Show that if we have on the same line OA + OB + OC = 0, PQ + PR + PS = 0, then AQ + BR + CS = 30P

The value of the function [tex]g^{-1[/tex](1) is -8, [tex]h^{-1[/tex](1) is -3/2 and h o [tex]h^{-1[/tex] is 1.

To find the requested values, let's work through each one step by step:

Finding [tex]g^{-1[/tex](1):

To find [tex]g^{-1[/tex](1), we need to determine the input value (x) that corresponds to an output of 1 in the function g. Looking at the given function g, we can see that the ordered pair (-8, 1) is the only pair where the output is 1. Therefore, [tex]g^{-1[/tex](1) is equal to -8.

Finding [tex]h^{-1[/tex](1):

To find [tex]h^{-1[/tex](1), we need to determine the input value (x) that gives an output of 1 in the function h(x) = 8x + 13. We can solve this equation for x by rearranging it:

h(x) = 8x + 13

1 = 8x + 13

8x = 1 - 13

8x = -12

x = -12/8

x = -3/2

So, [tex]h^{-1[/tex](1) is equal to -3/2.

Finding h o [tex]h^{-1[/tex]:

To find h o [tex]h^{-1[/tex], we need to evaluate the composition of functions h and [tex]h^{-1[/tex] at the input value of 5. First, we need to find [tex]h^{-1[/tex](5) by plugging in 5 into the inverse function we found earlier:

[tex]h^{-1[/tex](5) = -3/2

Next, we substitute the result back into the original function h(x) = 8x + 13:

h o [tex]h^{-1[/tex] = h([tex]h^{-1[/tex](5)) = h(-3/2) = 8(-3/2) + 13 = -12 + 13 = 1

Therefore, h o [tex]h^{-1[/tex] is equal to 1.

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The solutions to the equation

(

sin

⁡

�

−

1

)

(

3

tan

⁡

�

+

1

)

=

0

(sinx−1)(

3

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tanx+1)=0 in the interval

[

0

,

2

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)

[0,2π) are

�

=

�

2

x=

2

π

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 and

�

=

5

�

6

x=

6

5π

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.

To find the solutions, we need to set each factor of the equation equal to zero and solve for

�

x.

Setting

sin

⁡

�

−

1

=

0

sinx−1=0, we get

sin

⁡

�

=

1

sinx=1. The only solution in the interval

[

0

,

2

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[0,2π) is

�

=

�

2

x=

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.

Setting

3

tan

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+

1

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3

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tanx+1=0, we solve for

tan

⁡

�

=

−

1

3

tanx=−

3

​

1

​

. The value of

tan

⁡

�

tanx is negative in the second and fourth quadrants. The reference angle with

tan

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�

=

1

3

tanx=

3

​

1

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 is

�

6

6

π

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. In the second quadrant, the angle is

�

+

�

6

=

7

�

6

π+

6

π

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=

6

7π

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. In the fourth quadrant, the angle is

2

�

−

�

6

=

11

�

6

2π−

6

π

​

=

6

11π

​

.

Thus, the solutions in the interval

[

0

,

2

�

)

[0,2π) are

�

=

�

2

x=

2

π

​

,

�

=

7

�

6

x=

6

7π

​

, and

�

=

11

�

6

x=

6

11π

​

. However,

�

=

7

�

6

x=

6

7π

​

 falls outside the given interval. Hence, the solutions are

�

=

�

2

x=

2

π

​

 and

�

=

5

�

6

x=

6

5π

​

.

The solutions to the equation

(

sin

⁡

�

−

1

)

(

3

tan

⁡

�

+

1

)

=

0

(sinx−1)(

3

​

tanx+1)=0 in the interval

[

0

,

2

�

)

[0,2π) are

�

=

�

2

x=

2

π

​

 and

�

=

5

�

6

x=

6

5π

​

. These values were obtained by setting each factor of the equation equal to zero and solving for

�

x. It is important to note that

�

=

7

�

6

x=

6

7π

​

 is a solution as well, but it falls outside the given interval. Therefore, the final solutions are

�

=

�

2

x=

2

π

​

 and

�

=

5

�

6

x=

6

5π

.

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True. A control group is a group where no treatment is given. A control group serves as a comparison group for assessing the effects of treatment or exposure to a particular intervention.

The statement "A parameter is a numerical description of a sample's characteristics" is False.A parameter is a numerical value that describes a population characteristic. Parameters are typically unknown and estimated by calculating sample statistics. A statistic is a numerical description of a sample's characteristics. Statistic is the correct term for this statement. The waiting time at the Department of Motor Vehicles is an example of quantitative data. False, the waiting time at the Department of Motor Vehicles is an example of quantitative data, not qualitative data. Qualitative data refers to non-numerical data such as colors, textures, smells, tastes, and so on.Inferential statistics involves using a sample to draw a conclusion about a corresponding population.

True. Inferential statistics involves using sample data to draw conclusions about a larger population. It allows us to determine whether the observed differences between groups are statistically significant or if they are due to chance alone. The census is data from the whole population. True. Census data is data that has been collected from the entire population rather than a sample of that population. Finally, the statement "About 1/4 of the data lies below the first quartile, Q1" is true. The first quartile, Q1, marks the boundary below which one-quarter of the data lie.

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Researchers should be aware of internal and external validity because they impact the validity of the research findings. Internal validity refers to the extent to which the study design eliminates the influence of extraneous variables on the outcome. External validity refers to the extent to which the results can be generalized to the larger population.

To ensure that a research study is valid, both internal and external validity should be considered. The research question, research design, sample selection, and data collection methods all impact the internal and external validity of the study.

Researchers should be aware of internal validity to:

Researchers should be aware of external validity to:

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(a) The probability that the sample proportion is less than 0.80 is approximately 0.1867.

(b) The probability that the sample proportion is more than or equal to 0.82 is approximately 0.8133.

(c) The probability that the sample proportion is at most 0.85 is approximately 0.8577.

To calculate the probabilities related to the sample proportion in this scenario, we can use the properties of the sampling distribution of proportions and assume that the sample follows a binomial distribution.

(a) To calculate the probability that the sample proportion is less than 0.80, we can use the normal approximation to the binomial distribution. The mean of the sampling distribution is equal to the population proportion, which is 0.82 in this case, and the standard deviation is calculated as the square root of (p(1-p)/n), where p is the population proportion and n is the sample size.

