*NEED ANSWER ASAP*
Find the slope of (-4/6) and m=3/4.

Since none of the 13 classmates had been given math homework between the original survey and Kelly's second survey, the sum of the values in the second data set is the same as the sum of the values in the original data set. Therefore, the change in the means can be determined without calculating the mean of either data set by considering the number of data points in each set.

Since both data sets have the same number of data points, the change in the means will be zero. This is because the mean is calculated by dividing the sum of the values by the number of data points, and since the sum of the values is the same in both data sets, the means will also be the same.

In other words, if the mean of the first data set is x, then the sum of the values in the first data set is 13x (since there are 13 classmates), and the sum of the values in the second data set is also 13x (since none of the values have changed). Therefore, the mean of the second data set will also be x, and the change in the means will be zero.

Answer: 7x^3y^4z^2

Step-by-step explanation:

Answer:

h

Step-by-step explanation:

Answer:

14, 15, 16

Step-by-step explanation:

n + n+1  + n + 2 = 45

3n + 3 = 45

3n = 42

n = 14

14, 15, 16

Answer:

m= 70

s= 40

a= 110

d= 40

Step-by-step explanation:

Angle M is equal to 70 because it is an isoleases triangle.

The 3 sides of a triangle is equal yo 180. So if angle m is 70, then 70+70 is 140. 180-140 is 40. This makes angle s= 40.

Now we have to find angle a. Angle a and 70 make a straight line. So the line is 180. 180-70 is 110. Angle a is equal to 110.

Now we have to find angle d. As I said before All the angles of a triangle equal to 180. 30+110= 140 now 180 - 140= 40. This makes angle d= 40.

The expected value and variance of

E(Y) = 0

Var(Y) = 1

We can start by finding the expected value of Y:

E(Y) = E[(X - µx) / δx]

Using the linearity of expectation, we can rewrite this as:

E(Y) = (1 / δx) × E(X - µx)

Now, E(X - µx) is simply the expected deviation of X from its mean, which is 0. Therefore:

E(Y) = (1 / δx) × 0 = 0

So the expected value of Y is 0.

Next, let's find the variance of Y:

Var(Y) = Var[(X - µx) / δx]

Using the property Var(aX) = a2Var(X) for any constant a, we can rewrite this as:

Var(Y) = (1 / δx2) × Var(X - µx)

Expanding this expression, we get:

Var(Y) = (1 / δx2) × [Var(X) - 2Cov(X, µx) + Var(µx)]

Since Var(µx) = 0 (because µx is a constant), this simplifies to:

Var(Y) = (1 / δx2) ×[Var(X) - 2Cov(X, µx)]

Now, we know that Var(X) = δ2x (the square of the standard deviation), and Cov(X, µx) = 0 (because µx is a constant). Therefore:

Var(Y) = (1 / δx2) × [δ2x - 2(0)] = 1

So the variance of Y is 1.

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To find the expected value of Y, we use the linearity of expectation. The expected value of Y is 0 and the variance of Y is 1.

E(Y) = E(X - µx / δx)
    = E(X) - E(µx / δx)    (since E(aX) = aE(X))
    = µx - µx / δx         (since E(c) = c for any constant c)
    = µx(1 - 1/δx)

To find the variance of Y, we use the properties of variance:

Var(Y) = Var(X - µx / δx)
         = Var(X) + Var(µx / δx) - 2Cov(X, µx / δx)    (since Var(aX + bY) = a^2Var(X) + b^2Var(Y) + 2abCov(X, Y))
         = Var(X) + 0 - 2(µx/δx)Var(X) / δx    (since Cov(X, c) = 0 for any constant c)
         = δ^2x - 2µx(δ^2x) / δ^3x
         = δ^2x(1 - 2/δx)

Given a random variable X with expected value µx and variance δ^2x, the expected value and variance of Y = (X - µx) / δx are as follows:

Expected value of Y:
E(Y) = E((X - µx) / δx) = (E(X) - µx) / δx = (µx - µx) / δx = 0

Variance of Y:
Var(Y) = Var((X - µx) / δx) = (1/δ^2x) * Var(X) = (1/δ^2x) * δ^2x = 1

Therefore, the expected value of Y is 0 and the variance of Y is 1.

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The solution to the differential equation 3y^2 - xy dx using the method for solving homogeneous equations is y = kx^3, where k is a constant.

To solve the given differential equation using the method for solving homogeneous equations, we follow these steps:

Step 1: Rewrite the equation in the form dy/dx = f(x, y), where f(x, y) is a function of x and y.

The given differential equation is 3y^2 - xy dx. Let's rearrange it to the form dy/dx = f(x, y):

dy/dx = (3y^2 - xy) / dx

Step 2: Make the substitution y = vx, where v is a new variable.

Using the substitution y = vx, we can rewrite the equation as follows:

dy/dx = d(vx)/dx

dy/dx = v + x dv/dx

Step 3: Differentiate the substitution y = vx with respect to x.

To differentiate y = vx, we use the product rule:

dy/dx = v + x dv/dx

Step 4: Substitute the expressions obtained in steps 2 and 3 into the differential equation.

Substituting the expressions from steps 2 and 3 into the differential equation, we get:

v + x dv/dx = (3(vx)^2 - x(vx)) / dx

v + x dv/dx = 3v^2x^2 - vx^2

Step 5: Simplify and rearrange the equation.

Rearranging the terms, we have:

x dv/dx = 2v - v^2

Step 6: Separate the variables and integrate.

Separating the variables, we get:

(1 / (2v - v^2)) dv = (1 / x) dx

Integrating both sides, we have:

∫ (1 / (2v - v^2)) dv = ∫ (1 / x) dx

Step 7: Evaluate the integrals and solve for v.

Integrating both sides, we get:

(1/2) ln|2v - v^2| = ln|x| + C1

Step 8: Solve for v.

Taking the exponential of both sides, we have:

2v - v^2 = C2 * x

Step 9: Convert back to the original variable y.

Since y = vx, we substitute vx for y in the equation obtained in step 8:

2y - (y/x)^2 = C2 * x

This is the general solution to the given homogeneous differential equation.

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\(f(x)=4x+8\\y=4x+8\\x=4y+8\\\\\)

Proceed the inverse. Swap the x and y.

\(x-8=4y\\\frac{x}{4}-\frac{8}{4}=y\\y=h(x)\\h(x)=\frac{x}{4}-2\\h(x)=\frac{1}{4}x-2\)

So the answer is first one.

Answer:

no of girls - 300

no of boys - 450

no of girls - 6

no of boys - 9

A number, a variable, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by an equal sign form an equation. Word illustration: The product of 8 and 3. Word illustration: The product of 8 and 3 is 11

Given

Let boys be x and girls y

x = 2y-3

x - 2y = -3 ------(i)

x = 60/100(x+y)

10x = 6x + 6y

4x = 6y

2x = 3y

2x - 3y =0------(ii)

Subtract (ii) from (i)*2

no of girls - 6

no of boys - 9

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1) Probability that Exactly 2 have used their phones to access the Internet is; P(X = 2) = 0.2759

2) Probability that more than 2 have used their phones to access the Internet is; P(X > 2) = 0.1798

3) Probability that at least 8 have used their phones to access the Internet is; P(X ≥ 8) = 0.00001

4) Probability that no more than 5 have used their phones to access the Internet is; P(X ≤ 5) = 0.99862

The binomial probability is defined as the probability of exactly x successes on n repeated trials, with p probability.

The formula for binomial probability distribution is;

P(X = x) = nCx * p^(x) * (1 - p)^(n - x)

Where;

P(X = x) = binomial probability

x = number of times for a specific outcome within n trials

nCx = number of combinations

p = probability of success on a single trial

q = probability of failure on a single trial

n = number of trials

1) We are given;

p = 15% = 0.15

n = 10

Thus;

P(X = 2) = 10C2 * 0.15^(2) * (1 - 0.15)^(10 - 2)

P(X = 2) = 0.2759

2) Probability that more than 2 have used their phones to access the Internet is;

P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

Using binomial probability calculator, we have;

P(X > 2) = 0.1798

3) Probability that at least 8 have used their phones to access the Internet is;

P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10)

Using binomial probability calculator, we have;

P(X ≥ 8) = 0.00001

4) Probability that no more than 5 have used their phones to access the Internet is;

P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X ≤ 5) = 0.99862

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Answer:

B and C

Step-by-step explanation:

I took a test

Answer:

answer is b and c

Step-by-step explanation:

Answer:

417

Step-by-step explanation:

372+45 = 417

Just add both gram values

After assuming that A is a regular language recognized by a deterministic finite automaton, we find that A^2_1 is a regular language if A is a regular language.

To prove that the class of regular languages is closed under the operation A^2_1, where A^2_1 = {x | for some y, |x| = |y| and xy ∈ A}, we need to show that if A is a regular language, then A^2_1 is also a regular language.

Let's assume that A is a regular language recognized by a deterministic finite automaton (DFA) M = (Q, Σ, δ, q0, F), where:

- Q is the set of states,

- Σ is the input alphabet,

- δ is the transition function,

- q0 is the initial state,

- F is the set of final states.

We need to construct a DFA M' = (Q', Σ', δ', q0', F') that recognizes the language A^2_1.

The idea behind constructing M' is to simulate two copies of M in parallel, keeping track of the lengths of the input strings separately and ensuring that the lengths of the concatenated strings are equal.

Formally, the DFA M' = (Q', Σ', δ', q0', F') is defined as follows:

- Q' = Q × Q, representing pairs of states from M.

- Σ' = Σ, since the input alphabet remains the same.

- δ' is the extended transition function defined as:

 - For each (p, q) ∈ Q' and each a ∈ Σ, δ'((p, q), a) = (δ(p, a), δ(q, a)).

- q0' = (q0, q0), representing the initial states of M.

- F' = {(p, q) | p ∈ F}, where p and q are states from M.

Intuitively, the DFA M' keeps track of the current states of the two copies of M as it reads the input symbols. It transitions to the next pair of states based on the input symbol and the transitions of the individual copies of M. The final states of M' are the pairs of states where the first component comes from the final states of M.

Now, let's prove that M' recognizes the language A^2_1.

1. If x ∈ A^2_1, then there exist y and z such that |x| = |y| = |z| and xy ∈ A. Since A is recognized by M, there exists a path in M from q0 to a final state in F when reading xy. By simulating M' on input x, M' will reach a final state (p, q) ∈ F' where p comes from a final state in F. Therefore, M' accepts x.

2. If x ∉ A^2_1, then for any y and z with |x| = |y| = |z|, xy ∉ A. This implies that no matter how we split x into y and z, the concatenated string xy cannot be recognized by M. Hence, when simulating M' on input x, M' will not reach any final state. Therefore, M' rejects x.

Based on the above arguments, we have shown that M' recognizes the language A^2_1. Since A was assumed to be a regular language, we have proven that the class of regular languages is closed under the operation A^2_1.

Thus, A^2_1 is a regular language if A is a regular language.

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Answer:

chips

Step-by-step explanation:

see above

Answer:

Chips

Step-by-step explanation:

Answer:    The elimination method is a technique for solving systems of linear equations. Let's walk through a couple of examples.

Example 1

We're asked to solve this system of equations:

\begin{aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end{aligned}

2y+7x

5y−7x

​

=−5

=12

​

We notice that the first equation has a 7x7x7, x term and the second equation has a -7x−7xminus, 7, x term. These terms will cancel if we add the equations together—that is, we'll eliminate the xxx terms:

\begin{aligned} 2y+\redD{7x} &= -5 \\ +~5y\redD{-7x}&=12\\ \hline\\ 7y+0 &=7 \end{aligned}

2y+7x

+ 5y−7x

7y+0

​

=−5

=12

=7

​

​

Solving for yyy, we get:

\begin{aligned} 7y+0 &=7\\\\ 7y &=7\\\\ y &=\goldD{1} \end{aligned}

7y+0

7y

y

​

=7

=7

=1

​

Plugging this value back into our first equation, we solve for the other variable:

\begin{aligned} 2y+7x &= -5\\\\ 2\cdot \goldD{1}+7x &= -5\\\\ 2+7x&=-5\\\\ 7x&=-7\\\\ x&=\blueD{-1} \end{aligned}

2y+7x

2⋅1+7x

2+7x

7x

x

​

=−5

=−5

=−5

=−7

=−1

​

The solution to the system is x=\blueD{-1}x=−1x, equals, start color #11accd, minus, 1, end color #11accd, y=\goldD{1}y=1y, equals, start color #e07d10, 1, end color #e07d10.

We can check our solution by plugging these values back into the original equations. Let's try the second equation:

\begin{aligned} 5y-7x &= 12\\\\ 5\cdot\goldD{1}-7(\blueD{-1}) &\stackrel ?= 12\\\\ 5+7 &= 12 \end{aligned}

5y−7x

5⋅1−7(−1)

5+7

​

=12

=

?

12

=12

​

Yes, the solution checks out.

If you feel uncertain why this process works, check out this intro video for an in-depth walkthrough.

Example 2

We're asked to solve this system of equations:

\begin{aligned} -9y+4x - 20&=0\\\\ -7y+16x-80&=0 \end{aligned}

−9y+4x−20

−7y+16x−80

​

=0

=0

​

We can multiply the first equation by -4−4minus, 4 to get an equivalent equation that has a \purpleD{-16x}−16xstart color #7854ab, minus, 16, x, end color #7854ab term. Our new (but equivalent!) system of equations looks like this:

\begin{aligned} 36y\purpleD{-16x}+80&=0\\\\ -7y+16x-80&=0 \end{aligned}

36y−16x+80

−7y+16x−80

​

=0

=0

​

Adding the equations to eliminate the xxx terms, we get:

\begin{aligned} 36y-\redD{16x} +80&=0 \\ {+}~-7y+\redD{16x}-80&=0\\ \hline\\ 29y+0 -0&=0 \end{aligned}

36y−16x+80

+ −7y+16x−80

29y+0−0

​

=0

=0

=0

​

​

Solving for yyy, we get:

\begin{aligned} 29y+0 -0&=0 \\\\ 29y&=0 \\\\ y&=\goldD 0 \end{aligned}

29y+0−0

29y

y

​

=0

=0

=0

​

Plugging this value back into our first equation, we solve for the other variable:

\begin{aligned} 36y-16x+80&=0\\\\ 36\cdot 0-16x+80&=0\\\\ -16x+80&=0\\\\ -16x&=-80\\\\ x&=\blueD{5} \end{aligned}

36y−16x+80

36⋅0−16x+80

−16x+80

−16x

x

​

=0

=0

=0

=−80

=5

​

The solution to the system is x=\blueD{5}x=5x, equals, start color #11accd, 5, end color #11accd, y=\goldD{0}y=0y, equals, start color #e07d10, 0, end color #e07d10.

Want to see another example of solving a complicated problem with the elimination method? Check out this video.

Practice

PROBLEM 1

Solve the following system of equations.

\begin{aligned} 3x+8y &= 15\\\\ 2x-8y &= 10 \end{aligned}

3x+8y

2x−8y

​

​

Step-by-step explanation:

From 1976 until 1987, red M&Ms were removed from packets owing to a research associating the food dye used to color them to cancer.

