if h(x) = 6 5f(x) , where f(5) = 6 and f '(5) = 4, find h'(5). h'(5) =

(112)3 =336

(211)3 =633

sum of the numbers:

(112)3 +(211)3=

336+633=

969

product of the numbers:

(112)3 × (211)3=

336×633=

212688

The package weighs 6.86 102 N.

We must apply the formula for the mechanical advantage of a lever in order to get the weight of the crate:

Input force minus output force is the mechanical advantage.

When the weight of the crate acts as the output force and the force supplied to the lever acts as the input force.

If we rearrange the formula, we obtain:

Mechanical advantage times input force equals output force.

Inputting the values provided yields:

Output Force is equal to 3.43 x 2.0 102 N.

Force at output: 6.86 102 N.

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The derivative of x = 2 of the given function its value is -13/4.

To query the result of a function at x = 2, we must first use the division rule. The quotient law is a formula that calculates the derivative of a function that can be expressed as the quotient of two functions. Let

f(x) = h(x) and g(x) = x. We can express the function h(x)/x as f(x)/g(x). Now we can use the quotient rule like this:

d/dx(h(x)/x)) = d/dx(f(x)/g(x))

= [( g(x) * f '(x) )) - (f(x) * g'(x))] / (g(x))^2

= [(x * h'(x)) - (h (x) * 1) ] / x ^2

Now we can put the values ​​given as x = 2 and h(2) = 9 and h'(2) = -2 into the formula:

d /dx(h(x)/ x) ) x = 2 = [ (2 * (-2)) - (9 * 1)] / 2^2

= (-4 - 9) / 4

= -13/4

Therefore, the derivative of x = 2 of the given function its value is -13/4.

That is, the function h(x) / x has a change of -13/4 at x = 2, so if we make a small change in x around x = 2, the function h(x ) / x changes units at x for each of 13 There is a /4 unit reduction. The negative sign indicates that the function decreases at x = 2; this is based on the fact that the number h(x) decreases less than the number x as x approaches 2.

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The margin of error at a 90% confidence level is approximately 0.053.

To find the margin of error at a 90% confidence level, we first need to calculate the standard error, which is the standard deviation of the sampling distribution of the proportion:

SE = [tex]\sqrt{[p(1-p)/n]}[/tex]
where p is the estimated proportion (p-hat) and n is the sample size.

Plugging in the values, we get:

SE = sqrt[0.24(1-0.24)/340] ≈ 0.032

Next, we use the formula for margin of error at a 90% confidence level:

ME = z*SE
where z is the z-score corresponding to the desired confidence level.

Since we are looking for a 90% confidence level, we use z = 1.645:

ME = 1.645*0.032 ≈ 0.053

Therefore, the margin of error at a 90% confidence level is approximately 0.053. This means that if we were to repeat the sampling process many times, 90% of the intervals we construct would contain the true proportion of the population within 0.053 of our estimated proportion.

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Answer:

A: What are the coordinates of point W'?

The coordinates of point W' are (6, 6).

B: What are the coordinates of point V'?

The coordinates of point V' are (6, 2).

maybe i think so

The simplified expression for f(x+h) is, f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 8. This corresponds to option d in your list of choices.

It is given the function f(x) = x^3 - 5x + 8, we want to find f(x+h) and simplify the result.

1. Replace x with (x+h) in the function f(x) = x^3 - 5x + 8.
2. f(x+h) = (x+h)^3 - 5(x+h) + 8

Now, we will simplify the expression,

3. Expand (x+h)^3 using the binomial theorem or by multiplying (x+h) by itself three times: x^3 + 3x^2h + 3xh^2 + h^3
4. Distribute -5 to the terms inside the parenthesis: -5x - 5h
5. Combine the terms obtained in steps 3 and 4 with the constant 8: x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 8

So, the simplified expression for f(x+h) is,

f(x+h) = x^3 + 3x^2h + 3xh^2 + h^3 - 5x - 5h + 8

This corresponds to option d in your list of choices.

