Question 4 Two eigenvalues of a 3 x 3 matrix A are λ = 3,4. Determine the third (integer) eigenvalue if det(A) = 60 Question 5 A is a 2 x 2 real matrix. Choose the correct statement below. A can be diagonalized if it has 2 unit eigenvectors All two eigenvalues are distinct and the eigenvectors that correspond to distinct eigenvalues are orthogonal Eigenvectors that correspond to distinct eigenvalues are orthogonal Eigenvalues of A are the roots of a quadratic polynomial

1. What is the probability a randomly selected bag will have a positive test? Give your answer to four decimal places is 0.1032

2. Given a randomly selected bag has a positive test, what is the probability it actually contains a large amount of liquid is 0.3780

3. Given a randomly selected bag has a positive test, what is the probability it does not contain a large amount of liquid is  0.6219

Let's characterize the taking after occasions:

A: The pack contains huge sums of fluid.

B: The test is positive.

We are given the taking after probabilities:

P(A) = 0.04

P(B | A) = 0.98

P(B | not A) = 0.07

1. To discover the likelihood of a positive test, we are able to utilize the law of adding up to likelihood:

P(B) = P(B | A) P(A) + P(B | not A) P(not A)

= 0.98 * 0.04 + 0.07 * 0.96

= 0.1032

So the likelihood of a haphazardly chosen pack having a positive test is 0.1032 (adjusted to four decimal places).

2. To discover the likelihood that a sack really contains large amounts of fluid given a positive test, we are able to utilize Bayes' hypothesis:

P(A | B) = P(B | A) P(A) / P(B)

= 0.98 * 0.04 / 0.1032

= 0.3780

So the likelihood that a pack really contains expansive sums of fluid given a positive test is 0.3780 (adjusted to four decimal places).

3. To discover the likelihood that a sack does not contain expansive sums of fluid given a positive test, ready to utilize Bayes' hypothesis again:

P(not A | B) = P(B | not A) P(not A) / P(B)

= 0.07 * 0.96 / 0.1032

= 0.6219

So the likelihood that a pack does not contain expansive sums of fluid given a positive test is 0.6219 (adjusted to four decimal places). 

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The given relationship in the task content is not a function as more than one output value is paired with the same input value.

Recall that a relationship is said to be a function only if one output value is attached to each input value of the relationship.

On this note, by observation; the pair of coordinates (6, 3) and (6, 9) implies that two output values are assigned to the same input value. Consequently, the given relationship is not a function.

The complete and correct sentence is therefore; Since one input value is paired with two output values; the relationship is not a function.

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The simplified expression is given by (x² - 3x - 3) / ((x + 3)(x - 2)(x - 4)).

To simplify this expression, we need to find a common denominator for the two fractions and then combine them. To do this, we need to factor the denominators of both fractions.

Let's start with the first fraction's denominator:

x² + x - 6

We need to find two numbers that multiply to -6 and add to +1. These numbers are +3 and -2. Therefore, we can write:

x² + x - 6 = (x + 3)(x - 2)

Now let's factor the second fraction's denominator:

x² - 6x + 8

We need to find two numbers that multiply to 8 and add to -6. These numbers are -2 and -4. Therefore, we can write:

x² - 6x + 8 = (x - 2)(x - 4)

Now we can rewrite the original expression with a common denominator:

(x(x - 2) - (1)(x + 3)) / ((x + 3)(x - 2)(x - 4))

Next, we can simplify the numerator:

(x² - 2x - x - 3) / ((x + 3)(x - 2)(x - 4))

(x² - 3x - 3) / ((x + 3)(x - 2)(x - 4))

Finally, we can't simplify this expression any further. Therefore, the simplified expression is:

(x² - 3x - 3) / ((x + 3)(x - 2)(x - 4))

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Answer: The box is large enough to ship the coat.

15000cm to the power of 3>10000cm to the power of 3

Step-by-step explanation:

V box=25x30x20

         =750+20

         =15000cm to the power of 3

So the box is large enough to ship the coat

The characteristics of the visual representations are:

Dot plots :

Histogram :

Box Plot :

What are these graphs used for ?

Dot plot visually displays individual data points, while simultaneously providing frequency information for each interval. It enables one to easily visualize the median and is ideal for summarizing large data sets; additionally, it allows calculation of the mean.

