3. (a). Draw 10 Observations from a N(-2,5) as compute the sample mean, and variance. (b). Draw 100 Observations from a N(-2,5) as compute the sample mean, and variance. (c). Draw 1000 Observations from a N(-2,5) as compute the sample mean, and variance. (d). Draw 10,000 Observations from a N(-2,5) as compute the sample mean, and variance. (e). Draw 1,000,000 Observations from a N(-2,5) as compute the sample mean, and variance. (f). How do these values compare to the true mean and variance? Do you notice anything as the sample size gets larger.

Answers

Answer 1

(a) Ten observations drawn from N(-2, 5) and their sample mean, and variance are as follows:Observations from N(-2, 5) -7.174 -1.152 -5.209 -5.462 -2.745 -2.867 -2.322 -5.746 -7.559 -0.755Sample mean: -4.126

Sample variance: 7.107(b) A hundred observations drawn from N(-2, 5) and their sample mean, and variance are as follows:Sample mean: -1.802Sample variance: 4.225(c) A thousand observations drawn from N(-2, 5) and their sample mean, and variance are as follows:Sample mean: -2.109

Sample variance: 5.042(d) Ten thousand observations drawn from N(-2, 5) and their sample mean, and variance are as follows:Sample mean: -2.016Sample variance: 4.864(e) A million observations drawn from N(-2, 5) and their sample mean, and variance are as follows:Sample mean: -2.0002Sample variance: 5.0019

Summary:As the sample size increases, the sample variance decreases and becomes closer to the actual variance (5). In general, the sample means for all the samples (n = 10, n = 100, n = 1,000, n = 10,000, and n = 1,000,000) drawn from N(-2,5) are close to the actual mean (-2).

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Related Questions

Using the transformations u=x-y and v=x+y to evaluate ·JJ x-y/x+y dA over a square region with vertices (0.2): (1.1): (2.2) and (1,3), which ONE of the following values will be the CORRECT VALUE of the double integral?

Answers

The correct value of the double integral is 8.

For evaluate the integral ∫∫ x-y/x+y dA over the given square region, we can use the transformations u = x - y and v = x + y.

Then, the region of integration in the (x, y) plane maps to the region of integration in the (u, v) plane as follows:

(0, 2) → (-2, 2)

(1, 1) → (0, 2)

(2, 2) → (0, 4)

(1, 3) → (-2, 4)

The Jacobian of this transformation is given by:

∂(u, v)/∂(x, y) = 2

So, the integral becomes:

∫∫ x-y/x+y dA = ∫∫ (u+v)/2 dudv

Integrating this over the region in the (u, v) plane, we get: ·

∫∫ (u+v)/2 dudv = 1/2 ∫∫ u dudv + 1/2 ∫∫ v dudv

Integrating over the limits of integration, we get:

1/2∫∫ u dudv = 0

1/2 ∫∫ v dudv = (1/2) × [(2) - (-2)] × [(4-0)/2]

                   = 8

Therefore, the correct value of the double integral is 8.

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EXAM1-2 please show all the
[4 pts.] Resuelva: (x-2y+z= −4
2x + y - 2z = 4
x + 3y – 3z = 8
x+y-2z=3 .
[4 pts.] Resuelva: x + y -2z = 3
2x-y + 3z = 5
x- 2y + 5z = 7

Answers

The solution to the system of equations is x = 1, y = 8/3, and z = 1/3.

To solve the system of equations:

Equation 1: x - 2y + z = -4

Equation 2: 2x + y - 2z = 4

Equation 3: x + 3y - 3z = 8

Equation 4: x + y - 2z = 3

We can use the method of elimination or substitution to find the values of x, y, and z that satisfy all the equations.

Let's use the elimination method to solve this system of equations. We'll start by eliminating the variable x. To eliminate x between equations 2 and 3, we'll multiply equation 3 by 2 and equation 2 by -1:

Equation 2 (multiplied by -1): -2x - y + 2z = -4

Equation 3 (multiplied by 2): 2x + 6y - 6z = 16

Adding equations 2 and 3 eliminates x:

(-2x - y + 2z) + (2x + 6y - 6z) = (-4) + 16

-2x + 2x + (-y + 6y) + (2z - 6z) = 12

5y - 4z = 12   -----> Equation 5

Now let's eliminate x between equations 1 and 4. Multiply equation 4 by -1:

Equation 4 (multiplied by -1): -x - y + 2z = -3

Adding equations 1 and 4 eliminates x:

(x - 2y + z) + (-x - y + 2z) = -4 + (-3)

-3y + 3z = -7  -----> Equation 6

We now have two equations in terms of y and z: Equation 5 (5y - 4z = 12) and Equation 6 (-3y + 3z = -7). To eliminate y, multiply Equation 6 by 5 and Equation 5 by 3:

Equation 5 (multiplied by 3): 15y - 12z = 36

Equation 6 (multiplied by 5): -15y + 15z = -35

Adding equations 5 and 6 eliminates y:

(15y - 12z) + (-15y + 15z) = 36 + (-35)

-12z + 15z = 1

3z = 1

z = 1/3

Substitute the value of z back into Equation 6:

-3y + 3(1/3) = -7

-3y + 1 = -7

-3y = -8

y = 8/3

Substitute the values of y and z back into Equation 1:

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

x - 16/3 + 1/3 = -4

x - 15/3 = -4

x - 5 = -4

x = 1

Therefore, the solution to the system of equations is x = 1, y = 8/3, and z = 1/3.

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if f(x) = 19,x t^6 dt, then f'(x)=

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To find the derivative of the function f(x) = ∫[tex][x to t^6][/tex]19 dt, we can apply the Fundamental Theorem of Calculus.

According to the Fundamental Theorem of Calculus, if a function F(x) is defined as the integral of another function f(t) from a constant to x, i.e., F(x) = ∫[c to x] f(t) dt, then the derivative of F(x) with respect to x is equal to the integrand f(x), i.e., F'(x) = f(x).

In this case, we have f(x) = 19 * t^6 dt, where the integration is performed from x (a constant) to t^6.

Therefore, by applying the Fundamental Theorem of Calculus, we can conclude that:

f'(x) = d/dx ∫[x to t^6] 19 dt = 19 * d/dx (t^6)

Differentiating [tex]t^6[/tex] with respect to x, we obtain:

f'(x) = 19 * [tex]6t^{6-1}[/tex] * dt/dx

= 19 * 6[tex]t^5[/tex] * dt/dx

= 114[tex]t^5[/tex] * dt/dx

So, the derivative of f(x) is given by f'(x) = [tex]114t^5[/tex] * dt/dx, where dt/dx represents the derivative of t with respect to x.

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The differential equation for small deflections of a rotating string is of the form ) + pw²y = 0 dx Obtain the general solution of this equation under the following assumptions: T = T₁x", p = px"; T₁ = 1² p₂w²

Answers

The general solution of the given differential equation is

y = Acos(√(px"w²)x) + Bsin(√(px"w²)x)

To obtain the general solution of the given differential equation, let's go through the solution step by step.

The given differential equation is:

d²y/dx² + p*w²*y = 0

Let's substitute the given assumptions:

T = T₁x"

p = px"

T₁ = 1²p₂w²

Now, rewrite the equation with the substituted values:

d²y/dx² + px"w²*y = 0

Next, let's solve this differential equation. Assume that the solution is of the form y = e^(rx), where r is a constant to be determined.

Taking the first derivative of y with respect to x:

dy/dx = re^(rx)

Taking the second derivative of y with respect to x:

d²y/dx² = r²e^(rx)

Now, substitute these derivatives back into the differential equation:

r²e^(rx) + px"w²*e^(rx) = 0

Divide through by e^(rx) to simplify:

r² + px"w² = 0

Now, solve for r:

r² = -px"w²

r = ±i√(px"w²)

Since r is a constant, we can rewrite it as r = ±iω, where ω = √(px"w²).

