Help please. I would appreciate it.

The required probabilities are

a. P(X < 75) = 0.29, b. P(X < 28) = 0, and c. P(X = 75) = 0

A continuous probability distribution is a type of probability distribution that describes the probabilities of all possible values that a continuous random variable can take within a specific range.

In contrast to discrete probability distributions, which describe the probabilities of discrete outcomes, continuous probability distributions describe the probabilities of continuous outcomes.

Here we have

For a continuous random variable X,

P(28 ≤ X ≤ 75) = 0.15 and P(X > 75) = 0.14

Given probabilities can be calculated as follows

a. P(X < 75)

P(X < 75) = P(X ≤ 75) - P(X = 75)

= 0.15 + 0.14

= 0.29

b. P(X < 28)

P(X < 28) = P(X ≤ 28) = 0,

[ since X cannot be less than 28 if P(28 ≤ X ≤ 75) = 0.15 ]

c. P(X = 75)

P(X = 75) = P(X ≤ 75) - P(X < 75)

= 0.15 - 0.29

= -0.14.

However, this is not a valid probability since probabilities cannot be negative.

Therefore, P(X = 75) = 0.

Therefore,

The required probabilities are

a. P(X < 75) = 0.29, b. P(X < 28) = 0, and c. P(X = 75) = 0

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The null hypothesis would be H0: p = 0.05. The alternative hypothesis would be Ha: p < 0.05.

When forming hypotheses in this context, we typically state a null hypothesis (H0) and an alternative hypothesis (H1). Here's how you can fill in the blanks:

Null hypothesis (H0): The percentage of U.S. adults who suffer from a mental illness has not changed. H0: p = 0.05

Alternative hypothesis (H1): The percentage of U.S. adults who suffer from a mental illness has decreased. H1: p < 0.05

In this case, "p" represents the percentage of U.S. adults with a mental illness. The social worker will use statistical tests to analyze data and determine if there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

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The required answer is 2[(1/7) ln |sin(x)| - (1/6) ln|6sin(x)+7|] + C

To solve the integral 2∫cos(x)/(6sin^2(x) + 7sin(x)) dx, we will first make a substitution to express the integrand as a rational function.

A rational function is a polynomial divided by a polynomial. f(x) = x/x-3 is a rational function .Rational functions are used to approximate or model more complex equations . integrals (antiderivative functions) of rational functions. Any rational function can be integrated by partial fraction decomposition of the function into a sum of functions.  In this case, one speaks of a rational function and a rational fraction over K.

The integrand as a rational function and then evaluate the integral.

Step 1: Make a substitution
Let u = sin(x), so du = cos(x) dx.

The integral now becomes:

2∫(du) / (6u^2 + 7u)

Step 2: Express the integrand as a rational function
Since the integrand is already a rational function, no further simplification is needed.

Step 3: Evaluate the integral

2∫(1/7) du - ∫(1/(6u+7)) du] = 2[(1/7) ln |u| - (1/6) ln|6u+7|] + C

To evaluate the integral, we perform partial fraction decomposition on the integrand:
A rational function is any function that can be defined by a rational fraction, which is an algebraic fraction such that both the numerator and the denominator are polynomials. The coefficients of the polynomials need not be rational numbers; they may be taken in any field K. In this case, one speaks of a rational function and a rational fraction over K.

A constant function such as f(x) = π is a rational function since constants are polynomials. The function itself is rational, even though the value of f(x) is irrational for all x. Rational functions are used to approximate or model more complex equations

2∫(du) / (6u^2 + 7u) = 2∫(A/u + B/(6u+7)) du

By clearing the denominators, we get:

1 = A(6u+7) + B(u)

Now, we can solve for A and B:

When u = 0, 1 = 7A => A = 1/7
When u = -7/6, 1 = B(-1) => B = -1

So the integral becomes:

2∫((1/7)/u - 1/(6u+7)) du

Now, we can integrate each term:

2 [∫(1/7) du - ∫(1/(6u+7)) du] = 2[(1/7) ln |u| - (1/6) ln|6u+7|] + C

Step 4: Substitute back in terms of x

Finally, substitute u = sin(x) back into the equation:

2[(1/7) ln |sin(x)| - (1/6) ln|6sin(x)+7|] + C

This is the evaluated integral.

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The Cartesian product B × A is:
B × A = {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}

To find the Cartesian product B × A, where A = {a, b, c, d} and B = {y, z}, we need to create ordered pairs with the first element from set B and the second element from set A. The Cartesian product B × A is:

B × A = {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}

In this case, you have two sets: A = {a, b, c, d} and B = {y, z}. To find the Cartesian product B × A, you would take each element from set B and pair it with every element in set A. Let's go through it step by step:

Start with set B = {y, z}.

Take the first element from set B, which is y.

Pair y with each element in set A, which is {a, b, c, d}, to get the following ordered pairs: (y, a), (y, b), (y, c), and (y, d).

Next, take the second element from set B, which is z.

Pair z with each element in set A, which is {a, b, c, d}, to get the following ordered pairs: (z, a), (z, b), (z, c), and (z, d).

Collect all the ordered pairs obtained in the previous steps to get the Cartesian product B × A, which is {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}.

The Cartesian product B × A is indeed {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}, which consists of all possible ordered pairs with the first element from set B and the second element from set A.

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The Cartesian product B × A is:
B × A = {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}

To find the Cartesian product B × A, where A = {a, b, c, d} and B = {y, z}, we need to create ordered pairs with the first element from set B and the second element from set A. The Cartesian product B × A is:

B × A = {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}

In this case, you have two sets: A = {a, b, c, d} and B = {y, z}. To find the Cartesian product B × A, you would take each element from set B and pair it with every element in set A. Let's go through it step by step:

Start with set B = {y, z}.

Take the first element from set B, which is y.

Pair y with each element in set A, which is {a, b, c, d}, to get the following ordered pairs: (y, a), (y, b), (y, c), and (y, d).

Next, take the second element from set B, which is z.

Pair z with each element in set A, which is {a, b, c, d}, to get the following ordered pairs: (z, a), (z, b), (z, c), and (z, d).

Collect all the ordered pairs obtained in the previous steps to get the Cartesian product B × A, which is {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}.

The Cartesian product B × A is indeed {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}, which consists of all possible ordered pairs with the first element from set B and the second element from set A.

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OB. The set is a subspace of P₂. The set contains the zero vector of P₂, the set is closed under vector addition, and the set is closed under multiplication by scalars.

To show this, we need to verify the three conditions for a set to be a subspace:

The set contains the zero vector: The zero vector of P₂ is 0t² = 0, which is in the set since any real number multiplied by 0 is 0.

The set is closed under vector addition: Let p(t) = at² and q(t) = bt² be two polynomials in the set. Then p(t) + q(t) = (a +

Answer:

3.87cm

hope this helped

Mean of x for sample size is [28.34, 32.66]

The closest option is (c) 30.5.

The mean of the sample means will be the same as the population mean, which is 30.5.

The standard error of the mean, which is the standard deviation of the sampling distribution of the mean, can be calculated as:

SE = σ / sqrt(n)

where σ is the population standard deviation and n is the sample size.

SE = 3.5 / sqrt(10) = 1.108

The mean of x for sample size of 10 can be calculated as:

x = μ ± z*(SE)

where μ is the population mean, z is the z-score corresponding to the desired level of confidence (we'll assume 95% here), and SE is the standard error of the mean.

Using a z-score of 1.96 for a 95% confidence interval, we have:

x = 30.5 ± 1.96*(1.108) = [28.34, 32.66]

Therefore, the closest option is (c) 30.5.

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The moment of inertia of the cylinder about the given axis is (3/2)mr². Option d is correct.

The moment of inertia of a uniform cylinder of radius r and mass m rotating about its central axis (perpendicular to its length) is (1/2)mr². However, in this case, the cylinder is rotating about a horizontal axis that is parallel and tangent to the cylinder. This axis passes through the center of the cylinder, so we can use the parallel axis theorem to find the moment of inertia about the given axis.

The parallel axis theorem states that the moment of inertia of a rigid body about any axis parallel to its center of mass axis is equal to the moment of inertia about the center of mass axis plus the product of the mass of the body and the square of the distance between the two axes.

In this case, the distance between the center of the cylinder and the given axis is (1/2)l. Therefore, the moment of inertia of the cylinder about the given axis is:

Therefore, the moment of inertia of the cylinder about the given axis is (3/2)mr², which is option (d).

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Answer: x²+y²=25

Step-by-step explanation:

Formula for a circle is:

(x-h)²+(y-k)²=r²

where (h, k) is your center and r is radus.

