answer: Solution: Given system isx′=(−10)(0−1)xWe know that the characteristic equation of the above system is given statistical by |A-λI|=0.λ^2+2λ+1=0Solving the above equation we get the eigenvalues of Aλ1=-1,λ2=-1.
The eigenvectors corresponding to the eigenvalues λ1 and λ2 are defined as (A-λ1I)v1=0 and (A-λ2I)v2=0 respectively, where v1 and v2 are the eigenvectors corresponding to λ1 and λ2 respectively. From (A-λ1I)v1=0, we get(A+I)v1=0⇒v1=(−1,1)From (A-λ2I)v2=0, we getA−Iv2=0⇒v2=(1,0)Let P be the matrix whose columns are the eigenvectors of A, i.e.P=[−1 1 1 0]Using P, we can write A in Jordan form asA=PJP−1whereJ=diag(λ1,λ2)=diag(−1,−1).
Therefore, x′=Ax becomes y′=JP−1x′or, x′=Py′=PJP−1xLet Y=P−1x. Then y=P−1x satisfies y′=JP−1x′=Jy′.So, the system can be transformed into the following form by letting
[tex]y=P−1x:$$y'=\begin{bmatrix}-1&1\\0&-1\\\end{bmatrix}y$$[/tex]
The above system of equation has the general
[tex]y=c1e^(-t)+c2e^(-t)y=c1e^(-t)+c2e^(-t)[/tex]
twhere c1 and c2 are arbitrary constants.To plot the phase plane diagram we can use online websites or graphing software like MATLAB, Mathematica etc.
The phase plane diagram is given as follows.The critical point is (0,0) which is the only critical point of the system. The phase portrait has all trajectories moving towards the critical point and hence the critical point is asymptotically stable.
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Answer the following, show all necessary solutions. 1. Use any method to solve for the unknowns (5 points): 2x-y-3z=0 -x+2y-3z=0 x + y + 4z = 0 2.
Given the following matrices, verify that (5 points each): 4 A = B = c=1} 1 5 D= -1 0 #8 1 E= 1 2 a. C(A+B)=CA + CB b. (DT)¹=D c. B=(B²)¹=(B₁¹)² d. (A¹)¹=(A¹) ¹ 3. Find matrix A given the following expression (5points) -3 7 (7A)-¹ = [¯ 1 4. Compute for p(A) if p(x)=x²-2x+1 when using the matrix A in number 2 (5 points).
The solution to the matrix is 0 and matrix A=B=C
How to solve the matrix?In mathematics, a matrix (plural matrices) 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.
The given equations are
2x-y-3z=0
-x+2y-3z=0
x + y + 4z = 0
Expressing these in matrix form to have
[tex]\left[\begin{array}{ccc}2&-1&-3\\-1&2&-3\\1&1&4\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}0\\0\\0\end{array}\right][/tex]
The determinant of the matrix is given as
2[8+3] +1[-4+3] -3[-1-2]
This gives 2(11) -1(-1) -3(-3)
22+1+9 = 32
the determinant of the matrix is 32
Using Cramer's rule,
To find x,
[tex]\left[\begin{array}{ccc}0&-1&-3\\0&2&-3\\0&1&4\end{array}\right] / 32 , y = \left[\begin{array}{ccc}2&0&-3\\-1&0&-3\\1&0&4\end{array}\right] /32, z= \left[\begin{array}{ccc}2&-1&0\\-1&2&0\\1&1&0\end{array}\right] /32[/tex]
0[8+3] +1[0+0) -3[0+0] /32, y= 2[0-0]-0[-4+3] -3[0-0]/32, z = 2[0+0] +1[0-0] +0[-1-2]/32
0[11]+1[0]-3[0]/32, y = 2[0]-0[-1]0]/32, z = 2[0] +1[0] +0[-3]/32
= 0+0+0=0/32, y = 0+0+0 = 0/32, z = 0+0+0 = 0/32
Therefore in each case the values of x, y and z are 0
This implies that A=B-C
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"
4. The equation 2x + 3y = a is the tangent line to the graph of the function, $(x) = bx at I=2 Find the values of a and 8.
The values of a & b are a = 3y + 2x and b = (2x - 9y) / 2 for the equation 2x + 3y = a is the tangent-line to the graph of the function, f(x) = bx at I=2
Given that equation 2x + 3y = a is the tangent line to the graph of the function f(x) = bx at I = 2,
we can differentiate the equation f(x) = bx using the chain rule and find its slope at I = 2.
We know that the slope of the tangent line and the derivative of the function evaluated at x = 2 are the same slope of the tangent line at
x = 2
= f '(2)
f(x) = bx
f '(x) = b2x3y = (a - 2b)/2
Differentiate f(x) with respect to x.
b2x = 3y
f'(2) = b(2)
= 6y
Substitute f '(2) = b(2)
= 6y in the equation
3y = (a - 2b)/2.6y
= (a - 2b)/2
Multiply both sides by 2.
12y = a - 2b ----(1)
Also, substitute x = 2 and y = f(2) in 2x + 3y = a.2(2) + 3f(2) = a. .......(2)
Now, we need to eliminate the variable a from equations (1) and (2).
Substitute the value of a from equation (1) in (2).
2(2) + 3f(2) = 12y + 2b3f(2)
= 12y + 2b - 4
Multiply both sides by 1/3.
f(2) = 4y + 2/3 ----(3)
From equation (1), a = 12y + 2b.
Substitute this value of a in 2x + 3y = a.
2x + 3y = 12y + 2b2x + 3y - 12y
= 2b2x - 9y
= 2b
Therefore, a = 12y + 2b and
b = (2x - 9y) / 2.
Substitute b = (2x - 9y) / 2 in
a = 12y + 2b.
We get,a = 12y + 2((2x - 9y) / 2)
a = 12y + 2x - 9y
= 3y + 2x
Therefore, a = 3y + 2x and b = (2x - 9y) / 2.
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The vector r is twice as long as the vector δ. The angle between the vectors is 60°. The vector projection of δ on r is (-3, 0, 2). Determine r.
