Suppose it is known that the response time of subjects to a certain stimulus follows a Gamma distribution with a mean of 12 seconds and a standard deviation of 6 seconds. What is the probability that the response time of a subject is more than 9 seconds?

Step-by-step explanation:

A. Each point in the image has the same x-coordinate as the corresponding point

in the figure.

D. Each point in the image moves the same distance and direction from the

figure.

These two statements are true about a figure and an image created by a translation. When a figure is translated, every point in the figure is moved the same distance and direction. This means that each point in the image has moved the same way as its corresponding point in the figure. Additionally, since a translation only involves moving a figure without changing its shape or size, the image and figure are the same shape and size, but just in different positions. As such, statement B and C are not true for figures and images created by translation.

To graph the image G'R'A'M', we need to add 10 to each x-coordinate and subtract 2 from each y-coordinate:

G': (-9+10, -9-2) = (1,-11)

R': (-8+10, -6-2) = (2,-8)

A': (-4+10, -6-2) = (6,-8)

M': (-5+10, -9-2) = (5,-11)

Graphing these points and connecting them gives us parallelogram G'R'A'M'.

Answer:

Set your calculator to degree mode.

Draw a line from point O to a vertex of this octagon to form a right triangle.

tan(67.5°) = 17/x, so x = 17/tan(67.5°)

Area = (1/2)(34/tan(67.5°))(8)(17) = 957.7

[tex]\underset{ \textit{angle in degrees} }{\textit{area of a regular polygon}}\\\\ A=na^2\cdot \tan\left( \frac{180}{n} \right) ~~ \begin{cases} n=sides\\ a=apothem\\[-0.5em] \hrulefill\\ n=8\\ a=17 \end{cases}\implies A=(8)(17)^2\tan\left( \frac{180}{8} \right) \\\\\\ A=2312\tan(22.5^o)\implies A\approx 957.7[/tex]

Make sure your calculator is in Degree mode.

In this case, we can use a standard normal distribution table or a calculator to find that the value for P75 is approximately 12.1.

Here is a normal bell curve for a normal distribution with mean (μ) of 8 and standard deviation (σ) of 5:

The horizontal axis represents the range of possible values for the variable X, and the vertical axis represents the probability density of each value occurring.

The curve is symmetrical around the mean (μ=8), which is located at the peak of the curve. The standard deviation (σ=5) determines the width of the curve, with wider curves having larger standard deviations.

The shaded area under the curve represents the probability of a value falling within a certain range. For example, the area under the curve between X=3 and X=13 represents the probability of a value falling within 1 standard deviation of the mean, which is approximately 68%.

The percentile of a certain value can also be determined from the normal distribution. For example, the 75th percentile (represented as P75) is the value that separates the lowest 75% of values from the highest 25%. In this case, we can use a standard normal distribution table or a calculator to find that the value for P75 is approximately 12.1.

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Answer:

90 or 270

Step-by-step explanation:

To determine the bearing of the tree from the student, we need to use the reference direction of North. Typically, bearings are measured in degrees clockwise from North.

If the student stands 15 meters away directly west of the tree, then we can draw a line connecting the student and the tree, which would be a line going directly east to west.

Since we want to find the bearing of the tree from the student, we need to measure the angle between the line connecting the student and the tree and the North direction. Since the line between the student and the tree is going directly east to west, this angle will be either 90 degrees or 270 degrees, depending on which direction we consider as North.

If we consider North to be directly above the student, then the bearing of the tree from the student would be 90 degrees, since the line connecting the student and the tree is perpendicular to the North direction.

If we consider North to be directly below the student, then the bearing of the tree from the student would be 270 degrees, since the line connecting the student and the tree is perpendicular to the South direction (which is opposite to North).

Therefore, depending on the reference direction of North that we choose, the tree's bearing from the student will be either 90 degrees or 270 degrees.

Answer:lol it was 7

Step-by-step explanation:

The following parts can be answered by the concept of hypothesis test.

a. To reduce the risk of a Type I error, a researcher can control the significance level or alpha level of their hypothesis test. By setting a lower alpha level (such as 0.01 instead of 0.05), the researcher is decreasing the likelihood of rejecting the null hypothesis when it is actually true.

b. To reduce the standard error, a researcher can increase the sample size of their study. As the sample size increases, the standard error decreases because there is less variability in the sample means. Additionally, ensuring that the sample is representative of the population can also help reduce standard error.

