The answer to question is given below.
(i) To find the mean, E(N), use the following formula:
mean = E(N) = ∑[n · P(N = n)] for all values of n.
The distribution given above is geometric: P(N = n) = p^ n (1-p)
where n = 1,2,3,...
Therefore, E(N) = ∑[n · P(N = n)] = ∑ [n · p^ n (1-p)] for n = 1,2,3,...
Since this sum is infinite, we have to truncate the summation and compute the mean for a finite number of terms. We can use 40 since there are 40 independent observations. Therefore, we have:
E(N) ≈ ∑[n · P(N = n)] for n = 1 to 40
= 1 · p(1-p) + 2 · p^2(1-p) + 3 · p^3(1-p) + ... + 40 · p^40(1-p)
= (1-p) ∑[n · p^n] for n = 1 to 40
= (1-p) [p + 2p^2 + 3p^3 + ... + 40p^40]
= (1-p) p ∑[n · p^(n-1)] for n = 1 to 40
= (1-p) p [1 + 2p + 3p^2 + ... + 40p^39]
= p(1-p) [1p^(1-1) + 2p^(2-1) + 3p^(3-1) + ... + 40p^(40-1)]
= p(1-p) ∑[n · p^(n-1)] for n = 1 to 40
= p(1-p) ∑[(n-1+1) · p^(n-1)] for n = 1 to 40
= p(1-p) [∑[(n-1) · p^(n-1)] + ∑[1 · p^(n-1)]] for n = 1 to 40
= p(1-p) [∑[n · p^(n-1)] - ∑[p^(n-1)]] + p(1-p) ∑[p^(n-1)] for n = 1 to 40
= p(1-p) [d/dp ∑[p^n]] - p(1-p) (1/(1-p)) + p(1-p) (1/(1-p))
= p(1-p) [d/dp (1/(1-p)) ∑[(1-p)p^n]] + 1
= p(1-p) [d/dp (1/(1-p)) (1-p)/(1-p)^(40+1)] + 1
= p(1-p) [d/dp (1-p)^(-40)) + 1
= p(1-p) (40(1-p)^(-41)) + 1
= 40p/(1-p) - 40
(ii) Since this is a geometric distribution, the maximum likelihood estimator (MLE) for the unknown parameter p is given by:
MLE: ê = x/n
where x is the number of successes (species of insect caught in a trap) and n is the sample size (40). To find an equation that determines the MLE ê, differentiate the log-likelihood and equate it to zero to find the maximum of the likelihood function. The log-likelihood for a geometric distribution is given by:
L = ∑ [log(P(N = n))] for n = 1 to 40
= ∑ [log(p^n (1-p))] for n = 1 to 40
= ∑ [n · log(p) + log(1-p)] for n = 1 to 40
= (40 · log(p) + ∑ [n · log(p)] + ∑ [log(1-p)])
Use the formula MLE = x/n to replace p by x/n. This gives:
L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])
Differentiate L with respect to x/n and equate to zero to obtain the MLE ê:0 = dL/d(x/n) = (40/x) - (n/x) - ∑ [1/(1 - x/n)]
Solving this equation will give ê in terms of x and n.
(iii) The Fisher information, I(p), is defined as:
I(p) = -E[d^2/dp^2 L] = E[d/dp (d/dp L)]
where L is the log-likelihood function.
From part (ii), we have:
L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])
Therefore,
∂L/∂p = (40/x) - (n/x) and∂^2L/∂p^2 = -40/x^2.The Fisher information is therefore:
I(p) = -E[d^2/dp^2 L] = E[40/x^2] = 40E[x/n]^(-2) = 40/p^2.
Using the asymptotic normality of the MLE, the 95% confidence interval for p is approximately given by:
p ± 1.96 · sqrt(Var(p))
where Var(p) = 1/I(p) = p^2/40.
Using Ô = 0.75, we have ê = x/n = (N1 + N2 + ... + N40)/40 = (28 + 30 + ... + 22)/40 = 0.625.
Therefore, p ± 1.96 · sqrt(Var(p))= 0.625 ± 1.96 · sqrt(0.625^2/40)= (0.478, 0.772).
