In the diagram below of right triangle ABC, CD is the altitude to hypotenuse AB, CB = 6, and AD = 5.

What is the length of BD?
1) 5
2) 9
3) 3
4) 4

Given:

AE = AC + CE = 4 + 12 = 16

BE = BD + DE = 4⅔ + 14 = 14/3 + 14 = 56/3

To Prove:

AB || CD

Now,

By converse of ∆ proportionality theorem

EC/CA = ED/DB

12/4 = 14/4⅔

3 = 14 ÷ 14/3

3 = 14 × 3/14

3 = 3

L H S = R H S

HENCE PROVED!

The value of dot product (6+8)⋅(−10) is -140.

The value of (6u+v)⋅(u−10v) can be found using the distributive property and the dot product formula, which is (6u⋅u)+(v⋅u)-(60v⋅v). Substituting the given values, we get (6(6)²)+(8(6))-(60(4)²) = 92.

In the given problem, we are given the values of dot product, norms of two vectors u and v. We need to find the value of (6+8)⋅(−10) and (6u+v)⋅(u−10v). Using the formula for dot product, we get the value of the first expression as -140. For the second expression, we use the distributive property and the formula for dot product.

After substituting the given values, we simplify the expression to get the answer 92. The dot product is a useful tool in linear algebra and can be used to find angles between vectors, projections of vectors, and more.

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The repeated measures test, also known as a within-subjects or paired design, decreases the influence of individual differences compared to an independent t-test. The correct answer is c. paired/independent.

A repeated measures test decreases variability between subjects because it is a within-subjects design, meaning that each participant is measured multiple times under different conditions. This reduces the variability between participants and increases the power of the test. In contrast, an independent t-test is a between-subjects design and compares the means of two independent groups, resulting in more variability between subjects. The type of test statistic used depends on the design of the study - a within design uses a paired test statistic, while a between design uses an independent test statistic. Therefore, the answer is c. paired/independent.

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a) The value of R2 is (1,3).

b) The value of the discriminant D = 4.

c) There is no local maximum.

d) No saddle point.

e) The maximum value of f on R2 is 3.

f) The minimum value of f on R2 is also 3

In mathematics, a saddle point is a point on the surface of a function where there is a critical point in one direction, but a minimum or maximum point in another direction. In other words, it is a point on the surface of a function where the tangent plane in one direction is a minimum, and the tangent plane in another direction is a maximum.

(A) The partial derivatives of f(x,y) are:

fx = 2x - 2

fy = 2y - 6

Setting fx = 0 and fy = 0, we get:

2x - 2 = 0

2y - 6 = 0

Solving these equations, we get the critical point (1,3).

(E) To find the maximum value of f on R2, we need to compare the value of f at the critical point (1,3) with the values of f on the boundary of the region enclosed by R2. The boundary of R2 consists of three line segments:

The line segment from (0,0) to (3,3)

The line segment from (3,3) to (3,6)

The line segment from (3,6) to (0,0)

We can parametrize each line segment and substitute it into f to get its value along the boundary. Alternatively, we can use the fact that the maximum and minimum values of a continuous function on a closed, bounded region occur at critical points or at the boundary.

Since there is only one critical point and it is a local minimum, the maximum value of f on R2 occurs on the boundary. We can calculate the value of f at each vertex of the triangle:

f(0,0) = 3

f(3,3) = 3

f(3,6) = 3

The maximum value of f on R2 is 3.

(F) Similarly, the minimum value of f on R2 occurs on the boundary. Using the same calculations as above, we find that the minimum value of f on R2 is also 3.

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The PDF of of two random variable at x = a/2 is 4.851.

