Evaluate: S√2 1 (u⁷/2 - 1/u⁵)du

Required value of angle R is 21°.

What is the relationship between the angle of circumference and angle of centre of circles?

Following is the relationship between angle of an arc on that circle and the angle of the center of a circle,

1)We can say that the angle of an arc is half the angle at the center that it deleted.

2) Or we can say oppositely the angle at the center of a circle is twice the angle of the arc that it subtends.

The measure of an arc is proportional to the measure of the angle it subtends and that the angle at the center of a circle is always twice the angle formed by any two points on the circumference of the circle that are connected to the center This relationship is based on the fact that.

Here, angle of circumference is angle R and angle of centre is angle P.

According to their relationship,

angle P = 2 × angle R

15x+3 = 2 × ( 10x-3)

We are Simplifying,

15x+3 = 20x-6

20x-15x = 3+6

5x = 9

x = 9/5

So, required angle R = 10x + 3 = 10×(9/5)+3 = 21

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As a result, 16% of the time, the sample's mean weight will be higher than 613 grammes. We calculate the answer as 613 grammes by rounding to the closest gramme.

A mathematical equation is a formula that connects two claims and uses the equals symbol (=) to denote equivalence. An equation in algebra is a mathematical statement that establishes the equivalence of two mathematical expressions. For instance, in the equation 3x + 5 = 14, the equal sign places a space between the variables 3x + 5 and 14. The relationship between the two sentences that are written on each side of a letter may be understood using a mathematical formula. The symbol and the single variable are frequently the same. as in, 2x - 4 equals 2, for instance.

According to the Central Limit Theorem, a sample of 26 fruits will have a distribution of sample means that is likewise normal, with a mean of 598 grammes and a standard deviation of 34/sqrt(26) grammes.

The z-score for the 82nd percentile may be calculated using a conventional normal distribution table or calculator and is around 0.91.

We therefore have:

0.91 = (x - 598) / (34 / sqrt(26))

After finding x, we obtain:

x = 598 + 0.91 * (34 / sqrt(26)) ≈ 613

As a result, 16% of the time, the sample's mean weight will be higher than 613 grammes. We calculate the answer as 613 grammes by rounding to the closest gramme.

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A survey was recently conducted in which 650 BU students were asked about their web browser preferences. Of all the students asked, 461 said that they sometimes use Firefox. 72 of the students said that they never use Chrome or Firefox. 12 of the Firefox users stated that they never use Chrome. Given that a person uses Chrome, 20.67% is the probability that the person never uses Firefox.

To find the probability that a person never uses Firefox given they use Chrome, we first need to determine the number of students who use Chrome and then find the number of those students who never use Firefox. Let's follow these steps:
1. Find the total number of students who never use Chrome or Firefox: 72 students
2. Find the total number of students who use Firefox: 461 students
3. Of these Firefox users, find the number who never use Chrome: 12 students
4. Since there are 650 students in total, find the number of students who use Chrome: 650 - 72 (students who never use either browser) - 12 (Firefox users who never use Chrome) = 566 students
5. Subtract the number of Firefox users from the total number of students to find the number of students who never use Firefox: 650 - 461 = 189 students
6. Now, find the number of Chrome users who never use Firefox: 189 - 72 (students who never use either browser) = 117 students
7. Finally, calculate the probability of a Chrome user never using Firefox: divide the number of Chrome users who never use Firefox (117) by the total number of Chrome users (566): 117 / 566 ≈ 0.2067
Therefore, the probability that a person never uses Firefox given that they use Chrome is approximately 0.2067 or 20.67%.

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The area of the trapezoid is 3.2 m².

A trapezoid is a quadrilateral  that has exactly one pair of parallel sides. Therefore, let's find the area of the trapezoid as follows:

area of the trapezoid = 1 / 2 (a + b)h

where

Therefore,

a = 1.2 m

b = 2.8 m

sin 45 = h / 2.3

cross multiply

h = 2.3 sin 45

h = 2.3 × 0.70710678118

h = 1.62634559673

h = 1.6

Therefore,

area of the trapezoid = 1 / 2 (a + b)h

area of the trapezoid = 1 / 2 (1.2 + 2.8)1.6

area of the trapezoid = 1 / 2 (4.0)1.6

area of the trapezoid = 3.2 m²

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B = 5/2, and the particular solution is:

[tex]y_p = (5/2)x^2e^{(-3x)[/tex]

C = 15/8, and the particular solution is:

[tex]y_p = (15/8)x^2e^{(4x)[/tex]

E = 1, and the particular solution is:

[tex]y_p = e^{(2x)[/tex]

The particular solution is [tex]y = -3/4 + Ce^{(-3/2)x,[/tex] where C is a constant determined by the initial conditions.