Let's denote the length of vector δ as δ and the length of vector r as r. Since r is twice as long as δ, we have r = 2δ.
The vector projection of δ on r is given by the formula:
projδr = (δ · r / ||r||^2) * r,
where · denotes the dot product and ||r||^2 represents the squared length of r.
We are given that the vector projection of δ on r is (-3, 0, 2). So we have:
(-3, 0, 2) = (δ · r / ||r||^2) * r.
Since the angle between δ and r is 60°, we know that δ · r = ||δ|| ||r|| cos(60°) = δr/2, where δr represents the product of the lengths of δ and r.
Substituting this into the equation, we get:
(-3, 0, 2) = (δr/2 / ||r||^2) * r.
We can rewrite this as:
(-3, 0, 2) = (δr/2 ||r||^2) * 2δ.
Comparing the corresponding components, we have:
δr/2 = -3,
||r||^2 = 2^2 = 4.
From the first equation, we find δr = -6. Substituting this into the second equation, we get:
(-6)^2 = 4 ||r||^2.
Simplifying, we have:
36 = 4 ||r||^2.
Dividing both sides by 4, we get ||r||^2 = 9.
Taking the square root of both sides, we obtain ||r|| = 3.
Since we know that r = 2δ, we can express r as:
r = 2δ = 2 * 3 = 6.
Therefore, the vector r is (6, 6, 6).
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Consider again the functions from the questions above, namely 1 f(x) = 4√√x + 2x¹/2 - 8x-7/8 + x² +2 and f(x) - = ²³x³/² − 2x³/² + √3x³ − 2x² + x − 1. Find the indefinite integral [ f(x) dx for each function. Each item is worth 15 marks.
The indefinite integral for the given functions are :
(a) ∫ f(x) dx = (8/3)x^(3/4) + (4/3)x^(3/2) - (16/15)x^(1/8) + (1/3)x^3 + 2x + C
(b) ∫ f(x) dx = (4/5)x^(5/2) - (4/5)x^(5/2) + (2/3√3)x^(5/2) - (2/3)x^3 + (1/2)x^2 - x + C
To find the indefinite integral of each function, we will integrate term by term using the power rule and the properties of radicals.
(a) f(x) = 4√√x + 2x^(1/2) - 8x^(-7/8) + x^2 + 2
The indefinite integral of each term is as follows:
∫ 4√√x dx = (8/3)x^(3/4)
∫ 2x^(1/2) dx = (4/3)x^(3/2)
∫ -8x^(-7/8) dx = (-16/15)x^(1/8)
∫ x^2 dx = (1/3)x^3
∫ 2 dx = 2x
Therefore, the indefinite integral of f(x) is:
∫ f(x) dx = (8/3)x^(3/4) + (4/3)x^(3/2) - (16/15)x^(1/8) + (1/3)x^3 + 2x + C
(b) f(x) = 2³√x³/² - 2x^(3/2) + √3x³ - 2x² + x - 1
The indefinite integral of each term is as follows:
∫ 2³√x³/² dx = (4/5)x^(5/2)
∫ -2x^(3/2) dx = (-4/5)x^(5/2)
∫ √3x³ dx = (2/3√3)x^(5/2)
∫ -2x² dx = (-2/3)x^3
∫ x dx = (1/2)x^2
∫ -1 dx = -x
Therefore, the indefinite integral of f(x) is:
∫ f(x) dx = (4/5)x^(5/2) - (4/5)x^(5/2) + (2/3√3)x^(5/2) - (2/3)x^3 + (1/2)x^2 - x + C
Note: The "+ C" represents the constant of integration, which is added because indefinite integrals have an infinite family of solutions.
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Consider a random sample of size n from a normal distribution, X;~ N(μ, 2), suppose that o2 is unknown. Find a 90% confidence interval for uit = 19.3 and s2 = 10.24 with n = 16.
(_____, _____)
The 90% confidence interval for the population mean μ is (18.047, 20.553).
What is the 90% confidence interval for the population mean?A 90% confidence interval provides a range of values within which the true population mean is likely to fall. In this case, we have a random sample of size n = 16 from a normal distribution with unknown variance. The sample mean is 19.3, and the sample variance is 10.24.
To calculate the confidence interval, we use the t-distribution since the population variance is unknown. With a sample size of 16, the degrees of freedom is n - 1 = 15. From statistical tables or software, the critical value corresponding to a 90% confidence level and 15 degrees of freedom is approximately 1.753. The margin of error can be calculated as the product of the critical value and the standard error of the mean.
The standard error is the square root of the sample variance divided by the square root of the sample size, which yields approximately 0.806. Thus, the margin of error is 1.753 * 0.806 = 1.411. The lower bound of the confidence interval is the sample mean minus the margin of error, while the upper bound is the sample mean plus the margin of error. Therefore, the 90% confidence interval for the population mean μ is (19.3 - 1.411, 19.3 + 1.411), which simplifies to (18.047, 20.553).
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Given the two 3-D vectors a=[-5, 5, 3] and b=(-6, 4, 5), find the dot product and angle (degrees) between them. Also find the cross product (d = a cross b) and the unit vector in the direction of d. ans: 8 =
The dot product of vectors a and b is 8.
What is the scalar product of vectors a and b?It is possible to determine the dot product of two vectors by multiplying and adding the elements that make up each vector. In this instance, (-5*-6) + (5*4) + (3*5) = 30 + 20 + 15 = 65 is the dot product of vectors a=[-5, 5, 3] and b=(-6, 4, 5).
The equation = can be used to determine the angle between vectors a and b.
(a · b / (|a| * |b|))
The magnitudes of the vectors a and b are shown here as |a| and |b|, respectively. The magnitudes of a and b are ((-5)2 + 52 + 32) = 75 for a and ((-6)2 + 42 + 52) = 77 for b, respectively. When we enter these values into the formula, we obtain: =
47.17 degrees are equal to (65 / (75 * 77)).
Taking the determinant of the matrix generated yields the cross product of the vectors a and b.
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Consider the initial value problem for the function y,
y’ 6 cos(3t)/ y^4 -6 t^2/y^4=0
y(0) =1
(a) Find an implicit expression of all solutions y of the differential equation above, in the form y(t, y) = c, where c collects all constant terms. (So, do not include any c in your answer.) y(t, Ψ =___________ Σ
(b) Find the explicit expression of the solution y of the initial value problem above.
Ψ =___________ Σ
(a) The implicit expression of all solutions y is given by t^3 + 2 ln|y| - 2t^2 + 2ln|y|^3 = Ψ, where Ψ collects constant terms.
(b) The explicit expression of the solution y for the initial value problem y(0) = 1 is given by y(t) = [(2t^2 + 2ln|y(0)|^3 - Ψ)/2]^(-1/3).
(a) To find an implicit expression, we rearrange the terms and integrate both sides of the given differential equation. This leads to an equation that combines the terms involving t and y, resulting in an expression involving both variables. The constant terms are collected in Ψ.
(b) To obtain the explicit expression, we use the initial condition y(0) = 1 to determine the value of the constant term Ψ. Substituting this value back into the implicit expression gives the explicit solution, which provides a direct relationship between t and y.
The expression allows us to calculate the value of y for any given t within the valid domain. By plugging in specific values of t into the equation, we can obtain corresponding values of y.
The solution represents the function y(t) explicitly in terms of t, providing a clear understanding of how the function evolves with respect to the independent variable.
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Evaluate the indefinite integral. (Use C for the constant of integration.) √x³ sin(7 + x7/2) dx X
To evaluate the indefinite integral of √(x³) sin(7 + [tex]x^(7/2[/tex])) dx, we can use the substitution method. Let u = 7 + [tex]x^(7/2)[/tex], then differentiate u with respect to x to find du/dx.
Let's perform the substitution u =[tex]7 + x^(7/2)[/tex]. Taking the derivative of u with respect to x, we have du/dx = [tex](7/2) * x^(5/2[/tex]). Solving for dx, we get dx = [tex](2/7) * x^(-5/2)[/tex]du.
Substituting these expressions into the integral, we have ∫√(x³) sin(7 + [tex]x^(7/2)) dx = ∫√(x³) sin(u) * (2/7) * x^(-5/2)[/tex]du.
We can simplify this expression to [tex](2/7) ∫ x^(-5/2) * √(x³)[/tex] * sin(u) du. Rearranging the terms, we have (2/7) ∫[tex](sin(u) / x^(3/2))[/tex] du.
Now, we can integrate with respect to u, treating x as a constant. The integral of sin(u) is -cos(u), so the expression becomes (-2/7) * cos(u) / x^(3/2) + C, where C is the constant of integration.
Substituting u = 7 + x^(7/2) back into the expression, we have (-2/7) * cos([tex]7 + x^(7/2)) / x^(3/2)[/tex] + C.
Therefore, the indefinite integral of √(x³) sin(7 + x^(7/2)) dx is (-2/7) * cos(7 + [tex]x^(7/2)) / x^(3/2[/tex]) + C, where C is the constant of integration.
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in problem 5, for n = 3, if the coin is assumed fair, what are the probabilities associated with the values that x can take on?
The correct answer is probability is 1/8 for a coin is flipped n times, where n is some fixed positive integer.
Let x be the number of times that "heads" appears.
Let p denote the probability that "heads" appears on any individual flip, and assume that the coin is fair,
So that p = 0.5.
The probability that x = k, for k = 0, 1, 2, ..., n
For n = 3, if the coin is assumed fair, the probabilities associated with the values that x can take on are as follows:
Probability that x = 0:
This means that all of the coin flips resulted in tails.
Thus, the probability of this event is:P(x=0) = 1/2 * 1/2 * 1/2
= 1/8
Probability that x = 1:
This means that exactly one of the coin flips resulted in heads.
The probability of this event is:P(x=1) = 3(1/2 * 1/2 * 1/2)
= 3/8
Probability that x = 2:
This means that exactly two of the coin flips resulted in heads.
The probability of this event is:P(x=2) = 3(1/2 * 1/2 * 1/2)
= 3/8
Probability that x = 3:
This means that all of the coin flips resulted in heads.
Thus, the probability of this event is:P(x=3) = 1/2 * 1/2 * 1/2
= 1/8
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I don't see why (II) is false ??
Exercise 14
Let G be a group. Which of the following statement(s) is/are true:
I. If G is noncyclic, then there exists a proper non-cyclic subgroup of G.
II. If a, b € G and |a| and |b| are finite, then |ab| is finite.
III. naEG c(a) = G if and only if G is abelian.
(a) I and II only
(b) II and III only (c) III only (d) II only
(e) I and III only
The correct answer is option (a) "I and II only."
Statement (I) is true because a noncyclic group must have a proper non-cyclic subgroup. Statement (II) is also true as the product of two elements with finite orders has a finite order.
In the given exercise, we need to determine which of the statements are true for a group G.
Statement (I): This statement is true. If G is a noncyclic group, it means there is no element in G that generates the entire group. Therefore, there must exist a proper non-cyclic subgroup in G.
Statement (II): This statement is true. If a and b are elements of G with finite orders, then their product ab will also have a finite order. This is because the order of ab is the least common multiple of the orders of a and b, which is finite.
Statement (III): This statement is false. The condition na ∈ C(a) = G implies that a commutes with every element in G, but it does not necessarily make G an abelian group.
Based on the explanations, we can conclude that statement (I) and statement (II) are true, while statement (III) is false. Therefore, the correct answer is option (a) "I and II only."
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. Suppose that x is an exponential random variable with parameter λ = 2. Let Y₁, Y2, be two observation samples of a single variable x with attenuation factors h₁ =3,h₂=2 and noise N₁, N₁, respectively. Y₁ =h₁X + N₁ ; Y₂=h₂X + N₂₁
Given an exponential random variable x with parameter λ = 2, two observation samples Y₁ and Y₂ are obtained by attenuating x with factors h₁ = 3 and h₂ = 2 respectively, and adding independent noise terms N₁ and N₂₁.
