοn tο the differential equatiοn y' + 10y = [tex]\rm 8e^{(-10t)[/tex]u(t) with initial cοnditiοn y(0) = 0 is y(t) = (8/10)(1 - [tex]\rm e^{(-10t))[/tex].

e) The sοlutiοn tο the differential equatiοn 60y' + 8y = u(t) with initial cοnditiοns y(0) = 0 and y'(0) = 1 is y(t) = (1/60)(1 - [tex]\rm e^{(-8t))[/tex].

f) The sοlutiοn tο the differential equatiοn 6y' + 9y = sin(2t)u(t) with initial cοnditiοns y(0) = 0 and y'(0) = 0 is y(t) = (1/13)(1 - cοs(2t)).

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Solving the expressions, resulted to the value of the multiplication

x * y * z = 0.429 (to 3 dp)

To find the value of x * y * z we first need to solve the given equations individually

20ˣ = 16

take the ㏑ of both sides:

x ㏑ 20= ㏑ 16

x = ㏑ 16 / ㏑ 20

x = 0.926

10^y = 20

take ㏑ of both sides:

y ㏑ 10 = ㏑ 20

y = ㏑ 10 / ㏑ 20

y = 0.769

4^z = 10:

take the ㏑ of both sides:

z ㏑ 4 = ㏑ 10

z = ㏑ 4 / ㏑ 10

z = 0.602

solving for x * y * z

x * y * z = 0.926 * 0.769 * 0.602

x * y * z = 0.429

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The area of the land is 2 × 10⁸ m²

A rectangle is a type of quadrilateral that has its parallel sides equal to each other and all the four vertices are equal to 90 degrees.

The wild life refuge takes the shape of a rectangle and the area of a rectangle is expressed as;

A = l × w

where w is the width and l is the length

length = 5 × 10⁵

width = 4 × 10²

Area of the wild life refuge = 5 × 10⁵ × 4 × 10²

= 20× 10⁷

therefore the of the new wild life refuge is 20× 10⁷

= 2 × 10⁸ m²

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a) The curve that best fits the data is given as follows: Curve 1.

b) The predicted amount after 50 days is given as follows: 198.84 mg.

For a data-set, the definition of a residual is that it is the difference of the actual output value by the predicted output value, that is:

Residual = Observed - Predicted.

Item a:

The curve of best fit should have the lowest residuals, that is, the dots should be as close as possible to the curve, hence curve 1 is the best fit.

Item b:

The curve of best fit is defined as follows:

[tex]y = 546(0.98)^x[/tex]

Hence the expected amount after 50 days is given as follows:

[tex]y = 546(0.98)^{50}[/tex]

y = 198.84 mg.

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The completed cumulative frequency table is as follows:

Length (cm) Cumulative Frequency

<48.5            0

<53.5            7

<58.5           44

<63.5           77

<68.5           80

To complete the cumulative frequency table based on the given grouped frequency distribution, we can add up the frequencies as we move down the table.

Given:

Length, to the nearest cm:

49-53

54-58

59-63

64-68

Number of kettles:

7

37

33

3

Let's complete the cumulative frequency table:

Length (cm) Cumulative Frequency

<48.5 0

<53.5 7 (frequency of the first group)

<58.5 7 + 37 = 44 (cumulative frequency of the first two groups)

<63.5 44 + 33 = 77 (cumulative frequency of the first three groups)

<68.5 77 + 3 = 80 (cumulative frequency of all groups)

Therefore, the completed cumulative frequency table is as follows:

Length (cm) Cumulative Frequency

<48.5 0

<53.5 7

<58.5 44

<63.5 77

<68.5 80

In the last row, the cumulative frequency is 80, which represents the total number of kettles.

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The complete question:

The table gives the grouped frequency distribution for the lengths

of the electrical cords of 80 kettles.

Length, to the nearest cm

49-53

54-58

59-63

64-68

Number of kettles

7

37

33

3

Complete the cumulative frequency table.

Length (cm)

<48.5

<53.5

<58.5

<63.5

<68.5

Cumulative frequency

0

7

__

Answer:

Dave ate 3 potato chips with 5/6 pounds then:

3*5/6=5/2 pounds

Answer:

12 2/3 yards are equivalent to 38 feet.

