The Donaldson Game Co. orders boxes of collectable game cards for $60 per box. They plan to sell the boxes of cards for $100 per box in January. A new series of cards is expected at the end of February and the company plans to place boxes of the current cards on sale for $50 per box in February before the new series is released. Any unsold boxes after February are donated to a game convention. A review of historic sales of the cards at $100 per box yielded a normal distribution with a mean of 200 boxes and a standard deviation of 30. Sales at the discounted price of $50 in February are estimated to be a normal distribution of 15 boxes and a standard deviation of 5 boxesThe alternative order you can consider is listed in the following table:Alternatives 150 175 200 225 250 275Do not worry about calculating the standard deviation of the expected profit. Given the alternatives, which one has the highest expected profit? Solve using Excel-only formulas. Create 1,000 replications. Use a one-way data table to calculate the expected profit for each alternative order amounts.

Answer:

The answer is 11 to the nearest tenth

The approximate reciprocal of 0.72 is 1.3889.

The ✓1.7 depicted as a fraction is (√170)/10.

It should be noted that to calculate the reciprocal of 0.72, we simply divide 1 by 0.72:

1/0.72 = 1.388888888888889

Thus, the approximate reciprocal of 0.72 is 1.3889.

Also, to determine the fractional equivalent for √1.7, we may again rationalize the denominator through multiplying both numerator and denominator with the expression contained beneath the radical:

√1.7=√(17/10)=(√17)/(√10)

Multiplying every entity within the adjoined numerator and denominator with (√10) offers:

(√17)/(√10)*(√10)/(√10)= (√170)/10

Therefore, √1.7 depicted as a fraction is:

√1.7=(√170)/10

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The area of region d is 4.5π.

To find the area of region d between the circles of radius 4 and radius 5 centered at the origin in the second quadrant, we can use the following steps:

Find the area of the larger circle (radius 5) and subtract the area of the smaller circle (radius 4) to find the area of the annulus (ring-shaped region) between them:

Area of larger circle = π[tex](5)^2[/tex] = 25π

Area of smaller circle = π[tex](4)^2[/tex] = 16π

Area of annulus = (25π) - (16π) = 9π

Divide the annulus into two equal parts since we are only interested in the portion of the region in the second quadrant. This gives us:

Area of region d = 1/2 (9π) = 4.5π

Therefore, the area of region d is 4.5π.

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To find the volume of the solid, whose base is the region between the curve y=20 sin x.

We know that the base of the solid is the region between the curve y=20 sin x. We also know that the solid is bounded by the x-axis and the plane z=0.

Therefore, the height of the solid is the distance between the curve and the plane z=0. This distance is simply given by the function y=20 sin x.

To find the volume of the solid, we need to integrate the area of each cross-sectional slice of the solid as we move along the x-axis. The area of each slice is simply the area of the base times the height.

The area of the base is given by the integral of y=20 sin x over the region of interest. This integral is:

∫ y=20 sin x dx from x=0 to x=π

= -cos(x) * 20 from x=0 to x=π

= 40

Therefore, the area of the base is 40 square units.

The height of the solid is given by y=20 sin x. Therefore, the volume of each slice is:

dV = (area of base) * (height)

= 40 * (20 sin x) dx

Integrating this expression from x=0 to x=π, we get:

V = ∫ dV from x=0 to x=π

= ∫ 40 * (20 sin x) dx from x=0 to x=π

= 800 [cos(x)] from x=0 to x=π

= 1600

Therefore, the volume of the solid is 1600 cubic units.

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To find the volume of the solid, whose base is the region between the curve y=20 sin x.

We know that the base of the solid is the region between the curve y=20 sin x. We also know that the solid is bounded by the x-axis and the plane z=0.

Therefore, the height of the solid is the distance between the curve and the plane z=0. This distance is simply given by the function y=20 sin x.

To find the volume of the solid, we need to integrate the area of each cross-sectional slice of the solid as we move along the x-axis. The area of each slice is simply the area of the base times the height.

The area of the base is given by the integral of y=20 sin x over the region of interest. This integral is:

∫ y=20 sin x dx from x=0 to x=π

= -cos(x) * 20 from x=0 to x=π

= 40

Therefore, the area of the base is 40 square units.

