The distribution of actual weights of 8-ounce chocolate bars produced by a certain machine is Normal with mean 8.1 ounces and standard deviation 0.1 ounces. Company managers do not want the weight of a chocolate bar to fall below 8 ounces, for fear that consumers will complain. (a) Find the probability that the weight of a randomly selected candy bar is less than 8 ounces Forty candy bars are selected at random and their mean weight is computed. (b) Calculate the mean and standard deviation of the sampling distribution of (c) Find the probability that the mean weight of the forty candy bars is less than 8 ounces. (d) Would your answers to (a), (b), or (c) be affected if the weights of chocolate bars produced by this machine were distinctly non-Normal? Explain.

Answers

Answer 1

a. the probability that the weight of a randomly selected candy bar is less than 8 ounces is approximately 0.1587 or 15.87%. b. the standard deviation of the sampling distribution is 0.1 / sqrt(40) ≈ 0.0159 ounces. c. the probability that the mean weight of the forty candy bars is less than 8 ounces is almost 0 or very close to 0.

(a) The probability that the weight of a randomly selected candy bar is less than 8 ounces can be found by calculating the cumulative probability using the Normal distribution. Given that the distribution of weights is Normal with a mean of 8.1 ounces and a standard deviation of 0.1 ounces, we want to find P(X < 8), where X represents the weight of a candy bar.

Using the properties of the Normal distribution, we can standardize the variable X using the formula Z = (X - μ) / σ, where Z is the standard normal random variable, μ is the mean, and σ is the standard deviation.

For our case, we have Z = (8 - 8.1) / 0.1 = -1.

Using a standard normal distribution table or a calculator, we find that the cumulative probability for Z = -1 is approximately 0.1587. Therefore, the probability that the weight of a randomly selected candy bar is less than 8 ounces is approximately 0.1587 or 15.87%.

(b) The mean of the sampling distribution of the sample mean can be calculated as the same as the mean of the population, which is 8.1 ounces.

The standard deviation of the sampling distribution of the sample mean, also known as the standard error of the mean, can be calculated using the formula σ / sqrt(n), where σ is the standard deviation of the population and n is the sample size.

In our case, the standard deviation of the population is 0.1 ounces, and the sample size is 40 candy bars. Therefore, the standard deviation of the sampling distribution is 0.1 / sqrt(40) ≈ 0.0159 ounces.

(c) To find the probability that the mean weight of the forty candy bars is less than 8 ounces, we can again use the properties of the Normal distribution. Since the mean and standard deviation of the sampling distribution are known, we can standardize the variable using the formula Z = (X - μ) / (σ / sqrt(n)).

In this case, we have Z = (8 - 8.1) / (0.0159) ≈ -6.29.

Using a standard normal distribution table or a calculator, we find that the cumulative probability for Z = -6.29 is extremely close to 0. Therefore, the probability that the mean weight of the forty candy bars is less than 8 ounces is almost 0 or very close to 0.

(d) The answers to (a), (b), and (c) would not be affected if the weights of chocolate bars produced by this machine were distinctly non-Normal. This is because of the Central Limit Theorem, which states that regardless of the shape of the population distribution, as the sample size increases, the sampling distribution of the sample mean approaches a Normal distribution.

In our case, we have a sufficiently large sample size of 40, which allows us to rely on the Central Limit Theorem. As long as the sample size is large enough, the sampling distribution of the sample mean will still be approximately Normal, even if the population distribution is non-Normal.

Therefore, we can still use the Normal distribution to calculate probabilities and determine the mean and standard deviation of the sampling distribution, regardless of the population distribution being non-Normal.

Answer:

1 < -9

Step-by-step explanation:

A positive number can't be less than a negative

1 < -9
.......................

