Which expressions are equal to 8 × 105? Select all that apply
A. 8 x 1,000
B. 8 x 10,000
C. 8 x 1000,000
D. 8 x 10 x 10 x 10 x 10
E. 8 x 10 x 10 x 10 x 10 x 10

The maximum height of the ball is 2.437 meters.

What we have to change if the maximum height of the ball is more than 2.4 meters is the initial value of time.

We are given the quadratic function that represents the path of the ball kicked as; h(t) = -4.9t² + 6t + 0.6

Where;

t is the time in seconds and h is the height in meters above the ground.

This polynomial is a second degree polynomial that has a negative leading coefficient . Thus, this tells us that it is a downward parabola which gives the maximum value at vertex.

If the parabola is defined as;

f(x) = ax² + bx + c

Then the vertex of the parabola is;

(-b/2a, f(-b/2a))

x-coordinate of vertex = -6/(2 * - 4.9) = 0.612

y-coordinate of vertex = f(0.612) = 2.437

This tells us that the vertex is (0.612, 2.437) and the maximum height of the ball is 2.437 meters.

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Answer:

Step-by-step explanation:

The following is an example of a quadratic inequality: x^2 + 2x - 8 > 0.

This is an example of a quadratic inequality because it is an inequality involving a quadratic function (x^2 + 2x - 8) which is of the form ax^2 + bx + c where a, b, and c are constants.

Quadratic inequalities can be solved by factoring the quadratic expression, using the Quadratic Formula, or graphing the inequality. In order to solve a quadratic inequality, it is important to understand the types of solutions that can be expected.

Solutions may include one solution, two solutions, or no solutions. Additionally, when graphing a quadratic inequality, the graph may be a line or a region. To determine which type of graph to use, the type of inequality must be identified (greater than or less than). If the inequality is a greater than or less than inequality, then the graph will be a region. If the inequality is an equal to inequality, then the graph will be a line.

Solving and graphing quadratic inequalities is an important skill in algebra as it allows us to visualize and solve complex problems.

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The area of the region bounded by

\[y=\frac{7}{(4+x)^2}+\frac{5}{7+x^2},\ y=0,\ x\ge5\]is 0.0188 (rounded to four decimal places).

Here's how to get the solution:

We are asked to find the area of the region bounded by the two curves.

The curves intersect at (5, 0) because x can not be less than 5.

They meet again at the point x ≈ 1.281.

Now, we must find the integrals for both functions in the given range.

We'll call the first function "f (x)" and the second "g (x)."

f(x) = 7 / (4 + x)² + 5 / (7 + x²)

g(x) = 0

The area between the two curves is obtained by finding the integral of the difference of the two functions.

The area is given by:

\[\int_{5}^{1.281} [f(x) - g(x)] dx\]

Since there is no point of intersection beyond x ≈ 1.281, we will use this value for the limit of integration.

Integrating:

\[\begin{aligned} &\int_{5}^{1.281} [f(x) - g(x)] dx \\ =& \int_{5}^{1.281} \left[\frac{7}{(x+4)^2}+\frac{5}{x^2+7}-0\right] dx \\ =& -\left[\frac{7}{4+x}+\sqrt{7}\tan^{-1}\left(\frac{x}{\sqrt{7}}\right)\right]_{5}^{1.281} \\ =& 0.0188. \end{aligned}\]

Thus, the area of the region is 0.0188.

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The predicted number of runs for the player with only 86 hits can be calculated using the equation Runs = Hits + Walks - Home Runs. Plugging in the values given, we get: Runs = 86 + Walks - Home Runs. Therefore, the predicted number of runs for the player is dependent on the number of walks and home runs they have.

To solve for the predicted number of runs, we can use the following steps:

Therefore, the predicted number of runs for the player with only 86 hits can be calculated by plugging in the values of the number of walks and home runs into the equation Runs = Hits + Walks - Home Runs.

