I need help with this question

If the average annual stock return is 11. 3%, the average holds a portfolio of $8600 worth 30 years.

It's the ratio of two integers stated as a fraction of a hundred parts. It is a metric for comparing two sets of data, and it is expressed as a percentage using the percent symbol.

The usage of percentages is widespread and diverse. For instance, numerous data in the media, bank interest rates, retail discounts, and inflation rates are all reported as percentages. For comprehending the financial elements of daily life, percentages are crucial.

It is given that, the average annual stock return is 11. 3% and you begin your investment portfolio with $2,000,

Suppose the amount he earns in one year is x,

x=  11. 3%. of $2,000

x=220

The portfolio be worth 30 years if the average holds are,

=220 × 30

=$ 6600

The net cost is the sum of the return and the initial investment,

=$ 6600 + $ 2000

=$8600

Thus, if the average annual stock return is 11. 3%, the average holds a portfolio of $8600 worth 30 years.

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The constant average rate of change is the slope of the line which can be used to write the linear function.

What is a Data Table ?

A table of data is linear if there is a constant average rate of change between all pairs of points.

Testing for Linearity

Finding a Function from a Table that is Linear

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Answer:

y = 1/5 x

Step-by-step explanation:

Answer:

1

--

5

Step-by-step explanation

Can you give jasmynpurifoy branliest

Answer:

8

Step-by-step explanation:

if we take the 2 that is in the R.H.S and put it in L.H.S

it becomes 16÷2=8

The quadratic equation in one variable x² = 16, will have solutions at the values of x = 4, -4.

Any equation of the form ax² + bx + c = 0, where a, b, and c are constants, a ≠ 0, and x is a variable, is a quadratic equation in one variable, x. a ≠ 0, because if a = 0, x² term will be missing and the equation will become a linear equation in one variable, x.

We have been given a quadratic equation in one variable: x² = 16.

We have been asked for the values of x, which are the solutions to the given equation.

We solve the equation by following steps:

Subtract 16 from both sides of the equation to get,

x² = 16

or, x² - 16 = 16 - 16

or, x² - 16 = 0

or, x² - 4² = 0.

We factorize this using the formula a² - b² = (a + b)(a - b) taking a = x, and b = 4.

∴ x² - 4² = 0

or, (x + 4)(x - 4) = 0.

By the zero-product law, we know that if A*B = 0, then either A = 0, or B = 0, or both A and B = 0.

∴ Either, x + 4 = 0. ⇒ x = -4

Or, x - 4 = 0. ⇒ x = 4.

∴ The quadratic equation in one variable x² = 16, will have solutions at the values of x = 4, -4.

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Answer:

The correct option is;

b) 0 sq, units

Step-by-step explanation:

The vertices of the triangle are;

(4, 0), (2, 3), (8, -6)

The distance formula fr finding the length of a segment is given as follows;

\(l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}\)

Where, (x₁, y₁) and (x₂, y₂) are the coordinates of the end points of the line

For the points (4, 0) and (2, 3) , we have;

√((3 - 0)² + (2 -4)²) = √13

Distance from (4, 0) to (2, 3) = √13

For the points (4, 0) and (8, -6) , we have;

√((-6 - 0)² + (8 -4)²) = √13 =

Distance from (4, 0) to (8, -6) = 2·√13

For the points (2, 3) and (8, -6) , we have;

√((-6 - 3)² + (8 -2)²) = 3·√13 =

Distance from (2, 3) to (8, -6) = 3·√13

Therefore, the perimeter of the triangle = 6·√13

The semi perimeter s = 3·√13

The area of the triangle,  \(A = \sqrt{s\cdot \left (s-a \right )\cdot \left (s-b \right ) \cdot \left ( s-c \right )}\)

Where;

a, b, and c are the length of the sides of the triangle;

\(A = \sqrt{3\cdot \sqrt{3} \cdot \left (3\cdot \sqrt{3} -\sqrt{3} \right )\cdot \left (3\cdot \sqrt{3} -2 \cdot \sqrt{3} \right ) \cdot \left ( 3\cdot \sqrt{3} -3\cdot \sqrt{3} \right )} = 0\)

Therefore, the area = 0 sq, units.

Answer:

0 is 5

1 is 7

2 is 9

5 is 15

7 is 19

Answer:

Step-by-step explanation:

It usually works best for factoring to remove any common factors from the coefficients. Here, they are all even numbers, so have a common factor of 2. After taking that out, you have ...

  2(4x² -4x -3)

To factor this, you are looking for factors of (4)(-3) that have a sum of -4.

  -12 = 1(-12) = 2(-6) = 3(-4)

These pairs have sums of -11, -4, -1. So, we are interested in the factors 2 and -6. We can use those to rewrite the middle term, then factor by pairing.

  = 2(4x² +2x -6x -3) = 2((2x(2x+1) -3(2x+1)) = 2(2x -3)(2x +1)

Of these, the factors that are on your list of choices are ...

Answer:

35 inches

Step-by-step explanation:

Answer:

1615441

Step-by-step explanation:

Use BPEMDAS:

Multiply in the parenthesis 1st:

(1271)²

Then exponent:

1615441

Answer:

259.5

Step-by-step explanation:

8.1*12=97.2
Area of trapiezium = 1/2(b+a)h
(2.8+8.1)=10.9
10.9*3/2=16.35
16.35*2=32.7
2.8*12=33.6
33.6+32.7+97.2=163.5
4*12*2=96
163.5+96=259.5

Answer:

  see attached

Step-by-step explanation:

The dilation factor of 1/2 moves each point to half its previous distance from the origin.

The quadratic example is ax² + b x + c = 0 where x stands for an unknown value and where a, b, and c stand for known numbers is known as a quadratic equation in algebra.

Quadratic equations are second-degree polynomial equations with at least one squared term. It is also referred to as quadratic equations.

The numerical coefficients a, b, and c are known, while the unknown variable x is. For example , x2 + 2x + 1. It also goes by the name quadratic equation as in: b x +c=0    

The terms a, b, and c are also referred to as quadratic coefficients.    

'Examples of Quadratics

x² –x – 9 = 0

5x² – 2x – 6 = 0
3x² + 4x + 8 = 0

-x² +6x + 12 = 0  (an example of non-quadratic equation is x³ − x² − 5 = 0)

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Answer:

Her journey by car.

Step-by-step explanation:

3/5 is more than 3/10. I hope this helps :)

Answer:  3/10 of her journey by autorikshaw.

Step-by-step explanation:

The value of x and y using the substitution method are 5/4 and 5/4.

An equation is a mathematical statement that is made up of two expressions connected by an equal sign.

We have,

Two equations:

3x + 5y = 10 _____(1)

9y + 3x = 15 ______(2)

From (1) we get,

3x = 10 - 5y

x = (10 - 5y) / 3 ______(3)

Putting (3) in (2) we get,

9y + 3 x [(10 - 5y) / 3] = 15

9y + 10 - 5y = 15

4y = 15 - 10

4y = 5

y = 5/4

Putting y = 5/4 in (3) we get,

x = (10 - 5 x 5/4) / 3

x = (10 - 25/4) / 3

x = (40 - 25) / (4 x 3)

x = 15 / 12

x = 3x5 / 3x4

x = 5/4

Thus,

The value of x and y using the substitution method are 5/4 and 5/4.

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From the given figure the angle x° is quals to 38°.

Given triangle is a right-angled triangle,

In the right-angled triangle the opposite side of the triangle = 6.1

The hypotenuse of the triangle = 9.9

In a right-angled triangle, by using little big trigonometry we know that,

sin theta = opposite side of the triangle/hypotenuse side of the triangle

From the given figure sin x° = opposite side of x / hypotenuse side

sin x° = 6.1/9.9

x° = \(sin^{-1}\)  (6.1/9.9)

x° = 38.03°

From the above analysis, we can conclude that the angle of x° is equal to 38.03° ≅ 38°.

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Answer: Yes, because the results are unlikely to occur by chance.

Step-by-step explanation: apparently, the weight loss program has a statistical significance, because the results are likely to also occur by chance. because it occurrs by chance, this applits to a happening which has occurred without an intent, volition, or plan. an event or encounter that occurs by chance usually implies an occurrence of some importance.

(b) The test statistic for the t-test would be -29.96.

(f) The predicted average cob weight when there are 134 plants in the plot is 264.39 grams, and it may not be reliable due to the extrapolation beyond the observed range of plant density.

(8) The estimated change in cob weight for a five-unit increase in plant density is -3.60 grams, indicating a decrease.

(b) To perform the t-test, we need to calculate the t-statistic using the formula: t = (β1 - 0) / (SE(β1)), where β1 is the coefficient estimate for plant density and SE(β1) is its standard error. Here, the coefficient estimate for plant density is -0.720626 and its standard error is <0.0001, so the t-statistic is -29.96.

(f) To predict the average cob weight when there are 134 plants in the plot, we use the regression equation: Cob Weight = Intercept + (Plant Density x β1). Substituting the given values, we get Cob Weight = 316.37619 + (134 x -0.720626) = 264.39 grams. However, this estimate may not be reliable as it is an extrapolation beyond the observed range of plant density.

(8) The estimated change in cob weight for a five-unit increase in plant density can be calculated as: ΔCob Weight = 5 x -0.720626 = -3.60 grams, indicating a decrease. The negative sign indicates that as the plant density increases, the cob weight decreases.

(j) The intercept represents the predicted value of the response variable (cob weight) when the predictor variable (plant density) is zero. However, in this problem, it is not meaningful as it is not possible to have zero plant density in a cornfield. Therefore, interpreting the intercept is not appropriate in this context.

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Answer:

Possible

Step-by-step explanation:

Answer:

256

Step-by-step explanation:

Answer:

256 cm

Step-by-step explanation:

16 x 4 = 64

16 x 8 = 128

8 x 8 = 64

64 + 64 = 128  + 128 = 256

Answer:

Range = highest value - lowest value

Explanation:

This is the required formula to find range from a distribution table.

Answer:

try 45 its a great answer

Answer:

infinite solutions

Step-by-step explanation:

Multiply the second equation by 2

6x + 2y = 10

3x + y = 5

6x+2y = 10

This is identical to the first equation

This means there are infinite solutions

The solution of the nonlinear programming problem is non-negative.

To solve the given nonlinear programming problem using dynamic programming, we need to follow these steps:

We define a set of subproblems based on the constraints and the objective function. In this case, our subproblems can be defined as finding the maximum value of the objective function for different values of x₁, x₂, and x₃, while satisfying the constraint x₁ + 2x₂ + 3x₃ ≤ 7.

Next, we need to establish a recurrence relation that relates the optimal solution of a larger subproblem to the optimal solutions of its smaller subproblems. In our case, let's denote the maximum value of the objective function as F(x₁, x₂, x₃), where x₁, x₂, and x₃ are the variables that satisfy the constraint.

F(x₁, x₂, x₃) = max {5x₁ - x₁² + 3x₂ + x₃³ + F(x₁', x₂', x₃')},

where x₁ + 2x₂ + 3x₃ ≤ 7,

and x₁', x₂', x₃' satisfy the constraint x₁' + 2x₂' + 3x₃' ≤ 7.

Once the table is filled, the final entry in the table represents the maximum value of the objective function for the given problem. We can also backtrack through the table to determine the values of x₁, x₂, and x₃ that yield the maximum value.

Finally, we need to verify that the obtained solution satisfies all the constraints of the original problem. In our case, we need to ensure that x₁ + 2x₂ + 3x₃ ≤ 7 and that x₁, x₂, and x₃ are non-negative.

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The linearly independent solution of the non-homogeneous equation is y = y-c + y-p, y = c1×(x²(-1/2)cos(x)) + c2×(x²(-1/2)sin(x)) + (8/35)×x²(3/2) + (2/35)×x²(-1/2) where c1 and c2 are arbitrary constants.

The associated homogeneous equation is: x²2y'' + xy' + (x²2 - 1/4)y = 0

The complementary solution can be found by assuming y has the form y-c = c1y1 + c2y2, where c1 and c2 are constants, and y1 and y2 are the given linearly independent solutions.

y-c = c1×(x²(-1/2)cos(x)) + c2×(x²(-1/2)sin(x))

Now, the particular solution, denoted as y-p, of the non-homogeneous equation.

y-p has the form:

y-p = Ax²(3/2) + Bx²(-1/2)

where A and B are constants to be determined.

The first and second derivatives of y-p:

y-p' = A×(3/2)x²(1/2) - (1/2)Bx²(-3/2)

y-p'' = A(3/4)×x²(-1/2) + (3/4)Bx²(-5/2)

Substituting these into the non-homogeneous equation:

x²2y_-p'' + xy-p' + (x²2 - 1/4)×y-p = x²(3/2)

x²2×(A×(3/4)x²(-1/2) + (3/4)Bx²(-5/2)) + x(A×(3/2)x²(1/2) - (1/2)Bx²(-3/2)) + (x^2 - 1/4)(Ax²(3/2) + Bx²(-1/2)) = x²(3/2)

Simplifying and collecting like terms:

(3A/4)x²(3/2) + (3B/4)x²-1/2) + (3A/2)x²(3/2) - (1/2)Bx²(3/2) + (A - (1/4))x²(5/2) + (B/4)x²(1/2) - (A/4)x²(-1/2) + Bx²(-3/2) = x²(3/2)

Matching the coefficients of like powers of x:

[(3A/4) + (3A/2) - (1/2)B]x²(3/2) + [(3B/4) + (B/4)]x²(-1/2) + [(A - (1/4))]x²(5/2) + [(-A/4) + B]x²(-1/2) + [B/4]x²(-3/2) = x²(3/2)

Equating the coefficients of x²(3/2) on both sides:

(3A/4) + (3A/2) - (1/2)B = 1

(9A/4) - (1/2)B = 1

Equating the coefficients of x²(-1/2) on both sides:

[(3B/4) + (B/4)] - (A/4) = 0

(4B/4) - (A/4) = 0

Simplifying the equations:

(9A - 2B) = 4

4B - A = 0

Solving these equations simultaneously ,A = 8/35 and B = 2/35.

Therefore, the particular solution is: y-p = (8/35)×x²(3/2) + (2/35)×x²(-1/2)

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Step-by-step explanation:

Question:-

2x+3y+2=0

2x+y=-1

Answer:-

2x+y=-1

y= -1 - 2x

2x+3(-1-2x)=0

2x-3-6x=0

-4x-3=0

-4x=3

x=3/-4

2(3/-4)+y=-1

3/-2+y=-1

3+y=-1 x -2

y=2/3

Hope this helps you

have a good day

Answer:

D

Step-by-step explanation:

2x(x^4 + 6x^2 - 27)

x^4 + 6x^2 - 27 = 0

x^2 = y

y^2+ 6y - 27 = 0

(y+9)(y-3) = 0

(x^2+9)(x^2-3) = 0

2x(x^2-3)(x^2+9)

Answer:

The answer is B

Answer:

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

Initial height is the y-intercept of the line.

Yearly growth rate is the slope.

So we have:

The equation of the line with these constants is: