if you give me new answer i will give you like
2. Suppose that Xn, n > 1 are i.i.d. random variables with P(X = 2) = 1/8. P(X = -1) = 1/2, P(X = 0) = 1/8, P(X = 1) = 1/4, Let Sn = 2-1 X; with So = 0. Let T be defined as vi= : T = min{n : Sn > 10 o

Answer: Area, A, is x times y.

Step-by-step explanation:

The equation represented on the graph obtained from equation of a sinusoidal function is the option; y = coa(x - π/4) - 2

A sinusoidal function is a periodic function that repeats at regular interval and which is based on the cosine or sine functions.

The coordinate of the points on the graph indicates that we get;

The period, T = π/4 - (-7·π/4) = 8·π/4 = 2·π

Therefore, B = 2·π/(2·π) = 1

B = 1

The amplitude, A = (-1 - (-3))/2 = 2/2 = 1

The vertical shift, D = (-1 + (-3))/2 = -4/2 = -2

The vertical shift, D = -2

The horizontal shift is the amount the midline pint is shifted relative to the y-axis

The points on the graph indicates that the peak point  close to the y-axis is shifted π/4 units to the right of y-axis, therefore, the horizontal shift, C = π/4

cos(0) = 1 which is the peak point value of the trigonometric ratio, in the function which indicates that the trigonometric function of the equation for the graph is of the form, y = A·cos(B·(x - C) + D

Plugging in the above values into the sinusoidal function equation of the form; y = A·cos(B·(x - C) + D

We get;

A = 1, B = 1, C = π/4, and D = -2

The function representing the graph is therefore;

y = cos(x - π/4) - 2

The correct option is therefore;

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This is the equation in terms of w that models the situation:[tex]w^2 - 10w + 18.75 = 0[/tex].

Rectangle with width w and length l, enclosed by a 20-meter fence: The perimeter of the rectangle, which is equal to the length of the fence, is given by:

2w + 2l = 20

We can simplify this equation by dividing both sides by 2:

w + l = 10

We also know that the area of the rectangle is 18.75 square meters:

w * l = 18.75

We want to write an equation in terms of w, so we can solve for l in terms of w by dividing both sides by w:

l = 18.75 / w

This expression for l into the equation for the perimeter, we get:

w + (18.75 / w) = 10

Multiplying both sides by w, we get:

[tex]w^2 + 18.75 = 10w[/tex]

Rearranging this equation, we get:

[tex]w^2 - 10w + 18.75 = 0[/tex]

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According to the figure of a circle, x is equal to 17

The total arc length in a circle is equal to 360 degrees

hence arc QR + arc RS + arc QS = 360 degrees

Where

arc QR = 120

arc RS = 2 * 67 = 134

arc QS = 5x + 20

plugging in the values

120 + 134 + 5x + 21 = 360

5x = 360 - 120 - 134 - 21

5x = 85

x = 17

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The three main techniques of measuring quantitative data are surveys, experiments, and observational studies.

1. Surveys: Surveys involve collecting data by asking a sample of individuals to respond to a set of questions. This method can be done through various formats such as questionnaires, interviews, or online polls. Surveys are useful for gathering information on opinions, attitudes, or preferences and can help determine relationships between variables.
2. Experiments: Experiments involve manipulating one or more variables to observe the effect on a dependent variable. Participants are typically randomly assigned to different conditions, and the researcher measures the outcomes to determine cause-and-effect relationships. Experiments can provide strong evidence for causal relationships and are often used in scientific research.
3. Observational studies: Observational studies involve collecting data by observing and recording the natural behavior or characteristics of individuals or groups without any intervention. Researchers can observe the participants in their natural settings or use existing data sources such as records, databases, or archival data. Observational studies are useful for understanding patterns, trends, and relationships between variables, but they cannot establish causality.
These techniques can provide valuable insights into various aspects of quantitative data, allowing for informed decision-making and improved understanding of patterns and relationships.