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Answer:

One number is 6 and the other number is 4

Step-by-step explanation:

Helping in the name of Jesus.

Answer:

3x=12

divide boths by the coefficient of x

and x= 4

The MLE for u is the sample mean and the MLE for 02 is the sample variance.

To find the maximum likelihood estimators (MLEs) for both u and 02, we need to first write down the likelihood function.

The likelihood function for a normal distribution with unknown mean u and unknown variance 02 is given by:

L(u,02|X1,X2,...,Xn) = (2π02)^(-n/2) exp[-1/(2*02) Σ(Xi-u)^2]

Taking the natural logarithm of the likelihood function, we get:

log L(u,02|X1,X2,...,Xn) = -n/2 log(2π02) - 1/(2*02) Σ(Xi-u)^2

To find the MLE for u, we differentiate the log likelihood function with respect to u and set it equal to zero:

d/d u log L(u,02|X1,X2,...,Xn) = 1/(2*02) Σ(Xi-u) = 0

Solving for u, we get:

u = ΣXi / n

Therefore, the MLE for u is simply the sample mean.

To find the MLE for 02, we differentiate the log likelihood function with respect to 02 and set it equal to zero:

d/d(02) log L(u,02|X1,X2,...,Xn) = -n/(2*02) + 1/(2*02^2) Σ(Xi-u)^2 = 0

Solving for 02, we get:

02 = Σ(Xi-u)^2 / n

Therefore, the MLE for 02 is simply the sample variance.

In summary, the MLE for u is the sample mean and the MLE for 02 is the sample variance.

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Using the Central Limit Theorem, we have that:

a) Since there are at least 10 successes and 10 failures, the condition is met

b) Using the formula  [tex]$SE_{\hat{p}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]with n = 400 and p = 0.11, the standard error is of 0.0156.

(a) In order to apply the normal approximation to the binomial distribution, the sample size must be large enough such that np and n(1-p) are both greater than or equal to 10, where n is the sample size and p is the probability of success.

In this case, we have n=400 and the observed proportion of defective bulbs is [tex]$\hat{p}=44/400=0.11$[/tex].

Thus, np=4000.11=44 and n(1-p)=4000.89=356.6, which are both greater than 10. Therefore, it is reasonable to assume that the condition for obtaining an interval estimate for [tex]$p_x$[/tex]using the normal approximation is met.

(b) The standard error of the sample proportion [tex]$\hat{p}$[/tex]is given by:

[tex]$SE_{\hat{p}}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$[/tex]

Plugging in the values, we get:

[tex]$SE_{\hat{p}}=\sqrt{\frac{0.11(1-0.11)}{400}}\approx 0.0156$[/tex]

Therefore, the standard error of [tex]$p_x$[/tex]is approximately 0.0156.

It states that for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean  and standard deviation , as long as  and .

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Full Question: A large company produces an equal number of brand-mame lightbulbs and generic lightbulbs. The director of quality control sets guidelines that lightbulbs that are defect lightbulbs that are defective is not equal to the director wants to estimate the average of the two proportions. Production will be stopped if there is evidence that the proportion of all ve is greater than 0. 10. The director also believes that the proportion of brand-name proportion of generic lightbulbs that are defective. Therefore, the o estmate the proportion of brand-name lightbulbs that are defective, a simple random sample of 400 brand-name lightbulbs brand-name lightbulbs that are defective in a sample of 400, and let px represent the proportion of all brand-name lightbulbs that are defective. It is reasonable to assume that X is a binomial random variable.

(a) One condition for obtaining an interval estimate for px is that the distribution of px is approximately number of is taken and 44 are found to be defective. Let X represent the normal Is it reasonable to assume that the condition is met? Justify your answer.

(b) The suandard error of hr is approumately O 0156 Show how the value of the standard error is calculated

We need to check the success/failure condition to ensure that the sampling distribution is approximately normal.

