The standard deviation of the sample proportion will be 0.14. So, the correct option is option D) 0.14.
The formula for the standard deviation of a sample proportion is given by:
standard deviation = √[p(1-p)/n]
where p is the proportion of successes in the population (i.e. the proportion of students who cleared the exam), and n is the sample size.
In this case, p = 225/300 = 0.75, since in the school of 300 students 225 students cleared the exam. The sample size is n = 10, as sample of 10 of these students is taken.
Plugging these values into the formula, we get:
standard deviation = √[0.75(1-0.75)/10] = √[0.01875] = 0.1366
Rounding to two decimal places, the answer 0.14.
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Solve the following initial value problem. y' (t) - 2y = 6, y(2) = 2 Show your work for solving this problem and your answer on your own paper. y(t) = (Type an exact answer in terms of e.)
The solution to the initial value problem. y' (t) - 2y = 6, y(2) = 2 is y(t) = -3 + (5/e^4)e^(2t)..
To solve the initial value problem y'(t) - 2y = 6 with y(2) = 2,
we will use an integrating factor and the given initial condition. Here's the step-by-step solution:
1. Identify the integrating factor: The integrating factor is e^(-2t).
2. Multiply the equation by the integrating factor: e^(-2t)y'(t) - 2e^(-2t)y = 6e^(-2t).
3. Observe that the left side of the equation is the derivative of the product y(t)e^(-2t): (y(t)e^(-2t))' = 6e^(-2t).
4. Integrate both sides with respect to t: ∫(y(t)e^(-2t))' dt = ∫6e^(-2t) dt.
5. Integrate: y(t)e^(-2t) = -3e^(-2t) + C.
6. Solve for y(t): y(t) = -3 + Ce^(2t).
7. Apply the initial condition y(2) = 2: 2 = -3 + Ce^(4).
8. Solve for C: C = (5/e^4).
9. Substituting C back into the solution for y(t): y(t) = -3 + (5/e^4)e^(2t).
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A regression model made to conform to a sample set of data, compromising predictive power is called __________.
cross-validation
flooding
overfitting
binary choice
When a regression model is created to fit a sample set of data, its prediction ability is reduced overfitting. Thus, option C is correct.
What is the regression model?Overfitting is a phenomenon in machine learning where a regression model is trained too well on the sample data.
to the point where it starts to memorize the data instead of learning the underlying patterns or trends. As a result, the overfitted model may not generalize well to unseen data and may exhibit poor predictive power when used for making predictions on new data.
The term "compromising predictive power" in the question suggests that the model is not able to accurately predict outcomes on new, unseen data due to overfitting.
Essentially, the model becomes too specialized to the sample data it was trained on and loses its ability to generalize to new data points.
Flooding is not a term related to machine learning or regression modeling. Binary choice refers to a decision between two options and is not related to overfitting.
Therefore, When a regression model is created to fit a sample set of data, its prediction ability is reduced overfitting
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Using the Wronskian in Problems 15-18, verify that the given functions form a fundamental solution set for the given differential equation and find a general solution. y'" + 2y" - 11y' - 12y = 0; {e^3x, e^-x, e^-4x}
A general solution
[tex]y(x) = c1e^{3x} + c2e^{-x} + c3*e^{-4x}[/tex]
What is Wronskian?To verify that the given functions form a fundamental solution set for the differential equation y''' + 2y" - 11y' - 12y = 0, we can use the Wronskian. The Wronskian is defined as:
W(x) = | y1(x) y2(x) y3(x) |
| y1'(x) y2'(x) y3'(x) |
| y1''(x) y2''(x) y3''(x) |
where y1(x), y2(x), and y3(x) are the given functions.