In this scenario, x represents an exponential random variable with a rate parameter λ = 2. The exponential distribution is commonly used to model the time between events in a Poisson process, where events occur continuously and independently at a constant average rate. The parameter λ determines the average rate of event occurrences.
To obtain the observation sample Y₁, the random variable x is attenuated by a factor of h₁ = 3, which means the magnitude of x is reduced by a factor of 3. Additionally, the noise term N₁ is added to Y₁, representing random variations or errors in the measurement process. Similarly, for the observation sample Y₂, the attenuation factor is h₂ = 2, and the noise term N₂₁ is added.
The attenuation factors h₁ and h₂ can be used to adjust the magnitude or intensity of the observed samples relative to the original exponential random variable x. By attenuating the signal, the observed samples may have reduced amplitudes compared to x. The noise terms N₁ and N₂₁ introduce random variations or errors into the observations, which can be caused by measurement inaccuracies, environmental disturbances, or other sources of interference.Overall, the given observations Y₁ and Y₂ provide a modified representation of the original exponential random variable x, taking into account attenuation factors and added noise terms.
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What percentage of the global oceans are Marine Protected Areas
(MPA's) ?
a. 3.7% b. 15.2% c. 26.7% d. 90%
Option (c) 26.7% of the global oceans are Marine Protected Areas (MPAs). Marine Protected Areas (MPAs) are designated areas in the oceans that are set aside for conservation and management purposes.
They are intended to protect and preserve marine ecosystems, biodiversity, and various species. MPAs can have different levels of restrictions and regulations, depending on their specific objectives and conservation goals.
As of the current knowledge cutoff in September 2021, approximately 26.7% of the global oceans are designated as Marine Protected Areas. This means that a significant portion of the world's oceans has some form of protection and management in place to safeguard marine life and habitats. The establishment and expansion of MPAs have been driven by international agreements and initiatives, as well as national efforts by individual countries to conserve marine resources and promote sustainable practices.
It is worth noting that the percentage of MPAs in the global oceans may change over time as new areas are designated or existing MPAs are expanded. Therefore, it is important to refer to the most up-to-date data and reports from reputable sources to get the most accurate and current information on the extent of Marine Protected Areas worldwide.
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Question 2
Find the fourth order Taylor polynomial of f(x) = 3 / x³ - 7 at x = 2.
To find the fourth-order Taylor polynomial of the function f(x) = 3 / (x³ - 7) centered at x = 2, we need to compute the function's derivatives and evaluate them at x = 2.
Let's begin by finding the derivatives:
f(x) = 3 / (x³ - 7)
First derivative:
f'(x) = (-9x²) / (x³ - 7)²
Second derivative:
f''(x) = (18x(x³ - 7) + 18x²) / (x³ - 7)³
Third derivative:
f'''(x) = (18(x³ - 7)³ + 54x(x³ - 7)² + 54x²(x³ - 7)) / (x³ - 7)⁴
Fourth derivative:
f''''(x) = (72(x³ - 7)² + 54(3x²(x³ - 7)² + 3x(x³ - 7)(18x(x³ - 7) + 18x²))) / (x³ - 7)⁵
Now, we can evaluate these derivatives at x = 2:
f(2) = 3 / (2³ - 7) = 3 / (8 - 7) = 3
f'(2) = (-9(2)²) / (2³ - 7)² = -36 / (8 - 7)² = -36
f''(2) = (18(2)(2³ - 7) + 18(2)²) / (2³ - 7)³ = 0
f'''(2) = (18(2³ - 7)³ + 54(2)(2³ - 7)² + 54(2)²(2³ - 7)) / (2³ - 7)⁴ = 54
f''''(2) = (72(2³ - 7)² + 54(3(2)²(2³ - 7)² + 3(2)(2³ - 7)(18(2)(2³ - 7) + 18(2)²))) / (2³ - 7)⁵ = -432
Now, we can write the fourth-order Taylor polynomial:
P₄(x) = f(2) + f'(2)(x - 2) + (f''(2) / 2!)(x - 2)² + (f'''(2) / 3!)(x - 2)³ + (f''''(2) / 4!)(x - 2)⁴
Plugging in the values we calculated:
P₄(x) = 3 + (-36)(x - 2) + (0 / 2!)(x - 2)² + (54 / 3!)(x - 2)³ + (-432 / 4!)(x - 2)⁴
Simplifying further:
P₄(x) = 3 - 36(x - 2) + 9(x - 2)³ - 18(x - 2)⁴
Therefore, the fourth-order Taylor polynomial of f(x) = 3 / (x³ - 7) centered at x = 2 is P₄(x) = 3 - 36(x - 2) + 9(x - 2)³ - 18(x - 2)⁴.
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graph the cosecant function
y=1/2 csc 2x
please show ALL steps
To graph the cosecant function y = (1/2) csc(2x), we have to follow some steps.
Step 1: Determine the period
The period of the cosecant function is equal to 2π divided by the coefficient of x inside the trigonometric function. In this case, the coefficient is 2. Therefore, the period is 2π/2 = π.
Step 2: Identify key points
To graph the function, we need to identify some key points within one period. Since the cosecant function is the reciprocal of the sine function, we can look at the key points of the sine function and their reciprocals. The key points of the sine function in one period (0 to 2π) are as follows:
At x = 0, sin(2x) = sin(0) = 0.
At x = π/2, sin(2x) = sin(π) = 0.
At x = π, sin(2x) = sin(2π) = 0.
At x = 3π/2, sin(2x) = sin(3π) = 0.
At x = 2π, sin(2x) = sin(4π) = 0.
These key points will help us determine the x-values at which the cosecant function will have vertical asymptotes.
Step 3: Plot the key points and asymptotes
Plot the identified key points and draw vertical asymptotes at x-values where the cosecant function is undefined (i.e., where the sine function is equal to zero).
Step 4: Sketch the graph
Based on the key points, asymptotes, and the general shape of the cosecant function, sketch the graph by connecting the points and following the behavior of the function.