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well, the midpoint of segment AB is the center of the circle, so

\(~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{9}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 6 -2}{2}~~~ ,~~~ \cfrac{ 9 +1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ 10 }{2} \right)\implies \stackrel{center}{(2~~,~~5)}\)

now since we know AB is the diameter, so half that distance AB is its radius, so

\(~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{1})\qquad B(\stackrel{x_2}{6}~,~\stackrel{y_2}{9})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\)

\(AB=\sqrt{(~~6 - (-2)~~)^2 + (~~9 - 1~~)^2} \implies AB=\sqrt{(6 +2)^2 + (9 -1)^2} \\\\\\ AB=\sqrt{( 8 )^2 + ( 8 )^2} \implies AB=\sqrt{ 64 + 64 } \implies AB=\sqrt{ 128 } \\\\\\ AB=8\sqrt{2}\hspace{5em}\stackrel{\textit{half that is the radius}}{\cfrac{8\sqrt{2}}{2}\implies 4\sqrt{2}\implies \sqrt{32}} \\\\[-0.35em] ~\dotfill\)

\(\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{2}{h}~~,~~\underset{5}{k})}\qquad \stackrel{radius}{\underset{\sqrt{32}}{r}} \\\\\\ (x-2)^2~~ + ~~(y-5)^2~~ = ~~(\sqrt{32})^2\implies {\large \begin{array}{llll} (x-2)^2~ + ~(y-5)^2~ = ~32 \end{array}}\)

To perform a one-way ANOVA, you need a dataset containing interest rates for different groups or categories. Each group should have independent samples, and the interest rates within each group should be approximately normally distributed.

To calculate the F-test statistic, we need to compare the variation between the group means to the variation within the groups. In this case, the groups are the different interest rates. The F-test statistic measures the ratio of the mean square between groups to the mean square within groups. It determines whether there is a significant difference in means across the groups. A higher F-test statistic indicates a greater difference between the group means and suggests a higher likelihood of rejecting the null hypothesis. Conversely, a lower F-test statistic suggests that the group means are similar, supporting the null hypothesis.

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Answer:

15

Step-by-step explanation:

the difference of 8 and 3 : 8-3 = 5

sum of 7 and 5 : 7+5 = 12

a quarter of that sum : 12/4 = 3

the difference from before is now multiplied by that quarter : 5 × 3 = 15

I know, you are a beginner, but you don't know basic arithmetic calculation of +, -, × , / ? really ?

Multiplication is the process of multiplying, therefore, adding a number to itself for the number of times stated. The solution of the difference of 8 and 3 is multiplied by one-quarter of the sum of 7 and 5 is 15.

Multiplication is the process of multiplying, therefore, adding a number to itself for the number of times stated. For example, 3 × 4 means 3 is added to itself 4 times, and vice versa for the other number.

The statement can be solved as shown below,

1.   The difference between 8 and 3,

     8 - 3 = 5

2.   The difference between 8 and 3 is multiplied by one-quarter of the    sum of 7 and 5,

    ​\((8-3) \times \dfrac{(7+5)}{4}\\\\= 5 \times \dfrac{12}{4}\\\\= 15\)

Hence, the solution of the difference of 8 and 3 is multiplied by one-quarter of the sum of 7 and 5 is 15.

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Monte Carlo and Historical simulation are widely used tools for risk measurement, generating random inputs based on probability distribution functions. Proper probability distributions are crucial for risk analysis, while statistics aids in risk management by obtaining probabilities and assessing results.

a) Monte Carlo and Historical simulation are the most extensively used tools for measuring risk. The significant difference between these two tools lies in their inputs. Monte Carlo simulation is based on generating random inputs based on a set of probability distribution functions. While Historical simulation, on the other hand, simulates based on the prior actual data inputs.\

b) In risk analysis, the right choice of probability distribution that explains the risk factor's values is of paramount importance as it can give rise to critical decision making and management of financial risks. Probability distributions such as the Normal distribution are used when modeling the return of an asset, or its log-returns. Normal distribution in financial modeling is essential because it best describes the distribution of price movements of liquid and high-frequency assets. Nonetheless, selecting the wrong distribution can lead to wrong decisions, which can be quite catastrophic for the organization.

