The function f(x)=x^(2)-2,x>=0 is one -to-one (a) Find the inverse of f

Hello !

Answer:

[tex]\Large \boxed{\sf x=-11}[/tex]

Step-by-step explanation:

We want to find the value of x that satisfies the following equation :

[tex]\sf 5x+2=4x-9[/tex]

Let's isolate x !

First, substract 4x from both sides :

[tex]\sf 5x+2-4x=4x-9-4x\\x+2=-9[/tex]

Now let's substract 2 from both sides :

[tex]\sf x+2-2=-9-2\\\boxed{\sf x=-11}[/tex]

Have a nice day ;)

Hello!

5x + 2 = 4x - 9

5x - 4x = - 9 - 2

x = -11

The emphasis is on the Counting and Cardinality standard in the early childhood years according to the National Council of Teachers of Mathematics.

The National Council of Teachers of Mathematics emphasizes the following standards in the early childhood years:

- Counting and Cardinality

- Operations and Algebraic Thinking

- Number and Operations in Base Ten

- Measurement and Data

- Geometry

The National Council of Teachers of Mathematics recognizes that all five math standards are important in the early childhood years. However, they place a particular emphasis on the standards related to counting and cardinality. This includes developing skills in counting, understanding numbers, and recognizing numerical relationships.

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The Poisson distribution would be a good choice for the probability distribution of r if finding prehistoric artifacts is described as a rare occurrence.

The Poisson distribution is commonly used to model the number of rare events occurring in a fixed interval of time or space.

In the scenarios provided, the occurrence of finding prehistoric artifacts is described as either common or rare.

If finding prehistoric artifacts is a rare occurrence, it aligns with the characteristics of the Poisson distribution. The Poisson distribution is appropriate when events are infrequent and the probability of multiple events happening in a short interval is low.

The assumption of events being dependent or independent is not explicitly stated, so it cannot be used as a determining factor for choosing the Poisson distribution.

Therefore, based on the information given, the Poisson distribution would be a good choice for the probability distribution of the number of prehistoric artifacts found if the events are described as rare occurrences.

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If a boat is 80 miles away from the marina, sailing directly toward it at 20 miles per hour, then the equation for the distance of the boat from the marina, d, after t hours is d= 20t+ 80

To find the equation for the distance, follow these steps:

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In summary: a. The domain of f(x) is all real numbers except x = 6/1 and x = -1/6. b. There are no vertical asymptotes for f(x). c. There is no hole in the graph of f(x).

a. To find the domain of the function f(x), we need to determine the values of x for which the function is defined. In this case, the function f(x) is defined for all real numbers except where the denominator is equal to zero.

So, we set the denominator equal to zero and solve for x:

[tex]6x^2 - 35x - 6 = 0[/tex]

Using factoring or the quadratic formula, we can find the roots of this equation. The roots are x = 6/1 and x = -1/6.

b. To find the vertical asymptotes of f(x), we look for values of x where the function approaches positive or negative infinity as x approaches those values.

In this case, there are no vertical asymptotes for f(x) because the denominator [tex]6x^2 - 35x - 6[/tex] does not approach zero as x approaches any particular value. Hence, there are no vertical asymptotes.

c. To determine if there is a hole in the graph of f(x), we need to check if there are any common factors between the numerator [tex](x^2 - 4x - 12)[/tex] and the denominator [tex](6x^2 - 35x - 6).[/tex]

Factoring the numerator, we have:

[tex]x^2 - 4x - 12 = (x - 6)(x + 2)[/tex]

The denominator does not have any common factors with the numerator. Therefore, there is no hole in the graph of f(x).

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The observation of longer delay measurements in the previous hop compared to subsequent hops in traceroute can be attributed to factors such as network congestion, increased traffic, and variations in routing protocols. The delay between hop#10 and hop#11 is calculated to be -1ms, although this negative value could be due to measurement discrepancies.

The observation of longer delay measurements in the previous hop compared to subsequent hops in a traceroute can be attributed to factors like network congestion, routing changes, and variations in network infrastructure.

Each network node introduces its own processing and forwarding delays, which can vary based on factors like node load and network conditions. In the given example, hop #10 and hop #11 are part of the same network provider, but calculating the delay between them based on the provided measurements is not possible.

Accurately determining the delay between specific hops requires access to raw packet timestamps, network topology knowledge, and routing algorithms, which are not available in a regular traceroute.