(iv) When N1 = 100, the maximum likelihood estimate, ê(1), can be found iteratively as follows:
ê(1) = 0.70MLE = ê(1) = x/n
where x is the number of species of insect caught in a trap and n = 40. Therefore, ê(1) can be computed from the data. For example, if x = 25, then ê(1) = 25/40 = 0.625. To obtain ê(2), we need to solve the equation obtained in part (ii) for n = 100:0 = (40/x) - (100/x) - ∑ [1/(1 - x/100)]
We can use Newton's method to solve this equation numerically. Let ê(2) be the root obtained after one iteration of Newton's method. Then, we have:
ê(2) = ê(1) - f(ê(1))/f'(ê(1))where f(p) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] and f'(p) = -∑ [x/100(x - 100)^2].
For example, if x = 25 and ê(1) = 0.625, then:
ê(2) = 0.625 - f(0.625)/f'(0.625)= 0.625 - (-0.1155)/(0.1869)= 0.625 + 0.6171= 1.242.
This value is not valid since the MLE must lie between 0 and 1. Therefore, we need to use a different starting value of ê. Let ê(1) = 0.80. Then, we have:
f(0.80) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] = 0.0142f'(0.80) = -∑ [x/100(x - 100)^2] = -0.1026
Using Newton's method, we have:
ê(2) = 0.80 - f(0.80)/f'(0.80)= 0.80 - (0.0142)/(-0.1026)= 0.9367
ê(3) can be obtained in a similar manner by solving the equation obtained in part (ii) for n = 100 using ê(2) as the starting value.
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Question (5 points): Using Laplace transform to solve the IVP: y" + 5y = eat, y(0) = 0, y' (O) = 0, then, we have Select one: y(t) =(-1 1 93 – 4s2 + 5s – 20 O y(t) = 2-1 1 482 33 5s + 20 O None of these. Ο Ο 1 yt 3 – (t) = c{+60-60-20) {{+40 + 5a + 20 O 1 y(t) = 2-1
Given the differential equation is y''+5y = e^(at) with initial conditions y(0) = 0 and y'(0) = 0, the correct answer is: y(t) = (-1/5√(5)) cos (√(5)t)+(1/5) sin (√(5)t).
To solve the given initial value problem using Laplace transform, we need to apply Laplace transform on both sides of the differential equation.
y''+5y=e^(at) L{y''+5y} = L{e^(at)) s^2Y(s)-sy(0)-y'(0)+5Y(s)=1/(s-a) [by Laplace transform formula] s^2Y(s)+5Y(s)=1/(s-a) ... [i]
Applying Laplace transform on both sides, we get:
L{y''+5y}=L{e^(at))
Using the initial conditions, we get Y(s)=1/[(s-a)(s^2+5)] Y(s) = [A/(s-a)] + [(Bs+C)sin(t)+ (Ds+E)cos(t)]/√(5) (i)
To find the values of A, B, C, D, and E, we take the inverse Laplace transform of both sides of equation (i) using partial fraction expansion. Let's solve for A:
Y(s)=A/(s-a)+(Bs+C)sin(t)/√(5)+(Ds+E)cos(t)/√(5)
Multiplying by s-a on both sides: (s-a)Y(s)=A+Bssin(t)/√(5)+Csincos(t)/√(5)+Dscos(t)/√(5)+Esin(t)/√(5)
Taking the inverse Laplace transform: y(t)=Ae^(at)+(B/√(5))sin(√(5)t)+(C/√(5))cos(√(5)t)+(D/√(5))cos(√(5)t)+(E/√(5))sin(√(5)t)
Differentiating y(t) with respect to t, we get:
y'(t)=Aae^(at)+Bcos(√(5)t)-Csin(√(5)t)-Dsin(√(5)t)+Ecos(√(5)t)
Using the initial conditions, y(0)=0 and y'(0)=0 in equation (iii), we get:
0=A+E ...(iv)0=A+B/√(5)+D/√(5) ...(v)
Solving equations (iv) and (v) simultaneously, we get A=0, B=√(5)/5, D=-√(5)/5, and E=0
Substituting these values in equation (iii), we get: y(t)=(√(5)/5)sin(√(5)t)-(√(5)/5)cos(√(5)t)
Therefore, the correct answer is: y(t)= (-1/5√(5))cos(√(5)t)+(1/5)sin(√(5)t).