In your question, (h,k)=(0,0) because it says the center is at the origin and

r=5 because the hypotenuse of that triangle is the radius of the circle

You can substitute into the formula for a cirlcle with (0,0) and r=5

x²+y²=5²

x²+y²=25

The answer of the given question based on the triangle is (a) the coordinate is given wrong so it does not for any equilateral triangle, (b) the steps are given below to draw equilateral triangle.

A line segment is  part of  line that is bounded by two distinct endpoints. It can be measured by its length, which is the distance between its endpoints. A line segment is a straight line that extends between its endpoints, but it does not continue indefinitely in either direction.

Draw a segment (6,9,12) centimeters long. Then from one endpoint, draw a (30,60,90) ° angle.

Neither of these steps is correct for drawing an equilateral triangle with all sides equal to 6 centimeters.

To draw an equilateral triangle, we can follow these steps:

Draw a straight line segment of 6 centimeters.

At one endpoint of the segment, use a compass to draw a circle with a radius of 6 centimeters. This will be the circle that intersects the other endpoint of the segment.

Without changing the compass width, place the compass on the other endpoint of the segment and draw a second circle of radius 6 centimeters.

Draw a straight line segment connecting the two points where the circles intersect.

This will create an equilateral triangle with all sides equal to 6 centimeters.

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For the first equation, the value of x that does not satisfy the equation sin 2x + sin x = 0 is (c) 271.

For the second equation, all values of x satisfy the equation sin x + sin x = 0, and the answer is (e) All Satisfy.

For the first equation of Trigonometry, sin 2x + sin x = 0, we can use the identity sin 2x = 2 sin x cos x to rewrite it as:

2 sin x cos x + sin x = 0

Factoring out sin x, we get:

sin x (2 cos x + 1) = 0

So the equation is satisfied when sin x = 0 or 2 cos x + 1 = 0. Solving the second equation for cos x, we get:

2 cos x = -1

cos x = -1/2

So the equation is satisfied when sin x = 0 or cos x = -1/2.

The values of x that satisfy these conditions are x = nπ (where n is an integer) and x = (2n+1)π/3 (where n is an integer).

Therefore, the value of x that does not satisfy the equation sin 2x + sin x = 0 is (c) 271.

For the second equation, sin x + sin x = 0, we can simplify it to:

2 sin x = 0

This equation is satisfied when sin x = 0, which occurs at x = nπ (where n is an integer).

Therefore, all values of x satisfy the equation sin x + sin x = 0, and the answer is (e) All Satisfy.

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The derivative of f(x) is f'(x) = -2, so we can substitute these values into the formula to get δy = -2 * 0.2 = -0.4.

This problem involves using the concept of the derivative to calculate the change in the output variable y, given a small change in the input variable x.

Specifically, we are given the function y = f(x) = -2x, the value of x at which we want to evaluate the change, x = 2, and the size of the change in x, δx = 0.2.

To find the corresponding change in y, δy, we can use the formula δy = f'(x) * δx, where f'(x) is the derivative of f(x) evaluated at x.

In this case, the derivative of f(x) is f'(x) = -2, so we can substitute these values into the formula to get δy = -2 * 0.2 = -0.4.

This tells us that a small increase of 0.2 in x will result in a decrease of 0.4 in y, since the derivative of the function is negative.

This problem illustrates the concept of local linearization, which is the approximation of a nonlinear function by a linear function in a small region around a point.

The derivative of the function at a point gives us the slope of the tangent line to the function at that point, and this slope can be used to approximate the function in a small region around the point.

This approximation can be useful for estimating changes in the output variable given small changes in the input variable.

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(a) The taylor polynomial is 3(x) ≈ ln(1 + 6) + (2/7)(x - 3) - (1/21)(x - 3)² + (1/63)(x - 3)³

(b) The accuracy of the approximation is  |R3(x)| ≤ 0.000274

(c) Graphing |R3(x)| confirms the accuracy.

a)  To approximate f(x) = ln(1 + 2x) by a Taylor polynomial with degree n=3 at a=3, find the first three derivatives of f(x) and evaluate them at x=3. Then, use the formula T3(x) = f(3) + f'(3)(x - 3) + f''(3)(x - 3)²/2! + f'''(3)(x - 3)³/3!.

b) Use Taylor's Inequality to estimate the accuracy of the approximation for the given interval, 2.7 ≤ x ≤ 3.3. First, find the fourth derivative of f(x), then find its maximum value in the interval. Finally, use the formula |R3(x)| ≤ (M/4!)(x - 3)^4, where M is the maximum value of the fourth derivative.

c) To check the result from part (b), graph the remainder function |R3(x)| in the given interval. If the maximum value of the graph is close to the value found in part (b), this confirms the accuracy of the approximation.

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The  system of equations that represent the production time needed to build each doghouse and the production time needed for painting and assembling each doghouse.

5x + 2y = 50 (The production time for building)

1x + 2y = 30 (The production time for painting and assembling)

We shall make x be the number of vinyl doghouses made and y be the number of treated lumber doghouses made.

The time to build x vinyl doghouses = 5x hours,

The time to build y treated lumber doghouses = 2y hours.

Hence: the full time spent building doghouses is:  

5x + 2y hours

The time to paint x vinyl doghouses = 1x hour

The time to assemble y treated lumber doghouses = 2y hours.

Hence total time spent painting and assembling doghouses is:

1x + 2y hours

There is:

50 hours every week towards building doghouses

30 hours every week towards painting and assembling doghouses.

Hence the system of equations that stand for the production time needed to build all of the doghouse as well as painting and assembling each doghouse is:

5x + 2y = 50 (production time for building)

1x + 2y = 30 (production time for painting and assembling)

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The series is a divergent p-series with p = 1/6.

To determine whether the given series is convergent or divergent, we first need to understand the series itself. The series you've provided is:

Σ (n=1 to infinity) (1 / n√6)

This series is a p-series with p equal to the exponent of n in the denominator. In this case, p = 1/6 (since n√6 = n^(1/6)).

A p-series converges if p > 1 and diverges if p ≤ 1. In this case, p = 1/6, which is less than 1.

Your answer: The series is a divergent p-series with p = 1/6.

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(a) An example of a random variable X whose expected value is 5, but the probability that X = 5 is 0 is a dice roll, where X represents the number rolled. If the dice is fair, then the expected value of X is (1+2+3+4+5+6)/6 = 3.5. However, if we assign a probability of 0.1667 to each number from 1 to 4 and a probability of 0 to 5 and 6, then the expected value of X is still 5, but the probability that X = 5 is 0.

(b) An example of a random variable Y whose expected value is negative, but the probability that Y is positive is close to 1 is the temperature difference between two cities in winter. Let Y represent the temperature difference in Celsius degrees between City A and City B on a given day. We know that City A is colder than City B in winter, so the expected value of Y is negative. However, if we take the absolute value of Y, which represents the temperature difference regardless of direction, then the expected value of |Y| is positive. Moreover, if the temperature difference between the two cities is small, then the probability that Y is positive (i.e. City B is warmer) is close to 1.

(c) The statement E( 1/ Z ) = 1 / E(Z) is not always true. We can prove this by giving a counterexample. Let Z be a random variable that takes the value 1 with probability 1/2 and the value 2 with probability 1/2. Then, E(Z) = (1+2)/2 = 1.5. However, E(1/Z) = (1/1)(1/2) + (1/2)(1/2) = 3/4, and 1/E(Z) = 2/3. Therefore, E(1/Z) ≠ 1/E(Z) in this case.

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Answer:

Step-by-step explanation:

Find the linearization of the function z = x squareroot y at the point (6, 1). L(x, y) = Find the linearization of the function f(x, y) = squareroot 36 - 4x^2 - 4y^2 at the point (-1, 2). L(x, y) = Use the linear approximation to estimate the value of f(-1.1, 2.1) =

The recommended model for trend and seasonality is the Seasonal-Trend Decomposition using Loess (STL) regression model.

The variables in the model are time (t), trend (Tt), seasonality (St), and residual (Rt).

The number of independent variables depends on the frequency of data and degree of seasonality, and can be determined by the formula 2q x m.

To develop a model for trend and seasonality, we can use a regression model known as the Seasonal-Trend Decomposition using Loess (STL).

The variables in the model are:

The number of independent variables in the regression depends on the frequency of the data and the degree of seasonality. If the data has a daily frequency and exhibits daily seasonality, the regression model will have 365 independent variables (one for each day of the year). If the data has a monthly frequency and exhibits monthly seasonality, the regression model will have 12 independent variables (one for each month of the year).

The number of independent variables can be determined by the formula 2q × m, where q is the number of harmonics (usually set to 1 or 2) and m is the number of observations per season (e.g., 12 for monthly data).

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If the relative intensity of a quake is multiplied by 10t, the Richter scale reading increases by t units.

The Richter scale measures the intensity of earthquakes logarithmically. When the relative intensity of a quake is multiplied by 10t, it means that the earthquake's amplitude increases by a factor of 10t. Since the Richter scale is based on the logarithm (base 10) of the amplitude, the scale reading will increase by t units.

This is because the logarithm of a product (10t * original amplitude) is equal to the sum of the logarithms (log10(10t) + log10(original amplitude)). Since log10(10t) = t, the Richter scale reading increases by t units when the relative intensity is multiplied by 10t.

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If the relative intensity of a quake is multiplied by 10t, the Richter scale reading increases by t units.

The Richter scale measures the intensity of earthquakes logarithmically. When the relative intensity of a quake is multiplied by 10t, it means that the earthquake's amplitude increases by a factor of 10t. Since the Richter scale is based on the logarithm (base 10) of the amplitude, the scale reading will increase by t units.

This is because the logarithm of a product (10t * original amplitude) is equal to the sum of the logarithms (log10(10t) + log10(original amplitude)). Since log10(10t) = t, the Richter scale reading increases by t units when the relative intensity is multiplied by 10t.

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The regular expression "10" matches any string that starts with zero or more ones followed by zero or more zeros. The regular expression "(10 u 0)(1 u 10)" matches any string that starts with zero or more occurrences of "10" or "0", followed by zero or more occurrences of "1" or "10".

The regular expression 10 matches any string that contains zero or more ones followed by zero or more zeros. This includes empty strings as well as strings containing only ones or only zeros, as well as any combination of ones and zeros.

The regular expression (10 u 0)(1 u 10) matches any string that starts with zero or more occurrences of the string "10" or "0", followed by zero or more occurrences of either "1" or "10".

This regular expression matches strings such as "10", "1010", "0101", "001110", and so on. It allows for any number of occurrences of "0" between any pair of "1" or "10".

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The value of the matrix is in the given image below:

A matrix is a rectangular arrangement of numbers, symbols, or expressions that are organized into rows and columns.

Its size is described as m x n – where m stands for the number of rows while n denotes the number of columns.

Mathematicians, physicists, engineers, and even computer scientists frequently utilize matrices in order to manage data, fix equations, convert geometric shapes, and explore intricate systems.

Additionally, they are an essential tool when it comes to linear algebra and machine learning.

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The answer is b) 120.

To solve this problem, we can use the binomial probability formula:

P(X=k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n is the number of trials (in this case, 19)
- k is the number of successes we want (in this case, 5)
- p is the probability of getting a tail on a single coin toss (which is 0.5 for a fair coin)
- "n choose k" is a combination formula that gives us the number of ways to choose k items from a set of n items (it can be calculated as n!/(k!(n-k)!))

Plugging in the values, we get:
P(X=5) = (19 choose 5) * 0.5^5 * 0.5^(19-5)
P(X=5) = (19 choose 5) * 0.5^19
P(X=5) ≈ 0.2026

Finally, we need to multiply this probability by the total number of possible outcomes (which is 2^19, since there are 2 possible outcomes for each toss):

Total number of outcomes with 19 coin tosses = 2^19 = 524,288
Number of outcomes with exactly 5 tails = 0.2026 * 524,288 ≈ 106,288
Therefore, the answer is b) 120.

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The required answer is the number of cereals (10): 304 / 10 = 30.4

To find the mean amount of sugar in the sample of 10 different cereals, we need to add up all the amounts of sugar and divide by the number of cereals in the sample.
the mean (often simply described as the "average") is that it is not skewed by a small proportion of extremely large or small values, and therefore provides a better representation of the center. Median income,

Adding up the amounts of sugar:

24 + 30 + 47 + 43 + 7 + 47 + 13 + 44 + 39 + 10 = 304

Dividing by the number of cereals (which is 10):

304 / 10 = 30.4
So the mean amount of sugar in the sample is 30.4 centigrams.
If we need to round to the nearest hundredth place, the answer would be 30.40 centigrams.

To find the mean amount of sugar in the 10 different cereals, including Cheerios, Corn Flakes, Fruit Loops, and others, follow these steps:

1. Add up the amounts of sugar in each cereal: 24 + 30 + 47 + 43 + 7 + 47 + 13 + 44 + 39 + 10 = 304 centigrams
2. Divide the total amount of sugar (304 centigrams) by,

the number of cereals (10): 304 / 10 = 30.4

The arithmetic mean (or simply mean) of a list of numbers, is the sum of all of the numbers divided by the number of numbers. Similarly, the mean of a sample. , usually denoted by. , is the sum of the sampled values divided by the number of items in the sample.

the mean (often simply described as the "average") is that it is not skewed by a small proportion of extremely large or small values, and therefore provides a better representation of the center.