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we cannot determine a specific solution for the given differential equation with the given initial condition. Hence the correct answer is d) None of the above.

To solve the given differential equation (lny)y' = -x^2y, we can separate the variables and integrate both sides.

(lny)dy = -x^2ydx

Integrating both sides:

∫(lny)dy = ∫(-x^2y)dx

Integrating the left side using integration by parts:

[ ylny - ∫(1/y)dy ] = ∫(-x^2y)dx

Simplifying:

ylny - ∫(1/y)dy = -∫(x^2y)dx

Using the integral of 1/y and integrating the right side:

ylny - ln|y| = -∫(x^2y)dx

Simplifying further:

ln(y^y) - ln|y| = -∫(x^2y)dx

Combining the logarithmic terms:

ln(y^y/|y|) = -∫(x^2y)dx

Simplifying the expression inside the logarithm:

ln(|y|) = -∫(x^2y)dx

At this point, we cannot proceed to find a closed-form solution since the integral on the right side is not straightforward to evaluate. Additionally, the given initial condition y(0) = e cannot be directly incorporated into the solution process.

Therefore, we cannot determine a specific solution for the given differential equation with the given initial condition. Hence, the correct answer is d) None of the above.

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The statement "The function f(x) = x^2 - 2^x has a zero between x = 1.9 and x = 2.1" is true. To determine if the function f(x) = x^2 - 2^x has a zero between x = 1.9 and x = 2.1, we can evaluate the function at both endpoints and check if the signs of the function values differ.

Let's calculate the function values:

For x = 1.9:

f(1.9) = (1.9)^2 - 2^(1.9) ≈ -0.187

For x = 2.1:

f(2.1) = (2.1)^2 - 2^(2.1) ≈ 0.401

Since the function values at the endpoints have different signs (one negative and one positive), and the function f(x) = x^2 - 2^x is continuous, we can conclude that by the Intermediate Value Theorem, there must be at least one zero of the function between x = 1.9 and x = 2.1.

Therefore, the statement "The function f(x) = x^2 - 2^x has a zero between x = 1.9 and x = 2.1" is true.

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The test statistic (t) for this scenario is approximately 2.613.

To calculate the test statistic in this scenario, we'll use the provided information:

Sample size (n) = 29

Sample mean (x(bar)) = 6.9

Sample standard deviation (s) = 1.5

We also need the null hypothesis value for the population mean (μ₀). In this case, the null hypothesis is that the average number of items per customer order is 6 or less, so we'll use μ₀ = 6.

The formula for the test statistic (t) in a one-sample t-test is:

t = (x(bar) - μ₀) / (s / √(n))

Plugging in the values, we get:

t = (6.9 - 6) / (1.5 / √(29))

Calculating this expression, we find:

t ≈ 2.613

Therefore, the(t) for this scenario is approximately 2.613.

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The complete question is :

In this scenario, what is the test statistic?

A small business owner would like to test the claim that the average number of items per customer order is greater than 6 items.

Sample size =29 customers

Sample mean =6.9 items

Sample standard deviation =1.5 items

Calculate the test statistic using the formula:

t0=x¯−μ0sn√

The equation of the plane through the points (6,3,1),(4,0,2) and is perpendicular to the plane 2z=5x+4y is given by -2(x - 6) + 9(y - 3) + 22(z - 1) = 0.

Given that the two points are A(6, 3, 1) and B(4, 0, 2). First, we find the vector AB = B - A = (-2, -3, 1). We have a plane perpendicular to the plane 2z = 5x + 4y, which means that the normal vector to the plane is <5, 4, -2>.

Now let us find the equation of the plane containing A and is perpendicular to the given plane. We know that the normal vector to this plane is perpendicular to both the plane and AB.

Vector n × AB = <5, 4, -2> × <-2, -3, 1>

= <-2, 9, 22>.

The normal vector to the plane through A is given by <-2, 9, 22>.

The equation of the plane is -2(x - 6) + 9(y - 3) + 22(z - 1) = 0.

The equation of the plane through the points (6,3,1),(4,0,2) and is perpendicular to the plane 2z=5x+4y is given by -2(x - 6) + 9(y - 3) + 22(z - 1) = 0.

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There are 2 × 3 = 6 possible ordered pairs in A × B.(b, a) ∉ A2 since A2 is the Cartesian product of A with itself, and (b, a) is not a valid ordered pair in this product. The only possible ordered pairs are (a, a) and (b, b).

