Which of the following statements is (are) true?
a. The standard deviation is resistant to extreme values.
b. The interquartile range is resistant to extreme values.
c. The median is resistant to extreme values.
d. Both b and c.

False, the null distribution is the distribution of the test statistic assuming the null hypothesis is true; it is mound-shaped and symmetric about the null mean.

The null distribution is the distribution of the test statistic under the assumption that the null hypothesis is true. However, its shape and symmetry are not necessarily predetermined.

The null distribution can take various forms depending on the specific test and the underlying data. It may or may not be mound shaped or symmetric about the null mean. The shape and characteristics of the null distribution are determined by the specific hypothesis being tested, the sample size, and other factors.

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The correct order of integration and associated limits for the given triple integral I = ∫∫∫[f(x, y, z)] dx dy dz with limits: 0 ≤ x ≤ 1 - 2y, 0 ≤ y, 0 ≤ z ≤ 2y²

The correct order of integration and associated limits for the given triple integral, let's first examine the region of integration R and its slices in different planes.

Region R is defined by 0 ≤ z ≤ 2y² and 0 ≤ x ≤ 1 - 2y.

1.Slices in the zy-plane: In the zy-plane, z is restricted to 0 ≤ z ≤ 2y², and y is unrestricted. Therefore, the integral can be written as:

I = ∫∫∫ f(x, y, z) dV = ∫∫∫ f(x, y, z) dz dy dx

2.Slices in the zx-plane: In the zx-plane, z is unrestricted, and x is restricted to 0 ≤ x ≤ 1 - 2y. Therefore, the integral can be written as:

I = ∫∫∫ f(x, y, z) dV = ∫∫∫ f(x, y, z) dx dz dy

3.Slices in the yz-plane: In the yz-plane, y is unrestricted, and z is restricted to 0 ≤ z ≤ 2y². Therefore, the integral can be written as:

I = ∫∫∫ f(x, y, z) dV = ∫∫∫ f(x, y, z) dy dz dx

Considering the given hint, we can choose any of the above orders of integration as all of them are correct ways to write the integral. However, for simplicity, let's choose the order: I = ∫∫∫ f(x, y, z) dz dy dx.

Now, let's determine the limits of integration for each variable in this order:

∫∫∫ f(x, y, z) dz dy dx = ∫∫ [∫[f(x, y, z) dz] from z=0 to z=2y²] dy dx

The innermost integral with respect to z is evaluated from 0 to 2y². The next integral with respect to y is evaluated from 0 to a certain limit determined by the region R. Finally, the outermost integral with respect to x is evaluated from 0 to 1 - 2y.

Therefore, the order of integration and the associated limits for the triple integral are:

I = ∫∫∫ f(x, y, z) dz dy dx

I = ∫∫ [∫[f(x, y, z) dz] from z=0 to z=2y²] dy dx

I = ∫∫∫[f(x, y, z)] dx dy dz with limits: 0 ≤ x ≤ 1 - 2y, 0 ≤ y, 0 ≤ z ≤ 2y²

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Answer:

There is a 1/3 (or 0.33%) probability of rolling a number greater than or equal to 5.

Step-by-step explanation:

First, find what numbers are greater than or equal to 5:

5 and 6

Find what options you can get on a number cube:

1, 2, 3, 4, 5, and 6

Out of the 6 possible outcomes, there are only 2 that will get a number greater than or equal to 5. Write this as a fraction:

2/6

Simplify:

1/3

14. (1112 - 6vw - 3wa)-(-702 + vw + 13w) = 1814 - 7vw - 3wa - 13w

15. The equivalent of the expression (5a + 26 - 4c) is 25a2 + 10ab - 20ac + 482 - 86c + 1602 + c.

16.  (b + b)(4 - 5)(25 - 8) = -34

14. Simplify (1112 - 6vw - 3wa)-(-702 + vw + 13w).

Given expression is (1112 - 6vw - 3wa)-(-702 + vw + 13w)

⇒ 1112 - 6vw - 3wa + 702 - vw - 13w

⇒ 1814 - 7vw - 3wa - 13w

15. We are to find the equivalent of the expression (5a + 26 - 4c).

a. 25a2 + 20ab - 40ac +482 - 16bc + 1602

b. 25a2 + 10ab - 20ac + 482 - 86C + 1602

c. 25a2 + 482 + 1602

d. 10a + 4b-8c5a + 26 - 4c

= 5a - 4c + 26 = 25a2 - 20ac +482 - 4c2 + 52 - 8ac

= 25a2 - 20ac + 482 - 4c2 + 10a - 8c = Option (b)

⇒ 25a2 + 10ab - 20ac + 482 - 86c + 16c2 + c.

16. Expand and simplify. (b + b)(4 - 5)(25 - 8)

Given expression is (b + b)(4 - 5)(25 - 8) = 2b(-1)(17) = -34

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Answer:

Step-by-step explanation:

Given equation:

q is independent variable, we need to solve it for s:

Correct choice is

a) The differential equation is y = [tex]e^{(4x)/x}[/tex] + 3 b) The particular solution is y = [tex]e^{x^3 - x}[/tex] - 1 c) The particular solution to the differential equation is given by the equations is y = 2x or y = -2x.

a) To find the general solution to the differential equation:

x(dy/dx) + 3y = [tex](e^{4x})/{x^2}[/tex]

We can start by rearranging the equation:

dy/dx = [[tex](e^{4x})/{x^2}[/tex] - 3y]/x

This equation is linear, so we can use an integrating factor to solve it. The integrating factor is given by:

μ(x) = e^(∫(1/x) dx) = [tex]e^{ln|x|}[/tex] = |x|

Multiplying both sides of the equation by the integrating factor:

|x| * dy/dx - 3|xy| = [tex]e^{(4x)/x}[/tex]

Now, let's integrate both sides with respect to x:

∫(|x| * dy/dx - 3|xy|) dx = ∫([tex]e^{(4x)/x}[/tex]) dx

Using the properties of absolute values and integrating term by term:

∫(|x| * dy) - 3∫(|xy|) dx = ∫([tex]e^{(4x)/x}[/tex]) dx

Integrating each term separately:

∫(|x| * dy) = ∫([tex]e^{(4x)/x}[/tex]) dx + 3∫(|xy|) dx

To integrate ∫(|x| * dy), we need to know the form of y. Let's assume y = y(x). Integrating ∫[tex](e^{4x)/x}[/tex] dx gives us a natural logarithm term.

Integrating 3∫(|xy|) dx can be done using different cases for the absolute value of x.

By solving these integrals and rearranging the equation, you can find the general solution for y(x).

b) To find the particular solution to the differential equation:

dy/dx = (y + 1)(3x² - 1)

subject to the condition that y = 0 at x = 0.

We can solve this equation using separation of variables. Rearranging the equation:

dy/(y + 1) = (3x² - 1) dx

Now, let's integrate both sides:

∫(dy/(y + 1)) = ∫((3x² - 1) dx)

The left-hand side can be integrated using the natural logarithm function:

ln|y + 1| = x³ - x + C1

Solving for y, we have:

[tex]y + 1 = e^{x^3 - x + C1}\\y = e^{x^3 - x + C1} - 1[/tex]

Using the initial condition y = 0 at x = 0, we can find the particular solution. Substituting these values into the equation:

0 = [tex]e^{0 - 0 + C1}[/tex] - 1

1 = [tex]e^{C1}[/tex]

C1 = ln(1) = 0

Therefore, the particular solution is:

y = [tex]e^{x^3 - x}[/tex] - 1

c) To find the particular solution to the differential equation:

x(dy/dx) - y = y

subject to the condition that y = 2 at x = 1.

We can simplify the equation:

x(dy/dx) = 2y

Now, let's separate variables and integrate:

(1/y) dy = (1/x) dx

Integrating both sides:

ln|y| = ln|x| + C2

Simplifying further:

ln|y| = ln|x| + C2

ln|y| - ln|x| = C2

ln(|y/x|) = C2

|y/x| =  [tex]e^{C2}[/tex]

Since we are given the initial condition y = 2 at x = 1, we can substitute these values into the equation:

|2/1| = [tex]e^{C2}[/tex]

2 =    [tex]e^{C2}[/tex]

C2 = ln(2)

Therefore, the particular solution is:

|y/x| = [tex]e^{ln(2)}[/tex]

|y/x| = 2

Solving for y, we have two cases:

y/x = 2

y = 2x

y/x = -2

y = -2x

So, the particular solution to the differential equation is given by the equations:

y = 2x or y = -2x.

The complete question is:

a) Find the general solution to the differential equation

x dy/dx + 3y = (e⁴ˣ)/(x²)

b) Find the particular solution to the differential equation dy/dx = (y + 1)(3x² - 1)

subject to the condition that v = 0 at x = 0

c) Find the particular solution to the differential equation

x dy/dx (y) = y

subject to the condition that y = 2 at x = 1

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Answer: 40

Step-by-step explanation:

Answer: the second one/ \/36/6

Step-by-step explanation:

It cant be written any way else its already in its simplest form

Answer:

a=-2

Step-by-step explanation:

3a−6=−12

Step 1: Add 6 to both sides.

3a−6+6=−12+6

3a=−6

Step 2: Divide both sides by 3.

3a/ 3 = −6/ 3

a=−2

To prove that for any real number x, if  x²+ 7x < 0, then x < 0, we can use the properties of quadratic functions and inequalities.

By analyzing the quadratic expression, we can determine the conditions under which it is negative. This analysis shows that the inequality x²+ 7x < 0 holds true when x is less than 0. Consider the quadratic expression x² + 7x. To determine when this expression is negative, we can factor it as x(x + 7). According to the zero product property, this expression is equal to zero when either x or (x + 7) is equal to zero. Thus, the two critical points are x = 0 and x = -7.

Now, let's analyze the behavior of the quadratic expression in the intervals (-∞, -7), (-7, 0), and (0, +∞). Choose a test point from each interval, such as -8, -3, and 1, respectively. Evaluating the expression x²⁺7x for these test points, we find that for -8 and -3, the expression is positive, and for 1, it is positive as well.

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Answer:

Unbiased

Step-by-step explanation:

If b^ is equal to B this means that it is an unbiased estimator. When there is an absence of bias, we have an unbiased estimator. As an unbiased estimator it gives accurate information most of the time. The result it gives is not over estimated and also it is not underestimated.

Expected value = true value

Parameter estimates are correct on average

Thank you

Answer:

hi sh Sheet sh I monorailg Jericho improve Odom Ybor

Answer:

28.27

Step-by-step explanation:

A=πr2=π·32≈28.27433

Answer:

If you're using 3.14 for pi, it's 28.26

Step-by-step explanation:

Answer:

36.25 miles per gallon

Step-by-step explanation:

4/5 = .8

29/.8 = x/1

cross-multiply:

.8x = 29

x = 36.25

a. System in the matrix form is x' = Ax where A = [tex]\left[\begin{array}{ccc}0&1\\-5&6\end{array}\right][/tex] and x = [y, u].

b. The eigenvalues of the system are λ₁ = 5 and λ₂ = 1 and eigenvector are v₁ and v₂ = v₁, and v₁ is any non-zero value.

c. The general solution is equal to y(x) = c₁ × [tex]e^{(5x)[/tex] × [v₁] + c₂× [tex]e^{(x)[/tex]× [v₂].

a. Solution to the equation. y' - 5y + y = 32 is y(x) = c₁ × [tex]e^{(5x)[/tex] + c₂ × [tex]e^{(x)[/tex].

(a) To write the second order linear ODE as a system of two first order ODEs,

Introduce a new variable u = y'.

Then, we have,

u' = y'' - 5y + 6y

   = -5y + 6u

Now, write this as a system of two first order ODEs,

y' = u

u' = -5y + 6u

To express this system in matrix form,

Define the vector x = [y, u] and the matrix A = [tex]\left[\begin{array}{ccc}0&1\\-5&6\end{array}\right][/tex]

The system can then be written as,

x' = Ax

(b) To find the eigenvalues and eigenvectors of matrix A, solve the characteristic equation,

|A - λI| = 0

where I is the identity matrix.

Substituting the values of A, we have,

[tex]|\left[\begin{array}{ccc}0&1\\-5&6\end{array}\right][/tex] [tex]-\lambda\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]|[/tex] = 0

[tex]\left[\begin{array}{ccc}-\lambda&1\\-5&6-\lambda\end{array}\right][/tex] = 0

(-λ)(6-λ) - (-5)(1) = 0

λ²- 6λ + 5 = 0

Factoring the quadratic equation, we get,

(λ - 5)(λ - 1) = 0

So the eigenvalues are λ₁ = 5 and λ₂ = 1.

To find the corresponding eigenvectors,

solve the equation (A - λI)v = 0 for each eigenvalue.

Let us start with λ = 5

(A - 5I)v = 0

[tex]|\left[\begin{array}{ccc}1&1\\-5&6\end{array}\right]|[/tex] v = 0

v₁ + v₂ = 0

-5v₁ + v₂ = 0

From the first equation, we get v₂ = -v₁.