For the sampling distribution of S², we need to assume that the population follows a normal distribution in order to convert S² to a chi-square random variable.

To determine how many people we need to sample to reduce the standard deviation of the mean to 2.6, we found n=217.7515.

To use the t-distribution when finding P(Xˉ < some number), we need to assume that the population is normally distributed or approximately normal.

(c) In the Central Limit Theorem for 1 Proportion, we need to check the success/failure condition to ensure that the sampling distribution is approximately normal. This is because the theorem states that as the sample size increases, the sampling distribution of the proportion approaches a normal distribution, provided that the success/failure condition (np ≥ 10 and n(1-p) ≥ 10) is met. This allows us to make valid inferences about the population proportion.

(d) For the sampling distribution of S², we need to assume that the population follows a normal distribution in order to convert S² to a chi-square random variable. This is because the chi-square distribution is derived from the normal distribution, and using it assumes that the underlying population is normally distributed.

(e) To determine how many people we need to sample to reduce the standard deviation of the mean to 2.6 from a sample size of 92 with a standard deviation of 4, we found n=217.7515. Since we cannot sample a partial person, we need to round up to the nearest whole number, which is 218 people.

(f) To use the t-distribution when finding P(Xˉ < some number), we need to assume that the population is normally distributed or approximately normal. This is important because the t-distribution is derived from the normal distribution and is used when estimating population parameters, especially when the sample size is small and the population standard deviation is unknown.

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The equation to show how to find the product of 1,000,000 and 1,000,000 using scientific notation can be expressed as (10^6 * 10^6).

A number can be written in scientific notation  in a case whereby the number is greater than or equal to 1  however not up to 10 multiplied by a power of 10.

Given that 1,000,000 and 1,000,000 which can be written in scientific notation as 1.0 * 10^6 and 1*10^6, then th product can be written as (10^6 * 10^6) = 10^12.

Hence, the product is 10^12.

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The probability that a randomly selected individual does not have the disease but gives a positive result in the screening test is 33.8%.

The probability of having the disease is P(A) = 0.15, so the probability of not having the disease is P(~A) = 1 - P(A) = 0.85.

Using Bayes' theorem:

P(~A|B) = P(B|~A) * P(~A) / [P(B|A) * P(A) + P(B|~A) * P(~A)]

= 0.1 * 0.85 / [0.7 * 0.15 + 0.1 * 0.85]

= 0.338

Therefore, the probability that a randomly selected individual does not have the disease but gives a positive result in the screening test is 0.338, or about 33.8%.

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The volume of the empty portion of container B is 6104.2 ft³(nearest tenth)

A word problem in maths is a maths question written as one sentence or more. This statements are interpreted into mathematical equation or expression.

volume of empty space in container B = volume of B - volume of A

volume of A = πr²h

= 3.14 × 12² × 18

= 8138.88ft³

volume of B = πr²h

= 3.14 × 18² × 14

= 14243.04ft³

Therefore volume of empty space in B = 14243.04 - 8138.88

= 6104.2 ft³(nearest tenth)

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Answer: You would need to hold a membership for at least 17 months.

Step-by-step explanation:

The total number of prizes awarded in a year for each member is given by:

Monthly prizes = 12 x (1 + 1 + 1) = 36

Annual prizes = 2

Therefore, the total number of prizes awarded in a year is 38.

The probability of not winning any prize in a given month is (197/200) * (196/199) * (195/198) = 0.942

Therefore, the probability of winning at least one prize in a given month is 1 - 0.942 = 0.058.

The probability of not winning any prize in 12 months is (0.942)^12 = 0.399

Therefore, the probability of winning at least one prize in 12 months is 1 - 0.399 = 0.601.

To be confident of winning at least one prize, we want the probability to be greater than 0.5.

So, we want (1 - 0.942)^n < 0.5, where n is the number of months of membership.