The general solution can be expressed as a linear combination of the real and imaginary parts of the exponential function:

y = C₁e^(iωx) + C₂e^(-iωx)

Using Euler's formula, which states e^(ix) = cos(x) + isin(x), we can rewrite the general solution as:

y = C₁(cos(ωx) + isin(ωx)) + C₂(cos(ωx) - isin(ωx))

Simplifying further:

y = (C₁ + C₂)cos(ωx) + i(C₁ - C₂)sin(ωx)

Finally, we can combine C₁ + C₂ = A and i(C₁ - C₂) = B, where A and B are arbitrary constants, to obtain the general solution:

y = Acos(ωx) + Bsin(ωx)

Therefore, the general solution of the given differential equation, under the given assumptions, is:  y = Acos(√(px"w²)x) + Bsin(√(px"w²)x)

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7. An animal feed producer makes two types of grain: A and B. Each unit of grain A contains 2 grams of fat, 1 gram of protein, and 80 calories. Each unit of grain B contains 3 grams of fat, 3 grams of protein, and 60 calories. Suppose that the producer wants each unit of the final product to yield at least 18 grams of fat, at least 12 grams of protein, and at least 480 calories. If each unit of A costs 10 cents and each unit of B costs 12 cents, how many units of each type of grain should the producer use to minimize the cost?

Answers

The animal feed producer makes two types of grain, A and B. Each unit of grain A contains 2 grams of fat, 1 gram of protein, and 80 calories. Each unit of grain B contains 3 grams of fat, 3 grams of protein, and 60 calories.

Suppose that the producer wants each unit of the final product to yield at least 18 grams of fat, at least 12 grams of protein, and at least 480 calories.

If each unit of A costs 10 cents and each unit of B costs 12 cents, how many units of each type of grain should the producer use to minimize the cost?

First, let x be the number of units of grain A and y be the number of units of grain B, which are used to minimize the cost of the feed.

Let the function C(x, y) denote the cost of producing x units of grain A and y units of grain B.C(x,y) = 10x + 12y

where each unit of A costs 10 cents, and each unit of B costs 12 cents. The producer wants each unit of the final product to yield at least 18 grams of fat, at least 12 grams of protein, and at least 480 calories. Each unit of grain A contains 2 grams of fat, 1 gram of protein, and 80 calories; therefore, x units of grain A contain 2x grams of fat, x grams of protein, and 80x calories.

Similarly, y units of grain B contain 3y grams of fat, 3y grams of protein, and 60y calories.

Therefore, the following inequalities must be satisfied:2x + 3y >= 181x + 3y >= 12 80x + 60y >= 480 We use the graphing technique to solve this problem by finding the feasible region and using a corner point method. From the above inequalities, we plot the following equations on a graph and find the feasible region.

2x + 3y = 18,1x + 3y = 12,80x + 60y = 480

This is a plot of the feasible region. Now we need to find the corner points of the feasible region and evaluate C(x, y) at each point.(0, 4), (4.5, 1.5), (6, 0), (0, 12), and (9, 0) are the corner points of the feasible region.

We use these points to compute the minimum cost.

C(0,4) = 10(0) + 12(4)

= 48,C(4.5,1.5)

= 10(4.5) + 12(1.5)

= 57,C(6,0)

= 10(6) + 12(0)

= 60,C(0,12)

= 10(0) + 12(12)

= 144,C(9,0) = 10(9) + 12(0) = 90

Therefore, the minimum cost is 48 cents, which is obtained when 0 units of grain A and 4 units of grain B are used. The producer should use 0 units of grain A and 4 units of grain B to minimize the cost of producing the feed.

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or f (x) = 3x^4 - 12x^3 + 1 find the following. (A) f' (x) (B) The slope of the graph of f at x = 1 (C) The equation of the tangent line at x = 1 (D) The value(s) of x where the tangent line is horizontal (A) f'(x) = 12x^3 - 36x^2 (B) At x = 1, the slope of the graph of f is (C) At x = 1, the equation of the tangent line is y = (D) The tangent line is horizontal at x = (Use a comma to separate answers as needed.)

Answers

The tangent line is horizontal at x = 0 and x = 3.

(A) To find the derivative of the function f(x) = 3x^4 - 12x^3 + 1, we differentiate each term with respect to x using the power rule:

f'(x) = d/dx(3x^4) - d/dx(12x^3) + d/dx(1)

= 12x^3 - 36x^2 + 0

= 12x^3 - 36x^2

So, f'(x) = 12x^3 - 36x^2.

(B) To find the slope of the graph of f at x = 1, we evaluate f'(x) at x = 1:

f'(1) = 12(1)^3 - 36(1)^2

= 12 - 36

= -24

Therefore, the slope of the graph of f at x = 1 is -24.

(C) To find the equation of the tangent line at x = 1, we need both the slope and a point on the line. We already know the slope from part (B), which is -24. Now we can find the y-coordinate of the point on the graph of f(x) at x = 1 by substituting x = 1 into the original function:

f(1) = 3(1)^4 - 12(1)^3 + 1

= 3 - 12 + 1

= -8

So, the point (1, -8) lies on the graph of f(x) at x = 1. The equation of the tangent line can be written in point-slope form:

y - y1 = m(x - x1)

where (x1, y1) is the point on the line and m is the slope.

Using (1, -8) as the point and -24 as the slope, we have:

y - (-8) = -24(x - 1)

y + 8 = -24x + 24

y = -24x + 16

Therefore, the equation of the tangent line at x = 1 is y = -24x + 16.

(D) To find the value(s) of x where the tangent line is horizontal, we need to find where the derivative f'(x) = 0. Set f'(x) equal to zero and solve for x:

12x³ - 36x² = 0

Factor out common terms:

12x²(x - 3) = 0

Setting each factor equal to zero:

12x² = 0 => x² = 0 => x = 0

x - 3 = 0 => x = 3

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Please dont copy, solve it yourself, and explain it clearly, thank you 6.2.4 In the presence of a headwind of nor- malized intensity W, your speed on your bike is V = g(W) = 20 - 10W1/3 mi/hr. The wind intensity W is the continuous uni- form (-1,1) random variable. (Note: If W is negative, then the headwind is actually a tailwind.) Find the PDF fv(v)

Answers

To find the probability density function (PDF) of the speed v, we need to determine the cumulative distribution function (CDF) of v and then differentiate it with respect to v.

Let's denote the PDF of the wind intensity W as fw(w). Since W is a continuous uniform random variable over the interval (-1, 1), its PDF is constant within that interval and zero outside it. The CDF of v, denoted as Fv(v), can be calculated as follows: Fv(v) = P(V ≤ v) = P(g(W) ≤ v) = P(20 - 10W^(1/3) ≤ v).

To determine the probability, we need to find the range of W values that satisfy the inequality. Let's solve it: 20 - 10W^(1/3) ≤ v. -10W^(1/3) ≤ v - 20.

W^(1/3) ≥ (20 - v) / 10. W ≥ [(20 - v) / 10]^3. Since the wind intensity W is a continuous uniform random variable over (-1, 1), the probability that W falls within a certain range is equal to the length of that range. Therefore, the probability that W satisfies the inequality is: P(W ≥ [(20 - v) / 10]^3) = (1 - [(20 - v) / 10]^3) [since the length of (-1, 1) is 2]. Now, to find the PDF of v, we differentiate the CDF with respect to v: fv(v) = d/dv [Fv(v)] = d/dv [1 - [(20 - v) / 10]^3] = 3/10 [(20 - v) / 10]^2.  Therefore, the PDF of v, denoted as fv(v), is given by: fv(v) = 3/10 [(20 - v) / 10]^2.  Please note that this PDF is valid within the range of v where the inequality holds. Outside that range, the PDF is zero.

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Solve the equation 3|x-1|-1=11 Simplify the following expressions: Q.2.4.1 x²-4 x² + 4x +4 Q.2.4.2 9x²-25y² 3x² - 5xy Q.2.4.3 64a³-1256³ 4a²b-5ab² Q.2.4.4 √√4x²y√27x³y6 2,43 Q.2.4.5 [x₂² ]•Wx²y³] (4) (3) (3) (5) (4) (5)

Answers

An expression, which is used to indicate a mathematical relationship or computation, is a collection of numbers, variables, and mathematical operations (such as addition, subtraction, multiplication, and division).

1. Solve the equation 3|x-1|-1=11:

To solve this equation, we will isolate the absolute value term and then solve for x.

3|x-1| - 1 = 11

Add 1 to both sides:

3|x-1| = 12

Divide both sides by 3:

|x-1| = 4

Now we have two cases to consider, one where the expression inside the absolute value is positive and one where it is negative.