Using the given functions, we can compute the Wronskian as follows:
W(x) = |[tex]e^{3x} e^{-x} e^{-4x} || 3e^{3x} -e^{-x} -4e^{-4x} || 9e^{3x} e^{-x} 16e^{-4x}[/tex]|
Expanding the determinant, we get:
[tex]W(x) = e^{3x}(-e^{-x}*16e^{-4x} + e^{-4x}e^{-x}) - (-e^{-x}(-4e^{-4x}) - (-e^{3x})*16e^{-4x})e^{3x} + (3e^{3x}(-e^{-x}*e^{-4x}) - e^{-x}*9e^{3x}*16e^{-4x})[/tex]
Simplifying, we get:
W(x) = -23e^(-3x)
Since the Wronskian is nonzero everywhere, the functions {e^(3x), e^(-x), e^(-4x)} form a fundamental solution set for the differential equation.
To find the general solution of the differential equation, we can use the formula:
y(x) = c1y1(x) + c2y2(x) + c3*y3(x)
where c1, c2, and c3 are constants. Substituting the given functions, we get:
[tex]y(x) = c1e^{3x} + c2e^{-x} + c3*e^{-4x}[/tex]
This is the general solution of the given differential equation.
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Use the double line graph to answer the following questions
13a. How much combined money was in River
and Town Bank in 2000?
3b. How many years did Town Bank have
more money than River Bank?
c. Find the mean # of dollars per year River Bank had from 1998-2004.
The double line graph for the amount in the River Bank and Town Bank indicates;
13 a. $10,000
13 b. Two year
c. $4,000 per year
What is a graph of a function?A graph of a function shows the relationship that exists between the input and output values of the function.
The graph with the lines that have markings is the graph of the River Bank
The graph with the lines without markings is the graph of the Town Bank
13 a. In the year 2,000, the money in River Bank = $6,000
The money in Town Bank = $4,000
The combined amount in both banks in the year 2,000 is therefore;
Amount = $6,000 + $4,000 = $10,000
13 b. Town Bank had more money than Rivers Bank in the years; 1998, 2002
Therefore;
Towns Bank had more money that Rivers Bank in 2 years
c. The amount River Bank had between 1998 to 2004 are;
$5,000 + $3,000 + $6,000 + $4,000 + $1,000 + $7,000 + $2,000 = $28,000
The number of years between 1998 and 2004 = 7 years
The mean number of dollars per year River Bank had = $28,000/7 = $4,000 per year
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Use an appropriate test to determine whether the series converges. [infinity]∑k=1 k100/(k+5)! By the ______ this series _____.
By applying L'Hôpital's Rule 100 times or analyzing the degree of the polynomial, you'll find that L equals 0. Since L < 1, by the Ratio Test, this series converges.
To determine whether the series converges, you can use the Ratio Test. For the series Σk=1 to infinity (k100/(k+5)!), apply the Ratio Test by finding the limit as k approaches infinity of the absolute value of the ratio of consecutive terms:
L = lim (k→∞) |( (k+1)100 / (k+6)! ) / ( k100 / (k+5)! )|
Upon simplification, you'll find:
L = lim (k→∞) |(k+1)100 * (k+5)! / (k100 * (k+6)!)|
Cancel out the factorials and further simplify:
L = lim (k→∞) |(k+1)100 / (k100 * (k+6))|
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atch each third order linear equation with a basis for its solution space..1. y'''−5y''+y'−5y=02. y'''−y''−y'+y=03. y'''−7y''+12y'=04. y'''+3y''+3y'+y=05. ty'''−y''=06. y'''+y'=0A. et tet e−tB. 1 t t3C. 1 e4t e3tD. 1 cos(t) sin(t)E. e5t cos(t) sin(t)F. e−t te−t t2e−t
For each of the third-order linear equations, the basis for the solution space can be found by solving the characteristic equation and then finding the corresponding linearly independent solutions. The solutions for each equation are:
The characteristic equation is r³ - 5r² + r - 5 = 0, which has roots r = 4, 1±i. The basis for the solution space is {cos(t), sin(t), e^(4t)}.The characteristic equation is r³ - r² - r + 1 = 0, which has roots r = 1 (with multiplicity 3). The basis for the solution space is {e^(-t), te^(-t), t^2e^(-t)}.The characteristic equation is r³ - 7r² + 12r - 0 = 0, which has roots r = 0 (with multiplicity 2) and r = 7. The basis for the solution space is {e^(4t), e^(t), 1}.The characteristic equation is r³ + 3r² + 3r + 1 = 0, which has roots r = -1 (with multiplicity 3). The basis for the solution space is {e^(-t), e^(-t/2)cos((sqrt(3)/2)t), e^(-t/2)sin((sqrt(3)/2)t)}.The characteristic equation is r^3 - r^2 = 0, which has roots r = 0 (with multiplicity 2) and r = 1. The basis for the solution space is {t, 1}.The characteristic equation is r^3 + r = 0, which has roots r = 0 and r = ±i. The basis for the solution space is {e^(-t), cos(t), sin(t)}.To learn more about linear equation, here
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Select the rational number to help complete the circut
The rational number that helps complete the circuit is given as follows:
0.222...
