In the coordinate plane, the point A(-2,4) is translated to the point A’(-4,3). Under the same translation, the points B(-4,8) and C(-6,2) are translated to B’ and C’, respectively. What are the coordinates of B’ and C’?

The given statement "In order for a matrix B to be the inverse of A, the equations AB = I and BA = I must both be true." is true because of the definition of the inverse matrix.

An inverse matrix is obtained by dividing the adjugate of the given matrix by the determinant of the given matrix.

An inverse matrix is also known as a reciprocal matrix.

In order for a matrix B to be the inverse of A, both equations AB = I (Identity matrix) and BA = I must be true.

This is because the inverse of a matrix A, denoted as A⁻¹ (in this case, matrix B), should satisfy these conditions for it to be a true inverse.

When a matrix is multiplied by its inverse, the result is the identity matrix.

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The answer to the given statement is as follows:

It is inappropriate to apply the Empirical Rule to a population that is right-skewed

b. False.

The given statement is false because the rule of thumb, also known as the 68-95-99.7 rule, is a statistical rule that applies to the normal distribution. This rule was lost in our sample, with about 68% of the data falling within one standard deviation of the mean for a normal distribution, and about 95% of the data falling within two standard deviations from the mean, and about 99.7% of the data being lost in our sample. The deviation from the mean is the difference between the mean of the standard deviation.

Although the rule of thumb is most true for symmetric normal distributions, it can also be used for distributions, including right-skewed distributions.

However, as the distribution becomes more skewed, the rule of thumb may not be correct. In a right-skewed distribution, the mean is greater than the median and the tails of the distribution are to the right. In such a distribution, a rule of thumb might estimate the proportion of data that is one or two standard deviations from the mean.

Despite this limitation, the rule of thumb can be a useful tool for understanding the spread of data in right-skewed distributions. However, it is important to know that this law can predict the percentage of data in a given situation.

In such cases, other methods such as quartiles or percentages are more effective for analyzing the distribution of the data.

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The rational number that helps complete the circuit is given as follows:

0.222...

For this problem, we have two options to complete the circuit, as follows:

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The time of the rth arrival in a Poisson process follows a gamma distribution with parameters r and λ, where λ is the rate parameter.

The density function for the time of the rth arrival is: f(w) = λ^r * w^(r-1) * e^(-λw) / (r-1) where w is the time of the rth arrival. This density function gives the probability density of the time of the rth arrival occurring at a specific time w, given that there have been n arrivals up to time t.

The density function is derived from the fact that the time between successive arrivals in a Poisson process follows an exponential distribution with rate parameter λ, and the time of the rth arrival is the sum of r independent exponential random variables.

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The correct answer is 26%.

In an exponential regression model, the exact percentage of change can be calculated as: (exp(β1) - 1) × 100. Given that β1 = 0.23, let's calculate the percent increase in E(y):

Step 1: Calculate exp(β1): exp(0.23) ≈ 1.259
Step 2: Subtract 1: 1.259 - 1 = 0.259
Step 3: Multiply by 100: 0.259 × 100 = 25.9

The percent increase in E(y) is approximately 26%. So, the correct answer is 26%.

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The answer is True we use observed value not critical value

The answer is True. When computing the effect size, you use the observed value of t in the formula rather than the critical value. The critical value is used to determine statistical significance, while the effect size is calculated using the observed value to measure the strength or magnitude of the relationship between variables.

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The double line graph for the amount in the River Bank and Town Bank indicates;

13 a. $10,000

13 b. Two year

c. $4,000 per year

A graph of a function shows the relationship that exists between the input and output values of the function.

