(2) Consider the following LP. max s.t. z=2x1+3x2,,x1+2x2≤30, x1+x2≤20 ,x1,x2≥0 (a) Solve the problem graphically (follow the steps of parts (a)-(c) in problem (1)). (2.5 points) (b) Write the standard form of the LP. (c) Solve the LP via Simplex and write the optimal solution and optimal value.

Answers

Answer 1

The graphical solution and simplex method were used to solve the given linear programming problem. The optimal solution is (x1, x2) = (0, 2) with an optimal value of z = 70.0.

Given the LP, max z = 2x1 + 3x2

Subject to:

x1 + 2x2 ≤ 30

x1 + x2 ≤ 20

x1, x2 ≥ 0

(a) Solve the problem graphically:

Follow the steps of parts (a)-(c) in problem (1).

To solve the given problem graphically, follow these steps:

Step 1: Solve the equation x1 + 2x2 = 30.

This is the equation of the line passing through points (0, 15) and (30, 0). This line divides the feasible region into two parts - one on the upper side and one on the lower side.

Step 2: Solve the equation x1 + x2 = 20.

This is the equation of the line passing through points (0, 20) and (20, 0). This line divides the feasible region into two parts - one on the left side and one on the right side.

Step 3: Identify the feasible region.

The feasible region is the region that satisfies all the constraints of the given LP. It is the intersection of the two half-planes formed in Steps 1 and 2. The feasible region is shown below:

Step 4: Identify the objective function.

The objective function is z = 2x1 + 3x2. We need to maximize z.

Step 5: Draw the lines of constant z.

To maximize z, we need to draw lines of constant z. We can do this by selecting different values of z and then solving the equation 2x1 + 3x2 = z. The table below shows some values of z and their corresponding lines of constant z.

Step 6: Identify the optimal solution.

The optimal solution is the solution that maximizes the objective function z and lies on the boundary of the feasible region. In this case, the optimal solution is at the intersection of lines z = 12 and x1 + 2x2 = 30. The optimal solution is (12, 9). The optimal value is z = 39.

(b) Write the standard form of the LP:

The standard form of the LP is:

max z = 2x1 + 3x2

Subject to:

x1 + 2x2 ≤ 30

x1 + x2 ≤ 20

x1, x2 ≥ 0

(c) Solve the LP via Simplex and write the optimal solution and optimal value:

The initial simplex table is shown below:

BV x1 x2 s1 s2 RHS R

s1 1 2 1 0 30 0

s2 1 1 0 1 20 0

z -2 -3 0 0 0 0

The pivot column is x1, and the pivot row is R1. The pivot element is 1. We apply the following operations:

R1 → R1 - 2R2

s1 → s1 - 2s2

z → z - 2s2

The resulting simplex table is shown below:

BV x1 x2 s1 s2 RHS R

s1 -3/2 0 1 -1/2 10 6

s2 1/2 1 0 1/2 10 3

z -5 0 0 1 60 30

The pivot column is x2, and the pivot row is R2. The pivot element is 1/2. We apply the following operations:

R2 → 2R2

x1 → x1 + 3x2

s2 → s2 - (1/2)s1

z → z + 5x2 - (5/2)s1

The resulting simplex table is shown below:

BV x1 x2 s1 s2 RHS R

s1 -9/5 0 1/5 -1/5 4 6/5

x2 1/5 1 0 1/5 2 3/5

z 0 5 5/2 5/2 70 70

The optimal solution is (x1, x2) = (0, 2) and the optimal value is z = 70.

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Related Questions

Let f(x)=5x^2
(a) Use the limit process to find the slope of the line tangent to the graph of f at x=1. Slope at x=1 : (b) Find an equation of the line tangent to the graph of f at x=1. Tangent line: y=

Answers

Answer: Slope at x=1: 10Tangent line: y = 10x - 5

Let f(x)=5x^2

(a) Use the limit process to find the slope of the line tangent to the graph of f at x=1To find the slope of the line tangent to the graph of f at x=1, we will differentiate the function f(x) using the limit process.

We have the equation of the function f(x) as; f(x) = 5x^2To differentiate the equation of f(x) using the limit process, we need to follow the following steps;

Step 1: Let x → a, where a = 1, then h → 0

Step 2: Find the difference quotient of the function f(x)f(x + h) - f(x)/h = [5(x + h)^2 - 5x^2]/h

= [5(x^2 + 2xh + h^2) - 5x^2]/h

Step 3: Simplify the above expression(5x^2 + 10xh + 5h^2 - 5x^2)/h

= 10x + 5h

Step 4: Let h → 0, then the slope at x=1 is given by lim(h → 0) [10x + 5h]

= 10(1) + 5(0)

= 10

Therefore, the slope of the line tangent to the graph of f at x=1 is 10.

Slope at x=1: 10

(b) Find an equation of the line tangent to the graph of f at x=1.

Tangent line: y=To find an equation of the line tangent to the graph of f at x=1, we will use the point-slope form of the equation of the line.

The slope of the tangent line at x=1 is 10, and the point (1,5) lies on the tangent line.

Therefore, the equation of the line tangent to the graph of f at x=1 is; y - 5 = 10(x - 1)y - 5

= 10x - 10y

= 10x - 5

The required equation of the line tangent to the graph of f at x=1 is y = 10x - 5.