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Please help.
Is algebra.
Answer:
#4 is B
#5 is also B
Step-by-step explanation:
i big brain
Answer:
4. b) [tex]4x^2-20xy+25y^2[/tex]
5. b) [tex]x^2+14x+49[/tex]
Step-by-step explanation:
4. [tex](2x-5y)^2[/tex]
First, one must rewrite the exponential equation as a multiplication problem,
[tex](2x-5y)(2x-5y)[/tex]
Now distribute, multiply every term in one of the parenthesis by every term in the other parenthesis,
[tex]=(2x-5y)(2x-5y)\\\\=(2x)(2x)+(2x)(-5y)+(-5y)(2x)+(-5y)(-5y)[/tex]
Now simplify the given expression,
[tex]=(2x)(2x)+(2x)(-5y)+(-5y)(2x)+(-5y)(-5y)\\\\=4x^2-10xy-10xy+25y^2[/tex]
Combine like terms,
[tex]=4x^2-20xy+25y^2[/tex]
5.[tex](x+7)^2[/tex]
To solve this problem, one should follow the same series of steps as they did to solve the last expression. First, rewrite the exponential expression as a multiplication problem.
[tex](x+7)(x+7)[/tex]
Now distribute, multiply every term in one of the parenthesis by every term in the other parenthesis,
[tex]=(x)(x)+(7)(x)+(7)(x)+(7)(7)[/tex]
Simplify the expression,
[tex]=(x)(x)+(7)(x)+(7)(x)+(7)(7)\\\\=x^2 + 7x + 7x + 49[/tex]
Finally, combine like terms,
[tex]=x^2+14x+49[/tex]
Verify the equation: (tan x - 1)/(tan x + 1) = (1 - cot x)/(1 + cot x)
Answer:
True.
Step-by-step explanation:
given equation: (tan x - 1)/(tan x + 1) = (1 - cot x)/(1 + cot x)
1. manipulate the right side by using trigonometric identities
(tan(x) - 1)/(tan(x) + 1) = (-cos(x) + sin(x))/(cos(x) + sin(x))
2. manipulate the right side by using trigonometric identities
(-cos(x) + sin(x))/(cos(x) + sin(x)) = (-cos(x) + sin(x))/(cos(x) + sin(x))
Both sides of the equation are now equal -> (tan x - 1)/(tan x + 1) = (1 - cot x)/(1 + cot x) is true.
Answer? Please help!!
Answer:
true
Step-by-step explanation:
Greg and cc went to the burger stand and bought dinner. Greg had 2 cheeseburgers and 5 fries. CC bought 3 cheeseburgers and 2 fries. Greg paid $24.75. CC also paid $24.75. How much would 2 cheeseburgers and 3 fries cost?
Answer:
20.00
Step-by-step explanation:
Find the SUM of the perfect square roots that fall between the square root of 26 and the square root of 70.
Answer:
yo is your math class teacher named wicker? I think I'm ur classmate lol
IQs are known to be normally distributed with mean 100 and standard deviation 15. (a) What percentage of people have an IQ lower than 91? ? (b) Fill in the blank. 75% of the population have an IQ that is greater than Problem #7(a): Enter your answer as a percentage, correct to 2 decimals, without the % sign. e.g., 28.31 Problem #7(b): answer correct to 2 decimals
75% of the population has an IQ greater than 89.95.
(a)What percentage of people have an IQ lower than 91?The given distribution is the normal distribution, with the mean 100 and standard deviation 15. It is required to calculate the percentage of people having an IQ score lower than 91.
To calculate the percentage of people having an IQ score lower than 91, standardize the given IQ score of 91 using the formula of z-score.z=(x−μ)/σwherez is the standardized score,x is the raw score,μ is the mean, andσ is the standard deviation.
The values can be substituted as follows.z=(91−100)/15=−0.6Now, find the probability of having a z-score less than or equal to -0.6 using the standard normal distribution table.
