Solve for n using special segments of secants and tangents theoreom​

To solve the differential equation y'' - y = x^2/3 using the given form of yp (yp = ax^2 + bx + c), we first need to find the derivatives of yp.

yp = ax^2 + bx + c
yp' = 2ax + b
yp'' = 2a

Substituting yp, yp', and yp'' into the differential equation gives:

2a - (ax^2 + bx + c) = x^2/3

Simplifying this equation and comparing the coefficients of x^2, x, and the constant term gives:

-a = 1/3
b = 0
2a - c = 0

Solving for a, b, and c, we get:

a = -1/3
b = 0
c = -2a = 2/3

Therefore, the particular solution is yp = -x^2/3 + 2/3.

To find the general solution, we need to add the homogeneous solution, which is the solution to the associated homogeneous equation y'' - y = 0. The characteristic equation for this homogeneous equation is r^2 - 1 = 0, which has roots r = ±1.

Therefore, the general solution is:

y = c1e^x + c2e^-x - x^2/3 + 2/3

where c1 and c2 are constants determined by the initial or boundary conditions.

Final answer:

y = c1e^x + c2e^-x - x^2/3 + 2/3

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Pair 1: (2, 4), Pair 2: (4, 6), Pair 3: (6, 8), Pair 4: (8, 10), Pair 5: (10, 12)

To create a paired data set with the given conditions, we need to have five pairs of observations where the average of y1 and y2 are not equal, and the standard errors for y1 and y2 are greater than 0, but the standard error of the differences (sed) is 0.

Step 1: Invent five pairs of observations
Here's a paired data set that meets these requirements:

Pair 1: (2, 4)
Pair 2: (4, 6)
Pair 3: (6, 8)
Pair 4: (8, 10)
Pair 5: (10, 12)

Step 2: Verify that y¯1 and y¯2 are not equal
Calculate the average of y1 values and y2 values:

y¯1 = (2+4+6+8+10)/5 = 30/5 = 6
y¯2 = (4+6+8+10+12)/5 = 40/5 = 8

Since y¯1 ≠ y¯2 (6 ≠ 8), this condition is met.

Step 3: Verify that sey¯1 > 0 and sey¯2 > 0
As the data sets have different values, it's evident that the standard errors for y1 and y2 are greater than 0.

Step 4: Verify that sed = 0
Calculate the differences (d) for each pair:

d1 = 4 - 2 = 2
d2 = 6 - 4 = 2
d3 = 8 - 6 = 2
d4 = 10 - 8 = 2
d5 = 12 - 10 = 2

Since all differences are the same (2), the standard error of the differences (sed) is 0.
So, the paired data set provided meets all the given conditions.

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d) Least squares line ŷ = 521.2542 - 0.2349x.

e) Correlation coefficient r ≈ -0.9869

f) Estimated gold medal time ŷ for 1924 ≈ 71.26 sec

g) Estimated gold medal time ŷ for 1992 ≈ 53.14 sec

h) Estimate the gold medal time for the 2012 ŷ ≈ 52.12 sec

d) To find the least squares line, we need to calculate the mean and standard deviation of the year (x) and the time (y):

mean of x = (1912 + 1932 + 1952 + 1968 + 1976 + 1984 + 2000 + 2012)/8 = 1972

mean of y = (82.2 + 72.4 + 66.8 + 66.8 + 61.2 + 60.0 + 55.6 + 55.9 + 54.6 + 53.8 + 53.1)/11 = 63.2

standard deviation of x = √(((1912-1972)² + (1932-1972)² + ... + (2012-1972)²)/8) ≈ 44.54

standard deviation of y = √(((82.2-63.2)² + (72.4-63.2)² + ... + (53.1-63.2)²)/10) ≈ 10.53

Then, we can calculate the correlation coefficient r:

r = (1/10) * (((1912-1972)/44.54)(82.2-63.2)/10.53 + ((1932-1972)/44.54)(72.4-63.2)/10.53 + ... + ((2012-1972)/44.54)*(53.1-63.2)/10.53) ≈ -0.9869

