a three-quarter sector of a circle of radius 4 inches along with its interior is the 2-d net that forms the lateral surface area of a right circular cone by taping together along the two radii shown. what is the volume of the cone in cubic inches?

The limit is expressed as the integral from 0 to 4 of [(176x/3)² - 2(x/3)³]dx using the definition of a Riemann sum.

To express the given limit as an integral, we can use the definition of a Riemann sum

[tex]\lim_{n \to \infty}[/tex] ∑ i=1 to n  [(176i/n)^2 - 2((i/3)^³) * (4/n)] * (4/n)

This can be simplified as

[tex]\lim_{n \to \infty}[/tex] [(4/n) * ((176/3)² + (172/3)² + ... + (8/3)²) - (32/n) * ((1/3)³ + (2/3)³ + ... + (n/3)³)]

Taking the limit as n approaches infinity, this becomes:

∫₀⁴ [(176x/3)² - 2(x/3)³] dx

Therefore, the given limit can be expressed as the integral from 0 to 4 of [(176x/3)² - 2(x/3)³] dx.

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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? __

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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The tangent line to the curve at (1 ,- 1) is  found by using the point-slope formula: is, y + 1 = 0

Now, We can find the value of x and y values corresponding to t=1.

Hence, They are:

⇒ x= (1) = 1

And,  y = (1) - 2 = - 1

And, The slope of the tangent line to the graph is

⇒ dy/dx =   dy/dt / dx / dt = (2t - 2) / 1          

Thus, dy/dx (at t=1)

dy / dx = 0

3.  Now we have both a point (1, - 1) on the graph and the slope of the tangent line to the curve at that point: 0

Hence, The tangent line to the curve at (1 ,- 1) is  found by using the point-slope formula:

⇒ y - y₁ = m(x - x₁)

⇒ y - (-1) = 0 (x - 1),

or y + 1 = 0

Thus,  The tangent line to the curve at (1 ,- 1) is  found by using the point-slope formula: is, y + 1 = 0

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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 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 =.

Yes, it is appropriate to see if there is a significant effect due to either A or B. The P-value for B is P=0.0001, which rejects the null hypothesis that there is no effect due to B. (option d).

In this scenario, the question is whether there is a significant effect from the interaction of two variables, A and B, and whether there is a significant effect due to either A or B individually. The results show that the P-value for the interaction is 0.3958, which means we fail to reject the null hypothesis that there is no effect due to the interaction.

It is important to note that even though we reject the null hypothesis for B, we cannot conclude that there is a significant effect due to A, as we have not conducted a separate hypothesis test for A.

However, in scenario (d), we are told that it is appropriate to test for a significant effect due to either A or B, as we have significant evidence of an effect due to B. In this case, we can conduct a separate hypothesis test for A to determine whether there is a significant effect due to A as well.

Hence the correct option is (d).

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The missing lengths of each pair of similar figures are, respectively:

Case 10:

x = 50, y = 24 / 5, z = 400 / 3

Case 11:

x = 8√2, y = 32, z = 36

In this question we find two systems of similar figures, a pair of similar quadrilaterals and a pair of similar right triangles. Two figures are similar when both have congruent angles and each pair of corresponding sides are not congruent but proportional. Then, these systems can be described by proportion formulas:

Case 10

18 / 60 = y / 16 = 15 / x = 40 / z

Now we proceed to determine the missing lengths:

18 / 60 = y / 16

y = (18 / 60) × 16

y = 24 / 5

18 / 60 = 15 / x

x = (60 / 18) × 15

x = 50

18 / 60 = 40 / z

z = (60 / 18) × 40

z = 400 / 3

Case 11

z / 12 = 12 / 4

z = 12² / 4

z = 36

y + 4 = z

y = 36 - 4

y = 32

By Pythagorean theorem:

x = √(12² - 4²)

x = 8√2

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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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The Laplace transform of the solution. Y(s) = L {y(t)} = (24 / s⁵ + 5) / s and y(t) = 4t³ + t².

Consider the following initial value problem, defined for t ≥ 0: dy/dt = t³t, y(0) = 5.

a. Find the Laplace transform of the solution.                                                                                                                                                      

 Y(s) = L {y(t)} =b.

Obtain the solution y(t). y(t) =a.

Laplace transform of the solution:                                                                                                                                                                    First, let's solve the differential equation dy/dt = t³t.

We can rewrite it as dy/dt = t⁴.

Then, we'll take the Laplace transform of both sides.

Using the formula L{y'} = sY(s) - y(0),

we get:

sY(s) - y(0) = L{dy/dt}

= L{t⁴} = 4! / s⁵sY(s) - 5

= 24 / s⁵sY(s)

= 24 / s⁵ + 5Y(s)

= (24 / s⁵ + 5) / s

Therefore, the Laplace transform of the solution is Y(s) = (24 / s⁵ + 5) / s.

b. Solution: To obtain the solution y(t), we'll take the inverse Laplace transform of Y(s).

We can rewrite Y(s) as: Y(s) = (24 / s⁵ + 5) / s = 24 / s⁶ + 5s² / s⁶

By using the formula L⁻¹{1/sⁿ} = tⁿ⁻¹ / (n⁻¹)!, L⁻¹{sⁿ} = tⁿ⁺¹ / (n + 1)!, and L⁻¹{F(s)G(s)} = f(t) * g(t), we can find the inverse Laplace transform of Y(s).L⁻¹{Y(s)} = L⁻¹{24 / s⁶} + L⁻¹{5s² / s⁶}= 4t³ + t².

Therefore, the solution to the initial value problem is y(t) = 4t³ + t².

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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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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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The area inside one leaf of the rose is (9/8)π square units.

To find the area inside one leaf of the rose, we can use the formula for the area enclosed by a polar curve, which is:

[tex]A = (1/2) \int[\alpha ,\beta] r (\theta)^2 d\theta,[/tex]

where r(θ) is the equation of the curve in polar coordinates, and α and β are the angles that define the portion of the curve that we want to find the area of.

For the rose curve given by r = 6 sin(4θ), we can see that one leaf is traced out as θ varies from 0 to π/4.

So we can find the area inside one leaf by evaluating the integral:

[tex]A = (1/2) \int [0,\pi /4] (6 sin(4\theta ))^2 d\theta[/tex]

Simplifying the integrand, we get:

[tex]A = (1/2) \int [0,\pi /4] 36 sin^2(4\theta ) d\theta[/tex]

Using the identity [tex]sin^2(\theta ) = (1/2)(1 - cos(2\theta )),[/tex]we can rewrite this as:

A = (1/2) ∫[0,π/4] 18 - 18 cos(8θ) dθ

Integrating, we get:

A = [9θ - (9/8) sin(8θ)] [0,π/4]

A = [9(π/4) - (9/8) sin(2π)] - [0 - (9/8) sin(0)]

A = (9/8)π.

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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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Total area 59 square units.

To find the area under the split-domain function from x=-2 to x=3, we need to integrate each piece of the function separately over the given interval.

For x < 0, we have f(x) = 2x + 5. We can integrate this as follows:

∫(from -2 to 0) (2x + 5) dx = [x^2 + 5x] (from -2 to 0) = (0^2 + 5(0)) - (-2^2 + 5(-2)) = 4 + 10 = 14.

For x ≥ 0, we have[tex]f(x) = 5x^2[/tex] We can integrate this as follows:

∫(from 0 to 3) (5x^2) dx [tex]= [5/3 x^3][/tex] (from 0 to 3)

[tex]= 5/3 (3^3 - 0^3)[/tex]

= 45

Therefore, the total area under the function from x=-2 to x=3 is the sum of the areas of the two pieces

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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 estimated proportion (as a decimal) of the voting population that prefers Candidate A is 0.76, and the 95% confidence interval is (0.713, 0.807).

To estimate the proportion of the voting population (P) that prefers Candidate A based on the sample of 400 people, we can use the sample proportion:

[tex]p = 304/400 = 0.76[/tex]

We can construct a confidence interval for the true proportion P using the following formula:

[tex]p \± z\sqrt{(p(1-p)/n)[/tex]

where z* is the critical value from the standard normal distribution for a 95% confidence level [tex](z\times = 1.96)[/tex], sqrt is the square root function, and n is the sample size (n = 400).

Substituting the values, we get:

[tex]0.7 \± 1.96\sqrt{(0.76(1-0.76)/400)[/tex]

Simplifying this expression, we get:

[tex]0.76 \± 0.047[/tex]

The 95% confidence interval for the true proportion of the voting population that prefers Candidate A is:

(0.713, 0.807)

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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 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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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 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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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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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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Therefore, the cone's base radius is roughly 2.31 cm, and its vertical angle is roughly 70.8° (which was rounded to the closest degree).

A cone is a smooth-tapering, three-dimensional geometric object with a flat base and a pointed or rounded apex. A circle is used to create it by dragging it along an axis that is parallel to the circle's plane. The cone's apex is formed by the line's intersection with the circle, and the cone's base is formed by the circle.

a) The sidewall of the cone will be formed by the circle's sector. The following formula can be used to determine how long the sector's arc is:

Arc length is calculated as (angle/360) x 2r, where r is the circle's radius.