To solve for the particular solution of [tex](D+3)'y=15x^2e^{(-3x)[/tex], we first need to find the homogeneous solution by solving[tex](D+3)y_h = 0:[/tex]

[tex](D+3)y_h = 0[/tex]

[tex]y_h = Ae^{(-3x)[/tex]

The exponential shift method by assuming a particular solution of the form[tex]y_p = Bx^2e^{(-3x)[/tex]:

[tex](D+3)(Bx^2e^{(-3x)}) = 15x^2e^{(-3x)[/tex]

[tex](6B-15)x^2e^{(-3x)} = 15x^2e^{(-3x)[/tex]

B = 5/2, and the particular solution is:

[tex]y_p = (5/2)x^2e^{(-3x)[/tex]

So, the general solution is:

[tex]y = y_h + y_p = Ae^{(-3x)} + (5/2)x^2e^{(-3x)[/tex]

To solve for the particular solution of[tex](D-4)'y=15x^2e^{(4x)[/tex], we first need to find the homogeneous solution by solving[tex](D-4)y_h = 0:[/tex]

[tex](D-4)y_h = 0[/tex]

[tex]y_h = Be^{(4x)[/tex]

Now, we use the exponential shift method by assuming a particular solution of the form[tex]y_p = Cx^2e^{(4x)[/tex]:

[tex](D-4)(Cx^2e^{(4x)}) = 15x^2e^{(4x)[/tex]

[tex](8C-15)x^2e^{(4x)} = 15x^2e^{(4x)[/tex]

C = 15/8, and the particular solution is:

[tex]y_p = (15/8)x^2e^{(4x)[/tex]

So, the general solution is:

[tex]y = y_h + y_p = Be^{(4x)} + (15/8)x^2e^{(4x)[/tex]

To solve for the particular solution of[tex](D^2-2)^2y=16e^{(2x)[/tex], we first need to find the homogeneous solution by solving [tex](D^2-2)^2y_h = 0[/tex]:

[tex](D^2-2)^2y_h = 0[/tex]

[tex]y_h = (A+Bx)e^{\sqrt(2)x} + (C+Dx)e^{(-\sqrt(2)x)[/tex]

Now, we use the exponential shift method by assuming a particular solution of the form [tex]y_p = Ee^{(2x):[/tex]

[tex](D^2-2)^2(Ee^{(2x)}) = 16e^{(2x)[/tex]

[tex]16Ee^{(2x)} = 16e^{(2x)[/tex]

E = 1, and the particular solution is:

[tex]y_p = e^{(2x)[/tex]

So, the general solution is:

[tex]y = y_h + y_p = (A+Bx)e^{\sqrt(2)x} + (C+Dx)e^{(-\sqrt(2)x)} + e^{(2x)}[/tex]

To solve for the particular solution of[tex](D'D+3)y=9e^{(-3x),[/tex] we first need to find the homogeneous solution by solving [tex](D'D+3)y_h = 0:[/tex]

[tex](D'D+3)y_h = 0[/tex]

[tex]y_h = Acos(\sqrt(3)x) + Bsin(\sqrt(3)x)[/tex]

Now, we use the exponential shift method by assuming a particular solution of the form [tex]y_p = Ce^{(-3x)[/tex]:

[tex](D'D+3)(Ce^{(-3x)}) = 9e^{(-3x)[/tex]

[tex]0 = 9e^{(-3x)[/tex]

This is a contradiction, so we need to modify our assumption

[tex]Du + 3/2u = 9/2e^{(-3x)[/tex]

This is a first-order linear differential equation with integrating factor [tex]e^{(3/2)x[/tex]:

[tex]e^{(3/2)x}(Du + 3/2u) = e^{(3/2)x}(9/2)e^{(-3x)[/tex]

Integrating both sides:

[tex]e^{(3/2)x}u = -3/4e^{(-3x)} + C[/tex]

Multiplying both sides by [tex]e^{(-3/2)x[/tex]:

[tex]y = u = -3/4 + Ce^{(-3/2)x[/tex]

The particular solution is [tex]y = -3/4 + Ce^{(-3/2)x,[/tex]where C is a constant determined by the initial conditions.

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Since the calculated z-score (6.7) is greater than the critical z-score (2.33), the null hypothesis can be rejected, and it can be concluded that students who take the course score significantly higher than the general population.

The formula for Cohen’s d is (562-500)/100 = 0.62.

The effect size, estimated using Cohen’s d, was moderate (d = 0.62)."

a. To determine whether students who take the course score significantly higher than the general population, a one-tailed hypothesis test can be used with α = .01. The null hypothesis is that there is no difference between the sample mean and the population mean, and the alternative hypothesis is that the sample mean is significantly higher than the population mean. Using the formula for a z-score, the calculated z-score is (562-500)/(100/sqrt(20)) = 6.7. The critical z-score at α = .01 for a one-tailed test is 2.33.

b. Cohen’s d can be used to estimate the size of the effect of taking the course on SAT scores. Cohen’s d is calculated by taking the difference between the sample mean and the population mean, and dividing it by the standard deviation of the population. In this case, the formula for Cohen’s d is (562-500)/100 = 0.62. This indicates a medium effect size according to Cohen's guidelines.

c. In a research report, the results of the hypothesis test and the measure of effect size would be reported. For example, "The results of the one-tailed hypothesis test showed that students who took the course scored significantly higher on the SAT than the general population (z = 6.7, p < .01).

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We get that 636 voters should be sampled for a 90% confidence interval.

The Central Limit Theorem is a cornerstone of statistics, and it states that regardless of the population's underlying distribution, as sample size grows, the distribution of sample means of independent variables with similar distributions approaches a normal distribution. In other words, the Central Limit Theorem offers a method for approximating population behavior by examining sample behavior.

This theorem is significant because it enables statisticians to draw conclusions about a population from a sample, despite the fact that the population's distribution may be obscure or intricate.

The sample size for the distribution is given by:

[tex]n = (z* / E)^2 * p * (1 - p)[/tex]

Now, substituting the values we have:

[tex]n = (1.645 / 0.0334)^2 * 0.5 * (1 - 0.5) = 635.84[/tex]

Hence, We get that 636 voters should be sampled for a 90% confidence interval.

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If Caleb take 5/6 of hour and Seth took 9/10 of Caleb's time , then we can say that Caleb spent more time to complete the math homework.

The time taken by Caleb to complete his math's homework is = (5/6) of hour,

So, the Caleb's-time will be = (5/6)×60 = 50 minutes,

We also know that, Seth took (9/10) of Caleb's time, which means:

⇒ Seth's-Time is = (9/10) × 50 = 45 minutes.

On observing time taken by both Caleb and Seth,

We see that Caleb's time is greater than Seth's Time ,

Therefore, Caleb take more-time to complete the home work.

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The probability of the sample mean being between 21 and 22 inches is approximately 47.72%.

To find the probability of a sample mean between 21 and 22 inches, we'll use the z-score formula for sample means. The z-score is calculated as:

Z = (X - μ) / (σ / √n)

where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

First, we'll find the z-scores for the sample means of 21 and 22 inches:

Z₁ = (21 - 21) / (4.5 / √64) = 0
Z₂ = (22 - 21) / (4.5 / √64) ≈ 2.01

Now, we'll find the probability between these z-scores using a standard normal distribution table. The probability corresponding to Z₁=0 is 0.5, and for Z₂≈2.01, it's approximately 0.9772.

So, the probability of the sample mean being between 21 and 22 inches is:

P(21 ≤ X ≤ 22) = P(Z₂) - P(Z₁) ≈ 0.9772 - 0.5 = 0.4772

Therefore, the probability of the sample mean being between 21 and 22 inches is approximately 47.72%.

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The x-coordinate of the point of inflection is 9/4.

The interval on the left of the inflection point is 9/4 and on the function is concave down at (-∞, 9/4).

The interval on the right of the inflection point is 9/4 and on the function is concave up at (9/4, ∞).

In the given question we have to determine the intervals on which the given function is concave up or down and find the point of inflection.

The given function is:

f(x) = x(x−4√x)

Firstly finding the first and second derivatives.

f(x) = x^2−4x^{3/2}

f'(x) = 2x−4*3/2*x^{1/2}

f'(x) = 2x−6x^{1/2}

f''(x) = 2−6*(1/2)*x^{−1/2}

f''(x) = 2−3x^{−1/2}

Now finding the inflection point by equating the second derivative equal to zero.

f''(x) = 0

2−3x^{−1/2} = 0

After solving

x = 9/4

For left half of the number line of 9/4, f''(x)<0. So, the function is concave down in (-∞, 9/4).

For left right of the number line of 9/4, f''(x)>0. So, the function is concave up in (9/4, ∞).