The value in the table is 0.2743, which means the probability of having a z-score less than or equal to -0.6 is 0.2743.Thus, 27.43% of people have an IQ score lower than 91.
(a) 27.43% of people have an IQ lower than 91.(b)Fill in the blank. 75% of the population have an IQ that is greater than X.
In order to find X, the z-score can be calculated using the formula of z-score.z=(x−μ)/σwherez is the standardized score,x is the raw score,μ is the mean, andσ is the standard deviation.
The z-score for the given problem can be calculated as follows:z = (x - μ)/σ (standardized score formula)z = (x - 100)/15 (values substituted)To find the value of x for which 75% of the population have an IQ greater than x, we need to determine the z-score that corresponds to the 25th percentile.
This is because 75% of the population is above the 25th percentile and below the 100th percentile.Using a standard normal distribution table, we can find the z-score that corresponds to the 25th percentile. The z-score is approximately -0.67.
Now that we have the z-score, we can solve for x as follows.-0.67 = (x - 100)/15 (substitute z-score)-10.05 = x - 100 (multiply both sides by 15)-89.95 = x
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(2p²+7p-39)÷(2p-7)
Quotient ?
Answer:
Quotient: p + 7
Remainder: 10
Step-by-step explanation:
To find the quotient of the expression (2p² + 7p - 39) ÷ (2p - 7), we can use long division or synthetic division. Let's use long division:
____________________
2p - 7 | 2p² + 7p - 39
We start by dividing the first term of the dividend by the first term of the divisor, which gives us 2p² ÷ 2p = p. We then multiply p by the divisor (2p - 7) and subtract it from the dividend:
p
____________________
2p - 7 | 2p² + 7p - 39
- (2p² - 7p)
14p - 39
____________________
2p - 7 | 2p² + 7p - 39
- (2p² - 7p)
___________
14p - 39
We repeat the process by dividing the first term of the new dividend (14p - 39) by the first term of the divisor (2p - 7). This gives us (14p - 39) ÷ (2p - 7) = 7. We then multiply 7 by the divisor (2p - 7) and subtract it from the new dividend:
p + 7
____________________
2p - 7 | 2p² + 7p - 39
- (2p² - 7p)
___________
14p - 39
- (14p - 49)
___________
10
We are left with a remainder of 10. Therefore, the quotient is p + 7 with a remainder of 10.
Quotient: p + 7
Remainder: 10
Hope this helps!
p + 7 should be it.
I am not 100% sure?
Can you help me for these 3 questions.
Answer:
Step-by-step explanation:
1) a + a
2) 2n+4 = 6n
3) 1
4n = 4
n = 1
Let A(x)=∫x0f(t)dtA(x)=∫0xf(t)dt, with f(x)f(x) as in figure.
A(x)A(x) has a local minimum on (0,6)(0,6) at x=x=
A(x)A(x) has a local maximum on (0,6)(0,6) at x=x=
To determine the local minimum and local maximum of the function A(x) = ∫₀ˣ f(t) dt on the interval (0, 6), we need to analyze the behavior of A(x) and its derivative.
Let's denote F(x) as the antiderivative of f(x), which means that F'(x) = f(x).
To find the local minimum and maximum, we need to look for points where the derivative of A(x) changes sign. In other words, we need to find the values of x where A'(x) = 0 or A'(x) is undefined.
Using the Fundamental Theorem of Calculus, we have:
A(x) = ∫₀ˣ f(t) dt = F(x) - F(0)
Taking the derivative of A(x) with respect to x, we get:
A'(x) = (F(x) - F(0))'
Since F(0) is a constant, its derivative is zero, and we are left with:
A'(x) = F'(x) = f(x)
Now, let's analyze the behavior of f(x) based on the given figure to determine the local minimum and maximum of A(x) on the interval (0, 6). Without the specific information about the shape of the graph, it is not possible to determine the exact values of x that correspond to local minimum or maximum points.
To find the local minimum, we need to locate a point where f(x) changes from decreasing to increasing. This point would correspond to x = x_min.
To find the local maximum, we need to locate a point where f(x) changes from increasing to decreasing. This point would correspond to x = x_max.