The graph with the lines that have markings is the graph of the River Bank

The graph with the lines without markings is the graph of the Town Bank

13 a. In the year 2,000, the money in River Bank = $6,000

The money in Town Bank = $4,000

The combined amount in both banks in the year 2,000 is therefore;

Amount  = $6,000 + $4,000 = $10,000

13 b. Town Bank had more money than Rivers Bank in the years; 1998, 2002

Therefore;

Towns Bank had more money that Rivers Bank in 2 years

c. The amount River Bank had between 1998 to 2004 are;

$5,000 + $3,000 + $6,000 + $4,000 + $1,000 + $7,000 + $2,000 = $28,000

The number of years between 1998 and 2004 = 7 years

The mean number of dollars per year River Bank had = $28,000/7 = $4,000 per year

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When a regression model is created to fit a sample set of data, its prediction ability is reduced overfitting. Thus, option C is correct.

Overfitting is a phenomenon in machine learning where a regression model is trained too well on the sample data.

to the point where it starts to memorize the data instead of learning the underlying patterns or trends. As a result, the overfitted model may not generalize well to unseen data and may exhibit poor predictive power when used for making predictions on new data.

The term "compromising predictive power" in the question suggests that the model is not able to accurately predict outcomes on new, unseen data due to overfitting.

Essentially, the model becomes too specialized to the sample data it was trained on and loses its ability to generalize to new data points.

Flooding is not a term related to machine learning or regression modeling. Binary choice refers to a decision between two options and is not related to overfitting.

Therefore, When a regression model is created to fit a sample set of data, its prediction ability is reduced overfitting

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For each of the third-order linear equations, the basis for the solution space can be found by solving the characteristic equation and then finding the corresponding linearly independent solutions. The solutions for each equation are:

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The recursion, and subsequent terms are defined in terms of previous terms in the sequence

A) The recursive definition for the sequence An is:

An = (4n-2)An-1 + 4Ao, with A1 = 4Ao.

B) The recursive definition for the sequence An is:

An = n(n+1)An-1 + Ao, with A1 = Ao.

C) The recursive definition for the sequence An is:

An = 1 + (-1)nAn-2tAo, with A1 = Ao and A2 = 1 - Ao.

D) The recursive definition for the sequence An is:

An = n^2An-1 + Ao, with A1 = Ao.

These recursive definitions define each term of the sequence An as a function of one or more previous terms in the sequence, starting with a basis case. The basis case provides the starting point for the recursion, and subsequent terms are defined in terms of previous terms in the sequence.

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The alternating series error bound is (C) 1/26.7!, since 26.7! is the smallest factorial greater than [tex]120*M_5.[/tex]

The alternating series error bound for an alternating series of the form [tex]\sum (-1)^n b_n[/tex]is given by [tex]|R_n| < = b_{(n+1)}[/tex], where [tex]R_n[/tex] is the remainder term and [tex]b_n[/tex] is the absolute value of the (n+1)th term in the series.

In this case, the fourth degree Taylor polynomial for f about x = 0 is given by:

[tex]P_4(x) = f(0) + f'(0)x + (f''(0)/2)x^2 + (f'''(0)/6)x^3 + (f''''(0)/24)x^4[/tex]

The alternating series error bound for the approximation of f(x) by [tex]P_4(x)[/tex]is therefore:

[tex]|R_4(x)| < = |f(x) - P_4(x)| < = (M/5!) |x - 0|^5,[/tex]

where M is an upper bound for [tex]|f^{(5)}(c)[/tex]| on the interval [0,x] for some c between 0 and x.

Since the Taylor series for f about x=0 converges to f for all real x, we know that M is finite. Therefore, we can find an upper bound for [tex]|f^{(5)}(c)|[/tex]on [0,-1] using the Mean Value Theorem.

Let g(x) = f''''(x). Then, by the Mean Value Theorem, there exists some c between 0 and -1 such that:

g(c) = (g(0) - g(-1))/(-1 - 0) = g(0) - g(-1)

Since the fourth derivative of f is continuous, g is continuous on the interval [0,-1]. Therefore, by the Extreme Value Theorem, g attains its maximum and minimum values on [0,-1].