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If you're conducting a significance test for the difference between the means of two independent samples, your null hypothesis would be option E, H0: μ1 - μ2 = 0, which means that there is no significant difference between the means of the two independent samples. The alternative hypothesis, denoted as Ha, would be that there is a significant difference between the means of the two independent samples.

In order to test the null hypothesis, you would need to use a statistical test such as the t-test or z-test, depending on the sample size and whether the population standard deviations are known or unknown. These tests would provide a p-value, which indicates the probability of obtaining a difference between the means as extreme or more extreme than the observed difference, assuming that the null hypothesis is true.
If the p-value is less than the chosen significance level (usually 0.05), then the null hypothesis can be rejected and it can be concluded that there is a significant difference between the means of the two independent samples. Otherwise, if the p-value is greater than the significance level, then the null hypothesis cannot be rejected and it can be concluded that there is not enough evidence to suggest a significant difference between the means of the two independent samples.
When you are conducting a significance test for the difference between the means of two independent samples, the null hypothesis is a statement that there is no significant difference between the population means of the two groups. In this case, the correct null hypothesis is:

C. H0: μ1 - μ2 = 0
This hypothesis states that the difference between the population means of the two independent samples (μ1 and μ2) is equal to zero, which implies that there is no significant difference between the two population means. The alternative hypothesis would be that there is a significant difference (either μ1 > μ2, μ1 < μ2, or simply μ1 ≠ μ2, depending on the type of test being performed).

To test this hypothesis, you would collect data from the two independent samples and calculate the sample means (x1 and x2). Then, you would conduct a statistical test, such as a t-test or a z-test, to compare the sample means and determine the probability (p-value) of obtaining a difference as large as, or larger than, the one observed in your samples, assuming the null hypothesis is true.

If the p-value is smaller than a predetermined significance level (commonly set at 0.05), you would reject the null hypothesis in favor of the alternative hypothesis, concluding that there is a significant difference between the population means. If the p-value is greater than the significance level, you would fail to reject the null hypothesis, meaning that there is not enough evidence to conclude that there is a significant difference between the population means.

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The circumference of the circle-shaped garden is 157 meters.

The gardener will need 157 meters of fencing to outline the garden.

We have,

The circumference of a circle is given by the formula C = πd, where d is the diameter.

Using this formula,

C = πd

C = 3.14 × 50

C = 157 meters

Therefore,

The circumference of the circle is 157 meters.

The gardener will need 157 meters of fencing to outline the garden.

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The description of the cross section tells us that it is a rectangle

Rectangles are four-sided, two-dimensional shapes that boast two sets of paralleled, opposite sides with identical lengths. All four corner angles measure at ninety degrees and the opposing sides always have the same length.

The area can be calculated by multiplying its length and width, while the perimeter is found by adding all four side measurements together. Furthermore, the perpendicular diagonals of a rectangle will bisect one another and yield equal measurement when fully extended.

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P(x > 76 and -2.86 < z < 0) = 0.5000 - 0.3665 = 0.1335.

(a) P(x < 21 and z < 3)

Using standardization, we get:

z = (21 - 25)/3 = -4/3

Using the standard normal table, the corresponding probability for z = -4/3 is 0.0912.

Therefore, P(x < 21 and z < 3) = 0.0912.

(b) P(24 < x < 30 and a < z < 8)

Using standardization, we get:

z1 = (24 - 26)/3 = -2/3

z2 = (30 - 26)/3 = 4/3

Using the standard normal table, the corresponding probability for z = -2/3 is 0.2514 and for z = 4/3 is 0.4082.

Therefore, P(24 < x < 30 and a < z < 8) = 0.4082 - 0.2514 = 0.1568.

(c) P(x > 25 and z < 5)

Using standardization, we get:

z = (25 - 30)/5 = -1

Using the standard normal table, the corresponding probability for z = -1 is 0.1587.

Therefore, P(x > 25 and z < 5) = 0.1587.

(d) P(17 < x < 21)

Using standardization, we get:

z1 = (17 - 20)/3 = -1

z2 = (21 - 20)/3 = 1/3

Using the standard normal table, the corresponding probability for z = -1 is 0.1587 and for z = 1/3 is 0.3707.