Solving for n gives n > 16.4, which means we need to hold a membership for at least 17 months to be confident of winning at least one prize.

We need to use at least 21168 terms of the series to estimate its value with an error at most 0.000001.

Explanation: -

To estimate the value of the convergent series ∑n=1[infinity] 2 n^(-1.1) with an error at most 0.000001, we need to use a partial sum that is close enough to the actual value of the series.

One way to approach this is to use the error bound formula for a convergent series:

|S - Sn| ≤ a_n+1/(1 - r),

where S is the actual sum of the series, Sn is the sum of the first n terms of the series, an+1 is the (n+1)th term of the series, and r is the common ratio (in this case, r = 1/2^(1.1)).

We want to find the value of n such that the error |S - Sn| is at most 0.000001.

Plugging in the given values, we get:

0.000001 ≤ 2(n+1)^(-1.1)/(1 - 1/2^(1.1))

Solving for n using a calculator or computer algebra system, we get n ≈ 21168.

Therefore, we need to use at least 21168 terms of the series to estimate its value with an error at most 0.000001.

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Answer:

The plane should enter the clouds in about 3.3 hours; the plane should exit the cloud in about 6.0 hours

Step-by-step explanation:

I also made a table using your data in Desmos Graphic Calculator and the quadratic regression equation it gave me was

[tex]y=-15.3604x^2+141.912x+0.56701[/tex]

If your Desmos looks like mine, your table should be in box 1 (y1), and something like

[tex]y_{1}[/tex] ~ [tex]ax^2_{1}+bx_{1}+c[/tex] and it says STATISTICS, RESIDUALS, PARAMETERS (all this should be in the y2 box and it should graph the parabola itself)

First, under this equation, type y = 300 in the fourth box (y4)

Second, click on the wrench on the right and for your x axis, use -2.585 < 12.415 and for your y axis

Third, hover your mouse over the first spot where you parabola and your y = 300 line intersect.  You should see that the intersection coordinate is (3.261, 300)

Because the parabola points up at this first intersection, we know the plane is travelling upward, so this first coordinate is when the plane enters the clouds.  And rounding 3.261 to the nearest tenth gives us 3.3 hours

Fourth, hover your mouse over the second spot where your parabola and your y = 300 line intersect.  You should see that the intersection coordinates are (5.978, 300)

Because the parabola points down at this second intersection point, we know that plane is travelling downward, so this second coordinate is when the plane exits the clouds.  And rounding to the nearest tenth gives us 6.0 hours

If you can't figure out how to made the quadratic regression equation, I attached a picture of the Desmos graph I used for your question.

Question: what do negative exponents do?

Answer: (B)

Step-by-step explanation: I believe

a) The null space of A is a subspace of the 7-dimensional vector space R^7, and its dimension is 3.

b) The column space of A is a subspace of the 5-dimensional vector space R^5.

The null space of a matrix is the subspace of the vector space in which the matrix operates. In this case, since A is a 5x7 matrix, its null space is a subspace of the 7-dimensional vector space R^7.

a) The dimension of the null space can be found using the rank-nullity theorem, which states that the dimension of the null space plus the rank of the matrix equals the number of columns. Since the rank of A is 4 and it has 7 columns, we have:

dim(null space) + rank(A) = number of columns

dim(null space) + 4 = 7

dim(null space) = 3

Therefore, the null space of A is a subspace of R^7 with dimension 3.

b) The column space of a matrix is the subspace of the vector space generated by the columns of the matrix. In this case, since A is a 5x7 matrix, its column space is a subspace of the 5-dimensional vector space R^5. This is because the columns of A are vectors in R^5. Therefore, the column space of A is a subspace of R^5.

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It can be used to partition the nodes of the graph into two sets. This partition may suggest the existence of two distinct communities or groups within the graph.