Let's calculate the z-score for a sample proportion of 0.80:

z = (0.80 - 0.82) / sqrt(0.82 * (1 - 0.82) / 100) ≈ -0.8944

Using the z-table or a calculator, we can find the probability associated with this z-score. The probability that the sample proportion is less than 0.80 is approximately 0.1867.

(b) To calculate the probability that the sample proportion is more than or equal to 0.82, we can use the same approach. Let's calculate the z-score for a sample proportion of 0.82:

z = (0.82 - 0.82) / sqrt(0.82 * (1 - 0.82) / 100) ≈ 0

The probability that the sample proportion is more than or equal to 0.82 is equal to 1 minus the probability that it is less than 0.82. From part (a), we know that the probability of being less than 0.82 is approximately 0.1867. Therefore, the probability that the sample proportion is more than or equal to 0.82 is approximately 1 - 0.1867 = 0.8133.

(c) To calculate the probability that the sample proportion is at most 0.85, we can once again use the normal approximation. Let's calculate the z-score for a sample proportion of 0.85:

z = (0.85 - 0.82) / sqrt(0.82 * (1 - 0.82) / 100) ≈ 1.0645

Using the z-table or a calculator, we can find the probability associated with this z-score. The probability that the sample proportion is at most 0.85 is approximately 0.8577.

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The expression of the derivative of g(v) g′(v) is obtained by taking the derivative of v(1 − 7v^2) with respect to v.

That is, the derivative of g is:g′(v) = 1 * (1 − 7v^2) + v * [−14v]     = 1 − 7v^2 − 14v^2     = 1 − 21v^2The slope of g at v = 2 is given by g′(2) which can be computed as:g′(2) = 1 − 21(2)^2     = 1 − 84     = −83Therefore, the slope of g at v = 2 is −83.Let g(y) = −2y^3/8 − 3y^7/10 − 8y^6/8 be the given function of y.

We shall determine the derivative of g.Differentiating g(y) with respect to y gives the derivative of g as:g′(y) = [d/dy](-2y^3/8) + [d/dy](-3y^7/10) + [d/dy](-8y^6/8)     = (-6y^2)/8 − (21y^6)/10 − 6y^5     = -(3/4)y^2 - (21/10)y^6 - 6y^5Hence, the derivative of g is g′(y) = -(3/4)y^2 - (21/10)y^6 - 6y^5.The slope of g at y = 1 is given by g′(1) which can be calculated as:g′(1) = -(3/4)(1)^2 - (21/10)(1)^6 - 6(1)^5     = -3/4 - 21/10 - 6     = -67/20Therefore, the slope of g at y = 1 is -67/20.

In summary, we have seen how to compute the derivative of g(v) and determine its slope at v = 2. We have also determined the derivative of g(y) and its slope at y = 1.

Recall that the slope of a curve at any point is given by the derivative of that curve evaluated at that point. Therefore, to find the slope of a curve at a given point, we need to compute the derivative of the curve at that point.

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The correct statements are "There is a vertical stretch by 4" and "There is a horizontal shift left 7."Hence, option A is correct.

Given logarithmic function is [tex]y=4\log3(x+7)-2[/tex].

We need to select the correct statements from the given options. Let's discuss each statement one by one:

There is a vertical stretch by 4In general, the logarithmic function of the form y = logb(x) has the domain (0, ∞) and range (-∞, ∞). Here, we have [tex]y = 4\log3(x + 7) - 2[/tex].

Inside the log function, we have x + 7. This means that there is a horizontal shift to the left by 7 units.

Also, the coefficient 4 outside the log function will cause the vertical stretch.

Therefore, the statement "There is a vertical stretch by 4" is true.  There is horizontal shift right 2

Similarly, we can say that the logarithmic function y = logb(x) has the vertical asymptote x = 0 and the horizontal asymptote y = 0.

But in the given logarithmic function [tex]y=4\log3(x+7)-2[/tex], we have the inside function as x + 7, which means that there is a horizontal shift left by 7 units

.Therefore, the statement "There is horizontal shift right 2" is false.  There is a vertical shift up 7

The general form of the logarithmic function [tex]y = \log b(x)[/tex] has a y-intercept of (1, 0).

In the given logarithmic function [tex]y=4\log3(x+7)-2[/tex], we have y-intercept as (1, -2). This implies that there is a vertical shift down by 2 units.

Therefore, the statement "There is a vertical shift up 7" is false.  

There is a vertical stretch by 3

Here, we have [tex]y = 4\log3(x + 7) - 2[/tex].

Inside the log function, we have x + 7. This means that there is a horizontal shift to the left by 7 units. Also, the coefficient 4 outside the log function will cause the vertical stretch by a factor of 4.

Hence, the statement "There is a vertical stretch by 3" is false.

Therefore, the statement "There is a vertical stretch by 3" is false.  

There is a horizontal shift left 7

As we have already discussed that inside the log function, we have x + 7. This means that there is a horizontal shift to the left by 7 units.

Therefore, the statement "There is a horizontal shift left 7" is true.  Therefore, the correct statements are "There is a vertical stretch by 4" and "There is a horizontal shift left 7."Hence, option A is correct.

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There is a vertical stretch by 4 and a horizontal shift left 7 in the logarithmic function y = 4log3(x + 7)-2.

Consider the logarithmic function y = 4log3(x + 7)-2.

Explanation: In the logarithmic function, the equation is y = loga(x-h) + k where the logarithmic base is a, horizontal shift is h, and vertical shift is k. Logarithmic functions are the inverse of exponential functions so they have similarities in their transformations.

We can identify the true statements by comparing the given function with the standard form of the logarithmic function which is y = loga(x).

Given: y = 4log3(x + 7)-2

Comparing it to the standard form, there is a horizontal shift of 7 units to the left and a vertical shift of -2 units down. This makes the false statements:

There is a horizontal shift right 2.

There is a vertical shift up 7.

The vertical stretch or shrink is given by the coefficient of x. In this case, it is 4. So, there is a vertical stretch by 4. The coefficient of x can be either positive or negative so there is no horizontal stretch. Thus, the correct statements are:

There is a vertical stretch by 4.

There is a horizontal shift left 7.

Conclusion: There is a vertical stretch by 4 and a horizontal shift left 7 in the logarithmic function y = 4log3(x + 7)-2.