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Alberto invested $5,000 at 6% interest
compounded annually. What will be the
value of Alberto's investment after 8 years?
Answer:
$7969.24
Step-by-step explanation:
5000*1.06 to power of 8 = 7969.24037265
round to 2dp = $7969.24
Answer:
7400
Step-by-step explanation:
−x+9y=
\,\,-19
−19
3x-3y=
3x−3y=
\,\,9
9
Answer:
9!
Step-by-step explanation:
Answers this plz I need help
Answer:
y=2x+5
Step-by-step explanation:
u find the slope and since the y intercept is (0,5), that's the b value
hope this helps
Answer: 2x+5
Step-by-step explanation: calculate the slope of the points with the formula m= (y2-y1)/(x2-x1) then use the y-int to complete the equation
If a test is worth 90 points and I got 12/15 questions right then what would my grade be?
Answer: 80
Step-by-step explanation:
WILL GET BRAINLIEST The scatterplot shows the weight and miles per gallon of the nation's 40 best-selling cars.
Vehicle Weights and Miles
Per Gallon
Miles Per Gallon
Vehicle Weight
Which statement is most strongly supported by the scatterplot?
The last one as we can see that as the wight increases, the miles per gallon decreases.
Plz help! Due tonight!
During halftime of a football game, a slingshot launches T-shirts at the crowd. A T-shirt is launched from a height of 4 feet with an initial upward velocity of 80 feet per second. The T-shirt is caught 41 feet above the field. How long will it take the T-shirt to reach its maximum height? What is the maximum height? What is the range of the function that models the height of the T-shirt over time?
What is the area of one of the triangular faces? in 2 4 in. 3 in. 7 in. 7 in. 5 in. 4 in. 5 in.
Answer:
55
Step-by-step explanation:
Use the method of variation of parameters to find a particular solution of the differential equation 4y" – 4y +y = 16et/2 that does ' not involve any terms from the homogeneous solution. = Y(t) =
The particular solution that does not involve any terms from the homogeneous solution is given by:[tex]Y(t) = C3 + C4te^(-t/2).[/tex]
To find a particular solution of the given differential equation using the method of variation of parameters, we follow these steps:
Solve the associated homogeneous equation: 4y" - 4y + y = 0.
The characteristic equation is:
[tex]4r^2 - 4r + 1 = 0.[/tex]
Solving the quadratic equation, we find two repeated roots: r = 1/2.
Therefore, the homogeneous solution is given by: y_h(t) = C1[tex]e^(t/2)[/tex] + C2t[tex]e^(t/2),[/tex] where C1 and C2 are constants.
Find the particular solution using the variation of parameters.
Let's assume the particular solution has the form:
[tex]y_p(t) = u1(t)e^(t/2) + u2(t)te^(t/2).[/tex]
To find u1(t) and u2(t), we differentiate this expression:
[tex]y_p'(t) = u1'(t)e^(t/2) + u1(t)(1/2)e^(t/2) + u2'(t)te^(t/2) + u2(t)e^(t/2) + u2(t)(1/2)te^(t/2).[/tex]
We equate the coefficients of e^(t/2) and te^(t/2) on both sides of the original equation:
[tex](1/2)(u1(t) + u2(t)t)e^(t/2) = 16e^(t/2).[/tex]
From this, we can deduce that u1(t) + u2(t)t = 32.
Differentiating again:
[tex]y_p''(t) = u1''(t)e^(t/2) + u1'(t)(1/2)e^(t/2) + u1'(t)(1/2)e^(t/2) + u1(t)(1/4)e^(t/2) + u2''(t)te^(t/2) + u2'(t)e^(t/2) + u2'(t)(1/2)te^(t/2) + u2(t)e^(t/2) + u2(t)(1/2)te^(t/2).[/tex]
Setting the coefficient of [tex]e^(t/2)[/tex]equal to zero:
[tex](u1''(t) + u1'(t) + (1/4)u1(t))e^(t/2) = 0.[/tex]
Similarly, setting the coefficient of [tex]te^(t/2)[/tex]equal to zero:
[tex](u2''(t) + u2'(t) + (1/2)u2(t))te^(t/2) = 0.[/tex]
These two equations give us a system of differential equations for u1(t) and u2(t):
u1''(t) + u1'(t) + (1/4)u1(t) = 0,
u2''(t) + u2'(t) + (1/2)u2(t) = 0.