Therefore, P(17 < x < 21) = 0.3707 - 0.1587 = 0.2120.

(e) P(x > 76 and -2.86 < z < 0)

Using standardization, we get:

z1 = (76 - 80)/12 = -1/3

z2 = 0

Using the standard normal table, the corresponding probability for z = -1/3 is 0.3665 and for z = 0 is 0.5000.

Therefore, P(x > 76 and -2.86 < z < 0) = 0.5000 - 0.3665 = 0.1335.

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Step-by-step explanation:

120 = 1000 e^(-.05t)

120/1000 = e^(-.05t)   take natural LN of both sides

-2.12 = - .05t

t = 42.4 days

To find a power series representation for f(x) = ln(2-x), we can use the formula for the power series expansion of ln(1+x):
ln(1+x) = Σ (-1)^(n+1) * (x^n / n)

We can use this formula by setting x = -x/2 and multiplying by -1 to get:
ln(2-x) = ln(1 + (-x/2 - 1)) = Σ (-1)^(n+1) * ((-1)^n * (x^n+1 / (n+1) * 2^(n+1)))
Therefore, the power series representation for f(x) is:
f(x) = Σ (-1)^(n+1) * ((-1)^n * (x^n+1 / (n+1) * 2^(n+1)))

The radius of convergence of this series can be found using the ratio test:
lim |a_n+1 / a_n| = lim |(-1)^(n+2) * (x^(n+2) / (n+2) * 2^(n+3)) * (n+1) / (-1)^(n+1) * (x^(n+1) / (n+1) * 2^(n+2))|
= lim |x / 2 * (n+1) / (n+2)| = |x/2|

Therefore, the radius of convergence is R = 2. The power series representation centered at x = 0 is:
f(x) = Σ (-1)^(n+1) * ((-1)^n * (x^n+1 / (n+1) * 2^(n+1)))
To find a power series representation for the function f(x) = ln(2 - x), we can first rewrite the function as f(x) = ln(1 + (1 - x)). Now, we'll use the Taylor series formula for ln(1 + u) centered at x = 0:

ln(1 + u) = u - (1/2)u^2 + (1/3)u^3 - (1/4)u^4 + ... + (-1)^n(1/n)u^n + ...
In our case, u = (1 - x), so we can substitute it into the formula:
f(x) = (1 - x) - (1/2)(1 - x)^2 + (1/3)(1 - x)^3 - (1/4)(1 - x)^4 + ... + (-1)^n(1/n)(1 - x)^n + ...

This is the power series representation of the function f(x) = ln(2 - x) centered at x = 0.
Now, let's find the radius of convergence (R) using the ratio test:
lim (n -> ∞) |(-1)^{n+1}(1/n+1)(1 - x)^{n+1}| / |(-1)^n(1/n)(1 - x)^n|
Simplify the expression:
lim (n -> ∞) |(n/n+1)(1 - x)|
The limit depends on the value of (1 - x). To ensure convergence, the limit should be less than 1:
|(1 - x)| < 1

This inequality holds for -1 < (1 - x) < 1, which implies that the interval of convergence is 0 < x < 2. Therefore, the radius of convergence, R, is 1.

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(aH)(bH) = abH.

To prove that H is not a normal subgroup of G, we need to show that there exists an element g in G such that gHg^-1 is not a subset of H.

First, note that the left cosets of H in G are {1, b} and {a, ab}. Let g = a. Then we have:

gHg^-1 = a{1, b}a^-1 = {a, ab}

Since {a, ab} is not a subset of H, we have shown that H is not a normal subgroup of G.

Now, let's show the multiplication of the left coset

(aH)(bH) = {a, ab}{b, bb} = {ab, abb, b, bb}

To simplify this expression, we can use the fact that b^2 = 1 and ab = ba^-1. Then, we have:

(aH)(bH) = {ab, abb, b, bb} = {ba^-1, baa^-1, b, 1} = {b, a, ba, 1} = (abH)

Therefore, (aH)(bH) = abH.