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The general solution of the differential equation will be in the form: y(t) =[tex]C1 * e^(r1 * t) + C2 * e^(r2 * t) + C3 * e^(r3 * t) + C4 * e^(r4 * t) + C5 * e^(r5 * t),[/tex] where C1, C2, C3, C4, and C5 are arbitrary constants.

To find the general solution of the differential equation, we first need to find the characteristic equation by assuming a solution of the form y(t) = e^(rt). Plugging this into the differential equation, we get:

[tex]r^5 - 7r^4 + 13r^3 - 7r^2 + 12r = 0[/tex]

Factoring out an r term, we can simplify this to:

[tex]r(r^4 - 7r^3 + 13r^2 - 7r + 12) = 0[/tex]

We can solve for the roots of the polynomial using either factoring or the quadratic formula, but it turns out that there is only one real root, r = 1, with a multiplicity of 3, and two complex conjugate roots, r = 1 ± i. Therefore, the general solution is:

[tex]y(t) = C1 e^t + (C2 + C3 t + C4 t^2) e^(1+i)t + (C2 - C3 t + C4 t^2) e^(1-i)t + C5[/tex]

where C1, C2, C3, C4, and C5 are arbitrary constants to be determined by initial or boundary conditions. The last term, C5, represents the general solution to the homogeneous differential equation, since it contains no terms involving the roots of the characteristic equation.
To find the general solution of the given differential equation y(5) - 7y(4) + 13y'' - 7y' + 12y = 0, we first need to find the characteristic equation. The characteristic equation for this differential equation is:

[tex]r^5 - 7r^4 + 13r^3 - 7r^2 + 12r = 0.[/tex]
Now, we need to find the roots of this equation. Let's denote them as r1, r2, r3, r4, and r5.

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The derivative of f(x) = sin(x²) is f'(x) = 2x × cos(x²).

The chain rule is a rule of calculus used to find the derivative of a composition of functions. It allows us to differentiate a function that is constructed by combining two or more functions, where one function is applied to the output of another function.

To find the derivative of the function f(x) = sin(x²), we need to apply the chain rule of differentiation, which states that if f(x) = g(h(x)), then f'(x) = g'(h(x)) × h'(x).

Here, g(x) = sin(x) and h(x) = x². Therefore, g'(x) = cos(x) and h'(x) = 2x.

Applying the chain rule, we have

f'(x) = g'(h(x)) × h'(x) = cos(x²) × 2x

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Therefore, based on the information given, it seems that Bus Company A is doing slightly better in terms of on-time performance.

To determine which bus company seems to be doing better, we need to compare their on-time performance.

Bus Company A claims that it is typically on time 95% of the time. This means that out of 100 trips, it expects to be on time for 95 of them.

On the other hand, Bus Company B has a record of being on time 47 days out of the 50 days that it operates. This means that its on-time performance is:

47/50 = 0.94 or 94%

Comparing the two percentages,

we can see that Bus Company A claims to have a higher on-time performance (95%) than Bus Company B's actual on-time performance (94%).  Bus Company B's on-time performance is based on actual data from 50 days of operation.

Therefore, based on the information given, it seems that Bus Company A is doing slightly better in terms of on-time performance.

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Complete question:

Bus Company A claims that it is typically on time 95% of the time While Bus

Company B has a record of being on time 47 days out of the 50 days that it

operates. Which bus company seems to be doing better?

The probability that Diesel will win her 6th game on the 18th game is approximately 0.000568, or 0.0568%.

The probability that Diesel will win her 6th game on the 18th game is calculated using the binomial distribution formula.

Let's define the following terms:
- n = number of trials (games played) = 18
- k = number of successes (games won by Diesel) = 6
- p = probability of success (Diesel winning a game) = 1/4
- q = probability of failure (Daisy winning a game) = 1 - p = 3/4

The formula for the probability of exactly k successes in n trials is:

P(k) = (n choose k) * p^k * q^(n-k)

where (n choose k) is the binomial coefficient, which represents the number of ways to choose k successes out of n trials. It can be calculated as:

(n choose k) = n! / (k! * (n-k)!)

Plugging in the values, we get:

P(6) = (18 choose 6) * (1/4)^6 * (3/4)^12
= 18! / (6! * 12!) * (1/4)^6 * (3/4)^12
= 18564 * 0.00000244 * 0.0122
= 0.000568

Therefore, the probability that Diesel will win her 6th game on the 18th game is approximately 0.000568, or 0.0568%.

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Answer: its 12.92

first i multiplide 3.4 by 4.8

Answer: 120

Step-by-step explanation:

The factorial function multiplies all numbers below that number going to 1. For example,

   2! = 2 * 1 = 2

   3! = 3 * 2 * 1 = 6
   4! = 4 * 3 * 2 * 1 = 24

Thus, 5! would be 5 * 4 * 3 * 2 * 1 = 120.

Therefore, the complete homogeneous system solution is: x = [-0.1765, 0.6471, -0.0882] + t1[2, -1, 1] + t2[0, -1, 1]

The system, first we need to find the inverse of matrix A, which is:

A = [3 7 6]

[-3 11 2]

[4 0 -6]

det(A) = 3*(-611 - 20) - 7*(-3*-6) + 6*(-3*2) = -204

adj(A) = [72 18 42]

[54 6 33]

[14 28 14]

[tex]A^{(-1) }= adj(A)/det(A):[/tex]

[-0.3529 0.0882 -0.2059]

[-0.2647 -0.0294 -0.1618]

[-0.0686 -0.1373 -0.0686]

Next, we need to solve for Xp using Xp = [tex]A^{(-1)} * b:[/tex]

b = [1 2 2]

Xp = [tex]A^{(-1)} * b:[/tex]= [-0.1765, 0.6471, -0.0882]

To find Xn, we solve the associated homogeneous system Ax = 0:

[3 7 6][x1] [0]

[-3 11 2][x2] = [0]

[4 0 -6][x3] [0]

Writing this system in augmented form, we have:

[3 7 6 | 0]

[-3 11 2 | 0]

[4 0 -6 | 0]

We can use row reduction to solve for the reduced row echelon form:

[1 0 -2 | 0]

[0 1 1 | 0]

[0 0 0 | 0]

The general solution can be written as:

x = t1[2, -1, 1] + t2[0, -1, 1]

Here t1 and t2 are arbitrary constants.

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Correct Question:

Find the complete solution x= Xp + xn of the system Ax = b, A= b where 3 7 6 -3 11 2 4 0 -6 Xp stands for a particular solution and Xn the general solution of the associated homogeneous system. {1 2 3 3 10 2 2 -2 2 2 . }

The general solution of non-homogeneous system can be written as:

x = xh + xp = [2t + 1, t, -2s - 2] + [-1, -28, 1]

We can now write the augmented matrix of the system as:

[2    -4    5     8]

[-7   14   4   -28]

[3   -6    1      12]

We can use row reduction to determine whether the system is consistent and to find its solutions.

Performing the row reduction, we get:

[1  -2  0  2]

[0   0  1 -2]

[0   0  0 0]

From the last row of the row-reduced matrix, we can see that the system has a dependent variable, which means that there are infinitely many solutions. We can write the general solution as:

x = x1 = 2t + 1

y = y1 = t

z = z1 = -2s - 2

Here, t and s are arbitrary parameters.

To find a particular solution, we can use any method we like. One method is to use the method of undetermined coefficients. We can guess that xp is a linear combination of the columns of A, with unknown coefficients:

xp = k1[2 -7 3] + k2[-4 14 -6] + k3[5 4 1]

where k1, k2, and k3 are unknown coefficients.

We can substitute this into the system and solve for the coefficients. This gives:

k1 = -1

k2 = -2

k3 = 1

Therefore, a particular solution is:

xp = [-1 -28 1]

So the general solution can be written as:

x = xh + xp = [2t + 1, t, -2s - 2] + [-1, -28, 1]

where t and s are arbitrary parameters.

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One possible set of three equations that are all true and different for 1 2/3 is (5/3) + (1/3) = (8/3), (4/3) + (2/3) = (2), and (10/6) + (1/6) = (11/6).

Each of these equations represents a different way of expressing the same value of 1 2/3, which is equal to 5/3 or 1.6666... when expressed as a decimal.

Overall, these equations demonstrate the flexibility and versatility of mathematical expressions and show how different values can be represented in multiple ways through simple operations like addition and division.

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a) The summation notation of F₂ + F₄ + F₆ + ... + F₂ₙ is ∑ᵢ₌₁ᵗⁿ F₂ᵢ

b) Proved that F₂ + F₄ + F₆ + ... + F₂ₙ = F₂ₙ₊₁ - 1.

The Fibonacci sequence is defined as F₁ = 1, F₂ = 1, and Fₙ = Fₙ₋₁ + Fₙ₋₂ for n ≥ 3.

To express F₂ + F₄ + F₆ + ... + F₂ₙ in summation notation, we can observe that the terms are all even Fibonacci numbers. Thus, we can write:

∑ᵢ₌₁ᵗⁿ F₂ᵢ = ∑ᵢ₌₁ᵗⁿ₋₁ F₂ᵢ + F₂ₙ

where n is even.

We can then use the recurrence relation of the Fibonacci sequence to simplify this:

∑ᵢ₌₁ᵗⁿ F₂ᵢ = ∑ᵢ₌₁ᵗⁿ₋₁ F₂ᵢ + F₂ₙ

= (F₂₁ - 1) + F₂ₙ

= F₂ₙ₊₁ - 1

where we have used the fact that F₂₁ = F₁ = 1.

Therefore, we have shown that F₂ + F₄ + F₆ + ... + F₂ₙ = F₂ₙ₊₁ - 1.

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The distance that is being travelled by the flying squirrel in total would be = 18.2m.

To calculate the distance covered by the squirrel, the Pythagorean formula should be used. That is;

C² = a² + b²

a = 8m

b = 2m

c²= 8² + 2²

= 64+4

c = √68

= 8.2m

The total distance travelled by the flying squirrel is to find the perimeter of the triangle covered by the squirrel.

perimeter = length+width+height

= 8+2+8.2 = 18.2m.

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The inverse Laplace transform of F(s)=(8s^2-4s+12)/s(s^2+4) is 3 + 5cos(2t) - 2sin(2t).

Explanation:

To determine the inverse Laplace transform of(8s^2-4s+12)/s(s^2+4), follow these steps:

Step 1: To find the inverse Laplace transform of F(s)=(8s^2-4s+12)/s(s^2+4), use partial fraction decomposition:

F(s) = (A/s) + (Bs+C)/(s^2+4)

Step 2: Multiplying both sides by s(s^2+4) and equating coefficients, we get:

8s^2 - 4s + 12 = A(s^2+4) + (Bs+C)s

Simplifying, we get:

8s^2 - 4s + 12 = As^2 + 4A + B*s^2 + Cs

Equating coefficients, we get:

A + B = 8

C = -4

4A = 12

Solving for A, B, and C, we get:

A = 3

B = 5

C = -4

Therefore, we can write F(s) as:

F(s) = 3/s + (5s-4)/(s^2+4)

Step 3: Taking the inverse Laplace transform of each term separately, we get:

L^-1{3/s} = 3

L^-1{(5s-4)/(s^2+4)} = 5L^-1{s/(s^2+4)} - 2L^-1{1/(s^2+4)}

Using the table of Laplace transforms, we can find that:

L^-1{s/(s^2+4)} = cos(2t)

L^-1{1/(s^2+4)} = (1/2) sin(2t)

Therefore, the inverse Laplace transform of F(s) is:

L^-1{F(s)} = 3 + 5cos(2t) - 2sin(2t). where L^-1 denotes the inverse Laplace transform operator.

Therefore, the inverse Laplace transform of (8s^2-4s+12)/s(s^2+4) is 3 + 5cos(2t) - 2sin(2t).

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The given summation can be expanded as a series of terms, where each term is the product of two factors: one factor consists of the index variable, i or k+1, and the factorial i! or (k+1)!. The other factor consists of the sum of the first i or k terms of the corresponding factorial sequence, i.e., 1(1!), 2(2!), 3(3!), and so on.

Because i in the first term runs from 1 to k, the total is made up of the first k terms of the i! sequence multiplied by the appropriate value of i. Because k is the sole index variable in the second term, the total is composed of the first k terms of the k! sequence multiplied by each matching value of k.

The following terms have i ranging from k+1 to the summation's ultimate value, and the total is made up of the first i-1 terms of the i! sequence multiplied by each corresponding value of i. The first component, (k+1)!, accounts for the terms not included in the first two terms in the prior summations.

Overall, the summation represents a combination of factorials and their corresponding sum sequences, with the index variables determining the range of terms to include in each sum.

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The transformations of coordinates of A(2,2) is  A'' = (-1, -1) , B(1, -2) is  B'' = (0, 3) and C (4,-4) is  C'' = (-3, 5).

The coordinates are given in the figure as,

A's coordinates are (2, 2) ;

B's coordinates are (1, -2) ;

and C's coordinates are (4, -4)

The coordinates are to be transformed as A", B" and C'' as,

First invert the number's sign and then to add up 1.

Therefore,

For A" :

At x- axis, +2 becomes -2 and the by adding 1 we get, -1

Similarly at y- axis, +2 becomes -2 and the by adding 1 we get, -1

Thus, A'' = (-1, -1)

For B" :

At x- axis, +1 becomes -1 and the by adding 1 we get, 0

Similarly at y- axis, -2 becomes +2 and the by adding 1 we get, +3

Thus, B'' = (0, 3)

For C" :

At x- axis, +4 becomes -4 and the by adding 1 we get, -3

Similarly at y- axis, -4 becomes +4 and the by adding 1 we get, +5

Thus, C'' = (-3, 5)

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Answer:A(-1,5) B(0,1) C(-3,-1)

Step-by-step explanation:I had the question

To use the Integral Test to determine whether the series is convergent or divergent, we need to evaluate the integral ∫(xe^(-9x) dx) from 1 to ∞.

Let's evaluate the integral first:

∫(xe^(-9x) dx) from 1 to ∞

We use integration by parts for this, where:

u = x, dv = e^(-9x) dx
du = dx, v = -1/9 e^(-9x)

According to the integration by parts formula, ∫u dv = uv - ∫v du.

So, ∫(xe^(-9x) dx) = -1/9 x e^(-9x) - ∫(-1/9 e^(-9x) dx)

Now, we can integrate -1/9 e^(-9x) dx:

∫(-1/9 e^(-9x) dx) = (-1/9) * (-1/9) * e^(-9x) = 1/81 e^(-9x)

Thus, our integral becomes:

-1/9 x e^(-9x) - 1/81 e^(-9x)

Now we must evaluate the integral from 1 to ∞:

lim (x→∞) [ -1/9 x e^(-9x) - 1/81 e^(-9x) ] - [ -1/9 (1) e^(-9) - 1/81 e^(-9) ]

Since the exponential term e^(-9x) approaches 0 as x approaches ∞, the limit becomes:

0 - [ -1/9 e^(-9) - 1/81 e^(-9) ] = 1/9 e^(-9) + 1/81 e^(-9)

Since the integral is finite, the series is convergent.

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The matrix A of the rotation about the y-axis through an angle of π/2 clockwise as viewed from the positive y-axis is [tex]A=\left[\begin{array}{ccc}0 & 0 & -1 \\0 & 1 & 0 \\1 & 0 & 0\end{array}\right][/tex].

In mathematics, a matrix is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns, which is used to represent a mathematical object or a property of such an object.

To find the matrix A of the rotation about the y-axis through an angle of π/2 (90 degrees) clockwise as viewed from the positive y-axis, we can use the following rotation matrix:

[tex]A=\left[\begin{array}{ccc}\cos (\theta) & 0 & -\sin (\theta) \\0 & 1 & 0 \\\sin (\theta) & 0 & \cos (\theta)\end{array}\right][/tex]

Substitute θ with π/2, which is the angle of rotation.

[tex]A=\left[\begin{array}{ccc}\cos \left(\frac{\pi}{2}\right) & 0 & -\sin \left(\frac{\pi}{2}\right) \\0 & 1 & 0 \\\sin \left(\frac{\pi}{2}\right) & 0 & \cos \left(\frac{\pi}{2}\right)\end{array}\right][/tex]

Compute the trigonometric values for cos( π/2) and sin( π/2).

cos( π/2) = 0
sin( π/2) = 1

Substitute the computed values back into the matrix.

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

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The critical numbers of the function f(x) = [tex]x^{4} (x - 4)^{3}[/tex] are x = 0 and x = 4.

To find the critical numbers of the function f(x) = [tex]x^{4}(x - 4)^{3}[/tex], we need to find the values of x at which the derivative of f(x) is equal to zero or undefined.

First, we will find the derivative of f(x) using the product rule:

f'(x) = [tex]4x^{3} (x - 4)^{3} + x^{4} 3(x - 4)^{2}(1)[/tex]

Simplifying this expression, we get:

f'(x) = [tex]4x^{3} (x - 4)^{2} (4 - x)[/tex]

Now, we can set f'(x) equal to zero and solve for x:

[tex]4x^{3} (x - 4)^{2} (4 - x)[/tex] = 0

From this equation, we can see that the critical numbers are x = 0, x = 4, and x = 4.

To check if x = 4 is a critical number, we need to find the limit of f'(x) as x approaches 4 from the left and from the right:

lim x→4- f'(x) = lim x→4- 4[tex]x^{3}[/tex][tex](x - 4)^{2}[/tex](4 - x) = 0

lim x→4+ f'(x) = lim x→4+ 4[tex]x^{3}[/tex][tex](x - 4)^{2}[/tex](4 - x) = 0

Since both limits are equal to zero, x = 4 is a critical number.

Therefore, the critical numbers of the function f(x) = [tex]x^{4} (x - 4)^{3}[/tex] are x = 0 and x = 4.

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Yes, this is a valid confidence interval calculation for comparing the proportions of snowy days in Denver and Chicago.

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The rate of change of the area at the instant when the radius is 20 ft is -160π square feet per minute.

The area of a circle is given by the formula A = π [tex]r^2[/tex], where r is the radius of the circle.

We are given that the radius of the circle is decreasing at a rate of 4 ft/min. This means that the rate of change of the radius with respect to time is -4 ft/min (negative because the radius is decreasing).

At the instant when the radius is 20 ft, we can calculate the rate of change of the area by taking the derivative of the area formula with respect to time:

dA/dt = d/dt (π[tex]r^2)[/tex]

Using the chain rule, we get:

dA/dt = 2πr (dr/dt)

Substituting r = 20 ft and dr/dt = -4 ft/min, we get:

dA/dt = 2π(20)(-4) = -160π

Therefore, the rate of change of the area at the instant when the radius is 20 ft is -160π square feet per minute.

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If a student draws 1 counter out of the bag without looking, then the probability of drawing an orange counter is 0.25 or 25%.

Probability is a measure of the likelihood or chance of an event occurring. It is a number between 0 and 1, where 0 indicates that the event is impossible, and 1 indicates that the event is certain to occur.

Probability is usually expressed as a fraction, decimal, or percentage. For example, if the probability of an event occurring is 0.5, this means there is a 50% chance that the event will occur.

The probability of an event can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes. For example, if a fair six-sided die is rolled, the probability of rolling a 3 is 1/6, because there is one favorable outcome (rolling a 3) out of six possible outcomes (rolling a 1, 2, 3, 4, 5, or 6).

The probability of drawing an orange counter out of the bag can be calculated by dividing the number of orange counters by the total number of counters in the bag.

The total number of counters in the bag is:

6 + 2 + 10 + 6 = 24

The number of orange counters is:

6

Therefore, the probability of drawing an orange counter is:

6/24 = 1/4 = 0.25

So the probability of drawing an orange counter is 0.25 or 25%.

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On average, the metro train travels a distance of 90 kilometers in a single trip.

Since the metro train completes 4 round trips of 90 kilometres, the total distance traveled in a day would be 4290 = 720 kilometres (since a round trip is equivalent to two journeys of 90 kilometres).

To find the average distance traveled, we need to divide the total distance by the number of trips made. Since 4 round trips have been made, the number of trips made would be 4*2 = 8 (since each round trip is equivalent to 2 trips).

Therefore, the average distance traveled by the metro in a day would be 720/8 = 90 kilometres.

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Answer:

The answer is 5ft

Step-by-step explanation:

A=L×B

A=L×W

30=6×W

30=6W

divide both sides by 6

W=5ft

Answer:

40.82

Step-by-step explanation:

You need to use trig identities, which are sin(θ)=opposite length/hypotenuse length, cos(θ)=adjacent length/hypotenuse length, and tan(θ)=opposite length/adjacent length.

In your diagram, we see that the only available information is the length opposite of the angle x (19) and adjacent to angle x (22), so we will use the tan identity.

tan(x)=19/22

we need to solve for x, and so we need to get x alone. This can be done by using inverse tan: arctan or [tex]tan(x)^{-1}[/tex]. Note that we ARE NOT taking the equation to the exponent of -1, this is just notation for a trig identify.

arctan(x)tan(x)=x

x= arctan(19/22)

arctan(19/22)= 40.82

and so

x=40.82

The exponential Fourier series coefficients [tex]$D_n n$[/tex] for [tex]$x(t)$[/tex] are:

[tex]$D_n n=\{(p i-A) / 2 p i$[/tex] for [tex]$n=0$[/tex]

[tex]\left\{(-1)^{\wedge} n /(n \mathrm{~A})\right.[/tex] for [tex]$n=+/-1,+/-2, \ldots$[/tex]

To sketch [tex]$x(t)$[/tex] over two periods from [tex]$t=0$[/tex] to [tex]$4 \mathrm{pi}$[/tex], we first need to plot one period of [tex]$x(t)$[/tex], which is given as:

[tex]$$\begin{aligned}& \mathrm{x}(\mathrm{t})=(1 / \mathrm{A}) \mathrm{t} 0 < =\mathrm{t} < \mathrm{A} \\& =\mathrm{A} \mathrm{A} < =\mathrm{t} < \mathrm{pi} \\& =0 \mathrm{pi} < =\mathrm{t} < 2 \mathrm{pi}\end{aligned}$$[/tex]