Histograms present frequency by intervals which make them perfect also for analyzing larger data sets., This graphic allows calculating the mean value- a property that makes histograms an excellent tool in summarization tasks.

Box plots are incredibly useful when processing extensive amounts of data -They have visuals illustrating medians, split four ways. Box plots, similar to Dot plots and Histograms, allow computation of means as well.

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Answer: Dot Plot: individual data points

are seen, mean can be calculated

Histogram: frequency over each

interval is given, best used to summarize

large sets of data

Box Plot: breaks the data

into four equal

parts, best used to summarize

large sets of data, median can be seen

visually

"Got it right on Edmentum"

Explanation:

The median can be seen only on a box plot.

The data is broken into four equal parts on a box plot.

Box plots and histograms are best for large sets of data.

The individual data points are only seen on a dot plot. These points can be used to calculate the mean.

The frequency over each interval is given on a histogram

As we add more terms to the series, the plot approaches the original function f(x) = 1/x. Note that the series is only defined for x > 0, since f(x) is not defined at x = 0.

To sketch the Fourier cosine series of f(x) = 1/x, we need to first determine the Fourier coefficients. Recall that the Fourier cosine series is given by:

f(x) = a0/2 + ∑[n=1 to ∞] an cos(nπx/L)

where L is the period of the function (in this case, L = 2), and the Fourier coefficients are given by:

an = (2/L) ∫[0 to L] f(x) cos(nπx/L) dx

Using f(x) = 1/x, we can compute the Fourier coefficients as follows:

a0 = (2/L) ∫[0 to L] f(x) dx
  = (2/2) ∫[0 to 2] 1/x dx
  = ∞ (divergent)

an = (2/L) ∫[0 to L] f(x) cos(nπx/L) dx
  = (2/2) ∫[0 to 2] (1/x) cos(nπx/2) dx
  = (-1)^n π/2 (n ≠ 0)

Note that a0 is divergent, which means that the Fourier cosine series of f(x) will not have a constant term. Therefore, the Fourier cosine series of f(x) is given by:

f(x) = ∑[n=1 to ∞] (-1)^n π/2 cos(nπx/2)

To sketch this series, we can plot the partial sums of the series for a few values of n. For example, we can plot:

f1(x) = (-1)^1 π/2 cos(πx/2)
f2(x) = (-1)^1 π/2 cos(πx/2) + (-1)^2 π/2 cos(2πx/2)
f3(x) = (-1)^1 π/2 cos(πx/2) + (-1)^2 π/2 cos(2πx/2) + (-1)^3 π/2 cos(3πx/2)

and so on, up to some value of n. Here is what the plots look like for n = 1, 2, and 3:

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The evidence supports the claim that the average life of electric light bulbs is greater than 2000 hours.

a. Null Hypothesis: The average life of electric light bulbs is not greater than 2000 hours. Alternative Hypothesis: The average life of electric light bulbs is greater than 2000 hours.

b. The critical value for a one-tailed test at the 2% level of significance with 63 degrees of freedom is 2.33.

c. The relevant test statistic is:
t = (x - μ) / (s / √n)=[tex]= \frac{(2008 - 2000)}{\frac{12.31}{\sqrt{64}}}= 13.03[/tex]
Since the test statistic is greater than the critical value of 2.33, we can reject the null hypothesis and conclude that there is sufficient evidence to support the claim.

d. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed value, assuming the null hypothesis is true. Using a t-distribution table with 63 degrees of freedom, the p-value is less than 0.01. Since the p-value is less than the significance level of 0.02, we can reject the null hypothesis.

e. Yes, we reject the null hypothesis.

f. No, we do not reject the claim. The evidence supports the claim that the average life of electric light bulbs is greater than 2000 hours.

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To evaluate the given integral by changing to polar coordinates, we first need to determine the limits of integration in polar form. The region R is in the first quadrant and is bounded by the circles with the center of the origin and radii 2 and 3. In polar coordinates, the equation of a circle centered at the origin is given by r = a, where a is the radius.