Solving these equations, we obtain:
u1(t) = C3[tex]e^(-t/2)[/tex] + C4t[tex]e^(-t/2),[/tex]
u2(t) = -4C3[tex]e^(-t/2)[/tex] - 4C4t[tex]e^(-t/2).[/tex]
Substitute the values of u1(t) and u2(t) into the assumed particular solution:
[tex]y_p(t) = (C3e^(-t/2) + C4te^(-t/2))e^(t/2) - 4C3e^(-t/2) - 4C4te^(-t/2).[/tex]
Simplifying further:
[tex]y_p(t) = C3 + C4te^(-t/2) - 4C3e^(-t/2) - 4C4te^(-t/2).[/tex]
So, the particular solution that does not involve any terms from the homogeneous solution is given by:
[tex]Y(t) = C3 + C4te^(-t/2).[/tex]
Here, C3 and C4 are arbitrary constants that can be determined using initial conditions or boundary conditions if provided.
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is it A. B. Or C.
please help
Answer:c?
Step-by-step explanation:
Answer:
B
Step-by-step explanation:
Solve: |2x − 1| < 11.
Express the solution in set-builder notation.
{x|5 < x < 6}
{x|–5 < x < 6}
{x|x < 6}
{x|–6 < x < 6}
Answer:
the second one is the answer
Step-by-step explanation:
hope that helps
Answer:
B; {x|–5 < x < 6}
Step-by-step explanation:
The projection matrix is P = A(AT A)-1A". If A is invertible, what is e? Choose the best answer, e.g., if the answer is 2/4, the best answer is 1/2. The value of e varies based on A. Oe=b - Pb e = 0 Oe=AtAb
The value of e varies based on A. Oe=b - Pb e = 0 Oe=AtAb would be (AT A)-1 AT b.
The given projection matrix is P = A(AT A)-1A".
We have been asked to find the value of e if A is invertible. Let's proceed further and solve this problem. First, we need to find the product of A and its transpose, i.e., AT A.A.T.A = [a11 a12 ... a1n] [a21 a22 ... a2n] ... [an1 an2 ... ann] = [Σ(ai1)(aj1) Σ(ai1)(aj2) ... Σ(ai1)(ajn)] [Σ(ai2)(aj1) Σ(ai2)(aj2) ... Σ(ai2)(ajn)] ... [Σ(ain)(aj1) Σ(ain)(aj2) ... Σ(ain)(ajn)]
The inverse of AT A is (AT A)-1. Thus, (AT A)-1 AT A = I.Where I is the identity matrix. So we get P = A(AT A)-1 A".
Now, the value of e can be calculated as: Oe = b - Pe = b - A(AT A)-1 A" b = A x (AT A)-1 x AT b
This is the expression for the solution of the least square problem and if A is invertible, we can find the solution by directly calculating A-1 x b which is nothing but e. Thus, the value of e is e = A-1b.
Substituting the given expression of e, we get e = (AT A)-1 AT b.
Thus, the correct answer is e = (AT A)-1 AT b.
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Which of the following is NOT true for 6s + 25 + 5? *
A. Represents an algebraic expression
B. There is only one value for s
C. A phrase that simplifies to two terms
D. There is a solution for s = 5
Answer:
B. There is only one value for s
Step-by-step explanation:
I hope this works for u.. :3
Can I have a brainliest plz :))
Give the general solution of the linear system x+y-2z = 0 2x + 2y3z = 1 3x + 3y + z = 7.
Answer:
The general solution to the given linear system is x = 3z - 1, y = -z + 1, where z is a free variable. This means that the solution consists of infinitely many points that lie on a straight line in three-dimensional space.
To solve the linear system, we can use the method of elimination or Gaussian elimination. Here, we'll use Gaussian elimination to find the general solution.