Answer:

= 12  2/3 yds

Step-by-step explanation:

We know that 3 ft = 1 yd, so we divide 38 ft by 3 ft

38/3 = 12 with 2 left over

38 ft = 12  2/3 yds

Answer:

the value of b is 3√5

Step-by-step explanation:

To find the value of b, we can use the quadratic formula which is given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Here, a = 3, b = b and c = 10.

Let the roots of the equation be α and β, such that α > β.

We know that α - β = 4 1/3 or 13/3.

Using the sum and product of roots formula, we have:

α + β = -b/a ...(1)

αβ = c/a = 10/3 ...(2)

Squaring equation (1), we get:

(α + β)^2 = b^2/a^2

Substituting the values of (1) and (2), we get:

(b^2/9) = (13/3)^2 - 4(10/3)

Solving this equation, we get:

x = -21 or x = 5

Since b is a real number, we can discard the negative value of x. Therefore,

b = √(5*9) = 3√5

Therefore, the value of b is 3√5.

Based on the triple scalar product, u, v, and w are not coplanar and based on the definition of linear combination, u, v, and w are not coplanar and u, v, and p are linearly independent.

To show that u, v, and w are coplanar using their triple scalar product, we calculate the scalar triple product (u x v) · w.

If the scalar triple product is equal to zero, then the vectors are coplanar.

Let's calculate:

(u x v) · w = [(2, -1, 4) x (0, -9, 18)] · (0, -9, 18)

= (36, 0, 18) · (0, -9, 18)

= 36(0) + 0(-9) + 18(18)

= 0 + 0 + 324

= 324

Since the scalar triple product is not zero, u, v, and w are not coplanar.

To show that u, v, and w are coplanar using the definition that there exist two nonzero real numbers α and β such that w = αu + βv, we can write down the equations and solve for α and β.

We have:

w = (0, -9, 18)

u = (2, -1, 4)

v = (0, -9, 17)

Let's set up the equations:

0 = α(2, -1, 4) + β(0, -9, 17)

From the equations, we can see that the x-component equation is 2α = 0, which implies α = 0.

However, the y-component equation is -α - 9β = -9, which is inconsistent with α = 0.

Therefore, there are no nonzero real numbers α and β that satisfy the equations, and thus u, v, and w are not coplanar.

To show that u, v, and p are linearly independent, we need to demonstrate that there are no real numbers α, β, and γ, not all zero, such that αu + βv + γp = 0.

We have:

u = (2, -1, 4)

v = (0, -9, 17)

p = (0, -9, 17)

Let's set up the equations:

α(2, -1, 4) + β(0, -9, 17) + γ(0, -9, 17) = (0, 0, 0)

From the equations, we can see that the x-component equation is 2α = 0, which implies α = 0. However, the y-component equation is -α - 9β - 9γ = 0, which is inconsistent with α = 0.

Therefore, there are no nonzero real numbers α, β, and γ that satisfy the equations, and thus u, v, and p are linearly independent.

In summary, based on the results of the calculations, we conclude that u, v, and w are not coplanar, u, v, and w are not coplanar based on the definition of linear combination, and u, v, and p are linearly independent.

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The complete statement:

When the sectors of the circle are rearranged, they are close to forming a rectangle with a length of π units and a width of r units. The area of the rearranged figure is equal to that of the circle. So, the area of the circular base is πr² square units.

The area of the rearranged figure, which is the same as the area of the circle, remains unchanged. This is because the total area of the circle is still preserved when the sectors are rearranged.

Therefore, the area of the circular base of the cylinder is also equal to π square units. This is because the base of the cylinder is formed by the rearranged sectors of the circle, which have the same area as the original circle.

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The complete question:

Suppose the bottom surface, or base, of a cylinder is divided into 16 congruent sectors. Then the sectors are rearranged as shown. Open this GeoGebra activity to divide the circle into even more sectors and rearrange the sectors for yourself.

Complete the statements to compare the areas of the two figures and find the area of the base of the cylinder.

Select the correct answer from each drop-down menu.

When the sectors of the circle are rearranged, they are close to forming a____

with a length of ___▾ units and a width of r units.