Arithmetic is defined as working with numbers through addition, subtraction, multiplication, and division. Adding two and two to produce four is an example of arithmetic. Arithmetic is a fundamental branch of mathematics that studies the characteristics of classical operations on numbers such as addition, subtraction, multiplication, division, exponentiation, and root extraction. Arithmetic (derived from the Greek word arithmos, "number") relates to the fundamental components of number theory, arts of mensuration (measuring), and numerical computing (that is, the processes of addition, subtraction, multiplication, division, raising to powers, and extraction of roots).

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Answer:

D because when we multiply minus minus it became plus

A binomial is a factor of a polynomial has been determined by using the

polynomial division method.

To determine if a binomial is a factor of a polynomial, you can use the polynomial division method. The basic idea is to divide the polynomial by the binomial and check if the remainder is zero. If the remainder is zero, then the binomial is a factor of the polynomial. Here's the step-by-step process:

Write the polynomial and the binomial in standard form, with the terms arranged in descending order of their exponents.

Perform the long division of the polynomial by the binomial, similar to how you would divide numbers. Start by dividing the highest degree term of the polynomial by the highest degree term of the binomial.

Multiply the binomial by the quotient obtained from the division and subtract the result from the polynomial.

Repeat the division process with the new polynomial obtained from the subtraction step.

Continue dividing until you reach a point where the degree of the polynomial is lower than the degree of the binomial.

If the remainder is zero, then the binomial is a factor of the polynomial. If the remainder is non-zero, then the binomial is not a factor.

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The test statistic for the significance test is calculated as 3.6.

To determine if there is convincing evidence that the mean height of five-year-old children in this city is greater than 42.5 inches, we can perform a hypothesis test.

The null hypothesis, denoted as \(H_0\), assumes that the mean height is equal to 42.5 inches, while the alternative hypothesis, denoted as \(H_a\), assumes that the mean height is greater than 42.5 inches.

Using the given sample data, we can calculate the test statistic.

The sample mean height is 44.1 inches, and the standard deviation is 3.5 inches.

Since the population standard deviation is unknown, we can use a t-test.

The formula for the t-test statistic is given by (sample mean - hypothesized mean) / (sample standard deviation / √n).

Plugging in the values, we have (44.1 - 42.5) / (3.5 / √40) ≈ 3.6.

This test statistic measures how many standard deviations the sample mean is away from the hypothesized mean under the assumption of the null hypothesis.

To determine if the data provides convincing evidence, we compare the test statistic to the critical value corresponding to the significance level chosen for the test.

If the test statistic exceeds the critical value, we reject the null hypothesis in favor of the alternative hypothesis, providing evidence that the mean height of five-year-old children in this city is greater than 42.5 inches.

Without specifying the chosen significance level, we cannot definitively state if the data provides convincing evidence.

However, if the test statistic of 3.6 exceeds the critical value for a given significance level, we can conclude that the data provides convincing evidence at that specific level.

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Answer:

the second

Step-by-step explanation:

refers to well-founded standards of right and wrong that prescribe what humans should do, usually in terms of rights, obligations, benefits to society, justice

Answer:

255 added

Step-by-step explanation:

1700 x 55% = 935

1700 x 40% = 680

935 - 680 = 255

There are 255 more category of 1 tickets are added.

What is mean by Percentage?

A number or ratio that can be expressed as a fraction of 100 or a relative value indicating hundredth part of any quantity is called percentage.

To Calculate the percent of a number , divide the number by whole number and multiply by 100.

Given that;

Total number of seats in a theatre = 1700

And, 40% of the tickets to a concert at the theatre are category 1 tickets and the rest are of other categories.

The organizer decides to change some tickets from the other categories to category 1.

Now,

Number of category 1 tickets = 40% of 1700

                                             = 40/100 x 1700

                                            = 40 x 17

                                            = 680

And, The number of tickets of other categories = 55% of 1700

                                                                         = 55/100 x 1700

                                                                         = 55 x 17

                                                                         = 935

So, Number of added tickets of category 1 = 935 - 680

                                                                 = 255

Thus, There are 255 more category of 1 tickets are added.

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Answer:

y = $40 - ($3/day)x

Step-by-step explanation:

y = $40 - ($3/day)x represents the amount of money John has left after x days.

Marcus is using a plant food to fertilize his garden his garden is a rectangle Marcus uses 1 cup of plant food for every square yard of his garden.

To solve it, we need to first find the area of Marcus's garden in square yards, and then multiply it by 1 cup of plant food per square yard. Finally, we need to convert the total cups to ounces using the given conversion rate.

Let the length of Marcus's garden is l yards and the width is w yards. Then the area of his garden in square yards is

A = lw.

If Marcus uses 1 cup of plant food per square yard, then he will use A cups of plant food in total.

To convert cups to ounces, we can use the given conversion rate of 4 cups per 28 ounces.

1 cup = 28/4 ounces

Therefore, the total weight of plant food Marcus uses in ounces is

Total weight = A * (28/4) ounces

By substituting A = lw, we get

Total weight = lw * (28/4) ounces

Total weight = 7lw ounces

Hence, the total weight of plant food Marcus uses to fertilize his garden is 7 times the product of the length and width of his garden, measured in yards.

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(i) Context-Free Grammar for the set of odd-length strings in \( \{a, b\}^{*} \): S -> a | b | aSa | bSb

(ii) Context-Free Grammar for the set of even-length strings in \( \{a, b\}^{*} \): S -> ε | aSb | bSa | aSbS | bSaS

The above context-free grammar generates odd-length strings in the language \( \{a, b\}^{*} \). The start symbol S can produce a single 'a' or 'b' symbol as base cases. Additionally, S can generate strings of the form aSa or bSb, where S is enclosed by an 'a' and 'b'. This recursive rule allows for the generation of odd-length strings by adding pairs of 'a' and 'b' symbols around a central S symbol.

The above context-free grammar generates even-length strings in the language \( \{a, b\}^{*} \). The start symbol S can produce an empty string ε as a base case.

Additionally, S can generate strings of the form aSb or bSa, where an 'a' and 'b' are appended before and after the central S symbol. Furthermore, S can generate strings of the form aSbS or bSaS, where the central S symbol is surrounded by pairs of 'a' and 'b' symbols.

By using these context-free grammars, we can generate the desired sets of odd-length and even-length strings in \( \{a, b\}^{*} \) by following the production rules and recursively applying them to the start symbol.

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Since A and B are mutually exclusive, the probability of A occurring given that B has already occurred is 0. This means that P(A | B) = 0.

P(A ∩ B) = 0, since the events are mutually exclusive.

P(A | B) = 0, since the probability of A given B is 0 when A and B are mutually exclusive.

Since A and B are mutually exclusive, they cannot both occur at the same time. This means that the probability of them both happening at the same time is 0. As a result, P(A ∩ B) = 0.

Since A and B are mutually exclusive, the probability of A occurring given that B has already occurred is 0. This means that P(A | B) = 0.

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Answer:

Step-by-step explanation:

Triangle MLK

Answer:

x = \(\frac{5}{2}\) , y = \(\frac{27}{4}\)

Step-by-step explanation:

Equate the first 2 parts of the ratios on both sides of the equation and solve for x.

Expressing the ratios in fractional form, then

\(\frac{x}{3}\) = \(\frac{\frac{15}{4} }{\frac{9}{2} }\) = \(\frac{15}{4}\) × \(\frac{2}{9}\) = \(\frac{5}{6}\) ( cross- multiply )

6x = 15 ( divide both sides by 6 )

x = \(\frac{15}{6}\) = \(\frac{5}{2}\)

-----------------------------------------------------------------------------------

Equate the last 2 parts of the ratios on both sides and solve for y

\(\frac{\frac{9}{2} }{y}\) = \(\frac{3}{\frac{9}{2} }\) = 3 × \(\frac{2}{9}\) = \(\frac{2}{3}\) ( cross- multiply )

2y = \(\frac{9}{2}\) × 3 = \(\frac{27}{2}\) ( divide both sides by 2 )

y = \(\frac{27}{4}\)