To find the marginal PDF of X, we integrate the joint PDF with respect to Y over the range of possible values of Y:

[tex]f_X(x)[/tex]= ∫ f(x,y) dy from y=0 to y=8.15

= ∫ c dy from y=0 to y=8.15

= c * (8.15 - 0)

= 8.15c

Since the total area under the joint PDF must be equal to 1, we know that:

∫∫ f(x,y) dxdy = 1

We can use this to find the constant c:

∫∫ f(x,y) dxdy = ∫∫ c dxdy

= c * ∫∫ dxdy

= c * (a-0) * (8.15-0)

= c * a * 8.15

= 1

Therefore,

c = 1 / (a * 8.15)

Substituting this into our expression for [tex]f_X(x)[/tex], we get:

[tex]f_X(x)[/tex] = 8.15 / a

So, for x = a/2, we have:

[tex]f_X(a/2)[/tex] = 8.15 / (a/2)

= 16.3 / a

= 4.851

Therefore, the PDF of X at x = a/2 is 4.851.

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The test results suggest that there is not enough evidence to reject the null hypothesis.

A hypothesis test is a statistical method used to determine whether an assumption about a population parameter can be supported by sample data. In this case, the Small Business Administration hypothesized that 80% of franchisees own only one location. They collected a random sample of 120 franchisees and found that 85 owned only one store. To determine whether this sample result supports their hypothesis, they performed a hypothesis test with a significance level of 0.05.

The critical value for this test was determined based on the desired level of confidence, which in this case was 95%. The calculated test statistic was then compared to this critical value to determine whether the null hypothesis (that 80% of franchisees own only one location) can be rejected.

In this scenario, the calculated test statistic fell within the confidence interval, indicating that the null hypothesis cannot be rejected based on the sample data. This means that there is not enough evidence to support the claim that franchisers favor owners who operate multiple stores, at least not to the extent that it would significantly impact the distribution of franchise ownership.

It's important to that while the sample data may not support the hypothesis, it's possible that the true population parameter could still differ from the hypothesized value. However, based on the available data and the results of the hypothesis test, it appears that there is not enough evidence to support the claim that franchisers favor multi-store owners.

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The equation y has two outputs for each input of x, which proves that y²= 4x+16 is not a function.

A function is a relation between two sets of values such that each element of the first set is associated with a unique element of the second set.

In this case, y²= 4x+16 is an equation that is not a function as it does not satisfy the definition of a function.

It does not meet the criteria of having a unique output for each input. For example, when x = 0, the equation yields y²= 16.

Since y can be both positive and negative, there are two outputs for the same input. This violates the definition of a function and therefore this equation is not a function.

This can be proven mathematically by rearranging the equation to solve for y.

y²= 4x+16

y² -4x= 16

y² -4x+4= 16+4

(y-2)²= 20

y= ±√20 + 2

This equation shows that y has two outputs for each input of x, which proves that y²= 4x+16 is not a function.

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Answer: f(c) = 3.

Step-by-step explanation:

Since f is differentiable on the open interval (1,10), we can apply the Intermediate Value Theorem and Rolle's Theorem to draw some conclusions about the behavior of f on this interval.

I. f has at least 2 zeros.

This statement cannot be determined solely based on the given information. We know that f(2) = -5 and f(9) = -5, which means that f takes on the value of -5 at least twice on the interval (2, 9). However, we cannot conclude that f has at least 2 zeros without additional information. For example, consider the function f(x) = (x - 2)(x - 9), which satisfies the given conditions but has only 2 zeros.

II. The graph of f has at least one horizontal tangent.

This statement is true. Since f(2) = -5 and f(5) = 5, we know that f must cross the x-axis between x = 2 and x = 5. Similarly, since f(5) = 5 and f(9) = -5, we know that f must cross the x-axis between x = 5 and x = 9. Therefore, by the Intermediate Value Theorem, we know that f has at least one zero in the interval (2, 5) and at least one zero in the interval (5, 9). By Rolle's Theorem, we know that between any two zeros of f, there must be a point c where f'(c) = 0, which means that the graph of f has at least one horizontal tangent.