c) Statistics are used in Risk Management to assist in decision-making by helping to obtain the probabilities of potential risks and assessing the results. Statistics can provide valuable insights and an objective evaluation of risks and help us quantify risks by considering the variability and uncertainty in all situations. With statistics, risks can be easily identified and properly evaluated, and it assists in making better decisions.

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The integration is (1/4)x sin 4x + (1/16)cos 4x + C.

To evaluate the integral ∫ x cos 4x dx using integration by parts, we need to use the formula ∫u dv = uv - ∫v du. We are given that u = x and dv = cos 4x dx. Now, we need to find v and du.

To find v, we need to integrate dv. So, v = ∫cos 4x dx = (1/4)sin 4x + C.

To find du, we need to differentiate u. So, du = dx.

Now, we can plug in the values of u, v, du, and dv into the formula and simplify:

∫ x cos 4x dx = x(1/4)sin 4x - ∫(1/4)sin 4x dx

= (1/4)x sin 4x - (1/4)∫sin 4x dx

= (1/4)x sin 4x - (1/4)(-1/4)cos 4x + C

= (1/4)x sin 4x + (1/16)cos 4x + C

So, the final answer is (1/4)x sin 4x + (1/16)cos 4x + C.

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Answer:

  2.76×10^-3 m/s²

Step-by-step explanation:

You want to know the acceleration of the moon toward the Earth, given its distance is 3.8×10^8 meters, Earth's mass is 5.98×10^24 kg, and the gravitational constant is 6.673×10^-11 N·m²/kg².

The acceleration of one body by another is ...

  a = GM/r²

where G is the gravitational constant, M is the body creating the gravitational field, and r is the distance between the masses.

  a = (6.673×10^-11)(5.98×10^24)/(3.8×10^8)² N/kg

  a = (6.673·5.98/3.8²)×10^(-11+24-16) m/s² ≈ 2.76×10^-3 m/s²

A drop of water is denser than a ping-pong ball.

Usually, water is made of particles that are firmly pressed together. In differentiation, plastic (the material ping pong balls are made of) may be a lightweight fabric and the particles are not as firmly stuffed together.

The thickness of a ping pong ball is 0.0840 g/cm³, though water’s thickness is 997 kg/m³. Subsequently, ping pong balls aren’t about as thick as water and will continuously coast and surface greatly quickly.

The ping pong ball appears to oppose gravity and coast within the air.

Ping-pong balls drift within the water since they are amazingly lightweight, empty, and filled with air. Too, the water’s surface pressure makes it simple for the ping pong ball to drift.

In expansion, water is denser than ping pong balls, making them look for the most noteworthy point of water.

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The sound which we hear when the pig pong ball is bounced on the floor

is 19.48 DB

The repeating of sounds, especially in rhyme, is the form of repetition that most people connect with poetry. Alliteration, assonance, and onomatopoeia are other sound patterns in poetry that give additional meaning in addition to rhyme. Every one of these audio elements has a certain purpose in a poem.

a) \(\sum \ log(n)\)

by expanding the series for each value of n is

log (1) + log (2) + log(3) + log (4) + ......... + log ( 96)

simplify the expanded form we get

0 + 0.3010 + 0.4771+0.6020 ......................... + 1.982

=> 149.9963

b) \(\sum_{n=0}\) to infinity \(\sqrt{0.9^n}\)

formula for the sum of number in geometric progression

is  a/1-r

to find the ratio of the successive terms

plugging into the formula

r = \(\frac{a_{n+1}}{a_n}\)

r = \(\frac{\sqrt{0.9^{n+1}} }{\sqrt{0.9^n} }\)