Case 1: (x-1) is positive:

x-1 = 4

Add 1 to both sides:

x = 5

Case 2: (x-1) is negative:

-(x-1) = 4

Multiply both sides by -1 (to eliminate the negative sign):

x-1 = -4

Add 1 to both sides:

x = -3

Therefore, the solutions to the equation are x = 5 and x = -3.

2. Q 2.4.1 x²-4 x² + 4x +4:

combining similar terms

x² - 4x² + 4x + 4 = -3x² + 4x + 4

Q.2.4.2, "9x2-25y2 3x2 - 5xy," asks:

There are no similar terms to combine, thus the expression stays the same.

There are no similar terms to combine in Q.2.4.3 64a3-125b3 4a2b-5ab2, hence the expression is left alone.

Q.2.4.4: Separately simplify each square root in the following formula:

(27x3y6) = 3xy3 (y3) and ((4x2y) = 2xy

Add the condensed square roots together now:

√((4x2y)(27x3y6)) equals ((2xy * 3xy3(y3)).

Under the square root, multiply as follows: (2x * 3xy3 * (y3 * y)) = (6x2y4(y3 * y))

Q.2.4.5 [x²]•Wx²y³(4)(3)(3)(5)(4)(5):

Add the exponents together and multiply the coefficients:

[x²]•Wx²y³(4)(3)(3)(5)(4)(5) = x^(2 + 2) x = 4 * Wx2y7 * 14400 * Wx2y(3 + 4) * (4 * 3 * 3 * 5 * 4 * 5)

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Suppose the population of a particular endangered bird changes on a yearly basis as a discrete dynamic system. Suppose that initially there are 60 juvenile chicks and 30 60 breeding adults, that is xo = [\begin{array}{c}60\\30\end{array}\right]
Suppose also that the yearly transition matrix is
A = [\begin{array}{cc}0&1.25\\s&0.5\end{array}\right]
where s is the proportion of chicks that survive to become adults (note 9 S 0.5 that 0≤ s≤ 1 must be true because of what this number represents).

(a) Which entry in the transition matrix gives the annual birthrate of chicks per adult?
(b) Scientists are concerned that the species may become extinct. Explain why if 0 ≤ s < 0.4 the species will become extinct. (c) If s = 0.4, the population will stabilise at a fixed size in the long term. What will this size be?

Answers

(a) The annual birthrate of chicks per adult is represented by the entry which is 1.25.

b.  The species will become extinct if the total population decreases over time.

C. The populations stabilizes at s = 0.4

How to solve the matrix

(a) The annual birthrate of chicks per adult is represented by the entry which is 1.25.

(b) The species will become extinct if the total population decreases over time. The total population would be gotten at a given time that is given by multiplying the transition matrix A by the population vector at the previous time.

-λ (0.5 - λ) - 1.25 s

λ² - 0.5 λ - 1.25λ

when we solve this out we have the unknown

= 0.4

(c) If s = 0.4, the eigen values are

[tex]A = 1\left[\begin{array}{ccc}1.25\\1\\\end{array}\right][/tex]

The populations stabilizes at s = 0.4

which is a ratio of 1.25 : 1

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Find the Laplace transforms of the following functions: (a) y(t) = 14 (6) y(t) = 3t (c) y(t) = sin(2t) (d) y(t) = e-+43 (e) y(t) = (t - 4) u4(t).

Answers

Answer: The Laplace transform of

y(t) = (t - 4) u4(t) is

[tex]$\frac{4}{s} + \frac{1}{s^{2}}$[/tex]

Step-by-step explanation:

The Laplace transform can be obtained using the formula below:

[tex]$$F(s)=\int_{0}^{\infty} f(t) e^{-st} dt$$[/tex]

Let's use this formula to obtain the Laplace transforms of the given functions.

(a) y(t) = 14

Here, f(t)=14.

Substituting the value of f(t) in the above formula, we get:

\begin{align*}F(s) &=[tex]\int_{0}^{\infty} f(t) e^{-st} dt \\[/tex] &

= [tex]\int_{0}^{\infty} 14 \, e^{-st} dt \\[/tex] &

= [tex]\left[ \frac{14}{-s} \, e^{-st} \right]_{0}^{\infty} \\[/tex] &

=[tex]\frac{14}{s} \, [ 0 -1] \\[/tex] &

= [tex]\frac{-14}{s}\end{align*}[/tex]

Therefore, the Laplace transform of

y(t) = 14 is [tex]$\frac{-14}{s}$[/tex].

(b) y(t) = 3t

Here, f(t)=3t.

Substituting the value of f(t) in the above formula, we get:

\begin{align*}F(s) &=[tex]\int_{0}^{\infty} f(t) e^{-st} dt \\[/tex] &

= [tex]\int_{0}^{\infty} 3t \, e^{-st} dt \\[/tex]&

= [tex]\left[ \frac{3t}{-s} \, e^{-st} - \int_{0}^{\infty} \frac{3}{s} e^{-st} dt \right]_{0}^{\infty} \\[/tex] &

= [tex]\left[ \frac{3t}{-s} \, e^{-st} + \frac{3}{s^{2}} \, e^{-st} \right]_{0}^{\infty} \\[/tex] &

= [tex]\frac{3}{s^{2}}[/tex]end{align*}

Therefore, the Laplace transform of

y(t) = 3t is [tex]$\frac{3}{s^{2}}$[/tex].

(c) y(t) = sin(2t)

Here, f(t)=sin(2t).

Substituting the value of f(t) in the above formula, we get:

\begin{align*}F(s) &=[tex]\int_{0}^{\infty} f(t) e^{-st} dt \\[/tex] &

= [tex]\int_{0}^{\infty} \sin(2t) \, e^{-st} dt \\[/tex] &

=[tex]\int_{0}^{\infty} \frac{\sin(2t)}{s} \, s e^{-st} dt \\[/tex] &

= [tex]\frac{2}{s} \int_{0}^{\infty} \frac{\sin(2t)}{2} \, e^{-st} dt \\[/tex] &

=[tex]\frac{2}{s} \int_{0}^{\infty} \sin(x) \, e^{-\frac{s}{2}x} dx \qquad (\text{where } x=2t) \\[/tex]

&= [tex]\frac{2}{s} \cdot \frac{1}{1+(\frac{s}{2})^{2}}[/tex]end{align*}

Therefore, the Laplace transform of

y(t) = sin(2t) is [tex]$\frac{2}{s(1+(\frac{s}{2})^{2})}$[/tex].

(d) y(t) =[tex]e^(-4t)[/tex]

Here,

f(t)=[tex]e^{-4t}[/tex].

Substituting the value of f(t) in the above formula, we get:

\begin{align*}F(s) &

=[tex]\int_{0}^{\infty} f(t) e^{-st} dt \\[/tex] &

= [tex]\int_{0}^{\infty} e^{-4t} \, e^{-st} dt \\[/tex] &

= [tex]\int_{0}^{\infty} e^{-(s+4)t} dt \\[/tex] &

= [tex]\left[ \frac{1}{-(s+4)} \, e^{-(s+4)t} \right]_{0}^{\infty} \\[/tex] &

= [tex]\frac{1}{s+4}[/tex]end{align*}

Therefore, the Laplace transform of y(t) = [tex]e^(-4t) is \frac{1}{s+4}[/tex]

(e) y(t) = (t - 4) u4(t)

Here,

[tex]f(t)=(t-4)u_{4}(t)[/tex]

where [tex]u_{4}(t)[/tex] is the unit step function.

Substituting the value of f(t) in the above formula, we get:

\begin{align*}F(s) =[tex]\int_{0}^{\infty} f(t) e^{-st} dt \\[/tex]

= [tex]\int_{4}^{\infty} (t-4) \, e^{-st} dt \\[/tex] &

= [tex]\left[ -\frac{(t-4)}{s} \, e^{-st} \right]_{4}^{\infty} + \frac{4}{s} \\[/tex]

= [tex]\frac{4}{s} + \frac{1}{s^{2}}[/tex]end{align*}.

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Which of the following is not a graphical technique to display quantitative data? Group of answer choices
a. histogram
b. Stem-and-leaf
c. bar chart
d. scatterplot

Answers

The graphical technique that could be used to display quantitative data is Stem-and-leaf.Option B

What is Stem and leaf?

When displaying quantitative data in a tabular manner, stem-and-leaf divides each data point into a "stem" and "leaf." It is a way of quantitatively arranging and expressing data rather than a pictorial technique.

The stem-and-leaf plot is helpful for displaying data distribution and specific data points, but it is not a graphical method like the histogram, bar chart, or scatterplot, which directly depict data using graphical elements.

Hence, what we are going to use in the case of the data that we have here is the stem and leaf kind of plot.

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Use mathematical induction to prove that n(n+1) Σn,i=1 = [n(n+1)] / 2

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[(k+1)(k+2)] / 2 = RHS: By mathematical induction, equality is proven.

The following is the solution to the mathematical induction to prove that n(n+1) Σn,i=1 = [n(n+1)] / 2:

Step 1: Basis Step: Let’s check the equality for n=1.

LHS=1(1+1) Σ1,i=1=1 × 2/2=1 × 1=1.

RHS= [1(1+1)] / 2 = [2] / 2 = 1.

So, LHS=RHS =1 for n=1.

Step 2: Induction hypothesis: Suppose that the equality holds for any arbitrary positive integer k. That is,

k(k+1) Σk,i=1 = [k(k+1)] / 2.

This is the induction hypothesis.

Step 3: Induction Step: Let’s prove that equality holds for k+1 as well. i.e. (k+1)(k+2) Σk+1,i=1 = [(k+1)(k+2)] / 2.

The left-hand side of the equation is given by:(k+1)(k+2) Σk+1,i=1=k(k+1) + (k+1)(k+2).We know that k(k+1) Σk,i=1 = [k(k+1)] / 2 (Using Induction Hypothesis).

Therefore, (k+1)(k+2) Σk+1, i=1=k(k+1) + (k+1)(k+2)

= [k(k+1)] / 2 + (k+1)(k+2).

Taking the LCM of 2 in the numerator, we get

[k(k+1)] / 2 + 2(k+1)(k+2) / 2.= [k² + k + 2k + 2] / 2

= [(k+1)(k+2)] / 2 = RHS. Hence, by mathematical induction, equality is proven.

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2x² + 3x. 1 in the form fog. If g(x) = (x + 1), find the function f(x). 2+1 Let f(x) = 3x + 2 and g(x)= After simplifying, (fog)(x) = Question Help: Video Submit Question Question 7 Express the funct

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To express the function (fog)(x), we need to substitute the function g(x) into the function f(x) and simplify.

Given: f(x) = 3x + 2 ,g(x) = x + 1

To find (fog)(x), substitute g(x) into f(x): (fog)(x) = f(g(x))

Replace x in f(x) with g(x):(fog)(x) = f(x + 1)

Now substitute the function f(x) into (fog)(x): (fog)(x) = 3(x + 1) + 2

Simplify: (fog)(x) = 3x + 3 + 2

(fog)(x) = 3x + 5

So, the expression for (fog)(x) is 3x + 5.

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9. Let A = =[¹]. (15 points) a) Find the characteristic equation of A. b) Find the eigenvalues of A. c) Find bases for eigenspaces of A.

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a) The characteristic equation of matrix A is λ² - 4 = 0.

b) The eigenvalues of matrix A are λ = 2 and λ = -2.

c) The bases for the eigenspaces of matrix A are:

For eigenvalue λ = 2: v = [tex]\begin{bmatrix} 1 \\ -2 \end{bmatrix}[/tex]

For eigenvalue λ = -2: v = [tex]\begin{bmatrix} 1 \\ 2 \end{bmatrix}[/tex]

a) Finding the characteristic equation of matrix A:

The characteristic equation is obtained by finding the determinant of the matrix (A - λI), where λ is a scalar variable and I represents the identity matrix of the same size as A. In this case, A is a 2x2 matrix, so we subtract λI:

A - λI = [tex]\begin{bmatrix}0 & -1 \\4 & 0\end{bmatrix} - \begin{bmatrix}\lambda & 0 \\0 & \lambda\end{bmatrix} = \begin{bmatrix}-\lambda & -1 \\4 & -\lambda\end{bmatrix}[/tex]

Now, we find the determinant of this matrix:

det(A - λI) = (-λ)(-λ) - (-1)(4) = λ² - 4

Therefore, the characteristic equation of matrix A is:

λ² - 4 = 0

b) Finding the eigenvalues of matrix A:

To find the eigenvalues, we solve the characteristic equation we obtained in the previous step:

λ² - 4 = 0

We can factor this equation:

(λ - 2)(λ + 2) = 0

Setting each factor equal to zero, we have two cases:

λ - 2 = 0 or λ + 2 = 0

Solving each equation, we find two eigenvalues:

Case 1: λ - 2 = 0

λ = 2

Case 2: λ + 2 = 0

λ = -2

Therefore, the eigenvalues of matrix A are λ = 2 and λ = -2.

c) Finding bases for eigenspaces of matrix A:

To find the eigenspaces corresponding to each eigenvalue, we substitute the eigenvalues back into the equation (A - λI)v = 0, where v is the eigenvector. We solve for v to find the eigenvectors associated with each eigenvalue.

For the eigenvalue λ = 2:

(A - 2I)v = 0

Substituting the values, we have:

[tex]\begin{bmatrix}-2 & -1 \\4 & -2\end{bmatrix} \begin{bmatrix}v_1 \\v_2\end{bmatrix} = \begin{bmatrix}0 \\0\end{bmatrix}[/tex]

From the augmented matrix, we obtain the following equations:

-2v₁ - v₂ = 0 and 4v₁ - 2v₂ = 0

Simplifying each equation, we have:

-2v₁ = v₂ and 4v₁ = 2v₂

We can choose a convenient value for v₁, let's say v₁ = 1. Then, from the first equation, we find v₂ = -2.

Therefore, the eigenvector associated with λ = 2 is:

[tex]v = \begin{bmatrix}1 \\-2\end{bmatrix}[/tex]

For the eigenvalue λ = -2:

(A - (-2)I)v = 0

Substituting the values, we have:

[tex]\begin{bmatrix}2 & -1 \\4 & 2\end{bmatrix} \begin{bmatrix}v_1 \\v_2\end{bmatrix} = \begin{bmatrix}0 \\0\end{bmatrix}[/tex]

From the augmented matrix, we obtain the following equations:

2v₁ - v₂ = 0 and 4v₁ + 2v₂ = 0

Simplifying each equation, we have:

2v₁ = v₂ and 4v₁ = -2v₂

Again, we can choose a convenient value for v₁, let's say v₁ = 1. Then, from the first equation, we find v₂ = 2.

Therefore, the eigenvector associated with λ = -2 is:

[tex]v = \begin{bmatrix}1 \\2\end{bmatrix}[/tex]

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Complete Question:

9. Let A = [tex]\begin{bmatrix}0 &-1 \\ 4&0 \end{bmatrix}[/tex]. (15 points) a) Find the characteristic equation of A. b) Find the eigenvalues of A. c) Find bases for eigenspaces of A.

In how many ways can we arrange the integers 1, 2, 3, 4, 5 in a line so that there are no occurrence of the patterns 12, 23, 34, 45, 51?
a. 45
b. 40
C. 50
d. 60
e. None of the mentioned

Answers

To arrange the integers 1, 2, 3, 4, 5 in a line without any occurrence of the patterns 12, 23, 34, 45, 51, the number of possible arrangements can be determined. The options given are a) 45, b) 40, c) 50, d) 60, or e) None of the mentioned. correct answer is e) None of the mentioned.

To solve this problem, we can consider the given patterns as "forbidden" patterns. We need to count the number of arrangements where none of these forbidden patterns occur. One approach is to use complementary counting. There are 5! = 120 total possible arrangements of the integers 1, 2, 3, 4, 5. However, out of these, there are 5 arrangements where each forbidden pattern occurs once. Hence, the number of valid arrangements is 120 - 5 = 115. However, none of the given options matches this result, so the correct answer is e) None of the mentioned.

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Evaluate the following integrals below. Clearly state the technique you are using and include every step to illustrate your solution. Use of functions that were not discussed in class such as hyperbolic functions will not get credit.

(a)Why is this integral ∫4 1 /√3x-3 improper? If it converges, compute its value exactly (decimals are not acceptable) or show that it diverges.

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The integral ∫4 1 /√(3x-3) is improper because the integrand has a vertical asymptote at x = 1, resulting in an undefined value at that point. To determine if the integral converges or diverges, we need to evaluate its behavior as x approaches the endpoint of the interval.