III. For some c, c is greater than 2 but less than 5, f(c) = 3.

This statement is false. We know that f(2) = -5 and f(5) = 5, which means that f takes on all values between -5 and 5 on the interval (2, 5) by the Intermediate Value Theorem. Since the function is continuous on this interval, it must take on all values between its maximum and minimum. Therefore, there is no value of c between 2 and 5 for which f(c) = 3.

The resulting matrix using the specified row transformation to change the matrix; - 4 times row 1 added to row 2 is 0 -8

To apply the specified row transformation, we need to subtract 4 times the first row from the second row.

So the resulting matrix will be:

[  2              3

8 - 4 ( 2 )   4 - 4 ( 3 ) ]

which simplifies to:

[ 2   3

0   -8 ]

Therefore the resulting matrix for the specified row transformation is 0 and - 8.

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Answer:

The volume is 1 1/15 yards^3

Step-by-step explanation:

For the volume of a rectangular prism, you use this formula: L*W*H.

In this case, we're given 2/3, 4/5, and 2.

All you have to do here is 2/3 times 2 first, and you get 4/3.

But, we're not done yet.

We also have 4/5, so we also have to multiply 4/3 by 4/5, which gives you 16/15.

It says that we can do a proper fraction or mixed number, so the answer is 1  and 1/15, or 1 1/15.

The probability of selecting a college with in-state tuition less than $5,000 is 10%.

To calculate the probability that a randomly selected college will have an in-state tuition of less than $5,000, we first need to gather data on the number of colleges with in-state tuition less than $5,000 and the total number of colleges.

Let's assume that there are 500 colleges in the dataset, out of which 50 have in-state tuition less than $5,000.

So, the probability of selecting a college with in-state tuition less than $5,000 can be calculated as:

P(In-state tuition < $5,000) = Number of colleges with In-state tuition < $5,000 / Total number of colleges

P(In-state tuition < $5,000) = 50 / 500

P(In-state tuition < $5,000) = 0.1 or 10%

Therefore, the probability of selecting a college with in-state tuition less than $5,000 is 10%.

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Substituting in our expression for Σ, we get:

[tex]√n (r1)( . )( . )( rk) ~ Nk(0, (1/n) σ^2 Ik)[/tex]

This is the desired result.

If [tex]Xt[/tex]is an ARMA(p, q) process with [tex]p = q = 0,[/tex] then Xt is just white noise. In this case, Bartlett's formula gives us the asymptotic distribution of the sample autocorrelation coefficients, which can be written as:

[tex]√n (r1)( . )( . )( rk) ~ Nk(0, Ik)[/tex]

where [tex]r1, ..., rk[/tex] are the sample autocorrelation coefficients at lags 1 through [tex]k, √n[/tex] is the scaling factor, and Nk(0, Ik) is the multivariate normal distribution with mean 0 and identity covariance matrix.

To show this result, we can use the properties of white noise to derive the mean and covariance of the sample autocorrelation coefficients. For white noise, the sample mean is zero and the sample variance is constant. Therefore, we have:

E[tex](rj) = 0 for j = 1, ..., k[/tex]

Var [tex](rj) = 1/n for j = 1, ..., k[/tex]

To find the covariance between rj and rk, we use the fact that white noise has no autocorrelation at non-zero lags. Therefore, we have:

Cov [tex](rj, rk) = E[rjrk] - E[rj]E[rk][/tex]

Since Xt is white noise, we have:

E[tex][Xt] = 0[/tex] for all t

Cov [tex](Xt, Xs) = 0 for t ≠ s[/tex]

Therefore, we can write:

E[tex][rjrk] = E[(1/n) ∑(t=1)^(n-j) Xt Xt+j (1/n) ∑(t=1)^(n-k) Xt Xt+k]= (1/n^2) ∑(t=1)^(n-j) ∑(s=1)^(n-k) E[XtXt+jXsXs+k]= (1/n^2) ∑(t=1)^(n-j) E[XtXt+jXt+j+tXt+j+k]= (1/n^2) ∑(t=1)^(n-j) E[XtXt+j]E[Xt+j+tXt+j+k]= (1/n^2) ∑(t=1)^(n-j) Var(Xt)δ(t+k-j)[/tex]

where δ(i) is the Kronecker delta function, which is equal to [tex]1 if i = 0[/tex] and 0 otherwise. Using the fact that Var[tex](Xt) = σ^2[/tex] for all t, we can simplify this expression to:

E[tex][rjrk] = (1/n) σ^2 δ(k-j)[/tex]

Therefore, we have:

[tex]Cov(rj, rk) = E[rjrk] - E[rj]E[rk] = (1/n) σ^2 δ(k-j)[/tex]

Putting this together, we can write the covariance matrix of the sample autocorrelation coefficients as:

[tex]Σ = (1/n) σ^2 Ik[/tex]

where Ik is the k x k identity matrix. Therefore, the asymptotic distribution of the sample autocorrelation coefficients is:

[tex]√n (r1)( . )( . )( rk) ~ Nk(0, Σ)[/tex]

Substituting in our expression for Σ, we get:

[tex]√n (r1)( . )( . )[/tex]

[tex]( rk) ~ Nk(0, (1/n) σ^2 Ik)[/tex]

This is the desired result.

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The Answer is True.

The lognormal distribution is commonly used to model data that follows a log-transformed distribution. Taking the logarithm of a variable can often help to transform skewed or highly variable data into a more normal distribution, which can make it easier to analyze statistically.

Therefore, log10 transformation is a common technique used to create a log distribution for data that has a large range of values.

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a. The null hypothesis for this statement would be that the number of children the younger sister has is not dependent on the number of children the older sister has.

b. The null hypothesis for this statement would be that the distribution of family sizes for older and younger sisters is the same.

For a, The alternative hypothesis would be that there is a dependency between the two variables. This hypothesis can be tested using a chi-squared test for independence.

For b,The alternative hypothesis would be that the distributions are different. This hypothesis can be tested using a two-sample t-test for comparing means or a chi-squared test for comparing proportions.

Both hypotheses can be true or false independently. It is possible that the number of children the younger sister has is independent of the number of children the older sister has, but the distribution of family sizes could be different for older and younger sisters. Conversely, it is also possible that the number of children the younger sister has is dependent on the number of children the older sister has, but the distribution of family sizes is the same for both.

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Complete Question:  

A sociologist is studying influences on family size. He finds pairs of sisters, both of whom are married, and determines for each sister whether she has 0, 1, or 2 or more children. He wants to compare older and younger sisters. Explain what the following hypotheses mean and how to test them.

   a. The number of children the younger sister has is independent of the number of children the older sister has.

     b. The distribution of family sizes is the same for older and younger sisters. Could one hypothesis be true and the other false? Explain.

s = −81/4,t = −34 so its

Step-by-step explanation:

The quadratic function with the given features is defined as follows:

y = 0.86x² - 5.86x + 5.

The standard definition of a quadratic function is given as follows:

y = ax² + bx + c.

When x = 0, y = 5, hence the coefficient c is given as follows:

c = 5.

Hence:

y = ax² + bx + 5.

When x = 1, y = 0, hence:

a + b + 5 = 0

a + b = -5.

The discriminant is given as follows:

D = b² - 4ac.

Hence:

D = b² - 20a

The minimum value is of -4, hence:

-D/4a = -5

(b² - 20a)/4a = -5

b² - 20a = 20a

b² = 40a

Since a = -5 - b, we have that the value of b is obtained as follows:

b² = 40(-5 - b)

b² + 40b + 200 = 0.

b = -5.86.

Hence the value of a is of:

a = -5 + 5.86

a = 0.86.

Then the equation is:

y = 0.86x² - 5.86x + 5.