What are rational and irrational numbers?Rational numbers are numbers that can be represented by a ratio of two integers, which is in fact a fraction, such as numbers that have no decimal parts, or numbers in which the decimal parts are terminating or repeating. Examples are integers, fractions and mixed numbers.Irrational numbers are numbers that cannot be represented by a ratio of two integers, that is, they cannot be represented by fractions. They are non-terminating and non-repeating decimals, such as non-exact square roots.For this problem, we have two options to complete the circuit, as follows:
0.222..., which is a rational number, is it is a repeating decimal.the square root of 20, which is an irrational number, as the square root of 20 is non-exact.More can be learned about rational and irrational numbers at brainly.com/question/5186493
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Can you help me with this exercise
The coordinates of point P are (-3, -1).
What is the coordinate of point P?The coordinates of point P that divides the line segment AB in the ratio 1:4 is calculated as follows;
let the ratio = a : b = 1:4
P = ( (bx₂ + ax₁)/(b + a), (by₂ + ay₁)/( b + a) )
Where;
(x₁, y₁) and (x₂, y₂) are the coordinates of points A and BThe coordinate of point P is calculated as follows;
P = ( (4(-2) + 1 (-7))/(4 + 1), (4(0) + 1(-5) )/(4 + 1))
P = (-8 - 7)/(5), (0 - 5)/(5)
P = (-15/5), (-5/5)
P = (-3, - 1)
Thus, the coordinate of point P is determined by applying ratio formula on a line segment.
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Find the area of the triangle. Round your answer to the nearest tenth. A 58 yd 54° a. 1,360.8 yd² B 58 yd b. 1,682 yd² с c. 2,721.5 yd² d. 2,315.1 yd²
The Taylor series for a function f about x = 0 is given by Σ numbers x and converges to f for all real. If the fourth degree Taylor polynomial for f about x = 0 is used to approximate fl- , what is the alternating series error bound?
(A) 1/24 . 5!
(B) 1/25 . 6!
(C) 1/26.7!
(D) 1/27.8!
The alternating series error bound is (C) 1/26.7!, since 26.7! is the smallest factorial greater than [tex]120*M_5.[/tex]
How to find alternating series error bound?The alternating series error bound for an alternating series of the form [tex]\sum (-1)^n b_n[/tex]is given by [tex]|R_n| < = b_{(n+1)}[/tex], where [tex]R_n[/tex] is the remainder term and [tex]b_n[/tex] is the absolute value of the (n+1)th term in the series.
In this case, the fourth degree Taylor polynomial for f about x = 0 is given by:
[tex]P_4(x) = f(0) + f'(0)x + (f''(0)/2)x^2 + (f'''(0)/6)x^3 + (f''''(0)/24)x^4[/tex]
The alternating series error bound for the approximation of f(x) by [tex]P_4(x)[/tex]is therefore:
[tex]|R_4(x)| < = |f(x) - P_4(x)| < = (M/5!) |x - 0|^5,[/tex]
where M is an upper bound for [tex]|f^{(5)}(c)[/tex]| on the interval [0,x] for some c between 0 and x.