The given integral is improper because the denominator, √(3x-3), becomes zero at x = 1, which leads to division by zero. This indicates a vertical asymptote at x = 1, and the function is undefined at that point.

To analyze the convergence or divergence of the integral, we examine the behavior of the integrand as x approaches the endpoint of the interval, in this case, x = 1. Since the integrand approaches infinity as x approaches 1 from the left, and as x approaches negative infinity as x approaches 1 from the right, the integral diverges.

Therefore, the integral ∫4 1 /√(3x-3) diverges.

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Use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform. (Write your answer as a function of t.) ℒ−1 4s − 8 (s2 + s)(s2 + 1)

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The inverse Laplace transform is \mathcal{L}^{-1} \left[\frac{4s-8}{(s^2+s)(s^2+1)}\right] = 8 - 4e^{-t} - 4\cos(t).

We are to determine the inverse Laplace transform of the given function

ℒ−1 4s − 8 (s2 + s)(s2 + 1).

We are given that

ℒ−1 4s − 8 (s2 + s)(s2 + 1)

We know that Theorem 7.2.1 is defined as:\mathcal{L}^{-1}[F(s-a)](t)=e^{at}f(t)

By applying partial fraction decomposition, we get:

\frac{4s-8}{(s^2+s)(s^2+1)}

= \frac{As+B}{s(s+1)}+\frac{Cs+D}{s^2+1}\ implies 4s-8 = (As+B)(s^2+1)+(Cs+D)(s)(s+1)\ implies 4s-8 = As^3 + Bs + As + B + Cs^3 + Cs^2 + Ds^2 + Ds\ implies 0 = (A+C)s^3+C s^2+(A+D)s+B\ implies 0 = s^3(C+A)+s^2(C+D)+Bs+(AD-8)

Matching the coefficients, we get the following:

C+A=0

C+D=0

A=0

AD-8=-8

\implies A=0, D=-C

\implies C=-\frac{4}{5}

\implies B=\frac{8}{5}

Now the original function can be written as:

\frac{4s-8}{(s^2+s)(s^2+1)}

= \frac{8}{5}\cdot\frac{1}{s} - \frac{4}{5}\cdot\frac{1}{s+1} -\frac{4}{5}\cdot\frac{s}{s^2+1}\mathcal{L}^{-1}\left[\frac{4s-8}{(s^2+s)(s^2+1)}\right](t)

= \mathcal{L}^{-1}\left[\frac{8}{5}\cdot\frac{1}{s} - \frac{4}{5}\cdot\frac{1}{s+1} -\frac{4}{5}\cdot\frac{s}{s^2+1}\right](t)

= 8\mathcal{L}^{-1}\left[\frac{1}{s}\right](t) - 4\mathcal{L}^{-1}\left[\frac{1}{s+1}\right](t) - 4\mathcal{L}^{-1}\left[\frac{s}{s^2+1}\right](t)

= 8 - 4e^{-t} - 4\cos(t)

Therefore, the function is given by:\mathcal{L}^{-1} \left[\frac{4s-8}{(s^2+s)(s^2+1)}\right] = 8 - 4e^{-t} - 4\cos(t).

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There is a virus turning people into zombies who attack the living and never die.

No one knows where it came from, but when the virus was first detected, it was 2 days after a group of 16 archaeologists had opened up an ancient tomb.

Unfortunately, all 16 archaeologists had been turned to zombies.

Authorities believe the virus is spread when infected people bite someone who’s uninfected.

Each zombie bites three uninfected people each day.

a. How many zombies were there at day zero (i.e. t =0)?
b. If the number of zombies Z(t) takes the form , where A is the number of zombies at t = 0, what is k, the estimated growth rate of the virus?
c. How long will it take before the entire human population of the planet (which for this problem will be taken as 7 billion people) are turned into the undead?

Answers

(a) At day zero, the number of zombies, Z(0) = 16
Given that 16 archaeologists had opened up an ancient tomb, which is the cause of the virus. The given number of   zombies at day zero is 16.

(b) The number of zombies Z(t) takes the form
Z(t) = Ae^(kt), where A is the number of zombies at t=0 and k is the estimated growth rate of the virus.
At t=0, Z(0) = A
A = 16
Therefore, the number of zombies takes the form Z(t) = 16e^(kt)
To find k, we have to use the information provided. Each zombie bites three uninfected people each day. Thus, the number of newly infected people per day is 3Z(t).

The growth rate of the virus is given by dZ/dt. So we have,
dZ/dt = 3Z(t)
Separating the variables and integrating, we get
∫dZ/Z = ∫3dt
ln |Z| = 3t + C, where C is the constant of integration
At t = 0, Z = A = 16
ln |16| = C
C = ln 16
So the equation becomes
ln |Z| = 3t + ln 16
Taking the exponential of both sides, we get
|Z| = e^(3t+ln16)
|Z| = 16e^(3t)
Z = ±16e^(3t)
But since the number of zombies is always positive, we can ignore the negative sign. Hence,
Z(t) = 16e^(3t)
Comparing with Z(t) = Ae^(kt), we get
k = 3
Therefore, the estimated growth rate of the virus is 3.

(c)The entire human population of the planet is 7 billion.
Let P(t) be the number of uninfected people at time t.
Initially, P(0) = 7 billion
We know that each zombie bites three uninfected people each day.
So the number of newly infected people per day is 3Z(t)P(t).
The rate of change of uninfected people is given by dP/dt, which is negative since P is decreasing.
So we have,
dP/dt = -3Z(t)P(t)
Separating the variables and integrating, we get
∫dP/P = -∫3Z(t)dt
ln |P| = -3∫Z(t)dt + C, where C is the constant of integration
At t=0, Z = 16
So we have,
ln |7 billion| = -3(16t) + C
C = ln |7 billion| + 48t
Putting the value of C, we get
ln |P| = -3(16t) + ln |7 billion| + 48t
ln |P| = 32t + ln |7 billion|
Taking the exponential of both sides, we get
|P| = e^(32t+ln7billion)
|P| = 7 billione^(32t)
P = ±7 billione^(32t)
But since the number of uninfected people is always positive, we can ignore the negative sign. Hence,
P(t) = 7 billione^(32t)
When the entire population is infected, the number of uninfected people P(t) becomes zero.
So we have to solve for t in the equation P(t) = 0.
7 billione^(32t) = 0
e^(32t) = 0
Taking logarithms, we get
32t = ln 0
This is undefined, so the entire population will never be infected.

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1. (a) Without using a calculator, determine the following integral: x² - 8x + 52 6² dx. x² + 8x + 52 (Hint: First write the integrand I(x) as x² - 8x + 52 I(x) = 1+ ax + b x² + 8x + 52 x² + 8x + 52 where a and b are to be determined.) =

Answers

Substituting back u = x² + 8x + 52, the integral becomes: x² + 8x + 52 - 4 ln|x + 4| + C, where C is the constant of integration.

To determine the integral without using a calculator, we need to first find the values of a and b in the integrand. We can rewrite the integrand as:

I(x) = (x² - 8x + 52)/(x² + 8x + 52)

To find the values of a and b, we can perform polynomial division.

Dividing x² - 8x + 52 by x² + 8x + 52, we get:

         -16x + 0

     ------------

x² + 8x + 52 | x² - 8x + 52

           - (x² + 8x + 52)

            --------------

                  0

Therefore, the result of the division is -16x + 0.

Now, we can rewrite the integrand as:

I(x) = 1 - (16x/(x² + 8x + 52))

To evaluate the integral, we need to find the antiderivative of -16x/(x² + 8x + 52). This can be done by using substitution or partial fractions.

Let's use the substitution method. Let u = x² + 8x + 52, then du = (2x + 8) dx. Rearranging, we have dx = du/(2x + 8).

Substituting these values, the integral becomes:

∫ (1 - (16x/(x² + 8x + 52))) dx = ∫ (1 - (16/(2x + 8))) du/(2x + 8)

Simplifying, we have:

∫ (1 - 8/(2x + 8)) du = ∫ (1 - 4/(x + 4)) du

Integrating each term separately, we get:

u - 4 ln|x + 4| + C

Finally, substituting back u = x² + 8x + 52, the integral becomes:

x² + 8x + 52 - 4 ln|x + 4| + C

where C is the constant of integration.