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The argument of the complex number in this problem is given as follows:

arg(z) = 210º.

A complex number is a number that is composed by a real part and an imaginary part, as follows:

z = a + bi.

In which:

The argument of the number is given as follows:

arg(z) = arctan(b/a).

The number for this problem is given as follows:

[tex]z = -\frac{\sqrt{21}}{2} - \frac{\sqrt{7}}{2}i[/tex]

Hence the argument is given as follows:

arg(z) = [tex]\arctan{\left(\frac{1}{\sqrt{3}}\right)}[/tex]

arg(z) = [tex]\arctan{\left(\frac{\sqrt{3}}{3}}\right)}[/tex]

arg(z) = 180º + 30º

arg(z) = 210º.

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The domain of the function [tex]f(x) = \frac {3}{x^2 + 12x - 28}[/tex] is [tex](-\infty, -14) \cup (-14, 2) \cup (2, +\infty)[/tex], excluding the values -14 and 2 from the set of real numbers for which the function is defined.

To determine the domain of the function, we need to identify the values of x for which the function is defined. The function is defined for all real numbers except the values that make the denominator zero, as division by zero is undefined.

To find the values that make the denominator zero, we set the denominator equal to zero and solve the quadratic equation [tex]x^2 + 12x - 28 = 0[/tex]. Using factoring or the quadratic formula, we find that the solutions are x = -14 and x = 2.

Therefore, the function is undefined at x = -14 and x = 2. We exclude these values from the domain. The remaining real numbers, excluding -14 and 2, are included in the domain.

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The position vector v can be written as v = ⟨3, 4⟩.

The position vector v can be written using its components as follows:

v = 3i^ + 4j^

Here, i^ represents the unit vector in the x-direction (horizontal) and j^ represents the unit vector in the y-direction (vertical).

The coefficients 3 and 4 represent the components of the vector v along the x and y axes, respectively.

So, the position vector v can be written as v = ⟨3, 4⟩.

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(a) the interval is approximately [ 1.44, 12.85 ] (b) the 99% lower confidence bound for σ² is approximately 11.54 (c) the 95% lower confidence bound for σ² is approximately 9.38

In the given dataset of essential amino acid (Lysine) composition levels of soybean meals, we are tasked with constructing confidence intervals and bounds for the population variance (σ²). We will use a 99% confidence level for a two-sided interval and calculate both a lower confidence bound at 99% and a lower confidence bound at 95%.

(a) Construct a 99% two-sided confidence interval for σ²:

To construct a confidence interval for σ², we need to use the chi-square distribution. The formula for a two-sided confidence interval is:

[ (n - 1) * s² / χ²(α/2, n-1), (n - 1) * s² / χ²(1 - α/2, n-1) ]

Using the given dataset, we calculate the sample variance (s²) to be approximately 3.46. With a sample size of n = 10, the degrees of freedom (df) for the chi-square distribution is n - 1 = 9.

Using a chi-square distribution table or a statistical calculator, we find the critical values for α/2 = 0.005 and 1 - α/2 = 0.995 with 9 degrees of freedom. The critical values are approximately 2.700 and 21.666, respectively.

Substituting the values into the formula, we get the 99% two-sided confidence interval for σ² as [ (9 * 3.46) / 21.666, (9 * 3.46) / 2.700 ]. Simplifying, the interval is approximately [ 1.44, 12.85 ].

(b) Calculate a 99% lower confidence bound for σ²:

To calculate the lower confidence bound, we use the formula:

Lower bound = (n - 1) * s² / χ²(1 - α, n-1)

Using the same values as before, we substitute α = 0.01 into the formula and find the chi-square critical value χ²(0.99, 9) to be approximately 2.700.

Calculating the lower bound, we have (9 * 3.46) / 2.700 ≈ 11.54. Therefore, the 99% lower confidence bound for σ² is approximately 11.54.

(c) Calculate a 95% lower confidence bound for σ²:

Using a similar approach, we need to find the chi-square critical value for α = 0.05. The critical value χ²(0.95, 9) is approximately 3.325.

Calculating the lower bound, we have (9 * 3.46) / 3.325 ≈ 9.38. Hence, the 95% lower confidence bound for σ² is approximately 9.38.


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The answer is ,  the Taylor series (centered at c=0) for the function f(x) = e^(4x) is given by:

[tex]$$\large f(x) = \sum_{n=0}^{\infty} \frac{4^n}{n!}x^n$$[/tex]

The Taylor series expansion is a way to represent a function as an infinite sum of terms that depend on the function's derivatives.

The Taylor series of a function f(x) centered at c is given by the formula:

[tex]\large f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n[/tex]

Using the definition of Taylor series to find the Taylor series (centered at c=0) for the function f(x) = e^(4x), we have:

[tex]\large e^{4x} = \sum_{n=0}^{\infty} \frac{e^{4(0)}}{n!}(x-0)^n[/tex]

[tex]\large e^{4x} = \sum_{n=0}^{\infty} \frac{4^n}{n!}x^n[/tex]

Therefore, the Taylor series (centered at c=0) for the function f(x) = e^(4x) is given by:

[tex]$$\large f(x) = \sum_{n=0}^{\infty} \frac{4^n}{n!}x^n$$[/tex]

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The Taylor series for f(x) = e^(4x) centered at c = 0 is:

f(x) = 1 + 4x + 8x^2 + 32x^3/3 + ...

To find the Taylor series for the function f(x) = e^(4x) centered at c = 0, we can use the definition of the Taylor series. The general formula for the Taylor series expansion of a function f(x) centered at c is given by:

f(x) = f(c) + f'(c)(x - c) + f''(c)(x - c)^2/2! + f'''(c)(x - c)^3/3! + ...

First, let's find the derivatives of f(x) = e^(4x):

f'(x) = d/dx(e^(4x)) = 4e^(4x)

f''(x) = d^2/dx^2(e^(4x)) = 16e^(4x)

f'''(x) = d^3/dx^3(e^(4x)) = 64e^(4x)

Now, let's evaluate these derivatives at x = c = 0:

f(0) = e^(4*0) = e^0 = 1

f'(0) = 4e^(4*0) = 4e^0 = 4

f''(0) = 16e^(4*0) = 16e^0 = 16

f'''(0) = 64e^(4*0) = 64e^0 = 64

Now we can write the Taylor series expansion:

f(x) = f(0) + f'(0)(x - 0) + f''(0)(x - 0)^2/2! + f'''(0)(x - 0)^3/3! + ...