The height of the solid is given by y=20 sin x. Therefore, the volume of each slice is:

dV = (area of base) * (height)

= 40 * (20 sin x) dx

Integrating this expression from x=0 to x=π, we get:

V = ∫ dV from x=0 to x=π

= ∫ 40 * (20 sin x) dx from x=0 to x=π

= 800 [cos(x)] from x=0 to x=π

= 1600

Therefore, the volume of the solid is 1600 cubic units.

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Answer:

can be written as this in interval notation

[tex]( - 4 . \infty ) [/tex]

Step-by-step explanation:

since x is greater than -4 it is always going to be positive infinity on the right with -4 on the left.

if it is less than -4 then it is always going to be negative infinity on the left with -4 on the right

You can write the interval notation for the given set as:
(-4, ∞)

To write the set {x | x > -4} in interval notation, follow these steps:

1. Identify the lower limit of the interval: In this case, the lower limit is -4.
2. Identify the upper limit of the interval: Since x > -4, there is no upper limit, so we'll use infinity (∞) as the upper limit.
3. Determine whether the lower and upper limits are included in the set: In this case, x is strictly greater than -4, so -4 is not included. Therefore, we use the parenthesis "(" for the lower limit.

Interval notation is a way to describe continuous sets of real numbers by the numbers that bound them. Intervals, when written, look somewhat like ordered pairs. However, they are not meant to denote a specific point. Rather, they are meant to be a shorthand way to write an inequality or system of inequalities.

Now, you can write the interval notation for the given set as:

(-4, ∞)

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The equation that shows the relationship between x, the number of months, and y, the amount still owed on the loan, is y = -400x + 1200. The correct option is C.

The graph shows that the initial amount borrowed is 1200 and the loan payments reduce the amount owed by 400 pesos per month.

The amount still owed on the loan decreases linearly over time, so we can use the point-slope form of the equation for a line to express the relationship between x (the number of months) and y (the amount still owed on the loan):

y - y₁ = m(x - x₁)

where y₁ is the y-coordinate of a point on the line (in this case, the initial amount borrowed, which is 1200), m is the slope of the line (the rate at which the amount owed decreases, which is -400), and x₁ is the x-coordinate of the same point on the line (in this case, the first month, which is 1).

Substituting the values we have, we get:

y - 1200 = -400(x - 1)

Simplifying:

y = -400x + 1600

Therefore, the equation that shows the relationship between x, the number of months, and y, the amount still owed on the loan, is y = -400x + 1200. The correct option is C.

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The device is:

P(X > 7) = 1 - P(X ≤ 7) = 1 - F(7) = 1 - (-(1/7^2) + 1) = 0.0204

a) To confirm that f(x) is a probability density function, we need to check that it satisfies two properties: non-negativity and total area under the curve equal to 1.

Non-negativity: f(x) is non-negative for all x in its domain (1, infinity).

Total area under the curve:

∫1∞ f(x) dx = ∫1∞ 2x^(-3) dx

= [-x^(-2)] from 1 to ∞

= [-(1/∞) - (-1/1)]

= 1

Since f(x) satisfies both properties, it is a probability density function.

b) The cumulative distribution function (CDF) is given by:

F(x) = P(X ≤ x) = ∫1x f(t) dt

For x ≤ 1, F(x) = 0, since the smallest possible value of X is 1.

For x > 1, we have:

F(x) = ∫1x f(t) dt = ∫1x 2t^(-3) dt

= [-t^(-2)] from 1 to x

= -(1/x^2) + 1

So the CDF for this distribution is:

F(x) = {0 for x ≤ 1

-(1/x^2) + 1 for x > 1}

c) To find the mean, we use the formula:

E(X) = ∫1∞ x f(x) dx

= ∫1∞ x(2x^(-3)) dx

= 2 ∫1∞ x^(-2) dx

= 2 [-x^(-1)] from 1 to ∞

= 2(1-0)

= 2

So the mean of the distribution is 2.

d) The probability that the size of a random particle will be less than 5 micrometers is:

P(X < 5) = F(5) = -(1/5^2) + 1 = 0.96

e) The proportion of particles that will be detected by the device is:

P(X > 7) = 1 - P(X ≤ 7) = 1 - F(7) = 1 - (-(1/7^2) + 1) = 0.0204

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The event of rolling a sum of the numbers uppermost is 6 is E = {(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)}. The correct answer is b.