22) The Distributive law establishes that the two sets (A∩B)∪(A∩B) and A∩B are equal. The Distributive law states that (A∩B)∪C=(A-C)∩(BC) and (A∪B)∩C=(A-C)∪(BC).

This law describes the distribution of logical conjunctions and disjunctions and is true in both set theory and Boolean algebra.23) The false statement is A∩A2=∅.

This statement is not possible since A is the set {a, b}. It cannot be reduced to the empty set by taking its intersection with itself. Therefore, the statement is false.

The other options are all true:(b, 3) ∈ A × B since A × B is the set of all ordered pairs that can be formed by choosing one element from A and one element from B.

(b, 3) is one such ordered pair.|A × B| = 6 since there are 2 choices for the first element and 3 choices for the second element.

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a) To compute s for the given data set, we use the formula, where μ is the mean and N is the total number of data points.

b) If we multiply each data value by 3, the new data set will be as follows:33, 45, 51, 33, 24

The formula to compute s for this data set is similar to the one used in part a. We have

c) We can observe that the standard deviation changes if each data value is multiplied by a constant c.

If we multiply each data value by the same constant c, the standard deviation is |c| times larger.

For example, if we multiply each data value by 3, the standard deviation becomes 3 times larger than the original standard deviation.

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Pumping lemma is a technique that is used to show that certain languages are not regular. The pumping lemma states that every regular language has a pumping length such that every string in the language of length at least the pumping length can be divided into three pieces x, y, and z, such that the middle piece y is nonempty and the length of x and y is less than or equal to the pumping length, and the strings xyiz is also in the language. If a language fails to meet this condition, then it is not a regular language.

Now let's move on to the problem to prove that the following languages are not regular: 1. L1={0^n1^n2^n|n≥0,Σ={0,1,2}}.Pumping lemma: Let's assume that L1 is a regular language. Therefore, L1 satisfies the pumping lemma. Thus, there exists a positive integer p such that any string s ∈ L1 with length |s| ≥ p can be written as s = xyz, where:

|x y| ≤ p

|y| ≥ 1

xy i z ∈ L1 for all i ≥ 0

Let's select a string s ∈ L1 with length |s| ≥ p. Thus, s = 0p1p2p. Now let's divide the string into three parts:

x = 0k, y = 0m, z = 01p2p

Here, k + m ≤ p, and m > 0. Now let's try to pump y, which means that we repeat the middle section y i times where i is a positive integer. Therefore, the new string is

xyiz = 0k (0mi) 01p2p = 0k+mim 01p2p

Since we know that m > 0 and k+m ≤ p, then k+m+m ≤ p. Therefore, we can see that the number of 0's that come before 1's is less than the number of 1's that come before 2's. So, xyiz ∉ L1. This is a contradiction since xyiz should belong to L1 if L1 is a regular language. Thus, we can conclude that L1 is not a regular language.2. L2 = {ωωω|ω∈{a,b}∗}.Pumping lemma: Let's assume that L2 is a regular language. Therefore, L2 satisfies the pumping lemma. Thus, there exists a positive integer p such that any string s ∈ L2 with length |s| ≥ p can be written as s = xyz, where:

|x y| ≤ p

|y| ≥ 1

xy i z ∈ L2 for all i ≥ 0

Let's select a string s ∈ L2 with length |s| ≥ p. Since |s| ≥ p, the first three segments of s must be the same, say the segment "aaa". Therefore, s = aaax, where x is a string in {a,b}*. We can also write s as s = xyz, where

x = x1x2x3x4...xk

y = y1y2...yℓ, where ℓ ≤ p

z = z1z2z3...zq

where x1 = y1 = z1 = a, x2 = y2 = z2 = a, and x3 = y3 = z3 = a. Since y is nonempty, then ℓ > 1. Now let's try to pump y, which means that we repeat the middle section y i times where i is a positive integer. Therefore, the new string is

xyiz = x1 x2 x3 ... xi y1 y2 ... yℓ z1 z2 ... zq

For i = 0, we get xy0z = xyz = aaax ∈ L2.

For i = 2, we get xy2z = x1x2...xiy1y2...yℓx1x2...xiy1y2...yℓx3x4...xk ∈ L2.