Substituting the values we found:

f(x) = 1 + 4x + 16x^2/2! + 64x^3/3! + ...

Simplifying the terms:

f(x) = 1 + 4x + 8x^2 + 32x^3/3 + ...

Therefore, the Taylor series for f(x) = e^(4x) centered at c = 0 is:

f(x) = 1 + 4x + 8x^2 + 32x^3/3 + ...

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The osculating circle of the helix defined by y = x^3 - 3x + 1 at x = 1 has a center at (1, -1) and a radius of 1/6.

To determine the osculating circle of the helix defined by y = x^3 - 3x + 1 at x = 1, we need to find the center and radius of the osculating circle.

First, let's find the derivatives of the function y = x^3 - 3x + 1 with respect to x:

y' = 3x^2 - 3

y'' = 6x

Next, we evaluate the derivatives at x = 1 to find the slope and curvature of the helix at that point:

y'(1) = 3(1)^2 - 3 = 0

y''(1) = 6(1) = 6

The slope of the helix at x = 1 is 0, which means the helix is horizontal at that point.

Now, let's find the center and radius of the osculating circle.

Since the helix is horizontal at x = 1, the osculating circle will be centered at the point (1, y(1)). We need to find the y-coordinate of the helix at x = 1:

y(1) = (1)^3 - 3(1) + 1 = -1

Therefore, the center of the osculating circle is (1, -1).

The radius of the osculating circle is given by the reciprocal of the curvature at that point:

R = 1/|y''(1)| = 1/|6| = 1/6

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The odds against obtaining a 5 when a single fair die is tossed are 5 to 1.

The odds against obtaining a 5 when a single fair die is tossed are 5 to 1.

This means that there are 5 ways to not get a 5 (rolling a 1, 2, 3, 4, or 6) and only 1 way to get a 5.

Therefore, the probability of getting a 5 is 1/6.

The odds are a way of expressing probability.

In this case, the odds of obtaining a 5 when rolling a single fair die are 5 to 1.

This means that there are 5 ways of not getting a 5 and only 1 way of getting a 5 when rolling a die.

The probability of rolling a 5 is 1/6 since there are 6 possible outcomes when rolling a die and only 1 of those outcomes is a 5.

The formula for calculating odds is:

Odds Against = (Number of ways it won't happen) :

(Number of ways it will happen)In this case, the number of ways it won't happen is 5 (rolling a 1, 2, 3, 4, or 6) and the number of ways it will happen is 1 (rolling a 5).

So the odds against rolling a 5 are 5 to 1.

Another way to write this is as a fraction:5/1.

This can also be simplified to 5:1.

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Q3 = 63So, the lower Quartile (Q1) = 34, median (Q2) = 52.5, and upper quartile (Q3) = 63.

In order to solve the given problem, we first need to understand the concept of quartiles.

Quartiles are basically the values which divide a given set of data into four equal parts. In other words, we can say that a quartile is the value below which a given fraction of data falls.

The quartiles are of three types- Lower Quartile (Q1), Median (Q2), and Upper Quartile (Q3). These quartiles divide the data set into 4 equal parts where Q1, Q2, and Q3 are quartiles 1, 2 and 3, respectively.To solve the given problem, we will use the following formula: Q1 = L + [(N/4) - F]/fWhere, L is the lower limit of the quartile class is the total number of dataF is the cumulative frequency of the class interval preceding the quartile interval.f is the frequency of the quartile class intervalQ3 = L + [(3N/4) - F]/fWe are given the following data:

Content loaded43 32 68lower quartile median ==upper quartile62=10To solve for quartiles, we will first arrange the given data in ascending order.32, 43, 62, 68Since there are 4 numbers in the given data set, the median will be the mean of the two middle numbers. Therefore, the median of the data set = (43 + 62) / 2 = 52.5Now, to find the lower quartile (Q1), we need to follow the formula:Q1 = L + [(N/4) - F]/fFor this, we need to first find the class interval containing the lower quartile value. This class interval is 32 - 43.

The lower limit of this interval is 32. The frequency of this interval is 2. And the cumulative frequency of the interval preceding it is 0. Putting these values in the formula, we get:Q1 = 32 + [(4/4) - 0]/2Q1 = 32 + 2 = 34Therefore, Q1 = 34To find the upper quartile (Q3), we need to follow the formula:Q3 = L + [(3N/4) - F]/f

For this, we need to first find the class interval containing the upper quartile value. This class interval is 62 - 68. The lower limit of this interval is 62. The frequency of this interval is 1. And the cumulative frequency of the interval preceding it is 2. Putting these values in the formula, we get:Q3 = 62 + [(3*4/4) - 2]/1Q3 = 62 + 1 = 63

Therefore, Q3 = 63So, the lower quartile (Q1) = 34, median (Q2) = 52.5, and upper quartile (Q3) = 63.

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dot(u, v) = 144

hat(u) - hat(v) = 1/sqrt(145), 1/sqrt(145)>

3i = <3, 0>

3u - 4v = <12, -5>

u · v = 144

||u|| = sqrt(145)

Magnitude of vector u: The magnitude of a vector is calculated using the Pythagorean theorem. For vector u, the magnitude ||u|| is obtained by taking the square root of the sum of the squares of its components.

To determine the values of the given expressions involving the vectors u = <-8, 9> and v = <-9, 8>, we can perform the following calculations:

Dot product of u and v: The dot product is calculated by multiplying the corresponding components of the vectors and adding them together. In this case, the dot product of u and v is obtained by (-8)(-9) + (9)(8).

Difference of unit vectors u and v: To find the difference of unit vectors, we need to calculate the unit vectors of u and v first. The unit vector, also known as the direction vector, is obtained by dividing each component of a vector by its magnitude. Then, the difference of unit vectors is found by subtracting the corresponding components.

Vector 3i: The vector 3i is a vector that has a magnitude of 3 and points in the positive x-direction. Therefore, it can be represented as <3, 0>.

Linear combination of vectors 3u and -4v: A linear combination of vectors is obtained by multiplying each vector by a scalar and adding them together. In this case, we multiply 3u and -4v by their respective scalars and then add them.

