Select all ratios equivalent to 5:1

B. \( \sqrt{x} + \sqrt{x - 1} \) ✅

\(\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\orange{:}}}}}\)

\( \frac{1}{ \sqrt{x} - \sqrt{x - 1} } \\ \\= \frac{1}{ \sqrt{x} - \sqrt{x - 1} } \times \frac{ \sqrt{x} + \sqrt{x - 1} }{ \sqrt{x} + \sqrt{x - 1} } \\ \\ = \frac{ \sqrt{x} + \sqrt{x - 1} }{ ({ \sqrt{x} })^{2} - { (\sqrt{x - 1} })^{2} } \\\\ [∵(a + b)(a - b) = {a}^{2} - {b}^{2} ] \\ \\= \frac{ \sqrt{x} + \sqrt{x - 1} }{x - (x - 1)} \\ \\= \frac{ \sqrt{x} + \sqrt{x - 1} }{ x - x + 1} \\ \\= \sqrt{x} + \sqrt{x - 1} \)

\(\bold{ \green{ \star{ \orange{Mystique35}}}}⋆\)

The Special Product Formula for the "square of a sum" is \((A+B)^{2}\)= \(A^{2}\) + 2AB + \(B^{2}\) .

So

\((2x + 3)^{2}\)  = \((2x)^{2}\)   +2⋅(2x)⋅3+\(3^{2}\)  =4\(x^{2}\)  +12x+9.

​

The well-known for squaring the difference between two numbers or

\((a+b)^{2}\)=\(a^{2}\)+2ab+\(b^{2}\)  (1)

It can be calculated by multiplying the binomial a + b by one.

The square can also be obtained for a total of three sums:

\((a+b+c)^{2}\)=\(a^{2}\) + \(b^{2}\) + \(c^{2}\) + 2bc+ 2ca+ 2ab (2)

The contents of it might be described as the

Rule : The square of a sum is equal to the product of all the double products of the summands in twos plus the sum of all the squares of all the summands:        

(∑i ai)2=∑i \(a^{2}\)i+2∑i<j \(a_{i}\)\(a_{j}\)

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The given sequence consists of multiple patterns. In order to find the next term, we need to identify and apply the appropriate rules for each pattern separately.

The patterns observed include increments by a constant value, increments by alternate values, decrements by a constant value, and decrements by alternate values.

1. The first pattern consists of increments by a constant value of 4. Starting from 3, each term is obtained by adding 4 to the previous term: 3 + 4 = 7, 7 + 4 = 11, 11 + 4 = 15, and so on.

2. The second pattern involves increments by alternate values of 4 and 6. Starting from 7, the first increment is 4, and the second increment is 6. Therefore, the next term is obtained by adding 4, followed by adding 6 to the previous term: 19 + 4 = 23.

3. The third pattern includes increments by a constant value of 2. Starting from 2, each term is obtained by adding 2 to the previous term: 2 + 2 = 4, 4 + 2 = 6, and so on.

4. The fourth pattern consists of increments by alternate values of 3 and 4. Starting from 4, the first increment is 3, and the second increment is 4. Therefore, the next term is obtained by adding 3, followed by adding 4 to the previous term: 13 + 3 = 16.

5. The fifth pattern involves increments by a constant value of 2. Starting from 5, each term is obtained by adding 2 to the previous term: 5 + 2 = 7, 7 + 2 = 9, and so on.

6. The sixth pattern consists of increments by alternate values of 4 and 6. Starting from 5, the first increment is 4, and the second increment is 6. Therefore, the next term is obtained by adding 4, followed by adding 6 to the previous term: 17 + 4 = 21.

7. The seventh pattern includes decrements by a constant value of 1. Starting from 2, each term is obtained by subtracting 1 from the previous term: 2 - 1 = 1, 1 - 1 = 0, and so on.

8. The eighth pattern involves decrements by a constant value of 3. Starting from 15, each term is obtained by subtracting 3 from the previous term: 15 - 3 = 12, 12 - 3 = 9, and so on.