Substituting this into the second equation, we have -5v₁ - v₁ = 0,

which simplifies to -6v₁ = 0.

This implies v₁ = 0, and consequently, v₂ = 0.

So, for λ = 5, the eigenvector is v₁ = 0 and v₂ = 0.

Now, let us find the eigenvector for λ = 1.

(A - I)v = 0

[tex]|\left[\begin{array}{ccc}-1&1\\-5&5\end{array}\right][/tex] v = 0

-v₁ + v₂ = 0

-5v₁ + 5v₂ = 0

From the first equation, we get v₂ = v₁.

Substituting this into the second equation, we have -5v₁ + 5v₁ = 0,

which simplifies to 0 = 0.

This implies that v₁ can be any non-zero value.

So, for λ = 1, the eigenvector is v₁ and v₂ = v₁, where v₁ is any non-zero value.

(e) The general solution to the second order ODE can be expressed using the eigenvalues and eigenvectors as follows,

y(x) = c₁ ×[tex]e^{(\lambda_{1} x)[/tex] × v₁ + c₂ × [tex]e^{(\lambda_{2} x)[/tex]× v₂

Plugging in the values we found earlier, the general solution becomes,

y(x) = c₁ × [tex]e^{(5x)[/tex] × [v₁] + c₂× [tex]e^{(x)[/tex]× [v₂]

where [v₁] and [v₂] are the eigenvectors corresponding to the eigenvalues λ₁ = 5 and λ₂ = 1 respectively.

(a) To find the solution to the equation y' - 5y + y = 32,

Use the general solution obtained above.

Comparing the equation with the standard form y' - 5y + 6y = 0,

The equation corresponds to the case where λ₂ = 1.

Substitute λ = 1, v₁ = 1, and v₂ = 1 into the general solution.

y(x) = c₁ × [tex]e^{(5x)[/tex] × [1] + c₂ × [tex]e^{(x)[/tex] × [1]

Simplifying this expression, we have,

y(x) = c₁ × [tex]e^{(5x)[/tex] + c₂ × [tex]e^{(x)[/tex]

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Answer:

55

Step-by-step explanation:

To find this, simply plug 10 in for x.

6(10)-5

6*10=60

60-5=55

In a river bank

you can put the answers from the end

1.) 1/12

2.) 2/13

3.) 8/41

4.) 1/4

5.) 11/74

6.) 2/35

7.) 15/58

8.) 5/18

9.) 1/7

10.) 1/13

The exact distance between the points (8, -3) and (-2, 4) can be calculated using the distance formula in mathematics.

The formula for finding the distance between two points (x1, y1) and (x2, y2) in a two-dimensional Cartesian coordinate system is given by: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2). Using the coordinates (8, -3) and (-2, 4), we can substitute the values into the distance formula: Distance = sqrt((-2 - 8)^2 + (4 - (-3))^2) = sqrt((-10)^2 + (7)^2) = sqrt(100 + 49) = sqrt(149) ≈ 12.207

Therefore, the exact distance between the points (8, -3) and (-2, 4) is approximately 12.207 units.

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Answer:

D) 8

Step-by-step explanation:

We're looking for [tex]x[/tex], which means the 7x and 5x will have the SAME x.

7(8) = 56

For the C corner, it is cornered as a 90 degree angle, so that means the (7x) + 34 degrees NEEDS to equal 90.

7(8) = 56 + 34 = 90!!

Now we have to see if 5(8) is correct

5(8) = 40 + 50 [The degree mark] = WHICH EQUALS 90 TOO..

Therefore, the answer is D) 8

Answer:

x²+11x+30

Step-by-step explanation:

for this question, all we need to do is multiply (x+5) by (x+6)

we can use the distributive property.

(x+5)(x+6)

x² + 5x + 6x + 30

x² + 11x + 30

Answer:

True

Step-by-step explanation:

Brainliest?

Answer:

I Believe the answer is True

Step-by-step explanation:

Answer:

Im so sorry im late! The answer is A i just took the quiz!

Step-by-step explanation:

The correct dot plot is given in option 1.

Any data that may be shown as dots or tiny circles is called a dot plot. Given that the height of the bar created by the dots indicates the numerical value of each variable, it is comparable to a bar graph or a simple histogram. Little amounts of data are shown using dot plots.

As per the given data:

There are 4 options, with each option represented by a diagram also the number and frequency table is given

Number:     3 4 5 7 8

Frequency: 3 2 4 2 3

We can find the correct diagram of the dot plot by observing the number of cross against each value of the number on the line and then matching the obtained value with the given number and frequency table.