Using the formula for the least squares regression line, we have:

b = r * (standard deviation of y / standard deviation of x) ≈ -0.2349

a = mean of y - b * mean of x ≈ 521.2542

Therefore, the least squares line is ŷ = 521.2542 - 0.2349x.

f) To estimate the gold medal time for 1924, we substitute x = 1924 into the equation for the least squares line:

ŷ = 521.2542 - 0.2349(1924) ≈ 71.26 sec

g) To estimate the gold medal time for 1992, we substitute x = 1992 into the equation for the least squares line:

ŷ = 521.2542 - 0.2349(1992) ≈ 53.14 sec

h) To estimate the gold medal time for 2012, we substitute x = 2012 into the equation for the least squares line:

ŷ = 521.2542 - 0.2349(2012) ≈ 52.12 sec

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Answer:

Let x be the number of ounces of 70% fruit juice. Then 54 - x will be the number of ounces of 100% fruit juice.

.7x + 54 - x = .90(54)

54 - .30x = 48.6

.3x = 5.4, so x = 18

18 oz of 70% fruit juice, 36 ounces of 100% fruit juice.

We need 18 oz of the 70% fruit punch and 36 oz of the 100% fruit juice to make 54 oz of a mixture that is 90% fruit juice.

Mathematical expression is defined as the collection of the numbers variables and functions by using operations like addition, subtraction, multiplication, and division.

Let's call the amount of the 70% fruit punch we'll use "x" and the amount of 100% fruit juice we'll use "y".

To start, we can create two equations based on the information we have.

The first equation represents the total amount of liquid in the final mixture:

x + y = 54

The second equation represents the percentage of fruit juice in the final mixture:

0.7x + y = 0.9(54)

Simplifying the second equation:

0.7x + y = 48.6

We can now use the first equation to solve for one of the variables in terms of the other.

Let's solve for "y".

y = 54 - x

Substituting this into the second equation:

0.7x + (54 - x) = 48.6

Simplifying as:

0.7x - x = 48.6 - 54

-0.3x = -5.4

x = 18

So we need 18 oz of the 70% fruit punch.

Using the first equation to solve for "y":

y = 54 - x

y = 54 - 18

y = 36

So we need 36 oz of the 100% fruit juice.

Therefore, we need 18 oz of the 70% fruit punch and 36 oz of the 100% fruit juice to make 54 oz of a mixture that is 90% fruit juice.

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Answer: $24

Step-by-step explanation: (30*80)/100

The hexadecimal addition of CA10 and 4F57 directly in hexadecimal is B197.

Hi! I'd be happy to help you carry out the hexadecimal addition of CA10 and 4F57 directly in hexadecimal. Here are the steps:

1. Write down the two hexadecimal numbers one below the other, with the least significant digits (rightmost digits) aligned:
  CA10
+ 4F57

2. Perform column-wise addition starting from the rightmost column (least significant digit):
  0 + 7 = 7

3. Move to the next column and add the corresponding digits. If the sum exceeds 15 (F in hexadecimal), carry over the excess to the next column:
  1 + 5 = 6 (no carry over needed)

4. Continue the addition for the remaining columns, remembering to carry over when necessary:
  A + F = 19 (in hexadecimal, this is 13); carry over the 1
  1 + C + 4 = 11 (in hexadecimal, this is B)

5. Combine the results from each column to obtain the final sum: B197

So, the hexadecimal addition of CA10 and 4F57 directly in hexadecimal is B197.

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Based on the options provided, the roots of the quadratic equation associated with the graph are -6 and 3, as they are the values where the graph intersects the x-axis. Thus, option A is correct.

In a quadratic equation in the form of [tex]"ax^2 + bx + c = 0"[/tex]  , the roots (or solutions) can be determined from the x-intercepts of the associated graph of the quadratic equation.

which represent the points where the graph intersects the x-axis. These points are also known as the "zeros" or "roots" of the quadratic equation.

