The function f(x) = 2x^3 – 36x^2 + 192x + 5 has one local minimum and one local maximum. This function has a local minimum at x equals with value and a local maximum at requals with value

Answer:

171 fish

Step-by-step explanation:

512/3=170.667 (keeps going on but I just rounded it to add 1 to the total of 170) so that is why 171 is the answer.

B) The alpha value of the reliability coefficient is one way to establish the validity of a questionnaire.

In its most basic form, an equation is a mathematical statement that indicates that two mathematical expressions are equal. 3x + 5 = 14, for example, is an equation in which 3x + 5 and 14 are two expressions separated by a 'equal' sign.

Here,

The alpha value of the reliability coefficient, also known as Cronbach's alpha, measures the internal consistency of a questionnaire. It indicates how well the items on the questionnaire are related to each other and form a consistent measure of the concept being assessed. A high alpha value indicates that the items on the questionnaire are measuring the same underlying concept and that the questionnaire is reliable.

Correlation between the score on the questionnaire and another measure of the same concept (Option A) and consistency by which multiple subjects are able to complete the questionnaire without error (Option D) can also provide information about the reliability of a questionnaire.

The alpha value of the reliability coefficient is one way to establish the validity of a questionnaire.

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Answer:

first one, and the third one

Step-by-step explanation:

Answer:

Options 2 and 3

Step-by-step explanation:

If there are 2 boys and 5 altogether, means there are 3 girls. The ratio of boys to girls would be 2:3.

Look at all the first and second options as fractions to make things simpler. Simplify the fractions and see which ones are equivalent to 2/3

4/10 = 2/5 FALSE

8/12 = 2/3 TRUE

Options 3, 4, and 5 are comparing boys to all students, so whatever matches the ratio 2/5 is true.

10/25 = 2/5 TRUE

20/30 = 2/3 FALSE

9/15 = 3/5 FALSE

No . of girls in Ms. Jeffry's environmental club = 6

No . of boys in Ms. Jeffry's environmental club = 12

To find the maximum number of groups she can make , we have to find the highest common factor ( HCF ) of the number of boys and girls in her club .

Factors of 6 = 1 , 2 , 3 and 6

Factors of 12 = 2 , 3 , 4 , 6 and 12

Hence the highest common factor of 6 and 12 are 6 .

Therefore , the maximum number of groups Ms . Jeffry can make where the number of girls and boys are same are 6 .

The area of the remaining space in the park is \(4x(3x - y)\) square units.

To find the area of the remaining space in the park, we need to subtract the area of the two crossroads from the total area of the park.

The park has a length of 4x units and a width of 3x units. This gives us a total area of \((4x)(3x) = 12x^2\) square units.

Each crossroad has a width of y units, and since there are two crossroads, the total width of the crossroads is 2y units.

To find the area of the crossroads, we multiply the total width by the length of the park.

Since the crossroads run through the center of the park, the length of the park is divided equally on both sides of each crossroad.

Therefore, the length of each crossroad is \((4x)/2 = 2x\) units.

The area of each crossroad is \((2y)(2x) = 4xy\) square units.

To find the area of the remaining space in the park, we subtract the area of the crossroads from the total area of the park: \(2x^2 - 4xy = 4x(3x - y)\) square units.

So, the area of the remaining space in the park is \(4x(3x - y)\)  square units.

In conclusion, the area of the remaining space in the park is \(4x(3x - y)\) square units.

This formula takes into account the dimensions of the park and the width of the crossroads.

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The trapezoidal rule uses trapezoidal approximations to approximate the definite integral, whereas the midpoint rule uses rectangular regions to do so. By first approximating the original function with piecewise quadratic functions, Simpson's rule approximates the definite integral.

Since

"frac" (1-x) = (1-x)-1 = 1+x2+x3+x4+.... = "sum_n=0" infty" xn "hspace"

5mm for "left | x | 1"

How can a power-based integral be evaluated?

Power Rule for Integrals To apply this rule, simply add one to the exponent.The expression is then divided by the same number.The constant C is added in the final step. Now, we will solve some problems to determine the integrals of polynomials and root functions.

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The reference angle for 138 is 58.

Let's first sketch the angle and then find its reference angle using the terms you mentioned.