Since the Taylor series for f about x=0 converges to f for all real x, we know that M is finite. Therefore, we can find an upper bound for [tex]|f^{(5)}(c)|[/tex]on [0,-1] using the Mean Value Theorem.
Let g(x) = f''''(x). Then, by the Mean Value Theorem, there exists some c between 0 and -1 such that:
g(c) = (g(0) - g(-1))/(-1 - 0) = g(0) - g(-1)
Since the fourth derivative of f is continuous, g is continuous on the interval [0,-1]. Therefore, by the Extreme Value Theorem, g attains its maximum and minimum values on [0,-1].
Let[tex]M_5 = max{|g(x)| : x in [0,-1]}[/tex]. Then we have:
[tex]|R_4(x)| < = M_5/5! |x|^5 = M_5/120[/tex]
Therefore, the alternating series error bound is (C) 1/26.7!, since 26.7! is the smallest factorial greater than [tex]120*M_5.[/tex]
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In an exponential regression model, the exact percentage of change can be calculated as: (exp(β1 ) − 1) × 100. If β1 = 0.23, what is the percent increase in E(y)?
25%
26%
75%
22%
The correct answer is 26%.
In an exponential regression model, the exact percentage of change can be calculated as: (exp(β1) - 1) × 100. Given that β1 = 0.23, let's calculate the percent increase in E(y):
Step 1: Calculate exp(β1): exp(0.23) ≈ 1.259
Step 2: Subtract 1: 1.259 - 1 = 0.259
Step 3: Multiply by 100: 0.259 × 100 = 25.9
The percent increase in E(y) is approximately 26%. So, the correct answer is 26%.
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What is the area of the shaded segment shown in O below?
Segment area=degree/360 pie r 2sin(degree)
The area of the segment is 1.68 squared.
How to find area of the shaded segment?The area of the shaded segment is the subtraction of the area of the triangle from the area of the sector OMN.
Therefore,
area of the segment = ∅ / 360 πr² - 1 / 2r²sin(∅)
area of the segment = 30 / 360 π(12)² - 1 / 2 (12)² sin 30°
area of the segment = 1 / 12 π(144) - 1 / 2(144)0.5
area of the segment = 12π - 36
area of the segment = 12(3.14) - 36
area of the segment = 37.68 - 36
area of the segment = 1.68 inches squared.
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4-2/13? Help my i wont answer:)
Answer:
4 - (2 / 13) = 3.84615385.
But I understand that you don't want to answer.
Step-by-step explanation:
The question reads: Find dy/dx by implicit differentiation. ex/y = 6x − y
I believe this involves first taking the log of both sides, then using implict differentiation, but I can't get the math to work out.
The derivative dy/dx is (6y - eˣ - yeˣ)/y².
To find dy/dx for the given equation ex/y = 6x - y, you can use implicit differentiation without taking the log.
Given the equation [tex]e^\frac{x}{y}[/tex] = 6x - y, you do not need to take the log of both sides. Instead, start by applying implicit differentiation to both sides with respect to x:
1. Differentiate ex with respect to x: d(eˣ)/dx = ex
2. Differentiate y with respect to x: d(y)/dx = dy/dx
3. Differentiate 6x with respect to x: d(6x)/dx = 6
4. Apply the quotient rule to d([tex]e^\frac{x}{y}[/tex])/dx: d(ex/y)/dx = (y * d(eˣ)/dx - eˣ * d(y)/dx) / y² = (y * eˣ - eˣ * dy/dx) / y²
5. Set the derivatives equal: (y * eˣ - eˣ * dy/dx) / y² = 6 - dy/dx
Now, solve for dy/dx:
6. Multiply both sides by y²: y * eˣ - eˣ * dy/dx = 6y² - y² * dy/dx
7. Rearrange terms: dy/dx * (y² + eˣ) = 6y² - y * eˣ
8. Solve for dy/dx: dy/dx = (6y² - y *eˣ) / (y² + eˣ) = (6y - ex - yeˣ) / y²
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Each student in Mrs. Wimberly’s six science classes planted a bean in a Styrofoam cup. All beans came from the same source, were planted using the same bag of soil, and were watered the same amount. Mrs. Wimberly has 24 students in each of her six classes. In first period, 21 of the 24 bean seeds sprouted.