Use the table to determine a reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3? One-half One-fourth 3 DNE

Answers

The reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]

Given the limit of a function expressed as:

[tex]\lim_{x \to 3}\frac{2x^2-x+15}{x^3-5x-12}[/tex]

First, we need to substitute x = 3 into the function to have:

[tex]=\frac{2(3)^2-3+15}{3^3-5(3)-12}\\=\frac{18-3+15}{27-15-12}\\=\frac{0}{0} (indeterminate)[/tex]

Apply l'hospital rule on the function:

[tex]=\lim_{x \to 3}\frac{\frac{d}{dx} (2x^2-x+15)}{\frac{d}{dx} (x^3-5x-12)}\\=\lim_{x \to 3}\frac{4x-1}{3x^2-5}\\[/tex]

Subtitute x = 3 into the result

[tex]=\frac{4(3)-1}{3(3)^2-5}\\=\frac{12-1}{27-5}\\=\frac{11}{22}\\=\frac{1}{2}[/tex]

Hence the reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]

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Answer:

1/2

Step-by-step explanation:

i am in your walls

Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30.

a. Find the probability that the number of successes is greater than 310.
P(X ˃ 310) = _____ (round to four decimal places as needed and show work)

b. Find the probability that the number of successes is fewer than 250.
P(X ˂ 250) = _____ (round to four decimal places as needed and show work)

Answers

P(X < 250) = P(X ≤ 249) = 0 (approximately) Hence, P(X ˃ 310) = 0 and P(X ˂ 250) = 0.

Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30. The probability that the number of successes is greater than 310 and the probability that the number of successes is fewer than 250 are to be found.

Solution: a)We know that P(X > 310) can be found using normal approximation.

We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.

Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630. Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.

Using the normal approximation formula, z = (X - μ) / σwhere X = 310, μ = np and σ = √(npq), we getz = (310 - 270) / √(900*0.30*0.70)z = 4.25

Using the z-table, the probability of z being greater than 4.25 is almost zero.

Therefore, P(X > 310) = P(X ≥ 311) = 0 (approximately)

b)We know that P(X < 250) can be found using normal approximation. We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.

Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630.

Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.

Using the normal approximation formula,z = (X - μ) / σwhere X = 250, μ = np and σ = √(npq), we getz = (250 - 270) / √(900*0.30*0.70)z = -4.25Using the z-table, the probability of z being less than -4.25 is almost zero.

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Given data: n = 900, P = 0.30.

a. The probability that the number of successes is greater than 310 is 0.0000.

b. The probability that the number of successes is fewer than 250 is 0.0174.

a. The formula for finding probability of binomial distribution is:

P(X > x) = 1 - P(X ≤ x)

P(X > 310) = 1 - P(X ≤ 310)

Mean μ = np

= 900 × 0.30

= 270

Variance σ² = npq

= 900 × 0.30 × 0.70

= 189

Standard deviation

σ = √σ²

= √189

z = (x - μ) / σ

z = (310 - 270) / √189

z = 4.32

Using normal approximation,

P(X > 310) = P(Z > 4.32)

= 0.00001673

Using calculator, P(X > 310) = 0.0000(rounded to four decimal places)

b. P(X < 250)

Mean μ = np

= 900 × 0.30

= 270

Variance σ² = npq

= 900 × 0.30 × 0.70

= 189

Standard deviation

σ = √σ²

= √189

z = (x - μ) / σ

z = (250 - 270) / √189

z = -2.12

Using normal approximation, P(X < 250) = P(Z < -2.12) = 0.0174.

Using calculator, P(X < 250) = 0.0174(rounded to four decimal places).

Therefore, the probability that the number of successes is greater than 310 is 0.0000 and the probability that the number of successes is fewer than 250 is 0.0174.