The mean amount of sugar in the cereals is 30.4 centigrams. Since it's already rounded to the nearest hundredth place, there's no need for further rounding.

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The domain is (-2,10)

The total gain at the end of the second year for both accounts combined is $509.09.

We have,

Amount saved = $8500

40% of 8500 is saved in saving account = 0.4 x 8500 = $3400

Remainder amount in stock plan = 8500 - 3400 = 5100

Working for savings plan

A = P(1 + r/n[tex])^{nt[/tex]

A = 3400(1 + 0.042/1)²

A = $3691.60

So, we gain = 3691.6 - 3400 = $291.6

Working for stock plan:  

The stock plan decreases 3% in the first year

= 5100 x 0.97

= $4947

and increases 7.5% in the second year.

= 4947 x 1.75

= $5318.03

So, we gain = 5318.03 - 5100 = $218.03

Thus, the total gain is

= 291.06 + 218.03

= $509.09

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The population whose properties are analyzed by the data can be described as households in City A.

Given numerical information is,

58% of households in City A were online.

We have to describe what describes this numerical information.

Here, it is described that a certain percent of households in a city A are online.

So the description is about the households of the city A.

So this is the population.

Hence the best description is households in City A.

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The sample size needed to provide a margin of error of 2 or less with a 0.95 probability when the population standard deviation equals 11 is option C, 117.

The formula for the margin of error (E) is:

E = z * (σ / sqrt(n))

where z is the z-score for the desired level of confidence (0.95 corresponds to z = 1.96), σ is the population standard deviation, and n is the sample size.

We are given that σ = 11 and we want the margin of error to be 2 or less with a 0.95 probability, so we can write:

2 = 1.96 * (11 / sqrt(n))

Solving for n, we get:

n = (1.96 * 11 / 2)^2

n ≈ 116.36

Rounding up to the nearest integer, we get n = 117.

Therefore, the sample size needed to provide a margin of error of 2 or less with a 0.95 probability when the population standard deviation equals 11 is option C, 117.

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Answer:

Step-by-step explanation:

2x

The equation which shows the relationship between x and y is A: y = 2x.

An equation is the statement of two expressions located on two sides connected with an equal to sign. The two sides of an equation is usually called as left hand side and right hand side.

Given that,

Alisha's soccer team is having a bake sale.

Alisha decides to bring chocolate chip cookies to sell.

There is a proportional relationship between the number of cookies Alisha sells, x, and the total cost (in dollars), y.

Proportional relationships are relationships for which the equations are of the form y = kx, where k is a constant.

Alisha sells 6 cookies for $12.00.

That is, total cost, y = 12 when the number of cookies, x = 6.

The equation becomes,

12 = 6k

k = 12/6 = 2

Required equation is y = 2x.

Hence the required equation is y = 2x.

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The remaining sides and angles of triangle ABC are as follows: B = 29°, BC ≈ 17.19, and AC ≈ 9.97.

To determine the remaining sides and angles of triangle ABC:

Follow these steps:

A = 130° 50', C = 20° 10', and AB = 6,

Step 1: Determine angle B.
Since the sum of angles in a triangle is always 180°, you can find angle B by subtracting angles A and C from 180°.
B = 180° - (130° 50' + 20° 10')
B = 180° - 151°
B = 29°

Step 2: Use the Law of Sines to find sides BC and AC.
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.

We can write this as:

BC / sin(A) = AC / sin(B) = AB / sin(C)

Step 3: Solve for side BC.
Using the known values, we can set up an equation to find BC:
BC / sin(130° 50') = 6 / sin(20° 10')
BC = (6 * sin(130° 50')) / sin(20° 10')

Step 4: Solve for side AC.
Using the known values, we can set up an equation to find AC:
AC / sin(29°) = 6 / sin(20° 10')
AC = (6 * sin(29°)) / sin(20° 10')

Step 5: Calculate the values of BC and AC.
BC ≈ (6 * sin(130° 50')) / sin(20° 10') ≈ 17.19
AC ≈ (6 * sin(29°)) / sin(20° 10') ≈ 9.97

In conclusion, the remaining sides and angles of triangle ABC are as follows:
B = 29°, BC ≈ 17.19, and AC ≈ 9.97.

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