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Answer:

Tooo mny

Step-by-step explanation:

I think, you need to add al the sides then subtract by 180

Answer:

x < - 4

Step-by-step explanation:

the open circle at - 4 indicates that x cannot equal - 4

the arrow and shaded part of the line to the left of - 4 indicates values that are solutions of the inequality, then the inequality representing the graph is

x < - 4

The solution of the given initial value problem for r as a vector function of t is [tex]r(t) = 3(t + 1)^{(3/2)} i + (-6 e^{-t} + 6) j + (ln(t + 1) + 1) k[/tex].

A differential equation is an equation that contains one or more terms and the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable).

To solve the given differential equation, we will integrate each component of the differential equation and apply the initial condition.

Differential Equation: dr/dt = [tex]9/2 (t + 1)^{1/2} i + 6 e^{-t} j + 1/(t + 1) k[/tex]
Initial condition: r(0) = k

Step 1: Integrate each component of the differential equation with respect to t:
[tex]r(t) = \int(9/2 (t + 1)^{1/2}) dt \ i + \int(6 e^{-t}) dt \ j + \int(1/(t + 1)) dt \ k[/tex]

Step 2: Solve the integrals:
[tex]r(t) = [3(t + 1)^{(3/2)}] i - [6 e^{-t}] j + [ln(t + 1)] k + C[/tex]

Step 3: Apply the initial condition r(0) = k:
[tex]k = [3(0 + 1)^{(3/2)}] i - [6 e^0] j + [ln(0 + 1)] k + C[/tex]
k = 0 i - 6 j + 0 k + C
C = 6j + k

Step 4: Substitute C back into the expression for r(t):
[tex]r(t) = [3(t + 1)^{(3/2)}] i - [6 e^{-t}] j + [ln(t + 1)] k + (6j + k)[/tex]

So, the vector function r(t) is:
[tex]r(t) = 3(t + 1)^{(3/2)} i + (-6 e^{-t} + 6) j + (ln(t + 1) + 1) k[/tex].

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The probabilities for:

a. P(48 <X <66) = 0.978.

b. P(X> 69) = 0.0618.

c. P(X <= 65)  = 0.8051

d. P(X < 60) = 0.5

a. P(48 < X < 66) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation, σ = [tex]\sqrt{(np(1-p)) }[/tex]= [tex]\sqrt{(150 * 0.4 * 0.6)[/tex] = 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution between the z-scores corresponding to x = 48 and x = 66:

z1 = (48 - 60) / 5.81 = -2.06

z2 = (66 - 60) / 5.81 = 1.03

Using a standard normal distribution table, we find the area between these z-scores to be approximately 0.978. Therefore, P(48 < X < 66) ≈ 0.978.

b. P(X > 69) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation, σ = [tex]\sqrt{(np(1-p)) }[/tex]= [tex]\sqrt{(150 * 0.4 * 0.6)[/tex] = 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution to the right of the z-score corresponding to x = 69:

z = (69 - 60) / 5.81 = 1.55

Using a standard normal distribution table, we find the area to the right of this z-score to be approximately 0.0618. Therefore, P(X > 69) ≈ 0.0618.

c. P(X <= 65) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation,  σ = [tex]\sqrt{(np(1-p)) }[/tex]= [tex]\sqrt{(150 * 0.4 * 0.6)[/tex]= 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution to the left of the z-score corresponding to x = 65:

z = (65 - 60) / 5.81 = 0.86

Using a standard normal distribution table, we find the area to the left of this z-score to be approximately 0.8051. Therefore, P(X <= 65) ≈ 0.8051.

d. P(X < 60) can be approximated using the normal distribution as follows:

mean, μ = np = 150 × 0.4 = 60

standard deviation, σ = sqrt(np(1-p)) = sqrt(150 × 0.4 × 0.6) = 5.81

We can standardize using the formula z = (x - μ) / σ to find the area under the standard normal distribution to the left of the z-score corresponding to x = 60:

z = (60 - 60) / 5.81 = 0

Using a standard normal distribution table, we find the area to the left of this z-score to be 0.5. Therefore, P(X < 60) ≈ 0.5.