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9. A checker is placed on a checkerboard in the top right corner. The checker can move diagonally downward. Determine the number of routes to the bottom of the board.

Answers

So, in general, the number of routes for the checker to reach the bottom of the board in an m x n checkerboard is [tex]2^{(m-1)}.[/tex]

To determine the number of routes for the checker to reach the bottom of the board, we need to consider the dimensions of the checkerboard and the possible moves the checker can make.

Let's assume the checkerboard has dimensions of m rows and n columns. Since the checker starts at the top right corner, it needs to reach the bottom row. The checker can only move diagonally downward, either to the left or to the right.

To reach the bottom row, the checker must make m-1 moves. Since each move can be either diagonal-left or diagonal-right, there are two options for each move. Therefore, the total number of routes can be calculated as 2 raised to the power of (m-1).

In mathematical notation, the number of routes is given by:

Number of routes = [tex]2^{(m-1)}[/tex]

For example, if the checkerboard has 8 rows, the number of routes would be:

Number of routes = [tex]2^{(8-1)[/tex]

= [tex]2^7[/tex]

= 128

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What is the family wise error rate (FWER) and how can you control for it using the Bonferroni procedure when conducting post hoc test for a significant one-way ANOVA? (400 words)

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The family-wise error rate (FWER) is the chance of making at least one Type I error in a family of tests. When several post-hoc assessments are conducted in one ANOVA, the possibility of a type I error rises.

In other words, when conducting several pairwise comparisons in a one-way ANOVA, the probability of at least one type I error increases. In such situations, the Bonferroni correction may be employed to control the family-wise error rate.To account for multiple comparisons when conducting a post hoc test following a one-way ANOVA, the Bonferroni correction is often utilized.

The procedure includes a series of pairwise comparisons between all of the sample groupings. Bonferroni correction involves calculating a new alpha value that is smaller than the original alpha value. The new alpha value is then divided by the total number of tests. The new alpha value is calculated as:α = α / n Where, α = initial alpha level, n = number of pairwise comparisons. The p-value that is typically used to determine whether or not a null hypothesis is rejected can be changed using the Bonferroni correction.

This correction is accomplished by lowering the alpha level for each of the evaluations. For example, if the significance level is set to 0.05, and a Bonferroni correction is applied to three tests, the new alpha value will be 0.0167. This is done to make sure that the overall probability of a Type I error stays below the desired level. When utilizing the Bonferroni correction, the likelihood of committing a type I error is reduced. The results obtained after applying the Bonferroni correction to a one-way ANOVA post hoc comparison will be more accurate because they will be less prone to a Type I error.

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1. Class relative frequencies must be used, rather than class frequencies or class percentages, when constructing a Pareto diagram. 2. A Pareto diagram is a pie chart where the slices are arranged from largest to smallest in a counterclockwise direction. 3. The sample variance and standard deviation can be calculated using only the sum of the data and the sample size, n. 4. The conditions for both the hypergeometric and the binomial random variables require that the trials are independent. 5. The exponential distribution is sometimes called the waiting-time distribution, because it is used to describe the length of time between occurrences of random events. 6. A Type I error occurs when we accept a false null hypothesis. 7. A low value of the correlation coefficient r implies that x and y are unrelated. 8. A high value of the correlation coefficient r implies that a causal relationship exists between x and y.

Answers

1. Class relative frequencies must be used, rather than class frequencies or class percentages, when constructing a Pareto diagram. The relative frequency of each class is calculated by dividing the frequency of each class by the total number of data points.

2. A Pareto diagram is a chart where the slices are arranged in descending order of frequency in a counterclockwise manner. Pareto chart is a graphical representation that displays individual values in descending order of relative frequency.

3. The sample variance and standard deviation can be calculated using only the sum of the data and the sample size, n. The sample variance and standard deviation are calculated using the sum of squared deviations, which can be calculated using only the sum of the data and sample size.

4. The conditions for both the hypergeometric and the binomial random variables require that the trials are independent. The hypergeometric and binomial random variables require independence among the trials.

5. The exponential distribution is sometimes called the waiting-time distribution because it describes the time between events' occurrences. The exponential distribution is a continuous probability distribution that is used to model waiting times.

6. A Type I error occurs when we accept a false null hypothesis. A Type I error occurs when we reject a true null hypothesis.

7. A low value of the correlation coefficient r implies that x and y are unrelated. When the value of the correlation coefficient is close to zero, x and y are unrelated.

8. A high value of the correlation coefficient r implies that a causal relationship exists between x and y. When the value of the correlation coefficient is close to 1, a strong relationship exists between x and y. This indicates that a causal relationship exists between the two variables.

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Solve the following eigenvalue problem AX = 2X, 1-1 1 A= 1 1 1 1 1 1

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The eigenvalues and eigenvectors of matrix $A$ are,λ = 0, with eigenvector $X_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}$λ = 3, with eigenvectors $X_2 = \begin{bmatrix}1\\1\\1\end{bmatrix}$ and $X_3 = \begin{bmatrix}-1\\1\\0\end{bmatrix}$.

The given eigenvalue problem is, $AX=2X$,

where $A=\begin{bmatrix}1 & -1 & 1\\1 & 1 & 1\\1 & 1 & 1\end{bmatrix}$First, we need to find the eigenvalues of matrix $A$.

The characteristic equation of matrix $A$ is given by,|A-λI| = 0Where, λ is the eigenvalue and I is the identity matrix of order 3.

Substituting A, we get,$\begin{vmatrix}1-λ & -1 & 1\\1 & 1-λ & 1\\1 & 1 & 1-λ\end{vmatrix}=0$Expanding the above determinant,

we get,$\begin{aligned}&(1-λ)\begin{vmatrix}1-λ & 1\\1 & 1-λ\end{vmatrix}-\begin{vmatrix}-1 & 1\\1 & 1-λ\end{vmatrix}+\begin{vmatrix}-1 & 1-λ\\1 & 1\end{vmatrix}\\&=(1-λ)[(1-λ)^2-1]-[(-1)(1-λ)-(1)(1)]+[-1(1-λ)-1(1)]\\&=(λ-3)λ^2=0\end{aligned}$Hence, the eigenvalues of matrix $A$ are λ = 0, λ = 3.

Now, we need to find the eigenvectors corresponding to the eigenvalues of matrix $A$.For λ = 0,$(A-0I)X=0$Therefore, $\begin{bmatrix}1 & -1 & 1\\1 & 1 & 1\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

On solving, we get the eigenvector as,$X_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}$For λ = 3,$(A-3I)X=0$Therefore, $\begin{bmatrix}-2 & -1 & 1\\1 & -2 & 1\\1 & 1 & -2\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$On solving,

we get the eigenvectors as,$X_2 = \begin{bmatrix}1\\1\\1\end{bmatrix}$ and $X_3 = \begin{bmatrix}-1\\1\\0\end{bmatrix}$Therefore, the eigenvalues and eigenvectors of matrix $A$ are,λ = 0,

with eigenvector $X_1 = \begin{bmatrix}-1\\1\\0\end{bmatrix}$λ = 3, with eigenvectors $X_2 = \begin{bmatrix}1\\1\\1\end{bmatrix}$ and $X_3 = \begin{bmatrix}-1\\1\\0\end{bmatrix}$.

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(a) Let A = (x² - 4|: -1 < x < 1}. Find supremum and infimum and maximum and minimum for A.

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Supremum and infimum are known as the least upper bound and greatest lower bound respectively.Supremum of a set is the least element of the set that is greater than all other elements of the set. We use the symbol ∞ to represent the supremum.Infimum of a set is the greatest element of the set that is smaller than all other elements of the set. We use the symbol - ∞ to represent the infimum

A = {(x² - 4) / (x² + 2) : -1 < x < 1}.Now, we need to find the supremum and infimum and maximum and minimum for A. . Now, we will find the derivative of f(x) = (x² - 4) / (x² + 2). To differentiate the given function, we can use the Quotient Rule for the differentiation of two functions.Using Quotient Rule, we get;[f(x)]' = [ (x² + 2) . 2x - (x² - 4) . 2x ] / (x² + 2)²= [4x / (x² + 2)² ] . (x² - 1)Put [f(x)]' = 0∴ [4x / (x² + 2)² ] . (x² - 1) = 0Or, x = 0, ±1 When x = -1, then f(x) = (-3) / 3 = -1. When x = 0, then f(x) = -4 / 2 = -2When x = 1, then f(x) = (-3) / 3 = -1.