So, the equations of the two circles are:

r = 2  and  r = 3

Since the region R is between these two circles, the limits of integration for r are:

2 ≤ r ≤ 3

To determine the limits of integration for θ, we need to consider the quadrant in which the region R lies. Since R is in the first quadrant, we have:

0 ≤ θ ≤ π/2

Now, we can express the integrand sin(x^2 + y^2) in terms of polar coordinates:

sin(x^2 + y^2) = sin(r^2)

Therefore, the integral in polar coordinates is:

∫∫R sin(x^2 + y^2) dA = ∫ from 0 to π/2 ∫ from 2 to 3 sin(r^2) r dr dθ

This integral can be evaluated using standard techniques of integration.
To evaluate the integral using polar coordinates, we first need to express the given region R and the integrand in terms of polar coordinates. In polar coordinates, x = r*cos(θ) and y = r*sin(θ), so x^2 + y^2 = r^2.

The region R is in the first quadrant and is bounded by the circles with radii 2 and 3. In polar coordinates, this translates to 0 ≤ θ ≤ π/2, 2 ≤ r ≤ 3.

Now we can rewrite the integral as:

integral_integral_R sin(x^2 + y^2) dA
= integral (θ=0 to π/2) integral (r=2 to 3) sin(r^2) * r dr dθ

Now we can evaluate the integral step by step:

1. Integrate with respect to r:
integral (θ=0 to π/2) [(-1/2)cos(r^2)] (from r=2 to r=3) dθ
= integral (θ=0 to π/2) [(-1/2)(cos(9) - cos(4))] dθ

2. Integrate with respect to θ:
[(-1/2)(cos(9) - cos(4))]*(θ evaluated from 0 to π/2)
= [(-1/2)(cos(9) - cos(4))] * (π/2)

So the final answer is:

(π/2)(-1/2)(cos(9) - cos(4))

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The simplified difference is:

3y/8 - 5/6y = (-y)/24

Note that there are no non-permissible values in this case.

a. 4x/(2x+5) + 10/(2x+5)

To add these two expressions, we need to find a common denominator. In this case, the common denominator is (2x+5):

4x/(2x+5) + 10/(2x+5) = (4x+10)/(2x+5)

Now we can simplify the numerator by factoring out a 2:

(4x+10)/(2x+5) = 2(2x+5)/(2x+5)

And we can cancel out the common factor of (2x+5):

2(2x+5)/(2x+5) = 2

Therefore, the simplified sum is:

4x/(2x+5) + 10/(2x+5) = 2

Note that the non-permissible value is x = -2.5, since this value would make the denominator equal to zero.

b. 3y/8 - 5/6y

To subtract these two expressions, we also need a common denominator. In this case, the common denominator is 24y:

3y/8 - 5/6y = (9y^2)/(24y) - (20y)/(24y)

We can simplify the first term in the numerator by canceling out a common factor of 3:

([tex]9y^2[/tex])/(24y) = (3y)/8

So the subtraction becomes:

3y/8 - 5/6y = (3y)/8 - (10y)/12

Now we can find a common denominator of 24:

(3y)/8 - (10y)/12 = (9y)/24 - (10y)/24

Simplifying the numerator gives:

(9y)/24 - (10y)/24 = (-y)/24

Therefore, the simplified difference is:

3y/8 - 5/6y = (-y)/24

Note that there are no non-permissible values in this case.

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The Laplace transform is a mathematical tool used to transform a function of time into a function of complex frequency. The inverse Laplace transform does the opposite, transforming a function of complex frequency back into a function of time.

In MATLAB, you can use the "laplace" function to compute the Laplace transform of a given function. The syntax for the "laplace" function is: laplace(f), where f is the function you want to transform.

For example, in Example 1, the function f(t) = 5sin(3t) is defined using MATLAB's symbolic math toolbox by typing ">>symst" to activate symbolic math, followed by ">>f=5* sin(3*t);" to define the function. The Laplace transform of this function is then computed using the "laplace" function as follows: ">>laplace(f)".

Similarly, in Example 2, the function f(t) = (t - 2)^2U(t - 2) is defined using MATLAB's "heaviside" function to represent the unit step function. The Laplace transform of this function is then computed using the "laplace" function as follows: ">>F=laplace(f)".

