Mathematics High School

## Answers

**Answer 1**

**4% is the answer!!!!!!!**

## Related Questions

9. Draw a net of a rectangular prism that has a

height of 2 units and bases that are 3 units long

and 1 unit wide.

### Answers

The image of a **net **of a **rectangular prism **is attached and the dimensions is as follows

**height **of 2 units,**bases **of 3 units long, and 1 unit **wide**

What is a net of a rectangular prism?

A rectangular prism is a 3 dimensional object having the dimensions as

Lengthwidth and depth

A **net **of a** rectangular** **prism **is rather a **2 dimensional** representation of a 3 dimensional rectangular prism that has been unfolded along its edges so that it can lay flat.

The rectangular prism's faces and their connections to one another are displayed in a flattened form of the prism. In order to depict the six faces of the rectangular prism, a net is often made up of several rectangles connected along their edges.

The image of the prism is attached and the color codes used for the edges are

black for 3 unitsred for 2 units, andgreen for 1 unit

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For the given cost and demand functions, find the production level that will maximize profit. (Round your answer to the nearest whole number.) C(q) 690 + 39 + 0.0392, p = 13 – q/600 = X units

### Answers

The **production level **that will maximize profit is approximately** 323 units.**

To find the production level that will maximize profit, you need to determine the quantity (q) where the **marginal cost **(MC) is equal to marginal revenue (MR). Given the cost function C(q) = 690 + 39q + 0.0392q^2 and the demand function p = 13 - q/600, let's proceed as follows:

1. Find the** revenue function** (R):

R(q) = p * q = (13 - q/600) * q

2. Simplify the revenue function:

R(q) = 13q - (q^2/600)

3. Calculate the **marginal revenue** (MR) by taking the derivative of the revenue function with respect to q:

MR(q) = dR/dq = 13 - (2q/600)

4. Calculate the marginal cost (MC) by taking the **derivative** of the cost function with respect to q:

MC(q) = dC/dq = 39 + 0.0784q

5. Set MR equal to MC and solve for q:

13 - (2q/600) = 39 + 0.0784q

6. Solve for q:

0.0806q = -26

q = -26 / 0.0806 ≈ 322.60

Since the problem asks for the nearest whole number, the production level that will maximize profit is approximately 323 units.

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Using the Pythagorean Theorem, prove Theorem 5.3.4 (Law of Cosine): Given a triangle ABC in a plane, let the length of the side opposite vertex A be denoted by a, that opposite vertex B be denoted by b, and that opposite vertex C be denoted by c. Then c2=a+62 – 2ab cos C where cos C refers to the cosine of |ZC|. Hint. Consider triangles AABC with three cases: (1) when ZC is an acute angle, (2) when ZC is a right angle, and (3) when ZC is an obtuse angle.

### Answers

The theorem uses the **Pythagorean Theorem** and some basic trigonometry so the Law of Cosines is c² = a² + b² - 2ab cos(C).

To prove Theorem 5.3.4, we begin by constructing a perpendicular line from **vertex** C to side AB, intersecting at point D. This creates two right triangles: ACD and BCD. Using the Pythagorean Theorem, we can express the lengths of the sides of these right triangles as:

AC² = a² - CD²

BC² = b² - CD²

Adding these two equations, we get:

AC² + BC² = a² + b² - 2CD²

Now, we need to find a way to express CD in terms of a, b, and C. To do this, we notice that **triangle** ZCD is similar to triangle ZAB (they share angle C), so we can write:

CD/AB = CZ/ZB

CD = AB × CZ/ZB

Using the law of sines, we can express CZ and ZB in terms of a, b, and C:

CZ = a sin(C)

ZB = b sin(C)

Substituting these into the **equation** for CD, we get:

CD = ab sin(C) / c

Substituting this expression for CD into our previous equation for AC² + BC², we get:

AC² + BC² = a² + b² - 2ab cos(C)

Multiplying both **sides** by c² gives us the desired result:

c² = a² + b² - 2ab cos(C)

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A triangular pyramid has a base shaped like an equilateral triangle. The legs of the equilateral triangle are all 7 centimeters long, and the height of the equilateral triangle is 6.1 centimeters. The pyramid's slant height is 9 centimeters. What is its surface area?

### Answers

The **surface area **of the **pyramid **is 2005. 06 centimeters square

How to determine the surface area

The **surface area**(S) of a **regular pyramid** is given by,

S = 1/2pl + B

Such that the parameters are p represents the perimeter of the base , l the slant height and B the base area of the pyramid.