Now, let's make the sign chart for f(x).x -1 0 1f(x) -ve -ve -ve. Thus, we can observe that the function is decreasing from (-1, 0) and (0, 1).∴ Maximum = f(-1) = -1, Minimum = f(1) = -1.Both the maximum and minimum values are -1. Let's find the supremum and infimum.S = {f(x): -1 < x < 1}Let's consider f(x) as y.Now, y = (x² - 4) / (x² + 2) ⇒ y(x² + 2) = x² - 4 ⇒ xy² + 2y - x² + 4 = 0. Now, the discriminant of this equation is;D = (2)² - 4y(-x² + 4) = 4x² - 16y.The roots of the given equation are;y = [-2 ± √D ] / 2x²Since x ∈ (-1, 1), √D ≤ 4√(1) = 4. Also, since y < 0, we can take the negative root.

So, y = [-2 - 4] / 2x² = -3 / x². For x ∈ (-1, 0), y ∈ (-∞, -2/3]For x ∈ (0, 1), y ∈ [-2/3, -∞). Thus, we can observe that -2/3 is the supremum of S and -∞ is the infimum of S.Thus, the given set A is Maximum = f(-1) = -1, Minimum = f(1) = -1, Supremum = -2/3 and Infimum = -∞.Hence, the solution.

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The maximum value of the set A is -3.

The minimum value of the set A is -4.

The supremum of the set A is -3.

The infimum of the set A is -4.

Maximum and minimum values:

Taking the derivative of the function with respect to x, we have:

f'(x) = 2x

Setting f'(x) = 0 to find critical points:

2x = 0

x = 0

We evaluate the function at the critical points and the endpoints of the interval:

f(-1) = (-1)² - 4 = -3

f(0) = (0)² - 4 = -4

f(1) = (1)² - 4 = -3

We can see that the maximum value within the interval is -3, and the minimum value is -4.

The supremum is the least upper bound, which means the largest possible value that is still within the set A.

The supremum is -3, as there is no value greater than -3 within the set.

The infimum is the greatest lower bound, which means the smallest possible value that is still within the set A.

The infimum is -4, as there is no value smaller than -4 within the set.

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purchased a total of 11 novels and magazines that have a combined selling price of $20, how many novels did she purchase?

Answers

The number of novels purchased was 9 novels.

Let the number of novels purchased be x and the number of magazines purchased be y.

Hence, [tex]x + y = 11.[/tex]

Let the selling price of novels be a and that of magazines be b.

Therefore, [tex]ax + by = 20.[/tex]

Similarly, given the price of magazines and novels as shown below:

[tex]a=  2\\b = 1[/tex]

We can use the given equations above to find the number of novels purchased.

To find the value of x, we substitute the value of a and b into the equations,

[tex]ax + by = $20$2x + $1y \\= $20[/tex]

We can also use the equation we found from [tex]x + y = 11,[/tex] and solve for [tex]y:y = 11 - x[/tex]

We can now substitute this value of y into the equation[tex]2x + 1y = 202x + 1(11 - x) \\= 201x \\=9x \\= 9 novels[/tex]

Therefore, the number of novels purchased was 9 novels.

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The function f(x) = 2x³ − 27x² + 48x + 9 has one local minimum and one local maximum. This function has a local minimum at x = ___
with function value ____
and a local maximum at x = ____
with function value_____

Answers

To find the local minimum and local maximum of a function, we need to locate the critical points where the derivative of the function is equal to zero or undefined. In this case, we can start by finding the derivative of f(x). Taking the derivative of f(x) = 2x³ - 27x² + 48x + 9 gives us f'(x) = 6x² - 54x + 48.

Next, we set f'(x) equal to zero and solve for x to find the critical points. By solving the quadratic equation 6x² - 54x + 48 = 0, we can find the values of x that correspond to the critical points. The solutions to the equation will give us the x-coordinates of the local minimum and local maximum.

Once we have the critical points, we can evaluate the function f(x) at these points to find the corresponding function values. The point with the lower function value will be the local minimum, and the point with the higher function value will be the local maximum. By substituting the critical points into f(x), we can determine the specific values of x and the corresponding function values for the local minimum and local maximum of the given function.

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For the function f(x)=x4 +2x³-5x² +10, determine: all critical and inflection points, all local and global extrema, and be sure to give y-values as well as exact x-values

Answers

The critical points are (0, 10), (-2.19, -18.61), and (1.19, 9.06). The inflection points are (-0.57, 10.15) and (0.57, 9.82). The local maximum is at x = 0 with a y-value of 10, and the local minima are at x = -2.19 and x = 1.19 with y-values of -18.61 and 9.06, respectively. There are no global extrema.

The first derivative is f'(x) = 4x^3 + 6x^2 - 10x, and the second derivative is f''(x) = 12x^2 + 12x - 10.

To find critical points, we set f'(x) = 0 and solve for x:

4x^3 + 6x^2 - 10x = 0.

By factoring, we can simplify the equation to:

2x(x^2 + 3x - 5) = 0.

This gives us critical points at x = 0 and x = (-3 ± √29)/2.

To find the inflection points, we set f''(x) = 0 and solve for x:

12x^2 + 12x - 10 = 0.

Using the quadratic formula, we find two possible solutions:

x = (-1 ± √7)/3.

Now, let's analyze the nature of these points:

At x = 0, the second derivative is positive, indicating a local minimum.

At x = (-3 + √29)/2, the second derivative is positive, indicating a local minimum.

At x = (-3 - √29)/2, the second derivative is negative, indicating a local maximum.

At x = (-1 ± √7)/3, the second derivative changes sign, indicating inflection points.

To find the y-values at these points, substitute the x-values back into the original function f(x).

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find the local maximal and minimal of the Function give below in the interval (-π, π)
f(x) = sin²(x) cos 00

Answers

The function f(x) = sin²(x) cos(2x) has local maxima and minima in the interval (-π, π).  The critical points are local maxima or minima. If f''(x) > 0 at a critical point, it is a local minimum, and if f''(x) < 0, it is a local maximum.

To find the local maxima and minima of the function, we need to determine the critical points and analyze the behavior of the function around those points.

First, let's find the critical points by taking the derivative of f(x) and setting it equal to zero:

f'(x) = 2sin(x)cos(x)cos(2x) - sin²(x)(-sin(2x)) = 2sin(x)cos(x)cos(2x) + sin²(x)sin(2x)

Setting f'(x) = 0, we have:

2sin(x)cos(x)cos(2x) + sin²(x)sin(2x) = 0

Simplifying this equation is not straightforward, and it does not have a simple analytical solution. Therefore, we can use numerical methods or graphing tools to approximate the critical points.

Once we have the critical points, we can evaluate the second derivative, f''(x), to determine whether the critical points are local maxima or minima. If f''(x) > 0 at a critical point, it is a local minimum, and if f''(x) < 0, it is a local maximum.

However, since finding the critical points and evaluating the second derivative of the given function involves complex trigonometric calculations, it would be best to use numerical methods or graphing tools to find the local maxima and minima in the given interval (-π, π).

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find the value of the derivative (if it exists) at the indicated extremum. (if an answer does not exist, enter dne.) f(x) = x2 x2 2

Answers

The value of the derivative at the indicated extremum is 0. The given function has maximum extremum at x = 0.

The function is given by;f(x) = x² / (x² + 2)Let us find the derivative of the given function, using the quotient rule;dy/dx = [(x² + 2).(2x) - x².(2x)] / (x² + 2)²= [2x(x² + 2 - x²)] / (x² + 2)²= [2x.2] / (x² + 2)²= 4x / (x² + 2)²

For the given function to have extremum, dy/dx = 0We have,dy/dx = 4x / (x² + 2)² = 0 => 4x = 0=> x = 0At x = 0, the function has extremum.

Let's find what type of extremum the function has.

Second derivative test;d²y/dx² = [(d/dx) {4x / (x² + 2)²}] = [(8x³ - 24x) / (x² + 2)³]Let's find the value of second derivative at x = 0;d²y/dx² = (8*0³ - 24*0) / (0² + 2)³= -3/4

As the value of the second derivative is negative, the function has a maximum at x = 0.Now, let us find the value of the derivative at the indicated extremum.x = 0dy/dx = 4x / (x² + 2)²= 4(0) / (0² + 2)²= 0The value of the derivative at the indicated extremum is 0.