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In the case of discrete random variables, the expectation of a function is defined as the sum of the function's values multiplied by their probabilities:

E(aX + bY + c) = ∑(aX + bY + c)P(X,Y)

We can break down the sum using properties of summation:

= a∑XP(X,Y) + b∑YP(X,Y) + c∑P(X,Y)

Since the sum of probabilities over all events equals 1:

= aE(X) + bE(Y) + c

For the continuous case, the expectation of a function is defined as the integral of the function's values multiplied by the joint probability density function (PDF):

E(aX + bY + c) = ∫∫(aX + bY + c)f(X,Y)dXdY

We can break down the integral using properties of integration:

= a∫∫Xf(X,Y)dXdY + b∫∫Yf(X,Y)dXdY + c∫∫f(X,Y)dXdY

Again, since the integral of the joint PDF over all events equals 1:

= aE(X) + bE(Y) + c

Thus, we have shown that for both discrete and continuous cases, the linearity of expectation holds:

E(aX + bY + c) = aE(X) + bE(Y) + c

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Answer:

Step-by-step explanation:

The probability of a woman having four girls in a row is 6.25%.

To calculate the probability that if a woman has four children, they will all be girls, you should use the rule of multiplication. This rule states that to calculate the probability of two or more independent events occurring together, you multiply the probability of each individual event. In this case, the probability of each child being a girl is 0.5 (assuming an equal chance of having a boy or girl), so you would calculate the probability as 0.5 x 0.5 x 0.5 x 0.5 = 0.0625 or 6.25%. Therefore, the probability of a woman having four girls in a row is 6.25%.

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The simplified form of the given expression written into one power is 3⁰

From the question, we are to simplify the given expression into one power

The given expression is

[tex]\frac{(3^{2})^{2}}{3 \ \cdot \ 3^{3}}[/tex]

To simplify the expression, we will implore the laws of indices

Simplifying the expression

[tex]\frac{(3^{2})^{2}}{3 \ \cdot \ 3^{3}}[/tex]

[tex]\frac{(3^{2\times 2})}{3^{1+3}}[/tex]

[tex]\frac{3^{4}}{3^{4}}[/tex]

Applying the division law of indices

[tex]3^{4-4}[/tex]

[tex]3^{0}[/tex]

Hence, the simplified expression is 3⁰

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0 is a single digit whole number for y that supports Doni's claim.

2 is a single digit whole number for y that does not supports Doni's claim.

Doni claims that [tex]\frac{3^y}{2^y} \leq 1[/tex]

We have to find a single digit whole number for y that supports Doni's claim.

Let 0 be the single digit whole number for y that supports Doni's claim.

1/1≤1

Now let us find  single digit whole number for y that does not support Doni's claim.

2 be the whole number

9/4≤1

2.25≤1

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The solution of the given equation is x = 2.

Here we have the following equation that we want to solve, it is:

[tex]125^x = \frac{625}{5^{-x}}[/tex]

We want to solve this for x, remember that a negative exponent means that we need to take the inverse, then we can rewrite the right side as:

[tex]125^x = \frac{625}{5^{-x}} = 625*5^x[/tex]

Now we can divide both sides by 5^x to get:

[tex]125^x = \frac{625}{5^{-x}} = 625*5^x\\\\(125/5)^x = 625\\\\25^x = 625\\\\[/tex]

Now we can apply the natural logarithm in both sides, we will get:

[tex]ln(25^x) = ln(625)\\\\x*ln(25) = ln(625)\\x = ln(625)/ln(25)\\\\x = 2[/tex]

That is the solution.

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Answer:

A = 120; C = 95

Step-by-step explanation:

We will need a system of equations to solve for C, the number of child tickets and A, the number of adult tickets.

We know that the sum of the revenue earned from both the child tickets and the adult tickets = the total revenue

Thus, our first equation is 3C + 14A = 1965

We also know that the sum of the total number of child and adult tickets = the the total number of tickets

Thus, our other equation is C + A = 215

We can solve using substitution by first isolating c in the second equation:

[tex]C+A=215\\C=-A+215[/tex]

Now, we can plug in the equation we just made for C in the first equation in our system to solve for A:

[tex]3(-A+215)+14A=1965\\-3A+645+14A=1965\\11A+645=1965\\11A=1320\\A=120[/tex]

Finally, we can solve for C using the second equation in our system by plugging in 120 for A:

[tex]C+120=215\\C=95[/tex]

Where c1 and c2 are arbitrary constants determined by initial conditions.