The event of rolling a sum of the numbers uppermost is 6 can occur in different ways, for example, rolling a 1 on the first die and a 5 on the second, or rolling a 2 on the first die and a 4 on the second, and so on.

The sum of the numbers on the dice is 6 in each of these cases. The set of all possible outcomes of this experiment is the sample space S, which consists of all possible pairs of numbers on the dice, such as (1,1), (1,2), (1,3), ..., (6,5), (6,6).

The event E of rolling a sum of 6 is the set of all pairs of numbers on the dice that add up to 6, which is E = {(1,5), (2,4), (3,3), (4,2), (5,1), (6,0)}.

Option b is the only answer choice that includes all these pairs of numbers, so it is the correct answer.

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(a) The covariance of X and Y is 0, since X and Y are independent.

(b) The standard deviation of U is sqrt(2(3^2) + 4(-4^2)) = 2*sqrt(13).

(c) The expected value of V is 0, since E(V) = 6E(X)E(Y) + 3E(Y) = 0.

(a) Since X and Y are independent, the covariance between them is 0. The formula for covariance is Cov(X,Y) = E(XY) - E(X)E(Y). Since E(XY) = E(X)E(Y) when X and Y are independent, the covariance is 0.

(b) The formula for the standard deviation of U is SD(U) = sqrt(Var(3X) + Var(-4Y)). Since Var(aX) = a^2Var(X) for any constant a, we can calculate Var(3X) = 3^2Var(X) = 9(2) = 18 and Var(-4Y) = (-4)^2Var(Y) = 16(4) = 64. Thus, SD(U) = sqrt(18 + 64) = 2*sqrt(13).

(c) The expected value of V is E(V) = E(6XY + 3Y). Since X and Y are independent, we can calculate this as E(6XY) + E(3Y) = 6E(X)E(Y) + 3E(Y). Since E(X) = 3 and E(Y) = 0, we get E(V) = 6(3)(0) + 3(0) = 0. Therefore, the expected value of V is 0.

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44. The cumulative frequency graph from the histogram is option A

45. E. none of above

A cumulative frequency graph, also known as an ogive, is a type of graph used in statistics to represent the cumulative frequency distribution of a dataset.

The graph displays the running total of the frequency of each value in the dataset on the y-axis, while the x-axis shows the values in the dataset.

Given that x = 1/2, y = 2/3 and z = 3/4

To evaluate x + y + z we use addition of fraction as follows

1/2 + 2/3 + 3/4

we convert to have same base of 12

6/6 * 1/2 + 4/4 * 2/3 + 3/3 * 3/4

6/12 + 8/12 + 9/12

adding results to

23/12 OR 1 11/12

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Answer:

no answer

Step-by-step explanation:

Answer:

see below

Step-by-step explanation:

1) adjacent angles are 2 angles right next to each other and are labeled with 3 letters, not 2.  

Examples in the picture would include <ABE, <ABD

Vertical angles are angles opposite of each other, so 2 examples are ABE and DBC

2) adjacent angles: PQT and QTR

vertical angles: PQR and SQR

3) a) adjacent

b) neither

c) vertical

d) vertical

e) adjacent

f) neither

hope this helps!

The absolute maximum value of f(x) on the interval [1, 4] is 1.5, which occurs at x = 1, and the absolute minimum value is 0.176, which occurs at x = 4.

To find the absolute maximum and minimum values of the function f(x) = 3/(x^2 + 1) on the interval [1, 4], we need to first find the critical points and then evaluate the function at the endpoints of the interval.

Critical points occur where the derivative of the function is equal to 0 or is undefined.

First, find the derivative of f(x):
f'(x) = -6x / (x^2 + 1)^2

To find the absolute maximum and minimum values of the function f(x) = 3/(x^2 + 1) on the interval [1, 4], we need to first find the critical points and the endpoints of the interval.
f'(x) = -6x/(x^2 + 1)^2 = 0

Next, we evaluate the function at the endpoints of the interval:
f(1) = 3/(1^2 + 1) = 1.5
f(4) = 3/(4^2 + 1) = 0.176
Set f'(x) to 0 and solve for x:
-6x / (x^2 + 1)^2 = 0

Since the denominator can never be 0, the only way this equation can be true is if the numerator is 0:
-6x = 0
x = 0

However, x = 0 is not in the interval [1, 4], so there are no critical points in the interval.