Let[tex]M_5 = max{|g(x)| : x in [0,-1]}[/tex]. Then we have:

[tex]|R_4(x)| < = M_5/5! |x|^5 = M_5/120[/tex]

Therefore, the alternating series error bound is (C) 1/26.7!, since 26.7! is the smallest factorial greater than [tex]120*M_5.[/tex]

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When analyzing statistical data, it is important to consider the assumptions underlying the methods used. One such assumption is homogeneity of variance (HOV), which tests whether the variance of two groups is equal.

Why did we not have to test for hov with our paired data?

We do not need to test for the homogeneity of variance (HOV) assumption in paired data because the assumption only applies to independent samples. In paired data, the same individuals are measured twice, and the two sets of measurements are dependent. Therefore, the assumption of equal variances between two groups does not apply, and we do not need to test for it.

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Why did we not have to test for HOV with our paired data?

Answer:

4 - (2 / 13) = 3.84615385.

But I understand that you don't want to answer.

Step-by-step explanation:

A general solution

[tex]y(x) = c1e^{3x} + c2e^{-x} + c3*e^{-4x}[/tex]

To verify that the given functions form a fundamental solution set for the differential equation y''' + 2y" - 11y' - 12y = 0, we can use the Wronskian. The Wronskian is defined as:

W(x) = | y1(x) y2(x) y3(x) |

| y1'(x) y2'(x) y3'(x) |

| y1''(x) y2''(x) y3''(x) |

where y1(x), y2(x), and y3(x) are the given functions.

Using the given functions, we can compute the Wronskian as follows:

W(x) = |[tex]e^{3x} e^{-x} e^{-4x} || 3e^{3x} -e^{-x} -4e^{-4x} || 9e^{3x} e^{-x} 16e^{-4x}[/tex]|

Expanding the determinant, we get:

[tex]W(x) = e^{3x}(-e^{-x}*16e^{-4x} + e^{-4x}e^{-x}) - (-e^{-x}(-4e^{-4x}) - (-e^{3x})*16e^{-4x})e^{3x} + (3e^{3x}(-e^{-x}*e^{-4x}) - e^{-x}*9e^{3x}*16e^{-4x})[/tex]

Simplifying, we get:

W(x) = -23e^(-3x)

Since the Wronskian is nonzero everywhere, the functions {e^(3x), e^(-x), e^(-4x)} form a fundamental solution set for the differential equation.

To find the general solution of the differential equation, we can use the formula:

y(x) = c1y1(x) + c2y2(x) + c3*y3(x)

where c1, c2, and c3 are constants. Substituting the given functions, we get:

[tex]y(x) = c1e^{3x} + c2e^{-x} + c3*e^{-4x}[/tex]

This is the general solution of the given differential equation.

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The area of the segment is 1.68 squared.

The area of the shaded segment is the subtraction of the area of the triangle from the area of the sector OMN.

Therefore,

area of the segment  = ∅ / 360 πr² - 1 / 2r²sin(∅)

area of the segment  = 30 / 360 π(12)² - 1 / 2 (12)² sin 30°

area of the segment  = 1 / 12 π(144) - 1 / 2(144)0.5

area of the segment  = 12π  - 36

area of the segment  = 12(3.14) - 36

area of the segment  = 37.68 - 36

area of the segment  = 1.68 inches squared.

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The derivative dy/dx is (6y - eˣ - yeˣ)/y².

To find dy/dx for the given equation ex/y = 6x - y, you can use implicit differentiation without taking the log.