The differential equation given is:

y'' - 6y' + 25y = 0

To find the two linearly independent solutions of the differential equation, we assume a solution of the form:

y = e^(rt)

where r is a constant to be determined. We then substitute this into the differential equation and obtain:

r^2 e^(rt) - 6r e^(rt) + 25 e^(rt) = 0

Dividing both sides by e^(rt), we get:

r^2 - 6r + 25 = 0

This is the characteristic equation of the differential equation, and we can solve for r using the quadratic formula:

r = (6 ± sqrt(6^2 - 4*25)) / 2

r = 3 ± 4i

Therefore, the two linearly independent solutions of the differential equation are:

y1 = e^(3x) cos(4x)

y2 = e^(3x) sin(4x)

To verify that these solutions are linearly independent, we can take the Wronskian of the solutions:

W(y1, y2) = y1y2' - y1'y2

= e^(3x) cos(4x) (3e^(3x) sin(4x) + 4e^(3x) cos(4x)) - e^(3x) sin(4x) (3e^(3x) cos(4x) - 4e^(3x) sin(4x))

= 5e^(6x)

Since the Wronskian is nonzero, the two solutions are linearly independent.

The general solution to the differential equation is then a linear combination of the two linearly independent solutions:

y = c1 e^(3x) cos(4x) + c2 e^(3x) sin(4x)

where c1 and c2 are arbitrary constants determined by initial conditions.

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The value of the integral ∫ [tex]cos^7(x) dx[/tex] is[tex](3\pi /32).[/tex]

Wallis's formulas are used to evaluate integrals of the form:

∫ [tex]sin^{n(x)} cos^{m(x)} dx[/tex]

where n and m are non-negative integers. We can use the trigonometric identity[tex]cos^{2(x)] + sin^{2(x)} = 1[/tex] to convert the powers of cosine to powers of sine.

Here, we have m = 7, so we can use the identity [tex]cos^{2(x)} = 1 - sin^{2(x)}[/tex] to write:

[tex]cos^{7(x)} = cos^{6(x)}[/tex] × [tex]cos(x)[/tex]

[tex]= (1 - sin^2(x))^3[/tex] ×[tex]cos(x)[/tex]

Now, we can use a substitution of [tex]u = sin(x), du = cos(x) dx[/tex]to convert the integral to a form that can be evaluated using Wallis's formulas:

∫ [tex]cos^7(x) dx =[/tex] ∫ [tex](1 - sin^2(x))^3[/tex] × [tex]cos(x) dx[/tex]

= ∫ [tex](1 - u^2)^3 du[/tex]

Using Wallis's formulas, we have:

∫ [tex](1 - u^2)^3 du = (1/8)[/tex]× β[tex](4, 4)[/tex]

[tex]= (1/8)[/tex] ×[tex][(3\pi /4) / sin(3\pi /4)][/tex]

[tex]= (3\pi /32)[/tex]

Substituting [tex]u = sin(x)[/tex], we have:

∫ [tex]cos^7(x) dx =[/tex] ∫ [tex](1 - u^2)^3 du = (3π/32)[/tex]

Therefore, the value of the integral ∫ [tex]cos^7(x) dx[/tex] is [tex](3\pi /32).[/tex]

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The probability that the point lies on the square is P = 0.498

to find that probability, we need to take the quotient between the area of the square and the area of the circle.

We can see that the square has a side length of 5 units, then its area is.

A = 5*5 = 25 square units.

The circle has a radius of 4 units, then its area is:

A' = 3.14*4^2 = 50.24 square units

Then the probability is:

P = 25/50.24 = 0.498

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From multiplcation operation, Alex has enough paint to cover 1,000 ft² of his apartment. The true statement is 2 gallons of paint will cover 1068 ft², Alex have enough paint of quantity 2 gallons.