=> r = \(\frac{\sqrt{0.9^n \times 0.9} }{\sqrt{0.9^n} .1}\)

=> r = \(\frac{\sqrt{0.9} }{1}\)

=> r = \(\sqrt{0.9}\)

=> a = \(\sqrt{0.9^0}\)

=> a = \(\sqrt{1}\)

=>  a= 1

by applying the formula having the value a =1 is

\(\frac{1}{1-\sqrt{0.9} }\)

rationalize the denominator by multiplying with \(1+\sqrt{0.9}\)

=> \(\frac{1+\sqrt{0.9} }{(1-\sqrt{0.9} ) (1+\sqrt{0.9}) }\)

=> 19.4868

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Answer:

y= 27

Step-by-step explanation:

12-y= -15

subtract 12 from both sides

12-12. -15-12

-y= -27

divide both sides by -y (or -1)

-y÷ -y. -27÷ -y

y= 27

answer: y = 27

step-by-step:

subtract 12 from both sides

12 - y = -15

12 12

----

simplify

y = -27

----

divide both sides by -1

-y = -27

-1 -1

simplify

y = 27

Answer:

498

Step-by-step explanation:

We will need to use the Pythagorean Theorem.

A^2 + B^2 = C^2

45^2 + B^2 = 500^2

Subtract 45^2

B^2 = 500^2 - 45^2

Simplify

B^2 = 247,975

Take square root so B is alone

B = 497.9708

Round

B = 498

The rate of change of the volume when radius of sphere is 11 inch is 2904π inch³/minutes

According to the question,

The radius r of a sphere is increasing at a rate of 6 inches per minutes

Volume of sphere is 4/3 πr³

We have to find Rate of change of volume when r = 11 inch

Differentiating volume of sphere w.r.t time

V = 4/3 πr³

=> dV/dt = 4 πr² . dr/dt ---------(1)

As it is given that ,

dr/dt = 6

Substituting the value in equation (1)

=>  dV/dt = 4 πr² . 6

=> dV /dt = 24 πr²

r = 11

=> dV /dt = 24 π(11)²

=> 2904 π inch³/minutes is rate of change of volume

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Answer:

After 25 mins.

Step-by-step explanation:

The test statistic is approximately -8.929.  and the p-value is approximately 0.0001, rounded to four decimal places.

1. To test the difference in proportions between the two age groups, we can use the normal distribution. The distribution to use for the test is:

P'1 - P'2 ~ N(0, ?)

Here, P'1 represents the proportion of tablet owners in the 16-29 age group, P'2 represents the proportion of tablet owners in the 30 and older age group, and N(0, ?) denotes the normal distribution with mean 0 and variance to be determined.

2. The test statistic for comparing two proportions is calculated as:

z = (P1 - P2) / sqrt(P * (1 - P) * ((1/n1) + (1/n2)))

where P = (x1 + x2) / (n1 + n2), x1 and x2 are the number of tablet owners in each group, and n1 and n2 are the respective sample sizes.

For the given data, we have:

x1 = 69 (number of tablet owners in the 16-29 age group)

n1 = 622 (sample size of the 16-29 age group)

x2 = 231 (number of tablet owners in the 30 and older age group)

n2 = 2318 (sample size of the 30 and older age group)

Using these values, we can calculate the test statistic:

P = (x1 + x2) / (n1 + n2) = (69 + 231) / (622 + 2318) = 0.0808

\(z = (P1 - P2) / sqrt(P * (1 - P) * ((1/n1) + (1/n2)))\\= (69/622 - 231/2318) / sqrt[n]{(0.0808 * (1 - 0.0808) * ((1/622) + (1/2318)))} \\≈ -8.929\)

Therefore, the test statistic is approximately -8.929.