9. The ninth pattern consists of decrements by alternate values of 3 and 5. Starting from 1, the first decrement is 3, and the second decrement is 5. Therefore, the next term is obtained by subtracting 3, followed by subtracting 5 from the previous term: -8 - 3 = -11.

10. The tenth pattern involves decrements by a constant value of 10. Starting from 100, each term is obtained by subtracting 10 from the previous term: 100 - 10 = 90, 90 - 10 = 80, and so on.

By applying the identified rules to each pattern, we can find the next term in each sequence.

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Answer: Let's call Sarah's original amount of money as x. Then, Christine had 0.8x.

After they spent $684 altogether, Christine had 0.8x - 0.25(0.8x) = 0.8x - 0.2x = 0.6x left.

Sarah had x - 684.

Since Christine had thrice as much money left as Sarah, we can write an equation:

0.6x = 3(x - 684)

Expanding the right side of the equation:

0.6x = 3x - 2052

Solving for x:

2.6x = 2052

x = 787.69

So Sarah originally had 787.69 dollars and spent 787.69 - 684 = 103.69 dollars.

Step-by-step explanation:

Answer:

Both x and y have to be 90

Step-by-step explanation:

x and the 90 degree angle below it must be equal because the lines are parallel.

And y would also be the same as x

Answer: x = 90, y = 90

Step-by-step explanation:

         We can use vertical angles and corresponding angles to solve this question. Vertical angles are congruent to each other, so are corresponding angles. The little box shows that the angle is equal to 90 degrees.

   See attached.

Answer:

D: 63

Step-by-step explanation:

The value of ∠BAD would be as follows:

D). 63

Given that,

The measure of ∠ABC \(= 114\)

We know that,

BD bisects the ║gm ABCD

So,

∠ABD \(= 57\)

Since opposite angles of a ║gm give equal

∠ABC = ∠ADC

So,

∠ ADB \(= 57\)

Since the sum of all the angles of a ║gm is 360.

∠BAD can be found using this.

∵ ∠BAD \(= 63\)

Thus, option D is the correct answer.

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A polynomial is an equation made up of coefficients and in determinates that uses only the addition, subtraction, multiplication, and powers of positive-integer variables.

Multiplying Polynomials Find each product. 1) \(6v(2v + 3) 2) 7(-5v - 8)3) 2x(-2x -3)6v(2v + 3)\)

To distribute the 6v over the binomial 2v + 3, we multiply 6v by each term inside the parenthesis:

\(6v(2v) + 6v(3)\)

\(= 12v^2 + 18v7(-5v - 8)\)

To distribute the 7 over the binomial -5v - 8, we multiply 7 by each term inside the parenthesis:

\(7(-5v) + 7(-8)= -35v - 562x(-2x - 3)\)

To distribute the 2x over the binomial -2x - 3, we multiply 2x by each term inside the parenthesis:

\(2x(-2x) + 2x(-3)= -4x^2 - 6x\)

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Answer:

3x-8

Step-by-step explanation:

The researcher needs to use the critical value of t = 2.602 for constructing the 99% confidence interval around the mean.

To determine the critical value (t or z) for constructing a confidence interval, we need to consider two factors: the sample size and the desired level of confidence.

In this case, the researcher has 17 subjects and wants to construct a 99% confidence interval around a mean. The critical value depends on the distribution being used (t-distribution or standard normal distribution) and the degrees of freedom.

For small sample sizes (typically less than 30) or when the population standard deviation is unknown, the t-distribution is used. The degrees of freedom for the t-distribution are calculated as n - 1, where n is the sample size. In this case, the degrees of freedom would be 17 - 1 = 16.

To find the critical value for a 99% confidence interval using the t-distribution, we consult a t-table or use statistical software. Looking up the value for a 99% confidence level with 16 degrees of freedom, we find that the critical value is approximately 2.602.

Therefore, the researcher needs to use the critical value of t = 2.602 for constructing the 99% confidence interval around the mean.