We have Mr. Alex painted his apartment. Area of his apartment'walls = 178 ft²

Quantity of paint used by him to paint his apartment'walls with area 178 ft² =[tex] \frac{1}{3} \: \: gallons[/tex]

Total quantity of paint used in all

= 2 gallons

We have to check the provide paint is enough or not to cover 1,000 ft² of his apartment. Let the required paint for 1000 ft² be x gallons. Using multiplcation, 1/3 gallons quantity of paint will cover the area of apartment = 178 ft², so, 1 gallons quantity of paint will cover the area of apartment = 178 ×3 ft²= 534 ft²

Now, 2 gallons quantity of paint will cover the area of apartment = 2× 534 ft² = 1068 ft²> 1000 ft²

But he wants to paint 1000 ft² of his apartment in 2 gallons quantity (x=1.9 gal ). So, he has enough paint to paint his apartment.

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Complete question:

The above figure complete the question.

Alex painted 178 ft2 of his apartment’s walls with 1/3 gallon of paint. He has 2 gallons of paint in all. If he wants to cover 1,000 ft2 of his apartment, does he have enough paint? Complete a true statement

The value of variable x and y in the right triangle are 4 and 4√3 units respectively.

A right angle triangle is a triangle that has one of its angles as 90 degrees.

The variable x and y can be as follows:

cos 60° = adjacent / hypotenuse

cos 60° = x / 8

cross multiply

x = 8 cos 60°

x = 8 × 0.5

x = 4 units

Let's find the value of y as follows:

sin 60° = opposite / hypotenuse

Therefore,

sin 60° =  y / 8

cross multiply

y = 8 sin 60

y = 8 × √3 / 2

y = 4√3 units  

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(i) The transfer function of the system is:

H(z) = (0.4)^z / (0.2)^z - 0.3(0.4)^(z-1) / (0.2)^{z-1} - 2

(ii) The difference equation characterizing the system is:
y[n] = (0.4)^n x[n] - 0.3(0.4)^(n-1) x[n-1]

(i) To determine the transfer function of the system, we can take the Z-transform of both the input and output:

X(z) = (0.2)^z / (z - 0.4)
Y(z) = (0.4)^z / (z - 0.4) - 0.3(0.4)^(z-1) / (z - 0.4)

Then we can solve for the transfer function H(z) by dividing Y(z) by X(z):

H(z) = Y(z) / X(z)
    = (0.4)^z / (z - 0.4) - 0.3(0.4)^(z-1) / (z - 0.4) * (z - 0.4) / (0.2)^z
    = (0.4)^z / (0.2)^z - 0.3(0.4)^(z-1) / (0.2)^{z-1} - 2

So the transfer function of the system is H(z) = (0.4)^z / (0.2)^z - 0.3(0.4)^(z-1) / (0.2)^{z-1} - 2.

(ii) To determine the difference equation characterizing the system, we can use the formula for the output y[n] of a discrete-time LTI system with input x[n]:

y[n] = sum{k=0}{N-1} h[k] x[n-k]

where h[k] is the impulse response of the system. In this case, the impulse response can be found by setting x[n] = delta[n], the unit impulse function, and solving for y[n]:

h[n] = y[n] / delta[n]
    = (0.4)^n - 0.3(0.4)^(n-1)

So the difference equation characterizing the system is:

y[n] = (0.4)^n x[n] - 0.3(0.4)^(n-1) x[n-1]

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Answer

The result of 48 - 36 is 12. Then, if you divide 12 by 36, the result is 0.3333 or 1/3.

Step-by-step explanation:

Answer:

option B: x= 18

Step-by-step explanation:

To solve 1/3x - 1 = 5, we can start by adding 1 to both sides of the equation:

1/3x - 1 + 1 = 5 + 1

Simplifying:

1/3x = 6

Multiplying both sides by 3:

3(1/3x) = 3(6)

Simplifying:

x = 18

Therefore, the solution is x = 18, which is option B.

To the nearest tenth of a mile, Mr. Miller can see 9.8 miles of the horizon along the arc of his field of vision.

Based on the given information, we can use the formula for the arc length of a circle to find how much of the horizon Mr. Miller can see within his field of vision.

The formula for the arc length of a circle is:

arc length = (angle/360) x 2πr

where angle is the central angle of the arc in degrees, r is the radius of the circle, and 2πr is the circumference of the circle.

In this case, the central angle of the arc is 140 degrees, and the radius of the circle is the distance to the horizon, which is 4 miles. We can substitute these values into the formula:

arc length = (140/360) x 2π x 4

arc length = 0.388 x 8π

arc length = 9.8 miles

Therefore, to the nearest tenth of a mile, Mr. Miller can see 9.8 miles of the horizon along the arc of his field of vision.

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The exact values of the cosecant, secant, and cotangent ratios of 5π/6 are 2, -2/√3, and -√3, respectively.

To find the exact values of the cosecant, secant, and cotangent ratios of an angle of 5π/6, we need to use the definitions of these trigonometric functions and the values of the sine, cosine, and tangent of this angle.

First, we can find the sine and cosine of 5π/6 using the unit circle or reference angles:

sin(5π/6) = sin(π/6) = 1/2

cos(5π/6) = -cos(π/6) = -√3/2

Then, we can use the definitions of the cosecant, secant, and cotangent ratios:

cosec(5π/6) = 1/sin(5π/6) = 1/(1/2) = 2

sec(5π/6) = 1/cos(5π/6) = -2/√3

cot(5π/6) = cos(5π/6)/sin(5π/6) = (-√3/2)/(1/2) = -√3

Therefore, the exact values of the cosecant, secant, and cotangent ratios of 5π/6 are 2, -2/√3, and -√3, respectively.

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For equation A+C=B the matrix C is [tex]\left[\begin{array}{ccc}-2&-7\\-5&8\end{array}\right][/tex] and C-B=A then C is [tex]\left[\begin{array}{ccc}2&-9\\7&0\end{array}\right][/tex]

The given matrix A = [tex]\left[\begin{array}{ccc}2&-1\\6&-4\end{array}\right][/tex]

B=[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]

Now the equation is A+C=B

[tex]\left[\begin{array}{ccc}2&-1\\6&4\end{array}\right][/tex]+C  =[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]

C=[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]- [tex]\left[\begin{array}{ccc}2&-1\\6&-4\end{array}\right][/tex]

C=[tex]\left[\begin{array}{ccc}-2&-7\\-5&8\end{array}\right][/tex]

Now equation is C-B=A

C=A+B

= [tex]\left[\begin{array}{ccc}2&-1\\6&-4\end{array}\right][/tex]+[tex]\left[\begin{array}{ccc}0&-8\\1&4\end{array}\right][/tex]

C=[tex]\left[\begin{array}{ccc}2&-9\\7&0\end{array}\right][/tex]

Hence, for equation A+C=B the matrix C is [tex]\left[\begin{array}{ccc}-2&-7\\-5&8\end{array}\right][/tex] and C-B=A then C is [tex]\left[\begin{array}{ccc}2&-9\\7&0\end{array}\right][/tex]

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a. At (0,2] f is uniformly continuous.

b. At (2,∞) f is uniformly continuous.

c. At (0,∞) f is uniformly continuous.

A function connects an input with an output. It is analogous to a machine with an input and an output. And the output is somehow related to the input. The standard manner of writing a function is f(x) "f(x) =... "

(a) Let f(x) = x³ on (0,2]. Let ε > 0 be given. We need to find a δ > 0 such that |x - y| < δ implies |f(x) - f(y)| < ε for all x,y in (0,2]. Note that |f(x) - f(y)| = |x³ - y³| = |x - y||x² + xy + y²|. Since x,y ∈ (0,2], we have x² + xy + y² ≤ 12. Thus, if we choose δ = ε/12, then for any x,y ∈ (0,2] such that |x - y| < δ, we have |f(x) - f(y)| < ε. Hence, f is uniformly continuous on (0,2].

(b) Let f(x) = 1/2 on (2,∞). Let ε > 0 be given. We can choose any δ > 0 since for any x,y ∈ (2,∞), we have |f(x) - f(y)| = 0 < ε. Thus, f is uniformly continuous on (2,∞).

(c) Let f(x) = (x-1)/(x+1) on (0,∞). Let ε > 0 be given. We need to find a δ > 0 such that |x - y| < δ implies |f(x) - f(y)| < ε for all x,y in (0,∞). Note that |f(x) - f(y)| = |(x-1)/(x+1) - (y-1)/(y+1)| = |(x-y)(2/(x+1)(y+1))|. Thus, if we choose δ = ε/2, then for any x,y in (0,∞) such that |x - y| < δ, we have |f(x) - f(y)| = |(x-y)(2/(x+1)(y+1))| < ε. Hence, f is uniformly continuous on (0,∞).

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