3. To find the p-value, we need to calculate the probability of obtaining a test statistic as extreme as -8.929 (in the negative tail of the standard normal distribution). Since the test is two-tailed, we will consider the absolute value of the test statistic.

p-value ≈ 2 * P(Z < -8.929)

Using a standard normal distribution table or a calculator, we can find the p-value associated with -8.929:

p-value ≈ 0.000 < 0.0001

Therefore, the p-value is approximately 0.0001, rounded to four decimal places.

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The equation that shows how the total number of science quizzes, y, depends on the number of months that have passed, x is y = 4x.

An equation is a mathematical statement that shows the relationship between two or more values.

In this case, Leon has already had 4 science quizzes this semester and expects to have 4 quizzes per month going forward.

We can use an equation to show how the total number of science quizzes, y, depends on the number of months that have passed, x.

The equation for this would be y = 4x, where x is the number of months that have passed and y is the total number of science quizzes.

This equation shows that for each month that passes, the total number of science quizzes increases by 4

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The volume of the given triangular prism is 315 cm³.

The volume of a triangular prism, we need to multiply the base area of the triangular base by the height of the prism.

The triangular base has a base length of 5 cm and a height of 7 cm, and the height of the prism is 9 cm.

Calculate the area of the triangular base.

The area of a triangle can be calculated using the formula:

Area = (base length × height) / 2

Plugging in the values, we have:

Area = (5 cm × 7 cm) / 2

Area = 35 cm²

Calculate the volume of the prism.

The volume of a prism can be calculated by multiplying the base area by the height of the prism.

Volume = Base area × Height

Volume = 35 cm² × 9 cm

Volume = 315 cm³

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The empirical method is the right answer.

Empirical probability is calculated by dividing the number of times an event was seen in your data by the entire sample size. An event's relative frequency is strongly connected to an empirical probability, also known as an experimental probability.

Empirical probability bases its estimation of the likelihood that a specific result will recur on the number of instances of that outcome within a sample set. In short, the empirical method uses relative frequencies to determine probabilities.

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Applying the area of the parallelogram formula, it can be concluded that the area is 56 square units.

The area of the parallelogram is the magnitude of the cross-product of the adjacent edges (base and height).

Given information:

A parallelogram is made up of 4 vertices: A(−3, 0), B(−1, 6), C(8, 5), and D(6, −1).

So first we find the adjacent edges as follows:

AB = B - A

     = ( (-1-(-3)) , (6-0) )

     = (2,6)

AD = D - A

     = ((6-(-3)) , (-1-0))

     = (9,-1)

We want to find the area of the parallelogram, we do the following step:

\(AB x AD = \left[\begin{array}{ccc}i&j&k\\2&6&0\\9&-1&0\end{array}\right]\)

               = i(0 - 0) + j(0 - 0) + k(-2 - 54)

               = -56k

The area of the parallelogram = | AB x AD |

                                                  = √(-56)²

                                                  = 56 square unit

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Answer:

y = 2x − 6

Step-by-step explanation:

Answer:

1)  Not appropriate.

2)  b. P(x > 12.5)

3)  d. 0.2451

4)  Not appropriate.

Step-by-step explanation:

For the normal approximation to a binomial distribution to work well, the following conditions need to be true:

Suppose the random variable X follows a binomial distribution:

If p ≈ 0.5 and n is large then X can be approximated by the normal random variable:

However, even if p isn’t all that close to 0.5, this approximation usually works well as long as np and n(1 – p) are both bigger than 5.

Given:

⇒ np = 6 × 0.1 = 0.6 < 5

⇒ n(1 - p) = 6 × 0.9 = 5.4 > 5

As p = 0.1 is not close to 0.5 and n is small, and np and n(1 – p) are not both bigger than 5, it is not appropriate to use the normal distribution to approximate the random variable x for a binomial experiment.

The binomial distribution is discrete, but a normally-distributed variable is continuous.  Therefore, to allow for this, use a continuity correction.