The area of the rearranged figure is___

to that of the circle. So, the area of the circular base___

square units.

The test statistic for randomly interviewed a class of 16 students is given by 1.5.

The population mean is given by = µ = 2

The mean of the sample of class of 16 students is given by = 2.3

Length of size of sample is given by = (n) = 16

Standard deviation is given by = (s) = 0.8

So the test statistic for the test with randomly interviewed a class of 16 students is given by

= (mean of sample - µ)/(s/√n)

= (2.3 - 2)/(0.8/√16)

= 0.3/(0.8/4)

= 0.3/0.2

= 3/2

= 1.5

Hence the test statistic is given by 1.5.

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The correct label for the angle formed by line segments BC and BA measuring 65 degrees is ∠CBA.

In the given statement, line segments BC and BA are mentioned as the forming elements of the angle. When labeling an angle, the convention is to use three letters, with the middle letter representing the vertex of the angle. In this case, the vertex of the angle is point B. Therefore, the correct label for the angle would include B as the middle letter.

Among the options provided, ∠CBA correctly represents the angle formed by line segments BC and BA. It indicates that the vertex of the angle is B, with C and A being the endpoints of the two line segments. The order of the letters in the angle label follows a counter-clockwise direction, starting from the initial side (BC) and moving towards the terminal side (BA).

Options A (∠A), B (∠BCA), and C (∠b) do not accurately represent the given angle or follow the convention of labeling angles. Hence, the correct label for the angle formed by line segments BC and BA, measuring 65 degrees, is ∠CBA.

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The correct label for the angle formed by line segments BC and BA, measuring 65 degrees, is ∠CBA. The name of an angle starts at one ray, passes through the vertex, and ends at the other ray.

When we talk about angles, the label or name for an angle usually starts at one ray, passes through the vertex, and ends at the other ray. In this question, the angle is generated by the line segment BC (the one side of the angle), and line segment BA (the other side of the angle), with the vertex being at point B. Therefore, the correct label for this angle is ∠CBA or ∠ABC. The choice D.) ∠CBA is correct.

Note that the letters can be reversed (starting at A and ending at C) because angle measures are the same in both directions. The first and the last letters of the label name (C and A in this case) refer to the rays or line segments forming the angle, whereas the middle letter (B) refers to the vertex of the angle.

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The sinusoidal equations for each of the six scenarios are listed below:

Case 1: h = 60 - 20 · cos 15t

Case 2: h = 60 - 30 · cos 15t

Case 3: h = 40 - 30 · cos 15t

Case 4: h = 40 - 30 · cos 18t

Case 5: h = 40 - 20 · cos 18t

Case 6: h = 60 - 20 cos 18t

In this problem we need to determine six sinusoidal equations, each associated with a scenario. The sinusoidal model is introduced below:

h = H - 0.5 · D · cos (2π · f · t)

Where:

Case 1

h = 60 - 20 · cos (5π · t)

h = 60 - 20 · cos 15.708t

h = 60 - 20 · cos 15t

Case 2

h = 60 - 30 · cos (5π · t)

h = 60 - 30 · cos 15.708t

h = 60 - 30 · cos 15t

Case 3

h = 40 - 30 · cos (5π · t)

h = 40 - 30 · cos 15.708t

h = 40 - 30 · cos 15t

Case 4

h = 40 - 30 · cos (6π · t)

h = 40 - 30 · cos 18.849t

h = 40 - 30 · cos 18t

Case 5

h = 40 - 20 · cos (6π · t)

h = 40 - 20 · cos 18.849t

h = 40 - 20 · cos 18t

Case 6

h = 60 - 20 · cos (6π · t)

h = 60 - 20 · cos 18.849t

h = 60 - 20 cos 18t

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The value of x in the quadratic equation is 2 or 2

A quadratic equation is a second-order polynomial equation in a single variable x ax²+bx+c=0. with a ≠ 0 .

There are different methods of solving a quadratic equation but in this case we are using the formula method.

x = -b±√b² -4ac)/2a

for the equation x²-4x +4 = 0

a = 1, b = -4 and c = 4

-(-4) ±√ -4² -4×1 × 4)/2

x = 4 ± √ 16-16)/2

x = 4 ± √0)/2

x = 4 /2 or 4/2

x = 2 or 2

Therefore the value of x is 2 or 2

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The frequency-domain equivalent circuit equation obtained using phasor methods can be used to analyze the behavior of the circuit at different frequencies, and to predict the performance of the circuit under various operating conditions.  

To transform a circuit from the time domain to the frequency domain using phasor methods, follow these steps:

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The corresponding rectangular equation is: (y - 2)^2 (x-1)^2 - (x-1)^2 + 1 = 0

To eliminate the parameter θ, we need to find a way to express θ in terms of x and y. Let's start by rearranging the given equations:
x = 1 + sec θ    ->    sec θ = x - 1    ->    cos θ = 1/(x-1)
y = 2 + tan θ    ->    tan θ = y - 2

Now we can use the identity cos^2 θ = 1 - sin^2 θ to solve for sin θ:
cos^2 θ = 1 - sin^2 θ
(1/(x-1))^2 = 1 - sin^2 θ
sin^2 θ = 1 - (1/(x-1))^2
sin θ = ± sqrt(1 - (1/(x-1))^2)
Note that we take the square root of both sides, and since sin θ is positive in the first and second quadrants, we take the positive square root for 0 ≤ θ < π and the negative square root for π ≤ θ < 2π. Now we can substitute for sin θ and simplify:
If 0 ≤ θ < π:
x - 1 = 1/(cos θ) = 1/sqrt(1 - sin^2 θ) = (x-1)/sqrt((x-1)^2 - 1)
sqrt((x-1)^2 - 1) = x - 1
(x-1)^2 - 1 = (x-1)^2 - 2(x-1) + 1
y - 2 = tan θ = sin θ/cos θ = ± sqrt(1 - (1/(x-1))^2)/sqrt((x-1)^2 - 1)
y - 2 = ± sqrt((x-1)^2 - 1)/(x-1)
(y - 2)(x-1) = ± sqrt((x-1)^2 - 1)
(y - 2)^2 (x-1)^2 = (x-1)^2 - 1
(y - 2)^2 (x-1)^2 - (x-1)^2 + 1 = 0

If π ≤ θ < 2π:
x - 1 = 1/(cos θ) = -1/sqrt(1 - sin^2 θ) = -(x-1)/sqrt((x-1)^2 - 1)
sqrt((x-1)^2 - 1) = -(x - 1)
(x-1)^2 - 1 = (x-1)^2 + 2(x-1) + 1
y - 2 = tan θ = sin θ/cos θ = ± sqrt(1 - (1/(x-1))^2)/sqrt((x-1)^2 - 1)
y - 2 = ± sqrt((x-1)^2 - 1)/(x-1)
(y - 2)(x-1) = ± sqrt((x-1)^2 - 1)
(y - 2)^2 (x-1)^2 = (x-1)^2 - 1
(y - 2)^2 (x-1)^2 - (x-1)^2 + 1 = 0

Therefore, the corresponding rectangular equation is:
(y - 2)^2 (x-1)^2 - (x-1)^2 + 1 = 0

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Out of the given options, a cumulative frequency plot would not be a good display to show the clusters in a distribution.

A cumulative frequency plot is a graph that displays the cumulative frequency of a dataset on the y-axis and the values of the variable being measured on the x-axis. While this plot is useful in determining the proportion of data below a certain value, it does not provide clear information about the clusters or distribution of data. A stem and leaf plot, histogram, dotplot, and boxplot are all excellent displays to show the clusters in a distribution as they provide a visual representation of the data distribution, including its shape, center, and spread. These displays can help to identify any outliers, gaps, or patterns in the data, making them useful tools for data analysis.

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The scale factors are 5.5 and 4/3

The scale factors creates a reduction

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

(x, y) ⇒ (5.5x, 5.5y)

(x, y) ⇒ (4/3x, 4/3y)

In a dilation centered at the origin, the rule is represented as

(x, y) ⇒ (5.5k, 5.5k)

Where

k = scale factor

This means that the scale factors are 5.5 and 4/3

The general rule is that

Using the above as a guide, we have the following:

The scale factors creates a reduction because they are greater than 1

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As regards whether the conditions for inference were met, C. No, the Large Counts condition is not met.

First, the random condition is met since the health inspected selected a sample of workers in random order.