In the case of the values -6 and 3 that you mentioned, they are the x-intercepts or zeros of the graph. This means that when you plug in -6 and 3 into the equation or function that corresponds to the graph, the result is zero.

Based on the options provided, the roots of the quadratic equation associated with the graph are -6 and 3, as they are the values where the graph intersects the x-axis.

Therefore, option a.  [tex]-6[/tex] and  [tex]3 i[/tex]s the correct answer.

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The solutions manual for "Linear Algebra and its Applications" by Gilbert Strang can typically be found on online bookstores, such as Amazon or Barnes & Noble.

Alternatively, you may also be able to find a digital copy of the solutions manual through online forums or academic websites. It is important to note that obtaining unauthorized copies of copyrighted materials is illegal, so be sure to obtain the solutions manual through legitimate sources.

You can find the solutions manual for "Linear Algebra and its Applications" by Gilbert Strang on various online platforms, such as Amazon or the publisher's website. You can also check your local bookstore or academic library for a copy of the solutions manual.

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Average value of f on the closed interval 2≤x≤6 ≈ -1.848

The average value of a function f(x) on a closed interval [a,b] is given by:

1/(b-a) × integral from a to b of f(x) dx

So, in this case, we need to find:

1/(6-2) × integral from 2 to 6 of f(x) dx

First, let's find the integral of f(x):

integral of (x²+x)cos(5x) dx

= (1/5) × integral of (x²+x) d(sin(5x))   (integration by parts)

= (1/5) × [(x²+x)sin(5x) - integral of (2x+1)sin(5x) dx]

= (1/5) × [(x²+x)sin(5x) + (2x+1)(cos(5x))/5] + C

So, the average value of f on [2,6] is:

1/(6-2) * integral from 2 to 6 of f(x) dx

= 1/4 × [(6²+6)sin(30) + (2×6+1)(cos(30))/5 - (2²+2)sin(10) - (2×2+1)(cos(10))/5]

≈ -1.848

Therefore, the answer is (b) -1.848 (rounded to three decimal places)

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In the two functions as the value of V(x) increases, the value of W(x) also increases.

The value of functions, V(x) and W(x) is determined as follows;

for h(-2, 1/4); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁻²⁺³ = 2¹ = 2

w(x) = 2ˣ ⁻ ³ = 2⁻²⁻³ = 2⁻⁵ = 1/32

for h(-1, 1/2); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2² = 4

w(x) = 2ˣ ⁻ ³ = 2⁻⁴ = 1/16

for h(0, 1); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2³ = 8

w(x) = 2ˣ ⁻ ³ = 2⁻³ = 1/8

for h(1, 2); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁴ = 16

w(x) = 2ˣ ⁻ ³ = 2⁻² = 1/4

for h(2, 4); the value of the functions is calculated as follows;

v(x) = 2ˣ ⁺ ³ = 2⁵ = 32

w(x) = 2ˣ ⁻ ³ = 2⁻¹ = 1/2

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The line of best fit for the scatter plot is given as follows:

y = 2x.

The slope-intercept representation of a linear function is given by the equation presented as follows:

y = mx + b

The coefficients of the function and their meaning are described as follows:

The graph touches the y-axis at the origin, hence the intercept b is given as follows:

b = 0.

Hence:

y = mx.

When x = 20, y = 40, hence the slope m is given as follows:

20m = 40

m = 40/20

m = 2.

Hence the equation is:

y = 2x.

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As a result, the cable is **3317.7 CM⁵ in diameter in round mils.

A circular mil (CM) is a measurement of area that is particularly useful for describing the size of a wire or cable's cross section. It is equivalent to the surface area of a circle with a one mil (a thousandth of an inch, or 0.0254 mm²) diameter. It is equivalent to roughly 0.7854 millionths of an inch square.

This formula can be used to get the cable's diameter in circular mils:

- Each strand has a diameter of 5.63 mils.

Each strand has an area of (5.63/2)2 × π, or 24.9 circular mils (CM).

- 133 strands have a total area of 133× 24.9, or 3317.7 CM.

As a result, the cable is **3317.7 CM**⁵ in diameter in round mils.

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Answer:

C. A set of objects chosen from a larger set in which the order of the objects doesn't matter.

Step-by-step explanation:

That is the definition of a combination. Order does not matter

For example, selecting 3 students from a total of 10 students. The set selected is a smaller set from a larger set

Answer:

3 hours minus 15 minutes = 2 hours 45 minutes (2.75 hours)

(2.75 mph)(2.75 hours) = 7.5625 miles

For X following a χ²-distribution with 20 degrees of freedom, the point 'a' such that P(X < a) = 0.025 is approximately 8.26.

To find the point 'a' such that P(X < a) = 0.025, where X follows a χ²-distribution with 20 degrees of freedom, you need to use the inverse chi-square distribution function (also called the chi-square quantile function).

Here's the step-by-step explanation:

1. Identify the given parameters: X follows a χ²-distribution with 20 degrees of freedom, and we need to find a point 'a' such that P(X < a) = 0.025.

2. Use the inverse chi-square distribution function (quantile function) with the given probability and degrees of freedom. This function will give you the value of 'a' corresponding to the specified probability.

In most statistical software or calculators, you can find this function. For example, in R programming, you can use the "qchisq()" function:

a = qchisq(0.025, df = 20)

3. Calculate the value of 'a'.

In this case, a ≈ 8.26.

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a. The expected value of the sum of all incurred but as yet

b. Unreported claims at time t is λt times the expected value of a single claim amount.

Probability is a branch of mathematics that deals with measuring the likelihood of an event occurring. It involves quantifying the chances of different outcomes of a random experiment, such as flipping a coin, rolling a die, or drawing a card from a deck.

According to the given information:

(a) Let N(t) be the number of claims incurred up to time t, and let S be the set of times when claims are incurred but not yet reported. Then, the probability that there are exactly n incurred but as yet unreported claims at time t is given by:

P(N(t) - |S| = n) = P(N(t) = n + |S|) × P(|S|)

Since the occurrence of claims follows a Poisson process with rate λ, the probability of n + |S| claims in time t is:

P(N(t) = n + |S|) = ( + |S| / (n + |S|)!)

The distribution of the time until a claim is reported, G, gives the probability that a claim is reported within some time interval after it is incurred. The probability that a claim is incurred but not reported by time t is given by:

P(|S|) = P(G > t)

Putting all these pieces together, we get:

P(N(t) - |S| = n) = ( + |S| / (n + |S|)!) ×) × P(G > t)

(b) Let X_i denote the claim amount for the i-th incurred but as yet unreported claim. Then, the total claim amount for all incurred but as yet unreported claims at time t is:

Y(t) = Σ_i= - |S| X_i

We can find the expected value of Y(t) by using the law of total expectation:

E(Y(t)) = E[E(Y(t) | N(t), S)]

Given N(t) and S, the expected value of Y(t) is just the sum of the expected values of the claim amounts for the unreported claims:

E(Y(t) | N(t), S) = Σ_i= E(X_i)

Since the claim amounts are independent and identically distributed according to F, we have:

E(X_i) = E(F)

Thus, we get:

E(Y(t)) = E[E(Y(t) | N(t), S)] = E[(n + |S|)E(F)] = λt × E(F)

Therefore, the expected value of the sum of all incurred but as yet unreported claims at time t is λt times the expected value of a single claim amount.

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The simplified versions of the given circle equation are a. x^2 + (y-1)^2 = 81, b.(y-1)^2 + z^2 = 81, and c. x^2 + z^2 = 81

a. The circle of radius 9 centered at (0, 1, 0) lying in the xy-plane can be described with the equation:
(x-0)^2 + (y-1)^2 = 9^2
Simplified, this becomes: x^2 + (y-1)^2 = 81

b. The circle of radius 9 centered at (0, 1, 0) lying in the yz- plane can be described with the equation:
(y-1)^2 + (z-0)^2 = 9^2
Simplified, this becomes: (y-1)^2 + z^2 = 81

c. If the circle of radius 9 centered at (0, 1, 0) is lying in the plane y, it means that the y-coordinate is constant throughout the circle. In this case, the equation becomes:
x^2 + z^2 = 9^2
Simplified, this becomes: x^2 + z^2 = 81, with y = 1

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The volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, x = 1 about the y-axis is π/2 cubic units.

To find the volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, x = 1 about the y-axis, we can use the method of cylindrical shells.

Consider a thin strip of width dx at a distance x from the y-axis. When this strip is rotated about the y-axis, it generates a cylindrical shell with radius x and height y = x².

The volume of the cylindrical shell is given by:

dV = 2πx(y)dx

= 2πx(x²)dx

= 2πx³dx

To find the total volume, we need to integrate the volume of all such cylindrical shells from x = 0 to x = 1:

V = ∫₀¹ 2πx³ dx

= 2π ∫₀¹ x³ dx

= 2π [x⁴/4]₀¹

= 2π(1/4)

= π/2

Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x², y = 0, x = 1 about the y-axis is π/2 cubic units.

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The equation of the tangent plane to z = ln(2x y) at (1, 3) is x + (1/3)y - z + ln(2) + ln(3) = 0.

To find the equation of the tangent plane to the surface z = ln(2xy) at the point (1, 3), we need to find the partial derivatives of z with respect to x and y at (1, 3), which are:

∂z/∂x = 1/x

∂z/∂y = 1/y

Evaluating these partial derivatives at (1, 3), we get:

∂z/∂x(1, 3) = 1/1 = 1

∂z/∂y(1, 3) = 1/3

So the normal vector to the tangent plane at (1, 3) is given by:

N = <1, 1/3, -1>

Now we need to find the constant term in the equation of the plane. To do this, we substitute the coordinates of the point (1, 3) into the equation of the surface:

z = ln(2xy) = ln(2(1)(3)) = ln(6)

So the equation of the tangent plane is:

x + (1/3)(y - 3) - z + ln(6) = 0

Simplifying, we get:

x + (1/3)y - z + ln(2) + ln(3) = 0

So the equation of the tangent plane to the surface z = ln(2xy) at the point (1, 3) is x + (1/3)y - z + ln(2) + ln(3) = 0.

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Step-by-step explanation:

A trapezoid bases are the parallel sides

   the AVERAGE of the bases  X  the height    is the area

First one :   Height = 6cm    average of bases = (9.3+4.1) / 2 = 6.7 cm

    area =   6.7 cm * 6 cm = 40.2 c^2

The other three are similar...just different numbers...Using this method, you should be able to do them now .....

Answer:

A trapezoid bases are the parallel sides

  the AVERAGE of the bases  X  the height    is the area

First one :   Height = 6cm    average of bases = (9.3+4.1) / 2 = 6.7 cm

   area =   6.7 cm * 6 cm = 40.2 c^2

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Step-by-step explanation:

The maximum value of f(x, y, z) = xyz subject to the given constraints is f(4, 4, 2) = 32.

To maximize f(x, y, z) = xyz subject to constraints x y z = 32 and x - y + z = 12 using Lagrange multipliers, we need to define a function L(x, y, z, λ, μ) that includes the constraints:

L(x, y, z, λ, μ) = xyz + λ(xyz - 32) + μ(x - y + z - 12)

Now, we need to find the partial derivatives with respect to x, y, z, λ, and μ, and set them to zero:

∂L/∂x = yz + λyz + μ = 0
∂L/∂y = xz + λxz - μ = 0
∂L/∂z = xy + λxy + λ = 0
∂L/∂λ = xyz - 32 = 0
∂L/∂μ = x - y + z - 12 = 0

Solving this system of equations, we get the following:

x = 4
y = 4
z = 2

Now, we can find the maximum value of f:

f(4, 4, 2) = 4*4*2 = 32

Thus, the maximum value of f(x, y, z) = xyz subject to the given constraints is f(4, 4, 2) = 32.