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To determine if the normal approximation to the binomial is appropriate, we need to check if both np and n(1-p) are greater than or equal to 10.

(a) For n = 500 and p = 0.33, np = 165 and n(1-p) = 335, both of which are greater than 10. Therefore, the normal approximation to the binomial is appropriate.

(b) For n = 400 and p = 0.01, np = 4 and n(1-p) = 396, which are not both greater than 10. Therefore, the normal approximation to the binomial is not appropriate.

(c) For n = 100 and p = 0.61, np = 61 and n(1-p) = 39, both of which are greater than 10. Therefore, the normal approximation to the binomial is appropriate.
To determine if the normal approximation to the binomial is appropriate in each situation, we can use the following rule of thumb: the normal approximation is suitable when both np and n(1-p) are greater than or equal to 10.
A binomial is a polynomial that is the sum of two terms, each of which is a monomial .It is the simplest kind of a sparse polynomial after the monomials.

(a) n = 500, p = 0.33
np = 500 * 0.33 = 165
n(1-p) = 500 * (1 - 0.33) = 500 * 0.67 = 335
Since both values are greater than 10, the normal approximation is appropriate.
Normal distributions are important in statistics and are often used in the natural to represent real-valued random variables whose distributions are not known. Their importance is partly due to the central limit theorem.

(b) n = 400, p = 0.01
np = 400 * 0.01 = 4
n(1-p) = 400 * (1 - 0.01) = 400 * 0.99 = 396
Since np is less than 10, the normal approximation is not appropriate.

(c) n = 100, p = 0.61
np = 100 * 0.61 = 61
n(1-p) = 100 * (1 - 0.61) = 100 * 0.39 = 39
Since both values are greater than 10, the normal approximation is appropriate.

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Answer:

The answer is 0.

Step-by-step explanation:

[tex] log_{2}(32) = 5[/tex]

[tex] log_{5}(5) = 1[/tex]

[tex] log_{3}(1) = 0[/tex]

The given values modulo 45 are 192 (mod 45) = 12, 194 (mod 45) = 14, 198 (mod 45) = 18, and 1916 (mod 45) = 1.

To find the value of modulo of 192 (mod 45),

we can divide 192 by 45 and take the remainder

192 = 4 (45) + 12

So, 192 (mod 45) = 12.

To find 194 (mod 45),

we can divide 194 by 45 and take the remainder

194 = 4 (45) + 14

So, 194 (mod 45) = 14.

To find 198 (mod 45),

we can divide 198 by 45 and take the remainder

198 = 4 (45) + 18

So, 198 (mod 45) = 18.

To find 1916 (mod 45),

we can first reduce 1916 by reducing each digit

1916 = 1 (mod 45)

Therefore, 1916 (mod 45) = 1.

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You should originally invest $148.

To solve this problem, we need to use the formula for present value of a bond:

[tex]PV = C/(1+r)^t[/tex]

where PV is the present value, C is the annual coupon payment, r is the annual interest rate, and t is the time to maturity in years.

We know that the General Electric bonds had a time to maturity of 87 years and an annual rate of 5.25%. We also know that they paid a total of $6,630 at maturity. Let's call the original investment amount X.

Using the formula, we can set up the following equation:

[tex]6,630 = X/(1+0.0525)^{87[/tex]

Simplifying this equation, we get:

[tex]X = 6,630 * (1+0.0525)^{-87[/tex]

Using a calculator, we get:

X = $147.91

Rounding this to the nearest dollar, the answer is:

$148

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The terms of the sequence continue to increase without bound, we can say that the sequence is not bounded.

To determine whether the sequence is increasing or decreasing, we need to compare consecutive terms of the sequence.

For n = 1, a1 = 6(1) + 2 = 8

For n = 2, a2 = 6(2) + 2 = 14

For n = 3, a3 = 6(3) + 2 = 20

Since each term of the sequence is greater than the previous one, we can say that the sequence is increasing.

