One of the biggest factors in a credit score is credit age. The credit age is the average length of accounts. Higher credit scores are given to longer credit ages. Suppose we have 4 accounts open: Auto Loan: 1 year 4 months Credit Card: 4 years 1 month Credit Card: 1 year 11 months Credit Card: 1 year 8 months The credit card of 4 years and 1 month has the highest balance and interest rate. We payoff the credit card and close the account. Give the new credit age. (Enter as a decimal and round to the hundredths) Question 6 1 pts

The Laplace transform can be used to solve the initial-value problem y'' + 10y' + 9y = 0, y(0) = 1, y'(0) = 0. The solution is given by \(y(t) = \frac{1}{3} e^{-t} - \frac{1}{3}e^{-9t}\)

The Laplace transform of the differential equation y'' + 10y' + 9y = 0 is given by L(y'') + 10L(y') + 9L(y) = 0, where L denotes the Laplace transform. Using the property of the Laplace transform that \(L(y') = sY(s) - y(0)\), where Y(s) is the Laplace transform of y(t), and \(L(y'') = s^{2}Y(s) - sy(0) - y'(0)\), we can rewrite the transformed equation as \(s^{2}Y(s) - s + 10sY(s) - 10 + 9Y(s) = 0\)

Solving for Y(s), we get \(Y(s) = \frac{1}{(s^{2} + 10s + 9)}\).

To find the inverse Laplace transform of Y(s), we factor the denominator of Y(s) as (s+1)(s+9) and use partial fractions to write Y(s) as \(Y(s) = \frac{1}{((s+1)(s+9))} = \frac{(1/8)}{(s+1)} - \frac{(1/8)} {(s+9)}\).

Taking the inverse Laplace transform of each term, we get \(y(t) = \frac{1}{8}}e^{-t} - \frac{1}{8} e^{-9t}}\).

Using the initial conditions y(0) = 1 and y'(0) = 0, we can solve for the constants in the solution.

We have \(y(0) = \frac{1}{8}(1 - 1) = 0\), which is not equal to 1.

To fix this, we add the term \(\frac{1}{3} e^{-t}\), which has a derivative of \(-\frac{1}{3}e^{-t}\) and thus does not affect the value of y'(0).

Therefore, the solution to the initial-value problem is \(y(t) = \frac{1}{3} e^{-t} - \frac{1}{3}e^{-9t}\).

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Answer:

There's 3 answers: the third, fourth, and the sixth one. 1  2  3

                                                                                           4  5  6

Step-by-step explanation:

The expression in the question gets simplified to p-6, and so does the third one, fourth one is just itself, and the sixth one also simplifies to p-6.

The pοpulatiοn mean is apprοximately 84.93 and the pοpulatiοn standard deviatiοn is apprοximately 38.54.

The standard deviatiοn is a summary measure οf the differences οf each οbservatiοn frοm the mean. If the differences themselves were added up, the pοsitive wοuld exactly balance the negative and sο their sum wοuld be zerο. Cοnsequently, the squares οf the differences are added.

We can use the fοrmula fοr standardizing a variable using z-scοres:

z = (x - μ) / σ

where z is the z-scοre, x is the cοrrespοnding raw scοre, mu is the pοpulatiοn mean, and sigma is the pοpulatiοn standard deviatiοn.

We have twο pairs οf (x, z) values:

x = 83, z = -0.05

x = 0.93, z = 2.0

We can use these tο create twο equatiοns with twο unknοwns ( and sigma):

-0.05 = (83 - μ) / σ

2.0 = (0.93 - μ) / σ

We can sοlve this system οf equatiοns by first sοlving οne οf the equatiοns fοr οne οf the unknοwns, and then substituting that expressiοn intο the οther equatiοn. Fοr example, we can sοlve the first equatiοn fοr mu:

mu = 83 + 0.05 * σ

Then we substitute this expressiοn fοr mu intο the secοnd equatiοn:

2.0 = (0.93 - (83 + 0.05 * σ)) / σ

Simplifying this equatiοn, we get:

2.0 = (0.93 / σ) - (83 / σ) - 0.05

2.05 = 0.93 / σ - 83 / σ

2.05 = (0.93 - 83) / σ

σ = (0.93 - 83) / 2.05

σ = 38.5366

Nοw we can use this value tο find the pοpulatiοn mean μ:

μ = 83 + 0.05 * σ

μ = 83 + 0.05 * 38.5366

μ = 84.9278

Therefοre, the pοpulatiοn mean is apprοximately 84.93 and the pοpulatiοn standard deviatiοn is apprοximately 38.54.