Hence, the main answer is 0. Summary: The value of the derivative at the indicated extremum is 0. The given function has maximum extremum at x = 0.

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Solve the following initial-value problems and compare the numerical solutions obtained with the Euler's method using the values of h = 0.1 and h = 0.2. Compare the results to the actual values. (a) y'=1+x², 0≤x≤1, y(0) = 0, y(x) tan x. =

Answers

The numerical solution obtained when h = 0.2 is more accurate compared to the numerical solution obtained when h = 0.1. Therefore, Euler's method is more accurate when h is smaller.

Given differential equation is y' = 1 + x², with initial conditions y(0) = 0.To find the value of y, let's use Euler's method which is given by:yi+1 = yi + h * f(xi, yi)Where h is the step size which is equal to 0.1 and 0.2.f(xi, yi) = 1 + x²i. Now, let's find the numerical values of y using Euler's method and compare them to actual values.a) y'=1+x², 0≤x≤1, y(0) = 0, y(x) tan x.

Given differential equation is y' = 1 + x², with initial conditions y(0) = 0.So, y(0) = 0. Therefore, we have to find y(x) using Euler's method with h = 0.1 and h = 0.2.

The value of x lies in the range 0 to 1.h = 0.1

Using Euler's method, we get:yi+1 = yi + h * f(xi, yi)Where f(xi, yi) = 1 + x²i

Now,x0 = 0y0 = 0xi = x0 + ih = 0.1x1 = x0 + 2h = 0.2y1 = y0 + h * f(x0, y0)y1 = 0 + 0.1 * (1 + (0)²) = 0.1x2 = x0 + 3h = 0.3y2 = y1 + h * f(x1, y1)y2 = 0.1 + 0.1 * (1 + (0.2)²) = 0.130x3 = x0 + 4h = 0.4y3 = y2 + h * f(x2, y2)y3 = 0.130 + 0.1 * (1 + (0.3)²) = 0.1710x4 = x0 + 5h = 0.5y4 = y3 + h * f(x3, y3)y4 = 0.1710 + 0.1 * (1 + (0.4)²) = 0.2150x5 = x0 + 6h = 0.6y5 = y4 + h * f(x4, y4)y5 = 0.2150 + 0.1 * (1 + (0.5)²) = 0.2640x6 = x0 + 7h = 0.7y6 = y5 + h * f(x5, y5)y6 = 0.2640 + 0.1 * (1 + (0.6)²) = 0.3180x7 = x0 + 8h = 0.8y7 = y6 + h * f(x6, y6)y7 = 0.3180 + 0.1 * (1 + (0.7)²) = 0.3770x8 = x0 + 9h = 0.9y8 = y7 + h * f(x7, y7)y8 = 0.3770 + 0.1 * (1 + (0.8)²) = 0.4410x9 = x0 + 10h = 1.0y9 = y8 + h * f(x8, y8)y9 = 0.4410 + 0.1 * (1 + (0.9)²) = 0.5100So, the value of y at x = 1 is 0.5100 when h = 0.1.

Now,h = 0.2Using Euler's method, we get:yi+1 = yi + h * f(xi, yi)Where f(xi, yi) = 1 + x²iNow,x0 = 0y0 = 0xi = x0 + ih = 0.2x1 = x0 + 2h = 0.4y1 = y0 + h * f(x0, y0)y1 = 0 + 0.2 * (1 + (0)²) = 0.2x2 = x0 + 3h = 0.6y2 = y1 + h * f(x1, y1)y2 = 0.2 + 0.2 * (1 + (0.4)²) = 0.36x3 = x0 + 4h = 0.8y3 = y2 + h * f(x2, y2)y3 = 0.36 + 0.2 * (1 + (0.6)²) = 0.568x4 = x0 + 5h = 1.0y4 = y3 + h * f(x3, y3)y4 = 0.568 + 0.2 * (1 + (0.8)²) = 0.848

So, the value of y at x = 1 is 0.848 when h = 0.2.Now, let's find the actual value of y(x).y' = 1 + x²Integrating both sides w.r.t x, we get:y = x + (1/3) x³ + cNow, using initial condition y(0) = 0, we get c = 0Therefore,y = x + (1/3) x³Now, y(1) = 1 + (1/3)

Therefore, y(1) = 1.3333Now, compare the numerical solutions obtained with the Euler's method using the values of h = 0.1 and h = 0.2 and actual values. Value of y(1)Actual value of y at x = 1 is 1.3333.Value of y(1) when h = 0.1 is 0.5100Value of y(1) when h = 0.2 is 0.848So, we can see that the actual value of y(1) is 1.3333. Value of y(1) when h = 0.2 is closer to the actual value of y(1).

Hence, we can say that the numerical solution obtained when h = 0.2 is more accurate compared to the numerical solution obtained when h = 0.1. Therefore, Euler's method is more accurate when h is smaller.

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We see that Euler's method with h = 0.1 provides more accurate results compared to the Euler's method with h = 0.2. This is because when h is smaller, the step size becomes smaller and hence the approximation becomes better.

Given that y'=1+x² and 0 ≤ x ≤ 1 and y(0) = 0, we need to solve the initial value problem and compare the numerical solutions obtained with Euler's method using the values of h = 0.1 and h = 0.2.

Compare the results to the actual values. (a) y'=1+x², 0≤x≤1, y(0) = 0, y(x) tan x. Solution:Given, y'=1+x².Using Euler's method, we have:y1 = y0 + hf(x0, y0), where f(x, y) = 1 + x².From the given data, x0 = 0, y0 = 0.Using h = 0.1, we gety1 = y0 + hf(x0, y0) = 0 + 0.1(1 + 0²) = 0.1

Similarly, y2 = y1 + hf(x1, y1) = 0.1 + 0.1(1 + 0.1²) = 0.1201 and so on.

Now, let us tabulate the values of x and y(x) using h = 0.1. x y(x) Euler's method tan(x)

Absolute error 0 0 0 0 0.00 0.1 0.1 0.1 0.001 0.002 0.2 0.1201 0.2027 0.0826 0.0015 0.3 0.1513 0.3163 0.1650 0.0015 0.4 0.1941 0.4685 0.2744 0.0084 0.5 0.2507 0.6694 0.4188 0.0174 0.6 0.3233 0.9322 0.6089 0.0238 0.7 0.4158 1.2767 0.8609 0.0262 0.8 0.5330 1.7298 1.1941 0.0307 0.9 0.6819 2.3253 1.6434 0.0385 1.0 0.8701 3.1071 2.2370 0.0469

Now, using h = 0.2, we gety1 = y0 + hf(x0, y0) = 0 + 0.2(1 + 0²) = 0.2Similarly, y2 = y1 + hf(x1, y1) = 0.2 + 0.2(1 + 0.2²) = 0.248and so on.

Now, let us tabulate the values of x and y(x) using h = 0.2. x y(x) Euler's method tan(x)

Absolute error 0 0 0 0 0.00 0.2 0.248 0.2027 0.0453 0.0088 0.4 0.3875 0.4685 0.0809 0.0809 0.6 0.5655 0.9322 0.3667 0.1989 0.8 0.8082 1.7298 0.9216 0.1134 1.0 1.1592 3.1071 1.9479 0.1923

Comparing the actual values and the Euler's method values with h = 0.1 and h = 0.2, we get: x tan(x) Euler's method with h = 0.1 Euler's method with h = 0.2 Actual y(x) Absolute error with h = 0.1

Absolute error with h = 0.2 0 0 0 0 0 0 0 0 0 0.1 0.1003 0.1000 0.1003 0.0003 0.0003 0.2 0.2027 0.1201 0.2480 0.0826 0.0453 0.3 0.3163 0.1513 0.3493 0.1650 0.0330 0.4 0.4685 0.1941 0.3875 0.2744 0.0809 0.5 0.6694 0.2507 0.5217 0.4188 0.1484 0.6 0.9322 0.3233 0.5655 0.6089 0.1989 0.7 1.2767 0.4158 0.9998 0.8609 0.2769 0.8 1.7298 0.5330 1.1724 1.1941 0.5574 0.9 2.3253 0.6819 1.6149 1.6434 0.9336 1.0 3.1071 0.8701 2.2370 2.2370 1.2670

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