Vector product of u and v: The vector product, also known as the cross product, is calculated using a formula involving the components of the vectors. The cross product of u and v can be obtained using the formula: (u_y * v_z - u_z * v_y)i - (u_x * v_z - u_z * v_x)j + (u_x * v_y - u_y * v_x)k.

Magnitude of vector u: The magnitude of a vector is calculated using the Pythagorean theorem. For vector u, the magnitude ||u|| is obtained by taking the square root of the sum of the squares of its components.

By performing these calculations, we can determine the exact values for each of the given expressions involving the vectors u and v.

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Set up the partial fraction decomposition for the given function:

[tex]$f(x)=\frac{(b)\sqrt{16x^3+12r^2+10x+2}}{(x^4-4x^2)(x^2+x+1)^2(x^2-3x+2)(x+3x^2+2)}$[/tex]

is: [tex]=\frac{3}{26}x^2-\frac{21}{13}x+\frac{9}{13}\ln|x+4|-\frac{1}{13}\arctan\frac{x+1}{2}+\frac{5}{39}\ln[(x+1)^2+4]+C[/tex]

The denominator factors completely to:

[tex]x^4-4x^2=x^2(x^2-4)=x^2(x-\sqrt{2})(x+\sqrt{2})[/tex]

[tex]x^2+x+1=(x-(\frac{-1+i\sqrt{3}}{2}))(x-(\frac{-1-i\sqrt{3}}{2}))$$[/tex]

[tex]x^2-3x+2=(x-2)(x-1)[/tex]

[tex]x+3x^2+2=(x+\frac{3}{2})^2-\frac{1}{4}[/tex]

Therefore,

[tex]\begin{aligned}\frac{f(x)}{(b)\sqrt{16x^3+12r^2+10x+2}}&=\frac{A}{x}+\frac{B}{x-\sqrt{2}}+\frac{C}{x+\sqrt{2}}+\frac{Dx+E}{x^2+x+1}+\frac{Fx+G}{(x^2+x+1)^2}\\&+\frac{H}{x-2}+\frac{J}{x-1}+\frac{Kx+L}{x+\frac{3}{2}+\frac{1}{2}}+\frac{Nx+M}{(x+\frac{3}{2}+\frac{1}{2})^2}\end{aligned}[/tex]

(a) [tex]$\int\frac{3x^2-3x+1}{(x^2+2x+5)(x+4)}dx$[/tex] is

[tex]$\frac{1}{13}\int\frac{3x^2-3x+1}{x+4}dx-\frac{1}{13}\int\frac{3x^2-3x+1}{x^2+2x+5}dx$[/tex].

[tex]=\frac{3}{13}\int\frac{x^2-x}{x+4}dx+\frac{1}{13}\int\frac{1}{x+4}dx-\frac{1}{13}\int\frac{3x^2-3x+1}{(x+1)^2+4}dx[/tex]

[tex]=\frac{3}{13}\int\frac{x^2-x}{x+4}dx+\frac{1}{13}\ln|x+4|-\frac{1}{39}\int\frac{3x-7}{(x+1)^2+4}d(x+1)[/tex]

[tex]=\frac{3}{13}\int x-4+\frac{16}{x+4}dx+\frac{1}{13}\ln|x+4|-\frac{1}{39}\int\frac{3}{t^2+4}dt-\frac{1}{39}\int\frac{x+5}{(x+1)^2+4}d(x+1)[/tex]

[tex]=\frac{3}{26}x^2-\frac{21}{13}x+\frac{9}{13}\ln|x+4|-\frac{1}{13}\arctan\frac{x+1}{2}+\frac{5}{39}\ln[(x+1)^2+4]+C[/tex]

where t=x+1 is used. The answer is obtained by partial fraction decomposition and substitution. The value of the constant C is not given.

Conclusion: Partial fraction decomposition is a technique to decompose complex rational functions into simpler rational functions. In this technique, we decompose a complex rational function into simpler fractions whose denominators are linear factors or irreducible quadratic factors. This is the easiest way to solve integrals of rational functions that are complex.

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The partial fraction decomposition for the given function and evaluated the given indefinite integrals.

∫dx/[(3x²-3x+1)³+1] = -1/6[1/(3x² - 3x + 2)] + 1/6[1/[(3x² - 3x + 2)²]] + 1/2[arctan(6x² - 6x + 1)] + C

∫dx/[(x² + 2x + 5)(x + 4)] = 1/(x + 4) - 2x + 1/(x² + 2x + 5) + C

∫dx/√(2 + 2x + 5) = (1/2) ∫du/√u

The partial fraction decomposition for the given function and evaluated the given indefinite integrals.

The steps to set up the partial fraction decomposition for a given function

f(x) = (b) √ 16x³ +12r² + 10x +2 (x⁴-4x²)(x² + x + 1)²(x²-3x + 2)(x+3x²+2) are given below:

Step 1: Factorise the denominator: x⁴ - 4x² = x²(x² - 4)

= x²(x + 2)(x - 2) (x² + x + 1)²(x² - 3x + 2)(x + 3x² + 2)

Step 2: Determine the degree of the numerator, which is less than the degree of the denominator. In this case, the degree of the numerator is 1. So, the partial fraction decomposition will be of the form

A/x + B/(x² + x + 1) + C/(x² - 3x + 2) + D/(x + 3x² + 2) + E/[(x² + x + 1)²]

Step 3: Multiply the original function f(x) by the denominator, set the numerator of the result equal to the sum of the numerators of the partial fractions, and simplify. Then, equate the coefficients of the resulting polynomial equations in x with each other. Solve the system of equations for the unknown constants. Then, use partial fraction decomposition to break up a rational function into simpler fractions.

Explanation: Given function is f(x) = (b) √ 16x³ +12r² + 10x +2 (x⁴-4x²)(x² + x + 1)²(x²-3x + 2)(x+3x²+2). Partial fraction decomposition is the decomposition of a rational function into simpler fractions. The steps to set up the partial fraction decomposition for a given function f(x) = (b) √ 16x³ +12r² + 10x +2 (x⁴-4x²)(x² + x + 1)²(x²-3x + 2)(x+3x²+2) are as follows:

Step 1: Factorise the denominator: x⁴ - 4x² = x²(x² - 4) = x²(x + 2)(x - 2) (x² + x + 1)²(x² - 3x + 2)(x + 3x² + 2)

Step 2: Determine the degree of the numerator, which is less than the degree of the denominator. In this case, the degree of the numerator is 1. So, the partial fraction decomposition will be of the form A/x + B/(x² + x + 1) + C/(x² - 3x + 2) + D/(x + 3x² + 2) + E/[(x² + x + 1)²]

Step 3: Multiply the original function f(x) by the denominator, set the numerator of the result equal to the sum of the numerators of the partial fractions, and simplify. Then, equate the coefficients of the resulting polynomial equations in x with each other. Solve the system of equations for the unknown constants. Then, use partial fraction decomposition to break up a rational function into simpler fractions.

The answers for the given integrals are as follows:

(a) ∫dx/[(3x²-3x+1)³+1] = 1/3 ∫du/u³

(by substitution, let u = 3x² - 3x + 1)

= -1/6[1/(u² + u + 1)] + 1/6[1/[(u² + u + 1)²]] + 1/2[arctan(2u - 1)] + C

= -1/6[1/(3x² - 3x + 2)] + 1/6[1/[(3x² - 3x + 2)²]] + 1/2[arctan(6x² - 6x + 1)] + C

(b) ∫dx/[(x² + 2x + 5)(x + 4)] = A/(x² + 2x + 5) + B/(x + 4)

= [A(x + 4) + B(x² + 2x + 5)]/[(x² + 2x + 5)(x + 4)]

= [(A + B)x² + (4A + 2B)x + (5B)]/[(x² + 2x + 5)(x + 4)]

= 1/(x + 4) - 2x + 1/(x² + 2x + 5) + C

(d) ∫dx/√(2 + 2x + 5)

= ∫dx/√(7 + 2x) = (1/√2)arcsin((2x + 1)/√7) + C

(e) ∫xdx/(1 + x²) = (1/2)ln(1 + x²) + (1/2)arctan(x) + C(d) ∫dr/[(b) √ 16r³ + 12r² + 10r + 2]

= (1/2) ∫du/√u

(by substitution, let u = 16r³ + 12r² + 10r + 2)

= (1/2) √u + C

= (1/2) √(16r³ + 12r² + 10r + 2) + C.

Conclusion: Thus, we have set up the partial fraction decomposition for the given function and evaluated the given indefinite integrals.

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The differential system is:dx/dt = 3x - y dy/dt = 9x - 3y

With the initial conditions:

x (0) = 2, y (0) = 4

It is asked to find the solution of this system of differential equations.

First, let's find the eigenvalues of the matrix:| λ -3 | = (λ-3) (-3) - 9 = λ² - 3 λ - 18 = 0(λ - 6) (λ + 3) = 0

Then, the eigenvalues are λ₁ = 6, λ₂ = -3

Let's find the eigenvectors for λ₁ = 6:x(1) - 3x(2) = 0x(1) = 3x(2)x(1) = k 1 , x(2) = k 2

Then, the eigenvector is:(k 1 , k 2 ) = (3, 1)The eigenvector for λ₂ = -3 is:(k 1 , k 2 ) = (1, 3)

Therefore, the solution of the system is:

x(t) = c₁ e^(6t) (3, 1) + c₂ e^(-3t) (1, 3)

Where c₁ and c₂ are constants determined by the initial conditions.

The value of x (0) = (2, 4) gives:

c₁ (3, 1) + c₂ (1, 3) = (2, 4)

The previous system can be written as:3c₁ + c₂ = 2c₁ + 3c₂ = 4

Then, we can solve for c₁ and c₂ using the elimination method. From the first equation we get:c₂ = 2 - 3c₁Replacing c₂ in the second equation we get:2c₁ + 3 (2 - 3c₁) = 42c₁ = 5c₁ = 5/2

Substituting this in c₂ = 2 - 3c₁

we have: c₂ = -1/2

The constants are :c₁ = 5/2, c₂ = -1/2T

he solution to the system is:

x(t) = (5/2) e^(6t) (3, 1) - (1/2) e^(-3t) (1, 3

)Therefore, the real solution is:

x₁ = (5/2) e^(6t) (3) - (1/2) e^(-3t) (1)x₂ = (5/2) e^(6t) (1) - (1/2) e^(-3t) (3)

The solution to the system is:

x₁ = (5/2) e^(6t) (3) - (1/2) e^(-3t) (1)x₂ = (5/2) e^(6t) (1) - (1/2) e^(-3t) (3)

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Plan 2 is preferred because it requires a significantly lower sample size (50) compared to Plan 1 (5000), resulting in a more efficient inspection process. Plan 2 achieves a similar level of quality control while reducing the inspection effort and associated costs.

To calculate the Average Total Inspection (ATI) for both plans, we need to use the formula:

ATI = (Probability of acceptance) * (Sample size) + (Probability of rejection) * (Total population)

For Plan 1 (c = 0):

Probability of acceptance = 1 - P(reject | p = 0.005, c = 0) = 1 - 0.09539 = 0.90461

Probability of rejection = P(reject | p = 0.005, c = 0) = 0.09539

ATI for Plan 1 = (0.90461) * (50) + (0.09539) * (5000) ≈ 45.23 + 476.95 ≈ 522.18

For Plan 2 (c = 2):

Probability of acceptance = 1 - P(reject | p = 0.005, c = 2) = 1 - 0.00206 = 0.99794

Probability of rejection = P(reject | p = 0.005, c = 2) = 0.00206

ATI for Plan 2 = (0.99794) * (50) + (0.00206) * (5000) ≈ 49.897 + 10.3 ≈ 60.197

In terms of Average Total Inspection (ATI), Plan 1 has an ATI of approximately 522.18, while Plan 2 has an ATI of approximately 60.197.  Therefore, Plan 2 achieves a similar level of quality control while reducing the inspection effort and associated costs.

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The probability of having at least 3 defective phones in a random sample of 30 phones, given a defect rate of 14%, can be calculated using binomial probability.

By finding the probability of exactly 3, 4, 5, ..., up to 30 defective phones and summing them together, we can determine the overall probability. In this case, the probability is approximately 0.984, or 98.4%.

To calculate the probability of at least 3 defective phones, we need to consider all possible scenarios where 3 or more phones are defective. We can calculate the probability for each scenario separately and sum them up to obtain the overall probability.

The probability of having exactly k defective phones in a sample of n phones can be calculated using the binomial probability formula. The formula is given by P(X = k) = (nCk) * p^k * (1-p)^(n-k), where nCk is the number of combinations of n items taken k at a time, p is the probability of a phone being defective, and (1-p) is the probability of a phone being non-defective.

In this case, p = 0.14 (14%) and n = 30. We need to calculate the probabilities for k = 3, 4, 5, ..., up to 30 and sum them together to find the probability of at least 3 defective phones. This can be done using a statistical software, spreadsheet, or calculator with binomial probability functions.

By performing the calculations, we find that the probability of having at least 3 defective phones in a random sample of 30 phones, given a defect rate of 14%, is approximately 0.984, or 98.4%. This means that in most cases, we can expect a high likelihood of encountering at least 3 defective phones in a sample of 30.

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For a binomial distribution with n=8 and p=0.73, the probability of exactly 3 successes is 0.255. The expected value (mean) is 5.84, and the variance is approximately 1.3092.

a. The probability of exactly 3 successes:
Here, n = 8, p = 0.73, q = 0.27 and x = 3

We use the binomial distribution formula which is given as:
P(X=x) = C(n, x) * p^x * q^(n-x)
P(X=3) = C(8, 3) * 0.73^3 * 0.27^5
= [8! / (3! * 5!)] * 0.73^3 * 0.27^5
= 0.255 (approx) [Ans]

b. The expected value(mean):
The expected value of a binomial distribution is given by:
μ = np
μ = 8 * 0.73
μ = 5.84 [Ans]

c. The variance of the random variable:
The variance of a binomial distribution is given by:
σ^2 = npq
σ^2 = 8 * 0.73 * 0.27
σ^2 = 1.3092
σ = √1.3092
σ = 1.1437 (approx)

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Therefore, the probability that washing dishes tonight will take between 12 and 15 minutes is 0.33

The probability is given by the formula:

Probability (P) = (length of the interval) / (total length of the distribution)

Now, the length of the interval is 15 - 12 = 3 and the total length of the distribution is 16 - 7 = 9.

Thus, Probability (P) = 3 / 9

= 1 / 3 = 0.33 (rounded to two decimal places).

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To calculate the excess reserves of the bank after the customer's deposit, we first need to determine the required reserves.

Required Reserves = Checkable Deposits * Required Reserve Ratio

Required Reserves = $10,101 * 0.08

Required Reserves = $808.08

Next, we can calculate the excess reserves:

Excess Reserves = Reserves - Required Reserves

Excess Reserves = $1,205 - $808.08

Excess Reserves = $396.92

Therefore, the bank's excess reserves after the customer's deposit are approximately $396.92. None of the provided options ($580.92, $544.44, $415.32, $401.11) match this value.

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The given integral is ∫e^(-3x)sin(3x)dx. We can solve this using partial integration twice.

Let's define u = e^(-3x) and dv = sin(3x)dx. Taking the respective differentials, we have du = -3e^(-3x)dx and v = -cos(3x)/3.

Applying the partial integration formula, we have ∫u dv = uv - ∫v du. Substituting the values, we get ∫e^(-3x)sin(3x)dx = -e^(-3x)cos(3x)/3 + ∫(1/3)e^(-3x)cos(3x)dx.

Now, we perform partial integration again on the remaining integral. Let's define u = e^(-3x) and dv = cos(3x)dx. Taking the respective differentials, we have du = -3e^(-3x)dx and v = sin(3x)/3.

Applying the partial integration formula once more, we have ∫u dv = uv - ∫v du. Substituting the values, we get the integral as (1/9)e^(-3x)sin(3x) + (1/9)∫e^(-3x)sin(3x)dx.

Substituting this result back into the previous equation, we obtain ∫e^(-3x)sin(3x)dx = -e^(-3x)cos(3x)/3 + (1/9)e^(-3x)sin(3x) + (1/9)∫e^(-3x)sin(3x)dx.

Rearranging the equation, we have (8/9)∫e^(-3x)sin(3x)dx = -e^(-3x)cos(3x)/3 + (1/9)e^(-3x)sin(3x).

Finally, solving for the integral, we get ∫e^(-3x)sin(3x)dx = -(9/8)e^(-3x)cos(3x) + (1/8)e^(-3x)sin(3x) + C, where C is the constant of integration.

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John earned $530 in interest in one year.

13. The percent of decrease in mass is about 6.25%.

To solve the interest earned by John in one year,  the formula below is used:

Interest = Principal × Rate

Note that John invested $4000 at an interest rate of 13 1/4% (or 13.25% as a decimal), so:

Interest = $4000 × 0.1325

Interest = $530

So, John earned $530 in interest in one year.

a)  Percentage increase = (Difference / Original Value) × 100

= ((77 cm - 52 cm) / 52 cm) × 100

= (25 cm / 52 cm) × 100

= 48.08%

b)  Increase = $24.00 × (7% / 100)

 = $24.00 × 0.07

= $1.68

 = $24.00 + $1.68

 = $25.68

c.  Percentage decrease = (Difference / Original Value) × 100

 = (($210.00 - $45.00) / $210.00) × 100

 = ($165.00 / $210.00) × 100

= 78.57%

d)  Decrease = 15 kg × (14% / 100)

= 15 kg × 0.14

= 2.1 kg

= 15 kg - 2.1 kg

= 12.9 kg

13.  Percent decrease = (Decrease in mass / Original mass) × 100

Percent decrease = ((24 kg - 22.5 kg) / 24 kg) × 100

                             = (1.5 kg / 24 kg) × 100

                               =6.25%

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See correct question below

11. John invested $4000 at an interest rate of 13 1/4%

​How much interest did John earn in one year? 12. a) 77 cm exceeds 52 cm by what percent? b) $24.00 increased by 7%% is how much? c) $45.00 is what percent less than $210.00? d) 15 kg decreased by 14% is how much? 13. A rough casting with a mass of 24 kg, after having been finished on a lathe, has a mass of 221/2 kg. What is the percent of decrease in mass?

John earned $530 in interest in one year a) 77 cm exceeds 52 cm by about 48.08%.  b) $24.00 increased by 7% is equal to $25.68.  c) $45.00 is about 78.57% less than $210.00.  d) 15 kg decreased by 14% is equal to 12.9 kg.

The percent of decrease in mass is about 6.25%.

What is the Interest?