Given the equation [tex]e^\frac{x}{y}[/tex] = 6x - y, you do not need to take the log of both sides. Instead, start by applying implicit differentiation to both sides with respect to x:

1. Differentiate ex with respect to x: d(eˣ)/dx = ex
2. Differentiate y with respect to x: d(y)/dx = dy/dx
3. Differentiate 6x with respect to x: d(6x)/dx = 6
4. Apply the quotient rule to d([tex]e^\frac{x}{y}[/tex])/dx: d(ex/y)/dx = (y * d(eˣ)/dx - eˣ * d(y)/dx) / y² = (y * eˣ - eˣ * dy/dx) / y²
5. Set the derivatives equal: (y * eˣ - eˣ * dy/dx) / y² = 6 - dy/dx

Now, solve for dy/dx:

6. Multiply both sides by y²: y * eˣ - eˣ * dy/dx = 6y² - y² * dy/dx
7. Rearrange terms: dy/dx * (y² + eˣ) = 6y² - y * eˣ
8. Solve for dy/dx: dy/dx = (6y² - y *eˣ) / (y² + eˣ) = (6y - ex - yeˣ) / y²

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Answer:

58

Step-by-step explanation:

Put the numbers in chronological order and find the middle number

0, 0, 38, 47, 58, 65, 78, 99, 227

Helping in the name of Jesus.

Answer: x * y = 2x + 5y. Formula used: x * y = 2x + 5y. Calculation: When x = 3, and y = 5. ⇒ 2x + 5y = (2 × 3) + (5 × 5) = 6 + 25 = 31

Step-by-step explanation:

The correct answer is option B, 3360x6y4.

The term containing x6 in the expansion of (x+2y)10 will arise from selecting the x term exactly 6 times out of 10 terms. We can select the x term in different ways by using the binomial theorem.

The binomial theorem states that for any positive integers n and k, the coefficient of x^(n-k) in the expansion of (x+y)^n is given by the binomial coefficient (n choose k), which is written as nCk and can be calculated using the formula:

nCk = n! / (k! * (n-k)!)

where ! denotes the factorial function.

In our case, we need to find the coefficient of x^6 in the expansion of (x+2y)^10, which is given by:

10C6 * x^6 * (2y)^4

= 210 * x^6 * 16y^4

= 3360x^6y^4

Therefore, the correct answer is option B, 3360x6y4.

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By applying L'Hôpital's Rule 100 times or analyzing the degree of the polynomial, you'll find that L equals 0. Since L < 1, by the Ratio Test, this series converges.

To determine whether the series converges, you can use the Ratio Test. For the series Σk=1 to infinity (k100/(k+5)!), apply the Ratio Test by finding the limit as k approaches infinity of the absolute value of the ratio of consecutive terms:

L = lim (k→∞) |( (k+1)100 / (k+6)! ) / ( k100 / (k+5)! )|

Upon simplification, you'll find:

L = lim (k→∞) |(k+1)100 * (k+5)! / (k100 * (k+6)!)|

Cancel out the factorials and further simplify:

L = lim (k→∞) |(k+1)100 / (k100 * (k+6))|

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The absolute value equation is:

|x - 15| = 0

Here, we have,

to write the absolute value equation:

We want an absolute value equation that only has the solution x = 15.

So we must have something equal to zero (so we avoid the problem with the signs that we can have with other numbers)

So the equation will be something like:

|x - a| = 0

And the solution is 15, so:

|15 - a | = 0

then a = 15

The equation is:

|x - 15| = 0

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Answer:

|x - 15| = 0

Notes:

You're probably going to give the other person Brainliest, but could you maybe consider giving it to me?

The probabilities of

a) P(3) = 0.07

b) P(<1) = 0.67

c) P(≤2) = 0.93

d) P(>2) = 0.07

Probability is a measure of the likelihood of an event occurring. It is a numerical value between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.

a) The probability of getting a value of 3 is simply the value of P(3),

P(3) = 0.07

b) To find the probability of getting a value less than 1, we add the probabilities of getting 0 and 1,

P( < 1) = P(0) + P(1) = 0.02 + 0.65 = 0.67

c) To find the probability of getting a value less than or equal to 2, we add the probabilities of getting 0, 1, and 2,

P( ≤ 2) = P(0) + P(1) + P(2) = 0.02 + 0.65 + 0.26 = 0.93

d) To find the probability of getting a value greater than 2, we simply look at the probability of getting a value of 3, which is 0.07.