The plot of one period of [tex]x(t)[/tex] is shown below:

  |          /\

  |         /  \

A |        /    \

  |       /      \

  |      /        \

  |_____/          \_____

     0     A      pi    2pi

To sketch [tex]x(t)[/tex] over two periods, we need to repeat this pattern twice. Since [tex]x(t)[/tex] is a 2pi-periodic signal, we only need to sketch one period to represent the entire signal over any number of periods. Therefore, we can simply repeat the above plot twice to obtain the sketch of [tex]x(t)[/tex] over two periods from [tex]t = 0[/tex] to [tex]4pi[/tex], as shown below:

  |          /\          /\

  |         /  \        /  \

A |        /    \      /    \

  |       /      \    /      \

  |_____/        \__/        \_____

     0     A      pi         2pi  3pi

To find the exponential Fourier series coefficients [tex]D_n[/tex], we can use the formula:

[tex]$D_{\ldots} n=(1 / T) * \int[T] x(t) e^{\wedge}(-j n w 0 t) d t$[/tex]

where T is the period of [tex]$x(t)$[/tex], w0 is the fundamental angular frequency, and n is an integer. Since [tex]$x(t)$[/tex] is a 2pi-periodic signal, we have [tex]$T=2 p i$[/tex] and [tex]$\mathrm{wO}=2 \mathrm{pi} / \mathrm{T}=1$[/tex].

The Fourier series coefficients [tex]$D_n n$[/tex] for [tex]$n=0,+/-1,+/-2, \ldots$[/tex] are given by:

[tex]$D_{\ldots} n=(1 / 2 p i) * \int[2 \mathrm{pi}] x(\mathrm{t}) \mathrm{e}^{\wedge}(-j n t) d t$[/tex]

For [tex]$\mathrm{n}=0$[/tex], we have:

[tex]{ D_0 }$[/tex][tex]=(1 / 2 p i)^* \int[2 \mathrm{pi}] \times(t) d t$[/tex]

[tex]=(1 / 2 \mathrm{pi}) *\left[(1 / \mathrm{A}) * \int[\mathrm{A}] \mathrm{t} d \mathrm{dt}+\mathrm{A}^* \int[\mathrm{pi}] \mathrm{dt}+0\right] \\& =(1 / 2 \mathrm{pi}) *\left[(1 / \mathrm{A}) *\left(\mathrm{~A}^{\wedge} 2 / 2\right)+\mathrm{A}(\mathrm{pi}-\mathrm{A})\right] \\[/tex]

[tex]& =(1 / 2 \mathrm{pi}) *[(\mathrm{~A} / 2)+\mathrm{A}(\mathrm{pi}-\mathrm{A})] \\& =(\mathrm{pi}-\mathrm{A} / 2 \mathrm{pi})\end{aligned}$$[/tex]

For [tex]$n=+/-1,+/-2, \ldots$[/tex], we have:

[tex]$$\begin{aligned}& D_n n=(1 / 2 p i)^* \int[2 p i] x(t) e^{\wedge}(-j n t) d t \\& =(1 / 2 p i)^*\left[(1 / A) * \int[A] t e^{\wedge}(-j n t) d t+A^* \int[\text { pi }] e^{\wedge}(-j n t) d t+0\right] \\& =(1 / 2 \text { pi })^*\left[(1 / A)^*\left((-1)^{\wedge} n-1\right)+A^*\left(1-(-1)^{\wedge} n\right) /(j n)\right] \\& =(-1)^{\wedge} n /(n A)\end{aligned}$$[/tex]

Therefore, the exponential Fourier series coefficients [tex]$D_n n$[/tex] for [tex]$x(t)$[/tex] are:

[tex]$D_n n=\{(p i-A) / 2 p i$[/tex] for [tex]$n=0$[/tex]

[tex]$\left\{(-1)^{\wedge} n /(n \mathrm{~A})\right.$[/tex] for [tex]$n=+/-1,+/-2, \ldots$[/tex]

Using the formula for the inverse Fourier series, we can write the

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The exponential Fourier series coefficients [tex]$D_n n$[/tex] for [tex]$x(t)$[/tex] are:

[tex]$D_n n=\{(p i-A) / 2 p i$[/tex] for [tex]$n=0$[/tex]

[tex]\left\{(-1)^{\wedge} n /(n \mathrm{~A})\right.[/tex] for [tex]$n=+/-1,+/-2, \ldots$[/tex]

To sketch [tex]$x(t)$[/tex] over two periods from [tex]$t=0$[/tex] to [tex]$4 \mathrm{pi}$[/tex], we first need to plot one period of [tex]$x(t)$[/tex], which is given as:

[tex]$$\begin{aligned}& \mathrm{x}(\mathrm{t})=(1 / \mathrm{A}) \mathrm{t} 0 < =\mathrm{t} < \mathrm{A} \\& =\mathrm{A} \mathrm{A} < =\mathrm{t} < \mathrm{pi} \\& =0 \mathrm{pi} < =\mathrm{t} < 2 \mathrm{pi}\end{aligned}$$[/tex]

The plot of one period of [tex]x(t)[/tex] is shown below:

  |          /\

  |         /  \

A |        /    \

  |       /      \

  |      /        \

  |_____/          \_____

     0     A      pi    2pi

To sketch [tex]x(t)[/tex] over two periods, we need to repeat this pattern twice. Since [tex]x(t)[/tex] is a 2pi-periodic signal, we only need to sketch one period to represent the entire signal over any number of periods. Therefore, we can simply repeat the above plot twice to obtain the sketch of [tex]x(t)[/tex] over two periods from [tex]t = 0[/tex] to [tex]4pi[/tex], as shown below:

  |          /\          /\

  |         /  \        /  \

A |        /    \      /    \

  |       /      \    /      \

  |_____/        \__/        \_____

     0     A      pi         2pi  3pi

To find the exponential Fourier series coefficients [tex]D_n[/tex], we can use the formula:

[tex]$D_{\ldots} n=(1 / T) * \int[T] x(t) e^{\wedge}(-j n w 0 t) d t$[/tex]

where T is the period of [tex]$x(t)$[/tex], w0 is the fundamental angular frequency, and n is an integer. Since [tex]$x(t)$[/tex] is a 2pi-periodic signal, we have [tex]$T=2 p i$[/tex] and [tex]$\mathrm{wO}=2 \mathrm{pi} / \mathrm{T}=1$[/tex].

The Fourier series coefficients [tex]$D_n n$[/tex] for [tex]$n=0,+/-1,+/-2, \ldots$[/tex] are given by:

[tex]$D_{\ldots} n=(1 / 2 p i) * \int[2 \mathrm{pi}] x(\mathrm{t}) \mathrm{e}^{\wedge}(-j n t) d t$[/tex]

For [tex]$\mathrm{n}=0$[/tex], we have:

[tex]{ D_0 }$[/tex][tex]=(1 / 2 p i)^* \int[2 \mathrm{pi}] \times(t) d t$[/tex]

[tex]=(1 / 2 \mathrm{pi}) *\left[(1 / \mathrm{A}) * \int[\mathrm{A}] \mathrm{t} d \mathrm{dt}+\mathrm{A}^* \int[\mathrm{pi}] \mathrm{dt}+0\right] \\& =(1 / 2 \mathrm{pi}) *\left[(1 / \mathrm{A}) *\left(\mathrm{~A}^{\wedge} 2 / 2\right)+\mathrm{A}(\mathrm{pi}-\mathrm{A})\right] \\[/tex]

[tex]& =(1 / 2 \mathrm{pi}) *[(\mathrm{~A} / 2)+\mathrm{A}(\mathrm{pi}-\mathrm{A})] \\& =(\mathrm{pi}-\mathrm{A} / 2 \mathrm{pi})\end{aligned}$$[/tex]

For [tex]$n=+/-1,+/-2, \ldots$[/tex], we have:

[tex]$$\begin{aligned}& D_n n=(1 / 2 p i)^* \int[2 p i] x(t) e^{\wedge}(-j n t) d t \\& =(1 / 2 p i)^*\left[(1 / A) * \int[A] t e^{\wedge}(-j n t) d t+A^* \int[\text { pi }] e^{\wedge}(-j n t) d t+0\right] \\& =(1 / 2 \text { pi })^*\left[(1 / A)^*\left((-1)^{\wedge} n-1\right)+A^*\left(1-(-1)^{\wedge} n\right) /(j n)\right] \\& =(-1)^{\wedge} n /(n A)\end{aligned}$$[/tex]

Therefore, the exponential Fourier series coefficients [tex]$D_n n$[/tex] for [tex]$x(t)$[/tex] are:

[tex]$D_n n=\{(p i-A) / 2 p i$[/tex] for [tex]$n=0$[/tex]

[tex]$\left\{(-1)^{\wedge} n /(n \mathrm{~A})\right.$[/tex] for [tex]$n=+/-1,+/-2, \ldots$[/tex]

Using the formula for the inverse Fourier series, we can write the

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