We start by writing the augmented matrix of the system:
[1 1 -2 | 0]
[2 2 3 | 1]
[3 3 1 | 7]
To simplify the matrix, we perform row operations to create zeros in the first column below the first entry. We subtract twice the first row from the second row and subtract three times the first row from the third row:
[1 1 -2 | 0]
[0 0 7 | 1]
[0 0 7 | 7]
Next, we divide the second and third rows by 7 to create leading ones:
[1 1 -2 | 0]
[0 0 1 | 1/7]
[0 0 1 | 1]
Now, we perform row operations to create zeros in the second column below the second entry. We subtract the third row from the second row:
[1 1 -2 | 0]
[0 0 1 | 1/7]
[0 0 0 | 0]
From the last row, we can see that 0z = 0, which means that z is a free variable. We can assign a parameter to z, say t, and solve for x and y in terms of t. From the first row, we have x + y - 2z = 0. Plugging in the values for x and y, we get x = 3z - 1 and y = -z + 1. Therefore, the general solution to the linear system is x = 3z - 1, y = -z + 1, where z is a free variable.
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Which describes the graph of the inequality g(x)>2√3x+2?
A. Shading above a solid line
B. Shading above a dotted line
C. Shading below a dotted line
D. Shading below a solid line
Answer:
A. Shading above a solid line.
Step-by-step explanation:
Let [tex]f(x) = 2\sqrt{3}\cdot x + 2[/tex], whose domain is all real numbers, since it is a first order polynomial (linear function) and meaning that for all element of [tex]x[/tex] exists one and only one value for [tex]f(x)[/tex], meaning a solid line. If [tex]g(x) > f(x)[/tex], then the range of all possible results is a shade area above the solid line.
Hence, the correct answer is A.
For a given norm on Rņwe call the matrix A ∈ Rmxn mxn isometry if ||AX|| = |x|| for all x ER". = • Show that the isometry must be regular. • Show that the set of isometries forms a
An isometry on R^n must be regular and the set of isometries forms a group under matrix multiplication.
An isometry is a linear transformation that preserves distances, meaning the norm of the transformed vector is equal to the norm of the original vector. To show that an isometry must be regular (i.e., invertible), we can assume there exists a non-invertible isometry matrix A. In this case, there exists a nonzero vector x such that Ax = 0. However, this contradicts the property of an isometry since ||Ax|| = ||0|| = 0, but ||x|| ≠ 0. Thus, an isometry must be regular.
The set of isometries forms a group under matrix multiplication because it satisfies the group axioms: closure (the product of two isometries is an isometry), associativity (matrix multiplication is associative), identity (identity matrix is an isometry), and inverses (the inverse of an isometry is also an isometry).
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HELPP ME PLSSSS NO BOTS OR I WILL REPORT YOUU!!
Answer:
True
Step-by-step explanation:
It pases vertical line test but does not have an inverse
By using the method of variation of parameters to solve a nonhomogeneous DE with W = -3 W2 = e 112 and W = er, we have ---- Select one: 42 Ou= 41 O U2= O None of these. -4 Ou2 = O U =
By using the method of variation of parameters to solve a nonhomogeneous DE with W = -3 W2 = e 112 and W = er, we have