Answer: Slope at x=1: 10Tangent line: y = 10x - 5

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a statistics professor has 115 students in a statistics class and would like to estimate the number of hours each student studied for the last exam. a random sample of 41 students was found to study an average of 7.3 hours with a standard deviation of 1.9 hours. the 98% confidence interval to estimate the average number of hours studying for the exam would be..

a- 5.18 and 9.42

b- 6.72 and 7.88

c- 5.82 and 8.79

d- 6.11 and 8.49

Answers

The 98% confidence interval to estimate the average number of hours studying for the exam is approximately 6.61 to 7.99.

Hence option D is correct.

Given that,

Number of students in the statistics class: 115

Sample size: 41 students

Average number of hours studied by the sample: 7.3 hours

Standard deviation of the sample: 1.9 hours

Desired confidence level: 98%

To accurately the problem and calculate the 98% confidence interval,

Use the formula:

Confidence Interval = Sample Mean ± (Z * Standard Error)

Where:

Sample Mean is the average number of hours studied by the sample (7.3 hours).

Z is the critical value corresponding to the desired confidence level (98%). For a 98% confidence level, the Z-value is approximately 2.326.

Standard Error is calculated by dividing the standard deviation of the sample (1.9 hours) by the square root of the sample size (41 students).

Calculate the confidence interval: Standard Error = 1.9 / √41 ≈ 0.2965

Confidence Interval = 7.3 ± (2.326 x 0.2965)

Now, Calculate the upper and lower bounds of the confidence interval:

Upper Bound = 7.3 + (2.326 * 0.2965) ≈ 7.3 + 0.6895 ≈ 7.9895

Lower Bound = 7.3 - (2.326 * 0.2965) ≈ 7.3 - 0.6895 ≈ 6.6105

Therefore, the 98% confidence interval to estimate the average number of hours studying for the exam is approximately 6.61 to 7.99.

Based on the given options, the correct answer would be:

d- 6.11 and 8.49

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In the following frequency distribution table, locate and solve the missing value:
Classes Frequency Cumulative
Frequency Percent
5 - 9 4 10.000%
10 - 14 6 25.000%
15 - 19 3 32.500%
20 - 24 7 25 - 29 15 87.500%
30 - 34 5 100.000%

Answers

a negative frequency is not possible, it indicates an error in the given data. Please verify the data or provide additional information to rectify the issue.

To solve the missing value in the frequency distribution table, we need to find the frequency for the class interval "25 - 29."

Given that the cumulative frequency for the previous class interval "20 - 24" is 7 and the cumulative frequency for the class interval "30 - 34" is 5, we can calculate the missing frequency by subtracting the cumulative frequency of the previous class from the cumulative frequency of the next class.

Missing Frequency = Cumulative Frequency (30 - 34) - Cumulative Frequency (20 - 24)

Missing Frequency = 5 - 7

Missing Frequency = -2

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The solution of \( y^{\prime}=(\cos y)^{2} x^{i} \) The following problem \( y^{n} x d x+x^{2} y d y=0 \) is exact when \( n \) is

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The equation [tex]\(y^{n} x dx + x^{2} y dy = 0\[/tex]) is exact for different values of n, depending on the value of y.

The given differential equation is \(y^{n} x dx + x^{2} y dy = 0\[tex]\(y^{n} x dx + x^{2} y dy = 0\[/tex]

To determine when the equation is exact, we can check if the following condition is satisfied:

[tex]\(\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\)[/tex]

where M is the coefficient of dx and N is the coefficient of dy.

In this case, we have [tex]M = y^n x and N = x^2 y.[/tex]

Taking the partial derivatives, we get:

[tex]\(\frac{{\partial M}}{{\partial y}} = n y^{n-1} x\)\(\frac{{\partial N}}{{\partial x}} = 2x y\)[/tex]

For the equation to be exact, \(\frac{{\partial M}}{{\partial y}}\) should be equal to \(\frac{{\partial N}}{{\partial x}}\).

Therefore, we have the equation:

[tex]\(n y^{n-1} x = 2x y\)[/tex]

Simplifying, we can cancel out the common factors:

[tex]\(ny^{n-1} = 2\)[/tex]

From this equation, we can solve for n:

(ny^{n-1} = 2\)[tex]\(ny^{n-1} = 2\)[/tex]

The value of n that satisfies this equation depends on the specific value of y. It is not a fixed value but rather varies with y.

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Write an equation of the line passing through (−2,4) and having slope −5. Give the answer in slope-intercept fo. The equation of the line in slope-intercept fo is For the function f(x)=x2+7, find (a) f(x+h),(b)f(x+h)−f(x), and (c) hf(x+h)−f(x)​. (a) f(x+h)= (Simplify your answer.) (b) f(x+h)−f(x)= (Simplify your answer.) (c) hf(x+h)−f(x)​= (Simplify your answer.)

Answers

The equation of the line passing through (−2,4) and having slope −5 is y= -5x-6. For the function f(x)= x²+7, a) f(x+h)= x² + 2hx + h² + 7, b) f(x+h)- f(x)= 2xh + h² and c) h·[f(x+h)-f(x)]​= h²(2x + h)

To find the equation of the line and to find the values from part (a) to part(c), follow these steps:

The formula to find the equation of a line having slope m and passing through (x₁, y₁) is y-y₁= m(x-x₁). Substituting m= -5, x₁= -2 and y₁= 4 in the formula, we get y-4= -5(x+2) ⇒y-4= -5x-10 ⇒y= -5x-6. Therefore, the equation of the line in the slope-intercept form is y= -5x-6.(a) f(x+h) = (x + h)² + 7 = x² + 2hx + h² + 7(b) f(x+h)-f(x) = (x+h)² + 7 - (x² + 7) = x² + 2xh + h² + 7 - x² - 7 = 2xh + h²(c) h·[f(x+h)-f(x)]​ = h[(x + h)² + 7 - (x² + 7)] = h[x² + 2hx + h² + 7 - x² - 7] = h[2hx + h²] = h²(2x + h)

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Differentiate.
f(x) = 3x(4x+3)3
O f'(x) = 3(4x+3)²(16x + 3)
O f'(x) = 3(4x+3)³(7x+3)
O f'(x) = 3(4x+3)2
O f'(x) = 3(16x + 3)²

Answers

The expression to differentiate is f(x) = 3x(4x+3)³. Differentiate the expression using the power rule and the chain rule.

Then, show your answer.Step 1: Use the power rule to differentiate 3x(4x+3)³f(x) = 3x(4x+3)³f'(x) = (3)(4x+3)³ + 3x(3)[3(4x+3)²(4)]f'(x) = 3(4x+3)³ + 36x(4x+3)² .

Simplify the expressionf'(x) = 3(4x+3)²(16x + 3): The value of f'(x) = 3(4x+3)²(16x + 3).The process above was a  since it provided the method of differentiating the expression f(x) and the final value of f'(x). It was  as requested in the question.

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Does the equation specify a function with independent variable x ? If so, find the domain of the function. If not, find a value of x to which there corresponds more than one value of y. y(x+y)=4

Answers

The equation does not specify a function with independent variable x and the domain of the function is all real numbers.

The given equation is y(x + y) = 4. In the given equation, we have two variables, x and y. To check whether the equation specifies a function with independent variable x, let's assume y to be a function of x. Then we can write y as follows:

y = f(x)

Substituting this value of y in the given equation:

y(x + y) = 4x + f(x) + [f(x)]² = 4

This is a quadratic equation of f(x). The general form of a quadratic equation is:

ax² + bx + c = 0

where a, b, and c are constants.

In this case, we have:

x² + 2x f(x) + [f(x)]² - 4 = 0

Now let's find the discriminant of the above equation:

D = b² - 4ac

   = 4 - 4[f(x)]² - 4(-4)

   = 16 - 4[f(x)]²

The discriminant must be greater than or equal to zero for the equation to have real solutions. So we have:

16 - 4[f(x)]² ≥ 0[f(x)]² ≤ 4f(x) ≤ ±2

Let's take the positive value for simplicity:

      f(x) ≤ 2

If we draw the graph of this quadratic function, we'll find that it is a downward-facing parabola, which means that there will be a value of x for which there corresponds more than one value of y. So the equation does not specify a function with independent variable x. Now let's find that value of x:

Let's assume y = k (a constant). Then we can write:

y(x + k) = 4x + ky² + kx - 4 = 0

This is a quadratic equation of y. Let's find the discriminant of this equation:

D = b² - 4ac= k² - 4(x)(kx - 4)= k² - 4kx + 16

Let's make this discriminant zero:

16 - 4kx + k² = 0kx = (k² + 16)/4

For any value of k, we can find a value of x that satisfies this equation.

Therefore, there corresponds more than one value of y for this value of x. Hence, the equation does not specify a function with independent variable x. The domain of the function is all real numbers.

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given a function f : a → b and subsets w, x ⊆ a, then f (w ∩ x) = f (w)∩ f (x) is false in general. produce a counterexample.

Answers

Therefore, f(w ∩ x) = {0} ≠ f(w) ∩ f(x), which shows that the statement f(w ∩ x) = f(w) ∩ f(x) is false in general.

Let's consider the function f: R -> R defined by f(x) = x^2 and the subsets w = {-1, 0} and x = {0, 1} of the domain R.

f(w) = {1, 0} and f(x) = {0, 1}, so f(w) ∩ f(x) = {0}.

On the other hand, w ∩ x = {0}, and f(w ∩ x) = f({0}) = {0}.

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Consider the line in R3 which
goes through the points (1, 2, 5) and (4, −2, 3). Does this line
intersect the sphere with radius 3 centered at (0, 1, 2), and if
so, where?
2. [Intersections] Consider the line in bb{R}^{3} which goes through the points (1,2,5) and (4,-2,3) . Does this line intersect the sphere with radius 3 centered at (0,1,2)

Answers

To determine if the line in [tex]R^3[/tex], which goes through the points (1, 2, 5) and (4, -2, 3), intersects the sphere with radius 3 centered at (0, 1, 2), we can find the equation of the line and the equation of the sphere, and then check for their intersection.

1. Equation of the line:

Direction vector = (4, -2, 3) - (1, 2, 5) = (3, -4, -2)

x = 1 + 3t

y = 2 - 4t

z = 5 - 2t

2. Equation of the sphere:

[tex](x - a)^2 + (y - b)^2 + (z - c)^2 = r^2x^2 + (y - 1)^2 + (z - 2)^2 = 3^2[/tex]

3. Finding the intersection:

[tex](1 + 3t)^2 + (2 - 4t - 1)^2 + (5 - 2t - 2)^2 = 9[/tex]

Simplifying the equation:

[tex]9t^2 - 9t - 16 = 0[/tex]

Solving this quadratic equation, we find two values for t: t = 1 and t = -2/3.

Substituting these values:

For t = 1:

x = 1 + 3(1) = 4

y = 2 - 4(1) = -2

z = 5 - 2(1) = 3

For t = -2/3:

x = 1 + 3(-2/3) = -1

y = 2 - 4(-2/3) = 4

z = 5 - 2(-2/3) = 9/3 = 3

Therefore, the line intersects the sphere at the points (4, -2, 3) and (-1, 4, 3).

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which function has the same range as
f(x)=−5/7(3/5)x ?

answer choices:
g(x)= 5/7(3/5) -x

g(x)= -5/7(3/5) -x

g(x)= 5/7(3/5)x

g(x)= -(-5/7)(5/3)x

Answers

Answer:

The range of a function is the set of all possible output values. To find a function with the same range as f(x) = (-5/7)(3/5)x, we need to find a function g(x) such that the output values of g(x) are the same as the output values of f(x).

Notice that the function f(x) is a linear function with slope (-5/7)(3/5) = -3/7, and y-intercept of 0. Therefore, any function with the same slope and y-intercept of 0 will have the same range as f(x).

Out of the given answer choices, we can see that the function g(x) = 5/7(3/5)x has the same slope as f(x) but the y-intercept is different (it is also 0). Therefore, g(x) = 5/7(3/5)x has the same range as f(x).

So, the answer is g(x) = 5/7(3/5)x.

Consider the exponential distribution with probability density function (PDF) f(t)=ae
−at
where a>0 is some unknown constant. Compute the probability some arbitrary draw y is greater than 2 when a=3, i.e. p(y>2). Note that the exponential distribution is bounded below by 0 . Enter your answer as a probability to 4 decimal places.

Answers

The probability that an arbitrary draw y is greater than 2 when a=3, i.e. P(y>2) is 0.0025 (approx)

The exponential distribution with probability density function (PDF) f(t)=ae-at, where a>0 is an unknown constant. Here, we need to compute the probability that some arbitrary draw y is greater than 2 when a=3, i.e. P(y>2)

We can use the formula of the cumulative distribution function(CDF), which is given by:

[tex]$F_{X}(x)=\int_{0}^{x}f_{X}(t) dt$[/tex]

to solve the problem. Thus, the CDF for an exponential distribution with parameter a is given by:

[tex]$F_{X}(x)

= \int_{0}^{x} f_{X}(t) dt

= \int_{0}^{x} ae^{-at} dt

= [-e^{-at}]_{0}^{x}

= 1 - e^{-ax}$[/tex]

We need to calculate the probability that y is greater than 2, i.e.

[tex]P(y>2).Thus, P(y>2)

= 1 - P(y<2)

The, P(y>2)

= 1 - F(2)

= 1 - (1 - e-2a)

= e-2a[/tex]

Now, a=3, substitute a=3 in the above equation.

P(y>2) = e-6 = 0.0025 (approx.)

The probability that an arbitrary draw y is greater than 2 when a=3, i.e. P(y>2) is 0.0025 (approx).

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13% of all Americans live in poverty. If 34 Americans are randomly selected, find the probability that a. Exactly 3 of them live in poverty. b. At most 1 of them live in poverty. c. At least 33 of them live in poverty.

Answers

Given data:

13% of all Americans live in poverty, n = 34 Americans are randomly selected.

In probability, we use the formula: P(E) = n(E)/n(A)Where, P(E) is the probability of an event (E) happeningn(E) is the number of ways an event (E) can happen

(A) is the total number of possible outcomes So, let's solve the given problems.

a) Exactly 3 of them live in poverty.The probability of 3 Americans living in poverty is given by the probability mass function of binomial distribution:

P(X = 3) = (34C3) × (0.13)³ × (0.87)³¹≈ 0.1203Therefore, the probability that exactly 3 of them live in poverty is 0.1203.

b) At most 1 of them live in poverty. The probability of at most 1 American living in poverty is equal to the sum of the probabilities of 0 and 1 American living in poverty:

P(X ≤ 1) = P(X = 0) + P(X = 1)P(X = 0) = (34C0) × (0.13)⁰ × (0.87)³⁴P(X = 1) = (34C1) × (0.13)¹ × (0.87)³³≈ 0.1068Therefore, the probability that at most 1 of them live in poverty is 0.1068.

c) At least 33 of them live in poverty.The probability of at least 33 Americans living in poverty is equal to the sum of the probabilities of 33, 34 Americans living in poverty:

P(X ≥ 33) = P(X = 33) + P(X = 34)P(X = 33) = (34C33) × (0.13)³³ × (0.87)¹P(X = 34) = (34C34) × (0.13)³⁴ × (0.87)⁰≈ 5.658 × 10⁻⁵Therefore, the probability that at least 33 of them live in poverty is 5.658 × 10⁻⁵.

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What is the measure of angle4? mangle4 = 40° mangle4 = 48° mangle4 = 132° mangle4 = 140°

Answers

The measure of angle 4 is 48 degree.

We have,

measure of <1= 48 degree

Now, from the given figure

<1 and <4 are Vertical Angles.

Vertical angles are a pair of opposite angles formed by the intersection of two lines. When two lines intersect, they form four angles at the point of intersection.

Vertical angles are always congruent, which means they have equal measures.

Then, using the property

<1 = <4 = 48 degree

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he highest recorded temperaturein the world was 38.0\deg C in El Azizia , Libya, on September 13, 1922. Calculate in degrees farenheit.

Answers

The highest recorded temperature in the world, 38.0°C in El Azizia, Libya, on September 13, 1922, is equivalent to 100.4°F.

The Fahrenheit scale divides the temperature range between these two points into 180 equal divisions or degrees. Each degree Fahrenheit is 1/180th of the temperature difference between the freezing and boiling points of water.

To convert Celsius to Fahrenheit, we use the formula:

°F = (°C × 9/5) + 32

Given that the temperature is 38.0°C, we can substitute this value into the formula:

°F = (38.0 × 9/5) + 32

°F = (342/5) + 32

°F = 68.4 + 32

°F = 100.4

Therefore, the highest recorded temperature in El Azizia, Libya, on September 13, 1922, was 100.4°F.

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A researcher fits a linear regression model and finds that the correlation coefficient is 0.95. Which of the following is NOT correct.
O A significant linear relationship exists between the response variable and the explanatory variables.
O The high correlation indicates that the linear model is a good model.
O More data exploration should be performed to justify the linear model.
O The linear model might not be the best model.

Answers

The statement "The high correlation indicates that the linear model is a good model" is NOT correct.

While a high correlation coefficient (in this case, 0.95) suggests a strong linear relationship between the variables, it does not necessarily indicate that the linear model is a good model. Correlation measures the strength and direction of the linear relationship but does not account for other important factors such as model assumptions, goodness-of-fit measures, or the presence of influential outliers.

Therefore, it is possible that other considerations, such as further data exploration, assessing model assumptions, evaluating goodness-of-fit measures (e.g., R-squared, residual analysis), and considering alternative models, need to be performed to determine if the linear model is indeed a good model. So, the correct statement is "More data exploration should be performed to justify the linear model."

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Find the measure of the arc or central angle indicated. Assume that lines which appear to be.

Answers

The measure of angle ∠HKF is equal to 87°

A straight angle is that of 180° and is formed on a straight line.

Linear pair of angles are formed when two lines intersect with each other at a single point. The sum of angles of a linear pair is always equal to 180°.

In the given figure,

∠JKF + ∠GKF = 180° since they together form the straight line JG.

given that ∠JKF  = 135°

∠GKF = 180° - ∠JKF  = 180° -  135°  = 45°

Now,  ∠HKF =  ∠GKF +  ∠HKG

given, ∠HKG = 42°

and now we know that ∠GKF = 45°

So, ∠HKF = 87°

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Let B_{1}=\{1,2\}, B_{2}=\{2,3\}, ..., B_{100}=\{100,101\} . That is, B_{i}=\{i, i+1\} for i=1,2, \cdots, 100 . Suppose the universal set is U=\{1,2, ..., 101\} . Determine

Answers

The solutions are: A. $\overline{B_{13}}=\{1,2,...,12,15,16,...,101\}$B. $B_{17}\cup B_{18}=\{17,18,19\}$C. $B_{32}\cap B_{33}=\{33\}$D. $B_{84}^C=\{1,2,...,83,86,...,101\}$.

The given question is as follows. Let $B_1=\{1,2\}, B_2=\{2,3\}, ..., B_{100}=\{100,101\}$. That is, $B_i=\{i,i+1\}$ for $i=1,2,…,100$. Suppose the universal set is $U=\{1,2,...,101\}$. Determine. In order to find the solution to the given question, we have to find out the required values which are as follows: A. $\overline{B_{13}}$B. $B_{17}\cup B_{18}$C. $B_{32}\cap B_{33}$D. $B_{84}^C$A. $\overline{B_{13}}$It is known that $B_{13}=\{13,14\}$. Hence, $\overline{B_{13}}$ can be found as follows:$\overline{B_{13}}=U\setminus B_{13}= \{1,2,...,12,15,16,...,101\}$. Thus, $\overline{B_{13}}=\{1,2,...,12,15,16,...,101\}$.B. $B_{17}\cup B_{18}$It is known that $B_{17}=\{17,18\}$ and $B_{18}=\{18,19\}$. Hence,$B_{17}\cup B_{18}=\{17,18,19\}$

Thus, $B_{17}\cup B_{18}=\{17,18,19\}$.C. $B_{32}\cap B_{33}$It is known that $B_{32}=\{32,33\}$ and $B_{33}=\{33,34\}$. Hence,$B_{32}\cap B_{33}=\{33\}$Thus, $B_{32}\cap B_{33}=\{33\}$.D. $B_{84}^C$It is known that $B_{84}=\{84,85\}$. Hence, $B_{84}^C=U\setminus B_{84}=\{1,2,...,83,86,...,101\}$.Thus, $B_{84}^C=\{1,2,...,83,86,...,101\}$.Therefore, The solutions are: A. $\overline{B_{13}}=\{1,2,...,12,15,16,...,101\}$B. $B_{17}\cup B_{18}=\{17,18,19\}$C. $B_{32}\cap B_{33}=\{33\}$D. $B_{84}^C=\{1,2,...,83,86,...,101\}$.

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Evaluate (Standard Normal Distribution)

a) P(Z<1. 02)

b) P(Z>1. 98)

c)P(Z>-1. 26)

d) P(Z>-1. 52)

e)P(0. 38

f)P(-0. 91

g)P(-1. 97

h)P(0

Answers

a) P(Z<1.02) = 0.8461

b) P(Z>1.98) = 0.0239

c) P(Z>-1.26) = 0.8962

d) P(Z>-1.52) = 0.9357

e) P(Z<0.38) = 0.6497

f) P(Z<-0.91) = 0.1814

g) P(Z<-1.97) = 0.0242

h) P(Z<0) = 0.5

The standard normal distribution is a probability distribution that has a mean of 0 and a standard deviation of 1. It is commonly denoted as Z, and its values represent the number of standard deviations away from the mean.

In part (a), we are asked to find the probability that a random variable from the standard normal distribution is less than 1.02 standard deviations away from the mean. Using a standard normal distribution table or calculator, we find that this probability is 0.8461.

In part (b), we are asked to find the probability that a random variable from the standard normal distribution is greater than 1.98 standard deviations away from the mean. This can be rephrased as finding the probability that a random variable is less than -1.98 standard deviations away from the mean. Again, using a standard normal distribution table or calculator, we find that this probability is 0.0239.

In part (c), we are asked to find the probability that a random variable is greater than -1.26 standard deviations away from the mean. This can be rephrased as finding the probability that a random variable is less than 1.26 standard deviations away from the mean. Using a standard normal distribution table or calculator, we find that this probability is 0.8962.

In part (d), we are asked to find the probability that a random variable is greater than -1.52 standard deviations away from the mean. This can be rephrased as finding the probability that a random variable is less than 1.52 standard deviations away from the mean. Using a standard normal distribution table or calculator, we find that this probability is 0.9357.

In part (e), there seems to be some missing inputs or instructions. If we assume that the question is asking for the probability that a random variable is less than 0.38 standard deviations away from the mean, then using a standard normal distribution table or calculator, we find that this probability is 0.6497.

In part (f), there also seems to be some missing inputs or instructions. If we assume that the question is asking for the probability that a random variable is less than -0.91 standard deviations away from the mean, then using a standard normal distribution table or calculator, we find that this probability is 0.1814.

In part (g), we are asked to find the probability that a random variable is less than -1.97 standard deviations away from the mean. Using a standard normal distribution table or calculator, we find that this probability is 0.0242.

In part (h), we are asked to find the probability that a random variable is less than 0 standard deviations away from the mean, which is simply the probability of getting a value between negative and positive infinity. This probability is equal to 0.5.

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Determine whether the relation R on R, defined below, is reflexive, symmetric, transitive. Is it an equivalence relation? Mark and justify your answers.
Ry iff x-y=q for some q€ Q
R is reflexive / not reflexive because
R is symmetric / not symmetric because
R is transitive / not transitive because
R is an equivalence relation / not an equivalence relation

Answers

Given relation R on R, where Ry if and only if x-y=q for some q€ QTo determine whether the relation R on R, defined above, is reflexive, symmetric, transitive, and an equivalence relation or not;Reflexive Relation:An equivalence relation R on a non-empty set A is said to be reflexive if aRa holds for every aϵA.

Hence, in this relation, x-x=q for some qϵQ which is not possible. Hence, the relation is not reflexive. Symmetric Relation:An equivalence relation R on a non-empty set A is said to be symmetric if aRb implies bRa for any pair of elements a, bϵA.In this relation, x-y=q which is not same as y-x. Hence, the relation is not symmetric.

Transitive Relation:An equivalence relation R on a non-empty set A is said to be transitive if aRb, and bRc implies aRc for any a, b, cϵA. In this relation, x-y=q and y-z=q.

Substituting the value of q in both equations, we get x-y=y-z or x=2y-z. This value of x is not independent of y and z. Hence, the relation is not transitive.As the relation is neither reflexive nor symmetric nor transitive. Hence, it is not an equivalence relation.

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How do you find the solutions of a linear equation and linear inequalities in one variable?.

Answers

By isolating the variable in one side of the equation/inequality.

How do you find the solutions of a linear equation and linear inequalities in one variable?.

what we understand as solution, is the value that the variable takes when the equation/inequality are true.

To solve them, we need to isolate the variable in one of the sides by using logical operations that don't affect the equation/inequality, and once it is isolated, we can know the value (or values) that the variable can take.

for example in the equation

4 = 3x + 2

We isolate x, to do so we subtract 2 in both sides of the equation

4 - 2 = 3x + 2 -2

2 = 3x

Now divide both sides by 3, we will get:

2/3 = 3x/3

2/3 = x

That is the solution, for an inequality we would so a similar thing, but the symbol is different (and multipliying or dividing by negative numbers changes the direction of the sign).

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when using simple linear regression, we use confidence intervals for the _____ and prediction intervals for the ____ at a given level of x.

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When using simple linear regression, we use confidence intervals for the regression line and prediction intervals for the individual predicted values at a given level of x.

Confidence intervals for the regression line provide a range within which we are confident the true regression line lies. It helps us estimate the uncertainty associated with the regression coefficients (intercept and slope) and assess the significance of the relationship between the independent variable (x) and the dependent variable (y).

On the other hand, prediction intervals provide a range within which we expect individual future observations to fall, given a specific value of x. Prediction intervals account for both the uncertainty in estimating the regression line and the inherent variability of individual data points around the line.

In summary, confidence intervals provide information about the precision of the estimated regression line, while prediction intervals give an indication of the expected variability of individual observations around the line.

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Q3. [12 points ] Let A=\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 1 \\ -1 & 1 & α\end{array}\right] . Find all values of α for which a) {A} is Singular. b) \mat

Answers

a) Matrix A is singular when α = 0.

b) For matrix A:

    a) It is singular when α = 0.

    b) It is invertible for any value of α that is not equal to zero.

a) To find the values of α for which matrix A is singular, we need to determine when the determinant of A is equal to zero. The determinant of A can be calculated using cofactor expansion:

|A| = 1(1(α) - 1) - 0(0(α) - 1) + (-1)(0(1) - 1(1))

= α - 1 - (-1)

= α

For matrix A to be singular, the determinant |A| must be zero. Therefore, we have:

α = 0

So, matrix A is singular when α = 0.

b) To find the values of α for which matrix A is invertible, we need to determine when the determinant of A is non-zero. From the previous calculation, we know that the determinant of A is equal to α. Therefore, matrix A will be invertible for any value of α that is not equal to zero.

In summary, for matrix A:

a) It is singular when α = 0.

b) It is invertible for any value of α that is not equal to zero.

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Suppose that X+2y=1 and 2x+y=5. What is Y?
Problem 5. Suppose that x+2 y=1 and 2 x+y=5 . What is - A. 0 - B. -3 - C. 3 - D. -1 E. 1

Answers

Answer is D.  y = -1

Here, there are not a 2 separate questions, but their is only 1 question.

Given, x + 2y = 1 and 2x + y = 5

Now we have to find the value of y.

To solve for y, let's eliminate x by multiplying the first equation by 2 and subtracting it from the second linear equation:

2(x + 2y = 1) => 2x + 4y = 2.

Subtracting the equation from the 2nd equation:

2x + y = 5- (2x + 4y = 2)  -----> -3y = 3y = -1

Hence, y = -1

Hence, the value of y is -1.

Answer: D. -1

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g a search committee is formed to find a new software engineer. there are 66 applicants who applied for the position. 1) how many ways are there to select a subset of 1515 for a short list?

Answers

The number of ways to select a subset of 1515 for a short list is,

⇒ ⁶⁶C₁₅

We have to give that,

A search committee is formed to find a new software engineer.

And, there are 66 applicants who applied for the position.

Hence, a number of ways to select a subset of 15 for a short list is,

⇒ ⁶⁶C₁₅

Simplify by using a combination formula,

⇒ 66! / 15! (66 - 15)!

⇒ 66! / 15! 51!

Therefore, The number of ways to select a subset of 1515 for a shortlist

⇒ ⁶⁶C₁₅

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Compute the kernel for each of the following homomorphisms ϕ. (a) ϕ:Z→Z such that ϕ(1)=12 (b) ϕ:Z×Z→Z such that ϕ(1,0)=3 and ϕ(0,1)=6.

Answers

The kernel for the homomorphism ϕ: Z → Z with ϕ(1) = 12 is {0} and for the homomorphism ϕ: Z × Z → Z with ϕ(1, 0) = 3 and ϕ(0, 1) = 6 is the set of pairs (a, b) such that a = -2b.

(a) For the homomorphism ϕ: Z → Z such that ϕ(1) = 12, the kernel is the set of integers that map to the identity element in the codomain, which is 0. In other words, the kernel consists of all integers n such that ϕ(n) = 0. To find these integers, we can solve the equation ϕ(n) = 12n = 0. Since 12n = 0 implies n = 0, the kernel of ϕ is {0}.

(b) For the homomorphism ϕ: Z × Z → Z such that ϕ(1, 0) = 3 and ϕ(0, 1) = 6, the kernel is the set of pairs of integers that map to the identity element in the codomain, which is 0. We need to find all pairs (a, b) such that ϕ(a, b) = 0. From the given information, we have 3a + 6b = 0. Dividing both sides by 3, we get a + 2b = 0.

This equation implies that a = -2b. Therefore, the kernel of ϕ is the set of all pairs (a, b) such that a = -2b.

In conclusion, the kernel of the homomorphism ϕ in (a) is {0}, and the kernel of the homomorphism ϕ in (b) is the set of all pairs (a, b) such that a = -2b.

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The equation y=23.1x represents Arianys's earnings in dollars and cents, y, for working x hours.

Answers

Therefore, the equation y=23.1x represents Arianys's earnings in dollars and cents, y, for working x hours.

The equation y=23.1x represents Arianys's earnings in dollars and cents, y, for working x hours.

Here, the numerical coefficient of the equation 23.1 represents the amount earned per hour.

Thus, when Arianys works x hours, she earns 23.1x dollars.

For instance, if Arianys works 5 hours, she will earn 23.1*5= 115.5 dollars.

It should be noted that the equation y=23.1x is a linear equation with a slope of 23.1.

The slope of the line represents the rate of change of y with respect to x.

Here, it means that Arianys will earn 23.1 dollars for each additional hour worked.

This equation can also be used to determine the number of hours worked if the amount earned is known.

For example, if Arianys earned 231 dollars, we can find the number of hours worked by dividing the total earnings by the hourly rate. Thus, the number of hours worked will be:

x= 231/23.1

= 10 hours.

The coefficient 23.1 is the hourly rate of earnings, and the equation can be used to determine the number of hours worked or the amount earned for a given number of hours.

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A sculptor makes a miniature model before starting the final version. Her model is scaled so that (1)/(4) of an inch corresponds to 6 feet on the final version. The base of her model is (5)/(12) of an inch. How big will the base of the final be?

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A sculptor makes a miniature model before starting the final version. Her model is scaled so that (1)/(4) of an inch corresponds to 6 feet on the final version. The base of her model is (5)/(12) of an inch. The solution to this problem is that the length of the base of the final version is 72/5 inches.

Given: A sculptor makes a miniature model before starting the final version. Her model is scaled so that (1)/(4) of an inch corresponds to 6 feet on the final version. The base of her model is (5)/(12) of an inch.

Let's first calculate how many inches correspond to 1 foot in the final version.1/(4) inch corresponds to 6 feet. Therefore, 1 inch corresponds to 6/(1/(4)) feet= 6 × 4= 24 feet

So, 1 foot in the final version will be 1/24th of an inch. Let x be the length of the base of the final version. Then, according to the scale of the model, 1/4th of an inch represents 6 feet. On the model, the length of the base is (5)/(12) inches.

Therefore, x inches on the final version represent (6 × 1)/(4 × 5)/(12)= 72/5 feet. So, the length of the base of the final version is 72/5 inches.

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Suppose 20% of the students graduated from a technical university are not employed within 6 months after graduation. A random sample of 20 graduated students were selected.
(a) State the random variable, X and write the appropriate distribution. (2 Marks)
(b) Based on (a), find the probability that, after graduation
i) three students are not employed within 6 months. (1 Mark)
ii) more than five students are not employed within 6 months. (2 Marks)
iii) No students are not employed within 6 months. (1 Mark)
iv) What is the average students are not employed within 6 months. (2 Marks)

Answers

(a) X represents the number of students not employed within 6 months. The appropriate distribution is the binomial distribution.

(b) i) P(X = 3), ii) P(X > 5), iii) P(X = 0), iv) E(X) = 4.

(a) The random variable X represents the number of students in the random sample who are not employed within 6 months after graduation. The appropriate distribution for this scenario is the binomial distribution.

(b) Based on the binomial distribution:

i) The probability that three students are not employed within 6 months is given by:

  P(X = 3) = (20% of 20 choose 3) * (0.20)^3 * (0.80)^(20-3)

ii) The probability that more than five students are not employed within 6 months is given by:

  P(X > 5) = 1 - P(X ≤ 5)

           = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)]

iii) The probability that no students are not employed within 6 months is given by:

  P(X = 0) = (20% of 20 choose 0) * (0.20)^0 * (0.80)^(20-0)

iv) The average number of students not employed within 6 months can be calculated using the expected value of the binomial distribution, which is given by:

  E(X) = n * p

  In this case, E(X) = 20 * 0.20 = 4 students.

Please note that the actual calculations for the probabilities in (i), (ii), and (iii) may require numerical evaluation using a calculator or statistical software.

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Jennifer is building a post for her mailbox. To find the correct dimensions, she needs to expand this expression: (x-3)(x - 7)(x - 2) Select the equivalent expression written in the format ax^2 + bx+ cx+d. a.) x^3 + 6x^2 + 13x - 42 b.) x^3-12x^2 +41x-42 c.) x^3 - 6x^2–13x +42 d.) x^3 + 12x^2-41x +42

Answers

The equivalent expression written in the format ax^2 + bx + cx + d is (b) x^3 - 12x^2 + 41x - 42.

Jennifer is building a post for her mailbox. To find the correct dimensions, she needs to expand this expression: (x-3)(x - 7)(x - 2) Select the equivalent expression written in the format ax^2 + bx+ cx+d. a.) x^3 + 6x^2 + 13x - 42 b.) x^3-12x^2 +41x-42 c.) x^3 - 6x^2–13x +42 d.) x^3 + 12x^2-41x +42 EXPLAIN

To expand the expression (x-3)(x - 7)(x - 2), we can use the distributive property and multiply the first two factors, and then multiply the result by the third factor:

(x-3)(x - 7)(x - 2) = (x^2 - 7x - 3x + 21)(x - 2)

= (x^2 - 10x + 21)(x - 2)

= x^3 - 2x^2 - 10x^2 + 20x + 21x - 42

= x^3 - 12x^2 + 41x - 42

So the expanded form of the expression is x^3 - 12x^2 + 41x - 42.

Therefore, the equivalent expression written in the format ax^2 + bx + cx + d is (b) x^3 - 12x^2 + 41x - 42.

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Consider the differential equation u" + u = 0 on the interval (0,π). What is the dimension of the vector space of solutions which satisfy the homogeneous boundary conditions (a) u(0) = u(π), and (b) u(0) = u(π) = 0. Repeat the question if the interval (0,π) is replaced by (0, 1) and (0,2π).

Answers

Interval (0, π) with boundary condition u(0) = u(π):

Dimension of the vector space of solutions: 1.

Interval (0, π) with boundary condition u(0) = u(π) = 0:

Dimension of the vector space of solutions: 0.

Interval (0, 1) with boundary condition u(0) = u(1):

Dimension of the vector space of solutions: 0.

Interval (0, 2π) with boundary condition u(0) = u(2π):

Dimension of the vector space of solutions: 1.

For the differential equation u" + u = 0 on the interval (0, π), we can find the dimension of the vector space of solutions satisfying different homogeneous boundary conditions.

(a) If we have the boundary condition u(0) = u(π), it means that the solution must be periodic with a period of 2π. This condition implies that the solutions will be linear combinations of the sine and cosine functions.

The general solution to the differential equation is u(x) = A cos(x) + B sin(x), where A and B are constants. Since the solutions must satisfy the boundary condition u(0) = u(π), we have:

A cos(0) + B sin(0) = A cos(π) + B sin(π)

A = (-1)^n A

where n is an integer. This implies that A = 0 if n is odd and A can be any value if n is even. Thus, the dimension of the vector space of solutions is 1.

(b) If we impose the boundary condition u(0) = u(π) = 0, it means that the solutions must not only be periodic but also satisfy the additional condition of vanishing at both ends. This condition implies that the solutions will be linear combinations of sine functions only.

The general solution to the differential equation is u(x) = B sin(x). Since the solutions must satisfy the boundary conditions u(0) = u(π) = 0, we have:

B sin(0) = B sin(π) = 0

B = 0

Thus, the only solution satisfying the given boundary conditions is the trivial solution u(x) = 0. In this case, the dimension of the vector space of solutions is 0.

Now, let's consider the differential equation on different intervals:

For the interval (0, 1), the analysis remains the same as in case (b) above, and the dimension of the vector space of solutions with the given boundary conditions will still be 0.

For the interval (0, 2π), the analysis remains the same as in case (a) above, and the dimension of the vector space of solutions with the given boundary conditions will still be 1.

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