To solve the interest earned by John in one year,  the formula below is used:

Interest = Principal × Rate

Note that John invested $4000 at an interest rate of 13 1/4% (or 13.25% as a decimal), so:

Interest = $4000 × 0.1325

Interest = $530

So, John earned $530 in interest in one year.

a) Percentage increase = (Difference / Original Value) × 100

= ((77 cm - 52 cm) / 52 cm) × 100

= (25 cm / 52 cm) × 100

= 48.08%

b)  Increase = $24.00 × (7% / 100)

= $24.00 × 0.07

= $1.68

= $24.00 + $1.68

= $25.68

c.  Percentage decrease = (Difference / Original Value) × 100

= (($210.00 - $45.00) / $210.00) × 100

= ($165.00 / $210.00) × 100

= 78.57%

d)  Decrease = 15 kg × (14% / 100)

= 15 kg × 0.14

= 2.1 kg

= 15 kg - 2.1 kg

= 12.9 kg

13.  Percent decrease = (Decrease in mass / Original mass) × 100

Percent decrease = ((24 kg - 22.5 kg) / 24 kg) × 100

                            = (1.5 kg / 24 kg) × 100

                             =6.25%

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the temperature of the equipment is changing at a rate of 0.3 degree/s per second after the balloon is released.

The question is to find the temperature of the equipment per second after the balloon is released.

Given,  height is changing by 1.5 meter/s and temperature is changing 0.2deg/ meter

. We need to find the rate of change of temperature, that is dT/dt .From the question, we know the following data: dh/dt = 1.5 (m/s)dT/dh = 0.2 (degree/m)We need to find dT/dt. To find dT/dt, we can use the chain rule of differentiation, which states that the derivative of a composite function is the product of the derivative of the outer function and the derivative of the inner function multiplied together.

Let h(t) be the height of the balloon at time t, and let T(h) be the temperature at height h. Then we have T(h(t)) as the temperature of the balloon at time t. We can differentiate this with respect to time using the chain rule as follows: dT/dt = dT/dh × dh/dt Substitute the given values and we getdT/dt = 0.2 × 1.5 = 0.3 degree/s. Thus, the temperature of the equipment is changing at a rate of 0.3 degree/s per second after the balloon is released.

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The solution of equation after differentiate,

f' (x) = 1/3 (18x - 5) (9x² - 5x + 7)²

We have to given that,

Differential equation is,

f (x) = ∛(9x² - 5x + 7)

Now, We can differentiate the equation with respect to x as,

f (x) = ∛(9x² - 5x + 7)

f' (x) = 1/3 (9x² - 5x + 7)³⁻¹ (18x - 5)

f' (x) = 1/3 (18x - 5) (9x² - 5x + 7)²

Therefore, the solution of equation after differentiate,

f' (x) = 1/3 (18x - 5) (9x² - 5x + 7)²

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(a) Mean of the sampling distribution (μx) = 16

  Variance of the sampling distribution (σ²x) ≈ 0.1111

  Standard deviation of the sampling distribution (σx) ≈ 0.3333

(b) Mean of the sampling distribution (μx) = 550

   Variance of the sampling distribution (σ²x) ≈ 0.0581

   Standard deviation of the sampling distribution (σx) ≈ 0.2410

(c) Mean of the sampling distribution (μx) = 5

   Variance of the sampling distribution (σ²x) ≈ 0.0057

   Standard deviation of the sampling distribution (σx) ≈ 0.0756

(d) Mean of the sampling distribution (μx) = 81

   Variance of the sampling distribution (σ²x) ≈ 0.0049

   Standard deviation of the sampling distribution (σx) ≈ 0.0700

The mean, variance, and standard deviation of the sampling distribution of the sample mean, denoted as x, can be calculated using the following formulas:

Mean of the sampling distribution (μx) = μ (same as the population mean)

Variance of the sampling distribution (σ²x) = σ² / n (where σ is the population standard deviation and n is the sample size)

Standard deviation of the sampling distribution (σx) = σ / √n

Let's calculate the values for each situation:

a) μ = 16, σ = 2, n = 36

Mean of the sampling distribution (μx) = 16

Variance of the sampling distribution (σ²x) = (2²) / 36 = 0.1111

Standard deviation of the sampling distribution (σx) = √(0.1111) ≈ 0.3333

b) μ = 550, σ = 3, n = 155

Mean of the sampling distribution (μx) = 550

Variance of the sampling distribution (σ²x) = (3²) / 155 ≈ 0.0581

Standard deviation of the sampling distribution (σx) = √(0.0581) ≈ 0.2410

c) μ = 5, σ = 0.2, n = 7

Mean of the sampling distribution (μx) = 5

Variance of the sampling distribution (σ²x) = (0.2²) / 7 ≈ 0.0057

Standard deviation of the sampling distribution (σx) = √(0.0057) ≈ 0.0756

d) μ = 81, σ = 3, n = 1831

Mean of the sampling distribution (μx) = 81

Variance of the sampling distribution (σ²x) = (3²) / 1831 ≈ 0.0049

Standard deviation of the sampling distribution (σx) = √(0.0049) ≈ 0.0700

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Answer:

(a) $52,400

(b) $44,540.

Step-by-step explanation:

(b) $52,400 - $7,860 = $44,540

tax-deferred retirement plans like

401k

are great because

$7,860 wouldve been lost to taxes.

instead it goes to retirement

To raise the complex number [tex][2(cis 110°)]^3[/tex] to the given power, we can use De Moivre's theorem, which states that for any complex number in polar form (r cis θ), its nth power can be expressed as[tex](r^n) cis (nθ).[/tex]

In this case, we have[tex][2(cis 110°)]^3[/tex]. Let's simplify it step by step:

First, raise the modulus (2) to the power of 3: (2^3) = 8.

Next, multiply the argument (110°) by 3: 110° × 3 = 330°.

Therefore,[tex][2(cis 110°)]^3[/tex]simplifies to (8 cis 330°).

Expressing the result in rectangular form (a + bi), we can convert the polar coordinates to rectangular form using the relationships cos θ = a/r and sin θ = b/r, where r is the modulus:

cos 330° = a/8

sin 330° = b/8

Using trigonometric identities, we find:

cos 330° = √3/2 and sin 330° = -1/2

Substituting these values, we get:

[tex]a = (8)(√3/2) = 4√3[/tex]

[tex]b = (8)(-1/2) = -4[/tex]

Therefore, [2(cis 110°)]^3, expressed in rectangular form, is 4√3 - 4i.

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