Which statement about the seeds in the remaining five classes is NOT supported by this information?
Responses
A 87.5% of the bean seeds should sprout.87.5% of the bean seeds should sprout.
B More than 100 bean seeds should sprout.More than 100 bean seeds should sprout.
C 1 out of 8 bean seeds will not sprout.1 out of 8 bean seeds will not sprout.
D At least 20 bean seeds will not sprout.At least 20 bean seeds will not sprout.
HELP ME PLEASEE IS TIMED!!!
Answer: D
Explanation: Since 21 out of 24 bean seeds sprouted in the first class, the probability of a bean seed sprouting is 21/24, or 0.875. This information does not provide any information about the seeds in the other five classes, other than that they were all planted using the same method. Therefore, we cannot make a definitive statement about how many seeds will or will not sprout in the other classes. Option A is supported by the given information, since 87.5% of the seeds in the first class sprouted. Option B is not necessarily supported by the given information, as it depends on how many seeds were used in total. Option C is not directly supported by the given information, but is a possible conclusion based on the probability of a seed sprouting. Option D is contradicted by the given information, since at most 3 out of 24 seeds did not sprout in the first class.
why did we not have to test for hov with our paired data?
When analyzing statistical data, it is important to consider the assumptions underlying the methods used. One such assumption is homogeneity of variance (HOV), which tests whether the variance of two groups is equal.
Why did we not have to test for hov with our paired data?
We do not need to test for the homogeneity of variance (HOV) assumption in paired data because the assumption only applies to independent samples. In paired data, the same individuals are measured twice, and the two sets of measurements are dependent. Therefore, the assumption of equal variances between two groups does not apply, and we do not need to test for it.
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Why did we not have to test for HOV with our paired data?
Find the Taylor polynomial T3(x) for the function f centered at the number a. f(x) = xe?5x, a = 0
Find the Taylor polynomial Find the Taylor polynomial T(x) for the function fcentered at the number a. (x) for the function fcentered at the number a.
the Taylor polynomial T3(x) for the function f(x) = x[tex]e^{(-5x)}[/tex]centered at a = 0 is T3(x) = 1 + 0.25[tex]x^{2}[/tex] - 0.0083[tex]x^{3}[/tex]
How to find the Taylor polynomial T3(x) for the function f(x)?To find the Taylor polynomial T3(x) for the function f(x) = x[tex]e^{(-5x)}[/tex]centered at a = 0, we need to find the first four derivatives of f(x) and evaluate them at x = 0:
f(x) = x[tex]e^{(-5x)}[/tex]
f'(x) = [tex]e^{(-5x)}[/tex] - 5x[tex]e^{(-5x)}[/tex]
f''(x) = 25x[tex]e^{(-5x)}[/tex] - 20[tex]e^{(-5x)}[/tex]
f'''(x) = -125x[tex]e^{(-5x)}[/tex] + 75[tex]e^{(-5x)}[/tex]
f''''(x) = 625x[tex]e^{(-5x)}[/tex] - 500[tex]e^{(-5x)}[/tex]
Now, we can write the Taylor polynomial T3(x) as:
T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] + (f'''(0)/3!)[tex]x^{3}[/tex]
T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] + (f'''(0)/3!)[tex]x^{3}[/tex]
T3(x) = f(0) = 0 + f'(0)x = [tex]e^{0}[/tex] × 1 - 5 × 0 × [tex]e^{0}[/tex] = 1
T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] = 1 + 0.25[tex]x^{2}[/tex]
T3(x) = f(0) + f'(0)x + (f''(0)/2!)[tex]x^{2}[/tex] + (f'''(0)/3!)[tex]x^{3}[/tex] = 1 + 0.25[tex]x^{2}[/tex] - 0.0083[tex]x^{3}[/tex]
Therefore, the Taylor polynomial T3(x) for the function f(x) = x[tex]e^{(-5x)}[/tex]centered at a = 0 is T3(x) = 1 + 0.25[tex]x^{2}[/tex] - 0.0083[tex]x^{3}[/tex]
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13) rank in order, from largest to smallest, electric field strength at five points near an infinite plane of charge
The electric field strength decreases as the distance from the plane increases.
How to rank electric field strength at five points near an infinite plane of charge?The electric field strength near an infinite plane of charge is given by:
E = σ/2ε0
where E is the electric field strength, σ is the surface charge density of the plane, and ε0 is the electric constant.
The electric field strength at a point near the plane depends on the distance from the plane and the orientation of the point relative to the plane.
Assuming that the surface charge density is constant, we can rank the electric field strength from largest to smallest based on the distance from the plane:
Point closest to the plane
Point at a distance of 2 times the distance from point 1
Point at a distance of 3 times the distance from point 1
Point at a distance of 4 times the distance from point 1
Point at a distance of 5 times the distance from point 1
This is because the electric field strength decreases as the distance from the plane increases.
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use y = (x − x0)m to solve the given differential equation. (x 9)2y'' − 9(x 9)y' 16y = 0 y(x) =
The solution to the differential equation is: [tex]y(x) = c1 (x-9)^4 + c2 (x-9)[/tex]where c1 and c2 are constants of integration.
To solve this differential equation using the method of "reducing to a polynomial equation", we can make the substitution:
x - 9 = t,
so that x = t + 9 and y(x) = y(t+9).
We can then rewrite the differential equation in terms of t as follows:
[tex][(t+9)^2] y'' - 9(t+9) y' + 16y = 0[/tex]
We can now make the substitution [tex]y = (t+9)^m[/tex], where m is some constant to be determined.
Taking the first and second derivatives of y with respect to t, we get:
[tex]y'=m(t+9)^{(m-1)}[/tex]
[tex]y'' = m(m-1) (t+9)^{(m-2)}[/tex]
Substituting these expressions into the differential equation, we get:
[tex][(t+9)^2] m(m-1)(t+9)^{m-2} - 9(t+9) m(t+9)^{m-1} + 16(t+9)^m = 0[/tex]
Simplifying, we get:
m(m-1) - 9m + 16 = 0
Solving this quadratic equation for m, we get:
m = 4 or m = 1
Therefore, the general solution to the differential equation is given by:
[tex]y(t) = c1 (t+9)^4 + c2 (t+9)[/tex]
where c1 and c2 are constants of integration.
Substituting back to x, we have:
[tex]y(x) = c1 (x-9)^4 + c2 (x-9)[/tex]
where c1 and c2 are constants of integration.
Therefore, the solution to the differential equation is:
[tex]y(x) = c1 (x-9)^4 + c2 (x-9)[/tex]
where c1 and c2 are constants of integration.
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Find the term containing x6 in the expansion of (x+2y)10
A. 3470x6y6
B. 3360x6y4
C. 1680x6y4
D. 3360x6y3
The correct answer is option B, 3360x6y4.
The term containing x6 in the expansion of (x+2y)10 will arise from selecting the x term exactly 6 times out of 10 terms. We can select the x term in different ways by using the binomial theorem.
The binomial theorem states that for any positive integers n and k, the coefficient of x^(n-k) in the expansion of (x+y)^n is given by the binomial coefficient (n choose k), which is written as nCk and can be calculated using the formula:
nCk = n! / (k! * (n-k)!)
where ! denotes the factorial function.
In our case, we need to find the coefficient of x^6 in the expansion of (x+2y)^10, which is given by:
10C6 * x^6 * (2y)^4
= 210 * x^6 * 16y^4
= 3360x^6y^4
Therefore, the correct answer is option B, 3360x6y4.
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Give a recursive definition of the sequence An, n=1,2,3,... if: Recursive Form Basis A) An 4n-2 An = An-1+ 4 Ao B) An n(n+1) An = An-1+ Ao C) An = 1+(-1)" An An-2t Ao A1 = D) An = n2 An = An-1+ Ао
The recursion, and subsequent terms are defined in terms of previous terms in the sequence
A) The recursive definition for the sequence An is:
An = (4n-2)An-1 + 4Ao, with A1 = 4Ao.
B) The recursive definition for the sequence An is:
An = n(n+1)An-1 + Ao, with A1 = Ao.
C) The recursive definition for the sequence An is:
An = 1 + (-1)nAn-2tAo, with A1 = Ao and A2 = 1 - Ao.
D) The recursive definition for the sequence An is:
An = n^2An-1 + Ao, with A1 = Ao.
These recursive definitions define each term of the sequence An as a function of one or more previous terms in the sequence, starting with a basis case. The basis case provides the starting point for the recursion, and subsequent terms are defined in terms of previous terms in the sequence.
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It is inappropriate to apply the Empirical Rule to a population that is right-skewed a. True b. False
The answer to the given statement is as follows:
It is inappropriate to apply the Empirical Rule to a population that is right-skewed
b. False.
The given statement is false because the rule of thumb, also known as the 68-95-99.7 rule, is a statistical rule that applies to the normal distribution. This rule was lost in our sample, with about 68% of the data falling within one standard deviation of the mean for a normal distribution, and about 95% of the data falling within two standard deviations from the mean, and about 99.7% of the data being lost in our sample. The deviation from the mean is the difference between the mean of the standard deviation.
Although the rule of thumb is most true for symmetric normal distributions, it can also be used for distributions, including right-skewed distributions.
However, as the distribution becomes more skewed, the rule of thumb may not be correct. In a right-skewed distribution, the mean is greater than the median and the tails of the distribution are to the right. In such a distribution, a rule of thumb might estimate the proportion of data that is one or two standard deviations from the mean.
Despite this limitation, the rule of thumb can be a useful tool for understanding the spread of data in right-skewed distributions. However, it is important to know that this law can predict the percentage of data in a given situation.
In such cases, other methods such as quartiles or percentages are more effective for analyzing the distribution of the data.
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X 0 1 2 3
P(x) .02 .65 .26 .07
Find the probability that a family owns:
Exactly 2 refrigerators is: ___
P(3) = ____
P( < 1) = ____
P( ≤ 2) = ____
P (>2) = ____
The probabilities of
a) P(3) = 0.07
b) P(<1) = 0.67
c) P(≤2) = 0.93
d) P(>2) = 0.07
Probability is a measure of the likelihood of an event occurring. It is a numerical value between 0 and 1, where 0 represents an impossible event and 1 represents a certain event.
a) The probability of getting a value of 3 is simply the value of P(3),
P(3) = 0.07
b) To find the probability of getting a value less than 1, we add the probabilities of getting 0 and 1,
P( < 1) = P(0) + P(1) = 0.02 + 0.65 = 0.67
c) To find the probability of getting a value less than or equal to 2, we add the probabilities of getting 0, 1, and 2,
P( ≤ 2) = P(0) + P(1) + P(2) = 0.02 + 0.65 + 0.26 = 0.93
d) To find the probability of getting a value greater than 2, we simply look at the probability of getting a value of 3, which is 0.07.
