Question 7 (Drag&Drop 2pts): A system of equations is given. Identify the steps in the correct

order to explain how to eliminate the x in the system of equations.

STEPS

Step 1: 5x + 4y = -14

3x + 6y = 6

Step 2: -15x12y = -42

Step 3: 15x + 30y = 30

Step 4: -15x - 12y = -42

15x + 30y = 30

Equation 1: 5x + 4y = -14

Equation 2: 3x + 6y =6

EXPLANATION

Answer: 23/45 or 0.51 rounded to the nearest hundredth

Step-by-step explanation:

4 1/9 can be written as 37/9, and 3 3/5 can be written as 18/5.

37/9-18/5 = (185/45) - (162/45) = 23/45 or 0.51

The sum of the polynomial is solved to be

A. 5x^5 + 7x^3 + 7x^2 + 25x

To find the sum of the given polynomials, we simply add the like terms. Like terms in this case are terms with the same degree of x.

The given polynomials are:

(9x^5 + 7x^3 + 21x) and

(-4x^5 + 7x^2 + 4x)

Adding the like terms:

9x^5 + (-4x^5) = 5x^5

7x^3 + 0 = 7x^3

0 + 7x^2 = 7x^2

21x + 4x = 25x

Putting it all together, we get:

(9x^5 + 7x^3 + 21x) + (-4x^5 + 7x^2 + 4x) = 5x^5 + 7x^3 + 7x^2 + 25x

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We can use the sample proportion of tankers without spills (146/211 = 0.692) to estimate the population proportion of tankers without spills. To construct the confidence interval, we need to find the margin of error and the critical value for an 80% confidence level.

Follow these steps:

1. Calculate the sample proportion:
In the sample of 211 tankers, 146 did not have spills, so 211 - 146 = 65 tankers had spills. The sample proportion (p-hat) is the number of tankers with spills divided by the total sample size:
p-hat = 65/211 ≈ 0.308

2. Determine the z-score for an 80% confidence interval:
Using a z-table or calculator, the z-score for an 80% confidence interval is approximately 1.282.

3. Calculate the standard error:
The standard error (SE) can be calculated using the formula: SE = sqrt(p-hat*(1-p-hat)/n)
SE = sqrt(0.308*(1-0.308)/211) ≈ 0.030

4. Construct the confidence interval:
Lower limit = p-hat - (z-score * SE)
Upper limit = p-hat + (z-score * SE)

Lower limit = 0.308 - (1.282 * 0.030) ≈ 0.277
Upper limit = 0.308 + (1.282 * 0.030) ≈ 0.339

So, the 80% confidence interval for the population proportion of oil tankers that have spills each month is approximately (0.277, 0.339), rounded to three decimal places.

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The value of x and y are 27 and 47 unit.

We are given the hexagon shape which we need to find the angles.

5x -1 + 4x + 2 + 5x + 6 + 2x - 2 + 3x + 5 + x - 10 = 540

Combine the like terms;

5x -1 + 4x + 2 + 5x + 6 + 2x - 2 + 3x + 5 + x - 10 = 540

9x + 7x + 4x = 540

20x = 540

x = 27

Now solve for y;

3x + 5 + 2y = 180

3(27) + 5 + 2y = 180

2y = 180 - 5 - 81

2y = 94

y = 47

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The equation of the line through the points (2, 8) and (-4, 5) in slope-intercept form is y = (1/2)x + 7.

To determine the equation of the line through the points (2, 8) and (-4, 5) and express it in slope-intercept form, follow these steps:

1. Calculate the slope (m) of the line using the formula: m = (y2 - y1) / (x2 - x1)
  In our case, (x1, y1) = (2, 8) and (x2, y2) = (-4, 5).
  m = (5 - 8) / (-4 - 2) = (-3) / (-6) = 1/2

2. Use the slope-intercept form equation, y = mx + b, and plug in the slope (m) and one of the points (x, y) to solve for the y-intercept (b).
  Let's use the point (2, 8).
  8 = (1/2) * 2 + b
  8 = 1 + b
  b = 7

3. Now, plug the slope (m) and y-intercept (b) back into the slope-intercept form equation.
  y = mx + b
  y = (1/2)x + 7

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Geometric shapes are defined as shapes that have a clear and defined outline, uniformity in their angles, and consistent measurements. Examples of geometric shapes include squares, triangles, and stars. These shapes are typically man-made and are commonly found in architecture and design.