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The value of the variable that encloses the top 17 percent of area will be the 83rd percentile.

In any distribution of a continuous random variable, the value of the variable that encloses the top 17 percent of area will be the 83rd percentile.

To understand this, we can think of the percentile as a measure of relative position in a distribution. The nth percentile represents the value below which n percent of the data falls.

In this case, the top 17 percent of area corresponds to the upper tail of the distribution. Therefore, the 83rd percentile represents the value below which 83 percent of the data falls, leaving the top 17 percent above it.

Hence, the value of the variable that encloses the top 17 percent of area will be the 83rd percentile.

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The correct equation would be:

0b. A = 240 - 4h

Explanation:

If Clayton can plow 4 acres per hour, the number of acres he can plow in 'h' hours would be 4h.

Therefore, the number of acres left to plow would be the total number of acres (240) minus the number of acres he has already plowed (4h).

Hence, the equation that represents the number of acres left to plow would be A = 240 - 4h.

The average rate of change between x and y in the given data is -1.

We have,

To find the average rate of change between two variables, we need to calculate the difference in the values of the variables and divide it by the difference in their corresponding inputs.

In this case, we have the following data points:

x: -2, -1, 0, 1

y: 7, 6, 5, 4

To find the average rate of change, we'll consider the first and last data points.

Change in y: 4 - 7 = -3

Change in x: 1 - (-2) = 3

Average rate of change = Change in y / Change in x = -3 / 3 = -1

Therefore,

The average rate of change between x and y in the given data is -1.

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Answer:

L = V/wh

Step-by-step explanation:

you have to get L by its self so on the original move the letters around to get it by its self. Imagen it’s like soliving for X

The number of possible combinations of three coins from the bag of 16 coins with different dates can be calculated using the formula for combinations, which is nCr = n! / r!(n-r)!, where n is the total number of objects, r is the number of objects to be chosen, and ! denotes factorial (the product of all positive integers up to the given number).

In this case, we have n = 16 (the total number of coins in the bag) and r = 3 (the number of coins to be chosen for each combination). Using the formula for combinations, we can calculate the number of possible combinations as follows:

nCr = 16! / 3!(16-3)!
nCr = (16 x 15 x 14) / (3 x 2 x 1)
nCr = 560

Therefore, 560 possible combinations of three coins can be chosen from the bag of 16 coins with different dates. These combinations could represent different historical events, significant dates, or other symbolic meanings depending on the dates inscribed on the coins. The calculation of combinations is an important concept in combinatorics and probability theory, and it has many real-world applications in fields such as statistics, economics, and computer science.

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I don't know sorry.

Sorry!

The expression at the upper and lower limits and the difference is

∫[0,1]√(4-3\(x^2\)) dx \(=(2 - (3/2)(1)^2) - (2 - (3/2)(0)^2) = (2 - 3/2) - (2 - 0) = (4/2 - 3/2) = 1/2.\)

To evaluate the integral ∫√(4-3\(x^2\)) dx, where the interval of integration is 0≤x≤1, we can use various techniques such as substitution or integration by parts. Let's proceed with the method of substitution to simplify the integral and find its value.

First, let's identify a suitable substitution for the integral. Since the expression inside the square root contains a quadratic term, it is beneficial to let u be equal to the square root of the quadratic expression. Therefore, we set u = √(4-3\(x^2\)).

Next, we need to find the differential of u with respect to x. Taking the derivative of both sides with respect to x, we have du/dx = (-6x)/(2√(4-3\(x^2\))) = -3x/√(4-3\(x^2\)).

Now, we can rewrite the integral in terms of the new variable u. Substituting u = √(4-3\(x^2\)) and du = (-3x/√(4-3\(x^2\))) dx into the integral, we have:

∫√(4-3\(x^2\)) dx = ∫u du

Our new integral is now much simpler, as it reduces to the integral of u with respect to u. Integrating u, we get:

∫u du = (1/2)\(u^2\) + C,

where C is the constant of integration.

Now, we can substitute back for u in terms of x. Recall that we set u = √(4-3x^2). Therefore, the final result becomes:

∫√(4-3x^2) dx = (1/2)(√\((4-3x^2))^2 + C = (1/2)(4-3x^2) + C = 2 - (3/2)x^2 + C.\)

To find the definite integral over the interval [0, 1], we need to evaluate the expression at the upper and lower limits and find the difference:

∫[0,1]√(4-3\(x^2\)) dx\(= (2 - (3/2)(1)^2) - (2 - (3/2)(0)^2) = (2 - 3/2) - (2 - 0) = (4/2 - 3/2) = 1/2.\)

Therefore, the value of the definite integral ∫√(4-3\(x^2\)) dx over the interval [0, 1] is 1/2.

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Answer:

0.2 pounds

Step-by-step explanation:

You can use unitary method:

To make 15 loaves of bread, flour needed is 3 pounds

To make 1 loaf of bread, flour needed is 3/15 pounds = 0.2 pounds

A) the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = cos(x) is:

T5(x) = 1 - (1/2!)x² + (1/4!)x⁴

B) the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = sin(x) is:

T5(x) = x - (1/3!)x³ + (1/5!)x⁵

C) the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = eˣ is:

T5(x) = 1 + x + (1/2!)x² + (1/3!)x³ + (1/4!)x⁴ + (1/5!)x⁵

A) To find the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = cos(x), we need to compute the derivatives of f(x) at x = 0 and use them to construct the polynomial.

First, let's find the derivatives of f(x) = cos(x) up to the fifth degree:

f¹(x) = -sin(x)

f²(x) = -cos(x)

f³(x) = sin(x)

f⁴(x) = cos(x)

f⁵(x) = -sin(x)

Next, we evaluate these derivatives at x = 0:

f(0) = cos(0) = 1

f¹(0) = -sin(0) = 0

f²(0) = -cos(0) = -1

f³(0) = sin(0) = 0

f⁴(0) = cos(0) = 1

f⁵(0) = -sin(0) = 0

Using these values, we can construct the fifth-degree Taylor polynomial approximation T5(x) centered at a=0:

T5(x) = f(0) + f¹(0)x + (f²(0)/2!)x² + (f³(0)/3!)x³ + (f⁴(0)/4!)x⁴ + (f⁵(0)/5!)x⁵

T5(x) = 1 + 0x + (-1/2!)x² + (0/3!)x³ + (1/4!)x⁴ + (0/5!)x⁵

Simplifying the expression, we get:

T5(x) = 1 - (1/2!)x² + (1/4!)x⁴

Therefore, the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = cos(x) is:

T5(x) = 1 - (1/2!)x² + (1/4!)x⁴

B) For the function f(x) = sin(x), we follow the same steps as above to find the fifth-degree Taylor polynomial approximation T5(x) centered at a=0.

The derivatives of f(x) = sin(x) up to the fifth degree are:

f¹(x) = cos(x)

f²(x) = -sin(x)

f³(x) = -cos(x)

f⁴(x) = sin(x)

f⁵(x) = cos(x)

Evaluating these derivatives at x = 0:

f(0) = sin(0) = 0

f¹(0) = cos(0) = 1

f²(0) = -sin(0) = 0

f³(0) = -cos(0) = -1

f⁴(0) = sin(0) = 0

f⁵(0) = cos(0) = 1

Constructing the fifth-degree Taylor polynomial approximation T5(x) centered at a=0:

T5(x) = f(0) + f¹(0)x + (f²(0)/2!)x² + (f³(0)/3!)x³ + (f⁴(0)/4!)x⁴ + (f⁵(0)/5!)x⁵

T5(x) = 0 + 1x + (0/2!)x² + (-1/3!)x^3 + (0/4!)x⁴ + (1/5!)x⁵

Simplifying the expression, we get:

T5(x) = x - (1/3!)x³ + (1/5!)x⁵

Therefore, the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = sin(x) is:

T5(x) = x - (1/3!)x³ + (1/5!)x⁵

C) For the function f(x) = eˣ, we follow the same steps as before to find the fifth-degree Taylor polynomial approximation T5(x) centered at a=0.

The derivatives of f(x) = eˣ up to the fifth degree are:

f¹(x) = eˣ

f²(x) = eˣ

f³(x) = eˣ

f⁴(x) = eˣ

f⁵(x) = eˣ

Evaluating these derivatives at x = 0:

f(0) = e⁰ = 1

f¹(0) = e⁰ = 1

f²(0) = e⁰ = 1

f³(0) = e⁰ = 1

f⁴(0) = e⁰ = 1

f⁵(0) = e⁰ = 1

Constructing the fifth-degree Taylor polynomial approximation T5(x) centered at a=0:

T5(x) = f(0) + f¹(0)x + (f²(0)/2!)x² + (f³(0)/3!)x³ + (f⁴(0)/4!)x⁴ + (f⁵(0)/5!)x⁵