Putting it all together, the graph of y = (1/2) csc(2x) will have vertical asymptotes at x = π/2, x = 3π/2, and so on. It will also have zero crossings at x = 0, x = π, x = 2π, and so on. The graph will repeat itself every π units due to the period of the function.
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urgent
The following points are the vertices of the Feasible Region. (-1,-5), (0, -9), (1, 5), (2, 6), (3, 2) From these values, the maximum value of the objective function, 2x - 4y, is O 42 O -20 O 18 O 36
The required maximum value of the Feasible region is 36.
The given vertices are (-1,-5), (0, -9), (1, 5), (2, 6), and (3, 2).
To find the maximum value of the objective function, 2x - 4y, we need to evaluate this function at each of these vertices and then choose the largest value obtained.
2x - 4y at (-1,-5) = 2(-1) - 4(-5) = 22x - 4y
at (0, -9) = 2(0) - 4(-9) = 36 (largest so far)2x - 4y
at (1, 5) = 2(1) - 4(5) = -182x - 4y
at (2, 6) = 2(2) - 4(6) = -122x - 4y
at (3, 2) = 2(3) - 4(2) = 2
Thus, the maximum value of the objective function, 2x - 4y, is 36.
Therefore, option O 36 is the correct answer.
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A shareholders' group, in lodging a protest, claimed that the mean tenure for a chief executive officer (CEO) was at least nine years. A survey of companies reported in The Wall Street Journal found a sample mean tenure of ¯ x = 7.27 years for CEOs with a standard deviation of s = 6.38 years. Assume 85 companies were included in the sample. Formulate a hypotheses that can be used to challenge the validity of the claim made by the shareholders? group. At a level of significance α = 0.05 , what is your conclusion?
Null Hypothesis (H0): The mean tenure for CEOs is at least nine years.
Alternative Hypothesis (H1): The mean tenure for CEOs is less than nine years.
In the given scenario, the sample mean tenure (¯x) is 7.27 years, and the standard deviation (s) is 6.38 years. The sample size is 85 companies. To test the hypotheses, we calculate the test statistic using the formula:
t = (¯x - μ) / (s / √n). In this case, μ represents the hypothesized mean tenure, which is nine years. After calculating the test statistic, we compare it to the critical value obtained from the t-distribution table with (n-1) degrees of freedom and the given significance level (α = 0.05). If the test statistic falls in the critical region, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
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For a function y = (x² + 2) (x³ + x² + 1)², state the steps to find the derivative.
Using product rule and chain rule, the derivative of the function y = (x² + 2)(x³ + x² + 1)² is given by:
y' = 2x(x³ + x² + 1)² + 2(x² + 2)(x³ + x² + 1)(3x² + 2x)
What is the derivative of the function?To find the derivative of the function y = (x² + 2)(x³ + x² + 1)², we can use the product rule and the chain rule.
Let's denote the first factor (x² + 2) as u and the second factor (x³ + x² + 1)² as v.
Using the product rule (u * v)', the derivative of the function is given by:
y' = u' * v + u * v'
First, let's find the derivative of u (x² + 2):
u' = d/dx (x² + 2)
= 2x
Next, let's find the derivative of v (x³ + x² + 1)² using the chain rule:
v' = d/dx (x³ + x² + 1)²
= 2(x³ + x² + 1) * (d/dx (x³ + x² + 1))
= 2(x³ + x² + 1) * (3x² + 2x)
Now we can substitute the values of u, u', v, and v' into the derivative formula:
y' = (2x) * (x³ + x² + 1)² + (x² + 2) * [2(x³ + x² + 1) * (3x² + 2x)]
Simplifying further:
y' = 2x(x³ + x² + 1)² + (x² + 2) * 2(x³ + x² + 1) * (3x² + 2x)
y' = 2x(x³ + x² + 1)² + 2(x² + 2)(x³ + x² + 1)(3x² + 2x)
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If you evaluate the integral expression Blank 1 Add your answer 12x-1|dx 5 Points the result is Blank 1 (use fraction or decimal in 2 decimal places, no spaces)
3 Points √�
The result of evaluating the integral expression ∫(12x - 1) dx is 6x^2 - x + C, where C is the constant of integration.
To evaluate the integral, we use the power rule of integration, which states that the integral of x^n dx is (1/(n+1))x^(n+1) + C, where C is the constant of integration. Applying this rule to the integral of 12x - 1, we integrate each term separately.
The integral of 12x is (12/2)x^2 = 6x^2, and the integral of -1 is -x. Therefore, the result of the integral expression is 6x^2 - x + C, where C is the constant of integration.
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Assignment I
Height of students in statistics
Fall 2004, Height in Inches
63 62 70 74 68
62 67 70 72 65
73 60 65 69
69 67 65 62
70 64 63 75
72 60 67 63
64 67 65 68
Construct Tally Sheet
⚫ Frequency Distribution Table
o Class, absolute, relative, and percentage distribution
⚫ Histogram and Frequency Polygon
⚫ Cumulative distribution, less than and percentiles included
The height of students in statistics in Fall 2004 is distributed with a mean of 67 inches and a standard deviation of 2 inches. The most common height is 67 inches, followed by 65 inches and 68 inches.
The tally sheet shows that the most common height is 67 inches, with 7 students. This is followed by 65 inches and 68 inches, with 6 students each. The least common height is 60 inches, with 1 student.
The frequency distribution table shows that the absolute frequency of each height is the same as the tally sheet. The relative frequency of each height is calculated by dividing the absolute frequency by the total number of students, which is 20. The percentage distribution of each height is calculated by multiplying the relative frequency by 100%.
The histogram shows the distribution of the data in a graphical form. The frequency polygon is a line graph that connects the midpoints of the tops of the bars in the histogram.
The cumulative distribution shows the percentage of students who are less than or equal to a certain height. The percentiles show the percentage of students who are equal to or less than a certain height.
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Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers.
f(x)=x−−√+2, g(x)=x2+3
Reminder, to use sqrt(() to enter a square root.
f(g(x))=
__________
g(f(x))=
__________
The mathematical procedure known as the square root is the opposite of squaring a number. It is represented by the character "." A number "x"'s square root is another number "y" such that when "y" is squared, "x" results.
Given functions:f(x)=x−−√+2g(x)=x2+3.
We add g(x) to the function f(x) to find f(g(x)):
f(g(x)) = f(x^2 + 3)
Let's now make this expression simpler:
f(g(x)) = (x^2 + 3)^(1/2) + 2
f(g(x)) is therefore equal to (x2 + 3 * 1/2) + 2.
We add f(x) to the function g(x) to find g(f(x)):
g(f(x)) = (f(x))^2 + 3
Let's now make this expression simpler:
g(f(x)) = ((x - √(x) + 2))^2 + 3
G(f(x)) = (x - (x) + 2)2 + 3 as a result.
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Calculus question need help answering please show all work,
Starting with the given fact that the type 1 improper integral
[infinity]
∫ 1/x^p dx converges to 1/p-1
1
when p>1, use the substitution u = 1/x to determine the values of p for which the type 2 improper integral
1
∫ 1/x^p dx
0
converges and determine the value of the integral for those values of p.
The type 2 improper integral ∫(1/x^p) dx from 0 to 1 converges for p < 1, and its value is 1/(1 - p).
We start by substituting u = 1/x, which gives us du = -dx/x^2. We can rewrite the integral in terms of u as follows:
∫(1/x^p) dx = ∫u^p (-du) = -∫u^p du.
Now we need to consider the limits of integration. When x approaches 0, u approaches infinity, and when x approaches 1, u approaches 1. So our integral becomes:
∫(1/x^p) dx = -∫u^p du from 0 to 1.
To evaluate this integral, we use the antiderivative of u^p, which is u^(p+1)/(p+1). Applying the limits of integration, we have:
∫(1/x^p) dx = -[u^(p+1)/(p+1)] evaluated from 0 to 1.
When p+1 ≠ 0 (i.e., p ≠ -1), the integral converges. Thus, p must be less than 1. Plugging in the limits of integration, we obtain:
∫(1/x^p) dx = -(1^(p+1)/(p+1)) + 0^(p+1)/(p+1) = -1/(p+1) = 1/(1-p).
Therefore, the type 2 improper integral converges for p < 1, and its value is 1/(1 - p).
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The type 2 improper integral ∫(1/x^p)dx from 0 to 1 converges when p < 1. The value of the integral for those values of p is 1/(1 - p).
To determine the values of p for which the type 2 improper integral converges, we can use the substitution u = 1/x. As x approaches 0, u approaches positive infinity, and as x approaches 1, u approaches 1. We can rewrite the integral in terms of u as follows:
∫(1/x^p)dx = ∫(1/(u^(1-p))) * (du/dx) dx
= ∫(1/(u^(1-p))) * (-1/x^2) dx
= ∫(-1/(u^(1-p))) * (x^2) dx.
Now, when p > 1, the original integral converges to 1/(p - 1). Therefore, for the type 2 improper integral to converge, we need the same behavior when p < 1. In other words, the integral must converge as x approaches 0. Since the limits of integration for the type 2 integral are from 0 to 1, the convergence at x = 0 is crucial.
For the integral to converge, we require that the integrand becomes finite as x approaches 0. In this case, the integrand is (-1/(u^(1-p))) * (x^2). As x approaches 0, the factor x^2 becomes infinitesimally small, and for the integral to converge, the term (-1/(u^(1-p))) must compensate for the decrease in x^2. This is only possible when p < 1, as the power of u in the denominator ensures that the integral converges.When p < 1, the type 2 improper integral converges, and its value can be found using the formula 1/(1 - p). Therefore, the value of the integral for those values of p is 1/(1 - p).
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4. Ms. Levi recommended that Ms. Garrett use a random number table to select her sample of 10 students. How would you recommend Ms. Garrett assign numbers and select her random sample? TALK the TALK Lunching with Ms. Garrett Ms. Garrett wishes to randomly select 10 students for a lunch meeting to discuss ways to improve school spirit. There are 1500 students in the school.
Random number table is a list of random digits used to make random selections. When the individuals or objects to be studied are presented in a numbered list, then a random sample can be drawn by the use of random numbers.
To make random selections, it is useful to use a table of random numbers. The use of random number tables to select the sample is appropriate because all members of the population have an equal chance of being selected.
There are several ways to use random numbers to select a sample of 10 students from a school of 1500 students.
These include:
Assigning a number to each student and selecting the numbers randomly from a table of random numbers.
Firstly, assign a unique number to each student in the school. It is important that each student is assigned a unique number so that each student has the same probability of being selected as any other student in the school.
The numbers can be assigned in any order, but it is often helpful to use a systematic method, such as assigning numbers alphabetically by last name or sequentially by student ID number.
Next, use a table of random numbers to select the sample of 10 students. This is done by starting at a random point in the table of random numbers and selecting the first number in the table that falls within the range of student numbers (e.g., 001-1500).
This is repeated until a sample of 10 students has been selected.
The advantage of using random numbers is that it ensures that the sample is unbiased and representative of the population.
It also eliminates the possibility of researcher bias in selecting the sample, which can occur if the researcher selects the sample based on personal preference or convenience.
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What are the differences and the similarity between a short futures contract and a option?
The main difference between a short futures contract and an option is the obligation involved. In a short futures contract, the seller is obligated to deliver the underlying asset at a predetermined price and date, regardless of market conditions.
In contrast, an option provides the buyer with the right, but not the obligation, to buy (call option) or sell (put option) the underlying asset at a specified price and date. Both short futures contracts and options are derivative financial instruments that allow investors to speculate on price movements, but options provide more flexibility as they do not carry the same obligation as futures contracts.
Obligation: In a short futures contract, the seller (short position) is obligated to deliver the underlying asset at a specified price and date in the future.
Potential Profit/Loss: The seller profits if the price of the underlying asset decreases, but faces losses if the price increases.
Market Exposure: The seller is exposed to unlimited downside risk, as there is no cap on potential losses.
Margin Requirements: Sellers need to maintain margin accounts to cover potential losses and ensure contract performance. Futures contracts require the seller to deliver the asset, while options provide the buyer with the right, but not the obligation, to buy or sell. Options offer more flexibility but come with a premium cost, while futures contracts have unlimited downside risk and require margin accounts.
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Let h(x) = 25x² + 20x +4.
(a) Find the vertex of the parabola. (b) Use the discriminant to determine the number of x-intercepts the graph will have. Then determine the x-intercepts. (a) The vertex is
(Type an ordered pair, using integers or fractions.)
the graph of the parabola will have one x-intercept, and its x-coordinate is -2/5.
To find the vertex of the parabola represented by the quadratic function [tex]h(x) = 25x² + 20x + 4[/tex], we can use the formula for the x-coordinate of the vertex, given by x = -b / (2a), where a and b are the coefficients of the quadratic term and the linear term, respectively.
In this case, a = 25 and b = 20. Plugging these values into the formula, we get:
[tex]x = -20 / (2 * 25)[/tex]
x = -20 / 50
x = -2/5
To find the y-coordinate of the vertex, we substitute the x-coordinate we found into the original function:
[tex]h(-2/5) = 25(-2/5)² + 20(-2/5) + 4[/tex]
[tex]= 25(4/25) - 8/5 + 4[/tex]
[tex]= 4 - 8/5 + 4[/tex]
[tex]= 4 - 8/5 + 20/5[/tex]
[tex]= (4 + 20 - 8) / 5[/tex]
[tex]= 16/5[/tex]
Therefore, the vertex of the parabola is at (-2/5, 16/5).
Now let's move on to part (b) of the question.
The discriminant (Δ) can be used to determine the number of x-intercepts the graph will have. In the quadratic formula, the discriminant is the expression under the square root (√) sign, given by Δ = b² - 4ac.
For our quadratic function h(x) = 25x² + 20x + 4, we have a = 25, b = 20, and c = 4. Substituting these values into the discriminant formula:
[tex]Δ = (20)² - 4(25)(4)[/tex]
= 400 - 400
= 0
Since the discriminant is equal to 0, it means that there is only one x-intercept for the graph of this parabola.
To determine the x-intercept, we can set h(x) equal to 0 and solve for x:
25x² + 20x + 4 = 0
However, since the discriminant is 0, we already know that there is only one x-intercept, which is the x-coordinate of the vertex we found earlier: x = -2/5.
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Factor the given polynomial. Factor out
−1
if the leading coefficient is negative.
2x2y−6xy2+10xy
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes within your choice.
A.2 x squared y minus 6 xy squared plus 10 xy equals enter your response here
2x2y−6xy2+10xy=enter your response here
B.
The polynomial is prime.
The given polynomial 2x²y - 6xy² + 10xy cannot be factored further.the given polynomial does not have any common factors that can be factored out,
To determine if the given polynomial can be factored, we look for common factors among the terms. In this case, we have 2x²y, -6xy², and 10xy.
We can try factoring out the greatest common factor (GCF) from the terms. The GCF is the largest term that divides evenly into each term.
Taking a closer look at the terms, we can see that the GCF is 2xy. Factoring out 2xy from each term gives us: 2xy(1x - 3y + 5)
However, this is not a complete factorization. The expression 1x - 3y + 5 cannot be factored further since it does not have any common factors or simplifications.
Therefore, the polynomial 2x²y - 6xy² + 10xy cannot be factored any further.
In summary, the given polynomial does not have any common factors that can be factored out, and the expression 1x - 3y + 5 cannot be simplified or factored. Thus, the polynomial 2x²y - 6xy² + 10xy is considered to be prime.
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Find the general solution of the system of equations. ′=(5
1 -4 1)x
The general solution of the system of equations is given by: x(t) = c₁ + c₂t, y(t) = -5c₁ - 5c₂t. Where c₁ and c₂ are arbitrary constants.
Solving for General Solution of a SystemTo find the general solution of the system of equations:
X' = AX
where X = [x, y] and
A = [tex]\left[\begin{array}{ccc}5&1\\-4&1\end{array}\right][/tex]
we can proceed as follows:
Let's write the system of equations separately:
x' = 5x + y
y' = -4x + y
Taking the derivatives of x and y with respect to some variable (e.g., time), we obtain:
x'' = 5x' + y'
y'' = -4x' + y'
We can rewrite the system of equations in matrix form as:
X'' = AX'
Now, let's substitute X' with another variable, say V:
V = X'
We have:
X'' = AV
Therefore, we now have a new system of equations:
V = X'
X'' = AV
Substituting V back into the second equation, we get:
X'' = A(X')
This becomes:
X'' = AX'
This implies that X' is an eigenvector of A with eigenvalue 0.
Next, we need to find the eigenvectors of A. To do that, we solve the equation:
(A - 0I)V = 0
where I is the identity matrix and V is the eigenvector.
For A = [tex]\left[\begin{array}{ccc}5&1\\-4&1\end{array}\right][/tex] the matrix (A - 0I) becomes:
[tex]\left[\begin{array}{ccc}5&1\\-4&1\end{array}\right][/tex]V = [tex]\left[\begin{array}{ccc}5&1\\-4&1\end{array}\right][/tex][tex]\left[\begin{array}{ccc}v_{1} \\v_{2} \end{array}\right][/tex] = [tex]\left[\begin{array}{ccc}0\\0\end{array}\right][/tex]
This gives us the following system of equations:
5v₁ + v₂ = 0
-4v₁ + v₂ = 0
We can solve this system of equations to find the eigenvectors:
5v₁ + v₂ = 0 --> v₂ = -5v₁
-4v₁ + v₂ = 0 --> v₂ = 4v₁
From these equations, we can choose a value for v₁ (e.g., 1) and calculate the corresponding v₂:
v₂ = -5(1) = -5
So, one eigenvector is v = [1, -5].