Now, evaluate the function at the endpoints of the interval:
f(1) = 3/(1^2 + 1) = 3/2 = 1.5
f(4) = 3/(4^2 + 1) = 3/17 ≈ 0.2

Since there are no critical points in the interval, the absolute maximum and minimum values occur at the endpoints. Thus, the absolute maximum value is 1.5 at x = 1, and the absolute minimum value is approximately 0.2 at x = 4.

Answer: Absolute maximum = 1.5 at x = 1
Absolute minimum ≈ 0.2 at x = 4

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The volume of the solid enclosed by the hyperboloid [tex]\frac{9}{2\pi }[/tex]

The hyperboloid's equation must be expressed in terms of r,θ, z in order to use polar coordinates.

[tex]$-x^2 - y^2 + z^2 = 6$[/tex]

Since[tex]$x = r\cos\theta$ and $y = r\sin\theta$[/tex], we can substitute and get:

[tex]$-r^2\cos^2\theta - r^2\sin^2\theta + z^2 = 6$[/tex]

Simplifying, we get:

[tex]$r^2 = \frac{6}{1-z^2}$[/tex]

Now, we need to find the limits of integration for r,θ and z. We know that the plane z = 3 intersects the hyperboloid when:

[tex]$-x^2 - y^2 + 3^2 = 6$[/tex]

Simplifying, we get:

$x^2 + y^2 = 3$

This is the equation of a circle centered at the origin with radius [tex]$\sqrt{3}$[/tex]. Since we're using polar coordinates, we can express this as:

[tex]$r = \sqrt{3}$[/tex]

For [tex]$\theta$[/tex], we can use the full range[tex]$0\leq \theta \leq 2\pi$[/tex]. For z, we have[tex]$0\leq z \leq 3$.[/tex]

Now, we can set up the triple integral to find the volume:

[tex]$V = \iiint dV = \int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\int_{0}^{3} r,dz,dr,d\theta$[/tex]

Solving the integral, we get:

[tex]$V = \int_{0}^{2\pi}\int_{0}^{\sqrt{3}} 3r,dr,d\theta = 3\pi\int_{0}^{\sqrt{3}} r,dr = \frac{9}{2}\pi$[/tex]

Therefore, the volume of the solid enclosed by the hyperboloid [tex]$-x^2 - y^2 + z^2 = 6$[/tex]and the plane [tex]$z = 3$[/tex] is   [tex]\\\frac{9}{2\pi }[/tex]

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The size of the set |E ∩ B| is 2, and the size of the set |B| is 3.

There are six possible outcomes when two dice are thrown:
{(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1), (6,2), (6,3)}.

Out of these 18 outcomes, the following three satisfy the event E (both dice are odd): (1,3), (3,1), and (3,3).
The following outcomes satisfy event B (the blue die is odd): (1,1), (1,3), (2,1), (2,3), (3,1), and (3,3).

Therefore, the size of the set |E ∩ B| is 2 (the two outcomes that satisfy both events are (1,3) and (3,1)), and the size of the set |B| is 3 (three outcomes satisfy the event B).

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The absolute maximum value is 2 at t = -1, and the absolute minimum value is -12 at t = 6.

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The probability of exactly three bulbs being replaced before the end of March is approximately 0.0126 or 1.26%.