Hence, the correct dot plot is given in option 1.

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Answer:

Given ABCD ~ EFGH

FG = BC(EF/AB)

FG = 7(9/6)

FG = 63/6

FG = 10.5

GH = CD(EF/AB)

GH = 11(9/6)

GH = 99/6

GH = 16.5

EH = AD(EF/AB)

EH = 12(9/6)

EH = 108/6

EH = 18

The function y = sin(kt) satisfies the differential equation y'' - 64y = 0 for pospositiveypospositiveyitiveitive values of k that are multiples of 8.

To determine the values of k for which the function y = sin(kt) satisfies the given differential equation, we need to substitute y into the equation and solve for k. Let's start by finding the first and second derivatives of y with respect to t.
The first derivative of y with respect to t is y' = kcos(kt), and the second derivative is y'' = -k^2sin(kt). Substituting these derivatives into the differential equation gives us:
(-k^2sin(kt)) - 64sin(kt) = 0Simplifying the equation, we get:
sin(kt) = -64*sin(kt)/k^2
We can divide both sides of the equation by sin(kt) (assuming sin(kt) is not zero) to get:
1 = -64/k^2
Solving for k^2, we find k^2 = -64. Since k must be positive, there are no positive values of k that satisfy this equation. Therefore, there are no positive values of k for which the function y = sin(kt) satisfies the given differential equation y'' - 64y = 0.

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Answer:

3=20

4=15

5=12

Step-by-step explanation:

Answer

c

Step-by-step explanation:

It is not d because of the way that it is formed

Then, the width of the pool covered by the concrete walkway is 16 + 2c feet (2c feet on each side), and the length of the pool covered by the concrete walkway is 26 + 2c feet (2c feet on each end).

We have given a rectangular swimming pool with a length of 26 feet and a width of 16 feet. We need to find the perimeter of the wood deck that surrounds the concrete walkway of width c around the pool and extends w feet on all sides.

Let's solve the given problem as follows:Firstly, let's calculate the dimensions of the concrete walkway. Let the width of the concrete walkway be 'c' feet.

Then, the width of the pool covered by the concrete walkway is 16 + 2c feet (2c feet on each side), and the length of the pool covered by the concrete walkway is 26 + 2c feet (2c feet on each end).

So, the dimensions of the pool and concrete walkway are (26 + 2c) ft. x (16 + 2c) ft.The dimensions of the wood deck that surrounds the concrete walkway by w feet on all sides will be (26 + 2c + 2w) ft. x (16 + 2c + 2w) ft.Now, let's write the expression for the perimeter of the wood deck.P = 2(Length + Width)P = 2[(26 + 2c + 2w) + (16 + 2c + 2w)]P = 2[42 + 4c + 4w]P = 84 + 8c + 8wThe expression for the perimeter of the wood deck is 84 + 8c + 8w. Hence, the answer is 84 + 8c + 8w.

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a. The relation R defined on the set S = {1, 2, 3, ..., 18, 19} is an equivalence relation. It is reflexive, symmetric, and transitive, b. The equivalence class of 1, denoted [1], consists of the perfect squares in S: {1, 4, 9, 16}, c. The equivalence classes with more than one element are [1], [2], [3], ..., [18], and [19]. Each equivalence class represents a set of numbers that are squares of integers.

a. To show that the relation R is an equivalence relation, we need to demonstrate that it is reflexive, symmetric, and transitive.

i. Reflexive: For R to be reflexive, every element in S must be related to itself. Since the square of any integer is still an integer, xRx holds for all x in S, satisfying reflexivity.

ii. Symmetric: For R to be symmetric, if xRy holds, then yRx must also hold. Since multiplication is commutative, if xy is a square of an integer, then yx is also a square of an integer. Hence, R is symmetric.

iii. Transitive: For R to be transitive, if xRy and yRz hold, then xRz must also hold. Since the product of two squares of integers is itself a square of an integer, xz is also a square of an integer. Thus, R is transitive.

b. To find the equivalence class of 1, denoted [1], we determine all elements in S that are related to 1 under R. In this case, [1] consists of the perfect squares in S: {1, 4, 9, 16}.

c. The equivalence classes with more than one element are [1], [2], [3], ..., [18], and [19]. Each equivalence class represents a set of numbers that are squares of integers. The equivalence class [1] includes all perfect squares in S, while the other equivalence classes consist of a single element, which are non-square integers.

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