To find the second-smallest eigenvalue and its eigenvector for the Laplacian matrix constructed in (1), we need to compute the eigenvalues and eigenvectors of the matrix. Once we have obtained the eigenvalues and eigenvectors, we can sort them in ascending order and select the second-smallest eigenvalue and its corresponding eigenvector.

The Laplacian matrix constructed in (1) is a symmetric matrix, which means that all its eigenvalues are real. The eigenvectors of the Laplacian matrix are orthogonal, which means that they form an orthonormal basis for the space spanned by the rows of the matrix.

Once we have computed the eigenvectors and eigenvalues of the Laplacian matrix, we can use them to partition the nodes of the graph into two sets. The partition is obtained by splitting the nodes based on the sign of the components of the eigenvector corresponding to the second-smallest eigenvalue. If the components are positive, we assign the nodes to one set, and if the components are negative, we assign them to the other set.

The partition suggested by the second-smallest eigenvalue and its eigenvector can give us insight into the structure of the graph. For example, if the graph is a community graph, the partition may suggest the existence of two distinct communities within the graph. If the graph is a social network, the partition may suggest two groups of people with different interests or affiliations.

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Answer: 410

Step-by-step explanation: 2870/7 = 410

(a) The set {(x1,x2)T|x1 + x2 = 0} is a subspace of R2.

To check whether the given set is a subspace of R2, we need to check whether it is closed under vector addition and scalar multiplication. Let u = (u1,u2)T and v = (v1,v2)T be two arbitrary vectors in the set, and let c be an arbitrary scalar. Then:

u + v = (u1 + v1, u2 + v2)

Since u1 + v1 + u2 + v2 = (u1 + u2) + (v1 + v2) = 0 + 0 = 0 (since u and v are in the set), we see that u + v is also in the set.

c*u = (c*u1, c*u2)

Since c*u1 + c*u2 = c*(u1 + u2) = c*0 = 0 (since u is in the set), we see that c*u is also in the set.

Therefore, the set {(x1,x2)T|x1 + x2 = 0} is a subspace of R2.

(b) It is not a subspace of R2

To check whether the given set is a subspace of R2, we need to check whether it is closed under vector addition and scalar multiplication.

Let u = (u1,u2)T and v = (v1,v2)T be two arbitrary vectors in the set, and let c be an arbitrary scalar. Then:

u + v = (u1 + v1, u2 + v2)

Since u21 = u22 and v21 = v22 (since u and v are in the set), we see that (u1 + v1)2 = (u2 + v2)2. Therefore, u + v is in the set.

c*u = (c*u1, c*u2)

Since u21 = u22 (since u is in the set), we see that (c*u1)2 = (c*u2)2. Therefore, c*u is in the set.

However, the set {(x1,x2)T|x21 = x22} is not a subspace of R2 because it does not contain the zero vector (0,0)T, which is required for any set to be a subspace.

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The oscillator frequency is 0.5 MHz (megahertz).

In digital electronics, an oscillator is a circuit that generates a continuous and repetitive waveform at a specific frequency.

This frequency is usually determined by the machine cycle, which is the time it takes for a single machine cycle to execute in a computer system.

The oscillator frequency can be calculated as the reciprocal of the machine cycle time.

If the machine cycle is given as 2 μs (microseconds), then the oscillator frequency is:

f = 1 / T

where T is the machine cycle time.

Substituting the given value of T, we get:

f = 1 / (2 μs)

To simplify this expression, we can convert microseconds to seconds by dividing by [tex]10^6:[/tex]

[tex]f = 1 / (2 \times 10^-6 s)[/tex]

Simplifying further, we get:

[tex]f = 0.5 \times 10^6 Hz[/tex]

Therefore, the oscillator frequency is 0.5 MHz (megahertz).

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The coefficients on both sides of the equation do not match, hence the given integro-differential equation cannot have a solution.

a). ƒ(t) = inverse Laplace transform of ƒ(s) = 1/55

b). y(t) = ƒ(t).

c). There is no answer to the given equation.