T5(x) = 1 + 1x + (1/2!)x² + (1/3!)x³ + (1/4!)x⁴ + (1/5!)x⁵

Simplifying the expression, we get:

T5(x) = 1 + x + (1/2!)x² + (1/3!)x³ + (1/4!)x⁴ + (1/5!)x⁵

Therefore, the fifth-degree Taylor polynomial approximation T5(x) centered at a=0 for the function f(x) = eˣ is:

T5(x) = 1 + x + (1/2!)x² + (1/3!)x³ + (1/4!)x⁴ + (1/5!)x⁵

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The simplified expression of cos^2x sin^4x is cos^2x - 2cos^4x + cos^6x.

The expression cos^2x sin^4x can be simplified by using the trigonometric identity that states sin^2x + cos^2x = 1.

We can rewrite cos^2x sin^4x as cos^2x (sin^2x)^2. Then, using the trigonometric identity sin^2x + cos^2x = 1, we can substitute (1 - cos^2x) for sin^2x.

This gives us cos^2x (1 - cos^2x)^2, which can be expanded using the binomial theorem to give cos^2x (1 - 2cos^2x + cos^4x).

Finally, we can simplify this expression further by distributing the cos^2x term and collecting like terms to get cos^2x - 2cos^4x + cos^6x.

Therefore, the simplified expression of cos^2x sin^4x is cos^2x - 2cos^4x + cos^6x.

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The price that maximizes annual revenue is $17.86 per bushel, which should be charged by the farmers; The quantity of corn that can be sold per year at 67,786 bushels per year; the farmers' resulting revenue will be  $1,210,392.96 per year.

To find the price that maximizes annual revenue, we need to differentiate the revenue function with respect to the price and set it equal to zero:

Revenue = pq = (6,600,000q^1.3)q

= 6,600,000q^2.3

dRevenue/dp = q

Setting dRevenue/dp = 0, we get q = 0, which is not a valid solution. Therefore, we need to consider the endpoints of the feasible range, which is q >= 13,000.

At q = 13,000, we have p = 6,600,000*13,000^(-0.3) ≈ $17.86 per bushel.

At q → ∞, we have p → 0.

So, the price that maximizes annual revenue is $17.86 per bushel, which should be charged by the farmers.

The quantity of corn that can be sold per year at that price is given by

q = (p/6,600,000)^(1/1.3)

= (17.86/6,600,000)^(1/1.3)

≈ 67,786 bushels per year.

The farmers' resulting revenue will be Revenue = p*q

= $17.86 * 67,786

≈ $1,210,392.96 per year.

Therefore, the price that maximizes annual revenue is $17.86 per bushel, which should be charged by the farmers; The quantity of corn that can be sold per year at 67,786 bushels per year; the farmers' resulting revenue will be  $1,210,392.96 per year.

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The probability of rolling a 3 on a pair of standard six-sided dice is 1/36.

There is only one way to roll a 3 on a pair of standard six-sided dice: one die must show a 1, and the other die must show a 2.

The number of possible outcomes when rolling two dice is given by the number of combinations of the two dice, which is 6x6 = 36.

The number of outcomes that result in a sum of 3 is 1 (as explained above). Therefore, the probability of rolling a 3 is:

P(rolling a 3) = number of outcomes that result in a sum of 3 / total number of possible outcomes = 1/36.

So the probability of rolling a 3 on a pair of standard six-sided dice is 1/36.

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Answer:

x = 15

General Formulas and Concepts:

Pre-Alg

Geometry

Step-by-step explanation:

Step 1: Identify

We notice from the picture that the 2 angles are vertical.

Step 2: Set up equation

x + 12 = 5x - 48

Step 3: Solve for x

Answer:

The last one

Step-by-step explanation:

Answer:

thats to many questions to answer

Step-by-step explanation:

Answer:

the Answer will be

\( \frac{3(a + 5)}{(a {}^{2} - 9) } \)

The slope in the context of predicting verbal SAT scores based on math SAT scores indicates the change in the verbal score for each unit increase in the math score.

In this survey of statistics undergraduate students at Prosperity University, the slope represents the rate of change in the verbal SAT score for every one-point increase in the math SAT score. A positive slope suggests that as the math score increases, the verbal score also tends to increase. The magnitude of the slope indicates the strength of the relationship between the two variables. A larger slope indicates a stronger positive association, meaning that a one-point increase in the math score is associated with a larger increase in the verbal score. Conversely, a negative slope would imply an inverse relationship, where an increase in the math score is associated with a decrease in the verbal score.