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Suppose that X₁, X₂,..., X₂ form a random sample from an exponential distribution with an unknown parameter 3. (a) Find the M.L.E. 3 of 3. (b) Let m be the median of the exponential distribution, that is, 1 P(X₁ ≤m) = P(X₁ ≥ m) = 2 Find the M.L.E. m of m. ‹8 ||

Answers

(a) MLE of $\lambda$ is obtained by maximizing the log-likelihood. Suppose that X1,X2,…,XnX1,X2,…,Xn are independent and identically distributed exponential random variables with parameter λ, then the probability density function of XiXi is given by $$f(x_i;\lambda) =\lambda e^ {-\lambda x_i}, \quad x_i\geq0. $$

The log-likelihood function is given by$$\begin{aligned}\ln L(\lambda) &= \ln (\lambda^n e^{-\lambda(x_1+x_2+\cdots+x_n)}) \\&=n\ln \lambda-\lambda(x_1+x_2+\cdots+x_n).\end{aligned}$$

The first derivative of the log-likelihood function with respect to λλ is$$\frac {d\ln L(\lambda)} {d\lambda} = \frac{n}{\lambda}-x_1-x_2-\cdots-x_n.$$

The first derivative is zero when $$\frac{n}{\lambda}-\sum_{i=1} ^{n} x_i=0. $$Hence, the MLE of λλ is $$\hat{\lambda} =\frac{n}{\sum_{i=1} ^{n} x_i}. $$

Substituting the value of $\hat{\lambda} $ gives the maximum value of the log-likelihood. So, the MLE of $\lambda$ is given by $$\boxed{\hat{\lambda} =\frac{n}{\sum_{i=1} ^{n} x_i}}. $$

The MLE of $\lambda$ is $\frac {3} {\sum_{i=1} ^{n} x_i}$.

(b) The median of the exponential distribution is given by$$m = \frac {\ln (2)} {\lambda}. $$

Therefore, the log-likelihood function for median is given by$$\begin{aligned}\ln L(m) &= \sum_{i=1}^{n} \ln f(x_i;\lambda)\\&= \sum_{i=1}^{n} \ln \left(\frac{1}{\lambda}e^{-x_i/\lambda}\right)\\&= -n\ln\lambda-\frac{1}{\lambda}\sum_{i=1}^{n}x_i.\end{aligned}$$

The first derivative of the log-likelihood function with respect to mm is$$\frac {d\ln L(m)} {dm} = \frac {1} {\lambda}-\frac {1} {\lambda^2} \sum_{i=1} ^{n}x_i\ln 2. $$

The first derivative is zero when $$\frac {1} {\lambda} =\frac{1}{\lambda^2}\sum_{i=1}^{n}x_i\ln 2.$$Hence, the MLE of mm is $$\boxed{\hat{m} = \frac{\ln 2}{\bar{x}}}.$$where $\bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_i.$Therefore, the MLE of m is $\frac {\ln 2} {\bar{x}}. $

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In a normal distribution, 95% of the data falls within 1 standard deviation of
the mean.

True or False?

Answers

Answer:

False

Step-by-step explanation:

A P E X

Apply the properties of exponents to determine which of these numerical expressions
are equivalent to 5^12. Select all that apply.

Very confused and forgot the rules to figuring this out.

Answers

Answer:

Second One-

[tex] {5}^{14}. {5}^{ - 2} [/tex]

Fifth One-

[tex] {5}^{6} \: . \: {5}^{6} [/tex]

Sixth One-

[tex] \sqrt{ {5}^{24} } [/tex]

Seventh One-

[tex] {5}^{11} \: . \: 5 [/tex]

Introduction
Scientists have established a timeline of events after the Big Bang, based on astronomical observations and our understanding of the physical laws of the universe, such as gravity and the speed of light. In this lab activity, you will gather evidence to support the Big Bang theory.
Problem:
How can models demonstrate theories of our expanding universe?
Hypothesis:
Review the virtual lab demonstration in the lesson and stop the video when prompted to formulate a hypothesis. Hypothesize (or predict) what will happen to the distances between the labeled circles when you blow up the balloon ¼ full, ½ full, and ¾ full. Remember to include independent and dependent variables in your hypothesis.
The carbon dioxide represents how galaxies will spread out.
Materials:
Watch the virtual lab demonstration video within the lesson. No additional materials are needed.
Variables:
For this investigation:
List the independent variable(s):
List the dependent variable(s):
List the controlled variable(s):
Procedures:

1. Watch the virtual lab demonstration video within the lesson and record your observations in Table 1.
2. Using your expanding universe data from Table 1, construct a line graph using the volume of the below on the X axis and the distance between points on the Y axis. Be sure to include units and add titles to the graphs. Refer to the graph example and graphing tutorial in the lesson if needed.
3. Complete the Questions and Conclusion section of the lab report.
Data and Observations:
Table 1: Expanding Universe Observations

Galaxies Distance: Uninflated balloon (centimeters)
Distance: ¼ full (centimeters) Distance: ½ full (centimeters) Distance: ¾ full (centimeters)
A to B
A to C
A to D
B to C
B to D
C to D

Construct a line graph using the expanding the universe data from table 1. The volume will be plotted on the x-axis. The distance between the points will be plotted on the y-axis. Be sure to include units and add titles to the graph. Refer to the graph example and graphing tutorial in the lesson if needed.
Place your graph here.
Questions and Conclusion
1. How does the density and distribution of your “stars” change as the balloon expands?
2. How does your expanding balloon model represent an expanding universe?
3. What are some shortcomings of using this model as a replica of universe expansion?
4. How does the model you created help to show that the Steady State theory is inaccurate?
5. Suggest a way that a scientist could create an even more accurate model of universe expansion.
6. What will happen to the gravitational force between stars as the universe continues to expand?
In conclusion, how did your prediction of distances between points compare to your experimental results? All I truly need is the variables question 3 and 5 and I’m good :) thank you <3 this is for science but I didn’t know which one to pick so I picked a random one lol

Answers

For scientific tools to be measured at such a height...you must first begin multiplication and do the simple division.

Let k be a constant and consider the function f(x,y,z) = kx? - kry + y2 -2yz - 22. (Thus, for example, if k = 4, then f(xy.z) = 4x2 - 4xy + 2y2-2yz -22) For what values (if any) of the constant k does / have a (nondegenerate) local maximum at (0.0.0)? For what values of k does / have a (nondegenerate) local minimum at (0.0.0)? Be sure to explain your reasoning.

Answers

The values of k for which the function f(x, y, z) = kx² - kry + y² - 2yz - 22 has a nondegenerate local maximum at (0, 0, 0) are when k > 0.

To find the critical points of the function, we need to calculate the partial derivatives with respect to each variable:

∂f/∂x = 2kx ∂f/∂y = -kr + 2y - 2z ∂f/∂z = -2y

2kx = 0 => x = 0 (Equation 1) -kr + 2y - 2z = 0 => r = y - z (Equation 2) -2y = 0 => y = 0 (Equation 3)

From Equation 3, we can see that y = 0. Substituting this into Equation 2, we get:

r = 0 - z r = -z (Equation 4)

The Hessian matrix is given by:

Calculating the second-order partial derivatives:

∂²f/∂x² = 2k ∂²f/∂y² = 2 ∂²f/∂z² = 0 ∂²f/∂x∂y = 0 ∂²f/∂y∂z = -2 ∂²f/∂z∂x = 0

Thus, the Hessian matrix becomes:

H = | 2k 0 0 | | 0 2 -2 | | 0 -2 0 |

D = ∂²f/∂x² ∂²f/∂y² ∂²f/∂z² + 2∂²f/∂x∂y ∂²f/∂y∂z ∂²f/∂z∂x - (∂²f/∂x² ∂²f/∂y∂z ∂²f/∂z∂x + ∂²f/∂y² ∂²f/∂z∂x ∂²f/∂x∂y ∂²f/∂z²)

Substituting the partial derivatives we calculated earlier:

D = (2k)(2)(0) + 2(0)(-2)(0) - (2k)(-2)(0) - (2)(0)(0) D = 0

If the determinant D is zero, the second derivative test is inconclusive. In such cases, we need to consider the eigenvalues of the Hessian matrix.