P( > 2) = P(3) = 0.07

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The given question is incomplete, the complete question is:

X 0 1 2 3

P(x) .02 .65 .26 .07

Find the probabilities

a) P(3) =

b) P( < 1) =

c) P( ≤ 2) =

d) P (>2) =

The solution to the differential equation is: [tex]y(x) = c1 (x-9)^4 + c2 (x-9)[/tex]where c1 and c2 are constants of integration.

To solve this differential equation using the method of "reducing to a polynomial equation", we can make the substitution:

x - 9 = t,

so that x = t + 9 and y(x) = y(t+9).

We can then rewrite the differential equation in terms of t as follows:

[tex][(t+9)^2] y'' - 9(t+9) y' + 16y = 0[/tex]

We can now make the substitution [tex]y = (t+9)^m[/tex], where m is some constant to be determined.

Taking the first and second derivatives of y with respect to t, we get:

[tex]y'=m(t+9)^{(m-1)}[/tex]

[tex]y'' = m(m-1) (t+9)^{(m-2)}[/tex]

Substituting these expressions into the differential equation, we get:

[tex][(t+9)^2] m(m-1)(t+9)^{m-2} - 9(t+9) m(t+9)^{m-1} + 16(t+9)^m = 0[/tex]

Simplifying, we get:

m(m-1) - 9m + 16 = 0

Solving this quadratic equation for m, we get:

m = 4 or m = 1

Therefore, the general solution to the differential equation is given by:

[tex]y(t) = c1 (t+9)^4 + c2 (t+9)[/tex]

where c1 and c2 are constants of integration.

Substituting back to x, we have:

[tex]y(x) = c1 (x-9)^4 + c2 (x-9)[/tex]

where c1 and c2 are constants of integration.

Therefore, the solution to the differential equation is:

[tex]y(x) = c1 (x-9)^4 + c2 (x-9)[/tex]

where c1 and c2 are constants of integration.

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The Correct Option is C that 71 x 10-² of the following gives the value of the expression below written in scientific notation.

The basic objective of scientific notation is to make computations with unusually big or small numbers simpler. The following examples demonstrate how all of the digits in a number in scientific notation are relevant because zeros are no longer utilised to set the decimal point.

The statement for a number n in scientific notation is of the type a10b, where an is an integer such that 1|a|10. B is also an integer. Multiplication: To get the full amount in scientific notation, multiply the decimal values. Add the 10 power exponents after that.

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the Taylor polynomial T3(x) for the function f(x) = x[tex]e^{(-5x)}[/tex]centered at a = 0 is T3(x) = 1 + 0.25[tex]x^{2}[/tex] - 0.0083[tex]x^{3}[/tex]

To find the Taylor polynomial T3(x) for the function f(x) = x[tex]e^{(-5x)}[/tex]centered at a = 0, we need to find the first four derivatives of f(x) and evaluate them at x = 0:

f(x) = x[tex]e^{(-5x)}[/tex]

f'(x) = [tex]e^{(-5x)}[/tex] - 5x[tex]e^{(-5x)}[/tex]

f''(x) = 25x[tex]e^{(-5x)}[/tex] - 20[tex]e^{(-5x)}[/tex]

f'''(x) = -125x[tex]e^{(-5x)}[/tex] + 75[tex]e^{(-5x)}[/tex]

f''''(x) = 625x[tex]e^{(-5x)}[/tex] - 500[tex]e^{(-5x)}[/tex]

Now, we can write the Taylor polynomial T3(x) as:

T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] + (f'''(0)/3!)[tex]x^{3}[/tex]

T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] + (f'''(0)/3!)[tex]x^{3}[/tex]