Step 1: Sketch the angle 13π/8
To sketch the angle, first, we need to determine its position on the coordinate plane. Since 13/8 is greater than  (or 8/8), it's located in the third quadrant. As a full circle is 2 (or 16/8), we can determine the angle measured counterclockwise from the positive x-axis as:
Angle = 13π/8

Step 2: Find the reference angle
A reference angle is an acute angle formed between the terminal side of the angle and the x-axis. In the third quadrant, the reference angle is calculated as:
Reference Angle = Angle - π

Now we substitute the values:
Reference Angle = 13π/8 - 8π/8 = 5π/8

So, the reference angle for 138 is 58.

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Answer:

2(2x+15)

Step-by-step explanation:

x+(x+30)+2x=

=3x+(x+30)=

=3x+x+30=

=4x+30=

=2(2x+15)

Hello!

To solve for the y-values in the table:

 \(\hookrightarrow \text{ Must plug in the corresponding x's value into the equation}\)

             \(y = 2(-2) + 3 = -4 + 3 = -1\)

             \(y = 2(0) + 3 =3\)

            \(y = 2(1) +3=5\)

           \(y = 2(3) + 3=6 + 3=9\)

Hope that helps!

The rest allowance for chopping down a tree is 0.67 hours (rounded to two decimal places) or 40 minutes. In an 8-hour shift, approximately 40 minutes should be allowed for rest.

To calculate the rest allowance, we need to determine the energy expenditure for chopping down a tree and convert it into a time duration.

Given that the energy expenditure associated with chopping down a tree is 8.0 kcal/min, we can calculate the rest allowance using the following formula:

Rest allowance = Energy expenditure (kcal/min) * Time duration (min) / Energy content of food (kcal).

As the rest allowance is typically a fraction of the energy expenditure, we can use the value of 0.25 (25%) as the input for the rest allowance calculation.

Rest allowance = 8.0 kcal/min * Time duration (min) / Energy content of food (kcal) = 0.25.

Solving for the time duration, we find:

Time duration (min) = 0.25 * Energy content of food (kcal) / 8.0 kcal/min.

To determine the time duration in hours, we divide the time duration in minutes by 60:

Time duration (hours) = Time duration (min) / 60.

The specific energy content of food is not provided in the question. Therefore, without knowing the energy content, we cannot calculate the exact time duration for the rest allowance.

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Answer:

The correct option is the second option.

How to make a Box-and-whisker plot:

Solution explanation:

if you apply the rules of the boxplot above with both choices, the second lines up perfectly.

The hypergeometric probabilities for the given values of n and x are:

(a) P(X = 2) = 0.5143

(b) P(X = 3) = 0.1429

(c) P(X = 0) = 0.3571

(d) P(X = 3) = 0.0714

The hypergeometric probability mass function is given by:

P(X = x) = (rCx)(N-rCn-x) / (NCn)

where N is the total number of items in the population, r is the number of "successes" in the population, n is the sample size, and x is the number of "successes" in the sample.

(a) For n = 4 and x = 2:

P(X = 2) = (4C2)(8-4C4-2) / (8C4)

= (6)(6) / (70)

= 0.5143

(b) For n = 6 and x = 3:

P(X = 3) = (4C3)(8-4C6-3) / (8C6)

= (4)(1) / (28)

= 0.1429

(c) For n = 3 and x = 0:

P(X = 0) = (4C0)(8-4C3-0) / (8C3)

= (1)(20) / (56)

= 0.3571

(d) For n = 3 and x = 3:

P(X = 3) = (4C3)(8-4C3-3) / (8C3)

= (4)(1) / (56)

= 0.0714

Therefore, the hypergeometric probabilities for the given values of n and x are:

(a) P(X = 2) = 0.5143

(b) P(X = 3) = 0.1429

(c) P(X = 0) = 0.3571

(d) P(X = 3) = 0.0714

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Answer:

15

Step-by-step explanation:

The probability of obtaining five heads in a row when flipping a coin is 0.03125. The probability of this event is relatively low which means that it is an unlikely event. The probability can also be expressed in terms of percentages as:3.125% or 1 in 32. Here, only one outcome is favourable, which is obtaining five heads in a row.