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Answer:

So it would not be 6 cm because, that's like the size of a couple paper clips! it's not 6 millimeters either because that's very very small! 6 kilometers is ALOT! Do it would be 6 meters! A meter is about the length of a door frame! I hoped this helped! Good luck on that IXL!

To reparametrize the curve with respect to arc length, we need to find the arc length function s(t) and then solve for t in terms of s.

The arc length function is given by:

s(t) = ∫√[r'(t)·r'(t)] dt

where r'(t) is the derivative of r(t) with respect to t.

We can calculate r'(t) as:

r'(t) = (2e^(2t)cos(2t) - 4e^(2t)sin(2t))i + (12e^(2t)sin(2t))j + (2e^(2t)sin(2t) + 6e^(2t)cos(2t))k

Now we can substitute this into the arc length formula and integrate:

s(t) = ∫√[(2e^(2t)cos(2t) - 4e^(2t)sin(2t))^2 + (12e^(2t)sin(2t))^2 + (2e^(2t)sin(2t) + 6e^(2t)cos(2t))^2] dt

This integral looks quite complicated, so we will use a numerical integration method to approximate s(t).

We can use the trapezoidal rule to numerically integrate s(t) between t = 0 and some value t = T:

s(T) ≈ ∑[s(iΔt) + s((i+1)Δt)]/2 * Δt

where Δt = T/n is the step size, and n is the number of intervals we use.

Once we have approximated s(t), we can solve for t in terms of s using numerical methods such as the bisection method or Newton's method.

For example, if we want to find the value of t that corresponds to s = 10, we can solve:

s(t) = 10

for t using numerical methods. Once we have t, we can plug it back into r(t) to get the reparametrized curve in terms of arc length s.

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The simplex matrix using the linear programming problem gives optimal solution x = 12.5, y = 0, with objective function value f = 9(12.5) + 3(0) = 112.5

To set up the simplex matrix for the given linear programming problem, we need to introduce slack variables for each inequality constraint and form the initial tableau as follows:

Basic Variables x y s1 s2 RHS

s1 2 3 1 0 300

s2 1 4 0 1 200

z -9 -3 0 0 0

In this tableau, x and y are the decision variables, s1 and s2 are the slack variables, and z is the objective function.

We start with the coefficients of the decision variables in the objective function, which are -9 and -3. We choose the variable with the most negative coefficient to enter the basis, which is y in this case.

To determine which variable to exit the basis, we calculate the ratio of the right-hand side (RHS) value to the coefficient of the entering variable for each constraint. The smallest nonnegative ratio corresponds to the variable that will exit the basis.

For the y variable, we have the following ratios:

s1: 300/3 = 100

s2: 200/4 = 50

Since the ratio for s2 is smaller, we choose s2 to exit the basis. To pivot, we divide the second row by 4 and perform row operations to eliminate the y variable from the other rows:

Basic Variables x y s1 s2 RHS

s1 2 0 1 -3/4 50

y 1/4 1 0 1/4 50

z -9/4 0 0 9/4 225

The new entering variable is x, with coefficient -9/4 in the objective function. The ratios for x are:

s1: 50/2 = 25

y: 50/(1/4) = 200

Therefore, y exits the basis and we pivot on the (2,1) element:

Basic Variables x y s1 s2 RHS

s1 1/2 0 1/2 -3/8 25

x 1/8 1 -1/8 1/8 12.5

z -9/8 0 9/8 9/8 237.5

The optimal solution is x = 12.5, y = 0, with objective function value f = 9(12.5) + 3(0) = 112.5

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Answer:

See below

Step-by-step explanation:

The volume of the described solid is (490/7) π h^3.

The formula for the volume of a right circular cone is:

V = 1/3 πr^2h

In this case, the cone has height 4h and base radius 7r. We need to express the volume in terms of h and r. Since the base radius is 7r, we can write:

r = (1/7) b

where b is the radius of the base of the cone. To find b, we use the fact that the height of the cone is 4h and the base radius is 7r:

h^2 + r^2 = (4h)^2

Substituting r = (1/7) b, we get:

h^2 + (1/49) b^2 = 16h^2

Solving for b, we get:

b^2 = 49(16h^2 - h^2) = 49(15h^2) = 735h^2

Therefore, b = sqrt(735)h, and the volume of the cone is:

V = 1/3 πr^2h = 1/3 π(49/49) b^2 (1/7) 4h = (2/21) π (735h^3)

Simplifying, we get:

V = (490/7) π h^3

Therefore, the volume of the described solid is (490/7) π h^3.

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Answer:

2210in^3

Step-by-step explanation:

since volume= length *width *height

and area= length * width

u alr know area

so if area=l*w

and area= 130

then 130=l*w

so subsittute 130 for length and width

volume=(130*17)

=2210in^3

Answer:

Step-by-step explanation:

Volume  = Area x Height

Area = 130 in^2

Height = 17 in

Volume = 130 x 17

Volume = 2210

We can, probability of completing a given optional assignment is 0.16, expect about 4.16 students out of the 26 to complete the optional assignment.

The expected number of students that will complete the assignment can be found using the formula:

E(X) = np

where X is the random variable representing the number of students who complete the assignment, n is the sample size, and p is the probability of completing the assignment.

Substituting the given values, we get:

E(X) = 26 * 0.16

E(X) = 4.16

Therefore, we can expect about 4.16 students out of the 26 to complete the optional assignment. Since we can't have a fractional part of a student, we can round this up or down to the nearest whole number as needed.

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Answer: 508.9380099cm³ or 508.94cm³ (2 decimal places)

Step-by-step explanation:

Volume of a cylinder = [tex]V = \pi r^{2} h[/tex]   (Volume = pi x radius squared x height)

We have the diameter so we half it to get the radius

6 ÷ 2 =3

Then we do [tex]\pi[/tex] x 3² x 18 = 162[tex]\pi[/tex] or 508.9380099

1 decimal places = 508.9

2 decimal places = 508.94

(a) The value of β that makes fX(x) a valid pdf is β = 0.2.

(b) The cdf for the random variable X is FX(x) = (3[tex]x^3[/tex]/10), for 0 < x ≤ 1.

(c) The probability that the random variable P(X > 2) = 0, since the range of possible values for X is from 0 to 1.

To find the value of β that makes fX(x) a valid pdf, we need to ensure that the area under the pdf from 0 to 1 is equal to 1.

∫0¹ fX(x) dx = ∫0¹ 0.6(x²/β) dx = 0.6/β ∫0¹ x² dx = 0.6/β [[tex]x^3[/tex]/3]0¹ = 0.6/β * (1/3) = 1

Solving for β, we get:

0.6/β * (1/3) = 1

β = 0.6/3 = 0.2

Therefore, β = 0.2 makes fX(x) a valid pdf.

To find the cdf for the random variable X, we integrate the pdf from 0 to x:

FX(x) = ∫[tex]0^x[/tex] fX(t) dt = ∫[tex]0^x[/tex] 0.6(t²/0.2) dt = 3[tex]t^3[/tex]/10|[tex]0^x[/tex] = (3[tex]x^3[/tex]/10) - 0

Therefore, the cdf for X is:

FX(x) = (3[tex]x^3[/tex]/10), for 0 < x ≤ 1

To find the probability that X is greater than 2, we need to use the fact that the probability of an event outside the sample space is 0. Since the range of possible values for X is from 0 to 1, the probability that X is greater than 2 is 0.

P(X > 2) = 0

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