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What is the interpretation of the slope in the context of predicting the verbal SAT score based on the math SAT score, according to the survey conducted among undergraduate students majoring in statistics at Prosperity University?

Answer:

Step-by-step explanation:

The solution is a darkest area of a plane. All dots inside this area are the solutions of the system.

Answer:

see attachment for graph

Step-by-step explanation:

< or > : dashed line

≤ or ≥ : solid line

< or ≤ : shade below the line

> or ≥ : shade above the line

To graph the line y < 3x-2

Rewrite the equation as: y = 3x - 2

Find two points on the line:

when x = 0,  y = 3(0) - 2 = -2  →  (0, -2)

when x = 3,  y = 3(3) - 2 = 7  →  (3, 7)

Plots the found points (0, -2) and (3, 7).

Draw a straight, dashed line through the points.

To graph the line y ≥ -x + 3

Rewrite the equation as: y = -x + 3

Find two points on the line:

when x = 0,  y = -(0) + 3 = 3  →  (0, 3)

when x = 3,  y = -(3) + 3 = 0  →  (3, 0)

Plots the found points (0, 3) and (3, 0).

Draw a straight, solid line through the points.

Shade above the solid line and below the dashed line to the right of where the two lines intersect.  The shaded area is the area of all possible solutions.

The formula to find the radius (r) of a circle when you know the circumference (C) is r = C/(2π).

The formula presented derived from the formula for the circumference of a circle, which is C = 2πr. By rearranging the equation and isolating the radius (r) on one side, we get r = C/(2π).

So, if the circumference (C) of a circle is 18 centimeters, you can use the formula r = C/(2π) to find the radius (r). Plugging in the given value for the circumference (C), we get:

r = 18/(2π)

Simplifying the equation gives:

r = 9/π

Therefore, the radius (r) of the circle is 9/π centimeters.

In conclusion, the formula you can use to find the radius (r) of a circle when you know the circumference (C) is r = C/(2π).

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Answer:

A+c

Step-by-step explanation:

because common sense

The direct method of Lyapunov is a technique used in the analysis of the stability of a dynamical system. It involves the use of a Lyapunov function to determine whether a system is stable or not.

A Lyapunov function is a scalar function V(x) that is defined on the state space of a dynamical system, where x is the state of the system. The function is chosen such that it is positive definite, i.e., V(x) > 0 for all x except possibly at the origin, where V(x) = 0. The time derivative of the Lyapunov function along the trajectories of the system, denoted by V'(x), is also chosen to be negative definite or negative semi-definite, i.e., V'(x) < 0 or V'(x) ≤ 0 for all x except possibly at the origin.

The direct method of Lyapunov uses this Lyapunov function to determine the stability of the system. If a Lyapunov function can be found that satisfies the above conditions, then the system is said to be stable or asymptotically stable, depending on whether V'(x) is negative definite or negative semi-definite, respectively. If a Lyapunov function cannot be found, then the stability of the system cannot be determined using this method.

In addition to determining stability, the direct method of Lyapunov can also be used to determine instability. If a Lyapunov function can be found that satisfies the above conditions, but with V'(x) positive definite or positive semi-definite instead of negative definite or negative semi-definite, respectively, then the system is unstable.

Overall, the direct method of Lyapunov provides a powerful tool for analyzing the stability of a wide range of dynamical systems, including linear and nonlinear systems, time-invariant and time-varying systems, and continuous and discrete-time systems.

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Answer:

beta increases

Step-by-step explanation:

Answer:

y = 50 is answers and it is alternate angle

Answer:

B

Step-by-step explanation:

The two angles are the same

The equations' graphs are shown below for comparison.
(d) y=(4^x)-3, represents the graph in question.

What is graphs?

The link between lines and points is described by a graph, which is a mathematical description of a network. A graph is made up of some points and the connecting lines. The length of the lines and the placement of the points are irrelevant., A node is the name for each element in a graph.

A graph, or "G," is a collection of nodes, or "vertexes," that are connected by edges, or "links," or "e." Thus G= (v , e).

A graph's vertex (also known as a node) is where two lines cross. It designates a place, such as a city, a crossroads, or a transportation hub (stations, harbours, and airports).

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Answer:

FALSE

Step-by-step explanation:

DONE THIS

Answer:

The awnser is C. 4 hours

Step-by-step explanation:

It takes her 2 hours two walk 1 mile so 2*2 is 4