To determine if the sequence is bounded, we need to check if it approaches infinity or if it has a finite upper and lower bound. Since the terms of the sequence continue to increase without bound, we can say that the sequence is not bounded.

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To use the Chi Squared test on the resulting data we can use the following statements in order:

D. He takes a random sample of balloons.

B. He samples 36 balloons at most.

C. He expects each color to appear at least 5 times.

A statistical hypothesis test used to examine if a variable is likely to come from a specific distribution is the Chi-square goodness of fit test. To ascertain if sample data is representative of the total population, it is widely utilised.

To let you know if there is a correlation, the Chi-Square test provides a P-value. An assumption is being considered, which we can test later, that a specific condition or statement may be true. Consider this: The data collected and expected match each other quite closely, according to a very tiny Chi-Square test statistic. The data do not match very well, according to a very significant Chi-Square test statistic. The null hypothesis is disproved if the chi-square score is high.

Here in the question,

To use the Chi Squared test on the resulting data we can use the following statements in order:

D. He takes a random sample of balloons.

B. He samples 36 balloons at most.

C. He expects each color to appear at least 5 times.

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The complete question is:

Peter bought a big pack of

360

360360 party balloons. The balloons come in

6

66 different colors which are supposed to be distributed evenly in the pack.

Peter wants to test whether the distribution is indeed even, but he doesn't want to go over the entire pack. So, he plans to take a sample and carry out a

χ

2

χ

2

\chi, squared goodness-of-fit test on the resulting data.

Which of these are conditions for carrying out this test? Choose 3 options.

A. He observes each color at least 5 times.

B. He samples 36 balloons at most.

C. He expects each color to appear at least 5 times.

D. He takes a random sample of balloons.

Answer: the answer is A B D

Step-by-step explanation:

We can start by simplifying each factor separately and then combine them.

(a + b)/(a^2 + b^2) can be simplified by multiplying both the numerator and denominator by (a - b):

(a + b)/(a^2 + b^2) * (a - b)/(a - b) = (a^2 - b^2)/(a^3 - b^3)

Next, we simplify a/(a - b) by multiplying both the numerator and denominator by (a + b):

a/(a - b) * (a + b)/(a + b) = a(a + b)/(a^2 - b^2)

Lastly, we simplify (a^4 - b^4)/(a + b)^2 by factoring the numerator and expanding the denominator:

(a^4 - b^4)/(a + b)^2 = [(a^2)^2 - (b^2)^2]/(a + b)^2 = [(a^2 + b^2)(a^2 - b^2)]/(a + b)^2

Now we can combine all three simplified factors:

(a + b)/(a^2 + b^2) * a/(a - b) * (a^4 - b^4)/(a + b)^2 = [(a^2 - b^2)/(a^3 - b^3)] * [a(a + b)/(a^2 - b^2)] * [(a^2 + b^2)(a^2 - b^2)]/(a + b)^2

Simplifying further, we can cancel out the (a^2 - b^2) terms and the (a + b) terms:

= [a(a + b)/(a^3 - b^3)] * [(a^2 + b^2)/(a + b)]

= a(a + b)(a^2 + b^2)/(a + b)(a^3 - b^3)

= a(a^2 + b^2)/(a^3 - b^3)

Therefore, the simplified expression is a(a^2 + b^2)/(a^3 - b^3)

The greatest common divisor of the pair of integers 19 and 1919 is 19.

the greatest common divisor (GCD) of the pair of integers you provided. The pair of integers in question is 19 and 1919.

To find the GCD, you can use the Euclidean algorithm:

1. Divide the larger integer (1919) by the smaller integer (19) and find the remainder.
  1919 ÷ 19 = 101 with a remainder of 0.

2. Since there is no remainder, the smaller integer (19) is the greatest common divisor.

So, the greatest common divisor of the pair of integers 19 and 1919 is 19.

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Answer:

8) The distance the jet traveled is the area under the graph.

9) (1/2)(600)(20 + 25) = 13,500 miles

10) (1/2)(600)(5) = 1,500 miles