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Answer:

Step-by-step explanation:

A. can be used to test joint hypotheses.

In statistical analysis, F-statistics are used to compare the fit of two nested models, typically to test joint hypotheses. Maximum likelihood estimators are a popular method for estimating the parameters of a statistical model by maximizing the likelihood function. They are widely used in various fields due to their desirable properties, such as being consistent and asymptotically efficient.

When F-statistics are computed using maximum likelihood estimators, they can still be employed to test joint hypotheses. This involves comparing the difference in the log-likelihoods between two nested models, one being a restricted model and the other being an unrestricted model. The test statistic, in this case, follows an F distribution under the null hypothesis, which states that the restrictions imposed on the model are valid.

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We have shown that the given set of functions {-3x + 4, 1 - 8x^2 + 8x} is orthogonal with respect to the weight function w(x) = 1 on the interval [-1, 1].

To show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval, we need to show that the integral of the product of any two functions in the set, multiplied by the weight function, over the interval is equal to zero, except when the two functions are the same.

Let's consider two functions from the set: sin(mx) and cos(nx), where m and n are integers.

∫₀²π sin(mx) cos(nx) dx = 0

We can use the trigonometric identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b) to rewrite the integral as:

∫₀²π (1/2)[sin((m+n)x) + sin((m-n)x)] dx

Since m and n are integers, the two sine terms inside the integral have different frequencies and are orthogonal on the interval [0, 2π]. Therefore, their integral over this interval is zero. Thus, we have:

∫₀²π sin(mx) cos(nx) dx = 0, for any integers m and n

Similarly, we can show that the integral of the product of any two other functions in the set, multiplied by the weight function, over the interval is also equal to zero, except when the two functions are the same. Therefore, we have shown that the given set of functions {1, sin(x), cos(x), sin(2x), cos(2x), sin(3x), cos(3x), ...} is orthogonal with respect to the weight function w(x) = 1 on the interval [0, 2π].

To show that the given set of functions is orthogonal with respect to the given weight on the prescribed interval, we need to show that the integral of the product of any two functions in the set, multiplied by the weight function, over the interval is equal to zero, except when the two functions are the same.

Let's consider two functions from the set: -3x + 4 and 1 - 8x^2 + 8x.

∫₋₁¹ (-3x + 4)(1 - 8x^2 + 8x) dx = 0

Expanding the product and integrating, we get:

∫₋₁¹ (-3x + 4)(1 - 8x^2 + 8x) dx = ∫₋₁¹ (-3x + 4) dx - 8∫₋₁¹ x^3 dx + 8∫₋₁¹ x^2 dx

Evaluating the integrals, we get:

∫₋₁¹ (-3x + 4)(1 - 8x^2 + 8x) dx = 0

Therefore, we have shown that the given set of functions {-3x + 4, 1 - 8x^2 + 8x} is orthogonal with respect to the weight function w(x) = 1 on the interval [-1, 1].

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The differential   [tex]dy=2[/tex] when [tex]x = 4[/tex]  and [tex]dx = 0. 2.[/tex]

To find the change in [tex]y,Δy[/tex] , when x changes from, we can use the [tex]4 to 4 +Δx = 4 + 0.2 = 4.2[/tex] formula:

[tex]Δy = y(x + Δx) - y(x)[/tex]

where[tex]y(x) = 5x^2.[/tex]

So, plugging in[tex]x = 4[/tex] and[tex]x + Δx = 4.2[/tex] , we get:

[tex]Δy = y(4.2) - y(4)[/tex]

[tex]= 5(4.2)^2 - 5(4)^2[/tex]

[tex]= 44.2[/tex]

Therefore, the change in y is [tex]44.2[/tex] when x changes from [tex]4 to 4.2.[/tex]

To find the differential dy when [tex]x = 4[/tex] and [tex]dx = 0.2,[/tex]  we can use the formula:

[tex]dy = f'(x) × dx[/tex]

where the derivative of y with respect to x, which is:

[tex]f'(x) = 10x[/tex]

Plugging in [tex]x = 4[/tex]   we get:

[tex]= 2[/tex]

Therefore, the differential [tex]dy = 2[/tex] when [tex]x = 4[/tex] and [tex]dx = 0.2.[/tex]

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Answer:

246 boys like to play outdoors

Step-by-step explanation:

82% = 0.82

0.82 x 300 = 246

Answer:

246 boys like to play outdoors

Step-by-step explanation:

82% = 0.82

0.82 x 300 = 246

For problem 1, the solution is an = n.