From the information given, we have that;

Side of **equilateral triangle **= 7 centimeters

Slant height of the pyramid() = 9 centimeters

Then, for the perimeter of the triangle = 3(7) = 21 centimeters

Area of the equilateral triangle = √3/4 × 7² = 49√3/4 centimeters square

Now, substitute the values

Surface area = 1/2 × 21 × 9 × 49√3/4

Multiply the values

Surface area = 2005. 06 centimeters square

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Find the area of the shaded polygons

### Answers

The **area** of the **polygon** is 372 square units.

What's Trapezium?

A convex **quadrilateral** having exactly one set of opposite sides that are parallel to one another is called a trapezium. When drawn on a piece of paper, the trapezium is a two-dimensional shape that resembles a table. A quadrilateral is a polygon in **Euclidean geometry** that has four sides and four vertices. With four sides, four angles, and four vertices, a trapezium also has four of each.

There are several trapezium examples in everyday life. The trapezium rule, which divides the area under the curve into a number of trapeziums and then evaluates each trapezium's area, is a significant use of the trapezium.

Now the parallel sides in given questions are 7 and 24. And the perpendicular is 24.

so the formula to calculate the area of the shaded polygon is

A= 1/2(7 + 24)(24)

= 1/2(31)(24

= 31 x 12

= 372

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Find the coefficient of x^7y^4 in the binomial expansion of

(1/5x -25y)^11

### Answers

The coefficient of [tex]x^7y^4[/tex] in the** binomial expansion** of [tex](1/5x - 25y)^{11[/tex] is 3465.

We can use the binomial theorem to **expand** [tex](1/5x - 25y)^{11[/tex]:

[tex](1/5x - 25y)^{11[/tex] = ∑(k=0 to 11) [11 choose k] [tex](1/5x)^{(11-k)} (-25y)^k[/tex]

The** coefficient **of x^7y^4 will be the coefficient of [tex](1/5x)^{(11-7) } (-25y)^4[/tex], which is:

[11 choose 4] [tex](1/5)^4 (-25)^4[/tex]

=[tex](11!/(4!7!)) (1/5)^4 (-25)^4[/tex]

= [tex]3465 (1/5)^4 (25^4)[/tex]

= [tex]3465 (1/5)^4 (5^8)[/tex]

=[tex]3465 (1/5)^4 (5^2)^4[/tex]

= 3465 (1/625) (625)

= 3465

Therefore, the coefficient of [tex]x^7y^4[/tex] in the binomial expansion of [tex](1/5x - 25y)^{11[/tex]is 3465.

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Nayeli is a

75

7575 page manuscript. Filémon is a

40

4040 page manuscript.

Since proofreading is faster, Nayeli completes

11

1111 pages an hour, but Filémon only completes

4

44 pages an hour.

When Nayeli and Filémon have the same number of pages left, how many pages will that be

### Answers

**Answer:**

20 pages

**Step-by-step explanation:**

A clothing store had 30 jackets. If j represents the number of jackets the store then sold, which expression can be used to determine the total number of jackets that were NOT sold?

### Answers

**Expression** can be used to determine the total number of jackets that were NOT sold 30 - j.

**What in mathematics is a linear equation?**

An **algebraic equation **B. y=mx+b (where m is the slope and b is the y-intercept) containing simple constants and first-order (linear) components, such as the following, is called a **linear equation**. The above is sometimes called a "linear equation in two variables" where x and y are variables.

An equation with only one variable is called a univariate linear equation. It contains the expression Ax + B = 0. where A and B are any two real numbers and x is an ambiguous variable with only one possible value.

A clothing store had 30 jackets.

j represents the number of jackets

Then expression can be used to determine the total number of jackets that were NOT sold = 30 - j

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Cam someone help me do this? I don’t know what to divide and multiply

### Answers

**Answer:**

x = -35.375

**Step-by-step explanation:**

hope this helps! :)

Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point.

a) (9, ????/2 , 7)

(x, y, z) =

(b) (8, − ????/4 , −4)

(x, y, z) =

### Answers

a) The rectangular** **coordinates of the point with cylindrical coordinates (9, π/2, 7) are: **(0, 9, 7)**.

b) The rectangular coordinates of the point with cylindrical** **coordinates (8, -π/4, -4) are: **(4√2, -4√2, -4)**.

a) (9, π/2, 7)

To convert **cylindrical coordinates** (r, θ, z) to rectangular coordinates (x, y, z), use the formulas:

x = r*cos(θ)

y = r*sin(θ)

z = z

For point (9, π/2, 7):

x = 9*cos(π/2) = 9*0 = 0

y = 9*sin(π/2) = 9*1 = 9

z = 7

So, the rectangular coordinates of the point are (0, 9, 7).

b) (8, -π/4, -4)

For point (8, -π/4, -4):

x = 8*cos(-π/4) = 8*(1/√2) = 4√2

y = 8*sin(-π/4) = 8*(-1/√2) = -4√2

z = -4

So, the rectangular coordinates of the point are (4√2, -4√2, -4).