The total number of outcomes is 2^5, since there are two possible outcomes of each flip, heads or tails. Therefore, the probability can be calculated as:P(5 heads in a row) = (number of favourable outcomes)/(total number of possible outcomes)P(5 heads in a row) = 1/32If the event of a coin being flipped five times is repeated ten thousand different times, it is expected that the event would result in five heads about 312 times. This is because the probability of getting five heads in a row is 1/32.

P(A or B) = P(A) + P(B)P(A or B) = 0.24 + 0.25P(A or B) = 0.49b.

The probability that a student will get an A OR a B OR a C is 0.73. Here, we need to add the probability of getting an A, B and C:P(A or B or C) = P(A) + P(B) + P(C)P(A or B or C) = 0.24 + 0.25 + 0.24P(A or B or C) = 0.73c. Here, we need to subtract the probability of getting an A, B and C from 1:P(lower than C) = 1 - P(A) - P(B) - P(C)P(lower than C) = 1 - 0.24 - 0.25 - 0.24P(lower than C) = 0.51Therefore, the probability that a student will get an A OR a B is 0.49, the probability that a student will get an A OR a B OR a C is 0.73 and the probability that a student will get a grade lower than a C is 0.51.

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Answer:

he should save $70 each week

Step-by-step explanation:

1200 - 640 = 560

560/8 = 70

A point estimate is the value of a statistic that estimates the value of a parameter. Question 2 is false and question 3 is true.

Question 1: A point estimate is the value of a statistic that estimates the value of a parameter.A point estimate is a single number that is used to estimate the value of an unknown parameter of a population, such as a population mean or proportion

Question 2: False

Mu (μ) is not used to estimate X. Mu represents the population mean, while X represents the sample mean. The sample mean, X, is used as an estimate of the population mean, μ.

Question 3: True

Beta (β) is indeed used to estimate the population proportion (p) when conducting hypothesis testing on a sample. Beta represents the probability of making a Type II error, which occurs when we fail to reject a null hypothesis that is actually false. By calculating the probability of a Type II error, we indirectly estimate the population proportion, p, under certain conditions and assumptions.

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The expression csc²θ - 2cot²θ can be simplified to (1 - 2cos²θ) / sin²θ is obtained by using trignomentry expressions. This expression is equivalent to option (F) in the given choices.

To simplify the expression csc²θ - 2cot²θ, we can rewrite csc²θ and cot²θ in terms of sinθ and cosθ.

csc²θ = (1/sinθ)² = 1/sin²θ

cot²θ = (cosθ/sinθ)² = cos²θ/sin²θ

Substituting these values back into the expression:

csc²θ - 2cot²θ = 1/sin²θ - 2(cos²θ/sin²θ)

Now, we can combine the terms with a common denominator:

= (1 - 2cos²θ) / sin²θ

This simplification matches option (F) in the given choices.

Therefore, the expression csc²θ - 2cot²θ can be expressed as (1 - 2cos²θ) / sin²θ.

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The probability that at most three of the 15 observations exceed 7.779 from a chi-square distribution with four degrees of freedom is approximately 0.8946.

To find the probability that at most three of the 15 observations exceed 7.779 from a chi-square distribution with four degrees of freedom, we can use the cumulative distribution function (CDF) of the chi-square distribution.

The chi-square distribution with ν degrees of freedom (where ν is the number of degrees of freedom) is a continuous probability distribution that describes the sum of squares of ν independent standard normal random variables.

First, we need to calculate the cumulative probability of exceeding 7.779 for a single observation from the chi-square distribution with four degrees of freedom. Let's call this probability p₁.

Using a chi-square distribution table or a statistical software, we can find that p₁ is approximately 0.0972.

Next, we want to find the probability of at most three observations exceeding 7.779 out of the 15 independent observations. Let's call this probability p.

We can use the binomial distribution to calculate p. The binomial distribution describes the probability of a certain number of successes (exceeding 7.779) in a fixed number of independent Bernoulli trials (each observation being a trial) with a constant probability of success (p₁).

Using the binomial cumulative distribution function, we can calculate p as follows:

\(p = \sum(i=0 to 3) [ C(15, i) * p_1^i * (1 - p_1)^(15-i) ]\),

where C(15, i) represents the binomial coefficient, which is the number of ways to choose i observations out of 15.

Evaluating this expression, we find that p is approximately 0.8946.

Therefore, the probability that at most three of the 15 observations exceed 7.779 from a chi-square distribution with four degrees of freedom is approximately 0.8946.

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The following statements are true for all real numbers a, b, m, and n, where a > 0 and b > 0:

\((a^m)^n = a^{(mn)}\)

\(a^m/a^n = a^{(m-n)}\)

\((ab)^n = a^n * b^n\)

The first statement, \((a^m)^n = a^{(mn)}\), is true because when we raise a number to a power and then raise the result to another power, it is equivalent to raising the number to the product of the two powers.

The second statement, \(a^m/a^n = a^{(m-n)}\), is also true. When we divide two numbers with the same base, we subtract the exponents. This property of exponents allows us to simplify division of exponential expressions.

The third statement, \((ab)^n = a^n * b^n,\) is true because when we raise a product of two numbers to a power, we can distribute the power to each factor individually. This property is known as the power of a product rule.

The other statements in the given options are not true for all real numbers a, b, m, and n. For example, \((a^n)^m\)is not equal to \(a^{(mn)}\) in general, and\(a^m + a^n\) is not equal to \(a^{(m+n)}\) in general.

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Transfer function of the system is:X(s)/Y(s) = 10/(s^2+1.6s+4)We need to find the following:

a) Y(t); b) Percent overshoot; c) Ultimate value of Y(t); d) Maximum value of Y(t) and e) Period of oscillation.

(a) Calculation of Y(t):The transfer function of the system isX(s)/Y(s) = 10/(s^2+1.6s+4)Now, applying the Laplace inverse on both sides,Y(s) = 10/(s^2+1.6s+4) × X(s)Taking the inverse Laplace of Y(s),y(t) = L^-1 {10/(s^2+1.6s+4) × X(s)}Using partial fraction decomposition to find the inverse Laplace of Y(s), we get:y(t) = 1.2508{ 2.22 e^(-0.8t) - 0.22 e^(-3.2t)}Therefore, the value of Y(t) is 1.2508{ 2.22 e^(-0.8t) - 0.22 e^(-3.2t)}.

(b) Calculation of percent overshoot:The transfer function of the system isX(s)/Y(s) = 10/(s^2+1.6s+4)The damping ratio (ζ) can be given asζ = 1/2 √(ζ²-4)ζ = 1/2 √(1.6²-4)ζ = 0.6For a second-order system with a damping ratio of 0.6, the percent overshoot is given as:%OS = e^(-ζπ/√(1-ζ²)) × 100%OS = e^(-0.6π/√(1-0.6²)) × 100OS = 26.12%Hence, the percent overshoot is 26.12%.

(c) Calculation of the ultimate value of Y(t):The transfer function of the system isX(s)/Y(s) = 10/(s^2+1.6s+4)For the ultimate value of Y(t), we take the limit of sY(s) as s tends to 0.The value of Y(s) is given as:Y(s) = 10/(s^2+1.6s+4) × X(s)On simplifying the above equation, we get:sY(s) + 1.6 Y(s) + 4 Y(s) = 10 X(s)Now, taking the limit of sY(s) as s approaches 0,sY(s) = lim s→0 sY(s) = 0Therefore, 1.6 Y(s) + 4 Y(s) = 10 X(s)Taking the limit of Y(s) as s approaches 0,0 = 10 X(0)Y(0) = 2.5Hence, the ultimate value of Y(t) is 2.5.

(d) Calculation of the maximum value of Y(t):The maximum value of Y(t) is given as:Ymax = 2.5 + (1+ζ²)^0.5 e^(-ζπ/√(1-ζ²)) / (ζ√(1-ζ²))The value of ζ is 0.6Hence, substituting the value of ζ in the above equation, we get:Ymax = 2.5 + (1+0.6²)^0.5 e^(-0.6π/√(1-0.6²)) / (0.6√(1-0.6²))Ymax = 3.129

(e) Calculation of the period of oscillation: The period of oscillation is given as:T = 2π / ωnWhere,ωn = √(1-ζ²) / (2ζ)Therefore,ωn = √(1-0.6²) / (2 × 0.6)ωn = 1.302Therefore,T = 2π / ωnT = 4.830sHence, the period of oscillation is 4.830s.

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The number of ways to place 4 distinguishable balls in 3 indistinguishable boxes can be found using the technique of "stars and bars" and is equal to 15.

We can use the stars and bars technique to represent each of the four identifiable balls as a star and the three indistinguishable boxes as two bars (|).

Using this strategy, we can consider the challenge as arranging four stars and two bars in a row. The row has a total of six spots, and we must choose two of them to place the bars. As a result, the binomial coefficient gives the number of ways to arrange the stars and bars:

C(6,2) = 6! / (2! * 4!) = 15

So there are 15 different ways to arrange four identifiable balls in three indistinguishable compartments.

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Step-by-step explanation:

For the cubic function to be an increasing function, dy/dx > 0 for all real values of x.

dy/dx

= d/dx [4x³ + 4px² + 16x - 9]

= 12x² + 8px + 16.

We have 12x² + 8px + 16 > 0.

=> 3x² + 2px + 4 > 0,

where a = 3, b = 2p and c = 4.

For the quadratic function to have a range that is always positive, the discriminant must be less than 0.

(in order to have no real roots)

We have b² - 4ac < 0.

=> (2p)² - 4(3)(4) < 0

=> 4p² - 48 < 0

=> p² - 12 < 0

=> p² < 12

=> -√12 < p < √12

=> -2√3 < p < 2√3.

That was a very long process but hopefully you understand all of it!

Answer:

a=11/12

Step-by-step explanation:

1/3+a=5/4

a=5/4 - 1/3

a=11/12

Answer:

11/12

Step-by-step explanation:

If we make both fractions have the same denominator, then it would be 4/12 + a = 15/12, so 15 - 4 = 11. So the answer would be 11/12.

Answer: 164.53%

Step-by-step explanation:

105.3/64=1.6453125

1.6453125*100=164.53125

164.53125=164.53%

Notice the answer is greater than 100 percent. This is because 105.3 is greater than 64, meaning that 105.3 is greater than 100% of 64, so 105.3 will have a greater percent value.

The result of the evaluation of the radical expression ³√-3 × ( 1 / 375 )^⅓ as required using the laws of indices is; -1 / 5.

It follows from the task content that the expression ³√-3 × ( 1 / 375 )^⅓ is to be evaluated for its result in simplest form.

Since the given expression is; ³√-3 × ( 1 / 375 )^⅓.

Therefore, it follows from the rational exponent law of indices that we have;

³√-3 × ³√(1 / 375)

= ³√ ( -3 × 1 / 375 )

= ³√ ( -3 / 375 )

= ³√ (-1 / 125)

= - 1 / 5.

On this note, the result of the radical expression as required is; -1 / 5.

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Answer: y = 2x + 1

Step-by-step explanation:  

Rewrite the equation in standard form:

x+2y=5

2y = -x + 5

y = -(1/2)x + (5/2)

A line perpendicular to this would have a slope that is the negative inverse of the reference line slope, -(1/2), here.  The new slope would be 2.  The y-intercept, (5/2), will change, so we'll just use b for the new y-intercept, and calculate it later:

y = 2x +b

We know that point (3,7) lies on the new line (it is a solution to the equation), so we can use this point to find the new y-intercept, b.

7 = 2*3 + b

b = 1

The perpendicular line is y = 2x + 1, and goes through point (3,7)

If you want to create a new variable, called inter_age_pollution, that is equal to the product of age and pollution in Stata, the command you would use is gen inter_age_pollution = age * pollution.

Stata is an incredibly versatile and powerful software program that is widely used by researchers in many fields, including economics, political science, and epidemiology.

If you have a cross-sectional dataset that includes variables such as LC, age, and pollution, you can create a new variable called inter_age_pollution that is equal to the product of age and pollution by using the following Stata command: gen inter_age_pollution = age * pollution.

This command creates a new variable called inter_age_pollution and sets its value to the product of age and pollution. This variable is now included in the dataset and can be used in subsequent analyses or visualizations.

To ensure that the command worked as intended, you should use the command "browse" or "list" to display the dataset and check that the values in the inter_age_pollution variable are consistent with your expectations.

In conclusion, if you want to create a new variable, called inter_age_pollution, that is equal to the product of age and pollution in Stata, the command you would use is gen inter_age_pollution = age * pollution.

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Answer:

150°

Step-by-step explanation:

By inscribed angle theorem:

\(x = 2 \times 75 \degree \\ \\ x = 150 \degree\)