The general solution of the system of equations is given by:
X(t) = [tex]c_{1}e^{(\lambda_{1}t)v_{1}} + c_{2}e^{(\lambda_{2}t)v_{2}}[/tex]
where λ₁ and λ₂ are the eigenvalues and v₁ and v₂ are the corresponding eigenvectors.
In this case, since we have only one eigenvalue of 0 (due to X' being an eigenvector of A with eigenvalue 0), the general solution becomes:
X(t) = [tex]c_{1}e^{(0t)v_{1}} + c_{2}e^{(0t)v_{2}}[/tex]
Simplifying, we have:
X(t) = c₁v₁ + c₂tv₂
Substituting the values for v₁ and v₂, we get:
X(t) = c₁[1, -5] + c₂t[1, -5]
Expanding, we have:
x(t) = c₁ + c₂t
y(t) = -5c₁ - 5c₂t
where c₁ and c₂ are arbitrary constants.
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Complete the following statements in the blanks provided. (1 Point each).
i. Write the first five terms of the sequence { an}, if a₁ = 6, an+1 = an/n
ii. Find the value of b for which the geometric series converges 20 36 1+ e +e²0 +e³0 +... = 2 b=
The first five terms of the sequence {an} can be found using the recursive formula given: an+1 = an/n. Starting with a₁ = 6, we can calculate the next terms as follows.
i. a₂ = a₁/1 = 6/1 = 6
a₃ = a₂/2 = 6/2 = 3
a₄ = a₃/3 = 3/3 = 1
a₅ = a₄/4 = 1/4 = 0.25
Therefore, the first five terms of the sequence are 6, 6, 3, 1, and 0.25.
ii. To find the value of b for which the geometric series converges to the given expression, we need to consider the sum of an infinite geometric series. The series can be expressed as:
S = 20 + 36 + 1 + e + e²0 + e³0 + ...
In order for the series to converge, the common ratio (r) of the geometric progression must satisfy the condition |r| < 1. Let's analyze the terms of the series to determine the common ratio:
a₁ = 20
a₂ = 36
a₃ = 1
a₄ = e
a₅ = e²0
...
We can observe that the common ratio is e. Therefore, for the series to converge, |e| < 1. However, the value of e is approximately 2.71828, which is greater than 1. Thus, the series does not converge.
As a result, there is no value of b for which the given geometric series converges.
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(a) Prove the following statement: Vm, x € R, if m € Z and rZ, then [x] + [2m -x] = 2m + 1. Va, b = Z, if a #0 and b‡0 then ged(a, b) - lcm(a, b) = ab. (b) Disprove the following statement: (4 marks) (2 marks)
For all m and x in R, if m is an integer and x is a real number, then [x] + [2m - x] = 2m + 1. The statement "For all a and b in Z, if a # 0 and b # 0 then ged(a, b) - lcm(a, b) = ab" is false.
Let m be an integer and x be a real number. Then [x] is the greatest integer less than or equal to x, and [2m - x] is the greatest integer less than or equal to 2m - x. Since m is an integer, [2m - x] is also an integer. Therefore, [x] + [2m - x] is an integer.
Now, let y = [x] + [2m - x]. Then y is an integer and y <= 2m. Since x is a real number, there exists a non-integer real number z such that z < x <= z + 1. Therefore, [x] = z and [2m - x] = 2m - z - 1.
Substituting these values for [x] and [2m - x] into the equation y = [x] + [2m - x], we get y = z + (2m - z - 1) = 2m. Therefore, y = 2m + 1.
The statement is false because it is possible for ged(a, b) - lcm(a, b) to be equal to zero. For example, if a = 1 and b = 1, then ged(a, b) = lcm(a, b) = 1, so ged(a, b) - lcm(a, b) = 0.
Another way to disprove the statement is to find a counterexample. A counterexample is an example that shows that the statement is false. For example, the numbers a = 2 and b = 3 are a counterexample to the statement because ged(a, b) - lcm(a, b) = 1 - 6 = -5.
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6. An input of 251³ u(t) is applied to the input of a Type 3 unity feedback system, as shown in Figure P7.1,
where
G(s) = 210(s + 4)(s+6)(s + 11)(s +13)/s³ (s+7)(s+14)(s +19)
Find the steady-state error in position.
In a Type 3 unity feedback system with the transfer function G(s), where G(s) = 210(s + 4)(s+6)(s + 11)(s +13)/s³ (s+7)(s+14)(s +19), the steady-state error in position can be determined by evaluating the system's transfer function at s = 0.
The steady-state error in position can be found by evaluating the transfer function of the system at s = 0. In this case, the transfer function of the system is G(s) = 210(s + 4)(s+6)(s + 11)(s +13)/s³ (s+7)(s+14)(s +19).
To find the steady-state error, we substitute s = 0 into the transfer function. When s = 0, the denominator of the transfer function becomes non-zero, and the numerator evaluates to 210(4)(6)(11)(13) = 2,090,640.
The steady-state error in position (ess) is given by the formula ess = 1 / (1 + Kp), where Kp represents the position error constant.
Since the system is a Type 3 system, the position error constant is non-zero. Therefore, we can compute the steady-state error as ess = 1 / (1 + Kp).
In this case, the Kp value can be determined by evaluating the transfer function at s = 0. Substituting s = 0 into the transfer function, we get G(0) = 2,090,640.
Therefore, the steady-state error in position (ess) is ess = 1 / (1 + 2,090,640) = 1 / 2,090,641.
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Which of the following values cannot be probabilities? 0,5/3, 1.4, 0.09, 1, -0.51, √2, 3/5 Select all the values that cannot be probabilities. A. -0.51 B. √2 C. 5 3 D. 3 5 E. 1.4 F. 0.09 G. 0 H. 1
We can see here that the values that cannot be probabilities are:
A. -0.51
B. √2
C. 5/3
What is probability?Probability is a measure of the likelihood of an event to occur. It is expressed as a number between 0 and 1, where 0 indicates that the event is impossible and 1 indicates that the event is certain.
A probability is a number between 0 and 1, inclusive. The values -0.51, √2, and 5/3 are all outside of this range.
Please note that:
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