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If the second derivative of f is given by f"(x) = x(2 - x) (x + 3)^2, then the graph of f has inflection points at which x values? A) x = -3 only B) x = 0 only C) x = 2 only D) x = 0 and x = 2 E) x = -3 and x = 2

### Answers

**Inflection points **at are x = 0, x = 2, and x = -3.

**A more detailed explanation of the answer.**

An inflection point occurs when the **second derivative **changes sign, which means the second derivative is equal to zero. To find the inflection points for the given function, we need to find the x values when f"(x) = 0.

Given f"(x) = x(2 - x)(x + 3)², we set it to 0:

0 = x(2 - x)(x + 3)²

Now, we find the values of x that satisfy the equation:

1) x = 0

2) 2 - x = 0 => x = 2

3) (x + 3)² = 0 => x = -3

So, there are inflection points at x = 0, x = 2, and x = -3. Therefore, the correct answer is E) x = -3 and x = 2.

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I am an odd number. If

you multiply or divide a

number by me, it doesn't

change. What number am

I?

### Answers

**Answer:**

**Step-by-step explanation:**

If you divide something by you'll get 0 and if you multiply any number by 0 you'll get 0.

**Hope this Helps :)**

*Pls Brainliest...*

I think it’s 0 but I’m not sure

To find the distance from the edge of the

lake to the tree on the island in the lake, Hannah set up a triangular

configuration as shown in the diagram. The distance from location A to

location B is 85 meters. The measures of the angles at A and B are 51° and

83°, respectively. What is the distance from the edge of the lake at B to the

tree on the island at C?

### Answers

Therefore, the **distance** from the edge of the lake at B to the tree on the island at C is approximately 134.6 meters.

How far away is the edge?

In the manufacture of sheet metal aircraft, the phrase "**edge distance**" is used. It describes the separation between a rivet's centre and the edge of the substance it is holding together. Edge Separation.

We must determine the length of side BC in the provided **triangle** in order to determine the distance between the lake's edge at B and the tree on the island at C.

Let's start by identifying the triangle's edges and angles according to the diagram.

A is the **angle's** opposite half, or a.

We are looking for the side b that is opposing the angle at B.

C's side is the side across from the right corner.

A is at a 51° inclination.

B is at an inclination of 83°.

At C, the right angle is 90°.

The **length** of side b can be determined using the rule of sines. According to the rule of sines, a/sin A = b/sin B = c/sin C.

Plugging in the numbers yields the following result: a/sin 51° = b/sin 83°

By multiplying both parts by sin 83° and then dividing by sin 51°, we can find b:

b = (a x sin 83°) / sin 51°

We are informed that the **distance** between points A and B is 85 meters, so a must equal 85 metres. When we connect this, we get:

b = (85 x sin 83°) / sin 51°

Using a calculator, we get:

b ≈ 134.6 meters

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Question:

To find the distance from the edge of the

lake to the tree on the island in the lake, Hannah set up a triangular

configuration as shown in the diagram. The distance from location A to

location B is 85 meters. The measures of the angles at A and B are 51° and

83°, respectively. What is the distance from the edge of the lake at B to the

tree on the island at C?

find the best approximation to ~u = (3, −7, 2, 3) as a linear combination of ~v1 = (2, −1, −3, 1) and ~v2 = (1, 1, 0, −1). Namely, find the linear combination of V1 and V2 that is closest to u.

### Answers

The best approximation of vector u = (3,-7,2,3) as a **linear combination** of v₁=(2,-1,-3,1) and v₂=(1,1,0,-1) is -17/7 v₁ + 19/14 v₂.