T3(x) = f(0) = 0 + f'(0)x = [tex]e^{0}[/tex] × 1 - 5 × 0 × [tex]e^{0}[/tex] = 1

T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] = 1 + 0.25[tex]x^{2}[/tex]

T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] + (f'''(0)/3!)[tex]x^{3}[/tex] = 1 + 0.25[tex]x^{2}[/tex] - 0.0083[tex]x^{3}[/tex]

Therefore, the Taylor polynomial T3(x) for the function f(x) = x[tex]e^{(-5x)}[/tex]centered at a = 0 is T3(x) = 1 + 0.25[tex]x^{2}[/tex] - 0.0083[tex]x^{3}[/tex]

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The solution to the initial value problem. y' (t) - 2y = 6, y(2) = 2 is  y(t) = -3 + (5/e^4)e^(2t)..

To solve the initial value problem y'(t) - 2y = 6 with y(2) = 2,

we will use an integrating factor and the given initial condition. Here's the step-by-step solution:

1. Identify the integrating factor: The integrating factor is e^(-2t).

2. Multiply the equation by the integrating factor: e^(-2t)y'(t) - 2e^(-2t)y = 6e^(-2t).

3. Observe that the left side of the equation is the derivative of the product y(t)e^(-2t): (y(t)e^(-2t))' = 6e^(-2t).

4. Integrate both sides with respect to t: ∫(y(t)e^(-2t))' dt = ∫6e^(-2t) dt.

5. Integrate: y(t)e^(-2t) = -3e^(-2t) + C.

6. Solve for y(t): y(t) = -3 + Ce^(2t).

7. Apply the initial condition y(2) = 2: 2 = -3 + Ce^(4).

8. Solve for C: C = (5/e^4).

9. Substituting C back into the solution for y(t): y(t) = -3 + (5/e^4)e^(2t).
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Answer: D

Explanation: Since 21 out of 24 bean seeds sprouted in the first class, the probability of a bean seed sprouting is 21/24, or 0.875. This information does not provide any information about the seeds in the other five classes, other than that they were all planted using the same method. Therefore, we cannot make a definitive statement about how many seeds will or will not sprout in the other classes. Option A is supported by the given information, since 87.5% of the seeds in the first class sprouted. Option B is not necessarily supported by the given information, as it depends on how many seeds were used in total. Option C is not directly supported by the given information, but is a possible conclusion based on the probability of a seed sprouting. Option D is contradicted by the given information, since at most 3 out of 24 seeds did not sprout in the first class.

The coordinates of point P are (-3, -1).

The coordinates of point P that divides the line segment AB in the ratio 1:4 is calculated as follows;

let the ratio = a : b = 1:4

P = ( (bx₂ + ax₁)/(b + a), (by₂ +  ay₁)/( b + a) )

Where;

The coordinate of point P is calculated as follows;

P = ( (4(-2) + 1 (-7))/(4 + 1),  (4(0) + 1(-5) )/(4 + 1))

P = (-8 - 7)/(5), (0 - 5)/(5)

P = (-15/5), (-5/5)

P = (-3, - 1)

Thus, the coordinate of point P is determined by applying  ratio formula on a line segment.

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The electric field strength decreases as the distance from the plane increases.

The electric field strength near an infinite plane of charge is given by:

E = σ/2ε0

where E is the electric field strength, σ is the surface charge density of the plane, and ε0 is the electric constant.

The electric field strength at a point near the plane depends on the distance from the plane and the orientation of the point relative to the plane.

Assuming that the surface charge density is constant, we can rank the electric field strength from largest to smallest based on the distance from the plane:

Point closest to the plane

Point at a distance of 2 times the distance from point 1

Point at a distance of 3 times the distance from point 1

Point at a distance of 4 times the distance from point 1

Point at a distance of 5 times the distance from point 1

This is because the electric field strength decreases as the distance from the plane increases.

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