Thus, the new string xyiz is not in L2 for i = 0 and i = 2. This contradicts the statement that xyiz is in L2 for all i ≥ 0 if L2 is regular. Therefore, we can conclude that L2 is not a regular language.

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The p-value of the test is given as follows:

0.19.

The significance level is given as follows:

0.10.

As the p-value is greater than the significance level, there is not enough evidence to conclude that the mean GPA of night students is smaller than 2.3 at the 0.10 significance level.

The equation for the test statistic is given as follows:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which:

The parameters for this problem are given as follows:

[tex]\overline{x} = 2.28, \mu = 2.3, s = 0.14, n = 39[/tex]

Hence the test statistic is given as follows:

[tex]t = \frac{2.28 - 2.3}{\frac{0.14}{\sqrt{39}}}[/tex]

t = -0.89.

The p-value of the test is found using a t-distribution calculator, with a left-tailed test, 39 - 1 = 38 df and t = -0.89, hence it is given as follows:

0.19.

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The probability of A or B is 0.95, calculated as P(A) + P(B) = 0.65. The Venn diagram shows all possible regions for two events A and B, with their intersection being the empty set. The probability is 0.95.

If the events A and B are disjoint with P(A) = 0.65 and P(B) = 0.30, the probability of A or B can be found as follows:

Probability of A or B= P(A) + P(B) [Since A and B are disjoint events]

∴ Probability of A or B = 0.65 + 0.30 = 0.95

So, the probability of A or B is 0.95.

Now, let's construct the complete Venn diagram for this situation. The complete Venn diagram shows all the possible regions for two events A and B and how they are related.

Since A and B are disjoint events, their intersection is the empty set. Here is the complete Venn diagram for this situation:Please see the attached image for the Venn Diagram.

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A stands for 995.2 gallons of the 20% solution.

To determine the number of gallons of the 20% and 45% solutions needed to fulfill the order for 1,244 gallons of a 25% acid solution, we can set up a system of equations based on the acid concentration and total volume.

Let A be the number of gallons of the 20% solution (20% acid concentration).

Let B be the number of gallons of the 45% solution (45% acid concentration).

We can set up the following equations:

Equation 1: Acid concentration equation

0.20A + 0.45B = 0.25 * 1244

Equation 2: Total volume equation

A + B = 1244

Simplifying Equation 1:

0.20A + 0.45B = 311

To solve this system of equations, we can use various methods such as substitution or elimination. Here, we'll use substitution.

From Equation 2, we can express A in terms of B:

A = 1244 - B

Substituting A in Equation 1:

0.20(1244 - B) + 0.45B = 311

Simplifying and solving for B:

248.8 - 0.20B + 0.45B = 311

0.25B = 62.2

B = 62.2 / 0.25

B = 248.8

Therefore, B (the number of gallons of the 45% solution) is 248.8.

Substituting B in Equation 2:

A + 248.8 = 1244

A = 1244 - 248.8

A = 995.2

Therefore, A (the number of gallons of the 20% solution) is 995.2.

In conclusion:

A = 995 (rounded to 0 decimal place)

B = 249 (rounded to 0 decimal place)

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The null hypothesis (H0) states that the average time for a technician to arrive after a service call is 35 minutes (μ = 35). The alternative hypothesis (Ha) states that the average time for a technician to arrive is less than 35 minutes (μ < 35).


The null hypothesis assumes that there is no significant difference between the claim made by the cable company and the actual average time. It states that the average time for a technician to arrive is equal to 35 minutes. On the other hand, the alternative hypothesis assumes that there is a significant difference and that the average time is less than 35 minutes.


In this case, the null hypothesis is testing the company's claim that a technician will arrive within 35 minutes after a service call. The alternative hypothesis, on the other hand, challenges this claim, suggesting that the average time may be less than 35 minutes. By analyzing data and conducting statistical tests, we can determine if the claim is supported or rejected.