To find the best approximation to u as a linear combination of v₁ and v₂, we need to find the coefficients x and y that minimize the distance between u and the linear combination x v₁ + y v₂. This **distance **is given by

d = ||u - (x v₁ + y v₂)||

where ||.|| denotes the Euclidean norm. Expanding this out gives

d^2 = ||u||^2 - 2x(u.v₁) - 2y(u.v₂) + x^2 ||v₁||^2 + y^2 ||v₂||^2 + 2xy (v₁.v₂)

where u.v₁ denotes the dot product of u and v1, and similarly for the other dot products.

To minimize d^2, we take the **partial derivatives** with respect to x and y and set them equal to zero

d^2/dx = -2(u.v₁) + 2x ||v₁||^2 + 2y (v₁.v₂) = 0

d^2/dy = -2(u.v₂) + 2y ||v₂||^2 + 2x (v₁.v₂) = 0

Solving these equations for x and y, we get

x = (u.v₁)(||v₂||^2) - (u.v₂)(v₁.v₂) / (||v₁||^2)(||v₂||^2) - (v₁.v₂)^2

y = (u.v₂)(||v₁||^2) - (u.v₁)(v₁.v₂) / (||v₁||^2)(||v₂||^2) - (v₁.v₂)^2

Plugging in the given values, we get

x = (32 + (-7)1 + 2(-3) + 31)(1^2) - (31 + (-7)1 + 20 + 3(-1))(21) / (2^21^2) - (1*1)^2

= -17/7

y = (31 + (-7)1 + 20 + 3(-1))(2^2) - (32 + (-7)1 + 2(-3) + 31)(12) / (1^22^2) - (1*2)^2

= 19/14

Therefore, the best approximation to u as a **linear combination** of v₁ and v₂ is

-17/7 v₁ + 19/14 v₂

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please show your steps

find the equation of the tangent line at the specified point.

if f(x) = ln(x), find the equation of the tangent line when x

= 4.

### Answers

The equation of the** tangent line** at **x = 4 is y = (1/4)x + ln(4) - 1.**

To find the **equation **of the tangent line at x = 4, we need to first find the slope of the tangent line. We can do this by taking the derivative of f(x) and evaluating it at x = 4:

f(x) = ln(x)

f'(x) = 1/x (by the power rule of **differentiation**)

f'(4) = 1/4

So the slope of the tangent line at x = 4 is 1/4.

Next, we need to find the y-**coordinate** of the point where the tangent line intersects the graph of f(x) at x = 4. We can do this by plugging x = 4 into the equation of f(x):

f(x) = ln(x)

f(4) = ln(4)

So the point where the tangent line **intersects** the graph of f(x) at x = 4 is (4, ln(4)).

Now we can use the **point-slope** form of the equation of a line to write the equation of the tangent line:

y - y1 = m(x - x1)

where m is the slope we just found (m = 1/4), and (x1, y1) is the point where the tangent line intersects the graph of f(x) (x1 = 4, y1 = ln(4)).

Substituting in these values, we get:

y - ln(4) = (1/4)(x - 4)

Simplifying this equation, we get:

y = (1/4)x + ln(4) - 1

So the equation of the tangent line at x = 4 is y = (1/4)x + ln(4) - 1.

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Maximize B = 4xy?, where x and y are positive numbers such that x + y2 = 10. The maximum value of B is (Simplify your answer. Type an exact answer, using radicals as needed.)

### Answers

The **maximum** value of **B ** is 37.33.

How to calculate the value of B?

We can begin by solving for one of the **variables** in terms of the other using the stated constraint. When we rearrange the constraint** equation**, we get:

[tex]y^2 = 10 - x[/tex]

Taking the square **root** of both sides (keeping in mind that y is positive) yields:

y = [tex]\sqrt{10-x}[/tex]

This expression for y can now be substituted into the objective function B:

B = 4xy = 4x[tex]\sqrt{10-x}[/tex]

We can maximize B by taking the derivative of B with respect to x and setting it to zero:

[tex]\frac{dB}{dx} = 4((10 - x) - \frac{x}{2(10 -x)} = 0[/tex]

When we multiply both sides by 2(10 - x), we get:

4[tex]\sqrt{10 - x}[/tex] - 2x = 0

When we simplify, we get:

2[tex]\sqrt{10 - x}[/tex]= x

Squaring both sides yields:

4(10 - x) = [tex]x^{2}[/tex]

We get a quadratic equation in x by expanding and rearranging:

[tex]x^2 + 4x - 40 = 0[/tex]

We can solve for x using the quadratic formula:

[tex]x = (-4 ± \sqrt{ (4^2 + 4(40)))/2 } = (-4 ± 2\sqrt{29} )/2 = -2 ±\sqrt{29}[/tex]

We use the solution x = -2 + 29 since x is positive.