A mathematical statement that establishes the equality of two expressions is known as an equation. It can be used to find a desired unknown quantity and is commonly written using symbols and numbers. Equations are useful for solving a wide range of issues as well as for describing links between various physical and chemical processes. Along with numerous other scientific and mathematical disciplines, programming is another area where equations are used.

Utilising Laplace transforms, the given integro-differential equation can be solved,

Let ƒ(t) = Laplace transform of ƒ(t).

Then,

(1) + (1 - 55°(E)) dě = 1

⇒ (1) + (1 - 55ƒ(s)) ƒ(s) = 1

⇒ ƒ(s) = [1 + (1 - 55ƒ(s)]/55

⇒ ƒ(s) = 1/55

Therefore, ƒ(t) = inverse Laplace transform of ƒ(s) = 1/55

The integro-differential equation is transformed into an initial value issue.

Let y(t) = ƒ(t).

Then,

(1) + (1 - 55°(E)) dě = 1

(1) + (1 - 55y(t)) y′(t) = 1

Considering t differently for each side,

y′′(t) = (1 - 55y(t))/55

Differentiating again,

y′′′(t) = -55y′(t)/55

Differentiating once more,

y(4)(t) = -55y′′(t)/55

We require four beginning values to solve this fourth order differential equation because of its complexity. Therefore,

y(0) = 0, y′(0) = 0, y′′(0) = 0, y′′′(0) = 1

c).The starting value problem's resolution

By varying the settings, we can use this strategy to address the initial value problem.

Let y1(t) = e2t, y2(t) = te2t, y3(t) = t2e2t, y4(t) = t3e2t.

Then,

y′1(t) = 2e2t, y′2(t) = e2t + 2te2t, y′3(t) = 2te2t + t2e2t, y′4(t) = 3t2e2t + t3e2t

y′′1(t) = 4e2t, y′′2(t) = 2e2t + 4te2t, y′′3(t) = 4te2t + 2t2e2t, y′′4(t) = 6t2e2t + 3t3e2t

y′′′1(t) = 6e2t, y′′′2(t) = 2e2t + 6te2t, y′′′3(t) = 6te2t + 2t2e2t, y′′′4(t) = 12t2e2t + 3t3e2t

By including these in the calculation,

[6e2t + 2e2t + 6te2t] + [-55(e2t + 2te2t + t2e2t + t3e2t)] = 1

8e2t + (-55te2t - 110t2e2t - 55t3e2t) = 1

Putting like terms' coefficients on both sides in equal amounts,

8 + (-55) = 1

-47 = 1

This cannot be done. As a result, the following equation cannot be solved.

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(a) To be a probability density function, f(x) must satisfy two conditions: f(x) ≥ 0 for all x. The total area under the curve of f(x) must be equal to 1.

We have:[tex]f(x) = c/(1 + x^2)[/tex]

For f(x) to be non-negative, we need c > 0. To find the value of c such that the total area under the density function of f(x) is equal to 1, we integrate f(x) from −∞ to +∞ and set the result equal to 1:

∫(−∞ to +∞) f(x) dx = ∫(−∞ to +∞) c/(1 + x^2) dx = cπ = 1

Therefore, c = 1/π, and f(x) = 1/(π(1 + x^2)) is a probability density function.

(b) We want to find [tex]P(−9 < X < 9) for X ~ f(x) = 1/(π(1 + x^2))[/tex]

Using the cumulative distribution function (CDF), we have:

[tex]F(x) = P(X ≤ x) = ∫(−∞ to x) f(t) dt = ∫(−∞ to x) 1/(π(1 + t^2)) dt[/tex]

[tex]= (1/π) tan^−1(x) + (1/2)[/tex]

So, using the CDF, we have:

[tex]P(−9 < X < 9) = F(9) − F(−9) =[/tex] [tex][tan^−1(9)/π + 1/2] − [tan^−1(−9)/π + 1/2][/tex]