On the other hand, organic shapes are irregular and asymmetrical in nature, often resembling forms found in nature. Examples of organic shapes include leaves, hands, and clouds. These shapes are often found in art and can evoke a sense of movement and fluidity.
When teaching shapes to a kindergarten class, it is important to differentiate between geometric and organic shapes to help children understand the unique characteristics of each. This can help develop their cognitive and spatial skills and encourage creativity in their art and design projects.

Overall, the distinction between geometric and organic shapes is an important concept to introduce to young children, as it lays the foundation for future learning in math and design. By teaching them the differences between these two types of shapes, we can help them develop a deeper understanding of the world around them.

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The given statement "Quantitative Easing was used extensively in the aftermath of the(the late 1990s/2000s) dot com crisis is false because Quantitative Easing was not used extensively in the aftermath of the dot com crisis in the late 1990s/2000s.

Quantitative easing (QE) was not used extensively in the aftermath of the dot-com crisis in the late 1990s and early 2000s. In fact, QE as a monetary policy tool gained prominence after the global financial crisis of 2008. The dot-com crisis primarily affected the technology sector, causing a stock market downturn, but it did not lead to a widespread financial crisis that would have necessitated the use of QE.

Therefore, the given statement is false.

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Answer:112.475

Step-by-step explanation:

Answer:

Opens downward and is thinner than the parent function.

The a value is negative remember I told you ax^2+bx+x=0

if the a value is negative it opens down, but if it's positive it opens up.

This graph is also stretched because a is greater than 1.

Answer:41.5

Step-by-step explanation:

By arranging the data from smallest to largest, like this:

40, 41, 42, 45, we can take the average of the two middle values divided by 2 to find the median. This is done with an equation like this:

(41+42)/2

Which comes out to be 41.5.

The value of the expression is 295.

We can simplify the expression by grouping the terms that have the same value:

26 + (25 + 25) + (24 + 24) - (23 + 23) - (22 + 22) - (21 + 21) - (20 + 20) + (19 + 19) + (18 + 18) + 17 - (16 + 16) + (15 + 15) + (14)

= 26 + 50 + 48 - 46 - 44 - 42 - 40 + 38 + 36 + 17 - 32 + 30 + 14

= 295

The given expression involves a series of numbers where some of them are added and some of them are subtracted. To simplify this expression, we need to group the terms that have the same value. We can see that the expression has pairs of numbers that add up to the same value, such as (25 + 25), (24 + 24), and so on. We can combine these pairs and simplify the expression further.

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d. random and estimated. The number of tickets in the box labeled 1 represents the number of customers in the population who are satisfied with their internet service provider.

This quantity is not fixed or known, as we are using a sample to estimate the proportion of the population who are satisfied. The tickets in the box are randomly selected from the population, and the number in the box is estimated based on the proportion of satisfied customers in the sample. Therefore, the quantity is both random and estimated. we can calculate the sample proportion and construct a confidence interval to estimate the true proportion of satisfied customers in the population.

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To find the maximum possible area of a triangle with a given sum of base and height, we will use the formula for the area of a triangle:

Area = 1/2 × base × height

Given that the sum of base and height is 12cm, let's denote base as "b" and height as "h". We have:

b + h = 12

To maximize the area, we want to maximize the product of base and height. From the given equation, we can express height as:

h = 12 - b

Now, let's substitute this into the area formula:

Area = 1/2 × b × (12 - b)

To find the maximum area, we will find the critical points of this equation by taking the derivative with respect to b:

d(Area)/db = 1/2 × (12 - 2b)

Set the derivative equal to zero and solve for b:

0 = 1/2 × (12 - 2b)
0 = 12 - 2b
b = 6

Now that we have the base, we can find the height using the given equation:

h = 12 - b
h = 12 - 6
h = 6

So, the maximum possible area of the triangle is:

Area = 1/2 × 6 × 6 = 18 cm²

To summarize:

a) The maximum possible area of the triangle is 18 cm².
b) The base and height of the triangle found in (a) are both 6 cm.

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The time that it takes for the coconut to land on the net is given as follows:

t = 2.1 seconds.