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complete question:

Determine the intervals on which the given function is concave up or down and find the point of inflection. Let f(x) = x(x−4√x)

The x-coordinate of the point of inflection is: ____

The interval on the left of the inflection point is: ____ , and on this interval f is: __ concave up? or down? __

The interval on the right is: ____ , and on this interval f is: __ concave up? or down? __

Answer:

68

Step-by-step explanation:

54 + c2

substitute c = 7 in the equation

54 +(7)2

54+14

=68

The x-intercepts are (0, 0), (1, 0) and (-1, 0), the y-intercept is (0, 0), the stationary points are (1, 0), (-1, 0), (1/√5, 16/25√5) and (-1/√5, -16/25√5) and the inflection points are (0, 0), (√(3/5), 4/25√(3/5)) and (-√(3/5), -4/25√(3/5)).

Given that a function p(x) = x (x² - 1)², we need to find the coordinates of the intercepts, stationary points, and the inflection point,

x-intercept =

0 = x (x² - 1)²

x = 0,

x² - 1 = 0

x = ± 1

Thus, the x-intercepts are (0, 0), (1, 0) and (-1, 0)

y-intercept =

y = x (x² - 1)²

y = 0(0-1)²

y = 0

Thus, y-intercept is (0, 0)

Differentiate the function,

dy/dx = d/dx[x (x² - 1)²]

= (x² - 1)² + 4x²(x²-1)]

= (x²-1)(5x²-1)

Put dy/dx = 0

(x²-1)(5x²-1) = 0

x²-1 = 0

x = ±1

5x²-1 = 0

x = ±1/√5,

When, x = ±1 then y = 0

When x = 1/√5, then,
y = 1/√5((1/√5)²-1)²

= 16/25√5

Similarly, for x = -1/√5,

y = -1/√5((1/√5)²-1)²

= -16/25√5

Thus, the stationary points are (1, 0), (-1, 0), (1/√5, 16/25√5) and (-1/√5, -16/25√5)

Now, differentiate the function y = (x²-1)(5x²-1)

d²y/dx² = (5x²-1)2x + (x²-1)10x

= 20x³ - 12x

Put d²y/dx² = 0,

20x³ - 12x = 0

4x(5x²-3) = 0

4x = 0, x = 0

5x²-3 = 0

x = ±√(3/5)

When x = 0, y = 0,

When x = √(3/5)

y = √(3/5)((√(3/5))²-1)²

= 4/25(√(3/5))

When x = -√(3/5)

y = -√(3/5)((√(3/5))²-1)²

Thus, the inflection points are (0, 0), (√(3/5), 4/25√(3/5)) and (-√(3/5), -4/25√(3/5)).

Hence the x-intercepts are (0, 0), (1, 0) and (-1, 0), the y-intercept is (0, 0), the stationary points are (1, 0), (-1, 0), (1/√5, 16/25√5) and (-1/√5, -16/25√5) and the inflection points are (0, 0), (√(3/5), 4/25√(3/5)) and (-√(3/5), -4/25√(3/5)).

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The distance a patient will walk on the ramp is approximately 26.84 meters.

We can use trigonometry to solve this problem.

The ramp has a 25° angle of inclination, which means it makes a 25° angle with the horizontal. The ramp has a height of 12 meters, which translates to a vertical height of 12 meters from the ramp's bottom to its top.

To determine the length of the ramp (the distance a patient would travel), we can use the tangent function:

tan(25°) = adjacent/opposite

Where

Putting this equation in a different way, we get:

adjacent = opposite/tan(25°)

We obtain the following by substituting the above values:

nearby = 12/tan(25°) = 26.84 meters

Therefore, the distance a patient will walk on the ramp is approximately 26.84 meters.

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The price elasticity of demand at x=15 is -1.5, indicating elastic demand. The revenue function is R(x) = x(700 - 3x), with elastic intervals (0, 116.67) and inelastic intervals (116.67, ∞).

To find the price elasticity of demand, we first need to compute the derivative of the demand function with respect to x (p'(x)). The demand function is p = 700 - 3x, so p'(x) = -3.

Now, we can compute the price elasticity of demand (E) using the formula: E = (p'(x) * x) / p(x). At x=15, we have p(15) = 700 - 3(15) = 655. Plugging the values into the formula, we get E = (-3 * 15) / 655 = -1.5.

Since E < -1, the demand is elastic at x=15.

For the revenue function, we use R(x) = x * p(x), which gives R(x) = x(700 - 3x). To find the intervals of elasticity and inelasticity, we set |E| = 1 and solve for x: |-3x/ (700-3x)| = 1. Solving for x, we find x ≈ 116.67. Hence, the intervals are elastic for (0, 116.67) and inelastic for (116.67, ∞).

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A numerical measure of linear association between two variables is the

b. correlation coefficient.

A correlation coefficient is a numerical measure of the strength and direction of the linear relationship between two variables.

It ranges from -1 to +1, where a value of -1 indicates a perfect negative correlation (as one variable increases, the other decreases), a value of +1 indicates a perfect positive correlation (as one variable increases, the other also increases), and a value of 0 indicates no linear correlation between the variables.

The correlation coefficient is an important tool in statistical analysis as it allows researchers to determine whether and how strongly two variables are related.

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the focused deterrence strategies employed in the city are working to reduce crime.

To create probabilities to show the focused deterrence worked in the city, we can calculate the conditional probabilities of crime reduction given the neighborhood received focused deterrence and crime reduction given the neighborhood did not receive focused deterrence.

Let A be the event that the neighborhood received focused deterrence, and B be the event that crime was reduced. Then, the probabilities we can calculate are:

P(B|A) = 75/90 = 0.8333

P(B|A') = 5/30 = 0.1667

Where A' is the complement event of A (the neighborhood did not receive focused deterrence).

These probabilities show that crime reduction is much more likely when the neighborhood received focused deterrence, indicating that the focused deterrence strategies employed in the city are working to reduce crime.

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The probability that a student scores higher than a Credit (C) in this large statistics course can be calculated by adding the probabilities of getting a Distinction (D) or High Distinction (HD): P(D) + P(HD) = 0.15 + 0.35 = 0.50

In the given distribution for the large statistics course, the probabilities for each grade category are as follows:

- Fail (Z): 0.15
- D: 0.05
- C: 0.20
- P: 0.30
- HD: 0.30

To find the probability that a student scores higher than a Credit (C), you need to add the probabilities of the categories above C, which are D and HD.

Probability (Score > C) = Probability (D) + Probability (HD) = 0.05 + 0.30 = 0.35

Therefore, the probability that a student scores higher than a Credit (C) is 0.35.

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The differential of the given function when y = f(x) and f(x) = 4x² + 8x is given by dy = (8x + 8)dx.

Function is equal to,

y = f(x)

And f(x) = 4x² + 8x

The differential of the function y = f(x) = 4x² + 8x,

Take the derivative of y with respect to x, which is equal to,

dy/dx = 8x + 8

This gives us the rate of change of y with respect to x.

The differential of y by multiplying both sides by dx ,

dy = (8x + 8)dx

Therefore, the differential of the function y = 4x² + 8x is equal to

dy = (8x + 8)dx

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The above question is incomplete, the complete question is:

Let y = f(x), where f(x) = 4x² + 8x .Find the differential of the function dy =.

The total length of the curve swept out by the point (x, y) as θ ranges from 0 to 2π is 64π units.

We can start by finding the derivative of x and y with respect to θ:

dx/dθ = -16 sin θ

dy/dθ = 16 cos θ

Using the formula for arc length of a curve in polar coordinates, we have:

L = ∫(a to b) √(r² + (dr/dθ)²) dθ

where r is the distance from the origin to the point (x, y).

Substituting x and y into r, we get:

r = √(x² + y²) = √(256 [tex]cos^{2\theta[/tex] + 256 [tex]sin^{2\theta[/tex]) = 16

Substituting dx/dθ and dy/dθ into (dr/dθ), we get:

(dr/dθ) = √((-16 sin θ)² + (16 cos θ)²) = 16

Therefore, the total length of the curve swept out by the point (x, y) as θ ranges from 0 to 2π is:

L = [tex]\int\limits^{2 \pi}_0[/tex] √(r² + (dr/dθ)²) dθ

= [tex]\int\limits^{2 \pi}_0[/tex] √(256 + 256) dθ

= [tex]\int\limits^{2 \pi}_0[/tex] 32 dθ

= 32θ |o to (2π)

= 64π

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Answer:

Part A:

The comparison is "greater than".

Explanation: Since Richard has only a fraction (specifically, 34/1) of the number of blue marbles that Parker has, multiplying a whole number (Parker's marbles) by a fraction (34/1) results in a product that is less than the original whole number (Parker's marbles). Therefore, Parker must have more blue marbles than Richard.

Part B:

To determine how many blue marbles Richard has, we can multiply the number of blue marbles Parker has by the fraction representing the fraction of blue marbles Richard has compared to Parker, which is 34/1.

Richard has 12 (Parker's marbles) multiplied by 34/1 (the fraction representing the fraction of blue marbles Richard has compared to Parker):

Richard has 12 * 34/1 = 408 blue marbles.

Step-by-step explanation:

in the answer

The solution is x ≈ 0.305.

The equation you are referring to is:

5 = 31n x - In x

To solve for x, we need to use numerical methods since this equation

cannot be solved algebraically.

One common method is to use the Newton-Raphson method. Here are

the steps:

Choose an initial guess for x. Let's start with x=1.

Calculate the function and its derivative at the current guess:

f(x) = 31n x - In x - 5

f'(x) = 31n - 1/x

Use the formula x1 = x0 - f(x0)/f'(x0) to find a new guess for x:

x1 = x0 - (31n x0 - In x0 - 5)/(31n - 1/x0)

Repeat steps 2 and 3 with the new guess until the value of f(x) is very

close to zero.

In other words, keep iterating until |f(x)| < ε, where ε is a small positive

number that represents the desired accuracy.

After several iterations, we get the solution x ≈ 0.305. Rounded to the

nearest thousandths, the solution is x ≈ 0.305.

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a. The probability in a sample of 100 lipsticks that the proportion that are flawed is more than 0.054 is 0.3372.

b. Therefore, the probability in a sample of 100 lipsticks that the proportion that are flawed is less than 0.0542 is 0.6409.

c. The probability in a sample of 100 lipsticks that the proportion that are flawed is between 0.048 and 0.0522 is 0.3182.

To solve this problem, we can use the normal distribution since the sample size is large enough (n=100) and the proportion of flawed lipsticks (p=0.05) is not too small or too large.

(a) Let X be the number of flawed lipsticks in a sample of 100.

Then X follows a binomial distribution with n=100 and p=0.05.

We can use the normal approximation to the binomial distribution with mean np=5 and variance np(1-p)=4.75.

Let Z be the standard normal random variable, then

P(X > 0.054*100) = P(X > 5.4)

[tex]= P((X - 5)/\sqrt{(4.75) } > (5.4 - 5)/\sqrt{(4.75)} )[/tex]

= P(Z > 0.42)

= 1 - P(Z ≤ 0.42)

= 1 - 0.6628

= 0.3372.

(b) Using the same approach as in part (a), we have

P(X < 0.0542*100) = P(X < 5.42)

[tex]= P((X - 5)/\sqrt{(4.75)} < (5.42 - 5)/\sqrt{(4.75)} )[/tex]

= P(Z < 0.362)

= 0.6409.

(c) Using the same approach as in part (a), we have

P(0.048100 < X < 0.0522100) = P(4.8 < X < 5.22)

[tex]= P((4.8 - 5)/\sqrt{(4.75)} < (X - 5)/\sqrt{(4.75)} < (5.22 - 5)/\sqrt{(4.75)} )[/tex]

= P(-0.63 < Z < 0.21)

= P(Z < 0.21) - P(Z < -0.63)

= 0.5832 - 0.2650

= 0.3182.

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A z-test was conducted to determine if community college students who lived on their own took more time to earn an AA. The results showed that the sample of 36 students who lived on their own took an average of 3.8 years (SD = 0.453) to earn their AA, which was significantly longer than the population mean of 3.3 years, z = 2.58, p = 0.005 (one-tailed). Therefore, there is sufficient evidence to support the chancellor's theory that students who live on their own take longer to graduate.

For this scenario, the appropriate way to finish reporting the results in APA format depends on the type of hypothesis test used.
If a one-sample t-test was used to test the hypothesis, the appropriate way to finish reporting the results in APA format would be:
A one-sample t-test was conducted to determine if community college students who lived on their own took more time to earn an AA. The results showed that the sample of 36 students who lived on their own took an average of 3.8 years (SD = 0.453) to earn their AA, which was significantly longer than the population mean of 3.3 years,

t(35) = 3.26, p = 0.002 (one-tailed).

Therefore, there is sufficient evidence to support the chancellor's theory that students who live on their own take longer to graduate.
If a z-test was used to test the hypothesis, the appropriate way to finish reporting the results in APA format would be:
A z-test was conducted to determine if community college students who lived on their own took more time to earn an AA. The results showed that the sample of 36 students who lived on their own took an average of 3.8 years (SD = 0.453) to earn their AA,

which was significantly longer than

the population mean of 3.3 years,

z = 2.58, p = 0.005 (one-tailed).

Therefore, there is sufficient evidence to support the chancellor's theory that students who live on their own take longer to graduate.

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The steady state temperature distribution is:

u(x) = -4.5x + 18

Now, For determine the steady state temperature distribution u(x), we can start by assuming that the temperature of the rod is not changing with time, that is,

⇒ au/dt = 0.