=[tex][tan^−1(9) − tan^−1(−9)]/π[/tex]

=[tex](1/π) tan^−1(9/−1)[/tex]

= 0.499 (rounded to three decimal places)

Therefore, P[tex](−9 < X < 9) ≈ 0.499.[/tex]

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Based on the terms you provided, I can help you identify the true statement(s):

1. As the sample size n increases, the distribution of the sum of the observations approaches a normal distribution.
2. The sample mean varies from sample to sample.
3. If the underlying population is not normal, the Central Limit Theorem (CLT) states that the distribution of the sample mean X approaches a normal distribution as the sample size n increases.
These statements are true. Note that the statement "As the sample size n increases, the variance of the sample mean X also increases" is incorrect, as the variance of the sample mean actually decreases when the sample size increases. Additionally, the statement "The distribution of the mean X is never exactly normal" is not universally true, as the distribution of the mean can be exactly normal under specific circumstances.

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Answer:

A formula that uses one or more previous terms to find the next term is a recursive formula.

Step-by-step explanation:

A recursive formula is a formula that defines any term of a sequence in terms of its preceding term(s).

The leading coefficient of the term with the highest power (6y⁴) is 6.

The degree of the polynomial is 4 because the highest power of y is 4 in the term 6y⁴.

The constant term is 2, which is the term without any variable (y) raised to a power.

The degree of a polynomial is the highest power of its variable. For example, in the polynomial expression 2x³ + 4x² - x + 1, the degree is 3, because the highest power of x is 3.

The given expression is:

4y + 3y³ + 6y⁴ - 3y³ - 7y + 2

To find the coefficient, degree, and constant of this polynomial, we can simplify it by combining like terms:

-4y + (3y³ - 3y³) + 6y⁴ - 7y + 2

= -4y - 7y + 6y⁴ + 2

= 6y⁴ - 11y + 2

Therefore, the coefficient of the term with the highest power (degree) is 6, the degree of the polynomial is 4, and the constant term is 2.

Coefficients:

The coefficient of the term with the highest power (6y⁴) is 6.

The coefficient of the y-term (-11y) is -11.

The coefficient of the constant term (2) is 2.

Degree:

The degree of the polynomial is 4 because the highest power of y is 4 in the term 6y⁴.

Constant:

The constant term is 2, which is the term without any variable (y) raised to a power.

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a. To find the value of c, we need to ensure that the total area under the density function is equal to 1. Therefore, the variance is: Var(Y) =[tex]0.129 - (0.307)^2[/tex] ≈ 0.051 and the standard deviation is: SD(Y) = [tex]\sqrt{Var(Y) ≈ 0.226}[/tex]

Therefore, we need to solve for c:[tex]∫0^1 cy^2(1-y)^4 dy = 1[/tex]

Using integration by parts and simplifying, we get:

Therefore, the density function is:[tex]f(y) = 252y^2(1-y)^4, 0 < y < 1[/tex]

[tex]f(y) = 0[/tex], elsewhere.

b. To find E(Y), we use the formula:[tex]E(Y) = ∫0^1 yf(y) dy[/tex]

Substituting the density function, we get:[tex]E(Y) = ∫0^1 252y^3(1-y)^4 dy[/tex]

This integral is not easy to solve analytically, so we use numerical integration. Using a calculator or software, we get:[tex]E(Y) ≈ 0.307[/tex]

c. To find the standard deviation of Y, we first need to find the variance:

[tex]Var(Y) = E(Y^2) - [E(Y)]^2[/tex]

To find, we use the formula:[tex]E(Y^2) = ∫0^1 y^2 f(y) dy[/tex]

Substituting the density function, we get: [tex]E(Y^2) = ∫0^1 252y^4(1-y)^4 dy[/tex]

Again, we use numerical integration to get:[tex]E(Y^2) ≈ 0.129[/tex]

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