For problem 2, the solution is an = 3n - 1.

For problem 3 (the hard problem), we can solve for the values of a, b, and c in the quadratic equation: [tex]an^2 + bn + c = 0[/tex], where a = 5, b = 6n - 1, and c = -2.

Using the quadratic formula, we get:

[tex]n= \frac{-b±\sqrt{b^{2}-4ac }  }{2a}[/tex]

Substituting the values of a, b, and c, we get:

[tex]n= \frac{-(6n-1)±\sqrt{(6n-1)^{2}-4(5)(-2) }  }{2(5)}[/tex]

Simplifying, we get:

[tex]n = \frac{(-6n+1 ± \sqrt{36n^{2}-48n+49 } ) }{10}[/tex]

Therefore, the solution for problem 3 is:

[tex]an= 5n^{2} + \frac{-6n+1 + \sqrt{36n^{2}-48n+49 } }{10}[/tex]

or

[tex]an= 5n^{2} + \frac{-6n+1 - \sqrt{36n^{2}-48n+49 } }{10}[/tex]

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According to the given box plot:

Most of the response times were under 13 minutes is true

Fewer than 30 of the response times were over 13 minutes is true

More than half of the response times were 11 minutes or greater is true

About 75% of the response times were 13 minutes or less is true

In box plot the difference between a dataset's first and third quartiles is used to determine the interquartile range (IQR), a measure of variability.

The interquartile range (IQR) depicts the range of values centred on the data's median. Because it is more resistant to outliers than other measures of variability like the range or the standard deviation, it is valuable in statistics.

Observations that are more than 1.5 times the IQR from the adjacent quartile are considered probable outliers and can be located using the IQR.

Box plots and the IQR are frequently combined to compare and summarise data from several groups or samples.

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The probability that a randomly-selected student from this survey ordered Ranch is approximately 0.2383.

We have,

Let R be the event that a student ordered Ranch dressing.

We want to find P(R), the probability that a randomly-selected student from the survey ordered Ranch.

Out of the 3000 students surveyed, 185 never ordered any dressing, so the remaining 3000 - 185 = 2815 students ordered some kind of dressing. Of these, 2100 ordered Caesar but not Ranch, so the remaining

2815 - 2100 = 715 students ordered Ranch or both dressings.

Now,

P(R) is the proportion of students who ordered Ranch or both dressings out of the total number of students surveyed:

P(R) = 715 / 3000 = 0.2383 (rounded to four decimal places)

Thus,

The probability that a randomly-selected student from this survey ordered Ranch is approximately 0.2383.

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Answer:

The range of a set of data is the difference between the highest and lowest values. In this case, the highest value is 98 and the lowest value is 65, so the range is 98-65 = 33. This means that the scores on the test ranged from 65 to 98, a difference of 33 points.

The range is a measure of the spread of the data. In this case, the range is relatively large, which means that the scores were spread out over a wide range of values. This suggests that the test was challenging and that there was a wide range of student abilities.

(a) The given statement, "If A is linearly dependent, then Sp{u, v} = Sp{u, w}" is false because there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0.

(b) The given statement, "The set A is linearly independent if and only if {u+v,v-w, w+ 2u} is linearly independent" is true because A is linearly independent if and only if the determinant of the matrix formed by u, v, and w is nonzero. The determinant of the matrix formed by {u+v, v-w, w+2u} can be obtained by performing column operations on the original matrix. Since these operations do not change the determinant, the set {u+v, v-w, w+2u} is linearly independent if and only if A is linearly independent.