To find the eigenvalues, we solve the characteristic equation:

det(H - λI) = 0

where λ is the eigenvalue and I is the identity matrix. Substituting the values from the Hessian matrix:

| 2k-λ 0 0 | | 0 2-λ -2 | | 0 -2 -λ |

The characteristic equation becomes:

(2k - λ)((2 - λ)(-λ) - (-2)(0)) - (0)((2 - λ)(-2) - (0)(0)) = 0 (2k - λ)(λ² - 2λ) = 0

From this equation, we can see that one eigenvalue is (2k - λ) = 0, which implies λ = 2k.

For our case, we have one eigenvalue (λ = 2k). Thus, the sign of λ depends on the value of k.

When k < 0, the point (0, 0, 0) is a nondegenerate local minimum. When k = 0, the second derivative test is inconclusive, and further analysis would be required to determine the nature of the critical point.

To know more about function here

x/2+<14

Then, multiply both sides by 2 to get x alone

x<28

On a number line, there would be an open circle (not filled in dot) on 28, and the entire left side of the number line would be filled in

Answer:

x<28

Step-by-step explanation:

x/2+4<18

multiply the 2 on both sides to get rid of it

x+8<36

isolate the x

x<28

on the number line, it's an open circle with the arrow pointing to the left.

Compute the pooled variance given the following data:

N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8

Round to two decimal places

Answers

By computing the pooled variance given the following data N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8, the pooled variance is 436.40.

To compute the pooled variance given N_1 = 18, n_2 = 14, s_1 = 7, s_2 = 8, we can use the formula below;

S_p² = [(n₁ - 1)S₁² + (n₂ - 1)S₂²] / (N - 2),

where S_p² = pooled variance, n₁ = sample size of first group, n₂ = sample size of second group, S₁² = variance of first group, S₂² = variance of second group, and N = total sample size.

To plug in the values, we have: N₁ = 18n₂ = 14S₁ = 7S₂ = 8

Substituting the values into the formula above we get;

S_p² = [(18 - 1)(7²) + (14 - 1)(8²)] / (18 + 14 - 2)S_p² = (17 × 49 + 13 × 64) / 30S_p² = 436.4

Round off to two decimal places to get 436.40.

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Let f(x)= e = 1+x. - a) Show that f has at least one real root (i.e. a number c such that f(c) = 0). b) Show that f cannot have more than one real root.

Answers

The function f(x) = e^(1+x) has at least one real root. The function f(x) = e^(1+x) cannot have more than one real root.

To show that f(x) has at least one real root, we need to find a value of x for which f(x) equals zero. Let's set f(x) = 0 and solve for x:

e^(1+x) = 0

Since e^(1+x) is always positive for any real value of x, there is no value of x that makes f(x) equal to zero. Hence, f(x) = e^(1+x) does not have any real roots. Therefore, we cannot show that f(x) has at least one real root.

To show that f(x) cannot have more than one real root, we need to demonstrate that there cannot be two distinct real values, say c1 and c2, such that f(c1) = f(c2) = 0. Let's assume that f(x) = 0 at two distinct values, c1 and c2:

e^(1+c1) = e^(1+c2) = 0

However, this equation is not possible since e^(1+c1) and e^(1+c2) are always positive for any real values of c1 and c2. Therefore, f(x) = e^(1+x) cannot have more than one real root.

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What is the value of x in the equation 13−2(+4)=8+1 13 x − 2 ( x + 4 ) = 8 x + 1 ?

Answers

Answer: x= 12/103

alternate form x =0.116505

Step-by-step explanation:

Picsart , Inshot , Gandr , Photo lab and Viva video.

The radius of a circle is 8 inches. What is the area?

r=8 in

Give the exact answer in simplest form.

Answers

Answer:

Radius = 8 inches

Area = [tex]\pi {r}^{2} [/tex]

[tex]area = \pi( {8}^{2})[/tex]

[tex] = 64\pi \: square \: inches[/tex]

A = [tex]201.06 ^{2} \: inches[/tex]

Step-by-step explanation:

Hope it is helpful....