To solve this problem, we need to use the exponential distribution formula:
f(x) = (1/β) * e^(-x/β)
where β is the mean and x is the time period.
In this case, β = 6 months, and we need to find the probability of exactly three bulbs being replaced before the end of March, which is three months from New Year's Eve.
So, we need to find the probability of three bulbs being replaced within three months, which can be calculated as follows:
P(X = 3) = (1/6)^3 * e^(-3/6)
         = (1/216) * e^(-0.5)
         ≈ 0.011
Therefore, the probability that exactly three bulbs will be replaced before the end of March is approximately 0.011.
To answer this question, we will use the Poisson distribution since it deals with the number of events (in this case, lightbulb replacements) occurring within a fixed interval (the time until the end of March). The terms used in this answer include exponential lifetime, mean, Poisson distribution, and probability.
The mean lifetime of the lightbulb is 6 months, so the rate parameter (λ) for the Poisson distribution is the number of events per fixed interval. In this case, the interval of interest is the time until the end of March, which is 3 months.
Since the mean lifetime of the bulb is 6 months, the average number of bulb replacements in 3 months would be (3/6) = 0.5.
Using the Poisson probability mass function, we can calculate the probability of exactly three bulbs being replaced (k = 3) in the 3-month period:
P(X=k) = (e^(-λ) * (λ^k)) / k!
P(X=3) = (e^(-0.5) * (0.5^3)) / 3!
P(X=3) = (0.6065 * 0.125) / 6
P(X=3) = 0.0126
So the probability of exactly three bulbs being replaced before the end of March is approximately 0.0126 or 1.26%.

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The area of the figure which consists of two trapezoids is calculated as: 186.0 cm².

The figure is composed of two trapezoids. Therefore, the area of the figure would be the sum of the areas of both trapezoids.

Area of trapezoid 1 = 1/2 * (a + b) * h

a = 20.0 cm

b = 12.0 cm

h = 6.0 cm

Area of trapezoid 1 = 1/2 * (20.0 + 12.0) * 6.0 = 96.0 cm²

Area of trapezoid 2 = 1/2 * (a + b) * h

a = 20.0 cm

b = 10.0 cm

h = 6.0 cm

Area of trapezoid 2 = 1/2 * (20.0 + 10.0) * 6.0 = 90.0 cm²

Area of the figure = 96.0 + 90.0 = 186.0 cm²

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The 95% confidence interval for the mean price of regular gasoline in that region is $3.396 to $3.584.The 90% confidence interval for the mean price of regular gasoline in that region is $3.413 to $3.567 3and 95% confidence interval for the mean price of regular gasoline in that region with a sample size of 80 would be $3.427 to $3.55

a) The 95% confidence interval for the mean price of regular gasoline in that region can be calculated as:

[tex]x ± z(\frac{s}{\sqrt{n} } )[/tex]

where X is the sample mean, s is the sample standard deviation, n is the sample size, and z is the critical value for the desired confidence level. For a 95% confidence level, z is 1.96.

Plugging in the given values, we get:

[tex]3.149 ± 1.96(\frac{0.29}{\sqrt{40} } )[/tex]

= 3.49 ± 0.094

So the 95% confidence interval for the mean price of regular gasoline in that region is $3.396 to $3.584.

b) Similarly, the 90% confidence interval for the mean can be calculated by using z = 1.645 (the critical value for a 90% confidence level):

3.49 ± 1.645(0.29/√40)

= 3.49 ± 0.077

So the 90% confidence interval for the mean price of regular gasoline in that region is $3.413 to $3.567.

c) If we had the same statistics from 80 stations, the standard error would decrease because the sample size is larger. The new standard error would be:

s/√80 = 0.29/√80 ≈ 0.032

Using the same formula as in part (a), but with the new standard error and z = 1.96, we get:

3.49 ± 1.96(0.032)

= 3.49 ± 0.063

So the 95% confidence interval for the mean price of regular gasoline in that region with a sample size of 80 would be $3.427 to $3.553.

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Summary for elements is: Sole proprietorship, partnership, limited liability company, corporation. Factors are: Insider trading regulations, company policies, market impact, personal financial situation.

Let's start by summarizing the required elements for various business entities described.

1. Sole Proprietorship:
Required elements: Single owner, personal liability for business debts, no legal separation between the owner and the business.
Example: A small bakery run by an individual owner.

2. Partnership:
Required elements: Two or more partners, shared profits and losses, personal liability for business debts.
Example: A law firm with multiple partners working together.

3. Limited Liability Company (LLC):
Required elements: Legal separation between owners and business, limited liability for business debts, flexible management structure.
Example: A consulting firm organized as an LLC.

4. Corporation:
Required elements: Legal separation between owners and business, limited liability for business debts, formal management structure with directors and officers, shares issued to represent ownership.
Example: A technology company with shareholders and a board of directors.

Similarities and differences: Sole proprietorships and partnerships have personal liability, while LLCs and corporations offer limited liability. LLCs and corporations also have legal separation between the owners and the business, unlike sole proprietorships and partnerships.

Now, let's discuss factors considered when a director of a company makes a large trade of the company's stock:

1. Insider trading regulations: Directors must comply with securities laws, avoiding trading based on non-public information.
2. Company policies: The director should follow any internal policies regarding stock trading, like blackout periods or approval requirements.
3. Market impact: The director should consider the potential impact of their trade on the company's stock price and market perception.
4. Personal financial situation: The director might consider their own financial goals, tax implications, and diversification needs.

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We have shown that |g(x) - g(y)| < 12ε whenever |x - y| < δ. Since ε was arbitrary, this shows that g(x) is uniformly continuous on (0, 1).

A stronger version of continuity known as uniform continuity ensures that functions defined on metric spaces, such as the real numbers, only vary by a small amount when their inputs change by a small amount. Contrary to uniform continuity, continuity merely demands that the function act "locally" around each point. To clarify, this means that for any given point x, there exists a tiny neighbourhood around x such that the function behaves properly inside that neighbourhood.

For the function g(x) to be continuous we need to have any ε > 0, and  δ > 0 such that if |x - y| < δ, then |g(x) - g(y)| < ε for all x, y in (0, 1).

Now, g(x) is bounded as the parent function f(x) is bounded.

Suppose, (0, 1) such that  |x - y| < δ.

Thus, without generality we have:

|g(x) - g(y)| = |x(1-x)f(x) - y(1-y)f(y)|

= |x(1-x)(f(x) - f(y)) + y(f(y) - f(x)) + xy(f(x) - f(y))|

≤ x(1-x)|f(x) - f(y)| + y|f(y) - f(x)| + xy|f(x) - f(y)|

< x(1-x)4ε + y4ε + xy4ε (by the choice of δ)

= 4ε(x(1-x) + y + xy)

< 4ε(x + y + xy)

≤ 4ε(1 + 1 + 1) = 12ε

Hence, we have shown that |g(x) - g(y)| < 12ε whenever |x - y| < δ. Since ε was arbitrary, this shows that g(x) is uniformly continuous on (0, 1).

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The solution to the equation [tex]\sqrt{3r^2} = 3[/tex] is given as follows:

[tex]r = \pm \sqrt{3}[/tex]

The equation in the context of this problem is defined as follows:

[tex]\sqrt{3r^2} = 3[/tex]

To solve the equation, we must isolate the variable r. The variable r is inside the square root, hence to isolate, we must obtain the square of each side, as follows:

3r² = 9.

Now we solve it as a quadratic equation as follows:

r² = 3.

[tex]r = \pm \sqrt{3}[/tex]

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To find f(pi), we need to use the fundamental theorem of calculus which states that if the integral of f(x) is F(x), then the derivative of F(x) with respect to x is f(x).

Given that the integral of f(x) is xsin2x, we can use this theorem to find f(x).Taking the derivative of xsin2x with respect to x gives: f(x) = d/dx (xsin2x), f(x) = sin2x + 2xcos2x, Now, to find f(pi), we simply substitute pi for x in the expression we just found: f(pi) = sin2(pi) + 2(pi)cos2(pi) , f(pi) = 0 + 2(pi)(-1) , f(pi) = -2pi .Therefore, f(pi) = -2pi. To find f(π) when the integral of f(x) is x*sin(2x),

we need to differentiate the given integral with respect to x. So, let's find the derivative of x*sin(2x) using the product rule: f(x) = d/dx(x*sin(2x)), f(x) = x * d/dx(sin(2x)) + sin(2x) * d/dx(x), f(x) = x * (cos(2x) * 2) + sin(2x) * 1, f(x) = 2x * cos(2x) + sin(2x), Now, to find f(π), we simply substitute x with π: f(π) = 2π * cos(2π) + sin(2π), Since cos(2π) = 1 and sin(2π) = 0, f(π) = 2π * 1 + 0 = 2π.

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The equation ý - Bo + BIx represents the sample regression function in the regression equation y = B0 + BIxI + ua.

What is the sample regression function?

It shows the relationship between the dependent variable y and the independent variable x, with B0 being the y-intercept and BIx being the slope of the regression line.

The explained sum of squares (SSE) measures the variability in y that is explained by the regression equation, while the total sum of squares (SST) measures the total variability in y.