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To construct a Deterministic Finite Accepted M such that L(M) = L(G), the language generated by grammar G = ({S, A, B}, {a, b}, S , {S -> abS, S -> A, A -> baB, B -> aA, B -> bb} ), the following steps should be followed:

Step 1: Eliminate the Null productions from the grammar by removing productions containing S. The grammar, after removing null production, becomes as follows.{S -> abS, S -> A, A -> baB, B -> aA, B -> bb}

Step 2: Eliminate the unit productions. The grammar is as follows. {S -> abS, S -> baB, S -> bb, A -> baB, B -> aA, B -> bb}

Step 3: Now we will convert the given grammar to an equivalent DFA by removing the left recursion. By removing the left recursion, we get the following productions. {S -> abS | baB | bb, A -> baB, B -> aA | bb}

Step 4: Draw the state diagram for the DFA using the following rules: State diagram for L(G) DFA 1. The start state is the initial state of the DFA. 2. The final state is the final state of the DFA. 3. Label the edges with symbols on which transitions are made. 4. A table for the transition function is created. The table for the transition function of L(G) DFA is given below:{Q, a} -> P{Q, b} -> R{P, a} -> R{P, b} -> Q{R, a} -> Q{R, b} -> R

Step 5: Construct the DFA using the state diagram and transition function. The DFA for the given language is shown below. The starting state is shown in green and the final state is shown in blue. DFA for L(G) -> ({Q, P, R}, {a, b}, Q, {Q, P}) Where, Q is the starting state P is the first intermediate state R is the second intermediate state.

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Now, we can use the analytical method to calculate the resultant of the vectors in the x-direction. The x-component of the resultant vector is given by:

Rx = Ax + Bx + Cx

Where,

Rx = x-component of the resultant vector

Ax = x-component of vector A

Bx = x-component of vector B

Cx = x-component of vector C

Substitute the values of the vectors in the formula and get the sum of the x-component

Rx = Ax + Bx + Cx = (2 + 2 + 7) m = 11 m

Therefore, the sum of x components of the resultant vector is 11m.

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Vector A has components of 2m and 3m along the x and y-axis, vector B has 2m and 0m, and vector C has 7m and 1m

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The simplified answer is 21.727444444444444, rounded to 15 decimal places.

To solve the given equation, we need to use the order of operations (PEMDAS) rule. This rule tells us to perform the operations in the following order:

Parentheses Exponents Multiplication and Division (from left to right) Addition and Subtraction (from left to right) Now, let's apply the PEMDAS rule to the given equation:32.123 - 7.1 / 3 × 4.39

First, we perform the division operation within the parentheses.7.1 ÷ 3 = 2.366666666666667 Next, we perform the multiplication operation.2.366666666666667 × 4.39 = 10.395555555555556

Now, we subtract the product from the initial value.32.123 - 10.395555555555556 = 21.727444444444444

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The function g(x) = 1/(x - 1) + 3 can be graphed using translations. The graph is obtained by shifting the graph of the parent function 1/(x) to the right by 1 unit and vertically up by 3 units.

The parent function of g(x) is 1/(x), which has a vertical asymptote at x = 0 and a horizontal asymptote at y = 0. To graph g(x) = 1/(x - 1) + 3, we apply translations to the parent function.

First, we shift the graph 1 unit to the right by adding 1 to the x-coordinate. This causes the vertical asymptote to shift from x = 0 to x = 1. Next, we shift the graph vertically up by adding 3 to the y-coordinate. This moves the horizontal asymptote from y = 0 to y = 3.

By applying these translations, we obtain the graph of g(x) = 1/(x - 1) + 3. The graph will have a vertical asymptote at x = 1 and a horizontal asymptote at y = 3. It will be a hyperbola that approaches these asymptotes as x approaches positive or negative infinity. The shape of the graph will be similar to the parent function 1/(x), but shifted to the right by 1 unit and up by 3 units.

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The value of the line integral ∮C (x³)y dx - x dy, where C is the circle x² + y² = 1 with counter clockwise orientation, is -π/2

To evaluate the line integral ∮C (x³)y dx - x dy, where C is the circle x² + y² = 1 with counter clockwise orientation, parameterize the circle and then use the parameterization to compute the integral.

parameterize the circle C as follows:

x = cos(t)

y = sin(t)

where t ranges from 0 to 2π.

Now, let's compute the integral using this parameterization:

∮C (x³)y dx - x dy

= ∫(0 to 2π) [(cos(t)³)(sin(t))(-sin(t)) - cos(t)(cos(t))] dt

= ∫(0 to 2π) [-cos(t)²sin(t) - cos²(t)] dt

To evaluate this integral, we need to expand the terms and simplify the expression:

= -∫(0 to 2π) (cos²(t)sin(t) + cos²(t)) dt

= -∫(0 to 2π) (cos²(t)sin(t)) dt - ∫(0 to 2π) (cos²(t)) dt

The first integral on the right-hand side is zero since the integrand is an odd function integrated over a symmetric interval.

The second integral simplifies as follows:

= -∫(0 to 2π) (1 - sin²(t)) dt

= -∫(0 to 2π) (1 - (1 - cos²(t))) dt

= -∫(0 to 2π) cos²(t) dt

Using the trigonometric identity cos^2(t) = (1 + cos(2t))/2, the integral as:

= -∫(0 to 2π) (1 + cos(2t))/2 dt

= -[t/2 + sin(2t)/4] evaluated from 0 to 2π

= -(2π/2 + sin(4π)/4 - 0/2 - sin(0)/4)

= -π/2

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The margin of error, if you want a 90% confidence interval for the true population, the mean age is; 1.62 years.

We will use the formula for the margin of error:

Margin of error = z × (σ / √(n))

where, z is the z-score for the desired level of confidence, σ is the population standard deviation, n will be the sample size.

For a 90% confidence interval, the z-score = 1.645.

Substituting the values:

Margin of error = 1.645 × (9.84 / √(100))

Margin of error = 1.62

Therefore, the margin of error will be 1.62 years.

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When you run the above code in MATLAB, it will display the rent for a 1-bedroom house, a 2-bedroom house, and a 3-bedroom house based on the problem.

To solve the given problem using MATLAB, we can set up a system of linear equations based on the given information and then solve it using MATLAB's matrix operations. Let's proceed with the following steps:

Step 1: Define the variables:

Let x be the rent for a 1-bedroom house,

y be the rent for a 2-bedroom house,

z be the rent for a 3-bedroom house.

Step 2: Formulate the equations based on the given information:

Equation 1: x + 2y + 3z = 2760 (total rent for all houses is $2760)

Equation 2: 0.1x + 0.2(2y) + 0.3(3z) = 692 (total repair cost is $692)

Equation 3: x + y = z + 120 (sum of rent for 1-bedroom and 2-bedroom house is $120 more than the rent for a 3-bedroom house)

Step 3: Convert the equations into matrix form:

We can rewrite the system of equations as a matrix equation AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

A = [1 2 3; 0.1 0.4 0.9; 1 1 -1]

X = [x; y; z]

B = [2760; 692; 120]

Step 4: Solve the system of equations using MATLAB:

Use the command X = A\B to solve the system of equations.

Step 5: Display the solution:

Display the values of x, y, and z to get the rent for a 1-bedroom house, a 2-bedroom house, and a 3-bedroom house, respectively.

Here is the MATLAB code to solve the problem:

```matlab

% Define the variables

syms x y z;

% Define the equations

eq1 = x + 2*y + 3*z == 2760;

eq2 = 0.1*x + 0.2*(2*y) + 0.3*(3*z) == 692;

eq3 = x + y == z + 120;

% Solve the system of equations

sol = solve([eq1, eq2, eq3], [x, y, z]);

% Display the solution

rent_1bedroom = sol.x;

rent_2bedroom = sol.y;

rent_3bedroom = sol.z;

% Print the results

disp(['Rent for a 1-bedroom house: $', num2str(rent_1bedroom)]);

disp(['Rent for a 2-bedroom house: $', num2str(rent_2bedroom)]);

disp(['Rent for a 3-bedroom house: $', num2str(rent_3bedroom)]);

```

When you run the above code in MATLAB, it will display the rent for a 1-bedroom house, a 2-bedroom house, and a 3-bedroom house based on the given problem.

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Answer:

team A had the higher score

Step-by-step explanation:

maths

The Sample space is S={1,2,3,4,5,6,7,8,9}, A={1,3,5,7,9}, B={1,2,3,4,5}. Bc is the complement of B.

Bc={6, 7, 8, 9}3)

A∪B={1, 2, 3, 4, 5, 7, 9}4) A∩B={1, 3, 5}5) Sa) P(S)=1 as S is a sample space and hence the probability of an event occurring is 1.6) (Ac)c = A= {1, 3, 5, 7, 9}.

Therefore, Ac = {2, 4, 6, 8}.

And (Ac)c = A.7) P(A) = n(A)/n(S) = 5/9 = 0.556 or 55.6%.8) P(B) = n(B)/n(S) = 5/9 = 0.556 or 55.6%.9) Bc

Bc = {1, 2, 3, 4, 5}.