PLEASE PLEASE PLEASE HELP 7 points

Answers

Answer:

a) 2x+(x+36)=90

Step-by-step explanation:

b) A1+A2=90°. (A=angle)

2x+(x+36)=90

2x+x+36=90

3x+36=90

3x=90-36

x=54/3

x=18

then A1=2x=2*18=36°

A2=x+36=18+36=54°

Please help :) thank you to whoever does help!?

Answers

Answer:[tex]\boxed{\boxed{\sf{a=5}}}[/tex]Solution Steps:

____________________________

1.) Fill in the formula: F = [tex]B^2-C^2=A^2[/tex]F = [tex]12^2-13^2=A^2[/tex]F = [tex]144-169=A^2[/tex]

2.) Subtract:[tex]169-144=25[/tex]

3.) Find the square root of 25:[tex]\sqrt{25} =5[/tex]

So your answer would be C, [tex]\bold{a=5}[/tex].

____________________________

John is cutting 3 wooden sticks to build part of a kite frame. The part he is building must be a right triangle.
Select all the possible lengths, in inches, of the sticks John could cut to make a right triangle.
A. 6, 8, 10
B. 2, 5, 10
C. 2, 3, 5
D. 12, 16, 20
E. 3, 4, 13

Answers

Answer:

I found 2 answers for this one

Step-by-step explanation:

B- 2,5,10

D-12,16,20

The possible length in inches of the stick that will make a right angle triangle are as follows

6, 8, 1012, 16, 20

What is a right angle triangle?

A right angle triangle has one of its angles as 90 degrees.

The right angle triangle must obeys the Pythagorean theorem.

c² = a² + b²

where

c = hypotenusea and b are the other legs.

Therefore, the sides that makes a right angle are as follows:

6, 8 , 10 : 6² + 8² = 10²

12, 16, 20 : 12² + 16² = 20²

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Park trails and their elevation:
Sand trail has a -2 feet elevation
Cactus Trail has 15 feet elevation
Southern Trail has a -12 feet elevation
Rocky Trail has 42 feet elevation

Chi hiked the Rocky Trail What is the opposite of the elevation of the Rocky Trail?

Answers

Answer:

fjekwnkewgnelwnlgnendndj

Step-by-step explanation:

What are the solutions of the system?
Sy=2x+6
\y= r2 +5 +6
O (0, -3) and (6, 0)
O(-3,0) and (-2, 0)
O (-3, 0) and (0, 6)
O
(0, 6) and (-2, 0)
ہے
1 2

Answers

The solution to the system is {eq}(x, y) = (2, 1){/eq}. For the given system, we have the solutions (-3, 0) and (0, 6).

To solve a system of equations, we must find the value of each variable that satisfies both equations. One of the most common methods for solving systems of equations is called substitution.

In substitution, we solve for one variable in one equation and then plug that expression into the other equation.

For example, if we have the system {eq}2x + y = 5,

\quad 4x - y = 7 {/eq},

we can solve for y in the first equation: {eq}y = 5 - 2x {/eq} Then we substitute this expression for y into the second equation:

{eq}4x - (5 - 2x) = 7 {/eq}

Simplifying gives {eq}6x = 12 {/eq}, or {eq}x = 2 {/eq} Once we have a value for one variable, we can substitute it into either equation to find the value of the other variable.

Using {eq}y = 5 - 2x {/eq}, we have {eq}y = 5 - 2(2) = 1 {/eq}These solutions represent the values of x and y that satisfy both equations in the system. To check,

we can substitute each solution into both equations to ensure they are valid. If both equations are satisfied, then we have found the correct solutions.

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Find the equations of the images of the following lines when reflected in the x-axis. a.y= 3x b.y= -x c. x = 0.

Answers

The equations of the images are after the transformations are

a. y = -3x

b. y = x

c. x = 0

How to determine the equations of the images

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

a. y = 3x

b. y = -x

c. x = 0.

The rule of the lines when reflected in the x-axis is

(x, y) = (x, -y)

This means that the functions are negated

So, we have the images to be

a. y = -3x

b. y = x

c. x = 0

Hence, the equations of the images are

a. y = -3x

b. y = x

c. x = 0