Given: W1=-3, W2=e^t and W3=er.The general solution of the non-homogeneous differential equation, y" + p(t) y' + q(t) y = g(t) , where p(t) and q(t) are functions of t and g(t) is non-zero function is given by;{eq}y = y_c + y_p {/eq}Where {eq}y_c {/eq} is complementary function and {eq}y_p {/eq} is particular function obtained by using variation of parameters.The solution is as follows:The given differential equation is{eq}y''+3y'+2y=-3e^{-t}+e^{t}+re^t{/eq}Characteristic equation is{eq}m^2+3m+2=0{/eq}Solving above equation gives us, {eq}m=-1,-2{/eq}Therefore, complementary function {eq}y_c=c_1e^{-t}+c_2e^{-2t} {/eq}Now, we find the particular solution by using the method of variation of parameters.Let {eq}y_p=u_1e^{-t}+u_2e^{-2t}{/eq}be a particular solution where {eq}u_1{/eq} and {eq}u_2{/eq} are functions of {eq}t.{/eq}Here W is a Wronskian and is given as:{eq}W=\begin{vmatrix}W_1&W_2\\W_1'&W_2'\\\end{vmatrix}=\begin{vmatrix}-3&e^t\\-1&e^t\\\end{vmatrix}=2e^{2t}+3e^{t}{/eq}Now, we find {eq}u_1{/eq} and {eq}u_2{/eq} as follows:{eq}u_1=\frac{-\int W_2 g(t) dt}{W}=\frac{-\int e^t(-3e^{-t}+e^{t}+re^t)dt}{2e^{2t}+3e^{t}}=-\frac{r}{5}-\frac{7}{10}+\frac{3}{10}e^{t}{/eq}Similarly,{eq}u_2=\frac{\int W_1 g(t) dt}{W}=\frac{\int -3e^{-t}(-3e^{-t}+e^{t}+re^t)dt}{2e^{2t}+3e^{t}}=-\frac{r}{5}-\frac{1}{10}+\frac{3}{10}e^{-2t}{/eq}Hence, the general solution of the differential equation is {eq}y=y_c+y_p=c_1e^{-t}+c_2e^{-2t}-\frac{r}{5}-\frac{7}{10}+\frac{3}{10}e^{t}-\frac{r}{5}-\frac{1}{10}+\frac{3}{10}e^{-2t}{/eq}So, Option D, {eq}-4u_2=0,~u_1=-\frac{r}{5}-\frac{7}{10}+\frac{3}{10}e^{t},~u_2=-\frac{r}{5}-\frac{1}{10}+\frac{3}{10}e^{-2t}{/eq} is correct.
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Which statements are true based on the diagram?
Select three options.
The approximation of 1 = integral (x – 3)e** dx by composite Trapezoidal rule with n=4 is: -25.8387 4.7846 -5.1941 15.4505
The approximation of the integral I is -5.1941 using the composite Trapezoidal rule with n = 4.
We need to divide the interval [0, 2] into subintervals and apply the Trapezoidal rule to each subinterval.
The formula for the composite Trapezoidal rule is given by:
I = (h/2) × [f(x₀) + 2f(x₁) + 2f(x₂) + ... + 2f(xₙ₋₁) + f(xₙ)]
Where:
h = (b - a) / n is the subinterval width
f(xi) is the value of the function at each subinterval point
In this case, n = 4, a = 0, and b = 2. So, h = (2 - 0) / 4 = 0.5.
Now, let's calculate the approximation:
[tex]f\left(x_0\right)\:=\:f\left(0\right)\:=\:\left(0\:-\:3\right)e^{\left(0^2\right)}\:=\:-3[/tex]
[tex]f\left(x_1\right)\:=\:f\left(0.5\right)\:=\:\left(0.5\:-\:3\right)e^{\left(0.5^2\right)}\:=-2.535[/tex]
[tex]f\left(x_2\right)\:=\:f\left(1\right)\:=\:\left(1\:-\:3\right)e^{\left(1^2\right)}\:=\:-1.716[/tex]
[tex]f\left(x_3\right)\:=\:f\left(1.5\right)\:=\:\left(1.5\:-\:3\right)e^{\left(1.5^2\right)}\:=\:-1.051[/tex]
[tex]f\left(x_4\right)\:=\:f\left(2\right)\:=\:\left(2\:-\:3\right)e^{\left(2^2\right)}\:=\:-0.065[/tex]
Now we can plug these values into the composite Trapezoidal rule formula:
I = (0.5/2) × [-3 + 2(-2.535) + 2(-1.716) + 2(-1.051) + (-0.065)]
= (0.25)× [-3 - 5.07 - 3.432 - 2.102 - 0.065]
= -5.1941
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Pablo saves $85 per month for 6 months. Then he deposits the money in an account that earns 2.1% simple interest. How much interest will he earn over 4 years? (no links just answers)
Answer:
$42.84
Step-by-step explanation:
P = 85 * 6 = 510
Formula:
I = Prt
Given:
P = 510
r = 2.1% or 0.021
t = 4
Work:
I = Prt
I = 510(0.021)(4)
I = 42.84