The other condition is that there should not be more than 10 % condition of the sample size and this is achieved because 100 workers is less than the population.

The expected number of cases of strep throat in each department is :

Expected number of cases in special operations = (6 cases) / 3 departments = 2 cases

Expected number of cases in main floor = (13 cases) / 3 departments = 4 cases

Expected number of cases in management = (5 cases) / 3 departments = 1. 67 cases

The large counts condition is not met as the expected number of cases is not more than 5.

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Answer: V ≈ 113.04 cubic meters

Step-by-step explanation: We know that the formula for the surface area of a sphere is:

SA = 4πr²

We are given that the surface area of this sphere is 113.04 square meters, so we can set up an equation:

113.04 = 4πr²

To solve for r, we need to isolate it. First, we can divide both sides by 4π:

113.04 ÷ 4π = r²

Simplifying:

r² ≈ 9.013

To solve for r, we can take the square root of both sides:

r ≈ 2.999

Now that we know the radius of the sphere is approximately 3 meters, we can use the formula for the volume of a sphere to find the volume:

V = (4/3)πr³

Substituting our value for r:

V = (4/3)π(2.999)³

Simplifying and rounding to the nearest hundredth:

V ≈ 113.04 cubic meters

Hope that helped!

The domain of the graphed function is D. 0≤x≤ 12, where x is all non-negative integers

From the given context, it appears that you're trying to determine the domain of a function that represents the earnings of a Math Club Car Wash event, based on the number of cars washed.

So, the domain would be non-negative integers.

The range of a function encompasses all potential values that can be utilized as input (typically denominated as 'x') for the function.

Assuming that 'x' refers to the number of cars cleaned, 'x' must exclusively be a whole number greater than or equal to zero since washing a negative number or a fraction of a car is not possible.

In your options, choice D. "0≤x≤ 12, where x is all non-negative integers" seems to be the best fit. It suggests that the number of cars washed ranged from 0 to 12, and only whole numbers (non-negative integers) were considered.

Therefore, the answer would be option D.

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The bakery earn $81.65 more in sales of chocolate chips than sugar in 15 hours

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

The table of values

Where the total amount  of sales are

Sugar = 1.15 * (6h - 5)

Chocolate chips = 1.15 * (10h + 6)

In 15 hours, we have

h = 15

This means that

Sugar = 1.15 * (6 * 15 - 5) = 97.75

Chocolate chips = 1.15 * (10 * 15 + 6) = 179.4

The difference is calculated as

Difference = 179.4 - 97.75

Difference = 81.65

Hence, the bakery earn $81.65 more in sales of chocolate chips than sugar in 15 hours

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The distribution of -5X₁ and 2X₂, where X₁ and X₂ are independent normal random variables, is also a normal distribution.

When we multiply a normal random variable by a constant, it affects the mean and standard deviation of the resulting distribution.

In this case, we have -5X₁ and 2X₂, where X₁ and X₂ are independent normal random variables.

For -5X₁, the mean becomes -5 times the mean of X₁, and the standard deviation becomes 5 times the standard deviation of X₁.

Similarly, for 2X₂, the mean becomes 2 times the mean of X₂, and the standard deviation becomes 2 times the standard deviation of X₂.

Since the mean and standard deviation determine the shape of the normal distribution, the distribution of -5X₁ and 2X₂ will still be a normal distribution.

However, the mean and standard deviation will be adjusted according to the multiplication factors.

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The equation of the least squares regression line, also known as the line of best fit is y = 7.9 - 0.3x.

To find the equation of the least squares regression line, we need to determine the slope (b) and the y-intercept (a). The slope can be calculated using the formula b = r(sy / sx), where r is the correlation coefficient, sy is the standard deviation of the y-values, and sx is the standard deviation of the x-values.

Given r = -0.2, sy = 3, and sx = 2, we can calculate the slope as b = -0.2(3 / 2) = -0.3.

To find the y-intercept, we use the formula a = y-bar - b(x-bar), where y-bar is the mean of the y-values and x-bar is the mean of the x-values. Given y-bar = 7.1 and x-bar = 5, we can calculate the y-intercept as a = 7.1 - (-0.3)(5) = 7.9.

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