(c)The given statement, "Assume that A is linearly dependent. We define u₁ = 2u, v₁ = -3u + 4v, and W₁ = u + 2v - tw for some t E R. Then, there exists t E R such that {u₁, v₁, w₁} is linearly independent" is true because  A is linearly dependent, there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0.

Let us discuss this in detail.

(a) False. If A is linearly dependent, then there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0. Without loss of generality, assume α ≠ 0. Then we can solve for u: u = (-β/α)v + (-γ/α)w. Therefore, u is a linear combination of v and w, which means Sp{u, v} = Sp{u, w}.

(b) True. We can write each vector in {u+v,v-w, w+2u} as a linear combination of u, v, and w:
u + v = 1u + 1v + 0w
v - w = 0u + 1v - 1w
w + 2u = 2u + 0v + 1w
We can set up the equation α(u+v) + β(v-w) + γ(w+2u) = 0 and solve for α, β, and γ:
α + β + 2γ = 0 (from the coefficient of u)
α + β = 0 (from the coefficient of v)
-β + γ = 0 (from the coefficient of w)
Solving this system of equations, we get α = β = γ = 0, which means {u+v,v-w, w+2u} is linearly independent.

(c) True. Since A is linearly dependent, there exist scalars α, β, and γ, not all zero, such that αu + βv + γw = 0. Without loss of generality, assume α ≠ 0. Then we can solve for u: u = (-β/α)v + (-γ/α)w. Therefore, u is a linear combination of v and w, which means we can write u as a linear combination of u₁, v₁, and w₁:
u = (2/5)u₁ + (-3/5)v₁ + (1/5)w₁
Similarly, we can write v and w as linear combinations of u₁, v₁, and w₁:
v = (-2/5)u₁ + (4/5)v₁ + (1/5)w₁
w = u₁ + 2v₁ - t₁w₁
where t₁ = (α + 2β - γ)/(-t). We can set up the equation αu₁ + βv₁ + γw₁ = 0 and solve for α, β, and γ:
2α - 3β + γ = 0 (from the coefficient of u₁)
-3β + 4γ = 0 (from the coefficient of v₁)
-α + 2β - tγ = 0 (from the coefficient of w₁)
Solving this system of equations, we get α = β = γ = 0 if and only if t = -8/5. Therefore, if we choose any t ≠ -8/5, then {u₁, v₁, w₁} is linearly independent.

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Answer:

-13.57142857142857

Step-by-step explanation:

Answer: If ABC measures 122 degrees and ADC is an inscribed angle that intercepts the same arc as a central angle ABC, then ADC measures half of ABC which is 61 degrees

Step-by-step explanation:

Answer:

61 degrees

Step-by-step explanation

Angle extended to the circumference is half the angle at the centre

ADC = ½ ABC = ½ × 122 = 61

The radius of convergence comes out as 1/4. To get the radius of convergence, we can use the ratio test.

Step:1. Let's call the nth term in the series a_n, where a_n = 2^(2n)/n^2.
Step:2. Using the ratio test, we take the limit as n approaches infinity of |a_(n+1)/a_n|:
|a_(n+1)/a_n| = (2^(2(n+1))/(n+1)^2) * (n^2/2^(2n))
Step:3. Simplifying this expression, we can cancel out the 2^n terms and get: |a_(n+1)/a_n| = 4((n^2)/(n+1)^2)
Step:4. Taking the limit as n approaches infinity, we get:
lim n→∞ |a_(n+1)/a_n| = 4
Since this limit is less than 1, the series converges.
Step:5. Now we just need to find the radius of convergence, which is given by:
R = 1/lim sup n→∞ |a_n|^(1/n)
Step:6. Taking the limit superior of |a_n|^(1/n), we get:
lim sup n→∞ |a_n|^(1/n) = lim sup n→∞ (2^(2n)/n^(2n/n))^(1/n)
= lim sup n→∞ 2^2 = 4
So the radius of convergence is:
R = 1/lim sup n→∞ |a_n|^(1/n) = 1/4
Therefore, the radius of convergence is 1/4.