Bc = {6, 7, 8, 9}.

Disjoint events are two events that do not share any element. A and B have one common element, which is 1, hence A and B are not disjoint.11) Draw a Venn diagram representing A∩B.The diagram below represents the intersection of A and B. In this case, the intersection of A and B is {1, 3, 5}.Therefore, the Venn diagram of A∩B is shown below.

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The total cost of producing 30,000 items is $290,000.

Given:

The total cost to produce 10,000 items is $130,000 and the total cost to produce 20,000 items is $210,000.

Using the linear model C = F + V x for total cost C to produce x items in terms of the fixed cost F and the per-item cost V , find F and V. F = V = b

Formula used in this problem:

C = F + V x

For 10,000 items:

C = F + V x

C = F + 10,000 V ----(1)

Total cost to produce 10,000 items is $130,000

C = 130,000

Put the value of C in equation (1), we get:

130,000 = F + 10,000 V

F + 10,000 V = 130,000 --------------(2)

For 20,000 items:

C = F + V x

C = F + 20,000 V ----(3)

Total cost to produce 20,000 items is $210,000

C = 210,000

Put the value of C in equation (3), we get:

210,000 = F + 20,000 V

F + 20,000 V = 210,000 --------------(4)

Solving equation (2) and (4) by elimination method:

Multiplying equation (2) by -2, we get:-

2F - 20,000 V = -260,000

Multiplying equation (4) by 1, we get:

F + 20,000 V = 210,000

Adding above two equations:-

2F - 20,000 V = -260,000

F + 20,000 V = 210,000-----------------------

(-F) = -50,000

F = $50,000

Putting the value of F in equation (2)

F + 10,000 V = 130,000

50,000 + 10,000 V = 130,000

10,000 V = 130,000 - 50,000

10,000 V = 80,000

V = 8

Total cost equation is:

C = F + V x

C = 50,000 + 8x

Put the value of x=30,000 in above equation, we get:

C = 50,000 + 8(30,000)

C = 50,000 + 240,000

C = $290,000

Therefore, the total cost of producing 30,000 items is $290,000.

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The highest possible score a student who is dropped from the top university would have scored is approximately 1020.

To find the highest possible score for a student who is dropped from the top university, we need to determine the cutoff score corresponding to the bottom 4% of the distribution.

Since the test scores follow a normal distribution with a mean of 1,200 and a standard deviation of 120, we can use the Z-score formula to find the cutoff score.

The Z-score formula is given by:

Z = (X - μ) / σ

Where:

Z is the Z-score

X is the raw score

μ is the mean

σ is the standard deviation

To find the cutoff score, we need to find the Z-score corresponding to the bottom 4% (or 0.04) of the distribution.

Using a standard normal distribution table or a calculator, we can find that the Z-score corresponding to the bottom 4% is approximately -1.75.

Now, we can rearrange the Z-score formula to solve for the raw score (X):

X = Z * σ + μ

Plugging in the values:

X = -1.75 * 120 + 1200

Calculating this equation gives us:

X ≈ 1020

Therefore, the highest possible score a student who is dropped from the top university would have scored is approximately 1020.

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Skewness of the normal distribution. When it comes to normal distribution, the skewness is equal to zero.

Skewness is a measure of the distribution's symmetry. When a distribution is symmetric, the mean, median, and mode will all be the same. When a distribution is skewed, the mean will typically be larger or lesser than the median depending on whether the distribution is right-skewed or left-skewed. It is not appropriate to discuss mean or median in the case of normal distribution since it is a symmetric distribution.

Therefore, the answer is None of the above.

In normal distribution, the skewness is equal to zero, and it is not appropriate to discuss mean or median in the case of normal distribution since it is a symmetric distribution.

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The margin of error at a 98% confidence level is approximately 4.26.To find the margin of error (EBM - Error Bound for a Population Mean) at a 98% confidence level.

We need to use the formula:

Margin of Error = Z * (s / sqrt(n))

where Z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.

For a 98% confidence level, the corresponding z-score is 2.33 (obtained from the standard normal distribution table).

Plugging in the values into the formula:

Margin of Error = 2.33 * (10 / sqrt(30))

Calculating the square root and performing the division:

Margin of Error ≈ 2.33 * (10 / 5.477)

Margin of Error ≈ 4.26

Therefore, the margin of error at a 98% confidence level is approximately 4.26.

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