It's important to note that if the sample size is large (typically greater than 30) and the population standard deviation is known, the standard normal distribution (z-distribution) can be used instead of the t-distribution. In that case, the critical value would be obtained directly from the standard normal distribution table or using the corresponding quantile function in statistical software.

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Answer:

7664025600

Step-by-step explanation:

I basically just multiplied first then added the sums. Also you can use a calculator to make sure ur answer is right.

Answer:

OPTION B

Step-by-step explanation:

SEE THE IMAGE FOR SOLUTION

HOPE IT HELPS

HAVE A GREAT DAY

The final solution for the integral is:

∫(xe^(kx))/(1+kx)^2 dx = -xe^(1+kx)/(k(1+kx)) + (1/k)∫e^(1+kx)/(1+kx) dx + D

If k = 0, the term (1/k)∫e^(1+kx)/(1+kx) dx simplifies to e^x + E.

To find the integral ∫(xe^(kx))/(1+kx)^2 dx, we can use integration by parts. Let's denote u = x and dv = e^(kx)/(1+kx)^2 dx. Then, we can find du and v using these differentials:

du = dx

v = ∫e^(kx)/(1+kx)^2 dx

Now, let's find the values of du and v:

du = dx

v = ∫e^(kx)/(1+kx)^2 dx

To find v, we can use a substitution. Let's substitute u = 1+kx:

du = (1/k) du

dx = (1/k) du

Now, the integral becomes:

v = ∫e^u/u^2 * (1/k) du

 = (1/k) ∫e^u/u^2 du

This is a well-known integral. Its antiderivative is given by:

∫e^u/u^2 du = -e^u/u + C

Substituting back u = 1+kx:

v = (1/k)(-e^(1+kx)/(1+kx)) + C

 = -(1/k)(e^(1+kx)/(1+kx)) + C

Now, we can apply integration by parts:

∫(xe^(kx))/(1+kx)^2 dx = uv - ∫vdu

                         = x(-(1/k)(e^(1+kx)/(1+kx)) + C) - ∫[-(1/k)(e^(1+kx)/(1+kx)) + C]dx

                         = -xe^(1+kx)/(k(1+kx)) + Cx + (1/k)∫e^(1+kx)/(1+kx) dx - ∫C dx

                         = -xe^(1+kx)/(k(1+kx)) + Cx + (1/k)∫e^(1+kx)/(1+kx) dx - Cx + D

                         = -xe^(1+kx)/(k(1+kx)) + (1/k)∫e^(1+kx)/(1+kx) dx + D

Now, let's focus on the integral (1/k)∫e^(1+kx)/(1+kx) dx. This integral does not have a simple closed-form solution for all values of k. However, we can compute it separately for specific values of k.

If k = 0, the integral becomes:

(1/k)∫e^(1+kx)/(1+kx) dx = ∫e dx = e^x + E

For k ≠ 0, there is no simple closed-form solution, and the integral cannot be expressed using elementary functions. In such cases, numerical methods or approximations may be used to compute the integral.

Therefore, the final solution for the integral is:

∫(xe^(kx))/(1+kx)^2 dx = -xe^(1+kx)/(k(1+kx)) + (1/k)∫e^(1+kx)/(1+kx) dx + D

If k = 0, the term (1/k)∫e^(1+kx)/(1+kx) dx simplifies to e^x + E.

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The expected number of defective items and the probability of at least 4 are defective is equal to 3.6 and 0.370 or 37.0%.

Total number of items 'n' = 18

Probability of an item being defective 'p' =20%

                                                                    = 0.2  

Expected number of defective items,

Use the formula for the expected value of a binomial distribution,

E(X) = np

where X is the number of defective items.

Plug in the values we have,

E(X) = 18 x 0.2

      = 3.6

Expect average items out of 18 to be defective = 3.6  .

Probability that at least 4 items are defective,

Calculate the probability of 4, 5, 6, ..., 18 defective items

Use the complement rule to simplify it,

P(at least 4 defective)

= 1 - P(less than 4 defective)

Using the CDF function,

'binomcdf' is the binomial cumulative distribution function.

18 is the number of trials,

0.2 is the probability of success,

And 3 is the maximum number of successes

P(less than 4 defective)

= binomcdf (18, 0.2, 3)

= P(X <= 3)

=\(\sum_{x=0}^{3}\) ¹⁸Cₓ × (0.2)^x × (0.8)^(18-x)

= ¹⁸C₀× (0.2)^0 × (0.8)^(18-0) + ¹⁸C₁× (0.2)^1 × (0.8)^(18-1) + ¹⁸C₂× (0.2)^2 × (0.8)^(18-2) + ¹⁸C₃× (0.2)^3 × (0.8)^(18-3)

= (0.8)^(18) + 18× (0.2) × (0.8)^(17) + 153 × (0.04) × (0.8)^(16) + 1632× (0.008) × (0.8)^(15)

= 0.630

Plug in the values,

P(at least 4 defective)

= 1 - 0.630

= 0.370

Therefore, the expected items to be defective and probability that at least 4 items out of 18 are defective is equal to 3.6 and 0.370 or 37.0%.

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Answer:

(2, 11)

Step-by-step explanation:

Using the information you were give, plug it into this formula:

(x₁ + x₂)/2, (y₁ + y₂)/2

This should give you a new coordinate.

Answer:

300xy, hope this helps ya

Answer:

1/5, 1/6, 1/7

Step-by-step explanation:

Just add 1 for the denominator for each numbers.

Answer:

Dividend=Divisor×Quotient+Remainder

So, Applying it:−

Let q(x),k(x) be quotient when f(x) is divided by x−1 and x−2 respectively

⇒f(x)=(x−1)q(x)+5

∴f(1)=5 ..... (1)

Also,f(x)=(x−2)k(x)+7

∴f(2)=7 ..... (2)

Now, let ax+b be the remainder when f(x) is divided by (x−1)(x−2) and g(x) be the quotient.

f(x)=(x−1)(x−2)g(x)+(ax+b)

Using (1) and (2)

5=a+b ...... (3)

7=2a+b ...... (4)

Solving (3) and (4), we get

a=2 and b=3

∴2x+3 is the remainder when f(x) is divided by (x−1)(x−2).

Answer: y= (1/4)x

Step-by-step explanation: Best guess given screenshot, matched with desmos.

For n ≥ 459, Algorithm A is better than Algorithm B when B uses n√n operations.

To determine the value of n₀ for which Algorithm A is better than Algorithm B when B uses n√n operations, we need to find the point at which the number of operations for Algorithm A is less than the number of operations for Algorithm B.

Algorithm A: 10n log n operations

Algorithm B: n√n operations

Let's set up the inequality and solve for n₀:

10n log n < n√n

Dividing both sides by n gives:

10 log n < √n

Squaring both sides to eliminate the square root gives:

100 (log n)² < n

To solve this inequality, we can use trial and error or graph the functions to find the intersection point. After calculating, we find that n₀ is approximately 459. Therefore, For n ≥ 459, Algorithm A is better than Algorithm B when B uses n√n operations.

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R-1.3: For \($n \geq 14$\), Algorithm A is better than Algorithm B when B uses \($n^2$\) operations.

R-1.4: Algorithm A is always better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

R-1.3:

Algorithm A: \($10n \log n$\) operations

Algorithm B: \($n^2$\) operations

We want to determine the value of \($n_0$\) such that Algorithm A is better than Algorithm B for \($n \geq n_0$\).

We need to compare the growth rates:

\($10n \log n < n^2$\)

\($10 \log n < n$\)

\($\log n < \frac{n}{10}$\)

To solve this inequality, we can plot the graphs of \($y = \log n$\) and \($y = \frac{n}{10}$\) and find the point of intersection.

By observing the graphs, we can see that the two functions intersect at \($n \approx 14$\). Therefore, for \($n \geq 14$\), Algorithm A is better than Algorithm B.

R-1.4:

Algorithm A: \($10n \log n$\) operations

Algorithm B: \($n\sqrt{n}$\) operations

We want to determine the value of \($n_0$\) such that Algorithm A is better than Algorithm B for \($n \geq n_0$\).

We need to compare the growth rates:

\($10n \log n < n\sqrt{n}$\)

\($10 \log n < \sqrt{n}$\)

\($(10 \log n)^2 < n$\)

\($100 \log^2 n < n$\)

To solve this inequality, we can use numerical methods or make an approximation. By observing the inequality, we can see that the left-hand side \($(100 \log^2 n)$\) grows much slower than the right-hand side \($(n)$\) for large values of \($n$\).

Therefore, we can approximate that:

\($100 \log^2 n < n$\)

For large values of \($n$\), the left-hand side is negligible compared to the right-hand side. Hence, for \($n \geq 1$\), Algorithm A is better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

So, for R-1.4, the value of \($n_0$\) is 1, meaning Algorithm A is always better than Algorithm B when B uses \($n\sqrt{n}$\) operations.

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The most she should pay for uniform annual maintenance to make it worthwhile to buy the van instead of leasing it is 21%.The answer is 21.

Question 1

Annual equivalent cost (AEC) is used to find the equivalent amount of uniform payments over the life of the equipment. It can be calculated by using the following formula:

\(AEC = (P - S)/A(P/A, i, n) + G\)

Where,

P = initial cost of the equipment

S = salvage value of the equipment

A = annual worth factor

G = uniform gradientP/A, i,

n = present value of an annuity factorG is the annual incremental increase or decrease in cost (i.e., the rate of increase or decrease in maintenance costs) each year.

The maintenance cost is estimated to be a uniform gradient of 12% per year for five years.

The annual equivalent cost (AEC) for the truck is as follows:

\(G = 0.12A = (P - S)/((P/A, i, n) + G)A = (12927 - 4417)/((P/A, i, n) + 0.12)\)

Using the P/A factor table, we obtain

\(P/A = 3.992 (n = 5 years, i = 20%)A = (12927 - 4417)/(3.992 + 0.12)AEC = $3,260.24\)

So, the annual equivalent cost (AEC) for the truck is $3,260.24.

Question 2

The present value of all expenses associated with buying and maintaining a vehicle, including depreciation, is referred to as the annual equivalent cost (AEC). The MARR is the minimum acceptable interest rate that a project or investment must yield in order to justify its capital investment.Annual payment (P) for leasing the van is $3,621

The initial cost of buying a van (P) is $5,902

We neeed to find the Salvage value (S) is $1,268N is 3 years G = ?

The AEC for buying a van can be calculated as follows:

\(AEC (Buy) = (P - S)/A(P/A, i, n) + G\)

Using the P/A factor table for n = 3 and i = 20%, we get

\(P/A = 2.59AEC (Buy) = (5902 - 1268)/ (2.59 + G)\)

Similarly, for leasing the van, the AEC can be calculated as follows:

\(AEC (Lease) = P/A(P/A, i, n) = 3,621/2.60 = $1,393.85\)

We need to calculate the uniform annual maintenance that she should pay to make it worthwhile to buy the van instead of leasing it.Therefore,

\(AEC (Buy) = AEC (Lease) => (5902 - 1268)/(2.59 + G) = 3621/2.60 => 2.59(3621) = 2242 + 871.15G => G = 0.21 or 21%\)

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Answer:

Equation:  \(\frac{7}{3}t + \frac{5}{2}t = 87\)

Hours: 18

Step by Step:

(7/3)t + (5/2)t = 87

(14/6)t + (15/6)t = 87

(29/6)t = 87

29t=522

t=28

The equation of the line in slope-intercept form is: y = -4x - 23

To obtain the equation of a line parallel to the line y = -4x + 2 and passing through the point (-4, -7), we can use the fact that parallel lines have the same slope.

The provided line has a slope of -4. Therefore, the parallel line will also have a slope of -4.

1. Point-slope form:

The point-slope form of a linear equation is written as y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

Using the point (-4, -7) and the slope -4, we can write the equation in point-slope form:

y - (-7) = -4(x - (-4))

Simplifying:

y + 7 = -4(x + 4)

2. Slope-intercept form:

The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept.

Using the slope -4 and the point (-4, -7), we can substitute these values into the slope-intercept form:

y = -4x + b

Now, to obtain the value of b, substitute the coordinates of the provided point (-4, -7):

-7 = -4(-4) + b

Simplifying:

-7 = 16 + b

b = -7 - 16

b = -23

Therefore, the equation of the line in slope-intercept form is:

y = -4x - 23

Both the point-slope form and the slope-intercept form describe the same line passing through (-4, -7) and parallel to the line y = -4x + 2.

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Here are the expressions and their values when m = 6:

In mathematics, the term quotient refers to the result obtained when one quantity is divided by another quantity. The quotient is the number of times the divisor goes into the dividend exactly, or as close as possible without going over.

For example, if we divide 10 by 2, the quotient is 5, because 2 goes into 10 exactly 5 times with no remainder. The quotient is the whole number part of the result of division, as opposed to the remainder, which is the amount left over after the division is performed.

In symbols, we can represent the quotient of a dividend a and a divisor b as a/b, where the result is the number of times b goes into a without remainder.

Using m=6, we can evaluate each expression:

6 / (6 - 3) = 2

(6 + 3) / (6 - 3) = 3

3m + 4m = 7m = 7(6) = 42

3m - (m / 2) = 3(6) - (6 / 2) = 18 - 3 = 15

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The expression that is equivalent to 135−−−√5–√ is: 131.764.

An equivalent expression is one that has the same value as another. To evaluate the given expression, the three minus signs all equate to -. So, the question is asking that we subtract the root of 5 from 135.

Also, the root of minus 1 will be subtracted from the answer that we arrive at.

135−−−√5 = 132.764

132.764 - √1 = 131.764

So, the decimal equivalent of this expression is 131.764.

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It all depends on how you want your hamburger. If you're serving it as a main dish, 300 hamburger patties (each around 4 ounces) should be plenty to feed 300 people. If you utilize hamburger meat as a topping or filler in a meal, you'll need a different amount.

Arithmetic operations is a field of mathematics that studies numbers and the operations on numbers that are helpful in all other disciplines of mathematics. It consists mostly of operations like addition, subtraction, multiplication, and division. Arithmetic Operator is used to conduct mathematical operations on the supplied operands such as addition, subtraction, multiplication, division, modulus, and so on. The four basic arithmetic operations are addition, subtraction, multiplication, and division. Arithmetic mean is a number calculated by dividing the sum of a set's elements by the number of values in the set.

Here,

It all depends on how you want to eat the hamburger. If you're serving it as a main course, 300 hamburger patties, each weighing around 4 ounces, should plenty to satisfy 300 people. It will need a different quantity of hamburger meat if you use it as a topping or filler in a dish.

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Complete question:

How much hamburger will it take to make 300 three oz patties with a 20% shrinkage?

We constructed an LPP such that A has rank 3 and none of the variables are slack variables. We then observed that the M-method and the two-phase method are not required to solve this LPP since we have already ensured that it is feasible.

Linear Programming Problems (LPP) can be solved by various methods such as graphical method, simplex method, dual simplex method, and so on. However, some LPPs require different methods based on the characteristics of the problem. One such example is when the rank of matrix A is 3 and none of the existing variables are slack variables. This question asks us to construct an LPP by selecting a suitable c, A (a 5 x 7 matrix), and b such that it looks like:Max Z = cxSubject to Ax = bb ≥ 0 and x ≥ 0And with the conditions that A should have rank 3 and none of the existing variables are slack variables.Let's start by selecting a matrix A. Since A should have rank 3, we can select a 5x7 matrix with rank 3. Let A be the following 5x7 matrix:$$\begin{bmatrix}1 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 1\end{bmatrix}$$Note that we have selected a matrix A such that none of the columns are all zeros. This is important to ensure that none of the variables are slack variables.Now let's select a vector b. Since we have a 5x7 matrix A, b should be a 5x1 vector. Let b be the following vector:$$\begin{bmatrix}2\\ 3\\ 4\\ 5\\ 6\end{bmatrix}$$Finally, we need to select a vector c. Since we want to maximize Z, c should be a 1x7 vector. Let c be the following vector:$$\begin{bmatrix}1 & 1 & 1 & 1 & 1 & 1 & 1\end{bmatrix}$$Now we can write the LPP as follows:Max Z = x1 + x2 + x3 + x4 + x5 + x6 + x7Subject to:x1 + x3 ≥ 2x2 + x4 ≥ 3x5 ≥ 4x3 + x6 ≥ 5x4 + x7 ≥ 6x1, x2, x3, x4, x5, x6, x7 ≥ 0Note that none of the variables are slack variables. Also, the LPP is feasible since x = [2, 3, 0, 5, 4, 6, 0] satisfies all the constraints and has a non-negative value for each variable.Now, let's see what happens when we use the M-method and the two-phase method to solve this LPP.M-method:When we use the M-method, we first add artificial variables to the LPP to convert it to an auxiliary LPP. The auxiliary LPP is then solved using the simplex method. If the optimal value of the auxiliary LPP is zero, then the original LPP is feasible. Otherwise, the original LPP is infeasible.Note that we have already ensured that the LPP is feasible. Therefore, the M-method is not required in this case.Two-phase method:When we use the two-phase method, we first convert the LPP into an auxiliary LPP. The auxiliary LPP is then solved using the simplex method. If the optimal value of the auxiliary LPP is zero, then the original LPP is feasible. Otherwise, the original LPP is infeasible and the two-phase method fails.Note that we have already ensured that the LPP is feasible. Therefore, the two-phase method is not required in this case.

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A linear programming problem (LPP) can be constructed by selecting appropriate c, A (a 5 x 7 matrix), and b so that it appears as follows:

Max Z = cx

Subject to  Ax = bb ≥ 0 and x ≥ 0 with the constraint that A must have a rank of 3 and none of the existing variables are slack variables.  

LPP is a technique for optimizing a linear objective function that is subject to linear equality and linear inequality constraints.

A linear programming problem, as the name implies, requires a linear objective function and linear inequality constraints.

Methods: M-Method and Two-Phase Method:

M-method:M-method is a linear programming technique for generating a basic feasible solution for a linear programming problem.
For a variety of LPPs, the M-method may be used to produce an initial fundamental feasible solution. It works by reducing the number of constraints in the problem by adding artificial variables and constructing an auxiliary linear programming problem.

Two-phase Method:This method solves linear programming problems using an initial feasible basic solution.

Phase I of this technique entails adding artificial variables to the system and using simplex methods to determine a fundamental feasible solution.

Phase II involves determining the optimum fundamental feasible solution to the original problem using the simplex method based on the original problem's constraints and objective function.

Both the M-method and the two-phase approach are methods for generating an initial fundamental feasible solution in linear programming.

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We can conclude that the socioeconomic class of the offender is related to the likelihood of arrests.

Based on the information provided, we can conclude that there is a significant relationship between arrests and the socioeconomic class of the offender.
Identify the chi-square critical score: 9.488

Identify the chi-square test statistic: 12.2
Compare the test statistic to the critical score:

If the test statistic (12.2) is greater than the critical score (9.488), then there is a significant relationship between the variables.
In this case, 12.2 > 9.488, so there is a significant relationship between arrests and the socioeconomic class of the offender.
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Step-by-step explanation:

Heya mate the answer is 30 degree....

See the attachment

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