P( > 2) = P(3) = 0.07
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The given question is incomplete, the complete question is:
X 0 1 2 3
P(x) .02 .65 .26 .07
Find the probabilities
a) P(3) =
b) P( < 1) =
c) P( ≤ 2) =
d) P (>2) =
when computing the effect size, you use the observed value of t in the formula, not the critical value (cv). True or False?
The answer is True we use observed value not critical value
The answer is True. When computing the effect size, you use the observed value of t in the formula rather than the critical value. The critical value is used to determine statistical significance, while the effect size is calculated using the observed value to measure the strength or magnitude of the relationship between variables.
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In order for a matrix B to be the inverse of A, the equations AB = I and BA = I must both be true. true or false
The given statement "In order for a matrix B to be the inverse of A, the equations AB = I and BA = I must both be true." is true because of the definition of the inverse matrix.
An inverse matrix is obtained by dividing the adjugate of the given matrix by the determinant of the given matrix.
An inverse matrix is also known as a reciprocal matrix.
In order for a matrix B to be the inverse of A, both equations AB = I (Identity matrix) and BA = I must be true.
This is because the inverse of a matrix A, denoted as A⁻¹ (in this case, matrix B), should satisfy these conditions for it to be a true inverse.
When a matrix is multiplied by its inverse, the result is the identity matrix.
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Use the region in the first quadrant bounded by √x, y=2 and the y-axis to determine the volume when the region is revolved around the line y = -2. Evaluate the integral.
A. 18.667
B. 17.97
C. 58.643
D. 150.796
E. 21.333
F. 32.436
G. 103.323
H. 27.4
If B=x*y then 2x*5y =
Answer: x * y = 2x + 5y. Formula used: x * y = 2x + 5y. Calculation: When x = 3, and y = 5. ⇒ 2x + 5y = (2 × 3) + (5 × 5) = 6 + 25 = 31
Step-by-step explanation:
Which of the following gives the value of
the expression below written in scientific
notation?
(9.1 x 10-3) + (5.8 x 10-2)
A. 1.49 x 10-4
B.
6.71 x 10-²
C. 9.68 x 10-3
D.
14.9 x 10-5
VD 4TO6
02
The Correct Option is C that 71 x 10-² of the following gives the value of the expression below written in scientific notation.
Why does scientific notation employ the number 10?The basic objective of scientific notation is to make computations with unusually big or small numbers simpler. The following examples demonstrate how all of the digits in a number in scientific notation are relevant because zeros are no longer utilised to set the decimal point.
What does math scientific notation mean?The statement for a number n in scientific notation is of the type a10b, where an is an integer such that 1|a|10. B is also an integer. Multiplication: To get the full amount in scientific notation, multiply the decimal values. Add the 10 power exponents after that.
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Write the absolute value equation that has the following solution(s). One solution: x=15
The absolute value equation is:
|x - 15| = 0
Here, we have,
to write the absolute value equation:
We want an absolute value equation that only has the solution x = 15.
So we must have something equal to zero (so we avoid the problem with the signs that we can have with other numbers)
So the equation will be something like:
|x - a| = 0
And the solution is 15, so:
|15 - a | = 0
then a = 15
The equation is:
|x - 15| = 0
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Answer:
|x - 15| = 0
Notes:
You're probably going to give the other person Brainliest, but could you maybe consider giving it to me?
consider a poisson process with paaramter. given that x(t) = n occur at time t, find the density function for wr, time of the rth arrival
The time of the rth arrival in a Poisson process follows a gamma distribution with parameters r and λ, where λ is the rate parameter.
The density function for the time of the rth arrival is: f(w) = λ^r * w^(r-1) * e^(-λw) / (r-1) where w is the time of the rth arrival. This density function gives the probability density of the time of the rth arrival occurring at a specific time w, given that there have been n arrivals up to time t.
The density function is derived from the fact that the time between successive arrivals in a Poisson process follows an exponential distribution with rate parameter λ, and the time of the rth arrival is the sum of r independent exponential random variables.
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