We can now utilize our y in terms of the x expression to obtain the corresponding value of y:

y = √(10 - x) = √(10 - (-2 + √29)) = √(12 - √29)

Finally, we can plug these x and y values into the goal function B to get the maximum value:

B = 4xy = 4(-2 + √29)√(12 - √29) ≈ 37.33

As a result, the highest value of B is around 37.33.

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lim(x+y)^2/(x^2+y^2)(x,y)---(0,0). Does this limit exist? explain why.

### Answers

We cannot use polar coordinates to directly evaluate the limit. However, since the **limit** exists along the x-axis and y-axis, we can conclude that the limit exists and is equal to 1.

To determine if the limit exists, we can try to evaluate the limit along different paths approaching (0,0). If the limit is the same along all paths, then we can conclude that the limit **exists**.

Let's first consider the limit along the x-axis (y=0):

```

lim [tex](x+0)^2 / (x^2+0^2)[/tex] as x approaches 0

= lim [tex]x^2 / x^2[/tex] as x approaches 0

= 1

```

Now let's consider the limit along the y-axis (x=0):

```

lim [tex](0+y)^2 / (0^2+y^2)[/tex] as y approaches 0

= lim [tex]y^2 / y^2[/tex] as y approaches 0

= 1

```

Since the limit along both the x-axis and y-axis is 1, we can conclude that the limit exists and is equal to 1.

We can also verify this by using polar coordinates. Let x=r cos(theta) and y=r sin(theta). Then as (x,y) approaches (0,0), r approaches 0. Substituting x and y in the expression for the limit, we get:

```

lim [(r cos(theta) + r sin(theta))^2] / [(r cos(theta))^2 + (r sin(theta))^2] as r approaches 0

= lim [r^2 (cos^2(theta) + 2 cos(theta) sin(theta) + sin^2(theta))] / (r^2) as r approaches 0

= lim (cos^2(theta) + 2 cos(theta) sin(theta) + sin^2(theta)) as r approaches 0

= lim (cos^2(theta) + sin^2(theta) + cos(theta) sin(theta) + cos(theta) sin(theta)) as r approaches 0

= lim (1 + 2 cos(theta) sin(theta)) as r approaches 0

```

The limit above does not exist since the value of `2 cos(theta) sin(theta)` depends on the **direction** of approach (theta). Therefore, we cannot use polar **coordinates** to directly evaluate the limit. However, since the limit exists along the x-axis and y-axis, we can conclude that the limit exists and is equal to 1.

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is the function f(x)=x^2+3 even or odd?

### Answers

Since x23=x23 x 2 3 = x 2 3 , the function is

Let r = 2 sin and r = 2 sin(20). Find the area of the region inside the circle and outside the rose.

### Answers

The area of the region inside the circle and outside the rose is **approximately **9.42477796077 square units.

How to find the area of the region inside a circle and outside a rose curve represented by two equations?

The equation r = 2 sin(θ) represents a rose curve with two petals. The equation [tex]r = 2 sin(20 ^\circ)[/tex] represents a single point on the rose curve.

To find the area of the region inside the circle and outside the rose, we need to find the area of the circle and subtract the **area **of the region inside the rose.

The equation of the circle is r = 2. The **radius **of the circle is 2, so its area is[tex]A = \pi r ^2 = 4\pi.[/tex]

To find the area of the region inside the rose, we need to integrate the equation r = 2 sin(θ) over the interval [0, π]. This will give us the area of one petal, so we'll multiply the result by 2 to get the area of the entire rose.

[tex]\int [0,\pi] 2 sin(\theta) d\theta = [-2 cos(\theta)] [0,\pi] = 4[/tex]

Therefore, the area of one petal is 2, and the area of the entire rose is 4. So the area of the region inside the rose is 4.

Finally, the area of the region inside the circle and outside the rose is the area of the circle minus the area of the **region inside the rose**:

[tex]A = 4\pi - 4 \approx 9.42477796077.[/tex]

Therefore, the area of the region inside the circle and outside the rose is approximately 9.42477796077 square units.

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Let the random variable U=.32 be a realization of a Uniform on (0,1) distribution and the random variables (Z1...Z10) = (-3.03, .63, .37, -2.06, .52, -1.38, 1.16, -.64, .57, 1.31) be realizations of 10 independent N(0,1) random variables.

Express a single realization from each of the following distributions in terms of the given value of U and/or a subset of Z1...Z10 as you may not need all 11 values.

a) F-dist(df=8,2)

b) Exp(gamma=3.4)

c) N(23,81)

### Answers

a) F-dist(df=8,2): To obtain a realization from an F distribution with 8 and 2 **degrees **of freedom (df), you can use the following transformation:

F = (Z1^2/8 + Z2^2/8 + ... + Z8^2/8) / (Z9^2/2 + Z10^2/2)

Using the given values of Z1...Z10, we get:

F = ((-3.03)^2/8 + (.63)^2/8 + ... + (-.64)^2/8) / ((.57)^2/2 + (1.31)^2/2)

b) Exp(gamma=3.4):

To obtain a realization from an exponential distribution with gamma=3.4, you can use the following transformation:

Exp = -3.4 * log(1-U)

Using the given **value **of U = 0.32, we get:

Exp = -3.4 * log(1-0.32)

c) N(23,81):

To obtain a realization from a normal distribution with mean 23 and variance 81, you can use the following transformation:

N = 23 + sqrt(81) * Z1

Using the given value of Z1 = -3.03, we get:

N = 23 + sqrt(81) * (-3.03)

a) F-dist(df=8,2): Since U is a realization of a Uniform on (0,1) distribution, we can use the inverse transform method to obtain a realization from the F-distribution with degrees of freedom 8 and 2. Let F be the F-distributed random variable, then we have F = qf(U, 8, 2), where qf is the inverse cumulative distribution function (quantile function) of the F-distribution with degrees of freedom 8 and 2.

b) Exp(gamma=3.4): The exponential distribution with parameter gamma=3.4 has a cumulative distribution function (CDF) of F(x) = 1 - exp(-x/3.4). Using the inverse transform method, we can obtain a realization from this distribution as follows: let Y be the exponential random **variable**, then Y = -3.4 * ln(1 - U).

c) N(23,81): The normal distribution with mean 23 and standard deviation 9 has a CDF of F(x) = (1/2) * (1 + erf((x-23)/(9*sqrt(2)))), where erf is the error function. Using the inverse transform method, we can obtain a realization from this distribution as follows: let X be the normal random variable, then X = 23 + 9*sqrt(2)*erf^{-1}(2*U-1), where erf^{-1} is the inverse of the error **function**. We can approximate the inverse of the error function using a numerical algorithm such as Newton's method.

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The monthly demand function for a product sold by a monopoly is p = 1908-1x2 dollars, and the average cost is C = 900 + 8x + x2 dollars. Production is limited to 1000 units and x is in hundreds of units. (a) Find the quantity that will give maximum profit. units (b) Find the maximum profit. (Round your answer to the nearest cent.) Need Help? Read It Submit Answer Save Progress Practice Another Version

### Answers

(a) To find the quantity that will give maximum **profit**, we need to maximize the profit function. The profit function is given by:

Profit = Total Revenue - Total Cost

Profit = (Price per unit x Quantity) - (Average Cost per unit x Quantity)

Profit = (1908 - x^2) x - (900 + 8x + x^2) x

Profit = 1908x - x^3 - 900x - 8x^2 - x^3

Profit = -2x^3 - 8x^2 + 1008x

To find the maximum profit, we need to take the derivative of the profit function and set it equal to zero:

dProfit/dx = -6x^2 - 16x + 1008 = 0

Solving for x, we get:

x = (-(-16) ± sqrt((-16)^2 - 4(-6)(1008))) / 2(-6)

x = 14.37 or x = -23.63

Since x represents the quantity in hundreds of units, we can only use the positive value, which means the quantity that will give maximum profit is approximately 1437 units.

(b) To find the maximum profit, we need to substitute the value of x into the profit function:

Profit = -2(1437)^3 - 8(1437)^2 + 1008(1437)

Profit = $307,261.44

Therefore, the maximum profit is approximately $307,261.44.

To find the maximum profit, we need to first find the profit function, P(x), which is the difference between revenue, R(x), and cost, C(x). Since the demand function is given as p = 1908 - x^2, we can find the revenue function by multiplying the price with the quantity (x):

R(x) = p * x = (1908 - x^2) * x

The cost function is given as C(x) = 900 + 8x + x^2. Now, let's find the profit function P(x):

P(x) = R(x) - C(x) = (1908 - x^2) * x - (900 + 8x + x^2)

(a) To find the quantity that will give maximum profit, we need to find the critical points by taking the **derivative** of P(x) and setting it to zero:

P'(x) = d(P(x))/dx = 1908 - 6x^2 - 8x

Set P'(x) = 0:

0 = 1908 - 6x^2 - 8x

Solve for x, keeping in mind that x is in hundreds of units and production is limited to 1000 units (10 hundreds of units). The solution within this constraint is x ≈ 6.07 hundreds of units.

(b) To find the maximum profit, plug the value of x (6.07) into the profit function P(x):

P(6.07) ≈ (1908 - 6.07^2) * 6.07 - (900 + 8 * 6.07 + 6.07^2)

Round your answer to the nearest cent: P(6.07) ≈ $5,993.44

So, the maximum profit is approximately $5,993.44

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Find the missing numbers. 20 points

### Answers

1/9 / 3 = 1/9 * 1/3 = 1/9*3

if fis a differentiable function of rand g(x,y) = f(xy), show that

### Answers

g(x, y) = f(xy) is a **differentiable function.**

We'll show that g(x, y) = f(xy) is a differentiable function by finding its **partial derivatives **with respect to x and y.

Given that f is a differentiable function of r, we know that f'(r) exists for all values of r. Now, let's find the partial derivatives of g(x, y) with respect to x and y:

1. ∂g/∂x:

Using the **chain rule**, we get ∂g/∂x = (∂f/∂r) * (∂r/∂x). Since r = xy, ∂r/∂x = y. Thus, ∂g/∂x = f'(xy) * y.

2. ∂g/∂y:

Similarly, we get ∂g/∂y = (∂f/∂r) * (∂r/∂y). Since r = xy, ∂r/∂y = x. Thus, ∂g/∂y = f'(xy) * x.

Since both partial derivatives ∂g/∂x and ∂g/∂y exist, we can conclude that g(x, y) = f(xy) is a **differentiable function.**

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5 averaged at 14 and the sum of three was 48. What was the

sum of the other two?

### Answers

The **sum **of the other two numbers is 22.

Define average

The average, also known as the arithmetic **mean**, is a measure of **central tendency** that is calculated by adding up a set of values and then dividing the sum by the total number of values. In other words, the average is the value that is typical or representative of a set of **data**.

Let's assume the two numbers we need to find as x and y.

We know that the average of 5 numbers is 14, so the sum of all 5 numbers is:

5 × 14 = 70

We also know that the sum of three of these numbers is 48, so the sum of the other two numbers must be:

70 - 48 = 22

The other two digits add up to 22, therefore they are 22 in total.

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If 3 men and 3 women are to be randomly assigned to six seats in a row at a theater, find the probability that they will alternate by gender.

The probability is__

### Answers

To find the **probability** that the 3 men and 3 women will alternate by gender when **randomly **assigned to six seats in a row at a theater, follow these steps:

1. Calculate the total number of possible seating arrangements: Since there are 6 people, there are 6! (6 factorial) ways to arrange them. 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 possible arrangements.

2. Calculate the number of seating arrangements where they alternate by gender: Since there are 3 men and 3 women, there are two possible patterns (MWMWMW or WMWMWM). For each pattern, there are 3! ways to arrange the men and 3! ways to arrange the women. So, there are 2 × (3! × 3!) = 2 × (6 × 6) = 72 arrangements where they alternate by gender.

3. Calculate the probability: Divide the number of alternating gender arrangements (72) by the total number of possible arrangements (720). Probability = 72/720 = 1/10.

The** probability** that the 3 men and 3 women will alternate by gender when **randomly** assigned to six seats in a row at a theater is 1/10.

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Write each expression in the form 2kx or 3kx for a suitable constant k. 2x - 3x (a) () (b) 1) */2 81 (c) 9 16 2x (a) (H)* 81

### Answers

To write each **expression** in the form 2kx or 3kx, we have the following: (a) -x; (b) 2/9√(3)x; (c) 3/4√(2)x.

(a) The given expression is 2x - 3x. Combining like **terms**, we get -x, which is in the form 2kx with k = -1/2.

(b) The given expression is (1/2)(3/2)^4x. Simplifying, we get (1/2)81x, which is in the form 2kx with k = 2/9√(3).

(c) The given expression is (3/4)^2(2)x. Simplifying, we get 9/16x, which is in the **form** 3kx with k = 3/4√(2).