The quadratic function giving the height of the coconut after t seconds is defined as follows:

h(t) = -4.9t² + S.

In which S represents the height of the canopy.

Ellie uses a pole pruner to release a bunch of coconuts from a canopy 24 meters above the ground, hence the value of S is given as follows:

S = 24.

Thus the function is:

h(t) = -4.9t² + 24.

The coconuts hit the net when h(t) = 1.5, hence the time is obtained as follows:

1.5 = -4.9t² + 24

4.9t² = 22.5

t = sqrt(22.5/4.9)

t = 2.1 seconds.

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Answer:

what

Step-by-step explanation:

Note that the graph that best shows the estimates of the survey rounded to the nearest 500  is Graph D. See the attached image.

If we have a total of 10,000 people, and 240 people respond to a survey.

If 180 of them own cars and 60 don't, then the ratio of the respondent to the total population is:

Those that own car = (180/240) * 10,000

= 7,500  people

Those that don't own a car = (60/240) * 10,000

= 2,500 people

This is what is depicted in Graph D, hence option D is the correct answer.

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The policyholder is insured for $238,000, and the actual cash value of the damages is $110,000. Therefore, the insurer will pay the actual cash value, which is $110,000, so option A is correct.

The claim payment for a flood policy is based on the replacement cost value (RCV) of the property and the actual cash value (ACV) of the damage. The RCV represents the cost to replace the damaged property with new property of like kind and quality, while the ACV represents the RCV less depreciation.

Even though the duplex's replacement cost is $240,000, it is insured for $238,000 in this case. The cash value of the flood damage is $110000. Since the policyholder is only covered for a portion of the replacement cost, the claim payment will be determined by the damage's $110,000 actual cash value. Therefore, the answer is A.

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If a point (x,y) is equidist point from the point (2,3) and (6,1) then the equation of locus is given by 2x -y =6

Using distance formula, the distnace between point (x, y) and (2, 3) is,

d₁ = √[(2 - x)² + (3 - y)²]

And  the distnace between point (x, y) and (6, 1) is,

d₂ = √[(6 - x)² + (1 - y)²]

A point (x,y) is equidist point from the point (2,3) and (6,1)

⇒ d₁ = d₂

⇒  √[(2 - x)² + (3 - y)²] = √[(6 - x)² + (1 - y)²]

⇒ (2 - x)² + (3 - y)² = (6 - x)² + (1 - y)²

⇒ 4 - 4x + x² + 9 - 6y + y² = 36 - 12x + x² + 1 - 2y + y²

⇒ -4x + 12x - 6y + 2y = 37 - 13

⇒ 8x - 4y = 24

⇒ 2x -y = 6

Hence proved.

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The complete question is:

if a point (x,y) is equidist point from the point (2,3) and (6,1) show that the equation of locus is given by 2x -y = 6

a) there are 32 possible subsets of C.

b)The cardinality of set M is the sum of these numbers, which is 781.

C) there are an infinite number of possible strings, the cardinality of set K, which contains all possible strings using letters from C, is also infinite.

a) The cardinality of set C is 5, as there are 5 distinct elements in the set. The cardinality of the power set of C, denoted as P(C), is 2^5 = 32, as there are 32 possible subsets of C.

b) A tree showing all possible strings of letters of length 5 or less starting with the letter z would look like:

z

├── v

│   ├── v

│   ├── w

│   ├── x

│   └── y

├── w

│   ├── v

│   ├── w

│   ├── x

│   └── y

├── x

│   ├── v

│   ├── w

│   ├── x

│   └── y

├── y

│   ├── v

│   ├── w

│   ├── x

│   └── y

└── z

   ├── v

   ├── w

   ├── x

   └── y

The cardinality of set M, which contains all possible strings of length 5 or less with letters from C, is equal to the sum of the cardinalities of all sets of strings of each length. Thus,

Set of strings Number of strings

Length 1 1

Length 2 5

Length 3 5^2 = 25

Length 4 5^3 = 125

Length 5 5^4 = 625

The cardinality of set M is the sum of these numbers, which is 1 + 5 + 25 + 125 + 625 = 781.

c) A tree showing all possible strings of any length would have an infinite number of branches. Each node in the tree would represent a different string, and the branches emanating from each node would represent the next letter that could be added to the string. Since there are an infinite number of possible strings, the cardinality of set K, which contains all possible strings using letters from C, is also infinite.

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The rank correlation is 1, this means that the two variables being compared are monotonically related, even if their relationship is not linear.

Given the data

X, ...rankX.....Y.....rankY......d=rx-ry........d²

4.........4...........-5......4................0..............0

53.......3.........-1........3................0...............0

86.......2.........13.......2................0................0

162......1..........16.......1.................0...............0

Then, using the rank correlation formula

p = 1 — 6•Σd² / n(n²—1)

p = 1 - 6• 0 / 4(4²-1)

p = 1 - 0

p = 1

So, the rank correlation is 1, this means that the two variables being compared are monotonically related, even if their relationship is not linear.

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Full Question: Use the table of values to calculate the linear correlation coefficient r.

X Y

4 -5

53 -1

86 13

162 16

a) The parameter of interest is the percentage of mail items received at the dorm that are considered packages. This can be denoted as p, where p is the proportion of packages out of all mail items received at the dorm. A confidence interval would be most appropriate for this inference procedure.

b) The parameter of interest is the difference in the average number of packages sent from online retailers to a neighborhood in Champaign and the average number of packages sent from online retailers to a neighborhood in Urbana. This can be denoted as μ1 - μ2, where μ1 is the average number of packages sent to Champaign and μ2 is the average number of packages sent to Urbana. A hypothesis test would be most appropriate for this inference procedure.

c) The parameter of interest is the difference in store pick-up rates between all cookies and all cakes sold by the bakery. This can be denoted as p1 - p2, where p1 is the proportion of cookies that are picked up in store and p2 is the proportion of cakes that are picked up in store. A confidence interval would be most appropriate for this inference procedure.

d) The parameter of interest is the association between the weight of the package and the delivery time. This can be denoted as ρ, where ρ is the correlation coefficient between the weight of the package and the delivery time. A hypothesis test would be most appropriate for this inference procedure.

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The first four terms of the sequence are 5, -4, 14, and -22.

We have,

The first term is given as a_1 = 5.

To find the next term, we use the formula:

a_n = -2a_(n-1) + 6.

Now,

a_2

= -2a_1 + 6

= -2(5) + 6

= -4

For the third term:

a_3 = -2a_2 + 6 = -2(-4) + 6 = 14

And for the fourth term:

a_4

= -2a_3 + 6

= -2(14) + 6

= -22

Thus,

The first four terms of the sequence are 5, -4, 14, and -22.

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The surface area of this cube is 384 sq ft.

To find the surface area of this cube:

You can follow these steps:

STEP 1: Identify the number of faces on the cube: A cube has 6 faces.
STEP 2: Determine the area of each face: Each face measures 64 sq ft.
STEP 3: Calculate the surface area: Multiply the area of each face by the total number of faces.

Surface area = (Area of each face) x (Total number of faces)
Surface area = (64 sq ft) x (6)
Surface area = 384 sq ft

The surface area of this cube is 384 sq ft.

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The expected value of a $4 bet on the state's lottery is -$0.84 and the expected value of a $4 bet on the casino game is -1.7¢ (which is equivalent to -$0.017), the state's lottery is a better bet in terms of producing a higher expected value.

a. There are 10 possible choices for each of the three digits, so the total number of different selections is 10 x 10 x 10 = 1000.

b. Since there is only one winning combination out of the 1000 possible selections, the probability of winning is 1/1000.

c. If you win, your net profit would be $500 - $4 = $496.

d. The expected value is the sum of the products of each possible outcome and its probability. In this case, the expected value is (1/1000) x $500 + (999/1000) x (-$4) = -$0.84.

e. Since the expected value of a $4 bet on the state's lottery is -$0.84 and the expected value of a $4 bet on the casino game is -1.7¢ (which is equivalent to -$0.017), the state's lottery is a better bet in terms of producing a higher expected value.

This is because the expected loss for the state's lottery is smaller than the expected loss for the casino game.

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The automaton starts in the initial state q0, reads the input string w, and uses the stack to keep track of the non-terminals in the production rules.

To reduce the given CFG to Greibach Normal Form (GNF), we can follow these steps:

Step 1: Eliminate the start symbol from the right-hand side of any production by introducing a new non-terminal symbol S0 and a new production S0 → S.

S0 → S

S → CA

A → BB

B → SB

S → CBA

S → a

Step 2: Eliminate the productions with more than one non-terminal on the right-hand side.

S0 → S

S → CAA1

A1 → BA2

A2 → SB

B → SB

S → CBAA3

A3 → a

Step 3: Convert the remaining productions into GNF form by replacing the first non-terminal symbol on the right-hand side with a terminal symbol or a new non-terminal symbol.

S0 → S

S → CAZ

Z → AZ

A → BBY

Y → SB

B → SB

S → CBAW

W → A3

A3 → a

The resulting CFG is in Greibach Normal Form.

To define the condition for acceptability of strings using final state and null stack method, we need to consider the corresponding pushdown automaton. If the automaton reaches a final state and the stack is empty, then the input string is accepted.

Formally, let P = (Q, Σ, Γ, δ, q0, Z0, F) be a pushdown automaton, where Q is the set of states, Σ is the input alphabet, Γ is the stack alphabet, δ is the transition function, q0 is the initial state, Z0 is the initial stack symbol, and F is the set of final states.

A string w ∈ Σ* is accepted by P using final state and null stack method if there exists a sequence of configurations

(q0, w, Z0) ⇒* (qf, ε, ε)

such that qf ∈ F and the stack is empty.

In other words, the automaton starts in the initial state q0, reads the input string w, and uses the stack to keep track of the non-terminals in the production rules. If it reaches a final state qf and the stack is empty, then the input string w is accepted.

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Answer:

Speed= distance/ time

Speed= (2/3)/(3/4) = 0.88 miles/hr

A) tan(2θ) = 5/12

B) cos(2θ) = -31

C) tan(θ/2) = ±(6/5)√6 - 3i/5

Given tanθ = -5 and 3x < θ < 5x/2. We need to find:

A) tan(2θ)

B) cos(2θ)

C) tan(θ/2)

First, we can find the value of θ using the given inequality:

3x < θ < 5x/2

Multiplying all terms by 2, we get:

6x < 2θ < 5x

Dividing all terms by 2, we get:

3x < θ < 5x/2

Since we are given that tanθ = -5, we know that θ is in the third quadrant. In the third quadrant, tanθ is negative and sinθ is negative, while cosθ is positive.

Using the Pythagorean identity, we can find the value of cosθ:

[tex]cos^2θ + sin^2θ = 1[/tex]

[tex]cos^2θ + (-5)^2 = 1[/tex]

[tex]cos^2θ = 1 - 25[/tex]

cosθ = √(1 - 25) = √(-24) = 2i√6/6 (taking the positive root since cosθ is positive in the third quadrant)

Now, we can use the double angle identities to find A) and B):

A) tan(2θ) = 2tanθ/(1-tan^2θ)

= 2(-5)/(1-(-5)^2)

= 10/24

= 5/12

B) cos(2θ) = [tex]cos^2θ - sin^2θ[/tex]

= (2i√[tex]6/6)^2[/tex] - (-[tex]5)^2[/tex]

= -6/3 - 25

= -31

Finally, we can use the half-angle identity to find C):

C) tan(θ/2) = ±√((1-cosθ)/1+cosθ))

= ±√((1-2i√6/6)/(1+2i√6/6))

= ±√((1-2i√[tex]6/6)^2[/tex]/(1-24/36))

= ±√((1-2i√6/[tex]6)^2[/tex]/(5/36))

= ±(6/5)√6 - 3i/5

Therefore, the exact values are:

A) tan(2θ) = 5/12

B) cos(2θ) = -31

C) tan(θ/2) = ±(6/5)√6 - 3i/5

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The domain of the function is all real numbers and  range is  y ≥ -1.

Since the vertex is at (1,-1), the axis of symmetry is x = 1.

This means that the domain of the function is all real numbers.

To find the range, we need to consider the y-values of the graph. Since the vertex is the lowest point of the graph, the range must be all y-values greater than or equal to -1.

However, since the parabola opens upwards, there is no upper bound on the y-values.

Therefore, the range is given by y ≥ -1.

Hence, the domain of the function is all real numbers and  range is  y ≥ -1.

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