The population regression function is the regression equation that applies to the entire population, while the sample regression function applies only to the sample data.

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Answer: 272/3 OR 90.67

Step-by-step explanation:

First, turn the mixed fraction into an improper fraction. Using the times-addition method, you take the whole number (2) and multiply it by the denomitor (3). You get 6, and then add the numerator (2) to 6, getting 8, so th improper fraction of the first term is 8/3.

Then, you multiply 8/3 by 34. To do this, you do 8 times 34 divided by 3. 34 times 8 is 272, and then you divide it by 3. You don't get a whole number, so the answer could be written as 272/3 or 90.67

Answer:

Let's assume that the cost of a box of candy is "x" dollars.

According to the problem, Karen bought 3 boxes of candy and 2 small bags of popcorn, and paid $18.35. So we can write the equation:

3x + 2y = 18.35

Similarly, Holly bought 4 boxes of candy and 3 small bags of popcorn, and paid $26.05. So we can write the equation:

4x + 3y = 26.05

We want to find the cost of a box of candy, so we can solve for "x" using these two equations. One way to do this is to use elimination. If we multiply the first equation by 3 and the second equation by -2, we can eliminate the "y" term:

9x + 6y = 55.05

-8x - 6y = -52.10

Adding these two equations gives:

x = 2.95

So the cost of a box of candy is $2.95.

Answer:

$2.95

Step-by-step explanation:

Let x be the cost of a box of candy while y be the cost of a small bag of popcorn.

Out of the given data, two equations is formulated.

Equation 1

Equation 2

Multiply 3 to both sides of Eq.1 to derive Eq.1'

Multiply 2 to both sides of Eq.2 to derive Eq.2'

Elimination using Eq.1' and Eq.2' to derive x

A box of candy costs $2.95

The directional derivative of f(x, y) = xy at point p(5, 5) in the direction from p to q(8, 1) is -1.

To find the directional derivative of f(x, y) = xy at point p(5, 5) in the direction from p to q(8, 1), we need to first find the unit vector in the direction from p to q.

This can be done by subtracting the coordinates of p from those of q to get the vector v = <3, -4> and then dividing it by its magnitude, which is sqrt(3^2 + (-4)^2) = 5. So, the unit vector in the direction from p to q is u = v/|v| = <3/5, -4/5>.

Next, we need to compute the gradient of f at point p, which is given by the partial derivatives of f with respect to x and y evaluated at p: grad(f)(5, 5) =  evaluated at (5, 5) = <5, 5>.

Finally, we can compute the directional derivative of f at point p in the direction of u as follows:

D_u f(5, 5) = grad(f)(5, 5) · u = <5, 5> · <3/5, -4/5> = (5)(3/5) + (5)(-4/5) = -1.

Therefore, the directional derivative of f(x, y) = xy at point p(5, 5) in the direction from p to q(8, 1) is -1.

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The general expression for the (100p)th percentile of the distribution is :

x_p = -ln(1 - p)/λ

The median of an exponential distribution with parameter λ is :

ln(2)/λ.

An exponential distribution is a continuous probability distribution that describes the time between events in a Poisson process, where events occur continuously and independently at a constant average rate.

The probability density function (PDF) of an exponential distribution with parameter λ is given by:

f(x) = λe^(-λx)

where x ≥ 0 and λ > 0.

To derive the (100p)th percentile of the distribution, we need to find the value x_p such that P(X ≤ x_p) = p, where p is a given percentile (e.g. p = 0.5 for the median). In other words, x_p is the value of X that separates the bottom p% of the distribution from the top (100-p)%.

To find x_p, we can use the cumulative distribution function (CDF) of the exponential distribution, which is given by:

F(x) = P(X ≤ x) = 1 - e^(-λx)

Using this formula, we can solve for x_p as follows:

1 - e^(-λx_p) = p
e^(-λx_p) = 1 - p
-λx_p = ln(1 - p)
x_p = -ln(1 - p)/λ

This is the general expression for the (100p)th percentile of the exponential distribution. To obtain the median, we set p = 0.5 and simplify:

x_median = -ln(1 - 0.5)/λ = ln(2)/λ

Therefore, the median of an exponential distribution with parameter λ is ln(2)/λ.

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