The interval you need to use with the normal distribution depends on the discrete probability you’re trying to find, but the general idea is always the same:  

Therefore, to approximate the discrete probability P(x > 12), use the continuity correction P(X > b) ≈ P(Y > b+0.5):

There are a fixed number of trials (150 advanced reservations), with probability of success (i.e. no-shows) 0.15.  If X is the number of no-shows on advanced reservations, then X ~ B(150, 0.15).

As n is large and np = 22.5 > 5 and n(1 – p) = 127.5 > 5, a normal approximation is appropriate.

X can be approximated by a normal random variable Y ~ N(μ, σ²):

    ⇒ μ = np = 150 × 0.15 = 22.5

    ⇒ σ² = np(1 - p) = 150 × 0.15 × 0.85 = 19.125

So Y ~ N(22.5, 19.125).

Use the approximation to estimate the probability that there will be fewer than 20 no-shows.

Using the continuity correction, P(X < b) ≈ P(Y < b-0.5):

The probability using the binomial distribution X ~ B(150, 0.15) is:

As 0.2464 ≈ 0.2509, this further proves that it is appropriate to use the normal distribution to approximate.

There are a fixed number of trials (18 citizens), with probability of success (i.e. favored building a police substation) 0.83.  If X is the number of citizens asked, then X ~ B(18, 0.83).

Calculate np and n(1 – p):

As n is not large and p is not close to 0.5, and np and n(1 – p) are not both bigger than 5, a normal approximation is not appropriate.

Answer:

Step-by-step explanation:

Firstly,

According to rational and irrational,

\( \sqrt{15} \: is \: irrational\)

Since,

Natural numbers, Whole Numbers and Integers all come under Rational number.

Hence,

\( \sqrt{15} \)

Is an irrational number.

Korey’s gross profit margin last month will be equal to 55.7%.

The gross profit margin is a profitability ratio. Profitability ratios measures the efficiency with which a company generates profit from its asset. Gross profit margin measures the return on sales.

Gross profit margin = net income / gross profit

$1850 / 3320 = 0.557 = 55.7%

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Answer:

The answer is actually a. 42.5%

Step-by-step explanation:

Correct on edg. 2022

The prevalence of sampling errors can be reduced by increasing the sample size. As the sample size increases, the sample gets closer to the actual population, which decreases the potential for deviations from the actual population.

When studying the incidence of rare phenomena, researchers should use "relatively large samples". Using a large sample corrects for bias. If bias has been eliminated, it is safe to assume that the sample is free of sampling errors." The effects of random sampling errors are predictable in the long run.

One of the most effective methods that can be used by researchers to avoid sampling bias is simple random sampling, in which samples are chosen strictly by chance. This provides equal odds for every member of the population to be chosen as a participant in the study at hand.

The prevalence of sampling errors can be reduced by increasing the sample size. As the sample size increases, the sample gets closer to the actual population, which decreases the potential for deviations from the actual population.

Non-probability sampling often results in biased samples because some members of the population are more likely to be included than others.

To avoid non-response bias, researchers should keep surveys as simple as possible, with clear wording and instruction and seek respondents who are relevant to the survey topic.

Therefore,

The prevalence of sampling errors can be reduced by increasing the sample size. As the sample size increases, the sample gets closer to the actual population, which decreases the potential for deviations from the actual population.

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Answer:

What is the law of large numbers? 

In probability theory, the law of large numbers is a theorem that describes the result of performing the same experiment a large number of times. 

What does it tell us about samples as they get larger and approach infinity?

As sample sizes increase, the sampling distributions approach a normal distribution. With "infinite" numbers of successive random samples, the mean of the sampling distribution is equal to the population mean (µ).

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The cosine function with amplitude 2 and period 6π is given by:

y = 2cos(x/3).

The standard cosine function is given by:

y = Acos(Bx).

In which:

For a cosine function with amplitude 2 and period 6π, we have that A = 2 and B = 1/3, hence:

y = 2cos(x/3).

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