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find the 3rd degree taylor polynomial, centered at x = 0, for cos( x )

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The 3rd degree **Taylor polynomial **for cos(x) centered at x=0 is: P3(x) = 1 - x²/2 + x⁴/24

To find the Taylor polynomial for a function, we start with the function's **Taylor series** representation and truncate it after the desired degree. For cos(x), we know its Taylor series representation is:

[tex]cos(x) = \sum (-1)^n x^{(2n) / (2n)![/tex]

To find the 3rd degree** polynomial**, we only need to keep the terms up to x⁴ in the series:

cos(x) ≈ 1 - x²/2 + x⁴/24

This approximation will be more accurate closer to x=0, as **higher order terms **in the series become more significant farther from the center of the series.

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Consider the variable coefficient linear second order non-hom*ogeneous ODE x²+yⁿ – 2xy' + (x² +2)y = 3x³ for 1 > 0. 1. Write down the associated hom*ogeneous equation.

2. Show that y₁(x) = x sin x and y₂(x) = xcosx soulutions of the associated hom*ogeneous equation.

3. Use the Wronskian test to determine if y₁ and y₂ are linearly independent for x > 0.

### Answers

The associated **hom*ogeneous equation** is x^2 - 2xy' + (x^2 + 2)y = 0. y1(x) = xsin(x) and y2(x) = xcos(x) are solutions. The Wronskian test shows that they are linearly independent for x > 0.

The associated hom*ogeneous equation is x^2 - 2xy' + (x^2 + 2)y = 0. To show that y₁(x) = x sin x and y₂(x) = x cos x are solutions of the associated **hom*ogeneous equation**, we substitute them into the equation and simplify

For y₁(x) = x sin x:

x^2 - 2x(sin x + cos x) + (x^2 + 2)(sin x) = 0

This simplifies to:

x^2(sin x) - 2x(cos x) = 0

which is satisfied for all x. Therefore, y₁(x) = x sin x is a solution of the associated hom*ogeneous equation.

For y₂(x) = x cos x:

x^2 - 2x(-sin x + cos x) + (x^2 + 2)(cos x) = 0

This simplifies to:

x^2(cos x) + 2x(sin x) = 0

which is also satisfied for all x. Therefore, y₂(x) = x cos x is also a solution of the associated hom*ogeneous equation.

To determine if y₁ and y₂ are linearly independent for x > 0, we use the **Wronskian test**

W(y₁, y₂)(x) = det [y₁(x) y₂(x) ; y₁'(x) y₂'(x)]

= det [x sin x x cos x ; x cos x - sin x x sin x]

= x^2

Since the Wronskian is nonzero for all x > 0, y₁ and y₂ are linearly independent for x > 0.

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use differentials to approximate the value of the expression. compare your answer with that of a calculator. (round your answers to four decimal places.) 3√29

### Answers

Using a calculator, we find that 3√29 ≈ 3.0345 to four** decimal **places, which is the same as our approximation using differentials.

How 3√29 ≈ 3.0345 to four decimal places?

To use differentials to approximate the value of the expression 3√29, we can follow these steps:

Start with the equation for the differential of [tex]y = x^(1/3)[/tex], which is:

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

Substitute x = 29 into the **equation**, giving us:

[tex]dy = (1/3)(29)^(-2/3)dx[/tex]

Approximate dx as 0.001, which gives us:

dx = 0.001

Calculate dy using the equation we derived in step 2:

[tex]dy = (1/3)(29)^(-2/3)(0.001) ≈ 0.00176[/tex]

Approximate the **value **of 3√29 as:

[tex]3√29 ≈ 3√(29 - 0.001) + dy[/tex]

≈ 3√28.999 + 0.00176

≈ 3.0345

Using a calculator, we find that 3√29 ≈ 3.0345 to four decimal places, which is the same as our approximation using differentials.

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4/5 the students in a school are in sixth grade How many students are in the school if 60 of them are sixth graders?

### Answers

If 4/5 the **students** in a school are in sixth grade , there are 75 students in the school.

We can use a **proportion** to solve this problem. Since 4/5 of the students are in sixth grade, we can write:

4/5 = 60/x

where x is the total number of students in the school.

To **solve** for x, we can cross-multiply:

4x = 5(60)

4x = 300

x = 75

To **check** our answer, we can verify that 60 of the 75 students are in sixth grade:

60/75 = 4/5

So, 4/5 of the students in the school are in sixth **grade**, and there are 75 students in the school.

Alternatively, we can use a proportion to find the number of sixth graders in the school and then **multiply** by 5/4 to find the total number of students:

60/x = 4/5

x = (60*5)/4

x = 75

This method yields the same answer as before.

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