This implies that the left-hand side of the partial differential equation simplifies to 0.

Hence, We can then rearrange the equation and integrate twice to obtain:

u(x) = C₁ x + C₂

where C₁ and C₂ are constants of integration.

Thus, To determine these constants, we can use the boundary conditions:

u(0,t) = 18

C₂ = 18

And, u(4,t) = 0

C₁ = (4) + 18 = 0,

C₁ = -4.5

Therefore, the steady state temperature distribution is:

u(x) = -4.5x + 18

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Answer:

The volume of the 2 and 1/4 box is 729/64 cubic inches.

Step-by-step explanation:

Volume of a Box

What is the volume of a 2 and 1/4 box

To calculate the volume of a box, you need to multiply its length, width, and height together. If the box has dimensions of 2 and 1/4, we need to convert the mixed number to an improper fraction.

2 and 1/4 = 9/4

Let's assume that the dimensions of the box are in inches. So, if the length of the box is 2 and 1/4 inches, its length in inches would be 9/4 inches.

Similarly, let's assume that the width and height of the box are also 2 and 1/4 inches, which means their measurements in inches would also be 9/4 inches.

Now we can calculate the volume of the box as:

Volume = Length x Width x Height

Volume = (9/4) x (9/4) x (9/4)

Volume = 729/64 cubic inches

It will take about 1.28 years (or 15 months) for the money to double in this account with continuous compounding.

After 10 years $57,000 grows to $96,500 in an account withcontinuous compounding. How long will it take money to double inthis account? Report your answer to the nearest month.

We can use the continuous compounding formula to solve this problem:

A = Pe^(rt)

where:

A = the amount after time t

P = the initial amount (principal)

r = the annual interest rate (as a decimal)

t = time (in years)

We are given that P = $57,000, A = $96,500, and the interest is compounded continuously. Therefore, we can solve for t:

ln(A/P) = rt

ln(96,500/57,000) = rt

0.54077 = rt

To find the time it takes for the money to double, we need to find the value of t when A = 2P (i.e., $114,000). Therefore, we can set up the following equation:

ln(2) = rt

ln(2) = 0.54077*t

t = ln(2)/0.54077

t ≈ 1.28 years

Therefore, it will take about 1.28 years (or 15 months) for the money to double in this account with continuous compounding.

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To find the indefinite integrals for the given functions. Here are the solutions:

(i) ∫(t^2 - t + 1)dt:
Integration is performed term-by-term:
∫t^2dt - ∫tdt + ∫1dt = (t^3/3) - (t^2/2) + t + C

(ii) ∫x*cos(x^2)dx:
Use u-substitution: u = x^2, so du/dx = 2x, or du = 2xdx
Now rewrite the integral: (1/2)∫cos(u)du = (1/2)*sin(u) + C
Substitute x^2 back in for u: (1/2)*sin(x^2) + C

(iii) ∫tan(3x)dx:
Use u-substitution: u = 3x, so du/dx = 3, or du = 3dx
Now rewrite the integral: (1/3)∫tan(u)du
The integral of tan(u) is ln|sec(u)|, so:
(1/3)*ln|sec(3x)| + C

(iv) ∫cos(t)dt/sin^2(t):
Use u-substitution: u = sin(t), so du/dx = cos(t), or du = cos(t)dt
Now rewrite the integral: ∫du/u^2 = -1/u + C
Substitute sin(t) back in for u: -1/sin(t) + C

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To find the value of K, we need to integrate the pdf over all possible values of X and set the result equal to 1, since the pdf must sum to 1 over the entire domain. Integrating fx(x) from negative infinity to positive infinity, we get: 1 = ∫ (-∞ to ∞) Ke^(-x^2/2) dx Using the standard integral for the Gaussian function, we can simplify this to: 1 = K√(2π)

Therefore, K = 1/√(2π). Now, to determine the probability of X being negative, we need to integrate the pdf over all negative values of X and divide by the total probability (which is 1):

P(X < 0) = ∫ (-∞ to 0) fx(x) dx / ∫ (-∞ to ∞) fx(x) dx

Substituting in our value for K, this becomes:

P(X < 0) = ∫ (-∞ to 0) (1/√(2π))e^(-x^2/2) dx / ∫ (-∞ to ∞) (1/√(2π))e^(-x^2/2) dx

We can simplify this using the definition of the Q-function, which is defined as:

Q(x) = 1/√(2π) ∫ (x to ∞) e^(-t^2/2) dt Therefore: P(X < 0) = Q(∞) - Q(0) Since Q(∞) = 0 (the tail of the Gaussian function goes to zero as x approaches infinity), this simplifies to: P(X < 0) = 1 - Q(0)

We can evaluate Q(0) using the standard integral for the Gaussian function: Q(0) = 1/√(2π) ∫ (0 to ∞) e^(-t^2/2) dt

Making the substitution u = t^2/2, du/dt = t/√2, we get: Q(0) = 2/√(π) ∫ (0 to ∞) e^(-u) du

This is just the standard integral for the exponential function, so: Q(0) = 2/√(π) Substituting this back into our expression for P(X < 0),

we get: P(X < 0) = 1 - 2/√(π) Simplifying, this becomes: P(X < 0) = √(π)/√(π) - 2/√(π) P(X < 0) = (√(π) - 2)/√(π)

Therefore, the probability of X being negative in terms of the Q-function is (√(π) - 2)/√(π).

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Answer:

Step-by-step explanation:

You want to find the true statements describing the graph of f(x) = (x -3)² -1.

The function is written in vertex form:

  f(x) = a(x -h)² +k . . . . . . . vertex (h, k)

Comparing to this form, we see the vertex is

  (h, k) = (3, -1)

The value of "a" is 1, so the graph opens upward. This means it is decreasing for x-values less than 3, It also means the range is upward from -1.

∫[tex]cos^6 x[/tex] dx Remember the formula for the cube of a sum [tex](A + B)^3 = A^3 + 3A^2 b + 3AB^2 + B^3.[/tex]

Using the formula for the cube of a sum, we have

[tex](cos x)^6 = (cos^2 x)^3 = (1 - sin^2 x)^3[/tex]

Expanding the cube, we get

[tex](1 - sin^2 x)^3 = 1 - 3 sin^2 x + 3 sin^4 x - sin^6 x[/tex]

Now, we can integrate each term separately we get

∫[tex](1 - 3 sin^2 x + 3 sin^4 x - sin^6 x) dx[/tex]

= ∫dx - 3∫[tex]sin^2 x[/tex] dx + 3∫[tex]sin^4 x[/tex] dx - ∫[tex]sin^6 x[/tex] dx

= x + 3/2 ∫(1 - cos(2x)) dx - 3/4 ∫[tex](1 - cos(2x))^2[/tex] dx - 1/6 ∫[tex](1 - cos(2x))^3[/tex] dx

Using the power-reducing formula for [tex]cos^2 x[/tex], we have

∫[tex]cos^2 x[/tex] dx = 1/2 ∫(1 + cos(2x)) dx = 1/2(x + 1/2 sin(2x)) + C

Using this formula, we can evaluate the integrals for [tex]sin^2 x[/tex] and [tex]sin^4 x[/tex] we get

∫[tex]sin^2 x[/tex] dx = 1/2 ∫(1 - cos(2x)) dx = 1/2(x - 1/2 sin(2x)) + C

∫[tex]sin^4 x[/tex] dx = ∫[tex]sin^2 x * sin^2 x[/tex] dx = (∫[tex]sin^2 x dx)^2[/tex] = [tex](1/2(x - 1/2 sin(2x)))^2[/tex] = [tex]1/4(x - 1/2 sin(2x))^2[/tex] + C

Similarly, using the power-reducing formula for [tex]cos^2 x[/tex], we have

∫[tex]cos^2 x[/tex]dx = 1/2 ∫(1 + cos(2x)) dx = 1/2(x + 1/2 sin(2x)) + C

Using this formula, we can evaluate the integrals for [tex](1 - cos(2x))^2[/tex]and[tex](1 - cos(2x))^3[/tex]we get

∫[tex](1 - cos(2x))^2 dx[/tex]  = ∫(1 - 2cos(2x) + [tex]cos^2(2x)[/tex]) dx

= x - 1/2 sin(2x) +[tex]1/4 sin^2(2x)[/tex]+ C

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

= x - 3/4 sin(2x) + 3/8 [tex]sin^2(2x)[/tex] - 1/16 [tex]sin^3(2x)[/tex] + C

Putting everything together, we get

∫[tex]cos^6 x[/tex] dx = x + 3/2(∫dx - ∫cos(2x) dx) - 3/4(∫dx - ∫cos(2x) dx + ∫(1 + cos(4x)) dx) - 1/6(∫dx - ∫cos(2x) dx + ∫(1 + cos(4x) + [tex]cos^2(6x)) dx)[/tex]

= x + 3/2x - 3/4x + 3/8 sin(2x) - 1/6x + 1/24 sin(4x)

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