In this instance, the length of  radius is 7.2 cm, and the angle is 300°. So,

Arc length equals (300/360) x 2 x (7.2) = 15

The cone's lateral surface area will be equal to the length of the sector's arc, and its base will be a circle of radius r. We thus have:

cone's lateral surface area equals rl.

where l is the cone's slant height. Using the sector's angle and radius, we can calculate l as follows:

ℓ² = r² + h²

h = r cos(150°)

h = -3.6

ℓ² = r² + (-3.6)² ℓ = √(r² + 12.96)

Now we can relate the cone's lateral surface area to the sector's arc's length:

r2(r2 + 12.96) = 225 r4 + 12.96r2 - 225 = 0 rl = 15 r2(r2 + 12.96) = 15

This equation is quadratic in r2. The quadratic formula can be used to find the value of r2:

r² = (-12.96 ± √(12.96² + 4(225)))/2 r² = 5.313 or r² = 33.647

We have r = 5.313 2.31 cm since r should be positive.

Therefore, the cone's base radius is roughly 2.31 cm.

b) The following formula can be used to determine the cone's vertical angle:

tan(θ/2) = r/ℓ

where l is the slant height, r is its base radius, and is the cone's vertical angle. Since we already understand r and l, we may find :

tan(θ/2) = 2.31/√(2.31² + 12.96)

θ/2 = tan⁻¹(0.308) θ ≈ 35.4°

As a result, the cone's vertical angle is roughly 70.8° (with a to the nearest degree).

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(a)  The base radius of the cone formed is approximately 6.00 cm.

(b) The vertical angle of the cone is approximately 38° (to the nearest degree).

What is circle?

A circle is a geometric shape that consists of all points in a plane that are equidistant from a fixed point called the center.

a) To find the base radius of the cone, we need to first find the length of the arc that makes up the sector.

The circumference of the full circle is 2πr, where r is the radius. So the circumference of the circle with radius 7.2 cm is:

C = 2π(7.2) ≈ 45.19 cm

The sector we're interested in has an angle of 300°, which is 5/6 of the full circle. So the length of the arc that makes up the sector is:

(5/6)C = (5/6)(45.19) ≈ 37.66 cm

Now we can use this length as the circumference of the base of the cone, and the radius of the base of the cone (let's call it R) is what we're trying to find. The formula for the circumference of a circle is C = 2πR, so we can set these two equations equal to each other and solve for R:

2πR = 37.66

R = 37.66/(2π)

R ≈ 6.00 cm

So the base radius of the cone formed is approximately 6.00 cm.

b) To find the vertical angle of the cone, we can use the formula:

tan(θ) = (opposite/adjacent)

where θ is the vertical angle of the cone, opposite is half the diameter of the base of the cone (which is just the radius we found in part a), and adjacent is the height of the cone.

We already found that the radius of the base of the cone is approximately 6.00 cm, so the diameter is 12.00 cm, and the opposite is half of that, or 6.00 cm. We just need to find the height of the cone now.

To do this, we can use the fact that the sector we started with can be rolled up to form the lateral surface of the cone. The length of the lateral surface of a cone is given by:

L = πr√(r² + h²)

where r is the radius of the base of the cone and h is the height. We already found that the radius is approximately 6.00 cm, and we know that the length of the lateral surface is approximately 37.66 cm (which is the same as the length of the arc that made up the sector). So we can plug in these values and solve for h:

37.66 = π(6.00)√(6.00² + h²)

h ≈ 8.15 cm

Now we have all the information we need to find the vertical angle:

tan(θ) = (6.00/8.15)

θ ≈ 38°

So the vertical angle of the cone is approximately 38° (to the nearest degree).

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The probability that a random sample of 100 bulbs will have an average life of more than 1,870 hours is approximately 14.69%

To find the probability that a random sample of 100 bulbs will have an average life of more than 1,870 hours, we need to use the mean, standard deviation, and sample size provided.

First, we need to find the standard error (SE) for the sample:
SE = (standard deviation) / √(sample size) = 190 / √100 = 19 hours

Next, we need to find the z-score for the given average life (1,870 hours):
z = (sample mean - population mean) / SE = (1870 - 1850) / 19 ≈ 1.05

Finally, we use a z-table to find the probability associated with this z-score. For a z-score of 1.05, the probability is 0.8531, which represents the area to the left of the z-score. However, we need the probability to the right (more than 1,870 hours), so we subtract this value from 1:

Probability = 1 - 0.8531 ≈ 0.1469

Thus, the probability that a random sample of 100 bulbs will have an average life of more than 1,870 hours is approximately 14.69%.

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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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∫[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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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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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.