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Answer:

68 squer meter

Step-by-step explanation:

it is irregular shape so u have to give section as i draw it then rename it as A1 and A2

A1 = L×W

=13m × 2m

= 26m2

A2= L × W

=7m × 6m

= 42m2

so after weget each area then we will add them b/c we need the total area of the figur not the section

let At = area of totalAt = A1 + A2

= 26m2 + 42m2

=68m2 good luck..

Answer:

The selling price of the house is $175,000 and they need to make a 20% down payment, so the down payment amount is:

Down payment = 20% x $175,000 = $35,000

To find the amount financed with the mortgage, we need to subtract the down payment from the selling price:

Amount financed = Selling price - Down payment

Amount financed = $175,000 - $35,000

Amount financed = $140,000

Next, we need to calculate the closing costs. Let's assume the closing costs are 3% of the selling price:

Closing costs = 3% x $175,000 = $5,250

Since the bank allows them to finance the closing costs as part of the mortgage, we need to add the closing costs to the amount financed:

Total amount of the mortgage = Amount financed + Closing costs

Total amount of the mortgage = $140,000 + $5,250

Total amount of the mortgage = $145,250

Therefore, the actual amount financed with the mortgage is $140,000, the closing costs are $5,250, and the total amount of the mortgage if the closing costs are financed is $145,250.

The table should be completed with the correct key features as follows;

Axis of symmetry (1st graph): x = 1.

Vertex (1st graph): (1, -9).

Minimum (1st graph): -9.

y-intercept (1st graph): (0, -8).

Axis of symmetry (2nd graph): x = 2.

Vertex (2nd graph): (2, 16).

Maximum (2nd graph): 16.

y-intercept (2nd graph): (0, 12).

In Mathematics and Geometry, the graph of a quadratic function would always form a parabolic curve because it is a u-shaped. Based on the first graph of a quadratic function, we can logically deduce that the graph is an upward parabola because the coefficient of x² is positive and the value of "a" is greater than zero (0).

Based on the second graph of a quadratic function, we can logically deduce that the graph is a downward parabola because the coefficient of x² is negative and the value of "a" is less than zero (0).

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The sum of the probabilities is equal to 1 and all probabilities are between 0 and 1 (inclusive). Therefore, this is a probability distribution.

For a distribution to be a probability distribution, it must satisfy two conditions:

The sum of the probabilities for all possible values of X must be equal to 1.

The probability for each possible value of X must be between 0 and 1 (inclusive).

Let's check each distribution:

a) X 4 6 8 10 P(X) -0.6 0.2 0.7 1.5

This distribution does not satisfy the second condition, since the probability for X = 4 is negative (-0.6). Therefore, this is not a probability distribution.

b) X 8 9 12 P(X) 2 1 1 3 6 6

This distribution satisfies the first condition, since the sum of the probabilities is equal to 1. However, it does not satisfy the second condition, since the probability for X = 9 is 1, which is greater than 1. Therefore, this is not a probability distribution.

c) X P(X) 1 1 2 1 1 4 3 4 1 1 4 4 4 4

This distribution satisfies both conditions, since the sum of the probabilities is equal to 1 and all probabilities are between 0 and 1 (inclusive). Therefore, this is a probability distribution.

d) X P(X) 1 0.3 3 0.1 5 0.2

This distribution satisfies both conditions, since the sum of the probabilities is equal to 1 and all probabilities are between 0 and 1 (inclusive). Therefore, this is a probability distribution.

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Answer:

The system of equations y = x + 6 and y = -x + 2 has no solution.

To see why, we can set the equations equal to each other and solve for x: x + 6 = -x + 2 2x = -4 x = -2

However, if we substitute x = -2 back into the original equations, we get: y = x + 6 = -2 + 6 = 4 y = -x + 2 = -(-2) + 2 = 4

So we end up with the same value for y in both equations, which means that the system has a unique solution of (-2, 4). Therefore, the answer is that neither system of equations listed in the search results has no solution.

Step-by-step explanation: